Disclaimer: This manuscript, and the accompanying Lean formalisation of the proof, have been generated by AI. The validity of the proof has been extensively machine-checked, but the proof is not fully understood by the human author.
Introduction
A difference equation is globally periodic if all of its orbits are periodic with a common period. The celebrated example is the Lyness equation
x n + 2 = 1 + x n + 1 x n (1) x_{n+2} = \frac{1+x_{n+1}}{x_n} \tag*{(1)} x n + 2 = x n 1 + x n + 1 (1) every positive solution of which repeats with period 5 [27 ] . Global periodicity is a fragile property, and equations that have it are rare. In a survey of open problems in low-dimensional dynamics, Gasull [ [12 ] records five globally periodic rational difference equations with nonnegative coefficients,
x n + 1 = x n , x n + 1 = 1 x n , x n + 2 = x n + 1 x n , x n + 2 = 1 + x n + 1 x n , x n + 3 = 1 + x n + 1 + x n + 2 x n , (2) x_{n+1}=x_n,\qquad x_{n+1}=\frac{1}{x_n},\qquad x_{n+2}=\frac{x_{n+1}}{x_n},\qquad x_{n+2}=\frac{1+x_{n+1}}{x_n},\qquad x_{n+3}=\frac{1+x_{n+1}+x_{n+2}}{x_n}, \tag*{(2)} x n + 1 = x n , x n + 1 = x n 1 , x n + 2 = x n x n + 1 , x n + 2 = x n 1 + x n + 1 , x n + 3 = x n 1 + x n + 1 + x n + 2 , (2) of periods 1, 2, 6, 5 and 8 respectively. The 8-cycle of the last was found by H. Todd [22 , 28 ] . New examples can be made from these in two trivial ways: by rescaling , y n = α x n y_n=\alpha x_n y n = α x n α > 0 \alpha>0 α > 0 unfolding : multiplying every index in an equation of order k k k ℓ ≥ 1 \ell\geq1 ℓ ≥ 1 ℓ k \ell k ℓ k ℓ \ell ℓ k ≥ 1 k \ge1 k ≥ 1
x n + k = A 0 + A 1 x n + A 2 x n + 1 + ⋯ + A k x n + k − 1 B 0 + B 1 x n + B 2 x n + 1 + ⋯ + B k x n + k − 1 , (3) x_{n+k}=\frac{A_0+A_1x_n+A_2x_{n+1}+\cdots+A_kx_{n+k-1}}{B_0+B_1x_n+B_2x_{n+1}+\cdots+B_kx_{n+k-1}}, \tag*{(3)} x n + k = B 0 + B 1 x n + B 2 x n + 1 + ⋯ + B k x n + k − 1 A 0 + A 1 x n + A 2 x n + 1 + ⋯ + A k x n + k − 1 , (3) with A i , B i ≥ 0 A_i,B_i \ge0 A i , B i ≥ 0 ∑ i A i > 0 \sum_i A_i > 0 ∑ i A i > 0 ∑ i B i > 0 \sum_i B_i > 0 ∑ i B i > 0 A 1 2 + B 1 2 ≠ 0 A_1^2+B_1^2 \ne0 A 1 2 + B 1 2 = 0 x n x_n x n (3) obtained from one of (2) by rescaling and unfolding?
Here I show that the answer is yes at every order.
Theorem 1.1. Let (3) be globally periodic, that is, suppose there is a p ≥ 1 p \ge1 p ≥ 1 x n + p = x n x_{n+p}=x_n x n + p = x n (3) is one of the following equations, written from the second on with the denominator cleared:
the shift x n + k = x n x_{n+k}=x_n x n + k = x n
x n + k x n = a x_{n+k}x_n=a x n + k x n = a a > 0 a>0 a > 0
k = 2 m k=2m k = 2 m x n + 2 m x n = b x n + m x_{n+2m}x_n=bx_{n+m} x n + 2 m x n = b x n + m b > 0 b>0 b > 0
k = 2 m k=2m k = 2 m x n + 2 m x n = b 2 + b x n + m x_{n+2m}x_n=b^2+bx_{n+m} x n + 2 m x n = b 2 + b x n + m b > 0 b>0 b > 0
k = 3 m k=3m k = 3 m x n + 3 m x n = b 2 + b ( x n + m + x n + 2 m ) x_{n+3m}x_n=b^2+b(x_{n+m}+x_{n+2m}) x n + 3 m x n = b 2 + b ( x n + m + x n + 2 m ) b > 0 b>0 b > 0
These are, respectively, the k k k (2) and the m m m (3) is equivalent to one of the five equations of (2) .
Conversely, each equation in the list is globally periodic, with periods k k k 2 k 2k 2 k 6 m 6m 6 m 5 m 5m 5 m 8 m 8m 8 m (2) ; so Theorem 1.1 is a complete classification.
The requirement of a common period loses nothing: if every orbit in ( 0 , ∞ ) k (0,\infty)^k ( 0 , ∞ ) k [29 ] ; often it can even be linearised globally [7 ] . (Discontinuous maps can be pointwise periodic without being periodic [9 ] .) This remark is not used in the proof.
The question has a long history. Essentially the same question was asked by Graham et al. [ [14 ] ; the generalisation that Graham had in mind for his Problem 1343 in Mathematics Magazine asks for equations of much the same form [22 Remark 1] . A version of the question is posed by Grove and Ladas [ [15 ] ; at order two it is Conjecture 6.5.6 of Kulenővić and Ladas [ [24 ] . After partial results of Kocić et al. [ [22 ] , order two was settled by Barbeau et al. [ [1 ] , and Csörnyei and Laczkovich [ [10 ] settled, even for complex coefficients, orders two and three and the reciprocal equations x n + k x n = c 0 + ∑ j = 1 k − 1 c j x n + j x_{n+k}x_n=c_0+\sum_{j=1}^{k-1}c_jx_{n+j} x n + k x n = c 0 + ∑ j = 1 k − 1 c j x n + j c j c_j c j [6 ] reduced the problem to a reciprocal family with palindromic coefficients, and showed, using a continuum of 2-periodic points at odd order (Section 6), that the answer is yes for k ∈ { 1 , 2 , 3 , 4 , 5 , 7 , 9 , 11 } k \in\{1,2,3,4,5,7,9,11\} k ∈ { 1 , 2 , 3 , 4 , 5 , 7 , 9 , 11 } [8 ] . Over the complex numbers and with arbitrary coefficients, periodic linear fractional recurrences have been studied from the point of view of birational geometry by Bedford and Kim [ [2 , 4 ] at orders 2 and 3, and Bedford and Kim [ [3 ] show that for every k k k k k k 4 k 4k 4 k
The proof has two quite different halves. The first is analytic and short. A surjectivity argument, followed by a Lyapunov function and the derivative at a fixed point, shows that a globally periodic equation is either the shift or of the reciprocal form
x n + k x n = c 0 + ∑ j ∈ S c j x n + j , c 0 ≥ 0 , c j > 0 for j ∈ S ⊂ { 1 , … , k − 1 } , (4) x_{n+k}x_n=c_0+\sum_{j\in S}c_jx_{n+j},\qquad c_0\geq0,\quad c_j>0\ \text{for }j\in S\subset\{1,\ldots,k-1\}, \tag*{(4)} x n + k x n = c 0 + j ∈ S ∑ c j x n + j , c 0 ≥ 0 , c j > 0 for j ∈ S ⊂ { 1 , … , k − 1 } , (4) and linearising at the interior equilibrium shows that the coefficients are palindromic, c j = c k − j c_j=c_{k-j} c j = c k − j θ = 1 \theta=1 θ = 1 c 0 > 0 c_0>0 c 0 > 0 θ = 0 \theta=0 θ = 0 c 0 = 0 c_0=0 c 0 = 0 support S S S gcd ( S ∪ { k } ) = 1 \gcd(S\cup\{k\})=1 g cd( S ∪ { k }) = 1 S S S c 0 = 0 c_0=0 c 0 = 0 λ k + 1 − ϵ ( λ j + λ k − j ) \lambda^k+1-\epsilon(\lambda^j+\lambda^{k-j}) λ k + 1 − ϵ ( λ j + λ k − j )
The second half, which is most of the paper, deals with supports of three or more elements, and it is combinatorial. The key observation is that global periodicity survives degeneration. If the initial data are taken to be powers δ e i \delta^{e_i} δ e i δ \delta δ θ \theta θ x n ≈ δ e n x_n\approx\delta^{e_n} x n ≈ δ e n x n + 2 x n = 1 + x n + 1 x_{n+2}x_n=1+x_{n+1} x n + 2 x n = 1 + x n + 1 e n + 2 + e n = min ( 0 , e n + 1 ) e_{n+2}+e_n=\min(0,e_{n+1}) e n + 2 + e n = min ( 0 , e n + 1 ) δ \delta δ c 0 > 0 c_0>0 c 0 > 0 [32 ] , now called ultradiscretisation and familiar from integrable systems [16 , 17 , 30 ] ; for the full support it was already used by Csörnyei and Laczkovich [ [10 ] . Tropical versions also appear in the theory of cluster algebras, where Fomin and Zelevinsky [ [11 ] use them to prove the periodicity of Zamolodchikov’s Y Y Y A 2 A_2 A 2
For larger supports, global periodicity of an equation is refuted, for every choice of positive coefficients at once, by exhibiting an unbounded solution of the tangent recursion , the linearisation of the min-plus recursion along a periodic solution. The idea is to perturb a periodic orbit of the min-plus recursion slightly. Only the terms that attain the minimum see the perturbation, so it follows a recursion of the same kind, and if the perturbation can grow without bound, the min-plus recursion, and with it the equation, cannot be periodic. For the supports with at least three elements, the periodic orbits used have, except at order 6, a minimum that is always negative, so that the constant term never attains it, and both values of θ \theta θ
I construct the unbounded tangent solutions explicitly, case by case (Table 1). Apart from the full support and most supports at odd order, which have tangent solutions of their own, the workhorse is the σ \sigma σ k k k 2 k 2k 2 k Definition 3.11 ). The hardest case is even order with every element of S S S { y : ν y m o d 2 k ∈ [ 0 , k ) } \{y:\nu y\bmod2k\in[0,k)\} { y : ν y mod 2 k ∈ [ 0 , k )} ν \nu ν 2 k 2k 2 k 2 k 2k 2 k
Several steps finish with a finite computation, for instance at small orders where the general arguments do not yet apply; each is stated precisely in the text and listed in Table 1.
Every finite computation is carried out in exact integer arithmetic, and the whole proof, including these computations, has been formalised and kernel-checked in the Lean proof assistant (Appendix § 14 ).
As far as I know, the following ingredients are new: the tangent recursion and the two-scale pinning lemma (Lemma 3.5 ), which turn an unbounded solution of the linearised min-plus recursion into a proof that the equation is not periodic; the use of such tangent solutions to refute global periodicity for every choice of coefficients at once; the σ \sigma σ Theorem 9.1 ). Ultradiscretisation itself is not new, nor is the use of the leading coefficients at orders two and three, which goes back to Csörnyei and Laczkovich [ [10 ] . I am not aware of an earlier use of the derivative of a min-plus map along a periodic orbit to show that the map is not periodic (“linearised” means something different in the ultradiscrete literature, namely straight-line motion on a tropical Jacobian). A by-product is a purely tropical statement: for every palindromic support with at least three elements, and either θ \theta θ
Section 2 reduces the problem to primitive palindromic supports of the reciprocal form and Section 3 develops the min-plus machinery and the σ \sigma σ Theorem 9.1 and the bounds on Jacobsthal’s function, the most mechanical case checks, and a description of the formalisation. A reader who wants only the shape of the argument can read Sections 2 and 3, the statements of the main theorems of the later sections, and Section 10.
Reduction to the reciprocal family
Here I show that a globally periodic equation (3) is the shift or of the reciprocal form (8) with palindromic coefficients, and that it is enough to treat supports with gcd ( S ∪ { k } ) = 1 \gcd(S \cup\{k\}) = 1 g cd( S ∪ { k }) = 1
Most of this is known. The reduction to the reciprocal form and palindromy are the Reduction Theorem (Theorem 2.4) of Cima et al. [ [6 ] , and Csörnyei and Laczkovich [ [10 ] proved ∣ c j ∣ = ∣ c k − j ∣ |c_j| = |c_{k-j}| ∣ c j ∣ = ∣ c k − j ∣ [25 ] ; it was used by Kocić et al. [ [22 ] , Csörnyei and Laczkovich [ [10 ] and, with affine normal forms at order two, Rubió-Massagué and Mañosa [2007].
The state map
Throughout, the coefficients of (3) are nonnegative, with A 0 + ∑ i = 1 k A i > 0 A_0+\sum_{i=1}^{k} A_i>0 A 0 + ∑ i = 1 k A i > 0 B 0 + ∑ i = 1 k B i > 0 B_0+\sum_{i=1}^{k} B_i>0 B 0 + ∑ i = 1 k B i > 0 A i + 1 A_{i+1} A i + 1 B i + 1 B_{i+1} B i + 1 x n + i x_{n+i} x n + i 0 ≤ i ≤ k − 1 0\leq i\leq k-1 0 ≤ i ≤ k − 1 A 1 A_1 A 1 B 1 B_1 B 1 x n x_n x n Ω = ( 0 , ∞ ) k \Omega=(0,\infty)^k Ω = ( 0 , ∞ ) k u = ( u 1 , … , u k ) ∈ Ω u=(u_1,\ldots,u_k)\in\Omega u = ( u 1 , … , u k ) ∈ Ω
g ( u ) = A 0 + ∑ i = 1 k A i u i B 0 + ∑ i = 1 k B i u i , F ( u ) = ( u 2 , … , u k , g ( u ) ) . (5) g(u)=\frac{A_0+\sum_{i=1}^{k}A_i u_i}{B_0+\sum_{i=1}^{k}B_i u_i},\qquad F(u)=(u_2,\ldots,u_k,g(u)). \tag*{(5)} g ( u ) = B 0 + ∑ i = 1 k B i u i A 0 + ∑ i = 1 k A i u i , F ( u ) = ( u 2 , … , u k , g ( u )) . (5) For u ∈ Ω u\in\Omega u ∈ Ω g ( u ) g(u) g ( u ) F F F Ω \Omega Ω ( x 0 , … , x k − 1 ) (x_0,\ldots,x_{k-1}) ( x 0 , … , x k − 1 ) (3) , and the window ( x n , … , x n + k − 1 ) (x_n,\ldots,x_{n+k-1}) ( x n , … , x n + k − 1 ) F n ( x 0 , … , x k − 1 ) F^n(x_0,\ldots,x_{k-1}) F n ( x 0 , … , x k − 1 ) globally p p p p ≥ 1 p\geq1 p ≥ 1 x n + p = x n x_{n+p}=x_n x n + p = x n n n n F p = id F^p=\operatorname{id} F p = id
Table 1. The cases of the proof, the mechanism and results each uses, and the steps that rest on a finite computation (“none” means that the case is proved by hand for all parameters). Every case uses the reduction of the first row. Except in the rows for ∣ S ∣ ≤ 2 |S| \le2 ∣ S ∣ ≤ 2 ∣ S ∣ ≥ 3 |S| \ge3 ∣ S ∣ ≥ 3 S S S θ \theta θ k k k X X X S S S L = { s ∈ S : 2 s < k } L = \{s \in S : 2s < k\} L = { s ∈ S : 2 s < k }
on Ω \Omega Ω F F F Ω \Omega Ω Theorem 2.1 , uses only this surjectivity; the second, Theorem 2.2 , uses periodicity itself.
The dichotomy
The idea is to ask for a preimage of a single one-parameter family of states, ( 1 , … , 1 , t ) (1,\ldots,1,t) ( 1 , … , 1 , t ) t > 0 t>0 t > 0
Theorem 2.1 (dichotomy). Suppose F ( Ω ) = Ω F(\Omega)=\Omega F ( Ω ) = Ω Then one of the following holds:
(a) A 1 = 0 A_1=0 A 1 = 0 B 0 = 0 B_0=0 B 0 = 0 B i = 0 B_i=0 B i = 0 i ≥ 2 i\ge2 i ≥ 2 B 1 > 0 B_1>0 B 1 > 0 (3) reads B 1 x n + k x n = A 0 + A 2 x n + 1 + ⋯ + A k x n + k − 1 B_1x_{n+k}x_n=A_0+A_2x_{n+1}+\cdots+A_kx_{n+k-1} B 1 x n + k x n = A 0 + A 2 x n + 1 + ⋯ + A k x n + k − 1
(b) B 1 = 0 B_1=0 B 1 = 0 A 0 = 0 A_0=0 A 0 = 0 A i = 0 A_i=0 A i = 0 i ≥ 2 i\ge2 i ≥ 2 A 1 > 0 A_1>0 A 1 > 0 (3) reads x n + k = A 1 x n B 0 + B 2 x n + 1 + ⋯ + B k x n + k − 1 x_{n+k}=\frac{A_1x_n}{B_0+B_2x_{n+1}+\cdots+B_kx_{n+k-1}} x n + k = B 0 + B 2 x n + 1 + ⋯ + B k x n + k − 1 A 1 x n
Proof. Fix t > 0 t>0 t > 0 ( 1 , … , 1 , t ) (1,\ldots,1,t) ( 1 , … , 1 , t ) F F F ( u , 1 , … , 1 ) (u,1,\ldots,1) ( u , 1 , … , 1 ) u > 0 u>0 u > 0 g ( u , 1 , … , 1 ) = t g(u,1,\ldots,1)=t g ( u , 1 , … , 1 ) = t A ′ = ∑ i ≥ 2 A i A'=\sum_{i\ge2}A_i A ′ = ∑ i ≥ 2 A i B ′ = ∑ i ≥ 2 B i B'=\sum_{i\ge2}B_i B ′ = ∑ i ≥ 2 B i k = 1 k=1 k = 1 t > 0 t>0 t > 0 u > 0 u>0 u > 0
t ( B 0 + B 1 u + B ′ ) = A 0 + A 1 u + A ′ . (6) t(B_0+B_1u+B')=A_0+A_1u+A'. \tag*{(6)} t ( B 0 + B 1 u + B ′ ) = A 0 + A 1 u + A ′ . (6) There are four cases, according to the signs of A 1 A_1 A 1 B 1 B_1 B 1
If A 1 = B 1 = 0 A_1=B_1=0 A 1 = B 1 = 0 u u u (6) , and taking t = 1 t=1 t = 1 t = 2 t=2 t = 2 B 0 + B ′ = A 0 + A ′ = 2 ( B 0 + B ′ ) B_0+B'=A_0+A'=2(B_0+B') B 0 + B ′ = A 0 + A ′ = 2 ( B 0 + B ′ ) B 0 + B ′ = 0 B_0+B'=0 B 0 + B ′ = 0 B i B_i B i
If A 1 = 0 < B 1 A_1=0<B_1 A 1 = 0 < B 1 t B 1 u = A 0 + A ′ − t ( B 0 + B ′ ) tB_1u=A_0+A'-t(B_0+B') t B 1 u = A 0 + A ′ − t ( B 0 + B ′ ) u > 0 u>0 u > 0 t ( B 0 + B ′ ) < A 0 + A ′ t(B_0+B')<A_0+A' t ( B 0 + B ′ ) < A 0 + A ′ t > 0 t>0 t > 0 t → ∞ t\to\infty t → ∞ B 0 + B ′ ≤ 0 B_0+B'\le0 B 0 + B ′ ≤ 0 B 0 = 0 B_0=0 B 0 = 0 B i = 0 B_i=0 B i = 0 i ≥ 2 i\ge2 i ≥ 2
If B 1 = 0 < A 1 B_1=0<A_1 B 1 = 0 < A 1 A 1 u = t ( B 0 + B ′ ) − A 0 − A ′ A_1u=t(B_0+B')-A_0-A' A 1 u = t ( B 0 + B ′ ) − A 0 − A ′ u > 0 u>0 u > 0 A 0 + A ′ < t ( B 0 + B ′ ) A_0+A'<t(B_0+B') A 0 + A ′ < t ( B 0 + B ′ ) t > 0 t>0 t > 0 t → 0 + t\to0^+ t → 0 + A 0 + A ′ ≤ 0 A_0+A'\le0 A 0 + A ′ ≤ 0 A 0 = 0 A_0=0 A 0 = 0 A i = 0 A_i=0 A i = 0 i ≥ 2 i\ge2 i ≥ 2
Finally, suppose A 1 > 0 A_1>0 A 1 > 0 B 1 > 0 B_1>0 B 1 > 0 λ = A 1 / B 1 \lambda=A_1/B_1 λ = A 1 / B 1 t = λ t=\lambda t = λ u u u (6) , leaving λ ( B 0 + B ′ ) = A 0 + A ′ \lambda(B_0+B')=A_0+A' λ ( B 0 + B ′ ) = A 0 + A ′ (6) at t = λ + 1 t=\lambda+1 t = λ + 1 ( B 0 + B ′ ) + B 1 u = 0 (B_0+B')+B_1u=0 ( B 0 + B ′ ) + B 1 u = 0 u > 0 u>0 u > 0
In case (a) the denominator is B 1 x n B_1x_n B 1 x n B 1 > 0 B_1>0 B 1 > 0 A 1 x n A_1x_n A 1 x n A 1 > 0 A_1>0 A 1 > 0
Branch (b) is the shift
Genuine periodicity, rather than surjectivity, now removes the rest of branch (b). Two simple tools are used. The first is a Lyapunov function: the coordinate sum Σ ( u ) = ∑ i u i \Sigma(u)=\sum_i u_i Σ ( u ) = ∑ i u i
Σ ( F ( u ) ) − Σ ( u ) = g ( u ) − u 1 , (7) \Sigma(F(u))-\Sigma(u)=g(u)-u_1, \tag*{(7)} Σ ( F ( u )) − Σ ( u ) = g ( u ) − u 1 , (7) so if g ( u ) < u 1 g(u)<u_1 g ( u ) < u 1 u ∈ Ω u\in\Omega u ∈ Ω Σ ( F p ( u ) ) < Σ ( u ) \Sigma(F^p(u))<\Sigma(u) Σ ( F p ( u )) < Σ ( u ) p ≥ 1 p\ge1 p ≥ 1 F p = id F^p=\operatorname{id} F p = id g ( u ) > u 1 g(u)>u_1 g ( u ) > u 1 F p = id F^p=\operatorname{id} F p = id F ( u ∗ ) = u ∗ F(u^*)=u^* F ( u ∗ ) = u ∗ D F ( u ∗ ) p = I DF(u^*)^p=I D F ( u ∗ ) p = I
Theorem 2.2 (branch (b) is the shift). A globally periodic equation of type (b) has A 1 = B 0 A_1=B_0 A 1 = B 0 and B i = 0 B_i=0 B i = 0 for all $i\ge 2; that is, it is the shift x n + k = x n x_{n+k}=x_n x n + k = x n
Proof. Let the equation be globally p p p F p = id F^p=\operatorname{id} F p = id Ω \Omega Ω g ( u ) = A 1 u 1 / ( B 0 + ∑ i ≥ 2 B i u i ) g(u)=A_1u_1/(B_0+\sum_{i\ge2}B_i u_i) g ( u ) = A 1 u 1 / ( B 0 + ∑ i ≥ 2 B i u i ) β = ∑ i ≥ 2 B i \beta=\sum_{i\ge2}B_i β = ∑ i ≥ 2 B i
Suppose first that β = 0 \beta=0 β = 0 B 0 > 0 B_0>0 B 0 > 0 g ( u ) = ( A 1 / B 0 ) u 1 g(u)=(A_1/B_0)u_1 g ( u ) = ( A 1 / B 0 ) u 1 A 1 < B 0 A_1<B_0 A 1 < B 0 g ( u ) < u 1 g(u)<u_1 g ( u ) < u 1 Ω \Omega Ω A 1 > B 0 A_1>B_0 A 1 > B 0 g ( u ) > u 1 g(u)>u_1 g ( u ) > u 1 (7) . So A 1 = B 0 A_1=B_0 A 1 = B 0 β > 0 \beta> 0 β > 0 A 1 ≤ B 0 A_1 \le B_0 A 1 ≤ B 0 u ∈ Ω u \in\Omega u ∈ Ω B 0 ≥ A 1 B_0 \ge A_1 B 0 ≥ A 1 g ( u ) < u 1 g(u) < u_1 g ( u ) < u 1 Ω \Omega Ω
There remains the case β > 0 \beta> 0 β > 0 A 1 > B 0 A_1 > B_0 A 1 > B 0 x ∗ = ( A 1 − B 0 ) / β > 0 x^* = (A_1-B_0)/\beta> 0 x ∗ = ( A 1 − B 0 ) / β > 0 B 0 + β x ∗ = A 1 B_0+\beta x^* = A_1 B 0 + β x ∗ = A 1 u ∗ = ( x ∗ , … , x ∗ ) u^*=(x^*,\ldots,x^*) u ∗ = ( x ∗ , … , x ∗ ) F F F F F F Ω \Omega Ω F p = id F^p=\operatorname{id} F p = id D F ( u ∗ ) p = D ( F p ) ( u ∗ ) = I DF(u^*)^p=D(F^p)(u^*)=I D F ( u ∗ ) p = D ( F p ) ( u ∗ ) = I u ∗ u^* u ∗ g g g A 1 A_1 A 1 ∂ g / ∂ u 1 = 1 \partial g/\partial u_1=1 ∂ g / ∂ u 1 = 1 ∂ g / ∂ u i = − x ∗ B i / A 1 \partial g/\partial u_i=-x^*B_i/A_1 ∂ g / ∂ u i = − x ∗ B i / A 1 i ≥ 2 i\ge2 i ≥ 2 d i = B i x ∗ / A 1 ≥ 0 d_i=B_i x^*/A_1\ge0 d i = B i x ∗ / A 1 ≥ 0
D F ( u ∗ ) w = ( w 2 , … , w k , w 1 − ∑ i ≥ 2 d i w i ) . DF(u^*)w=(w_2,\ldots,w_k,w_1-\sum_{i\ge2}d_iw_i). D F ( u ∗ ) w = ( w 2 , … , w k , w 1 − i ≥ 2 ∑ d i w i ) . Put ϕ ( λ ) = λ k + ∑ i ≥ 2 d i λ i − 1 \phi(\lambda)=\lambda^k+\sum_{i\ge2}d_i\lambda^{i-1} ϕ ( λ ) = λ k + ∑ i ≥ 2 d i λ i − 1 ϕ ( 0 ) = 0 \phi(0)=0 ϕ ( 0 ) = 0 ϕ ( 1 ) = 1 + β x ∗ / A 1 > 1 \phi(1)=1+\beta x^*/A_1>1 ϕ ( 1 ) = 1 + β x ∗ / A 1 > 1 λ ∈ ( 0 , 1 ) \lambda\in(0,1) λ ∈ ( 0 , 1 ) ϕ ( λ ) = 1 \phi(\lambda)=1 ϕ ( λ ) = 1 v = ( 1 , λ , … , λ k − 1 ) v=(1,\lambda,\ldots,\lambda^{k-1}) v = ( 1 , λ , … , λ k − 1 ) D F ( u ∗ ) v = λ v DF(u^*)v=\lambda v D F ( u ∗ ) v = λ v D F ( u ∗ ) v DF(u^*)v D F ( u ∗ ) v 1 − ∑ i ≥ 2 d i λ i − 1 = λ k 1-\sum_{i\ge2}d_i\lambda^{i-1}=\lambda^k 1 − ∑ i ≥ 2 d i λ i − 1 = λ k v = D F ( u ∗ ) p v = λ p v v=DF(u^*)^pv=\lambda^pv v = D F ( u ∗ ) p v = λ p v λ p = 1 \lambda^p=1 λ p = 1 0 < λ < 1 0<\lambda<1 0 < λ < 1
The reciprocal form and its normalisation
Theorem 2.3 (reduction). Every globally periodic equation of the form (3) is either the shift x n + k = x n x_{n+k}=x_n x n + k = x n (that is, A 1 = B 0 > 0 A_1=B_0>0 A 1 = B 0 > 0 A 1 = B 0 = 0 A_1=B_0=0 A 1 = B 0 = 0 B i = 0 B_i=0 B i = 0 i ≥ 2 i\ge2 i ≥ 2 B 1 > 0 B_1>0 B 1 > 0 B 1 B_1 B 1
x n + k x n = c 0 + ∑ j ∈ S c j x n + j , c j > 0 for j ∈ S ⊆ { 1 , … , k − 1 } , c 0 ≥ 0 , (8) x_{n+k}x_n=c_0+\sum_{j\in S}c_jx_{n+j},\qquad c_j>0\ \text{for }j\in S\subseteq\{1,\ldots,k-1\},\qquad c_0\ge0, \tag*{(8)} x n + k x n = c 0 + j ∈ S ∑ c j x n + j , c j > 0 for j ∈ S ⊆ { 1 , … , k − 1 } , c 0 ≥ 0 , (8) where
c 0 = A 0 B 1 , c j = A j + 1 B 1 ( 1 ≤ j ≤ k − 1 ) , S = { j ∈ { 1 , … , k − 1 } : A j + 1 > 0 } , (9) c_0=\frac{A_0}{B_1},\qquad c_j=\frac{A_{j+1}}{B_1}\quad(1\le j\le k-1),\qquad S=\{j\in\{1,\ldots,k-1\}:A_{j+1}>0\}, \tag*{(9)} c 0 = B 1 A 0 , c j = B 1 A j + 1 ( 1 ≤ j ≤ k − 1 ) , S = { j ∈ { 1 , … , k − 1 } : A j + 1 > 0 } , (9) and S ≠ ∅ S\ne\varnothing S = ∅ c 0 > 0 c_0>0 c 0 > 0
Proof. A globally periodic F F F Ω \Omega Ω Theorem 2.1 applies. In case (b), Theorem 2.2 gives the shift. In case (a) the equation is B 1 x n + k x n = A 0 + ∑ j = 1 k − 1 A j + 1 x n + j B_1x_{n+k}x_n=A_0+\sum_{j=1}^{k-1}A_{j+1}x_{n+j} B 1 x n + k x n = A 0 + ∑ j = 1 k − 1 A j + 1 x n + j B 1 > 0 B_1>0 B 1 > 0 (8) with the data (9) , after dropping the terms with A j + 1 = 0 A_{j+1}=0 A j + 1 = 0 (3) is not identically zero (recall A 1 = 0 A_1=0 A 1 = 0
From now on an equation of the form (8) is described by its normalised data : the order k k k S S S c 0 c_0 c 0 c j c_j c j j ∈ S j\in S j ∈ S c j = 0 c_j=0 c j = 0 j ∈ { 1 , … , k − 1 } ∖ S j\in\{1,\ldots,k-1\}\setminus S j ∈ { 1 , … , k − 1 } ∖ S θ ∈ { 0 , 1 } \theta\in\{0,1\} θ ∈ { 0 , 1 } c 0 > 0 c_0>0 c 0 > 0 (9) is a division of numerator and denominator by B 1 B_1 B 1 k = 1 k=1 k = 1 (8) is x n + 1 x n = c 0 x_{n+1}x_n=c_0 x n + 1 x n = c 0
The linearisation
The reciprocal family always has an interior equilibrium. Write s = ∑ j ∈ S c j s=\sum_{j\in S}c_j s = ∑ j ∈ S c j s + c 0 > 0 s+c_0>0 s + c 0 > 0 x 2 = c 0 + s x x^2=c_0+sx x 2 = c 0 + s x
x ∗ = 1 2 ( s + s 2 + 4 c 0 ) , (10) x^*=\frac{1}{2}\left(s+\sqrt{s^2+4c_0}\right), \tag*{(10)} x ∗ = 2 1 ( s + s 2 + 4 c 0 ) , (10) and the constant sequence x n = x ∗ x_n=x^* x n = x ∗ (8) . Put
ϵ j = c j x ∗ ( 1 ≤ j ≤ k − 1 ) , so that ∑ j = 1 k − 1 ϵ j = s x ∗ = 1 − c 0 x ∗ 2 ≤ 1 , (11) \epsilon_j=\frac{c_j}{x^*}\quad(1\le j\le k-1),\qquad\text{so that}\qquad\sum_{j=1}^{k-1}\epsilon_j=\frac{s}{x^*}=1-\frac{c_0}{x^{*2}}\le1, \tag*{(11)} ϵ j = x ∗ c j ( 1 ≤ j ≤ k − 1 ) , so that j = 1 ∑ k − 1 ϵ j = x ∗ s = 1 − x ∗ 2 c 0 ≤ 1 , (11) with equality exactly when c 0 = 0 c_0 = 0 c 0 = 0 linearisation of (8) at x ∗ x^* x ∗
y n + k + y n = ∑ j = 1 k − 1 ϵ j y n + j (12) y_{n+k} + y_n = \sum_{j=1}^{k-1} \epsilon_j y_{n+j} \tag*{(12)} y n + k + y n = j = 1 ∑ k − 1 ϵ j y n + j (12) whose characteristic polynomial is
χ ( λ ) = λ k + 1 − ∑ j = 1 k − 1 ϵ j λ j . (13) \chi(\lambda) = \lambda^k + 1 - \sum_{j=1}^{k-1} \epsilon_j \lambda^j. \tag*{(13)} χ ( λ ) = λ k + 1 − j = 1 ∑ k − 1 ϵ j λ j . (13) Every initial window ( y 0 , … , y k − 1 ) (y_0,\ldots,y_{k-1}) ( y 0 , … , y k − 1 ) (12) , and the window advances by the companion matrix Γ \Gamma Γ
Γ ( w 1 , … , w k ) = ( w 2 , … , w k , − w 1 + ∑ j = 1 k − 1 ϵ j w j + 1 ) . \Gamma(w_1,\ldots,w_k) = \left(w_2,\ldots,w_k,-w_1+\sum_{j=1}^{k-1}\epsilon_j w_{j+1}\right). Γ ( w 1 , … , w k ) = ( w 2 , … , w k , − w 1 + j = 1 ∑ k − 1 ϵ j w j + 1 ) . Theorem 2.4 (linearisation). Suppose (8) is globally p p p
(i) Γ p = I \Gamma^p = I Γ p = I (12) is p p p
(ii) every complex root λ \lambda λ χ \chi χ λ p = 1 \lambda^p = 1 λ p = 1
Proof. (i) By Theorem 2.3 the state map of (8) is F ( u ) = ( u 2 , … , u k , g ( u ) ) F(u) = (u_2,\ldots,u_k,g(u)) F ( u ) = ( u 2 , … , u k , g ( u )) g ( u ) = ( c 0 + ∑ j c j u j + 1 ) / u 1 g(u) = (c_0+\sum_j c_j u_{j+1})/u_1 g ( u ) = ( c 0 + ∑ j c j u j + 1 ) / u 1 Ω \Omega Ω u ∗ = ( x ∗ , … , x ∗ ) u^* = (x^*,\ldots,x^*) u ∗ = ( x ∗ , … , x ∗ ) ( x ∗ ) 2 = c 0 + ∑ j c j x ∗ (x^*)^2 = c_0+\sum_j c_jx^* ( x ∗ ) 2 = c 0 + ∑ j c j x ∗ g g g u ∗ u^* u ∗ ∂ g / ∂ u 1 = − ( x ∗ ) 2 / ( x ∗ ) 2 = − 1 \partial g/\partial u_1 = -(x^*)^2/(x^*)^2 = -1 ∂ g / ∂ u 1 = − ( x ∗ ) 2 / ( x ∗ ) 2 = − 1 ∂ g / ∂ u j + 1 = c j / x ∗ = ϵ j \partial g/\partial u_{j+1} = c_j/x^* = \epsilon_j ∂ g / ∂ u j + 1 = c j / x ∗ = ϵ j D F ( u ∗ ) = Γ DF(u^*) = \Gamma D F ( u ∗ ) = Γ F p = id F^p = \operatorname{id} F p = id Ω \Omega Ω Γ p = D ( F p ) ( u ∗ ) = I \Gamma^p = D(F^p)(u^*) = I Γ p = D ( F p ) ( u ∗ ) = I
(ii) If χ ( λ ) = 0 \chi(\lambda) = 0 χ ( λ ) = 0 v = ( 1 , λ , … , λ k − 1 ) v = (1,\lambda,\ldots,\lambda^{k-1}) v = ( 1 , λ , … , λ k − 1 ) Γ v = λ v \Gamma v = \lambda v Γ v = λ v Γ v \Gamma v Γ v − 1 + ∑ j ϵ j λ j = λ k -1+\sum_j\epsilon_j\lambda^j = \lambda^k − 1 + ∑ j ϵ j λ j = λ k v = Γ p v = λ p v v = \Gamma^p v = \lambda^p v v = Γ p v = λ p v λ p = 1 \lambda^p = 1 λ p = 1
Palindromy
The first structural consequence of Theorem 2.4 is that the coefficients are palindromic. The reason is that a linear map of finite order preserves a positive definite quadratic form, so its inverse is its adjoint. Here the form can be written down explicitly: on p p p period sum
⟨ u , v ⟩ = ∑ n = 0 p − 1 u n v n \langle u,v\rangle= \sum_{n=0}^{p-1} u_n v_n ⟨ u , v ⟩ = n = 0 ∑ p − 1 u n v n is positive definite, and it is unchanged by shifting both sequences, ∑ n < p u n + a v n + a = ⟨ u , v ⟩ \sum_{n<p}u_{n+a}v_{n+a}=\langle u,v\rangle ∑ n < p u n + a v n + a = ⟨ u , v ⟩ n ↦ u n v n n\mapsto u_nv_n n ↦ u n v n p p p (12) must therefore be the same when it is run backwards, and that is palindromy.
Theorem 2.5 (palindromy). If (8) is globally periodic, then c j = c k − j c_j=c_{k-j} c j = c k − j for 1 ≤ j ≤ k − 1 1\leq j\leq k-1 1 ≤ j ≤ k − 1 In particular the support is palindromic, S = k − S S=k-S S = k − S in the notation of (3) , A i = A k + 2 − i A_i=A_{k+2-i} A i = A k + 2 − i for 2 ≤ i ≤ k 2\leq i\leq k 2 ≤ i ≤ k
Proof. Let the equation be globally p p p (12) is p p p v v v
ρ n = v n + v n + k − ∑ j = 1 k − 1 ϵ j v n + k − j . \rho_n = v_n + v_{n+k} - \sum_{j=1}^{k-1}\epsilon_j v_{n+k-j}. ρ n = v n + v n + k − j = 1 ∑ k − 1 ϵ j v n + k − j . Let u u u ∑ n < p u n ρ n \sum_{n<p}u_n\rho_n ∑ n < p u n ρ n k k k j j j
∑ n < p u n v n = ∑ n < p u n + k v n + k , ∑ n < p u n v n + k − j = ∑ n < p u n + j v n + k , \sum_{n<p}u_nv_n=\sum_{n<p}u_{n+k}v_{n+k},\qquad\sum_{n<p}u_nv_{n+k-j}=\sum_{n<p}u_{n+j}v_{n+k}, n < p ∑ u n v n = n < p ∑ u n + k v n + k , n < p ∑ u n v n + k − j = n < p ∑ u n + j v n + k , so that
∑ n < p u n ρ n = ∑ n < p u n + k ( u n + k + u n − ∑ j = 1 k − 1 ϵ j u n + j ) = 0 , \sum_{n<p}u_n\rho_n=\sum_{n<p}u_{n+k}\left(u_{n+k}+u_n-\sum_{j=1}^{k-1}\epsilon_j u_{n+j}\right)=0, n < p ∑ u n ρ n = n < p ∑ u n + k ( u n + k + u n − j = 1 ∑ k − 1 ϵ j u n + j ) = 0 , because u u u (12) . On the other hand, using (12) for v v v v n + v n + k v_n+v_{n+k} v n + v n + k j ↦ k − j j\mapsto k-j j ↦ k − j
ρ n = ∑ j = 1 k − 1 ( ϵ j − ϵ k − j ) v n + j . \rho_n=\sum_{j=1}^{k-1}(\epsilon_j-\epsilon_{k-j})v_{n+j}. ρ n = j = 1 ∑ k − 1 ( ϵ j − ϵ k − j ) v n + j . This is a linear combination of shifts of v v v ρ \rho ρ (12) . Taking u = ρ u=\rho u = ρ ∑ n < p ρ n 2 = 0 \sum_{n<p}\rho_n^2=0 ∑ n < p ρ n 2 = 0 ρ n = 0 \rho_n=0 ρ n = 0 n < p n<p n < p n n n 1 ≤ j ≤ k − 1 1\le j\le k-1 1 ≤ j ≤ k − 1 v v v v j = 1 v_j=1 v j = 1 v i = 0 v_i=0 v i = 0 i < k i<k i < k 0 = ρ 0 = ϵ j − ϵ k − j 0=\rho_0=\epsilon_j-\epsilon_{k-j} 0 = ρ 0 = ϵ j − ϵ k − j x ∗ x^* x ∗ c j = c k − j c_j=c_{k-j} c j = c k − j S S S S = { j : c j > 0 } S=\{j:c_j>0\} S = { j : c j > 0 } A i A_i A i (9) .
Known types and primitive supports
The five equations (2) have a simple description in terms of normalised data. Recall that the ℓ \ell ℓ x n + k ′ = f ( x n , x n + 1 , … , x n + k ′ − 1 ) x_{n+k'}=f(x_n,x_{n+1},\ldots,x_{n+k'-1}) x n + k ′ = f ( x n , x n + 1 , … , x n + k ′ − 1 ) k ′ k' k ′ x n + ℓ k ′ = f ( x n , x n + ℓ , … , x n + ( k ′ − 1 ) ℓ ) x_{n+\ell k'}=f(x_n,x_{n+\ell},\ldots,x_{n+(k'-1)\ell}) x n + ℓ k ′ = f ( x n , x n + ℓ , … , x n + ( k ′ − 1 ) ℓ ) ℓ k ′ \ell k' ℓ k ′ ℓ \ell ℓ m ↦ x m ℓ + r m\mapsto x_{m\ell+r} m ↦ x m ℓ + r 0 ≤ r < ℓ 0\le r<\ell 0 ≤ r < ℓ unfold for the rescaled family y n = α x n y_n=\alpha x_n y n = α x n
Definition 2.6. The normalised data ( k , S , c ) (k,S,c) ( k , S , c ) (8) is of known type if one of the following holds:
(K1) S = ∅ S=\varnothing S = ∅
(K2) k = 2 m k=2m k = 2 m S = { m } S=\{m\} S = { m } c 0 = 0 c_0=0 c 0 = 0 c 0 = c m 2 c_0=c_m^2 c 0 = c m 2
(K3) k = 3 m k=3m k = 3 m S = { m , 2 m } S=\{m,2m\} S = { m , 2 m } c m = c 2 m c_m=c_{2m} c m = c 2 m c 0 = c m 2 c_0=c_m^2 c 0 = c m 2
Lemma 2.7. If the normalised data of an equation of type (a) is of known type, then the equation is equivalent, under a rescaling y n = x n / λ y_n=x_n/\lambda y n = x n / λ λ > 0 \lambda>0 λ > 0 (2) . In terms of the coefficients of (3) , with m m m A i = 0 A_i=0 A i = 0 i ≥ 1 i\ge1 i ≥ 1 A i = 0 A_i=0 A i = 0 i ∉ { 0 , m + 1 } i\notin\{0,m+1\} i ∈ / { 0 , m + 1 } A 0 = 0 A_0=0 A 0 = 0 A 0 B 1 = A m + 1 2 A_0B_1=A_{m+1}^2 A 0 B 1 = A m + 1 2 A i = 0 A_i=0 A i = 0 i ∉ { 0 , m + 1 , 2 m + 1 } i\notin\{0,m+1,2m+1\} i ∈ / { 0 , m + 1 , 2 m + 1 } A m + 1 = A 2 m + 1 A_{m+1}=A_{2m+1} A m + 1 = A 2 m + 1 A 0 B 1 = A m + 1 2 A_0B_1=A_{m+1}^2 A 0 B 1 = A m + 1 2
Proof. If x x x (8) , then y n = x n / λ y_n=x_n/\lambda y n = x n / λ y n + k y n = c 0 / λ 2 + ∑ j ∈ S ( c j / λ ) y n + j y_{n+k}y_n=c_0/\lambda^2+\sum_{j\in S}(c_j/\lambda)y_{n+j} y n + k y n = c 0 / λ 2 + ∑ j ∈ S ( c j / λ ) y n + j c 0 > 0 c_0>0 c 0 > 0 λ = c 0 \lambda=\sqrt{c_0} λ = c 0 y n + k = 1 / y n y_{n+k}=1/y_n y n + k = 1/ y n k k k y n + 1 = 1 / y n y_{n+1}=1/y_n y n + 1 = 1/ y n λ = c m \lambda=c_m λ = c m y n + 2 m y n = y n + m y_{n+2m}y_n=y_{n+m} y n + 2 m y n = y n + m y n + 2 m y n = 1 + y n + m y_{n+2m}y_n=1+y_{n+m} y n + 2 m y n = 1 + y n + m m m m y n + 2 = y n + 1 / y n y_{n+2}=y_{n+1}/y_n y n + 2 = y n + 1 / y n y n + 2 = ( 1 + y n + 1 ) / y n y_{n+2}=(1+y_{n+1})/y_n y n + 2 = ( 1 + y n + 1 ) / y n λ = c m \lambda=c_m λ = c m y n + 3 m y n = 1 + y n + m + y n + 2 m y_{n+3m}y_n=1+y_{n+m}+y_{n+2m} y n + 3 m y n = 1 + y n + m + y n + 2 m m m m y n + 3 = ( 1 + y n + 1 + y n + 2 ) / y n y_{n+3}=(1+y_{n+1}+y_{n+2})/y_n y n + 3 = ( 1 + y n + 1 + y n + 2 ) / y n (9) : c 0 = c m 2 c_0=c_m^2 c 0 = c m 2 A 0 / B 1 = A m + 1 2 / B 1 2 A_0/B_1=A_{m+1}^2/B_1^2 A 0 / B 1 = A m + 1 2 / B 1 2
The last step of the reduction removes common factors from the support. If every element of S S S k k k ℓ > 1 \ell>1 ℓ > 1 ℓ \ell ℓ ℓ \ell ℓ k / ℓ k/\ell k / ℓ [6 ] observe at order four, where they split a solution into its odd and even subsequences. Call S S S primitive (at order k k k gcd ( S ∪ { k } ) = 1 \gcd(S\cup\{k\})=1 g cd( S ∪ { k }) = 1
Lemma 2.8 (folding). Let (8) have normalised data ( k , S , c ) (k,S,c) ( k , S , c ) ℓ = gcd ( S ∪ { k } ) \ell=\gcd(S\cup\{k\}) ℓ = g cd( S ∪ { k }) ℓ = k \ell=k ℓ = k S = ∅ S=\varnothing S = ∅ k ′ = k / ℓ k'=k/\ell k ′ = k / ℓ S ′ = { j / ℓ : j ∈ S } S'=\{j/\ell:j\in S\} S ′ = { j / ℓ : j ∈ S } c 0 ′ = c 0 c'_0=c_0 c 0 ′ = c 0 c j ′ ′ = c j ′ ℓ c'_{j'}=c_{j'\ell} c j ′ ′ = c j ′ ℓ j ′ ∈ S ′ j'\in S' j ′ ∈ S ′
(i) equation (8) is the ℓ \ell ℓ
x n + k ′ x n = c 0 ′ + ∑ j ′ ∈ S ′ c j ′ ′ x n + j ′ (14) x_{n+k'}x_n=c'_0+\sum_{j'\in S'}c'_{j'}x_{n+j'} \tag*{(14)} x n + k ′ x n = c 0 ′ + j ′ ∈ S ′ ∑ c j ′ ′ x n + j ′ (14) of order k ′ k' k ′ θ \theta θ
(ii) if (8) is globally p p p (14) ;
(iii) S ′ S' S ′ k ′ k' k ′ c j = c k − j c_j=c_{k-j} c j = c k − j j j j c j ′ ′ = c k ′ − j ′ ′ c'_{j'}=c'_{k'-j'} c j ′ ′ = c k ′ − j ′ ′ j ′ j' j ′ S ′ S' S ′
(iv) if the data ( k ′ , S ′ , c ′ ) (k',S',c') ( k ′ , S ′ , c ′ ) ( k , S , c ) (k,S,c) ( k , S , c )
Proof. (i) Every element of S S S ℓ \ell ℓ k k k n = m ℓ + r n=m\ell+r n = m ℓ + r 0 ≤ r < ℓ 0\le r<\ell 0 ≤ r < ℓ j = j ′ ℓ j=j'\ell j = j ′ ℓ (8) at n n n
x ( m + k ′ ) ℓ + r x m ℓ + r = c 0 ′ + ∑ j ′ ∈ S ′ c j ′ ′ x ( m + j ′ ) ℓ + r , x_{(m+k')\ell+r}x_{m\ell+r}=c'_0+\sum_{j'\in S'}c'_{j'}x_{(m+j')\ell+r}, x ( m + k ′ ) ℓ + r x m ℓ + r = c 0 ′ + j ′ ∈ S ′ ∑ c j ′ ′ x ( m + j ′ ) ℓ + r , which is (14) at m m m m ↦ x m ℓ + r m\mapsto x_{m\ell+r} m ↦ x m ℓ + r
(ii) Let z z z (14) , and put x n = z ⌊ n / ℓ ⌋ x_n=z_{\lfloor n/\ell\rfloor} x n = z ⌊ n / ℓ ⌋ m ↦ x m ℓ + r m\mapsto x_{m\ell+r} m ↦ x m ℓ + r z z z x x x (8) , and it is therefore p p p p ℓ p\ell p ℓ z m + p = x m ℓ + p ℓ = x m ℓ = z m z_{m+p}=x_{m\ell+p\ell}=x_{m\ell}=z_m z m + p = x m ℓ + p ℓ = x m ℓ = z m m m m
(iii) and (iv) are immediate from the definitions: gcd ( S ′ ∪ { k ′ } ) = gcd ( S ∪ { k } ) / ℓ = 1 \gcd(S'\cup\{k'\})=\gcd(S\cup\{k\})/\ell=1 g cd( S ′ ∪ { k ′ }) = g cd( S ∪ { k }) / ℓ = 1 m m m m ℓ m\ell m ℓ Definition 2.6 . □
Corollary 2.9. Suppose that every globally periodic equation (8) whose support S S S c j = c k − j c_j=c_{k-j} c j = c k − j j j j (3) is either the shift x n + k = x n x_{n+k}=x_n x n + k = x n (2) , with coefficients as in Lemma 2.7.
Proof. Let (3) be globally p p p (8) , which is globally p p p c j = c k − j c_j=c_{k-j} c j = c k − j p p p (14) with primitive support and palindromic coefficients. By hypothesis its data is of known type, so by Lemma 2.8(iv) the data of (8) is of known type, and Lemma 2.7 finishes the proof. □
The remaining sections therefore work with a primitive palindromic support S S S c j = c k − j > 0 c_j=c_{k-j}>0 c j = c k − j > 0 S S S θ \theta θ
The min-plus map and the σ \sigma σ
Throughout this section the equation is (8) , with support S ⊆ { 1 , … , k − 1 } S\subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } θ \theta θ S ≠ ∅ S\ne\varnothing S = ∅ θ = 1 \theta=1 θ = 1 σ \sigma σ σ \sigma σ § 3.5 ) and the mirror lemma (Lemma 3.15), both used in the later sections.
The min-plus recursion and its periodicity
It is convenient to treat the constant term as a slot. Let
S θ = S ∪ { 0 } if θ = 1 , S θ = S if θ = 0 , S_\theta= S \cup\{0\}\ \text{if }\theta=1,\qquad S_\theta=S\ \text{if }\theta=0, S θ = S ∪ { 0 } if θ = 1 , S θ = S if θ = 0 , so that S θ ≠ ∅ S_\theta\ne\varnothing S θ = ∅ ( f n ) n ≥ 0 (f_n)_{n\ge0} ( f n ) n ≥ 0 n ≥ 0 n\ge0 n ≥ 0 j ∈ S θ j\in S_\theta j ∈ S θ
v n ( f , j ) = { 0 , j = 0 , f n + j , j ∈ S , m n ( f ) = min j ∈ S θ v n ( f , j ) . v_n(f,j)=
\begin{cases}
0, & j=0,\\
f_{n+j}, & j\in S,
\end{cases}
\qquad
m_n(f)=\min_{j\in S_\theta}v_n(f,j). v n ( f , j ) = { 0 , f n + j , j = 0 , j ∈ S , m n ( f ) = j ∈ S θ min v n ( f , j ) . Thus v n ( f , j ) v_n(f,j) v n ( f , j ) j j j n n n j ∈ S j\in S j ∈ S taps .
Definition 3.1. An integer sequence ( e n ) n ≥ 0 (e_n)_{n\ge0} ( e n ) n ≥ 0 solution of the min-plus recursion (for k k k S S S θ \theta θ
e n + k + e n = m n ( e ) = min ( { 0 if θ = 1 } ∪ { e n + j : j ∈ S } ) (15) e_{n+k}+e_n=m_n(e)=\min\bigl(\{0\text{ if }\theta=1\}\cup\{e_{n+j}:j\in S\}\bigr) \tag*{(15)} e n + k + e n = m n ( e ) = min ( { 0 if θ = 1 } ∪ { e n + j : j ∈ S } ) (15) for all n ≥ 0 n\ge0 n ≥ 0
Every integer window ( e 0 , … , e k − 1 ) (e_0,\ldots,e_{k-1}) ( e 0 , … , e k − 1 ) min-plus map T : Z k → Z k T:\mathbb{Z}^k\to\mathbb{Z}^k T : Z k → Z k T ( e 0 , … , e k − 1 ) = ( e 1 , … , e k − 1 , m 0 ( e ) − e 0 ) T(e_0,\ldots,e_{k-1})=(e_1,\ldots,e_{k-1},m_0(e)-e_0) T ( e 0 , … , e k − 1 ) = ( e 1 , … , e k − 1 , m 0 ( e ) − e 0 ) attaining sets of a solution e e e
A n = A n ( e ) = { j ∈ S θ : v n ( e , j ) = m n ( e ) } ( n ≥ 0 ) ; A_n=A_n(e)=\{j\in S_\theta:v_n(e,j)=m_n(e)\}\qquad(n\ge0); A n = A n ( e ) = { j ∈ S θ : v n ( e , j ) = m n ( e )} ( n ≥ 0 ) ; ties are allowed, so A n A_n A n
Each term of the orbit of (8) started from x i = δ e i x_i=\delta^{e_i} x i = δ e i C n δ e n C_n\delta^{e_n} C n δ e n δ → 0 + \delta\to0^+ δ → 0 + e n e_n e n C n C_n C n (8) , only the terms whose exponent attains the minimum.
Theorem 3.2 (leading coefficients). Let e e e (15) . For δ > 0 \delta>0 δ > 0 x ( δ ) = ( x n ( δ ) ) x(\delta)=(x_n(\delta)) x ( δ ) = ( x n ( δ )) be the solution of (8) with initial window x i ( δ ) = δ e i x_i(\delta)=\delta^{e_i} x i ( δ ) = δ e i ( 0 ≤ i < k ) (0\le i<k) ( 0 ≤ i < k ) Define C n C_n C n by C n = 1 C_n=1 C n = 1 for n < k n<k n < k and
C n + k C n = c 0 [ 0 ∈ A n ( e ) ] + ∑ j ∈ A n ( e ) ∩ S c j C n + j ( n ≥ 0 ) , (16) C_{n+k}C_n=c_0[0\in A_n(e)]+\sum_{j\in A_n(e)\cap S}c_jC_{n+j}\qquad(n\ge0), \tag*{(16)} C n + k C n = c 0 [ 0 ∈ A n ( e )] + j ∈ A n ( e ) ∩ S ∑ c j C n + j ( n ≥ 0 ) , (16) where [ 0 ∈ A n ( e ) ] [0\in A_n(e)] [ 0 ∈ A n ( e )] is 1 if the constant slot attains the minimum and 0 otherwise. Then C n > 0 C_n>0 C n > 0 and
lim δ → 0 + x n ( δ ) δ − e n = C n for every n ≥ 0. \lim_{\delta\to0^+}x_n(\delta)\delta^{-e_n}=C_n\qquad\text{for every }n\ge0. δ → 0 + lim x n ( δ ) δ − e n = C n for every n ≥ 0. Proof. First, x ( δ ) x(\delta) x ( δ ) (8) is a sum of nonnegative terms, at least one of them positive. Likewise C n > 0 C_n>0 C n > 0 A n ( e ) A_n(e) A n ( e ) (16) is a positive coefficient times a positive number.
The limit is proved by induction on n n n n < k n<k n < k x n ( δ ) δ − e n = 1 x_n(\delta)\delta^{-e_n}=1 x n ( δ ) δ − e n = 1 n , n + 1 , … , n + k − 1 n,n+1,\ldots,n+k-1 n , n + 1 , … , n + k − 1 M = m n ( e ) M=m_n(e) M = m n ( e ) (8) at n n n δ − M \delta^{-M} δ − M
x n + k ( δ ) δ − e n + k ⋅ x n ( δ ) δ − e n = c 0 δ − M + ∑ j ∈ S c j x n + j ( δ ) δ − e n + j ⋅ δ e n + j − M , x_{n+k}(\delta)\delta^{-e_{n+k}}\cdot x_n(\delta)\delta^{-e_n}
=c_0\delta^{-M}+\sum_{j\in S}c_jx_{n+j}(\delta)\delta^{-e_{n+j}}\cdot\delta^{e_{n+j}-M}, x n + k ( δ ) δ − e n + k ⋅ x n ( δ ) δ − e n = c 0 δ − M + j ∈ S ∑ c j x n + j ( δ ) δ − e n + j ⋅ δ e n + j − M , because e n + k + e n = M e_{n+k}+e_n=M e n + k + e n = M j ∈ S j\in S j ∈ S e n + j − M e_{n+j}-M e n + j − M j ∈ A n j\in A_n j ∈ A n c j C n + j c_jC_{n+j} c j C n + j j ∈ A n j\in A_n j ∈ A n θ = 1 \theta=1 θ = 1 M ≤ 0 M\le0 M ≤ 0 c 0 δ − M c_0\delta^{-M} c 0 δ − M c 0 c_0 c 0 M = 0 M=0 M = 0 0 ∈ A n 0\in A_n 0 ∈ A n θ = 0 \theta=0 θ = 0 (16) , and dividing by x n ( δ ) δ − e n → C n > 0 x_n(\delta)\delta^{-e_n}\to C_n>0 x n ( δ ) δ − e n → C n > 0 x n + k ( δ ) δ − e n + k → C n + k x_{n+k}(\delta)\delta^{-e_{n+k}}\to C_{n+k} x n + k ( δ ) δ − e n + k → C n + k
Theorem 3.3 (periodicity of the degeneration). Suppose the equation (8) is globally p p p periodic. Then every integer solution e e e of the min-plus recursion (15) satisfies e n + p = e n e_{n+p}=e_n e n + p = e n for all n ≥ 0 n \ge0 n ≥ 0 equivalently, T p = id T^p=\operatorname{id} T p = id on Z k \mathbb{Z}^k Z k Moreover the leading coefficients of Theorem 3.2 satisfy C n + p = C n C_{n+p}=C_n C n + p = C n for all n ≥ 0 n \ge0 n ≥ 0
Proof. Let x ( δ ) x(\delta) x ( δ ) C n C_n C n x n + p ( δ ) = x n ( δ ) x_{n+p}(\delta)=x_n(\delta) x n + p ( δ ) = x n ( δ ) δ > 0 \delta>0 δ > 0
δ e n + p − e n = x n ( δ ) δ − e n x n + p ( δ ) δ − e n + p ⟶ C n C n + p ( δ → 0 + ) . \delta^{e_{n+p}-e_n}=\frac{x_n(\delta)\,\delta^{-e_n}}{x_{n+p}(\delta)\,\delta^{-e_{n+p}}}\longrightarrow\frac{C_n}{C_{n+p}}\qquad(\delta\to0^+). δ e n + p − e n = x n + p ( δ ) δ − e n + p x n ( δ ) δ − e n ⟶ C n + p C n ( δ → 0 + ) . The limit is positive and finite, whereas δ a \delta^a δ a a > 0 a>0 a > 0 a < 0 a<0 a < 0 e n + p = e n e_{n+p}=e_n e n + p = e n C n + p = C n C_{n+p}=C_n C n + p = C n
The first part of Theorem 3.3 is a condition on ( S , θ ) (S,\theta) ( S , θ ) c j c_j c j [10 ] pass to the same piecewise-linear recursion (with max in place of min), taking initial data of the form x c i x^{c_i} x c i c i c_i c i Q \mathbb{Q} Q [10 ] , and Theorem 3.2 is its general form.
Tangent solutions
To use Theorem 3.3 one needs an integer solution of (15) that is not p p p periodic orbit and look at how nearby orbits move. Since T T T
Definition 3.4. Let e e e (15) , with attaining sets A n = A n ( e ) A_n=A_n(e) A n = A n ( e ) ( d n ) n ≥ 0 (d_n)_{n\ge0} ( d n ) n ≥ 0 tangent solution along e e e
d n + k + d n = min j ∈ A n v n ( d , j ) for all n ≥ 0. (17) d_{n+k}+d_n=\min_{j\in A_n}v_n(d,j)\qquad\text{for all }n\ge0. \tag*{(17)} d n + k + d n = j ∈ A n min v n ( d , j ) for all n ≥ 0. (17) The right-hand side of (17) is the one-sided derivative of the minimum: for small τ > 0 \tau>0 τ > 0
m n ( e + τ d ) = m n ( e ) + τ min j ∈ A n v n ( d , j ) , m_n(e+\tau d)=m_n(e)+\tau\min_{j\in A_n}v_n(d,j), m n ( e + τ d ) = m n ( e ) + τ j ∈ A n min v n ( d , j ) , because slots outside A n A_n A n
Lemma 3.5 (two-scale pinning). Let e e e integer solution of (15) and d d d tangent solution along e e e N ≥ 0 N\ge0 N ≥ 0 D ≥ 0 D\ge0 D ≥ 0 ∣ d n ∣ ≤ D |d_n|\le D ∣ d n ∣ ≤ D n < N n<N n < N κ > 2 D \kappa>2D κ > 2 D g g g (15) with initial window g i = κ e i + d i g_i=\kappa e_i+d_i g i = κ e i + d i 0 ≤ i < k 0\le i<k 0 ≤ i < k
g n = κ e n + d n for all 0 ≤ n < N . g_n=\kappa e_n+d_n\qquad\text{for all }0\le n<N. g n = κ e n + d n for all 0 ≤ n < N . Proof. By induction on n n n n < k n<k n < k g g g k ≤ n ≤ N k\le n\le N k ≤ n ≤ N n = m + k n=m+k n = m + k m , m + 1 , … , m + k − 1 m,m+1,\ldots,m+k-1 m , m + 1 , … , m + k − 1 N N N j ∈ S θ j\in S_\theta j ∈ S θ
v m ( g , j ) = κ v m ( e , j ) + v m ( d , j ) , v_m(g,j)=\kappa v_m(e,j)+v_m(d,j), v m ( g , j ) = κ v m ( e , j ) + v m ( d , j ) , and ∣ v m ( d , j ) ∣ ≤ D |v_m(d,j)|\le D ∣ v m ( d , j ) ∣ ≤ D M = m m ( e ) M=m_m(e) M = m m ( e ) j ∈ A m j\in A_m j ∈ A m κ M + v m ( d , j ) ≤ κ M + D \kappa M+v_m(d,j)\le\kappa M+D κ M + v m ( d , j ) ≤ κ M + D j ∉ A m j\notin A_m j ∈ / A m v m ( e , j ) v_m(e,j) v m ( e , j ) M M M v m ( e , j ) ≥ M + 1 v_m(e,j)\ge M+1 v m ( e , j ) ≥ M + 1 κ M + κ − D > κ M + D \kappa M+\kappa-D>\kappa M+D κ M + κ − D > κ M + D S θ S_\theta S θ A m A_m A m
m m ( g ) = κ M + min j ∈ A m v m ( d , j ) = κ ( e m + k + e m ) + ( d m + k + d m ) , m_m(g)=\kappa M+\min_{j\in A_m}v_m(d,j)=\kappa(e_{m+k}+e_m)+(d_{m+k}+d_m), m m ( g ) = κ M + j ∈ A m min v m ( d , j ) = κ ( e m + k + e m ) + ( d m + k + d m ) , by (15) for e e e (17) for d d d g m + k = m m ( g ) − g m g_{m+k}=m_m(g)-g_m g m + k = m m ( g ) − g m g m = κ e m + d m g_m=\kappa e_m+d_m g m = κ e m + d m g n = κ e n + d n g_n=\kappa e_n+d_n g n = κ e n + d n
Theorem 3.6 (tangent solutions). Let T T T min-plus map for ( S , θ ) (S,\theta) ( S , θ )
(i) If T p = id T^p = \operatorname{id} T p = id Z k \mathbb{Z}^k Z k d d d e e e (15) satisfies d n + p = d n d_{n+p} = d_n d n + p = d n n ≥ 0 n \ge0 n ≥ 0
(ii) Consequently, if some integer solution of (15) carries an unbounded integer tangent solution, then no power of T T T Z k \mathbb{Z}^k Z k (8) with support S S S θ \theta θ
Proof. (i) Fix n 0 ≥ 0 n_0 \ge0 n 0 ≥ 0 N = n 0 + p + 1 N = n_0 + p + 1 N = n 0 + p + 1 D ≥ 0 D \ge0 D ≥ 0 ∣ d n ∣ \lvert d_n\rvert ∣ d n ∣ n < N n < N n < N κ = 2 D + 1 \kappa= 2D + 1 κ = 2 D + 1 g g g κ e i + d i \kappa e_i + d_i κ e i + d i g n = κ e n + d n g_n = \kappa e_n + d_n g n = κ e n + d n n < N n < N n < N T p = id T^p = \operatorname{id} T p = id e e e g g g p p p
d n 0 + p = g n 0 + p − κ e n 0 + p = g n 0 − κ e n 0 = d n 0 . d_{n_0+p} = g_{n_0+p} - \kappa e_{n_0+p} = g_{n_0} - \kappa e_{n_0} = d_{n_0}. d n 0 + p = g n 0 + p − κ e n 0 + p = g n 0 − κ e n 0 = d n 0 . A p p p
(ii) By (i) no power of T T T S S S θ \theta θ p p p T p = id T^p = \operatorname{id} T p = id Z k \mathbb{Z}^k Z k
Definition 3.7. A tangent solution d d d ray if there are an integer q ≥ 1 q \ge1 q ≥ 1 q q q u u u d n + q = d n + u n d_{n+q} = d_n + u_n d n + q = d n + u n n ≥ 0 n \ge0 n ≥ 0
A ray grows linearly, since d i + m q = d i + m u i d_{i+mq} = d_i + m u_i d i + m q = d i + m u i i , m ≥ 0 i,m \ge0 i , m ≥ 0 σ \sigma σ q = 2 k q = 2k q = 2 k § 5 , with q = 8 k q = 8k q = 8 k
Certificates
Theorem 3.6 becomes a finite check when e e e w t ∈ A t w_t \in A_t w t ∈ A t
f t + k + f t = v t ( f , w t ) , (18) f_{t+k} + f_t = v_t(f,w_t), \tag*{(18)} f t + k + f t = v t ( f , w t ) , (18) which is a linear recursion: f f f ( w t ) (w_t) ( w t ) (18) is a tangent solution provided w t w_t w t v t ( f , ⋅ ) v_t(f,\mathord{\cdot}) v t ( f , ⋅ ) A t A_t A t margin condition need only be checked over one period, on two vectors.
Definition 3.8. A tangent certificate for ( k , S , θ ) (k,S,\theta) ( k , S , θ ) q ≥ 1 q \ge1 q ≥ 1 E , V , W ∈ Z k E,V,W \in\mathbb{Z}^k E , V , W ∈ Z k W ≠ 0 W \ne0 W = 0 ( w 0 , … , w q − 1 ) (w_0,\ldots,w_{q-1}) ( w 0 , … , w q − 1 ) S θ S_\theta S θ e e e (15) with initial window E E E d d d u u u (18) for 0 ≤ t < q 0 \le t < q 0 ≤ t < q V V V W W W
for every 0 ≤ t < q 0 \le t < q 0 ≤ t < q w t ∈ A t ( e ) w_t \in A_t(e) w t ∈ A t ( e ) v t ( d , w t ) ≤ v t ( d , j ) v_t(d,w_t) \le v_t(d,j) v t ( d , w t ) ≤ v t ( d , j ) v t ( u , w t ) ≤ v t ( u , j ) v_t(u,w_t) \le v_t(u,j) v t ( u , w t ) ≤ v t ( u , j ) j ∈ A t ( e ) j \in A_t(e) j ∈ A t ( e )
the windows return:
( e q , … , e q + k − 1 ) = E , ( d q , … , d q + k − 1 ) = V + W , ( u q , … , u q + k − 1 ) = W . (e_q,\ldots,e_{q+k-1}) = E,\qquad(d_q,\ldots,d_{q+k-1}) = V + W,\qquad(u_q,\ldots,u_{q+k-1}) = W. ( e q , … , e q + k − 1 ) = E , ( d q , … , d q + k − 1 ) = V + W , ( u q , … , u q + k − 1 ) = W . Checking validity means running q q q q q q
Proposition 3.9. If ( k , S , θ ) (k,S,\theta) ( k , S , θ ) (8) with support S S S θ \theta θ
Proof. Extend the word q q q w t = w t m o d q w_t = w_{t \bmod q} w t = w t mod q d d d u u u t ≥ 0 t \ge0 t ≥ 0 (18) . Since e e e ( e q , … , e q + k − 1 ) = E (e_q,\ldots,e_{q+k-1}) = E ( e q , … , e q + k − 1 ) = E e e e q q q A t + q = A t A_{t+q} = A_t A t + q = A t ( u t + q ) t (u_{t+q})_t ( u t + q ) t (18) (the word is q q q W W W u t + q = u t u_{t+q} = u_t u t + q = u t t t t ( d t + q ) t (d_{t+q})_t ( d t + q ) t ( d t + u t ) t (d_t+u_t)_t ( d t + u t ) t (18) with initial window V + W V+W V + W d t + q = d t + u t d_{t+q} = d_t+u_t d t + q = d t + u t
d t + m q = d t + m u t ( t , m ≥ 0 ) . d_{t+mq} = d_t + m u_t \qquad(t,m \ge0). d t + m q = d t + m u t ( t , m ≥ 0 ) . Now let n = t + m q n = t+mq n = t + m q 0 ≤ t < q 0 \le t < q 0 ≤ t < q A n = A t A_n = A_t A n = A t w n = w t w_n = w_t w n = w t j j j v n ( d , j ) = v t ( d , j ) + m v t ( u , j ) v_n(d,j) = v_t(d,j) + m v_t(u,j) v n ( d , j ) = v t ( d , j ) + m v t ( u , j ) 0 0 0 u u u t t t m ≥ 0 m \ge0 m ≥ 0 d d d v n ( d , w n ) ≤ v n ( d , j ) v_n(d,w_n) \le v_n(d,j) v n ( d , w n ) ≤ v n ( d , j ) j ∈ A n j \in A_n j ∈ A n w n ∈ A n w_n \in A_n w n ∈ A n (18) this says that d d d (17) : it is a tangent solution along e e e
Finally, W ≠ 0 W \ne0 W = 0 u i ≠ 0 u_i \ne0 u i = 0 i < k i < k i < k d i + m q = d i + m u i d_{i+mq} = d_i + m u_i d i + m q = d i + m u i m m m Theorem 3.6 completes the proof.
In this paper certificates are needed only once, for the support { 1 , 3 , 5 } \{1,3,5\} { 1 , 3 , 5 } 6 6 6 σ \sigma σ § 8 ). A single certificate serves both values of θ \theta θ
Proposition 3.10. Let k = 6 k = 6 k = 6 S = { 1 , 3 , 5 } S = \{1,3,5\} S = { 1 , 3 , 5 } θ = 0 \theta= 0 θ = 0 θ = 1 \theta= 1 θ = 1 q = 15 q = 15 q = 15
E = ( − 1 , 0 , − 1 , 0 , 0 , − 1 ) , V = ( − 1 , − 1 , 0 , 0 , 0 , 1 ) , W = ( 0 , − 1 , − 1 , 0 , − 1 , 1 ) , E = (-1,0,-1,0,0,-1), \qquad V = (-1,-1,0,0,0,1), \qquad W = (0,-1,-1,0,-1,1), E = ( − 1 , 0 , − 1 , 0 , 0 , − 1 ) , V = ( − 1 , − 1 , 0 , 0 , 0 , 1 ) , W = ( 0 , − 1 , − 1 , 0 , − 1 , 1 ) , and
( w t ) t = 0 14 = 5 , 1 , 5 , 1 , 3 , 5 , 1 , 5 , 5 , 3 , 5 , 1 , 5 , 1 , 3. (w_t)_{t=0}^{14} = 5,1,5,1,3,5,1,5,5,3,5,1,5,1,3. ( w t ) t = 0 14 = 5 , 1 , 5 , 1 , 3 , 5 , 1 , 5 , 5 , 3 , 5 , 1 , 5 , 1 , 3. Consequently no equation (8) of order 6 6 6 { 1 , 3 , 5 } \{1,3,5\} { 1 , 3 , 5 } θ \theta θ
The orbit e e e 5 5 5 − 1 , 0 , − 1 , 0 , 0 -1,0,-1,0,0 − 1 , 0 , − 1 , 0 , 0 E E E 15 15 15
Proof. The consequence follows from Proposition 3.9, so only validity needs checking. The run is listed in Table 2 . Each row t t t e t e_t e t d t d_t d t u t u_t u t t t t m t ( e ) m_t(e) m t ( e ) j ∈ S j \in S j ∈ S w t w_t w t θ = 0 \theta= 0 θ = 0 e e e d d d u u u 15 15 15 E E E V + W = ( − 1 , − 2 , − 1 , 0 , − 1 , 2 ) V+W = (-1,-2,-1,0,-1,2) V + W = ( − 1 , − 2 , − 1 , 0 , − 1 , 2 ) W W W
Table 2. The certificate of Proposition 3.10 for k = 6 k = 6 k = 6 S = { 1 , 3 , 5 } S = \{1,3,5\} S = { 1 , 3 , 5 } q = 15 q = 15 q = 15 15 , … , 20 15,\ldots,20 15 , … , 20 e = ( − 1 , 0 , − 1 , 0 , 0 , − 1 ) e = (-1,0,-1,0,0,-1) e = ( − 1 , 0 , − 1 , 0 , 0 , − 1 ) d = ( − 1 , − 2 , − 1 , 0 , − 1 , 2 ) d = (-1,-2,-1,0,-1,2) d = ( − 1 , − 2 , − 1 , 0 , − 1 , 2 ) u = ( 0 , − 1 , − 1 , 0 , − 1 , 1 ) u = (0,-1,-1,0,-1,1) u = ( 0 , − 1 , − 1 , 0 , − 1 , 1 ) m t = 0 m_t = 0 m t = 0 θ = 1 \theta= 1 θ = 1
Now let θ = 1 \theta= 1 θ = 1 m t ( e ) m_t(e) m t ( e ) 0 0 0 0 0 0 e e e t = 3 , 8 , 13 t = 3,8,13 t = 3 , 8 , 13 m t ( e ) = 0 m_t(e) = 0 m t ( e ) = 0 0 0 0 ( d t + w t , u t + w t ) = ( 0 , − 1 ) , ( − 2 , − 1 ) (d_t+w_t,u_t+w_t) = (0,-1),(-2,-1) ( d t + w t , u t + w t ) = ( 0 , − 1 ) , ( − 2 , − 1 ) ( − 3 , − 2 ) (-3,-2) ( − 3 , − 2 ) 0 0 0
Carriers and the σ \sigma σ
Certificates have to be found one support at a time. For most supports a single explicit construction does the job instead, and it reduces the dynamics to a combinatorial condition on residues: the periodic orbit takes only the values − 1 -1 − 1 0 0 0 2 k 2k 2 k k k k
Definition 3.11. Let k ≥ 2 k \ge2 k ≥ 2 S ⊆ { 1 , … , k − 1 } S \subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 }
A set P ⊆ Z / 2 k P \subseteq\mathbb{Z}/2k P ⊆ Z /2 k n + k ∈ P ⟺ n ∉ P n+k \in P \Longleftrightarrow n \notin P n + k ∈ P ⟺ n ∈ / P n n n Q = P ∩ { 0 , … , k − 1 } Q = P \cap\{0,\ldots,k-1\} Q = P ∩ { 0 , … , k − 1 } 0 ≤ i < k 0 \leq i < k 0 ≤ i < k i ∈ P i \in P i ∈ P i ∈ Q i \in Q i ∈ Q k + i ∈ P k+i \in P k + i ∈ P i ∉ Q i \notin Q i ∈ / Q n ≥ 0 n \geq0 n ≥ 0 n ∈ P n \in P n ∈ P n n n 2 k 2k 2 k P P P P P P Q Q Q carrier .
The orbit of P P P E n = − 1 E_n = -1 E n = − 1 n ∈ P n \in P n ∈ P E n = 0 E_n = 0 E n = 0
Let σ ∈ S \sigma\in S σ ∈ S ( σ , P ) (\sigma,P) ( σ , P ) σ \sigma σ bump for S S S P P P
(B1) P ∩ ( { 0 } ∪ S ) = { σ } P \cap(\{0\} \cup S) = \{\sigma\} P ∩ ({ 0 } ∪ S ) = { σ }
(B2) for every 0 < i < k 0 < i < k 0 < i < k j , j ′ ∈ S j,j' \in S j , j ′ ∈ S i + j ∈ P i+j \in P i + j ∈ P i + j ′ ∉ P i+j' \notin P i + j ′ ∈ / P P P P 2 k 2k 2 k i + j i+j i + j i + j ′ i+j' i + j ′ 0 0 0 σ \sigma σ k k k
When such a pair exists, S S S carry a σ \sigma σ bump . The two taps j j j j ′ j' j ′ i + j i+j i + j i + j ′ i+j' i + j ′ P P P i = 1 i=1 i = 1 Lemma 3.13 .
The bump with parameter σ \sigma σ n = ℓ k + i n = \ell k+i n = ℓ k + i ℓ ≥ 0 \ell\geq0 ℓ ≥ 0 0 ≤ i < k 0 \leq i < k 0 ≤ i < k
d ℓ k = { − ℓ , ℓ even , ℓ + 1 , ℓ odd , d ℓ k + σ = { 2 , ℓ even , − 2 , ℓ odd , (19) d_{\ell k} =
\begin{cases}
-\ell, & \ell\ \text{even},\\
\ell+1, & \ell\ \text{odd},
\end{cases}
\qquad
d_{\ell k+\sigma} =
\begin{cases}
2, & \ell\ \text{even},\\
-2, & \ell\ \text{odd},
\end{cases} \tag*{(19)} d ℓ k = { − ℓ , ℓ + 1 , ℓ even , ℓ odd , d ℓ k + σ = { 2 , − 2 , ℓ even , ℓ odd , (19) and d ℓ k + i = 0 d_{\ell k+i} = 0 d ℓ k + i = 0 i ≠ 0 , σ i \ne0,\sigma i = 0 , σ
The orbit of a carrier is a solution of the min-plus recursion as soon as every window of taps meets the carrier, and then the constant slot plays no role.
Lemma 3.12. Let P ⊆ Z / 2 k P \subseteq\mathbb{Z}/2k P ⊆ Z /2 k n ≥ 0 n \geq0 n ≥ 0 j ∈ S j \in S j ∈ S n + j ∈ P n+j \in P n + j ∈ P E E E P P P (15) for both values of θ \theta θ m n ( E ) = − 1 m_n(E) = -1 m n ( E ) = − 1
A n ( E ) = { j ∈ S : n + j ∈ P } for every n ≥ 0. A_n(E) = \{j \in S : n+j \in P\}
\qquad\text{for every } n \geq0. A n ( E ) = { j ∈ S : n + j ∈ P } for every n ≥ 0. In particular the constant slot never attains the minimum, and the tangent recursion (17) along E E E θ = 0 \theta= 0 θ = 0 θ = 1 \theta= 1 θ = 1
Proof. Since E E E − 1 -1 − 1 0 0 0 − 1 -1 − 1 m n ( E ) = − 1 m_n(E)=-1 m n ( E ) = − 1 n n n 0 0 0 n n n n + k n+k n + k P P P E n + k + E n = − 1 = m n ( E ) E_{n+k}+E_n=-1=m_n(E) E n + k + E n = − 1 = m n ( E )
Conditions (B1) and (B2) are constraints on the k k k Q Q Q k k k E E E σ \sigma σ σ \sigma σ
Lemma 3.13. Let ( σ , P ) (\sigma,P) ( σ , P ) σ \sigma σ S S S n ≥ 0 n \ge0 n ≥ 0 j ∈ S j \in S j ∈ S n + j ∈ P n+j \in P n + j ∈ P E E E P P P d d d E E E θ \theta θ d 2 m k = − 2 m d_{2mk}=-2m d 2 mk = − 2 m m ≥ 0 m \ge0 m ≥ 0
Proof. First, every window meets P P P n = ℓ k + i n=\ell k+i n = ℓ k + i 0 ≤ i < k 0 \le i < k 0 ≤ i < k k k k P P P i = 0 i=0 i = 0 ℓ \ell ℓ n + σ ∈ P n+\sigma\in P n + σ ∈ P i = 0 i=0 i = 0 ℓ \ell ℓ j ∈ S ∖ { σ } j \in S \setminus\{\sigma\} j ∈ S ∖ { σ } i = 1 i=1 i = 1 j ∉ P j \notin P j ∈ / P n + j ∈ P n+j \in P n + j ∈ P 0 < i < k 0<i<k 0 < i < k j j j i + j ∈ P i+j \in P i + j ∈ P j ′ j' j ′ i + j ′ ∉ P i+j' \notin P i + j ′ ∈ / P n + j ∈ P n+j \in P n + j ∈ P ℓ \ell ℓ n + j ′ ∈ P n+j' \in P n + j ′ ∈ P ℓ \ell ℓ
Next, the left-hand side of (17) . For n = ℓ k + i n=\ell k+i n = ℓ k + i
d n + k + d n = { 2 , i = 0 and ℓ even , 0 , otherwise . d_{n+k}+d_n=
\begin{cases}
2, & i=0\text{ and }\ell\text{ even},\\
0, & \text{otherwise}.
\end{cases} d n + k + d n = { 2 , 0 , i = 0 and ℓ even , otherwise . Indeed, at i = 0 i=0 i = 0 d ( ℓ + 1 ) k + d ℓ k = ( ℓ + 2 ) − ℓ = 2 d_{(\ell+1)k}+d_{\ell k}=(\ell+2)-\ell=2 d ( ℓ + 1 ) k + d ℓ k = ( ℓ + 2 ) − ℓ = 2 ℓ \ell ℓ − ( ℓ + 1 ) + ( ℓ + 1 ) = 0 -(\ell+1)+(\ell+1)=0 − ( ℓ + 1 ) + ( ℓ + 1 ) = 0 ℓ \ell ℓ i = σ i=\sigma i = σ − 2 + 2 = 0 -2+2=0 − 2 + 2 = 0 2 − 2 = 0 2-2=0 2 − 2 = 0
Then, the negative entries of d d d P P P d ℓ k d_{\ell k} d ℓ k ℓ ≥ 2 \ell\ge2 ℓ ≥ 2 ≡ 0 ( m o d 2 k ) \equiv0 \pmod{2k} ≡ 0 ( mod 2 k ) d ℓ k + σ d_{\ell k+\sigma} d ℓ k + σ ℓ \ell ℓ ≡ k + σ ( m o d 2 k ) \equiv k+\sigma\pmod{2k} ≡ k + σ ( mod 2 k ) 0 ∉ P 0 \notin P 0 ∈ / P σ ∈ P \sigma\in P σ ∈ P k + σ ∉ P k+\sigma\notin P k + σ ∈ / P
d n ≥ 0 whenever n ∈ P . (20) d_n \ge0 \qquad\text{whenever } n \in P. \tag*{(20)} d n ≥ 0 whenever n ∈ P . (20) Now compute the right-hand side of (17) , using the attaining sets A n = { j ∈ S : n + j ∈ P } A_n=\{j \in S:n+j \in P\} A n = { j ∈ S : n + j ∈ P } θ \theta θ n = ℓ k + i n=\ell k+i n = ℓ k + i
i = 0 i=0 i = 0 ℓ \ell ℓ n + j ∈ P n+j \in P n + j ∈ P j ∈ P j \in P j ∈ P A n = { σ } A_n=\{\sigma\} A n = { σ } d ℓ k + σ = 2 d_{\ell k+\sigma}=2 d ℓ k + σ = 2
i = 0 i=0 i = 0 ℓ \ell ℓ n + j ∈ P n+j \in P n + j ∈ P j ∉ P j \notin P j ∈ / P A n = S ∖ { σ } A_n=S\setminus\{\sigma\} A n = S ∖ { σ } d ℓ k + j d_{\ell k+j} d ℓ k + j j ≠ 0 , σ j\ne0,\sigma j = 0 , σ 0 0 0
0 < i < k 0<i<k 0 < i < k j ∈ A n j \in A_n j ∈ A n d n + j ≥ 0 d_{n+j}\ge0 d n + j ≥ 0 j ∈ A n j \in A_n j ∈ A n i + j ≢ 0 , σ ( m o d k ) i+j\not\equiv0,\sigma\pmod{k} i + j ≡ 0 , σ ( mod k ) j j j ℓ \ell ℓ j ′ j' j ′ ℓ \ell ℓ d n + j = 0 d_{n+j}=0 d n + j = 0 n + j ≡ i + j ( m o d k ) n+j\equiv i+j\pmod{k} n + j ≡ i + j ( mod k ) 0 0 0
In every case the right-hand side equals the left-hand side. Finally d 2 m k = − 2 m d_{2mk}=-2m d 2 mk = − 2 m
Theorem 3.14. If S ⊂ { 1 , … , k − 1 } S\subset\{1,\ldots,k-1\} S ⊂ { 1 , … , k − 1 } σ \sigma σ σ ∈ S \sigma\in S σ ∈ S S S S θ \theta θ
Proof. By Lemma 3.13 the orbit E E E P P P d d d θ = 0 \theta=0 θ = 0 θ = 1 \theta=1 θ = 1 d 2 m k = − 2 m d_{2mk}=-2m d 2 mk = − 2 m d d d Theorem 3.6 applies.
The full support never carries a σ \sigma σ § 5 .
Reading a carrier
Conditions (B1) and (B2) are checked many times in Sections 7 and 8, so I fix a vocabulary for them. Let S ⊆ { 1 , … , k − 1 } S \subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } σ ∈ S \sigma\in S σ ∈ S P P P Q = P ∩ { 0 , … , k − 1 } Q=P\cap\{0,\ldots,k-1\} Q = P ∩ { 0 , … , k − 1 } 0 < y < k 0<y<k 0 < y < k
y ∈ P ⟺ y ∈ Q , y + k ∈ P ⟺ y ∉ Q . (21) y\in P \Longleftrightarrow y\in Q,\qquad y+k\in P \Longleftrightarrow y\notin Q. \tag*{(21)} y ∈ P ⟺ y ∈ Q , y + k ∈ P ⟺ y ∈ / Q . (21) Since (B1) forces 0 ∉ Q 0\notin Q 0 ∈ / Q Q Q Q ( σ , Q ) (\sigma,Q) ( σ , Q ) Q ∩ S = { σ } Q\cap S=\{\sigma\} Q ∩ S = { σ } index is an integer i i i 0 < i < k 0<i<k 0 < i < k tap is an element j ∈ S j\in S j ∈ S i i i j j j position p = i + j p=i+j p = i + j 0 < p < 2 k 0<p<2k 0 < p < 2 k residue is y = p m o d k y=p\bmod k y = p mod k deleted at i i i y = 0 y=0 y = 0 y = σ y=\sigma y = σ survives ; a surviving tap reads 1 if p ∈ P p\in P p ∈ P i i i clause at i holds. The tap, and the position it reads, are unwrapped if p < k p<k p < k y = p y=p y = p wrapped if p > k p>k p > k y = p − k y=p-k y = p − k p = k p=k p = k
1 [ y ∈ Q ] if it is unwrapped , 1 [ y ∉ Q ] if it is wrapped . (22) \mathbf{1}[y\in Q]\text{ if it is unwrapped},\qquad\mathbf{1}[y\notin Q]\text{ if it is wrapped}. \tag*{(22)} 1 [ y ∈ Q ] if it is unwrapped , 1 [ y ∈ / Q ] if it is wrapped . (22) A position p p p y y y 0 < y < k 0<y<k 0 < y < k pin if its residue lies in S S S free if its residue lies outside S ∪ { 0 } S\cup\{0\} S ∪ { 0 } S ∖ { σ } S\setminus\{\sigma\} S ∖ { σ } Q Q Q (22) gives
an unwrapped pin reads 0 , a wrapped pin reads 1 . (23) \text{an unwrapped pin reads 0},\qquad\text{a wrapped pin reads 1}. \tag*{(23)} an unwrapped pin reads 0 , a wrapped pin reads 1 . (23) The reads of pins are therefore fixed by S S S σ \sigma σ Q Q Q
As an example, take k = 6 k=6 k = 6 S = { 1 , 2 , 4 , 5 } S=\{1,2,4,5\} S = { 1 , 2 , 4 , 5 } σ = 1 \sigma=1 σ = 1 Q = { 1 , 3 } Q=\{1,3\} Q = { 1 , 3 } P = { 1 , 3 , 6 , 8 , 10 , 11 } P=\{1,3,6,8,10,11\} P = { 1 , 3 , 6 , 8 , 10 , 11 } Q ∩ ( { 0 } ∪ S ) = { 1 } Q\cap(\{0\}\cup S)=\{1\} Q ∩ ({ 0 } ∪ S ) = { 1 } Q Q Q
Table 1 (PDF p. 17).
The unwrapped reads of 1 are of the free residue 3 ∈ Q 3\in Q 3 ∈ Q i = 5 i=5 i = 5 3 ∈ Q 3\in Q 3 ∈ Q ( 1 , Q ) (1,Q) ( 1 , Q ) σ \sigma σ Figure 1 shows the orbit and the bump in space and time.
Figure 1. The σ-bump of the example at k = 6 k=6 k = 6 S = { 1 , 2 , 4 , 5 } S=\{1,2,4,5\} S = { 1 , 2 , 4 , 5 } σ = 1 \sigma=1 σ = 1 Q = { 1 , 3 } Q=\{1,3\} Q = { 1 , 3 } ℓ \ell ℓ n = 6 ℓ + i n=6\ell+i n = 6 ℓ + i 0 ≤ i < 6 0\le i<6 0 ≤ i < 6 P P P E E E − 1 -1 − 1 E = 0 E=0 E = 0 2 k = 12 2k=12 2 k = 12 d d d (19) ; blank cells have d n = 0 d_n=0 d n = 0 − 2 -2 − 2 d 12 m = − 2 m d_{12m}=-2m d 12 m = − 2 m (20) requires.
Consequently, by Theorem 3.14 , no equation x n + 6 x n = c 0 + c 1 x n + 1 + c 2 x n + 2 + c 4 x n + 4 + c 5 x n + 5 x_{n+6}x_n=c_0+c_1x_{n+1}+c_2x_{n+2}+c_4x_{n+4}+c_5x_{n+5} x n + 6 x n = c 0 + c 1 x n + 1 + c 2 x n + 2 + c 4 x n + 4 + c 5 x n + 5 c 1 , c 2 , c 4 , c 5 c_1,c_2,c_4,c_5 c 1 , c 2 , c 4 , c 5
The mirror
Let S ⊆ { 1 , … , k − 1 } S\subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } Q Q Q σ \sigma σ Q ′ = k − Q Q'=k-Q Q ′ = k − Q k − σ k-\sigma k − σ k − i k-i k − i ( k − σ , Q ′ ) (k-\sigma,Q') ( k − σ , Q ′ ) i i i ( σ , Q ) (\sigma,Q) ( σ , Q ) ( σ , Q ) (\sigma,Q) ( σ , Q ) ( σ , Q ) (\sigma,Q) ( σ , Q ) ( k − σ , Q ′ ) (k-\sigma,Q') ( k − σ , Q ′ )
Lemma 3.15 (the mirror). Let S S S σ ∈ S \sigma\in S σ ∈ S Q , Q ′ ⊆ { 1 , … , k − 1 } Q,Q'\subseteq\{1,\ldots,k-1\} Q , Q ′ ⊆ { 1 , … , k − 1 } y ∈ Q ′ ⟺ k − y ∈ Q y\in Q'\Longleftrightarrow k-y\in Q y ∈ Q ′ ⟺ k − y ∈ Q 0 < y < k 0<y<k 0 < y < k k − i k-i k − i j ′ j' j ′ ( k − σ , Q ′ ) (k-\sigma,Q') ( k − σ , Q ′ ) v v v i i i k − j ′ k-j' k − j ′ ( σ , Q ) (\sigma,Q) ( σ , Q ) 1 − v 1-v 1 − v
Proof. Let P P P P ′ P' P ′ Q Q Q Q ′ Q' Q ′ k − j ′ k-j' k − j ′ S S S i i i p = i + k − j ′ p=i+k-j' p = i + k − j ′ k − i k-i k − i j ′ j' j ′ p ′ = k − i + j ′ p'=k-i+j' p ′ = k − i + j ′ p + p ′ = 2 k p+p'=2k p + p ′ = 2 k p ≡ − p ′ p\equiv-p' p ≡ − p ′ k k k σ ≡ − ( k − σ ) \sigma\equiv-(k-\sigma) σ ≡ − ( k − σ ) k − j ′ k-j' k − j ′ i i i j ′ j' j ′ k − i k-i k − i r = p ′ m o d k r=p'\mathbin{\bmod} k r = p ′ mod k j ′ j' j ′ p ′ = r p'=r p ′ = r p = ( k − r ) + k p=(k-r)+k p = ( k − r ) + k p ∈ P ⟺ k − r ∉ Q ⟺ r ∉ Q ′ ⟺ p ′ ∉ P ′ p\in P\Longleftrightarrow k-r\notin Q\Longleftrightarrow r\notin Q'\Longleftrightarrow p'\notin P' p ∈ P ⟺ k − r ∈ / Q ⟺ r ∈ / Q ′ ⟺ p ′ ∈ / P ′ p ′ = r + k p'=r+k p ′ = r + k p = k − r p=k-r p = k − r p ∈ P ⟺ k − r ∈ Q ⟺ r ∈ Q ′ ⟺ p ′ ∉ P ′ p\in P\Longleftrightarrow k-r\in Q\Longleftrightarrow r\in Q'\Longleftrightarrow p'\notin P' p ∈ P ⟺ k − r ∈ Q ⟺ r ∈ Q ′ ⟺ p ′ ∈ / P ′
Supports with at most two elements
This section deals with the supports with at most two elements. By Theorem 2.5 a globally periodic equation of the form (8) has a palindromic support, and by Lemma 2.8 it suffices to treat primitive supports. A palindromic support with at most two elements is ∅ \varnothing ∅ { k / 2 } \{k/2\} { k /2 } { j , k − j } \{j,k-j\} { j , k − j } 1 ≤ j < k / 2 1\le j<k/2 1 ≤ j < k /2 k / 2 k/2 k /2
S = ∅ S=\varnothing S = ∅ k = 1 k=1 k = 1 x n + 1 x n = c 0 x_{n+1}x_n=c_0 x n + 1 x n = c 0 c 0 > 0 c_0>0 c 0 > 0
S = { k / 2 } S=\{k/2\} S = { k /2 } k = 2 k=2 k = 2 S = { 1 } S=\{1\} S = { 1 } Theorem 4.1 ).
S = { j , k − j } S=\{j,k-j\} S = { j , k − j } gcd ( j , k ) = 1 \gcd(j,k)=1 g cd( j , k ) = 1 k = 3 j k=3j k = 3 j j = 1 j=1 j = 1 { 1 , 2 } \{1,2\} { 1 , 2 } Theorem 4.2 ); every other case is settled by Theorem 4.3 .
At orders two and three with c 0 > 0 c_0>0 c 0 > 0 λ 2 − ϵ 1 λ + 1 \lambda^2-\epsilon_1\lambda+1 λ 2 − ϵ 1 λ + 1 ϵ 1 = 2 cos ( 2 π q / p ) \epsilon_1=2\cos(2\pi q/p) ϵ 1 = 2 cos ( 2 π q / p ) q q q p ≥ 1 p\ge1 p ≥ 1 θ = 1 \theta=1 θ = 1 [10 Theorem 9] . What decides it is the second half of Theorem 3.3 : once round a cycle of the min-plus recursion the leading coefficients are multiplied by a fixed factor, which must be 1.
Order two
At k = 2 k = 2 k = 2 S = { 1 } S = \{1\} S = { 1 } (8) reads
x n + 2 x n = c 0 + c 1 x n + 1 , c 1 > 0 , c 0 ≥ 0. (24) x_{n+2}x_n = c_0 + c_1x_{n+1}, \qquad c_1 > 0,\qquad c_0 \ge0. \tag*{(24)} x n + 2 x n = c 0 + c 1 x n + 1 , c 1 > 0 , c 0 ≥ 0. (24) The substitution x n = c 1 y n x_n = c_1y_n x n = c 1 y n
y n + 2 y n = α + y n + 1 , α = c 0 c 1 2 ≥ 0 , (25) y_{n+2}y_n = \alpha+ y_{n+1}, \qquad\alpha= \frac{c_0}{c_1^2} \ge0, \tag*{(25)} y n + 2 y n = α + y n + 1 , α = c 1 2 c 0 ≥ 0 , (25) and it preserves positivity and periodicity of solutions. For α = 1 \alpha= 1 α = 1 [27 ] , of period 5; for α = 0 \alpha= 0 α = 0 [22 ] showed that the periods 5 and 6 occur only for α = 1 \alpha= 1 α = 1 α = 0 \alpha= 0 α = 0 [1 ] on the whole real plane, by Csörnyei and Laczkovich [ [10 ] with complex coefficients, and by Cima et al. [ [6 ] . I sketch the proof, which is essentially that of Csörnyei and Laczkovich [ [10 ] and of a remark of Barbeau et al. [ [1 ] , because it is the model for order three, where the details are given. The formalisation (Appendix § 14 ) contains a complete proof along the lines of the sketch.
Theorem 4.1 (order two). If (24) is globally periodic , then c 0 = 0 c_0 = 0 c 0 = 0 c 0 = c 1 2 c_0 = c_1^2 c 0 = c 1 2
Sketch of proof. Let c 0 > 0 c_0 > 0 c 0 > 0 α > 0 \alpha> 0 α > 0 θ = 1 \theta= 1 θ = 1 e n + 2 + e n = min ( 0 , e n + 1 ) e_{n+2} + e_n = \min(0,e_{n+1}) e n + 2 + e n = min ( 0 , e n + 1 ) ( e 0 , e 1 ) = ( 1 , 0 ) (e_0,e_1) = (1,0) ( e 0 , e 1 ) = ( 1 , 0 ) 1 , 0 , − 1 , − 1 , 0 1,0,-1,-1,0 1 , 0 , − 1 , − 1 , 0 (16) of Theorem 3.2 once round this cycle, with c 1 = 1 c_1 = 1 c 1 = 1 α \alpha α C 5 s + 1 C_{5s+1} C 5 s + 1 α \alpha α C 5 s + 4 = 1 / C 5 s + 1 C_{5s+4} = 1/C_{5s+1} C 5 s + 4 = 1/ C 5 s + 1 n = 5 s + 4 n = 5s + 4 n = 5 s + 4 C 5 s + 6 = α / C 5 s + 4 = α C 5 s + 1 C_{5s+6} = \alpha/C_{5s+4} = \alpha C_{5s+1} C 5 s + 6 = α / C 5 s + 4 = α C 5 s + 1 C 5 s + 1 = α s C_{5s+1} = \alpha^s C 5 s + 1 = α s Theorem 3.3 the period p p p C p + 1 = C 1 C_{p+1} = C_1 C p + 1 = C 1 α p / 5 = 1 \alpha^{p/5} = 1 α p /5 = 1 α = 1 \alpha= 1 α = 1
Order three
At k = 3 k = 3 k = 3 S = { 1 , 2 } S = \{1,2\} S = { 1 , 2 } Theorem 2.5 gives c 1 = c 2 c_1 = c_2 c 1 = c 2
x n + 3 x n = c 0 + c 1 ( x n + 1 + x n + 2 ) , c 1 > 0 , c 0 ≥ 0. (26) x_{n+3}x_n = c_0 + c_1(x_{n+1} + x_{n+2}), \qquad c_1 > 0,\qquad c_0 \ge0. \tag*{(26)} x n + 3 x n = c 0 + c 1 ( x n + 1 + x n + 2 ) , c 1 > 0 , c 0 ≥ 0. (26) The substitution x n = c 1 y n x_n = c_1y_n x n = c 1 y n
y n + 3 y n = α + y n + 1 + y n + 2 , α = c 0 c 1 2 ≥ 0. (27) y_{n+3}y_n = \alpha+ y_{n+1} + y_{n+2}, \qquad\alpha= \frac{c_0}{c_1^2} \ge0. \tag*{(27)} y n + 3 y n = α + y n + 1 + y n + 2 , α = c 1 2 c 0 ≥ 0. (27) For α = 1 \alpha= 1 α = 1 (2) . The following theorem is also known. With complex coefficients it follows from Theorems 1 and 4 of Csörnyei and Laczkovich [ [10 ] , and with nonnegative coefficients it is the case m = 1 m = 1 m = 1 [6 ] ; Kocić et al. [ [22 ] had shown that the period 8 occurs exactly when c 0 = c 1 2 c_0 = c_1^2 c 0 = c 1 2 c 0 > 0 c_0 > 0 c 0 > 0 [10 ] , written in the language of Theorem 3.2.
Theorem 4.2 (order three). If x n + 3 x n = c 0 + c 1 x n + 1 + c 2 x n + 2 x_{n+3}x_n = c_0 + c_1x_{n+1} + c_2x_{n+2} x n + 3 x n = c 0 + c 1 x n + 1 + c 2 x n + 2 c 1 , c 2 > 0 c_1,c_2 > 0 c 1 , c 2 > 0 c 0 ≥ 0 c_0 \ge0 c 0 ≥ 0 globally periodic , then c 1 = c 2 c_1 = c_2 c 1 = c 2 c 0 = c 1 2 c_0 = c_1^2 c 0 = c 1 2 In particular c 0 > 0 c_0 > 0 c 0 > 0
Proof. Suppose the equation is globally p p p p ≥ 1 p \ge1 p ≥ 1 Theorem 2.5 , c 1 = c 2 c_1 = c_2 c 1 = c 2 (26) . The two values of θ \theta θ
First suppose c 0 > 0 c_0 > 0 c 0 > 0 (27) is globally p p p α > 0 \alpha> 0 α > 0 e n + 3 + e n = min ( 0 , e n + 1 , e n + 2 ) e_{n+3} + e_n = \min(0,e_{n+1},e_{n+2}) e n + 3 + e n = min ( 0 , e n + 1 , e n + 2 ) ( e 0 , e 1 , e 2 ) = ( 1 , 0 , − 1 ) (e_0,e_1,e_2) = (1,0,-1) ( e 0 , e 1 , e 2 ) = ( 1 , 0 , − 1 )
Table 6.
The period is exactly 8, so Theorem 3.3 gives p = 8 r p=8r p = 8 r r ≥ 1 r\ge1 r ≥ 1 C 8 s = ξ C_{8s}=\xi C 8 s = ξ C 8 s + 1 = η C_{8s+1}=\eta C 8 s + 1 = η C 8 s + 2 = ζ C_{8s+2}=\zeta C 8 s + 2 = ζ n = 8 s , … , 8 s + 7 n=8s,\ldots,8s+7 n = 8 s , … , 8 s + 7 (16) give in turn
C 8 s + 3 = ζ ξ , C 8 s + 4 = ζ ξ η , C 8 s + 5 = η + 1 ξ η , C 8 s + 6 = 1 η , C 8 s + 7 = η + 1 ζ , C_{8s+3}=\frac{\zeta}{\xi},\quad C_{8s+4}=\frac{\zeta}{\xi\eta},\quad C_{8s+5}=\frac{\eta+1}{\xi\eta},\quad C_{8s+6}=\frac{1}{\eta},\quad C_{8s+7}=\frac{\eta+1}{\zeta}, C 8 s + 3 = ξ ζ , C 8 s + 4 = ξ η ζ , C 8 s + 5 = ξ η η + 1 , C 8 s + 6 = η 1 , C 8 s + 7 = ζ η + 1 , C 8 s + 8 = ζ ( α η + 1 ) η + 1 , C 8 s + 9 = α η , C 8 s + 10 = α ζ . C_{8s+8}=\frac{\zeta(\alpha\eta+1)}{\eta+1},\quad C_{8s+9}=\alpha\eta,\quad C_{8s+10}=\alpha\zeta. C 8 s + 8 = η + 1 ζ ( α η + 1 ) , C 8 s + 9 = α η , C 8 s + 10 = α ζ . For example, at n = 8 s + 2 n=8s+2 n = 8 s + 2 C 8 s + 5 C 8 s + 2 = C 8 s + 3 + C 8 s + 4 C_{8s+5}C_{8s+2}=C_{8s+3}+C_{8s+4} C 8 s + 5 C 8 s + 2 = C 8 s + 3 + C 8 s + 4 C 1 = 1 C_1=1 C 1 = 1 C 8 s + 1 = α s C_{8s+1}=\alpha^s C 8 s + 1 = α s Theorem 3.3 gives α r = C p + 1 = C 1 = 1 \alpha^r=C_{p+1}=C_1=1 α r = C p + 1 = C 1 = 1 α = 1 \alpha=1 α = 1 c 0 = c 1 2 c_0=c_1^2 c 0 = c 1 2
Now suppose c 0 = 0 c_0=0 c 0 = 0 x n + 3 x n = c 1 ( x n + 1 + x n + 2 ) x_{n+3}x_n=c_1(x_{n+1}+x_{n+2}) x n + 3 x n = c 1 ( x n + 1 + x n + 2 ) x ∗ = 2 c 1 x^*=2c_1 x ∗ = 2 c 1 ϵ 1 = ϵ 2 = 1 2 \epsilon_1=\epsilon_2=\frac{1}{2} ϵ 1 = ϵ 2 = 2 1 (11) , and the characteristic polynomial of Theorem 2.4 is
χ ( λ ) = λ 3 − 1 2 λ 2 − 1 2 λ + 1 = ( λ + 1 ) ( λ 2 − 3 2 λ + 1 ) . \chi(\lambda)=\lambda^3-\frac{1}{2}\lambda^2-\frac{1}{2}\lambda+1=(\lambda+1)\left(\lambda^2-\frac{3}{2}\lambda+1\right). χ ( λ ) = λ 3 − 2 1 λ 2 − 2 1 λ + 1 = ( λ + 1 ) ( λ 2 − 2 3 λ + 1 ) . By Theorem 2.4 every root of χ \chi χ p p p λ \lambda λ λ \lambda λ λ ˉ \bar{\lambda} λ ˉ λ λ ˉ = 1 \lambda\bar{\lambda}=1 λ λ ˉ = 1 λ + λ − 1 = 3 2 \lambda+\lambda^{-1}=\frac{3}{2} λ + λ − 1 = 2 3 λ \lambda λ λ − 1 \lambda^{-1} λ − 1 3 2 \frac{3}{2} 2 3 c 0 = 0 c_0=0 c 0 = 0
Two-term supports
The remaining small supports are the two-term supports { j , k − j } \{j,k-j\} { j , k − j }
Theorem 4.3 (two-term supports). Let 1 ≤ j 1\le j 1 ≤ j 2 j < k 2j<k 2 j < k gcd ( j , k ) = 1 \gcd(j,k)=1 g cd( j , k ) = 1 k ≠ 3 j k\ne3j k = 3 j S = { j , k − j } S=\{j,k-j\} S = { j , k − j } (8) with support S S S θ \theta θ
Note that gcd ( S ∪ { k } ) = gcd ( j , k ) \gcd(S\cup\{k\})=\gcd(j,k) g cd( S ∪ { k }) = g cd( j , k ) gcd ( j , k ) = 1 \gcd(j,k)=1 g cd( j , k ) = 1 S S S S S S k ≠ 3 j k\ne3j k = 3 j ( k , j ) = ( 3 , 1 ) (k,j)=(3,1) ( k , j ) = ( 3 , 1 ) Theorem 4.2 .
Throughout this subsection, for t ∈ C t\in\mathbb{C} t ∈ C
χ t ( λ ) = λ k + 1 − t ( λ j + λ k − j ) , (28) \chi_t(\lambda)=\lambda^k+1-t(\lambda^j+\lambda^{k-j}), \tag*{(28)} χ t ( λ ) = λ k + 1 − t ( λ j + λ k − j ) , (28) which is the characteristic polynomial of Theorem 2.4 for the support { j , k − j } \{j,k-j\} { j , k − j } ϵ j = ϵ k − j = t \epsilon_j=\epsilon_{k-j}=t ϵ j = ϵ k − j = t η 1 , … , η k \eta_1,\ldots,\eta_k η 1 , … , η k s r = ∑ i = 1 k η i r s_r=\sum_{i=1}^k\eta_i^r s r = ∑ i = 1 k η i r
The first step is to compute a few power sums. Because χ t \chi_t χ t
Lemma 4.4 (power sums). Let 1 ≤ j 1\le j 1 ≤ j 2 j < k 2j<k 2 j < k
for 1 ≤ r < k − j 1\le r<k-j 1 ≤ r < k − j s r = j t r / j s_r=jt^{r/j} s r = j t r / j j ∣ r j\mid r j ∣ r s r = 0 s_r=0 s r = 0
s k − j = ( k − j ) t + t s k − 2 j s_{k-j}=(k-j)t+ts_{k-2j} s k − j = ( k − j ) t + t s k − 2 j
s k = t ( s k − j + s j ) − k s_k=t(s_{k-j}+s_j)-k s k = t ( s k − j + s j ) − k
Proof. Write χ t ( λ ) = λ k + a 1 λ k − 1 + ⋯ + a k \chi_t(\lambda)=\lambda^k+a_1\lambda^{k-1}+\cdots+a_k χ t ( λ ) = λ k + a 1 λ k − 1 + ⋯ + a k 1 ≤ j < k − j < k 1\le j<k-j<k 1 ≤ j < k − j < k a j = a k − j = − t a_j=a_{k-j}=-t a j = a k − j = − t a k = 1 a_k=1 a k = 1 s r + a 1 s r − 1 + ⋯ + a r − 1 s 1 + r a r = 0 s_r+a_1s_{r-1}+\cdots+a_{r-1}s_1+ra_r=0 s r + a 1 s r − 1 + ⋯ + a r − 1 s 1 + r a r = 0 1 ≤ r ≤ k 1\le r\le k 1 ≤ r ≤ k
s r = t s r − j [ j < r ] + t s r − ( k − j ) [ k − j < r ] + r t [ r = j ] + r t [ r = k − j ] − k [ r = k ] . s_r=ts_{r-j}[j<r]+ts_{r-(k-j)}[k-j<r]+rt[r=j]+rt[r=k-j]-k[r=k]. s r = t s r − j [ j < r ] + t s r − ( k − j ) [ k − j < r ] + r t [ r = j ] + r t [ r = k − j ] − k [ r = k ] . where [ ⋅ ] [\cdot] [ ⋅ ] r < k − j r<k-j r < k − j s r = t s r − j s_r=ts_{r-j} s r = t s r − j j < r < k − j j<r<k-j j < r < k − j s j = j t s_j=jt s j = j t s r = 0 s_r=0 s r = 0 r < j r<j r < j r r r r = k − j r=k-j r = k − j r = k r=k r = k k − ( k − j ) = j k-(k-j)=j k − ( k − j ) = j
The power sums replace a delicate analysis of where the roots of χ t \chi_t χ t ∣ η i ∣ = 1 |\eta_i|=1 ∣ η i ∣ = 1 ∣ s r ∣ ≤ k |s_r|\le k ∣ s r ∣ ≤ k r r r r r r s r s_r s r t t t
Lemma 4.5 (interval bound). Let 1 ≤ j 1\le j 1 ≤ j 2 j < k 2j<k 2 j < k gcd ( j , k ) = 1 \gcd(j,k)=1 g cd( j , k ) = 1 k ≠ 3 j k\ne3j k = 3 j t ∈ C t\in\mathbb{C} t ∈ C χ t \chi_t χ t 1 1 1 t t t ∣ t ∣ ≤ 2 |t|\le2 ∣ t ∣ ≤ 2 t ≤ 2 t\le\sqrt{2} t ≤ 2
Proof. First, t t t χ t \chi_t χ t λ k χ t ( 1 / λ ) = χ t ( λ ) \lambda^k\chi_t(1/\lambda)=\chi_t(\lambda) λ k χ t ( 1/ λ ) = χ t ( λ ) χ t ( 0 ) = 1 \chi_t(0)=1 χ t ( 0 ) = 1
χ t ( λ ) = λ k χ t ( 1 / λ ) = ∏ i ( 1 − η i λ ) = ∏ i ( − η i ) ∏ i ( λ − η i − 1 ) = ∏ i ( λ − η i − 1 ) . \chi_t(\lambda)=\lambda^k\chi_t(1/\lambda)=\prod_i(1-\eta_i\lambda)=\prod_i(-\eta_i)\prod_i(\lambda-\eta_i^{-1})=\prod_i(\lambda-\eta_i^{-1}). χ t ( λ ) = λ k χ t ( 1/ λ ) = i ∏ ( 1 − η i λ ) = i ∏ ( − η i ) i ∏ ( λ − η i − 1 ) = i ∏ ( λ − η i − 1 ) . since ∏ i ( − η i ) = χ t ( 0 ) = 1 \prod_i(-\eta_i)=\chi_t(0)=1 ∏ i ( − η i ) = χ t ( 0 ) = 1 η i − 1 \eta_i^{-1} η i − 1 χ t \chi_t χ t ∣ η i ∣ = 1 |\eta_i|=1 ∣ η i ∣ = 1 η i − 1 = η i ‾ \eta_i^{-1}=\overline{\eta_i} η i − 1 = η i s j = ∑ i η i − j = ∑ i η i ‾ j = s j ‾ s_j=\sum_i\eta_i^{-j}=\sum_i\overline{\eta_i}^{\,j}=\overline{s_j} s j = ∑ i η i − j = ∑ i η i j = s j s j = j t s_j=jt s j = j t t = t ‾ t=\overline{t} t = t
Second, ∣ s r ∣ ≤ ∑ i ∣ η i ∣ r = k |s_r|\le\sum_i|\eta_i|^r=k ∣ s r ∣ ≤ ∑ i ∣ η i ∣ r = k r r r j ≥ 2 j\ge2 j ≥ 2 0 < k − 2 j < k − j 0<k-2j<k-j 0 < k − 2 j < k − j j ∤ k − 2 j j\nmid k-2j j ∤ k − 2 j j ∣ k j\mid k j ∣ k j = gcd ( j , k ) = 1 j=\gcd(j,k)=1 j = g cd( j , k ) = 1 s k − 2 j = 0 s_{k-2j}=0 s k − 2 j = 0 s k − j = ( k − j ) t s_{k-j}=(k-j)t s k − j = ( k − j ) t s j = j t s_j=jt s j = j t
s k = t ( ( k − j ) t + j t ) − k = k ( t 2 − 1 ) . s_k=t\bigl((k-j)t+jt\bigr)-k=k(t^2-1). s k = t ( ( k − j ) t + j t ) − k = k ( t 2 − 1 ) . So k ∣ t 2 − 1 ∣ ≤ k k|t^2-1|\le k k ∣ t 2 − 1∣ ≤ k t 2 ≤ 2 t^2\le2 t 2 ≤ 2 ∣ t ∣ ≤ 2 |t|\le\sqrt{2} ∣ t ∣ ≤ 2
Suppose instead j = 1 j=1 j = 1 k > 2 k>2 k > 2 k ≠ 3 k\ne3 k = 3 k ≥ 4 k\ge4 k ≥ 4 s k − 2 = t k − 2 s_{k-2}=t^{k-2} s k − 2 = t k − 2 s k − 1 = ( k − 1 ) t + t s k − 2 = ( k − 1 ) t + t k − 1 s_{k-1}=(k-1)t+ts_{k-2}=(k-1)t+t^{k-1} s k − 1 = ( k − 1 ) t + t s k − 2 = ( k − 1 ) t + t k − 1 ∣ t ∣ k − 2 ≤ k ≤ 2 k − 2 |t|^{k-2}\le k\le2^{k-2} ∣ t ∣ k − 2 ≤ k ≤ 2 k − 2 k ≥ 4 k\ge4 k ≥ 4 ∣ t ∣ ≤ 2 |t|\le2 ∣ t ∣ ≤ 2 t > 1 t>1 t > 1 s k − 1 > ( k − 1 ) + 1 = k s_{k-1}>(k-1)+1=k s k − 1 > ( k − 1 ) + 1 = k t ≤ 1 ≤ 2 t\le1\le\sqrt{2} t ≤ 1 ≤ 2
The last ingredient is Kronecker’s theorem [23 ] : if L L L ζ ∈ L \zeta\in L ζ ∈ L ∣ φ ( ζ ) ∣ ≤ 1 |\varphi(\zeta)|\le1 ∣ φ ( ζ ) ∣ ≤ 1 φ : L → C \varphi:L\to\mathbb{C} φ : L → C ζ \zeta ζ [5 ] .
Proof of Theorem 4.3 . Suppose that an equation (8) with support S = { j , k − j } S=\{j,k-j\} S = { j , k − j } p p p c j = c k − j c_j=c_{k-j} c j = c k − j ϵ j = ϵ k − j = : ϵ \epsilon_j=\epsilon_{k-j}=:\epsilon ϵ j = ϵ k − j =: ϵ (11) gives 2 ϵ = ϵ j + ϵ k − j ≤ 1 2\epsilon=\epsilon_j+\epsilon_{k-j}\le1 2 ϵ = ϵ j + ϵ k − j ≤ 1
0 < ϵ ≤ 1 2 . 0<\epsilon\le\frac{1}{2}. 0 < ϵ ≤ 2 1 . and by Theorem 2.4 every root η i \eta_i η i χ ϵ \chi_\epsilon χ ϵ p p p ϵ = ζ + ζ − 1 \epsilon=\zeta+\zeta^{-1} ϵ = ζ + ζ − 1 ζ \zeta ζ 0 < ϵ ≤ 1 2 0<\epsilon\le\frac{1}{2} 0 < ϵ ≤ 2 1 ζ \zeta ζ 9 9 9 ϵ \epsilon ϵ 2 \sqrt{2} 2
Step 1: ϵ \epsilon ϵ ζ + ζ − 1 \zeta+\zeta^{-1} ζ + ζ − 1 ζ \zeta ζ The coefficient of λ k − j \lambda^{k-j} λ k − j χ ϵ = ∏ i ( λ − η i ) \chi_\epsilon=\prod_i(\lambda-\eta_i) χ ϵ = ∏ i ( λ − η i ) − ϵ -\epsilon − ϵ ( − 1 ) j (-1)^j ( − 1 ) j j j j η 1 , … , η k \eta_1,\ldots,\eta_k η 1 , … , η k ϵ \epsilon ϵ ζ \zeta ζ z 2 − ϵ z + 1 z^2-\epsilon z+1 z 2 − ϵz + 1 ζ \zeta ζ ζ ≠ 0 \zeta\ne0 ζ = 0 ϵ = ζ + ζ − 1 \epsilon=\zeta+\zeta^{-1} ϵ = ζ + ζ − 1 L = Q ( ζ , η 1 , … , η k ) L=\mathbb{Q}(\zeta,\eta_1,\ldots,\eta_k) L = Q ( ζ , η 1 , … , η k ) ϵ \epsilon ϵ φ : L → C \varphi:L\to\mathbb{C} φ : L → C φ \varphi φ χ ϵ = ∏ i ( λ − η i ) \chi_\epsilon=\prod_i(\lambda-\eta_i) χ ϵ = ∏ i ( λ − η i ) χ φ ( ϵ ) = ∏ i ( λ − φ ( η i ) ) \chi_{\varphi(\epsilon)}=\prod_i(\lambda-\varphi(\eta_i)) χ φ ( ϵ ) = ∏ i ( λ − φ ( η i )) φ ( η i ) \varphi(\eta_i) φ ( η i ) p p p 1 1 1 φ ( ϵ ) \varphi(\epsilon) φ ( ϵ ) ∣ r ∣ ≤ 2 |r|\le2 ∣ r ∣ ≤ 2 φ ( ζ ) \varphi(\zeta) φ ( ζ ) z 2 − r z + 1 z^2-rz+1 z 2 − r z + 1 r 2 − 4 ≤ 0 r^2-4\le0 r 2 − 4 ≤ 0 ± 1 \pm1 ± 1 1 1 1 ∣ φ ( ζ ) ∣ = 1 |\varphi(\zeta)|=1 ∣ φ ( ζ ) ∣ = 1 ζ \zeta ζ
Step 2: the order of ζ \zeta ζ Let n 0 n_0 n 0 ζ \zeta ζ ∣ ζ ∣ = 1 |\zeta|=1 ∣ ζ ∣ = 1 ζ − 1 = ζ ‾ \zeta^{-1}=\overline{\zeta} ζ − 1 = ζ ϵ = 2 cos ( 2 π b / n 0 ) \epsilon=2\cos(2\pi b/n_0) ϵ = 2 cos ( 2 π b / n 0 ) b b b n 0 n_0 n 0 0 ≤ b ≤ n 0 / 2 0\le b\le n_0/2 0 ≤ b ≤ n 0 /2 cos \cos cos [ 0 , π ] [0,\pi] [ 0 , π ] cos ( 5 π / 12 ) = ( 6 − 2 ) / 4 > 1 4 \cos(5\pi/12)=(\sqrt{6}-\sqrt{2})/4>\frac14 cos ( 5 π /12 ) = ( 6 − 2 ) /4 > 4 1 cos ( π / 2 ) = 0 \cos(\pi/2)=0 cos ( π /2 ) = 0 0 < ϵ ≤ 1 2 0<\epsilon\le\frac12 0 < ϵ ≤ 2 1
5 24 < b n 0 < 1 4 , (29) \frac{5}{24}<\frac{b}{n_0}<\frac14, \tag*{(29)} 24 5 < n 0 b < 4 1 , (29) that is, 5 n 0 < 24 b < 6 n 0 5n_{0}<24b<6n_{0} 5 n 0 < 24 b < 6 n 0 n 0 ≤ 8 n_{0}\le8 n 0 ≤ 8 24 b < 48 24b<48 24 b < 48 b = 1 b=1 b = 1 b = 0 b=0 b = 0 5 n 0 < 24 < 6 n 0 5n_{0}<24<6n_{0} 5 n 0 < 24 < 6 n 0 n 0 ≥ 9 n_{0}\ge9 n 0 ≥ 9
Step 3: a large conjugate. The cyclotomic polynomial Φ n 0 \Phi_{n_{0}} Φ n 0 Q \mathbb{Q} Q ζ \zeta ζ e 2 π i / n 0 e^{2\pi i/n_{0}} e 2 π i / n 0 Q ( ζ ) \mathbb{Q}(\zeta) Q ( ζ ) C \mathbb{C} C ζ \zeta ζ e 2 π i / n 0 e^{2\pi i/n_{0}} e 2 π i / n 0 ζ ∈ L \zeta\in L ζ ∈ L φ : L → C \varphi:L\to\mathbb{C} φ : L → C
φ ( ϵ ) = φ ( ζ ) + φ ( ζ ) − 1 = e 2 π i / n 0 + e − 2 π i / n 0 = 2 cos ( 2 π / n 0 ) , \varphi(\epsilon)=\varphi(\zeta)+\varphi(\zeta)^{-1}=e^{2\pi i/n_{0}}+e^{-2\pi i/n_{0}}=2\cos(2\pi/n_{0}), φ ( ϵ ) = φ ( ζ ) + φ ( ζ ) − 1 = e 2 π i / n 0 + e − 2 π i / n 0 = 2 cos ( 2 π / n 0 ) , and, as in Step 1, Lemma 4.5 gives 2 cos ( 2 π / n 0 ) ≤ 2 2\cos(2\pi/n_{0})\le\sqrt{2} 2 cos ( 2 π / n 0 ) ≤ 2 n 0 ≥ 9 n_{0}\ge9 n 0 ≥ 9 0 < 2 π / n 0 < π / 4 0<2\pi/n_{0}<\pi/4 0 < 2 π / n 0 < π /4 2 cos ( 2 π / n 0 ) > 2 cos ( π / 4 ) = 2 2\cos(2\pi/n_{0})>2\cos(\pi/4)=\sqrt{2} 2 cos ( 2 π / n 0 ) > 2 cos ( π /4 ) = 2
The full support
The full support S = { 1 , … , k − 1 } S=\{1,\ldots,k-1\} S = { 1 , … , k − 1 } σ \sigma σ Q = { σ } Q=\{\sigma\} Q = { σ } i = k − 1 i=k-1 i = k − 1 i + j i+j i + j j ∈ S j\in S j ∈ S { k , … , 2 k − 2 } \{k,\ldots,2k-2\} { k , … , 2 k − 2 } P P P k + σ k+\sigma k + σ [10 ] . They also passed to its piecewise-linear shadow and showed, by exhibiting a non-periodic real solution at every order other than these, that it is periodic only at order 2 and, with c 0 > 0 c_{0}>0 c 0 > 0 c 0 > 0 c_{0}>0 c 0 > 0 [26 ] , and, with c 0 > 0 c_{0}>0 c 0 > 0 3 k − 1 3k-1 3 k − 1 [13 ] . The carrier Q = { 1 } Q=\{1\} Q = { 1 } σ = 1 \sigma=1 σ = 1
Theorem 5.1. Let k ≥ 4 k\ge4 k ≥ 4 (8) with full support S = { 1 , … , k − 1 } S=\{1,\ldots,k-1\} S = { 1 , … , k − 1 } θ \theta θ
The bound k ≥ 4 k\ge4 k ≥ 4 k = 2 k=2 k = 2 k = 3 k=3 k = 3 { 1 } \{1\} { 1 } { 1 , 2 } \{1,2\} { 1 , 2 } (2) .
The carrier
Throughout this section k ≥ 4 k\ge4 k ≥ 4 S = { 1 , … , k − 1 } S=\{1,\ldots,k-1\} S = { 1 , … , k − 1 } P ⊆ Z / 2 k P\subseteq\mathbb{Z}/2k P ⊆ Z /2 k Q = { 1 } Q=\{1\} Q = { 1 } Definition 3.11 ), that is
P = { 1 , k } ∪ { k + 2 , k + 3 , … , 2 k − 1 } , P=\{1,k\}\cup\{k+2,k+3,\ldots,2k-1\}, P = { 1 , k } ∪ { k + 2 , k + 3 , … , 2 k − 1 } , and let E E E n ≥ 0 n\ge0 n ≥ 0 n = ℓ k + i n=\ell k+i n = ℓ k + i ℓ ≥ 0 \ell\ge0 ℓ ≥ 0 0 ≤ i < k 0\le i<k 0 ≤ i < k ℓ \ell ℓ block and i i i column of n n n
Lemma 5.2. The orbit E E E θ \theta θ m n ( E ) = − 1 m_n(E)=-1 m n ( E ) = − 1
A ℓ k + i ( E ) = { { 1 } , ℓ even , i = 0 , { k − i } ∪ { k + 2 − i , … , k − 1 } , ℓ even , i ≥ 1 , { 2 , … , k − 1 } , ℓ odd , i = 0 , { 1 , … , k − 1 − i } ∪ ( { k + 1 − i } ∩ S ) , ℓ odd , i ≥ 1. A_{\ell k+i}(E)=
\begin{cases}
\{1\}, & \ell\ \text{even},\ i=0,\\
\{k-i\}\cup\{k+2-i,\ldots,k-1\}, & \ell\ \text{even},\ i\ge1,\\
\{2,\ldots,k-1\}, & \ell\ \text{odd},\ i=0,\\
\{1,\ldots,k-1-i\}\cup(\{k+1-i\}\cap S), & \ell\ \text{odd},\ i\ge1.
\end{cases} A ℓ k + i ( E ) = ⎩ ⎨ ⎧ { 1 } , { k − i } ∪ { k + 2 − i , … , k − 1 } , { 2 , … , k − 1 } , { 1 , … , k − 1 − i } ∪ ({ k + 1 − i } ∩ S ) , ℓ even , i = 0 , ℓ even , i ≥ 1 , ℓ odd , i = 0 , ℓ odd , i ≥ 1. Proof. Let 1 ≤ j ≤ k − 1 1 \le j \le k-1 1 ≤ j ≤ k − 1 ℓ \ell ℓ n + j ≡ i + j ( m o d 2 k ) n+j \equiv i+j \pmod{2k} n + j ≡ i + j ( mod 2 k ) 1 ≤ i + j ≤ 2 k − 2 1 \le i+j \le2k-2 1 ≤ i + j ≤ 2 k − 2 i + j < k i+j<k i + j < k P P P i + j = 1 i+j=1 i + j = 1 i = 0 i=0 i = 0 j = 1 j=1 j = 1 i + j ≥ k i+j \ge k i + j ≥ k P P P i + j = k i+j=k i + j = k i + j ≥ k + 2 i+j \ge k+2 i + j ≥ k + 2 ℓ \ell ℓ n + j ≡ k + i + j ( m o d 2 k ) n+j \equiv k+i+j \pmod{2k} n + j ≡ k + i + j ( mod 2 k ) i + j < k i+j<k i + j < k P P P i + j ≠ 1 i+j \ne1 i + j = 1 j ≥ 1 j \ge1 j ≥ 1 i ≥ 1 i \ge1 i ≥ 1 j ≥ 2 j \ge2 j ≥ 2 i + j ≥ k i+j \ge k i + j ≥ k i + j − k ∈ { 0 , … , k − 2 } i+j-k \in\{0,\ldots,k-2\} i + j − k ∈ { 0 , … , k − 2 } P P P i + j = k + 1 i+j=k+1 i + j = k + 1 { j ∈ S : n + j ∈ P } \{j \in S:n+j \in P\} { j ∈ S : n + j ∈ P } 1 1 1 k − i k-i k − i 2 2 2 1 1 1 2 2 2 □ \square □
In words, a tap j j j n = ℓ k + i n=\ell k+i n = ℓ k + i i + j i+j i + j ℓ \ell ℓ i + j < k i+j<k i + j < k i + j − k i+j-k i + j − k ℓ + 1 \ell+1 ℓ + 1 i = 0 i=0 i = 0 2 , … , i − 1 2,\ldots,i-1 2 , … , i − 1 i ≥ 1 i \ge1 i ≥ 1 i + 1 , … , k − 1 i+1,\ldots,k-1 i + 1 , … , k − 1 2 , … , k − 1 2,\ldots,k-1 2 , … , k − 1 i = 0 i=0 i = 0 i ≥ 2 i \ge2 i ≥ 2 k = 5 k=5 k = 5
Here the middle columns are 2 and 3.
The ray
Write the block as ℓ = 8 m + r \ell=8m+r ℓ = 8 m + r m ≥ 0 m \ge0 m ≥ 0 0 ≤ r < 8 0 \le r<8 0 ≤ r < 8 n = 8 k m + r k + i n=8km+rk+i n = 8 k m + r k + i d n d_n d n Table 3 : its value depends on r r r m m m i = 0 i=0 i = 0 i = 1 i=1 i = 1 2 ≤ i ≤ k − 2 2 \le i \le k-2 2 ≤ i ≤ k − 2 middle columns ) and i = k − 1 i=k-1 i = k − 1 k ≥ 4 k \ge4 k ≥ 4 a r a_r a r b r b_r b r c r c_r c r z r z_r z r r r r m m m f − m f-m f − m f + m f+m f + m f f f a 8 = a 0 a_8=a_0 a 8 = a 0 b 8 = b 0 b_8=b_0 b 8 = b 0 z 8 = z 0 z_8=z_0 z 8 = z 0 c 8 = − ( m + 1 ) c_8=-(m+1) c 8 = − ( m + 1 ) d d d q = 8 k q=8k q = 8 k m m m m + 1 m+1 m + 1 − 1 -1 − 1 + 1 +1 + 1
Table 3. The ray of Lemma 5.3. Left: the value of d n d_n d n n = 8 k m + r k + i n=8km+rk+i n = 8 k m + r k + i d n + k + d n d_{n+k}+d_n d n + k + d n r = 7 r=7 r = 7 m m m m + 1 m+1 m + 1
Lemma 5.3. The sequence d d d Table 3 is a tangent solution along E E E θ \theta θ d 8 k m + 2 k + 2 = − 1 − m d_{8km+2k+2}=-1-m d 8 k m + 2 k + 2 = − 1 − m m ≥ 0 m \ge0 m ≥ 0
Proof. Let n = ℓ k + i n=\ell k+i n = ℓ k + i ℓ = 8 m + r \ell=8m+r ℓ = 8 m + r d n + k + d n d_{n+k}+d_n d n + k + d n r r r r + 1 r+1 r + 1 i i i m m m Table 3 .
Now the right-hand side, the minimum of d n + j d_{n+j} d n + j j ∈ A n ( E ) j \in A_n(E) j ∈ A n ( E ) c r ≥ a r c_r \ge a_r c r ≥ a r c r ≥ z r c_r \ge z_r c r ≥ z r m ≥ 0 m \ge0 m ≥ 0 ℓ \ell ℓ i ≥ 3 i \ge3 i ≥ 3 ℓ + 1 \ell+ 1 ℓ + 1 i ≤ k − 3 i \le k - 3 i ≤ k − 3 k − 1 k - 1 k − 1
for r r r b r b_r b r i = 0 i = 0 i = 0 a r + 1 a_{r+1} a r + 1 i ≥ 1 i \ge1 i ≥ 1
for r r r z r z_r z r i ∈ { 0 , 1 } i \in\{0,1\} i ∈ { 0 , 1 } min ( z r , b r + 1 ) \min(z_r,b_{r+1}) min ( z r , b r + 1 ) 2 ≤ i ≤ k − 2 2 \le i \le k - 2 2 ≤ i ≤ k − 2 b r + 1 b_{r+1} b r + 1 i = k − 1 i = k - 1 i = k − 1
The tangent recursion (17) therefore reduces to the identities
r even : a r + a r + 1 = b r , b r + b r + 1 = c r + c r + 1 = z r + z r + 1 = a r + 1 ; r odd : a r + a r + 1 = b r + b r + 1 = z r , c r + c r + 1 = min ( z r , b r + 1 ) , z r + z r + 1 = b r + 1 . \begin{aligned}
r\ \text{even}: \quad&a_r + a_{r+1} = b_r,\qquad b_r + b_{r+1} = c_r + c_{r+1} = z_r + z_{r+1} = a_{r+1};\\
r\ \text{odd}: \quad&a_r + a_{r+1} = b_r + b_{r+1} = z_r,\qquad c_r + c_{r+1} = \min(z_r,b_{r+1}),\qquad z_r + z_{r+1} = b_{r+1}.
\end{aligned} r even : r odd : a r + a r + 1 = b r , b r + b r + 1 = c r + c r + 1 = z r + z r + 1 = a r + 1 ; a r + a r + 1 = b r + b r + 1 = z r , c r + c r + 1 = min ( z r , b r + 1 ) , z r + z r + 1 = b r + 1 . These are read off Table 3 . For the even rows r = 0 , 2 , 4 , 6 r = 0,2,4,6 r = 0 , 2 , 4 , 6 ( b r ; a r + 1 ) (b_r; a_{r+1}) ( b r ; a r + 1 ) ( 0 ; 1 ) (0;1) ( 0 ; 1 ) ( 0 ; 0 ) (0;0) ( 0 ; 0 ) ( 1 ; 0 ) (1;0) ( 1 ; 0 ) ( 1 ; 1 ) (1;1) ( 1 ; 1 ) r = 1 , 3 , 5 , 7 r = 1,3,5,7 r = 1 , 3 , 5 , 7 ( z r ; min ( z r , b r + 1 ) ; b r + 1 ) (z_r; \min(z_r,b_{r+1}); b_{r+1}) ( z r ; min ( z r , b r + 1 ) ; b r + 1 ) ( 1 ; 0 ; 0 ) (1;0;0) ( 1 ; 0 ; 0 ) ( 1 ; 1 ; 1 ) (1;1;1) ( 1 ; 1 ; 1 ) ( 0 ; 0 ; 1 ) (0;0;1) ( 0 ; 0 ; 1 ) ( 0 ; 0 ; 0 ) (0;0;0) ( 0 ; 0 ; 0 ) θ \theta θ
Finally, column 2 is a middle column because k ≥ 4 k \ge4 k ≥ 4 d 8 k m + 2 k + 2 = c 2 = − 1 − m d_{8km+2k+2} = c_2 = -1 - m d 8 k m + 2 k + 2 = c 2 = − 1 − m
Proof of Theorem 5.1 . By Lemma 5.2 the orbit E E E θ \theta θ Theorem 3.6 applies.
Odd orders
At odd order the palindromic supports have a rigid shape, and this section uses it to dispose of every one of them except the full support.
Theorem 6.1 (odd orders). Let k k k S ⊆ { 1 , … , k − 1 } S \subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } ∣ S ∣ ≥ 3 |S| \ge3 ∣ S ∣ ≥ 3 S ≠ { 1 , … , k − 1 } S \ne\{1,\ldots,k-1\} S = { 1 , … , k − 1 } S S S θ \theta θ
Write k = 2 h + 1 k = 2h + 1 k = 2 h + 1 { j , k − j } \{j,k-j\} { j , k − j } X ⊆ { 1 , … , h } X \subseteq\{1,\ldots,h\} X ⊆ { 1 , … , h }
S = { 2 x , k − 2 x : x ∈ X } . (30) S = \{2x, k - 2x : x \in X\}. \tag*{(30)} S = { 2 x , k − 2 x : x ∈ X } . (30) Two quite different mechanisms are needed, according to whether or not X X X dilated interval , that is X = μ ⋅ { 1 , … , M } X = \mu\cdot\{1,\ldots,M\} X = μ ⋅ { 1 , … , M } μ , M ≥ 1 \mu, M \ge1 μ , M ≥ 1 X X X § 6.4 ), an unbounded tangent solution along a fixed periodic orbit, which alternates between low values at odd positions and high values at even ones, can be written down in closed form; whether it exists turns on rigidity statements about translates and reflections of X X X k k k X X X § 6.5 ), the support carries a σ \sigma σ k = 7 k = 7 k = 7 X X X { 1 , 2 , 5 , 6 } \{1,2,5,6\} { 1 , 2 , 5 , 6 } { 1 , 3 , 4 , 6 } \{1,3,4,6\} { 1 , 3 , 4 , 6 } σ \sigma σ σ \sigma σ 2 7 2^7 2 7 σ \sigma σ
The alternating orbit has a precedent in the work of Cima et al. [ [6 ] . At odd order a palindromic equation has a one-parameter family of 2-periodic solutions, equal to t t t ( s t + c 0 ) / ( t − s ) (st + c_0)/(t - s) ( s t + c 0 ) / ( t − s ) t > s = ∑ j ∈ S , j e v e n c j t > s = \sum_{j \in S,\ j\ \mathrm{even}} c_j t > s = ∑ j ∈ S , j even c j [6 ] Proposition 2.6, which gives polynomial conditions on the coefficients that they solve at k = 5 , 7 , 9 , 11 k=5, 7, 9, 11 k = 5 , 7 , 9 , 11 t − s t-s t − s δ 3 \delta^{3} δ 3 δ → 0 + \delta\to0^{+} δ → 0 + − 3 -3 − 3 2 k 2k 2 k S S S
Throughout, k = 2 h + 1 k=2h+1 k = 2 h + 1 k k k { 0 , … , k − 1 } \{0,\ldots,k-1\} { 0 , … , k − 1 } n n n X X X n + X n+X n + X n − X n-X n − X X X X k k k X X X { 1 , … , h } \{1,\ldots,h\} { 1 , … , h } x , y ∈ X x,y\in X x , y ∈ X
x + y ≢ 0 ( m o d k ) , and − x ≡ k − x ∈ { h + 1 , … , 2 h } lies outside X ∪ { 0 } , (31) x+y\not\equiv0 \pmod{k}, \qquad\text{and} \qquad-x\equiv k-x\in\{h+1,\ldots,2h\}\text{ lies outside }X\cup\{0\}, \tag*{(31)} x + y ≡ 0 ( mod k ) , and − x ≡ k − x ∈ { h + 1 , … , 2 h } lies outside X ∪ { 0 } , (31) because 2 ≤ x + y ≤ 2 h < k 2\leq x+y\leq2h<k 2 ≤ x + y ≤ 2 h < k 1 [ ⋅ ] \mathbf{1}[\cdot] 1 [ ⋅ ] 1 U ( y ) = 1 [ y ∈ U ] \mathbf{1}_{U}(y)=\mathbf{1}[y\in U] 1 U ( y ) = 1 [ y ∈ U ] U U U
Lemma 6.2 (coordinates). Let S ⊆ { 1 , … , k − 1 } S\subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } X = { s / 2 : s ∈ S even } X=\{s/2:s\in S\text{ even}\} X = { s /2 : s ∈ S even } X ⊆ { 1 , … , h } X\subseteq\{1,\ldots,h\} X ⊆ { 1 , … , h } (30) holds, and ∣ S ∣ = 2 ∣ X ∣ |S|=2|X| ∣ S ∣ = 2∣ X ∣ ∣ S ∣ ≥ 3 |S|\geq3 ∣ S ∣ ≥ 3 ∣ X ∣ ≥ 2 |X|\geq2 ∣ X ∣ ≥ 2 S S S
Proof. By palindromy S S S { j , k − j } \{j,k-j\} { j , k − j } 2 x ≤ 2 h 2x\leq2h 2 x ≤ 2 h k − 2 x k-2x k − 2 x 2 ∣ X ∣ 2|X| 2∣ X ∣
The alternating orbit and the reduced recursion
The periodic orbit used below alternates, low on odd positions and high on even ones, except for a single raised value at one odd position of each period 2 k 2k 2 k n ≥ 0 n\geq0 n ≥ 0
E n = { − 2 n odd , n ≡ 1 ( m o d 2 k ) , − 3 n odd , n ≢ 1 ( m o d 2 k ) , − 1 n even , n ≡ k + 1 ( m o d 2 k ) , 0 n even , n ≢ k + 1 ( m o d 2 k ) . (32) E_n=
\begin{cases}
-2 & n\text{ odd},\ n\equiv1\pmod{2k},\\
-3 & n\text{ odd},\ n\not\equiv1\pmod{2k},\\
-1 & n\text{ even},\ n\equiv k+1\pmod{2k},\\
0 & n\text{ even},\ n\not\equiv k+1\pmod{2k}.
\end{cases} \tag*{(32)} E n = ⎩ ⎨ ⎧ − 2 − 3 − 1 0 n odd , n ≡ 1 ( mod 2 k ) , n odd , n ≡ 1 ( mod 2 k ) , n even , n ≡ k + 1 ( mod 2 k ) , n even , n ≡ k + 1 ( mod 2 k ) . (32) I call E E E
Lemma 6.3 (the alternating orbit). Let S S S E E E 2 k 2k 2 k S S S θ \theta θ E n + k + E n = − 3 E_{n+k}+E_n=-3 E n + k + E n = − 3 n n n n n n − 3 -3 − 3
A n = { j ∈ S : n + j odd , n + j ≢ 1 ( m o d 2 k ) } ; (33) A_n=\{j\in S:n+j\text{ odd},\ n+j\not\equiv1\pmod{2k}\}; \tag*{(33)} A n = { j ∈ S : n + j odd , n + j ≡ 1 ( mod 2 k )} ; (33) the constant slot, when θ = 1 \theta=1 θ = 1
Proof. As k k k n n n n + k n+k n + k n ≡ 1 n\equiv1 n ≡ 1 n + k ≡ k + 1 ( m o d 2 k ) n+k\equiv k+1\pmod{2k} n + k ≡ k + 1 ( mod 2 k ) − 2 -2 − 2 − 1 -1 − 1 E n + k + E n = − 3 E_{n+k}+E_n=-3 E n + k + E n = − 3 E E E − 3 -3 − 3 ≡ 1 ( m o d 2 k ) \equiv1\pmod{2k} ≡ 1 ( mod 2 k ) 0 > − 3 0>-3 0 > − 3 A n A_n A n n n n A n A_n A n j j j n + j ≢ 1 n+j\not\equiv1 n + j ≡ 1 n + j ≡ 1 ( m o d 2 k ) n+j\equiv1\pmod{2k} n + j ≡ 1 ( mod 2 k ) n n n
The tangent recursion along E E E
d n + k + d n = min j ∈ A n d n + j ( n ≥ 0 ) , (34) d_{n+k}+d_n=\min_{j\in A_n}d_{n+j}\qquad(n\geq0), \tag*{(34)} d n + k + d n = j ∈ A n min d n + j ( n ≥ 0 ) , (34) and every tap in A n A_n A n d d d d d d v t = d 2 t + 1 v_t=d_{2t+1} v t = d 2 t + 1 (30) . At the odd time n = 2 t + 1 n=2t+1 n = 2 t + 1 j = 2 x j=2x j = 2 x d 2 ( t + x ) + 1 = v t + x d_{2(t+x)+1}=v_{t+x} d 2 ( t + x ) + 1 = v t + x 2 ( t + x ) + 1 ≡ 1 ( m o d 2 k ) 2(t+x)+1\equiv1\pmod{2k} 2 ( t + x ) + 1 ≡ 1 ( mod 2 k ) t + x ≡ 0 ( m o d k ) t+x\equiv0\pmod{k} t + x ≡ 0 ( mod k ) n = 2 t + 1 + k n=2t+1+k n = 2 t + 1 + k j = k − 2 x j=k-2x j = k − 2 x v t + k − x v_{t+k-x} v t + k − x t − x ≡ 0 ( m o d k ) t-x\equiv0\pmod{k} t − x ≡ 0 ( mod k ) read sets , of the time 2 t + 1 2t+1 2 t + 1 2 t + 1 + k 2t+1+k 2 t + 1 + k
R + ( t ) = { t + x : x ∈ X , t + x ≢ 0 } , R − ( t ) = { t + k − x : x ∈ X , t − x ≢ 0 } , (35) R_{+}(t)=\{t+x:x\in X,\ t+x\not\equiv0\},\qquad R_{-}(t)=\{t+k-x:x\in X,\ t-x\not\equiv0\}, \tag*{(35)} R + ( t ) = { t + x : x ∈ X , t + x ≡ 0 } , R − ( t ) = { t + k − x : x ∈ X , t − x ≡ 0 } , (35) with congruences modulo k k k k k k R + ( t ) R_{+}(t) R + ( t ) t + X t+X t + X R − ( t ) R_{-}(t) R − ( t ) t − X t-X t − X ∣ X ∣ ≥ 2 |X|\ge2 ∣ X ∣ ≥ 2 x ∈ X x\in X x ∈ X t ± x ≡ 0 t\pm x\equiv0 t ± x ≡ 0 k k k reduced recursion is
v t + k − v t = min y ∈ R − ( t ) v y − min y ∈ R + ( t ) v y ( t ≥ 0 ) . (36) v_{t+k}-v_t=\min_{y\in R_{-}(t)}v_y-\min_{y\in R_{+}(t)}v_y\qquad(t\ge0). \tag*{(36)} v t + k − v t = y ∈ R − ( t ) min v y − y ∈ R + ( t ) min v y ( t ≥ 0 ) . (36) Proposition 6.4 (reduction to the odd positions). Let S S S ∣ S ∣ ≥ 3 |S|\ge3 ∣ S ∣ ≥ 3 X X X (36) has an unbounded integer solution ( v t ) t ≥ 0 (v_t)_{t\ge0} ( v t ) t ≥ 0 (8) with support S S S θ \theta θ
Proof. Define d n d_n d n n ≥ 0 n\ge0 n ≥ 0 d 2 t + 1 = v t d_{2t+1}=v_t d 2 t + 1 = v t
d n = min j ∈ A n d n + j − d n + k ( n even ) ; d_n=\min_{j\in A_n}d_{n+j}-d_{n+k}\qquad(n\ \text{even}); d n = j ∈ A n min d n + j − d n + k ( n even ) ; the right-hand side involves only odd positions (n + j n+j n + j j j j n + k n+k n + k (34) holds at every even n n n n = 2 t + 1 n=2t+1 n = 2 t + 1 n + k n+k n + k
d n + k = min j ∈ A n + k d n + k + j − d n + 2 k = min y ∈ R − ( t ) v y − v t + k , d_{n+k}=\min_{j\in A_{n+k}}d_{n+k+j}-d_{n+2k}=\min_{y\in R_{-}(t)}v_y-v_{t+k}, d n + k = j ∈ A n + k min d n + k + j − d n + 2 k = y ∈ R − ( t ) min v y − v t + k , and hence d n + k + d n = min y ∈ R − ( t ) v y − v t + k + v t = min y ∈ R + ( t ) v y = min j ∈ A n d n + j d_{n+k}+d_n=\min_{y\in R_{-}(t)}v_y-v_{t+k}+v_t=\min_{y\in R_{+}(t)}v_y=\min_{j\in A_n}d_{n+j} d n + k + d n = min y ∈ R − ( t ) v y − v t + k + v t = min y ∈ R + ( t ) v y = min j ∈ A n d n + j (36) . So d d d E E E ∣ X ∣ ≥ 2 |X|\ge2 ∣ X ∣ ≥ 2 d d d v v v Theorem 3.6 . □
Rays
The unbounded solutions of the reduced recursion used below are rays of the reduced recursion : they satisfy v t + k = v t + Δ ( t m o d k ) v_{t+k}=v_t+\Delta(t\bmod k) v t + k = v t + Δ ( t mod k ) Δ \Delta Δ Z / k \mathbb{Z}/k Z / k k k k (35) : the residues of R + ( t ) R_{+}(t) R + ( t ) R − ( t ) R_{-}(t) R − ( t )
R ˉ + ( r ) = ( r + X ) ∖ { 0 } , R ˉ − ( r ) = ( r − X ) ∖ { 0 } , r = t m o d k , (37) \bar{R}_{+}(r)=(r+X)\setminus\{0\},\qquad\bar{R}_{-}(r)=(r-X)\setminus\{0\},\qquad r=t\bmod k, \tag*{(37)} R ˉ + ( r ) = ( r + X ) ∖ { 0 } , R ˉ − ( r ) = ( r − X ) ∖ { 0 } , r = t mod k , (37) the punctured translate and the punctured reflection of X X X U ⊆ Z / k U\subseteq\mathbb{Z}/k U ⊆ Z / k r r r reflection-blocked if R ˉ − ( r ) ⊆ U \bar{R}_{-}(r)\subseteq U R ˉ − ( r ) ⊆ U translate-blocked if R ˉ + ( r ) ⊆ U \bar{R}_{+}(r)\subseteq U R ˉ + ( r ) ⊆ U
Lemma 6.5 (indicator rays). Let ∣ X ∣ ≥ 2 |X|\ge2 ∣ X ∣ ≥ 2 U ⊆ Z / k ∖ { 0 } U\subseteq\mathbb{Z}/k\setminus\{0\} U ⊆ Z / k ∖ { 0 } Δ : Z / k → Z \Delta:\mathbb{Z}/k\to\mathbb{Z} Δ : Z / k → Z
Δ ( r ) = 1 [ r reflection-blocked ] − 1 [ r translate-blocked ] . (38) \Delta(r)=\mathbf{1}[r\ \text{reflection-blocked}]-\mathbf{1}[r\ \text{translate-blocked}]. \tag*{(38)} Δ ( r ) = 1 [ r reflection-blocked ] − 1 [ r translate-blocked ] . (38) Suppose that 0 0 0 r ≠ 0 r\ne0 r = 0 r r r U U U R ˉ − ( r ) \bar{R}_{-}(r) R ˉ − ( r ) r r r R ˉ + ( r ) \bar{R}_{+}(r) R ˉ + ( r ) r r r Δ \Delta Δ
v t = 1 U ( t m o d k ) − 1 [ t ≡ 0 ] + ⌊ t / k ⌋ Δ ( t m o d k ) ( t ≥ 0 ) v_t=\mathbf{1}_U(t\bmod k)-\mathbf{1}[t\equiv0]+\lfloor t/k\rfloor\Delta(t\bmod k)\qquad(t\ge0) v t = 1 U ( t mod k ) − 1 [ t ≡ 0 ] + ⌊ t / k ⌋ Δ ( t mod k ) ( t ≥ 0 ) solves the reduced recursion (36) , and v m k = − 1 − m v_{mk}=-1-m v mk = − 1 − m v v v
Proof. The hypotheses say that Δ ( 0 ) = − 1 \Delta(0)=-1 Δ ( 0 ) = − 1 Δ ≥ 0 \Delta\ge0 Δ ≥ 0 Δ = 0 \Delta=0 Δ = 0 U ∪ { 0 } U\cup\{0\} U ∪ { 0 } v t + k − v t = Δ ( t m o d k ) v_{t+k}-v_t=\Delta(t\bmod k) v t + k − v t = Δ ( t mod k ) v m k = − 1 − m v_{mk}=-1-m v mk = − 1 − m 0 ∉ U 0\notin U 0 ∈ / U v v v y y y ρ = y m o d k \rho=y\bmod k ρ = y mod k 0 0 0 (35) . If ρ ∉ U \rho\notin U ρ ∈ / U Δ ( ρ ) = 0 \Delta(\rho)=0 Δ ( ρ ) = 0 v y = 0 v_y=0 v y = 0 ρ ∈ U \rho\in U ρ ∈ U v y = 1 + ⌊ y / k ⌋ Δ ( ρ ) ≥ 1 v_y=1+\lfloor y/k\rfloor\Delta(\rho)\ge1 v y = 1 + ⌊ y / k ⌋ Δ ( ρ ) ≥ 1 Δ ( ρ ) = 0 \Delta(\rho)=0 Δ ( ρ ) = 0 v v v 0 0 0 U U U 1 1 1 U U U Δ ( ρ ) = 0 \Delta(\rho)=0 Δ ( ρ ) = 0 (36) is 1 [ r reflection-blocked ] − 1 [ r translate-blocked ] \mathbf{1}[r\text{ reflection-blocked}]-\mathbf{1}[r\text{ translate-blocked}] 1 [ r reflection-blocked ] − 1 [ r translate-blocked ] r = t m o d k r=t\bmod k r = t mod k Δ ( r ) \Delta(r) Δ ( r )
The whole difficulty is to choose U U U X X X X X X
Rigidity of translates and reflections
Say that X X X symmetric about j ∈ Z / k j\in\mathbb{Z}/k j ∈ Z / k j − X ⊆ X j-X\subseteq X j − X ⊆ X j − X = X j-X=X j − X = X W = X ∪ { 0 } W=X\cup\{0\} W = X ∪ { 0 } symmetric about z z z z ∈ X z\in X z ∈ X z − x ∈ X z-x\in X z − x ∈ X x ∈ X ∖ { z } x\in X\setminus\{z\} x ∈ X ∖ { z } z − W = W z-W=W z − W = W X X X
Lemma 6.7 (translate rigidity). Let ∅ ≠ X ⊆ { 1 , … , h } \varnothing\ne X\subseteq\{1,\ldots,h\} ∅ = X ⊆ { 1 , … , h } ζ ≢ 0 ( m o d k ) \zeta\not\equiv0\pmod{k} ζ ≡ 0 ( mod k ) ζ + X ⊆ X ∪ { 0 } \zeta+X\subseteq X\cup\{0\} ζ + X ⊆ X ∪ { 0 } k k k X X X
Proof. Take 0 < ζ < k 0<\zeta<k 0 < ζ < k ζ ≤ h \zeta\le h ζ ≤ h x max = max X x_{\max}=\max X x m a x = max X x max + ζ x_{\max}+\zeta x m a x + ζ ( x max , 2 h ] (x_{\max},2h] ( x m a x , 2 h ] k k k 0 0 0 X X X ζ > h \zeta>h ζ > h μ = k − ζ ∈ { 1 , … , h } \mu=k-\zeta\in\{1,\ldots,h\} μ = k − ζ ∈ { 1 , … , h } ζ \zeta ζ x ↦ x − μ x\mapsto x-\mu x ↦ x − μ k k k x ∈ X x\in X x ∈ X x < μ x<\mu x < μ x − μ + k x-\mu+k x − μ + k { h + 1 , … , 2 h } \{h+1,\ldots,2h\} { h + 1 , … , 2 h } 0 0 0 X X X x ∈ X x\in X x ∈ X x ≥ μ x\ge\mu x ≥ μ x − μ ∈ X ∪ { 0 } x-\mu\in X\cup\{0\} x − μ ∈ X ∪ { 0 } X X X μ \mu μ max X = M μ \max X=M\mu max X = M μ μ \mu μ X X X 0 0 0 X = μ ⋅ { 1 , … , M } X=\mu\cdot\{1,\ldots,M\} X = μ ⋅ { 1 , … , M }
Lemma 6.8 (translates of a symmetric set). Let ∅ ≠ X ⊆ { 1 , … , h } \varnothing\ne X\subseteq\{1,\ldots,h\} ∅ = X ⊆ { 1 , … , h } j j j ζ ≠ 0 \zeta\ne0 ζ = 0 ζ + X ⊆ X ∪ { 0 , j } \zeta+X\subseteq X\cup\{0,j\} ζ + X ⊆ X ∪ { 0 , j } X X X
Proof. The translate can leave X X X 0 0 0 j j j x 1 + ζ ≡ 0 x_1+\zeta\equiv0 x 1 + ζ ≡ 0 x 2 + ζ ≡ j x_2+\zeta\equiv j x 2 + ζ ≡ j x 1 , x 2 ∈ X x_1,x_2\in X x 1 , x 2 ∈ X y = j − x 2 ∈ X y=j-x_2\in X y = j − x 2 ∈ X y ≡ ζ ≡ − x 1 y\equiv\zeta\equiv-x_1 y ≡ ζ ≡ − x 1 j j j Lemma 6.7 applies directly. If it leaves only at j j j y ∈ X y\in X y ∈ X x = j − y ∈ X x=j-y\in X x = j − y ∈ X x + ζ ≡ j − ζ ∈ X x+\zeta\equiv j-\zeta\in X x + ζ ≡ j − ζ ∈ X x + ζ ≡ j x+\zeta\equiv j x + ζ ≡ j y − ζ ≡ 0 y-\zeta\equiv0 y − ζ ≡ 0 − ζ + X ⊆ X ∪ { 0 } -\zeta+X\subseteq X\cup\{0\} − ζ + X ⊆ X ∪ { 0 } Lemma 6.7 applies to − ζ -\zeta − ζ
Lemma 6.9 (uniqueness of the centre of W W W ∅ ≠ X ⊆ { 1 , … , h } \varnothing\ne X\subseteq\{1,\ldots,h\} ∅ = X ⊆ { 1 , … , h } W = X ∪ { 0 } W=X\cup\{0\} W = X ∪ { 0 } z z z z ′ z' z ′ X X X
Proof. Put ζ = z ′ − z ≠ 0 \zeta=z'-z\ne0 ζ = z ′ − z = 0 ζ + X ⊆ X ∪ { 0 } \zeta+X\subseteq X\cup\{0\} ζ + X ⊆ X ∪ { 0 } Lemma 6.7 . For x = z x=z x = z x + ζ = z ′ ∈ X x+\zeta=z'\in X x + ζ = z ′ ∈ X x ≠ z x\ne z x = z y = z − x y=z-x y = z − x X X X x + ζ = z ′ − y x+\zeta=z'-y x + ζ = z ′ − y 0 0 0 y = z ′ y=z' y = z ′ X X X W W W z ′ z' z ′
Finally, reflection blocking against X X X
Lemma 6.10 (reflection blocking). If ( r − X ) ∖ { 0 } ⊆ X (r-X)\setminus\{0\}\subseteq X ( r − X ) ∖ { 0 } ⊆ X X X X r r r W W W r r r
Proof. If r ∉ X r\notin X r ∈ / X r − x r-x r − x r − X ⊆ X r-X\subseteq X r − X ⊆ X r ∈ X r\in X r ∈ X r − x ≠ 0 r-x\ne0 r − x = 0 x ∈ X ∖ { r } x\in X\setminus\{r\} x ∈ X ∖ { r } r − x ∈ X r-x\in X r − x ∈ X x x x W W W r r r
The non-dilated sets
Theorem 6.11 (non-dilated sets). Let S S S ∣ S ∣ ≥ 3 |S|\ge3 ∣ S ∣ ≥ 3 X X X (8) with support S S S θ \theta θ
Proof. By Proposition 6.4 and Lemma 6.5 it suffices to exhibit U U U Lemma 6.5 . Here ∣ X ∣ ≥ 2 |X|\ge2 ∣ X ∣ ≥ 2 X X X j j j j ≠ 0 j\ne0 j = 0 − X -X − X X X X (31) , and j ∉ X j\notin X j ∈ / X 0 = j − j ∈ X 0=j-j\in X 0 = j − j ∈ X U = X ∪ { j } U=X\cup\{j\} U = X ∪ { j } U = X U=X U = X 0 0 0 R ˉ + ( 0 ) = X \bar R_+(0)=X R ˉ + ( 0 ) = X r ≠ 0 r\ne0 r = 0 r + X ⊆ U ∪ { 0 } r+X\subseteq U\cup\{0\} r + X ⊆ U ∪ { 0 } Lemma 6.7 or Lemma 6.8 would make X X X
U = X ∪ { j } U=X\cup\{j\} U = X ∪ { j } r = j r=j r = j U U U R ˉ − ( j ) = j − X = X ⊆ U \bar R_-(j)=j-X=X\subseteq U R ˉ − ( j ) = j − X = X ⊆ U r = 0 r=0 r = 0 R ˉ − ( 0 ) = − X \bar R_-(0)=-X R ˉ − ( 0 ) = − X X X X j j j r ∉ { 0 , j } r\notin\{0,j\} r ∈ / { 0 , j } ζ = r − j ≠ 0 \zeta=r-j\ne0 ζ = r − j = 0 y ∈ X y\in X y ∈ X x = j − y x=j-y x = j − y X X X y + ζ = r − x y+\zeta=r-x y + ζ = r − x r r r ζ + X ⊆ X ∪ { 0 , j } \zeta+X\subseteq X\cup\{0,j\} ζ + X ⊆ X ∪ { 0 , j } Lemma 6.8 would make X X X Δ = 1 { j } − 1 { 0 } \Delta=1_{\{j\}}-1_{\{0\}} Δ = 1 { j } − 1 { 0 } R ˉ − ( j ) = X \bar R_-(j)=X R ˉ − ( j ) = X R ˉ + ( 0 ) = X \bar R_+(0)=X R ˉ + ( 0 ) = X Δ \Delta Δ
U = X U=X U = X − X -X − X X X X r r r Lemma 6.10 makes X X X W W W r r r W W W r r r r r r X = U X=U X = U Lemma 6.9 there is at most one such r r r Δ \Delta Δ X X X R ˉ + ( 0 ) = X \bar R_+(0)=X R ˉ + ( 0 ) = X Δ \Delta Δ z z z x ∈ X x\in X x ∈ X x ≠ z x\ne z x = z z − x ∈ R ˉ − ( z ) z-x\in\bar R_-(z) z − x ∈ R ˉ − ( z ) 0 0 0 z z z Δ ( z − x ) = 0 \Delta(z-x)=0 Δ ( z − x ) = 0 x x x ∣ X ∣ ≥ 2 |X|\ge2 ∣ X ∣ ≥ 2 R ˉ − ( z ) \bar R_-(z) R ˉ − ( z )
:::
Dilated intervals
When X = μ ⋅ { 1 , … , M } X=\mu\cdot\{1,\ldots,M\} X = μ ⋅ { 1 , … , M }
S = { 2 μ i , k − 2 μ i : 1 ≤ i ≤ M } , 2 μ M ≤ 2 h < k , (39) S=\{2\mu i,\ k-2\mu i:1\le i\le M\},\qquad2\mu M\le2h<k, \tag*{(39)} S = { 2 μ i , k − 2 μ i : 1 ≤ i ≤ M } , 2 μ M ≤ 2 h < k , (39) and the rigidity lemmas of § 6.3 no longer rule out blocking. These supports carry a σ \sigma σ σ \sigma σ k k k 2 k 2k 2 k 2 u 2u 2 u 2 u + k 2u+k 2 u + k u ∈ Z / k u\in\mathbb{Z}/k u ∈ Z / k Z / k \mathbb{Z}/k Z / k X X X
For s ∈ Z / k s\in\mathbb{Z}/k s ∈ Z / k s + X s+X s + X s − X s-X s − X s s s
Lemma 6.12 (carriers on Z / k \mathbb{Z}/k Z / k S S S X X X x 0 ∈ X x_0\in X x 0 ∈ X σ = 2 x 0 \sigma=2x_0 σ = 2 x 0 γ : Z / k → { 0 , 1 } \gamma:\mathbb{Z}/k\to\{0,1\} γ : Z / k → { 0 , 1 } P γ P_\gamma P γ 2 k 2k 2 k 2 u 2u 2 u γ ( u ) = 1 \gamma(u)=1 γ ( u ) = 1 2 u + k 2u+k 2 u + k γ ( u ) = 0 \gamma(u)=0 γ ( u ) = 0 P γ P_\gamma P γ P γ P_\gamma P γ γ \gamma γ ( σ , P γ ) (\sigma,P_\gamma) ( σ , P γ ) σ \sigma σ S S S
(F1) γ ( 0 ) = 0 , γ ( x 0 ) = 1 , γ ( x ) = 0 \gamma(0)=0,\ \gamma(x_0)=1,\ \gamma(x)=0 γ ( 0 ) = 0 , γ ( x 0 ) = 1 , γ ( x ) = 0 x ∈ X ∖ { x 0 } x\in X\setminus\{x_0\} x ∈ X ∖ { x 0 } γ ( − x ) = 1 \gamma(-x)=1 γ ( − x ) = 1 x ∈ X x\in X x ∈ X
(F2) for every s ∈ Z / k ∖ { 0 } s\in\mathbb{Z}/k\setminus\{0\} s ∈ Z / k ∖ { 0 } γ ( u ) \gamma(u) γ ( u ) u u u s + X s+X s + X 1 − γ ( u ) 1-\gamma(u) 1 − γ ( u ) u u u s − X s-X s − X u ∉ { 0 , x 0 } u\notin\{0,x_0\} u ∈ / { 0 , x 0 } 0 0 0 1 1 1
Proof. As k k k u ↦ 2 u u \mapsto2u u ↦ 2 u Z / k \mathbb{Z}/k Z / k 2 k 2k 2 k u ↦ 2 u + k u \mapsto2u+k u ↦ 2 u + k g ( n ) = 1 [ n ∈ P ] g(n)=1[n\in P] g ( n ) = 1 [ n ∈ P ] P P P 2 k 2k 2 k P = P γ P=P_\gamma P = P γ g ( 2 u ) = γ ( u ) g(2u)=\gamma(u) g ( 2 u ) = γ ( u ) g ( 2 u + k ) = 1 − γ ( u ) g(2u+k)=1-\gamma(u) g ( 2 u + k ) = 1 − γ ( u ) P γ P_\gamma P γ P P P γ ( u ) = g ( 2 u ) \gamma(u)=g(2u) γ ( u ) = g ( 2 u ) g ( 2 u + k ) = 1 − γ ( u ) g(2u+k)=1-\gamma(u) g ( 2 u + k ) = 1 − γ ( u ) P = P γ P=P_\gamma P = P γ γ \gamma γ P P P
(B1). The position 0 = 2 ⋅ 0 0=2\cdot0 0 = 2 ⋅ 0 P γ P_\gamma P γ γ ( 0 ) = 1 \gamma(0)=1 γ ( 0 ) = 1 2 x 2x 2 x P γ P_\gamma P γ γ ( x ) = 1 \gamma(x)=1 γ ( x ) = 1 k − 2 x = 2 ( − x ) + k k-2x=2(-x)+k k − 2 x = 2 ( − x ) + k P γ P_\gamma P γ γ ( − x ) = 0 \gamma(-x)=0 γ ( − x ) = 0 P γ ∩ ( { 0 } ∪ S ) = { σ } P_\gamma\cap(\{0\}\cup S)=\{\sigma\} P γ ∩ ({ 0 } ∪ S ) = { σ }
(B2). Let 0 < i < k 0<i<k 0 < i < k s ∈ Z / k s\in\mathbb{Z}/k s ∈ Z / k 2 s ≡ i ( m o d k ) 2s\equiv i\pmod{k} 2 s ≡ i ( mod k ) s ≠ 0 s\ne0 s = 0 i ↦ s i\mapsto s i ↦ s { 1 , … , k − 1 } \{1,\ldots,k-1\} { 1 , … , k − 1 } Z / k ∖ { 0 } \mathbb{Z}/k\setminus\{0\} Z / k ∖ { 0 } i ≡ 2 s ( m o d 2 k ) i\equiv2s\pmod{2k} i ≡ 2 s ( mod 2 k ) i i i i ≡ 2 s + k ( m o d 2 k ) i\equiv2s+k\pmod{2k} i ≡ 2 s + k ( mod 2 k ) i i i x ∈ X x\in X x ∈ X 2 x 2x 2 x i + 2 x ≡ 2 ( s + x ) i+2x\equiv2(s+x) i + 2 x ≡ 2 ( s + x ) 2 ( s + x ) + k 2(s+x)+k 2 ( s + x ) + k k − 2 x k-2x k − 2 x i + k − 2 x ≡ 2 ( s − x ) + k i+k-2x\equiv2(s-x)+k i + k − 2 x ≡ 2 ( s − x ) + k 2 ( s − x ) 2(s-x) 2 ( s − x ) i i i γ ( s + x ) \gamma(s+x) γ ( s + x ) 1 − γ ( s − x ) 1-\gamma(s-x) 1 − γ ( s − x ) i i i 2 x 2x 2 x i + 2 x ≡ 0 i+2x\equiv0 i + 2 x ≡ 0 i + 2 x ≡ σ = 2 x 0 ( m o d k ) i+2x\equiv\sigma=2x_0\pmod{k} i + 2 x ≡ σ = 2 x 0 ( mod k ) s + x ≡ 0 s+x\equiv0 s + x ≡ 0 x 0 x_0 x 0 2 2 2 k k k k − 2 x k-2x k − 2 x s − x ≡ 0 s-x\equiv0 s − x ≡ 0 x 0 x_0 x 0 i i i s s s
A unit modulo k k k μ \mu μ k k k X ′ = μ X X'=\mu X X ′ = μ X x 0 ′ = μ x 0 x'_0=\mu x_0 x 0 ′ = μ x 0 γ ′ ( u ) = γ ( μ − 1 u ) \gamma'(u)=\gamma(\mu^{-1}u) γ ′ ( u ) = γ ( μ − 1 u ) u ↦ μ − 1 u u\mapsto\mu^{-1}u u ↦ μ − 1 u 0 0 0 x 0 ′ x'_0 x 0 ′ x 0 x_0 x 0 X ′ X' X ′ X X X s ± X ′ s\pm X' s ± X ′ μ − 1 s ± X \mu^{-1}s\pm X μ − 1 s ± X μ − 1 s ≠ 0 \mu^{-1}s\ne0 μ − 1 s = 0 s ≠ 0 s\ne0 s = 0 γ \gamma γ ( X , x 0 ) (X,x_0) ( X , x 0 ) γ ′ \gamma' γ ′ ( X ′ , x 0 ′ ) (X',x'_0) ( X ′ , x 0 ′ ) X = { 1 , … , M } X=\{1,\ldots,M\} X = { 1 , … , M } x 0 = 1 x_0=1 x 0 = 1
Theorem 6.13 (the interval carrier). Let k = 2 h + 1 k=2h+1 k = 2 h + 1 M ≥ 2 M\ge2 M ≥ 2 2 M + 3 ≤ k 2M+3\le k 2 M + 3 ≤ k k k k u ∈ [ − h , h ] u\in[-h,h] u ∈ [ − h , h ] γ : Z / k → { 0 , 1 } \gamma:\mathbb{Z}/k\to\{0,1\} γ : Z / k → { 0 , 1 }
γ ( 0 ) = 0 , γ ( 1 ) = 1 , γ ( u ) = 0 ( 2 ≤ u ≤ M ) , γ ( − u ) = 1 ( 1 ≤ u ≤ M ) , γ ( u ) = 1 [ u ≡ M + 1 ( m o d 2 ) ] , γ ( − u ) = 1 − γ ( u ) ( M < u ≤ h ) . \begin{aligned}
\gamma(0)&=0,\qquad\gamma(1)=1,\qquad\gamma(u)=0\ (2\le u\le M),\qquad\gamma(-u)=1\ (1\le u\le M),\\
\gamma(u)&=1[u\equiv M+1\pmod{2}],\qquad\gamma(-u)=1-\gamma(u)\quad(M<u\le h).
\end{aligned} γ ( 0 ) γ ( u ) = 0 , γ ( 1 ) = 1 , γ ( u ) = 0 ( 2 ≤ u ≤ M ) , γ ( − u ) = 1 ( 1 ≤ u ≤ M ) , = 1 [ u ≡ M + 1 ( mod 2 )] , γ ( − u ) = 1 − γ ( u ) ( M < u ≤ h ) . Then γ \gamma γ X = { 1 , … , M } X=\{1,\ldots,M\} X = { 1 , … , M } x 0 = 1 x_0=1 x 0 = 1
Proof. Here h ≥ M + 1 ≥ 3 h\ge M+1\ge3 h ≥ M + 1 ≥ 3 { 0 } \{0\} { 0 } [ 1 , M ] [1,M] [ 1 , M ] [ − M , − 1 ] [-M,-1] [ − M , − 1 ] I + = [ M + 1 , h ] I_+=[M+1,h] I + = [ M + 1 , h ] I − = [ − h , − M − 1 ] I_-=[-h,-M-1] I − = [ − h , − M − 1 ] [ − h , h ] [-h,h] [ − h , h ] γ \gamma γ 0 0 0 1 1 1 excluded ; they are the residues removed in (F2). The function γ \gamma γ
γ ( − u ) = 1 − γ ( u ) \gamma(-u)=1-\gamma(u) γ ( − u ) = 1 − γ ( u ) 2 ≤ u ≤ h 2\le u\le h 2 ≤ u ≤ h u ≤ M u\le M u ≤ M 1 1 1 u > M u>M u > M
γ \gamma γ M M M M + 1 M+1 M + 1 − M − 1 -M-1 − M − 1 − M -M − M h h h h + 1 = − h h+1=-h h + 1 = − h γ ( M ) = 0 \gamma(M)=0 γ ( M ) = 0 γ ( M + 1 ) = 1 \gamma(M+1)=1 γ ( M + 1 ) = 1 γ ( − M ) = 1 \gamma(-M)=1 γ ( − M ) = 1 γ ( − M − 1 ) = 0 \gamma(-M-1)=0 γ ( − M − 1 ) = 0 γ ( − h ) = 1 − γ ( h ) \gamma(-h)=1-\gamma(h) γ ( − h ) = 1 − γ ( h )
γ \gamma γ I + I_+ I + ≡ M + 1 ( m o d 2 ) \equiv M+1\pmod{2} ≡ M + 1 ( mod 2 ) I − I_- I −
Also γ ( − 1 ) = 1 ≠ 0 = γ ( 2 ) \gamma(-1)=1\ne0=\gamma(2) γ ( − 1 ) = 1 = 0 = γ ( 2 )
Going round Z / k \mathbb{Z}/k Z / k [ 2 , M ] [2,M] [ 2 , M ] I + I_+ I + I − I_- I − [ − M , − 1 ] [-M,-1] [ − M , − 1 ] { 0 , 1 } \{0,1\} { 0 , 1 } [ 2 , M ] [2,M] [ 2 , M ] γ \gamma γ − 1 -1 − 1 2 2 2 γ \gamma γ R R R M M M R R R
than k k k I ± I_{\pm} I ± M ≥ 2 M \ge2 M ≥ 2 R ⊆ [ − M , 1 ] R \subseteq[-M,1] R ⊆ [ − M , 1 ] R ⊆ [ 0 , M ] R \subseteq[0,M] R ⊆ [ 0 , M ]
Suppose now that (F2) fails at s ≠ 0 s \ne0 s = 0 γ \gamma γ s + { 1 , … , M } s+\{1,\ldots,M\} s + { 1 , … , M } [ − M , 1 ] [-M,1] [ − M , 1 ] [ 0 , M ] [0,M] [ 0 , M ] s ∈ [ − M − 1 , 1 − M ] s \in[-M-1,1-M] s ∈ [ − M − 1 , 1 − M ] s ∈ { − 1 , 0 } s \in\{-1,0\} s ∈ { − 1 , 0 } s − { 1 , … , M } s-\{1,\ldots,M\} s − { 1 , … , M } s ∈ { 0 , 1 , 2 } s \in\{0,1,2\} s ∈ { 0 , 1 , 2 } s ∈ { M , M + 1 } s \in\{M,M+1\} s ∈ { M , M + 1 } [ − M − 1 , M + 1 ] [-M-1,M+1] [ − M − 1 , M + 1 ] 2 M + 3 ≤ k 2M+3 \le k 2 M + 3 ≤ k k k k [ − M − 1 , 0 ] [-M-1,0] [ − M − 1 , 0 ] { 0 } ∪ [ 1 , M + 1 ] \{0\} \cup[1,M+1] { 0 } ∪ [ 1 , M + 1 ] s ≠ 0 s \ne0 s = 0
Proof of Theorem 6.1
Proof of Theorem 6.1. Let k = 2 h + 1 k=2h+1 k = 2 h + 1 S S S X X X ∣ X ∣ ≥ 2 |X| \ge2 ∣ X ∣ ≥ 2 X X X Theorem 6.11 applies. Otherwise X = μ ⋅ { 1 , … , M } X=\mu\cdot\{1,\ldots,M\} X = μ ⋅ { 1 , … , M } S S S M = ∣ X ∣ ≥ 2 M=|X| \ge2 M = ∣ X ∣ ≥ 2 2 μ M ≤ 2 h 2\mu M \le2h 2 μ M ≤ 2 h S S S 2 μ i 2\mu i 2 μ i k − 2 μ i k-2\mu i k − 2 μ i gcd ( μ , k ) \gcd(\mu,k) g cd( μ , k ) S ∪ { k } S \cup\{k\} S ∪ { k } gcd ( μ , k ) = 1 \gcd(\mu,k)=1 g cd( μ , k ) = 1 μ = 1 \mu=1 μ = 1 M = h M=h M = h S S S μ = 1 \mu=1 μ = 1 M ≤ h − 1 M \le h-1 M ≤ h − 1 2 M + 3 ≤ k 2M+3 \le k 2 M + 3 ≤ k μ ≥ 2 \mu\ge2 μ ≥ 2 4 M ≤ 2 μ M < k 4M \le2\mu M < k 4 M ≤ 2 μ M < k 2 M + 3 ≤ 4 M < k 2M+3 \le4M < k 2 M + 3 ≤ 4 M < k M ≥ 2 M \ge2 M ≥ 2 Theorem 6.13 gives a function satisfying (F1) and (F2) for { 1 , … , M } \{1,\ldots,M\} { 1 , … , M } μ − 1 \mu^{-1} μ − 1 k k k X X X x 0 = μ x_0=\mu x 0 = μ Lemma 6.12 turns it into a 2 μ 2\mu 2 μ S S S Theorem 3.14 now shows that no equation with support S S S θ \theta θ
Even orders: supports with an even element
Throughout this section the order k k k Theorem 3.14 , the following theorem disposes of every support at even order, other than the full support, that is not all-odd. Reads are described in the vocabulary of § 3.5 , in particular (22) and (23), and Lemma 3.15 turns reads of 1 into reads of 0.
Theorem 7.1. Let k k k S ⊆ { 1 , … , k − 1 } S \subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } S S S σ \sigma σ ( σ , P ) (\sigma,P) ( σ , P )
The natural carriers at even order are the parity carriers : σ \sigma σ σ \sigma σ k k k y ↦ k − y y \mapsto k-y y ↦ k − y Q Q Q σ \sigma σ k − σ k-\sigma k − σ i i i S S S S S S i i i a = min S a=\min S a = min S S S S σ \sigma σ Theorem 7.7 ). If a a a σ \sigma σ Theorem 7.12 ), which carries one of two explicit carriers at σ = a \sigma=a σ = a 7.14 ). On a first reading one can take Lemmas 7.9 , 7.10 and 7.11 on trust.
Parity carriers
Write S 0 S_0 S 0 S 1 S_1 S 1 S S S class . Since k k k s ≡ k − s ( m o d 2 ) s \equiv k-s \pmod2 s ≡ k − s ( mod 2 ) s s s S 0 S_0 S 0 S 1 S_1 S 1 S c S_c S c k − min S c k-\min S_c k − min S c S S S S 1 ≠ ∅ S_1 \ne\varnothing S 1 = ∅ S ∪ { k } S \cup\{k\} S ∪ { k }
Definition 7.2. For σ ∈ S \sigma\in S σ ∈ S parity carrier of σ \sigma σ
Q σ = { σ } ∪ { y ∈ { 1 , … , k − 1 } : y ∉ S , y ≡ σ ( m o d 2 ) } . Q_\sigma=\{\sigma\}\cup\{y\in\{1,\ldots,k-1\}: y\notin S,\ y\equiv\sigma\pmod2\}. Q σ = { σ } ∪ { y ∈ { 1 , … , k − 1 } : y ∈ / S , y ≡ σ ( mod 2 )} . An index i i i silent for Q σ Q_{\sigma} Q σ i i i i i i c c c C c ( i ) C_c(i) C c ( i )
j ∈ S , i + j < k , i + j ≡ c ( m o d 2 ) ⟹ i + j ∈ S . (40) j \in S,\quad i+j<k,\quad i+j\equiv c\pmod{2}\quad\Longrightarrow\quad i+j\in S. \tag*{(40)} j ∈ S , i + j < k , i + j ≡ c ( mod 2 ) ⟹ i + j ∈ S . (40) Lemma 7.3 (reads of a parity carrier). Let k k k S S S σ ∈ S \sigma\in S σ ∈ S c c c σ \sigma σ
(a) Q σ ∩ S = { σ } Q_{\sigma}\cap S=\{\sigma\} Q σ ∩ S = { σ } ( B 1 ) (B1) ( B 1 )
(b) y ∈ Q k − σ ⟺ k − y ∈ Q σ y\in Q_{k-\sigma}\Longleftrightarrow k-y\in Q_{\sigma} y ∈ Q k − σ ⟺ k − y ∈ Q σ 0 < y < k 0<y<k 0 < y < k
(c) A surviving tap whose position has residue y y y y y y y ≡ c y\equiv c y ≡ c y ∈ S y\in S y ∈ S y ≢ c y\not\equiv c y ≡ c
(d) If ( σ , Q σ ) (\sigma,Q_{\sigma}) ( σ , Q σ ) σ \sigma σ σ ′ = σ \sigma'=\sigma σ ′ = σ σ ′ = k − σ \sigma'=k-\sigma σ ′ = k − σ σ ′ ∈ S \sigma'\in S σ ′ ∈ S σ ′ ≡ σ \sigma'\equiv\sigma σ ′ ≡ σ Q σ ′ Q_{\sigma'} Q σ ′
(e) If i i i Q σ Q_{\sigma} Q σ C c ( i ) C_c(i) C c ( i ) i i i Q σ Q_{\sigma} Q σ
Proof. (a) is the definition. (b) holds because y = k − σ ⟺ k − y = σ y=k-\sigma\Longleftrightarrow k-y=\sigma y = k − σ ⟺ k − y = σ y y y k − y k-y k − y y ≡ k − y y\equiv k-y y ≡ k − y k − σ ≡ σ k-\sigma\equiv\sigma k − σ ≡ σ (22) , using that a surviving tap has y ≠ σ y\ne\sigma y = σ
(d) By (a), the pair fails ( B 2 ) (B2) ( B 2 ) i i i i i i σ ′ = σ \sigma'=\sigma σ ′ = σ i i i Lemma 3.15 no surviving tap reads 1 at the index k − i k-i k − i Q k − σ Q_{k-\sigma} Q k − σ σ ′ = k − σ \sigma'=k-\sigma σ ′ = k − σ σ ′ ≡ σ \sigma'\equiv\sigma σ ′ ≡ σ k k k
(e) Let j ∈ S j\in S j ∈ S i + j < k i+j<k i + j < k i + j ≡ c i+j\equiv c i + j ≡ c i + j = σ i+j=\sigma i + j = σ i + j ∈ S i+j\in S i + j ∈ S j j j i + j i+j i + j y y y y = σ y=\sigma y = σ y ∈ Q σ y\in Q_{\sigma} y ∈ Q σ y y y y ≡ c y\equiv c y ≡ c
For the rest of the section, S S S k k k a = min S a=\min S a = min S b b b a a a max S = k − a \max S=k-a max S = k − a b b b k − b k-b k − b b ≤ k − b b\le k-b b ≤ k − b a < b a<b a < b
2 a < k , a + b < k . (41) 2a<k,\qquad a+b<k. \tag*{(41)} 2 a < k , a + b < k . (41) The next three lemmas say where a silent index can be. Near the start, the taps k − a k-a k − a k − min S c k-\min S_c k − min S c
Lemma 7.4 (the top taps). Let σ ∈ S \sigma\in S σ ∈ S c c c i i i Q σ Q_{\sigma} Q σ
(i) i < a i<a i < a i ≡ a + c ( m o d 2 ) i\equiv a+c\pmod{2} i ≡ a + c ( mod 2 )
(ii) i i i i < min S c i<\min S_c i < min S c
Proof. (i) The tap k − a k-a k − a y = i + k − a y=i+k-a y = i + k − a k − a < y < k k-a<y<k k − a < y < k y y y max S \max S max S y ≡ i + a ≡ c y\equiv i+a\equiv c y ≡ i + a ≡ c y ≠ σ y\ne\sigma y = σ Lemma 7.3 (c). (ii) Put m = min S c m=\min S_c m = min S c k − m k-m k − m y = i + k − m y=i+k-m y = i + k − m k − m < y < k k-m<y<k k − m < y < k y ≡ m ≡ c y\equiv m\equiv c y ≡ m ≡ c i i i k k k y y y c c c max S c = k − m \max S_c=k-m max S c = k − m
Beyond a a a Q σ Q_{\sigma} Q σ σ \sigma σ k − a k-a k − a k − b k-b k − b ( a , b ] (a,b] ( a , b ]
Lemma 7.5 (the parity dichotomy). Let σ ∈ S \sigma\in S σ ∈ S c c c i > a i>a i > a Q σ Q_{\sigma} Q σ i ≡ a + c ( m o d 2 ) i\equiv a+c\pmod{2} i ≡ a + c ( mod 2 ) i ≤ b i\le b i ≤ b
Proof. All elements of Q σ Q_\sigma Q σ c c c k − a k-a k − a i i i i > a i>a i > a i − a i-a i − a Lemma 7.3 (e) i − a ≡ c i-a\equiv c i − a ≡ c i ≡ a + c i\equiv a+c i ≡ a + c i > b i>b i > b k − b k-b k − b i ≡ b + c i\equiv b+c i ≡ b + c b ≠ a b\ne a b = a
Finally, a containment C c ( i ) C_c(i) C c ( i )
Lemma 7.6 (the gap). Let i i i C c ( i ) C_c(i) C c ( i ) S c ≠ ∅ S_c\ne\varnothing S c = ∅ S S S 1 − c 1-c 1 − c ( i , i + min S c ) (i,i+\min S_c) ( i , i + min S c )
Proof. Let s ∈ S s\in S s ∈ S 1 − c 1-c 1 − c s > i s>i s > i k − s ∈ S k-s\in S k − s ∈ S 1 − c 1-c 1 − c i + ( k − s ) < k i+(k-s)<k i + ( k − s ) < k i + ( k − s ) ≡ c i+(k-s)\equiv c i + ( k − s ) ≡ c i i i C c ( i ) C_c(i) C c ( i ) i + k − s ∈ S c i+k-s\in S_c i + k − s ∈ S c i + k − s ≤ max S c = k − min S c i+k-s\le\max S_c=k-\min S_c i + k − s ≤ max S c = k − min S c s ≥ i + min S c s\ge i+\min S_c s ≥ i + min S c
Even minimum
Theorem 7.7 (parity carriers at even minimum). Let k k k S S S a = min S a=\min S a = min S ( σ 1 , Q σ 1 ) (\sigma_1,Q_{\sigma_1}) ( σ 1 , Q σ 1 ) σ \sigma σ σ 1 ∈ S \sigma_1\in S σ 1 ∈ S ( σ 0 , Q σ 0 ) (\sigma_0,Q_{\sigma_0}) ( σ 0 , Q σ 0 ) σ \sigma σ σ 0 ∈ S \sigma_0\in S σ 0 ∈ S S S S σ \sigma σ
Proof. Since S S S S 1 ≠ ∅ S_1\ne\varnothing S 1 = ∅ b = min S 1 b=\min S_1 b = min S 1 a < b a<b a < b Lemma 7.3 (d) some index i 1 i_1 i 1 Q σ ′ Q_{\sigma'} Q σ ′ σ ′ ∈ S \sigma'\in S σ ′ ∈ S Lemma 7.4 with c = 1 c=1 c = 1 i 1 < a i_1<a i 1 < a a + 1 a+1 a + 1 i 1 ≤ a i_1\le a i 1 ≤ a min S 1 = b > a \min S_1=b>a min S 1 = b > a i 1 > a i_1>a i 1 > a Lemma 7.5 gives i 1 i_1 i 1 i 1 ≤ b i_1\le b i 1 ≤ b Lemma 7.3 (e) C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) Lemma 7.6 gives
S 0 ∩ ( i 1 , i 1 + b ) = ∅ . (42) S_0\cap(i_1,i_1+b)=\varnothing. \tag*{(42)} S 0 ∩ ( i 1 , i 1 + b ) = ∅ . (42) Now fix an even σ ∈ S \sigma\in S σ ∈ S i i i Q σ Q_\sigma Q σ
i < a i<a i < a i i i Lemma 7.4 (i) applies, as i ≡ a i\equiv a i ≡ a
i < a i<a i < a i i i b b b i + b < a + b < k i+b<a+b<k i + b < a + b < k i + b i+b i + b i 1 ≤ b < i + b < a + b < i 1 + b i_1\le b<i+b<a+b<i_1+b i 1 ≤ b < i + b < a + b < i 1 + b (42) it is not in S S S σ \sigma σ
i = a i=a i = a a a a C 0 ( a ) C_0(a) C 0 ( a ) Lemma 7.3 (e). Let s s s Z = { y ∈ S 0 : y + i 1 < k } Z=\{y\in S_0:y+i_1<k\} Z = { y ∈ S 0 : y + i 1 < k } a a a a + i 1 ≤ a + b < k a+i_1\le a+b<k a + i 1 ≤ a + b < k C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) s + i 1 ∈ S s+i_1\in S s + i 1 ∈ S s + i 1 ≤ max S 1 = k − b s+i_1\le\max S_1=k-b s + i 1 ≤ max S 1 = k − b a + s ≤ k − b − ( i 1 − a ) < k a+s\le k-b-(i_1-a)<k a + s ≤ k − b − ( i 1 − a ) < k C 0 ( a ) C_0(a) C 0 ( a ) a + s ∈ S 0 a+s\in S_0 a + s ∈ S 0 a + s + i 1 ≤ a + k − b < k a+s+i_1\le a+k-b<k a + s + i 1 ≤ a + k − b < k a + s ∈ Z a+s\in Z a + s ∈ Z s s s
i > a i>a i > a i i i Lemma 7.5 i i i i ≤ b i\le b i ≤ b i < b i<b i < b b b b C 0 ( i ) C_0(i) C 0 ( i ) x x x S 0 S_0 S 0 x ≤ i 1 x\le i_1 x ≤ i 1 a < i 1 a<i_1 a < i 1 x + i < i 1 + b ≤ 2 b ≤ k x+i<i_1+b\le2b\le k x + i < i 1 + b ≤ 2 b ≤ k C 0 ( i ) C_0(i) C 0 ( i ) x + i ∈ S 0 x+i\in S_0 x + i ∈ S 0 x x x x + i > i 1 x+i>i_1 x + i > i 1 x + i x+i x + i (42) , a contradiction.
This proves the claim. The same claim for the even element k − σ k-\sigma k − σ Lemma 7.3 (b) and Lemma 3.15 , gives a surviving tap reading 0 at every index for Q σ Q_\sigma Q σ Lemma 7.3 (a), ( σ , Q σ ) (\sigma,Q_\sigma) ( σ , Q σ ) σ \sigma σ
For the last sentence: either ( σ , Q σ ) (\sigma,Q_\sigma) ( σ , Q σ ) σ \sigma σ σ \sigma σ σ 1 = b \sigma_1=b σ 1 = b ( a , Q a ) (a,Q_a) ( a , Q a )
Odd minimum: where the failures sit
From now on a = min S a=\min S a = min S S 0 ≠ ∅ S_0\ne\varnothing S 0 = ∅ b = min S 0 b=\min S_0 b = min S 0 a < b ≤ k − b a<b\le k-b a < b ≤ k − b max S 1 = k − a \max S_1=k-a max S 1 = k − a max S 0 = k − b \max S_0=k-b max S 0 = k − b a + b < k a+b<k a + b < k b b b b b b
Lemma 7.8 (localisation). (a) If σ ∈ S \sigma\in S σ ∈ S i i i Q σ Q_\sigma Q σ C 0 ( i ) C_0(i) C 0 ( i ) i = a i=a i = a i i i a < i < b a<i<b a < i < b
(b) If σ ∈ S \sigma\in S σ ∈ S i i i Q σ Q_\sigma Q σ C 1 ( i ) C_1(i) C 1 ( i ) i i i i ≤ a i \le a i ≤ a i i i a < i ≤ b a < i \le b a < i ≤ b
Proof. The containments are Lemma 7.3 (e). (a) An even i < a i < a i < a Lemma 7.4 (ii), as i < a < b = min S 0 i < a < b = \min S_0 i < a < b = min S 0 i < a i < a i < a Lemma 7.4 (i), as i ≡ a i \equiv a i ≡ a i > a i > a i > a Lemma 7.5 makes i i i i ≤ b i \le b i ≤ b i < b i < b i < b i < a i < a i < a Lemma 7.4 (i), as i ≡ a + 1 i \equiv a + 1 i ≡ a + 1 i = a i = a i = a i > a i > a i > a Lemma 7.5 makes i i i i ≤ b i \le b i ≤ b
So a silent index of the even class sits at the end i = a i = a i = a ( a , b ) (a,b) ( a , b ) i ≤ a i \le a i ≤ a ( a , b ] (a,b] ( a , b ]
Lemma 7.9 (no inner failure of the even class). Let i 0 i_0 i 0 a < i 0 < b a < i_0 < b a < i 0 < b C 0 ( i 0 ) C_0(i_0) C 0 ( i 0 ) i 1 i_1 i 1 C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) i 1 i_1 i 1 i 1 ≤ a i_1 \le a i 1 ≤ a i 1 i_1 i 1 a < i 1 ≤ b a < i_1 \le b a < i 1 ≤ b
Proof. By Lemma 7.6 with c = 0 c = 0 c = 0 S S S ( i 0 , i 0 + b ) (i_0, i_0 + b) ( i 0 , i 0 + b )
If i 1 i_1 i 1 i 1 ≤ a i_1 \le a i 1 ≤ a b + i 1 < 2 b ≤ k b + i_1 < 2b \le k b + i 1 < 2 b ≤ k C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) S S S i 0 < b < b + i 1 < b + i 0 i_0 < b < b + i_1 < b + i_0 i 0 < b < b + i 1 < b + i 0
If i 1 i_1 i 1 x x x S S S x ≤ i 0 x \le i_0 x ≤ i 0 a < i 0 a < i_0 a < i 0 x + i 1 < 2 b ≤ k x+i_1 < 2b \le k x + i 1 < 2 b ≤ k x + i 1 ∈ S x+i_1 \in S x + i 1 ∈ S C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) x + i 1 > i 0 x+i_1 > i_0 x + i 1 > i 0 x x x x + i 1 ≥ i 0 + b x + i_1 \ge i_0 + b x + i 1 ≥ i 0 + b x ≤ i 0 x \le i_0 x ≤ i 0 i 1 = b i_1 = b i 1 = b a + b ∈ S a + b \in S a + b ∈ S i 0 < a + b < i 0 + b i_0 < a + b < i_0 + b i 0 < a + b < i 0 + b
The pairing of the two end failures is excluded by primitivity, and this is where the full support appears as the one obstruction. This proof and the next use an elementary observation. Say that a finite set W W W w w w closed under the step r > 0 r > 0 r > 0 y + r ∈ W y + r \in W y + r ∈ W y ∈ W y \in W y ∈ W y ≠ w y \ne w y = w y ∈ W y \in W y ∈ W y , y + r , y + 2 r , … y, y + r, y + 2r, \ldots y , y + r , y + 2 r , … W W W w w w W W W w w w r ∣ w − y r \mid w-y r ∣ w − y y , y + r , … , w y, y+r, \ldots, w y , y + r , … , w W W W
Lemma 7.10 (two end failures). Let S S S C 0 ( a ) C_0(a) C 0 ( a ) C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) i 1 i_1 i 1 i 1 ≤ a i_1 \le a i 1 ≤ a S S S
Proof. Put r = a + i 1 r = a + i_1 r = a + i 1
The odd class has step r r r Let y ∈ S 1 y \in S_1 y ∈ S 1 y < k − a y < k - a y < k − a a + y < k a + y < k a + y < k a + y ∈ S 0 a + y \in S_0 a + y ∈ S 0 C 0 ( a ) C_0(a) C 0 ( a ) a + y ≤ k − b a + y \le k - b a + y ≤ k − b a + y + i 1 < k a + y + i_1 < k a + y + i 1 < k i 1 < b i_1 < b i 1 < b y + r ∈ S 1 y + r \in S_1 y + r ∈ S 1 C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) S 1 S_1 S 1 r r r k − a k-a k − a y ∈ S 1 y \in S_1 y ∈ S 1 r ∣ k − a − y r \mid k-a-y r ∣ k − a − y r ∣ k − 2 a r \mid k-2a r ∣ k − 2 a
b = r b = r b = r k − b + i 1 < k k - b + i_1 < k k − b + i 1 < k S 1 S_1 S 1 C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) k − b + i 1 ≤ k − a k - b + i_1 \le k - a k − b + i 1 ≤ k − a C 0 ( a ) C_0(a) C 0 ( a ) a + i 1 + k − b < k a + i_1 + k - b < k a + i 1 + k − b < k S 0 S_0 S 0 max S 0 = k − b \max S_0 = k-b max S 0 = k − b k − b + i 1 = k − a k-b+i_1 = k-a k − b + i 1 = k − a b = r b = r b = r
The even class has step b b b Let y ∈ S 0 y \in S_0 y ∈ S 0 y < k − b y < k-b y < k − b y + i 1 < k y+i_1 < k y + i 1 < k y + i 1 ∈ S 1 y+i_1 \in S_1 y + i 1 ∈ S 1 C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) y + i 1 < k − b + i 1 = k − a y+i_1 < k-b+i_1 = k-a y + i 1 < k − b + i 1 = k − a a + y + i 1 < k a+y+i_1 < k a + y + i 1 < k S 0 S_0 S 0 C 0 ( a ) C_0(a) C 0 ( a ) y + b ∈ S 0 y+b \in S_0 y + b ∈ S 0 S 0 S_0 S 0 b b b k − b k-b k − b y ∈ S 0 y \in S_0 y ∈ S 0 b ∣ k − b − y b \mid k-b-y b ∣ k − b − y y = b y=b y = b b ∣ k b \mid k b ∣ k
Collapse. Now b ∣ k b \mid k b ∣ k b = r ∣ k − 2 a b = r \mid k-2a b = r ∣ k − 2 a b ∣ 2 a b \mid2a b ∣ 2 a b > a b > a b > a b = 2 a b = 2a b = 2 a a ∣ k a \mid k a ∣ k s ∈ S s \in S s ∈ S 2 a ∣ k − a − s 2a \mid k-a-s 2 a ∣ k − a − s s ∈ S s \in S s ∈ S 2 a ∣ k − 2 a − s 2a \mid k-2a-s 2 a ∣ k − 2 a − s a a a S ∪ { k } S \cup\{k\} S ∪ { k } a = 1 a = 1 a = 1 r = 2 r = 2 r = 2 a = 1 a = 1 a = 1 [ 1 , k − 1 ] [1,k-1] [ 1 , k − 1 ] S S S y ∈ ( 0 , k ) y \in(0,k) y ∈ ( 0 , k ) y − 1 ∈ S y-1 \in S y − 1 ∈ S 1 + ( y − 1 ) < k 1 + (y-1) < k 1 + ( y − 1 ) < k y ∈ S y \in S y ∈ S C 0 ( 1 ) C_0(1) C 0 ( 1 ) S S S
Lemma 7.11 (end against inner range). Suppose C 0 ( a ) C_0(a) C 0 ( a ) C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) i 1 i_1 i 1 a < i 1 ≤ b a < i_1 \le b a < i 1 ≤ b i 1 = b i_1 = b i 1 = b b ∣ k − 2 a b \mid k-2a b ∣ k − 2 a S 1 = { a , a + b , a + 2 b , … , k − a } S_1 = \{a, a+b, a+2b, \ldots, k-a\} S 1 = { a , a + b , a + 2 b , … , k − a }
Proof. By Lemma 7.6 with i = a i=a i = a c = 0 c=0 c = 0 S S S ( a , a + b ) (a,a+b) ( a , a + b ) a + i 1 ≤ a + b < k a+i_1\le a+b<k a + i 1 ≤ a + b < k S S S C 1 ( i 1 ) C_1(i_1) C 1 ( i 1 ) a + i 1 ≥ a + b a+i_1\ge a+b a + i 1 ≥ a + b i 1 = b i_1=b i 1 = b y ∈ S 1 y\in S_1 y ∈ S 1 y < k − a y<k-a y < k − a k − y ∈ S 1 k-y\in S_1 k − y ∈ S 1 a a a k − y ≥ a + b k-y\ge a+b k − y ≥ a + b y + b ≤ k − a y+b\le k-a y + b ≤ k − a y + b y+b y + b S 1 S_1 S 1 C 1 ( b ) C_1(b) C 1 ( b ) S 1 S_1 S 1 b b b k − a k-a k − a S 1 S_1 S 1 ≡ k − a ( m o d b ) \equiv k-a\pmod b ≡ k − a ( mod b ) a a a k − a k-a k − a S 1 S_1 S 1 b ∣ k − 2 a b\mid k-2a b ∣ k − 2 a S 1 S_1 S 1
Theorem 7.12 (the both-failing structure). Let k k k S S S a = min S a=\min S a = min S S 0 ≠ ∅ S_0\ne\varnothing S 0 = ∅ b = min S 0 b=\min S_0 b = min S 0 ( σ , Q σ ) (\sigma,Q_\sigma) ( σ , Q σ ) σ \sigma σ σ ∈ S \sigma\in S σ ∈ S
k = 2 a + M b for an integer M ≥ 1 , a < b < 2 a , k=2a+Mb\quad\text{for an integer }M\ge1,\qquad a<b<2a, k = 2 a + M b for an integer M ≥ 1 , a < b < 2 a , S 1 = { a + m b : 0 ≤ m ≤ M } , 2 a + m b ∈ S for 0 ≤ m < M . S_1=\{a+mb:0\le m\le M\},\qquad2a+mb\in S\quad\text{for }0\le m<M. S 1 = { a + mb : 0 ≤ m ≤ M } , 2 a + mb ∈ S for 0 ≤ m < M . Proof. By Lemma 7.3 (d) applied at σ = b \sigma=b σ = b i 0 i_0 i 0 Q σ 0 Q_{\sigma_0} Q σ 0 σ 0 \sigma_0 σ 0 σ = a \sigma=a σ = a i 1 i_1 i 1 Q σ 1 Q_{\sigma_1} Q σ 1 σ 1 \sigma_1 σ 1 Lemma 7.8 locates both, and Lemma 7.9 rules out i 0 ≠ a i_0\ne a i 0 = a C 0 ( a ) C_0(a) C 0 ( a )
Next, let σ ∈ S \sigma\in S σ ∈ S i i i Q σ Q_\sigma Q σ Lemma 7.8 (b) and Lemma 7.10 (with C 0 ( a ) C_0(a) C 0 ( a ) S S S i i i a < i ≤ b a<i\le b a < i ≤ b Lemma 7.11 gives
i = b , b ∣ k − 2 a , S 1 = { a , a + b , … , k − a } . (43) i=b,\qquad b\mid k-2a,\qquad S_1=\{a,a+b,\ldots,k-a\}. \tag*{(43)} i = b , b ∣ k − 2 a , S 1 = { a , a + b , … , k − a } . (43) Applying this to i 1 i_1 i 1 S 1 S_1 S 1 M = ( k − 2 a ) / b M=(k-2a)/b M = ( k − 2 a ) / b M ≥ 1 M\ge1 M ≥ 1 2 a < k 2a<k 2 a < k (41) . For 0 ≤ m < M 0\le m<M 0 ≤ m < M a + m b a+mb a + mb a + ( a + m b ) ≤ 2 a + ( M − 1 ) b < k a+(a+mb)\le2a+(M-1)b<k a + ( a + mb ) ≤ 2 a + ( M − 1 ) b < k C 0 ( a ) C_0(a) C 0 ( a ) 2 a + m b 2a+mb 2 a + mb S S S m = 0 m=0 m = 0 2 a ∈ S 0 2a\in S_0 2 a ∈ S 0 b ≤ 2 a b\le2a b ≤ 2 a
It remains to exclude b = 2 a b=2a b = 2 a M = 1 M=1 M = 1 k = 4 a k=4a k = 4 a S 1 ⊆ { a , 3 a } S_1\subseteq\{a,3a\} S 1 ⊆ { a , 3 a } S 0 ⊆ [ b , k − b ] = { 2 a } S_0\subseteq[b,k-b]=\{2a\} S 0 ⊆ [ b , k − b ] = { 2 a } a a a S ∪ { k } S\cup\{k\} S ∪ { k } a = 1 a=1 a = 1 k = 4 k=4 k = 4 S ⊇ { a , b , a + b } = { 1 , 2 , 3 } S\supseteq\{a,b,a+b\}=\{1,2,3\} S ⊇ { a , b , a + b } = { 1 , 2 , 3 } M ≥ 2 M\ge2 M ≥ 2 3 a = a + b ∈ S 1 3a=a+b\in S_1 3 a = a + b ∈ S 1 Lemma 7.3 (d) some index i 3 i_3 i 3 Q σ 3 Q_{\sigma_3} Q σ 3 σ 3 ∈ { 3 a , k − 3 a } \sigma_3\in\{3a,k-3a\} σ 3 ∈ { 3 a , k − 3 a } (43) i 3 = b = 2 a i_3=b=2a i 3 = b = 2 a k − a k-a k − a k + a k+a k + a a a a a ≠ σ 3 a\ne\sigma_3 a = σ 3 k − 3 a = a k-3a=a k − 3 a = a k = 4 a k=4a k = 4 a M = 1 M=1 M = 1 i 3 i_3 i 3
Carriers for the both-failing supports
The structure of Theorem 7.12 leaves the even elements of S S S ( b , k − b ) (b,k-b) ( b , k − b )
Lemma 7.13 (the case M = 1 M=1 M = 1 a a a b b b a < b < 2 a a<b<2a a < b < 2 a k = 2 a + b k=2a+b k = 2 a + b S ⊆ { 1 , … , k − 1 } S\subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } a , b ∈ S a,b\in S a , b ∈ S { a , a + b } \{a,a+b\} { a , a + b } [ b , 2 a ] [b,2a] [ b , 2 a ] σ = a \sigma=a σ = a
Q = { a , 3 a , b − a } ∪ { y ∈ { 1 , … , k − 1 } : y even , y ∉ S } Q=\{a,3a,b-a\}\cup\{y\in\{1,\ldots,k-1\}:y\text{ even},\ y\notin S\} Q = { a , 3 a , b − a } ∪ { y ∈ { 1 , … , k − 1 } : y even , y ∈ / S } is a σ \sigma σ
Proof. By palindromy a + b = k − a a+b=k-a a + b = k − a 2 a = k − b 2a=k-b 2 a = k − b S S S 3 a < k 3a<k 3 a < k b − a > 0 b-a>0 b − a > 0 a a a a + b a+b a + b b ≠ 2 a b\ne2a b = 2 a Q ∩ S = { a } Q\cap S=\{a\} Q ∩ S = { a } Q Q Q a a a 3 a 3a 3 a b − a b-a b − a Q Q Q ( 0 , b ) ∪ ( 2 a , k ) (0,b)\cup(2a,k) ( 0 , b ) ∪ ( 2 a , k ) Q Q Q
For (B2) it suffices, at each index i i i a a a b b b 2 a 2a 2 a a + b a+b a + b a a a b b b 2 a 2a 2 a a + b a+b a + b 2 b 2b 2 b 3 a 3a 3 a
For M ≥ 2 M \ge2 M ≥ 2 band carrier : a a a a a a k − a k-a k − a
Lemma 7.14 (the case M ≥ 2 M \ge2 M ≥ 2 S S S Theorem 7.12 with M ≥ 2 M \ge2 M ≥ 2
B = { y ∉ S : a < y < k − a } . \mathcal{B}=\{y\notin S:a<y<k-a\}. B = { y ∈ / S : a < y < k − a } . Then σ = a \sigma=a σ = a Q = { a } ∪ B Q=\{a\}\cup\mathcal{B} Q = { a } ∪ B σ \sigma σ
The idea is that at an index i > a i>a i > a k − a k-a k − a i − a i-a i − a i − a i-a i − a B \mathcal{B} B S S S a a a j j j S S S i i i k − j k-j k − j B \mathcal{B} B
Proof. (N1) For 1 ≤ m ≤ M 1\le m\le M 1 ≤ m ≤ M m b = k − ( 2 a + ( M − m ) b ) ∈ S mb=k-(2a+(M-m)b)\in S mb = k − ( 2 a + ( M − m ) b ) ∈ S
(N2) An odd y ∈ [ a , k − a ] y\in[a,k-a] y ∈ [ a , k − a ] y − a y-a y − a b b b S 1 = { a + m b } S_1=\{a+mb\} S 1 = { a + mb }
(N3) If a < x ≤ k − a a<x\le k-a a < x ≤ k − a j ∈ S j\in S j ∈ S x ≤ j < x + a x\le j<x+a x ≤ j < x + a
For (N3), I first show that every s ∈ S s\in S s ∈ S s < k − a = a + M b s<k-a=a+Mb s < k − a = a + M b S S S ( s , s + a ] (s,s+a] ( s , s + a ] s = q b + r s=qb+r s = q b + r 0 ≤ r < b 0\le r<b 0 ≤ r < b r < a r<a r < a q ≤ M q\le M q ≤ M q b ≤ s < a + M b < ( M + 1 ) b qb\le s<a+Mb<(M+1)b q b ≤ s < a + M b < ( M + 1 ) b a + q b ∈ S a+qb\in S a + q b ∈ S ( s , s + a ] (s,s+a] ( s , s + a ] r ≥ a r\ge a r ≥ a q b + a ≤ s < a + M b qb+a\le s<a+Mb q b + a ≤ s < a + M b q < M q<M q < M ( q + 1 ) b ∈ S (q+1)b\in S ( q + 1 ) b ∈ S ( s , s + a ] (s,s+a] ( s , s + a ] ( q + 1 ) b − s = b − r ≤ b − a < a (q+1)b-s=b-r\le b-a<a ( q + 1 ) b − s = b − r ≤ b − a < a x x x s s s S S S x x x a a a s < k − a s<k-a s < k − a j ∈ S j\in S j ∈ S ( s , s + a ] (s,s+a] ( s , s + a ] j ≥ x j\ge x j ≥ x s s s j ≤ s + a < x + a j\le s+a<x+a j ≤ s + a < x + a
The set B \mathcal{B} B Q ∩ S = { a } Q\cap S=\{a\} Q ∩ S = { a } B \mathcal{B} B y ↦ k − y y\mapsto k-y y ↦ k − y Q ′ = { k − a } ∪ B Q'=\{k-a\}\cup\mathcal{B} Q ′ = { k − a } ∪ B Q Q Q
Reads of 1. Let z ∈ { a , k − a } z\in\{a,k-a\} z ∈ { a , k − a } Q z = { z } ∪ B Q_z=\{z\}\cup\mathcal{B} Q z = { z } ∪ B i i i ( z , Q z ) (z,Q_z) ( z , Q z ) z ∈ S z\in S z ∈ S z ∉ B z\notin\mathcal{B} z ∈ / B B \mathcal{B} B
If i < a i<a i < a i i i a a a i + a i+a i + a ( a , 2 a ) ⊂ ( a , k − a ) (a,2a)\subset(a,k-a) ( a , 2 a ) ⊂ ( a , k − a ) ( i + a ) − a = i ∈ ( 0 , b ) (i+a)-a=i\in(0,b) ( i + a ) − a = i ∈ ( 0 , b ) B \mathcal{B} B
If i < a i<a i < a i i i b b b i + b i+b i + b ( b , a + b ) ⊂ ( a , k − a ) (b,a+b)\subset(a,k-a) ( b , a + b ) ⊂ ( a , k − a ) ( i + b ) − a ∈ ( b − a , b ) ⊂ ( 0 , b ) (i+b)-a\in(b-a,b)\subset(0,b) ( i + b ) − a ∈ ( b − a , b ) ⊂ ( 0 , b ) B \mathcal{B} B
If i = a i=a i = a 2 a ∈ S 2a\in S 2 a ∈ S 3 a 3a 3 a 3 a − a = 2 a ∈ ( b , 2 b ) 3a-a=2a\in(b,2b) 3 a − a = 2 a ∈ ( b , 2 b ) ( a , k − a ) (a,k-a) ( a , k − a ) 2 a < 2 b ≤ M b 2a<2b\le Mb 2 a < 2 b ≤ M b B \mathcal{B} B
If i > a i>a i > a k − a k-a k − a y = i − a ∈ ( 0 , k − a ) y=i-a\in(0,k-a) y = i − a ∈ ( 0 , k − a ) y ∉ B y\notin\mathcal{B} y ∈ / B y ≠ z y\ne z y = z
If i > a i>a i > a i − a = z i-a=z i − a = z z ≠ k − a z\ne k-a z = k − a z = a z=a z = a i = 2 a i=2a i = 2 a k − b k-b k − b 2 a > b 2a>b 2 a > b 2 a − b ∈ ( 0 , a ) 2a-b\in(0,a) 2 a − b ∈ ( 0 , a ) B \mathcal{B} B z z z
If i > a i>a i > a i − a ∈ B i-a\in\mathcal{B} i − a ∈ B a < i − a < k − a a<i-a<k-a a < i − a < k − a x = i − a x=i-a x = i − a j ∈ S j\in S j ∈ S i − a ≤ j < i i-a\le j<i i − a ≤ j < i k − j k-j k − j y = i − j ∈ ( 0 , a ) y=i-j\in(0,a) y = i − j ∈ ( 0 , a ) B \mathcal{B} B y = z y=z y = z z ≤ a < k − a z\le a<k-a z ≤ a < k − a z = a z=a z = a j = i − a ∈ B j=i-a\in\mathcal{B} j = i − a ∈ B j ∈ S j\in S j ∈ S y ≠ z y\ne z y = z
With z = a z=a z = a ( a , Q ) (a,Q) ( a , Q ) z = k − a z=k-a z = k − a k − i k-i k − i Lemma 3.15 gives a surviving read of 0 at i i i ( a , Q ) (a,Q) ( a , Q ) □ \square □
Corollary 7.15 (the both-failing supports). Under the hypotheses of Theorem 7.12 , S S S σ \sigma σ σ = a \sigma=a σ = a
Proof. If M ≥ 2 M\ge2 M ≥ 2 Lemma 7.14 . If M = 1 M=1 M = 1 k = 2 a + b k=2a+b k = 2 a + b S S S a a a a + b a+b a + b s s s b ≤ s b\le s b ≤ s b ≤ k − s b\le k-s b ≤ k − s s ∈ [ b , 2 a ] s\in[b,2a] s ∈ [ b , 2 a ] □ \square □
Proof of Theorem 7.1
Proof of Theorem 7.1 . Let a = min S a = \min S a = min S a a a Theorem 7.7 gives a σ \sigma σ a a a ( σ , Q σ ) (\sigma,Q_\sigma) ( σ , Q σ ) σ \sigma σ σ ∈ S \sigma\in S σ ∈ S S S S Theorem 7.12 , since it is primitive, palindromic, not full and has an even element; and then Corollary 7.15 gives a σ \sigma σ σ = a \sigma= a σ = a
By Theorem 3.14 , no equation (8) whose support satisfies the hypotheses of Theorem 7.1 is globally periodic, for either value of θ \theta θ
Even orders: all-odd supports
This section treats the supports at even order all of whose elements are odd, the hardest supports for the σ \sigma σ Definition 3.11 then split by parity (§ 8.1 ).
Theorem 8.1 (all-odd supports). Let k k k S ⊆ { 1 , … , k − 1 } S \subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } ∣ S ∣ ≥ 3 |S| \ge3 ∣ S ∣ ≥ 3 ( k , S ) ≠ ( 6 , { 1 , 3 , 5 } ) (k,S) \ne(6,\{1,3,5\}) ( k , S ) = ( 6 , { 1 , 3 , 5 }) S S S σ \sigma σ
The support { 1 , 3 , 5 } \{1,3,5\} { 1 , 3 , 5 } k = 6 k = 6 k = 6 σ \sigma σ σ ∈ { 1 , 3 , 5 } \sigma\in\{1,3,5\} σ ∈ { 1 , 3 , 5 } Q Q Q Proposition 3.10 .
Supports containing 1 are handled in § 8.2 by small explicit carriers. For the others, with a = min S ≥ 3 a = \min S \ge3 a = min S ≥ 3 half-arc : a free position y y y Q Q Q ν y \nu y ν y 2 k 2k 2 k t ∈ { 0 , k } t \in\{0,k\} t ∈ { 0 , k } y y y [ 0 , k ) [0,k) [ 0 , k ) ν \nu ν i i i ν i + ν S \nu i+\nu S ν i + ν S i i i S S S z ↦ k − z z \mapsto k-z z ↦ k − z ν \nu ν S S S k / 2 k/2 k /2 § 8.3 ). At even indices, for a unit ν \nu ν t t t ν S \nu S ν S Z / 2 k \mathbb{Z}/2k Z /2 k § 8.5 ); the one gap is the index 2 a 2a 2 a ν \nu ν Theorem 9.1 , supplies (§ 8.6 ). The supports { a , k / 2 , k − a } \{a,k/2,k-a\} { a , k /2 , k − a } k / 2 k/2 k /2 ν \nu ν § 8.4 ), and seventeen supports at orders k ≤ 42 k \le42 k ≤ 42
The parity split
Throughout this section k k k S ⊆ { 1 , … , k − 1 } S \subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } § 3.5 , in particular (22) and (23).
Suppose now that every element of S S S Q Q Q σ \sigma σ k − i k-i k − i S S S k k k ≡ 0 ( m o d k ) \equiv0 \pmod{k} ≡ 0 ( mod k ) σ \sigma σ σ + k \sigma+k σ + k
Supports containing 1
When 1 ∈ S 1 \in S 1 ∈ S k − 1 k-1 k − 1
Lemma 8.2 (a three-point carrier). Let k k k S S S 1 ∈ S 1 \in S 1 ∈ S b ∈ S b \in S b ∈ S 5 ≤ b ≤ k / 2 5 \le b \le k/2 5 ≤ b ≤ k /2 σ = b \sigma=b σ = b Q = { b − 1 , b , b + 1 } Q=\{b-1,b,b+1\} Q = { b − 1 , b , b + 1 } σ \sigma σ S S S
Proof. (B1) holds because b ± 1 b \pm1 b ± 1 0 ∉ Q 0 \notin Q 0 ∈ / Q k ≥ 2 b ≥ 10 k \ge2b \ge10 k ≥ 2 b ≥ 10 1 , b , k − b , k − 1 ∈ S 1,b,k-b,k-1 \in S 1 , b , k − b , k − 1 ∈ S
Even indices. Here every position is odd, and the only odd element of Q Q Q b = σ b=\sigma b = σ k − 1 k-1 k − 1 i + k − 1 i+k-1 i + k − 1 i + k − 1 = k + b i+k-1=k+b i + k − 1 = k + b i = b + 1 i=b+1 i = b + 1 i = b + 1 i=b+1 i = b + 1 k − b k-b k − b k + 1 k+1 k + 1 i + 1 < k i+1<k i + 1 < k i + 1 = b i+1=b i + 1 = b i = b − 1 i=b-1 i = b − 1 b b b 2 b − 1 < k 2b-1<k 2 b − 1 < k
Odd indices. Here every position is even and the only deleted position is k k k i ≥ 3 i \ge3 i ≥ 3 k − 1 k-1 k − 1 i + k − 1 i+k-1 i + k − 1 i − 1 i-1 i − 1 i − 1 = b ± 1 i-1=b \pm1 i − 1 = b ± 1 i = 1 i=1 i = 1 b b b b + 1 ∈ Q b+1 \in Q b + 1 ∈ Q i = b i=b i = b b + 1 b+1 b + 1 i = b + 2 i=b+2 i = b + 2 k − b k-b k − b k + 2 k+2 k + 2 2 ∉ Q 2 \notin Q 2 ∈ / Q b − 1 ≥ 4 b-1 \ge4 b − 1 ≥ 4 i = k − 1 i=k-1 i = k − 1 i + 1 i+1 i + 1 i + 1 = b ± 1 i+1=b \pm1 i + 1 = b ± 1 i = k − 1 i=k-1 i = k − 1 i = b i=b i = b b b b k − 1 k-1 k − 1 k + b − 1 k+b-1 k + b − 1 b − 1 ∈ Q b-1 \in Q b − 1 ∈ Q i = b − 2 i=b-2 i = b − 2 b b b 2 b − 2 < k 2b-2<k 2 b − 2 < k Q Q Q b ≥ 5 b \ge5 b ≥ 5
Lemma 8.3 (a carrier when 3 ∈ S 3 \in S 3 ∈ S k ≥ 8 k \ge8 k ≥ 8 S S S 1 , 3 ∈ S 1,3 \in S 1 , 3 ∈ S σ = 3 \sigma=3 σ = 3 Q = { 2 , 3 , 6 } Q=\{2,3,6\} Q = { 2 , 3 , 6 } σ \sigma σ S S S
Proof. Since k ≥ 8 k \ge8 k ≥ 8 Q ⊆ { 1 , … , k − 1 } Q \subseteq\{1,\ldots,k-1\} Q ⊆ { 1 , … , k − 1 } k − 3 k-3 k − 3 k − 1 k-1 k − 1
Even indices. As in Lemma 8.2 , every position is odd and the only odd element of Q Q Q σ \sigma σ k − 1 k-1 k − 1 i + k − 1 i+k-1 i + k − 1 i − 1 i-1 i − 1 i = 4 i=4 i = 4 k − 3 k-3 k − 3 k + 1 k+1 k + 1 i + 1 < k i+1<k i + 1 < k i = 2 i=2 i = 2 5 < k 5<k 5 < k
Odd indices. Every position is even, the only deleted position is k k k Q Q Q { 2 , 6 } \{2,6\} { 2 , 6 } { 2 , 6 } \{2,6\} { 2 , 6 } i = 1 i=1 i = 1 i = 3 i=3 i = 3 i = 7 i=7 i = 7 k − 3 k-3 k − 3 k + 4 k+4 k + 4 i ≥ 5 i \ge5 i ≥ 5 i ≠ 7 i \ne7 i = 7 k − 1 k-1 k − 1 k + i − 1 k+i-1 k + i − 1 i − 1 ∉ { 0 , 2 , 6 } i-1 \notin\{0,2,6\} i − 1 ∈ / { 0 , 2 , 6 } i = 1 i=1 i = 1 i = 5 i=5 i = 5 i = k − 1 i=k-1 i = k − 1 k − 3 k-3 k − 3 k + 2 k+2 k + 2 i ≤ k − 3 i \le k-3 i ≤ k − 3 i ≠ 1 , 5 i \ne1,5 i = 1 , 5 i + 1 < k i+1<k i + 1 < k i + 1 ∉ { 0 , 2 , 6 } i+1 \notin\{0,2,6\} i + 1 ∈ / { 0 , 2 , 6 }
Proposition 8.4 (all-odd supports containing 1). Let k k k S ⊆ { 1 , … , k − 1 } S \subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } 1 ∈ S 1 \in S 1 ∈ S ∣ S ∣ ≥ 3 |S| \ge3 ∣ S ∣ ≥ 3 ( k , S ) ≠ ( 6 , { 1 , 3 , 5 } ) (k,S) \ne(6,\{1,3,5\}) ( k , S ) = ( 6 , { 1 , 3 , 5 }) S S S σ \sigma σ
Proof. If 3 ∈ S 3 \in S 3 ∈ S k ≥ 8 k \ge8 k ≥ 8 k ≤ 6 k \le6 k ≤ 6 { 1 , … , k − 1 } \{1,\ldots,k-1\} { 1 , … , k − 1 } ∣ S ∣ ≥ 3 |S| \ge3 ∣ S ∣ ≥ 3 ( 6 , { 1 , 3 , 5 } ) (6,\{1,3,5\}) ( 6 , { 1 , 3 , 5 }) Lemma 8.3 applies. If 3 ∉ S 3 \notin S 3 ∈ / S k − 3 ∉ S k-3 \notin S k − 3 ∈ / S S S S k − 1 k-1 k − 1 b ∈ S b \in S b ∈ S 5 ≤ b ≤ k / 2 5 \le b \le k/2 5 ≤ b ≤ k /2 Lemma 8.2 applies. □
Half-arc carriers and the odd indices
From here on 1 ∉ S 1 \notin S 1 ∈ / S a = min S a=\min S a = min S a ≥ 3 a \ge3 a ≥ 3 σ = a \sigma=a σ = a L = { s ∈ S : 2 s < k } L=\{s \in S: 2s<k\} L = { s ∈ S : 2 s < k } S S S S S S S = L ∪ ( k − L ) S=L \cup(k-L) S = L ∪ ( k − L ) k / 2 k/2 k /2 k / 2 ∈ S k/2 \in S k /2 ∈ S a ∈ L a \in L a ∈ L S = { k / 2 } S=\{k/2\} S = { k /2 }
For integers v v v y y y z z z
H ( z ) = 1 [ z m o d 2 k < k ] , H v ( y ) = H ( v y ) , (44) H(z)=\mathbf{1}[z \bmod2k<k], \qquad H_v(y)=H(vy), \tag*{(44)} H ( z ) = 1 [ z mod 2 k < k ] , H v ( y ) = H ( v y ) , (44) where z z z 2 k 2k 2 k { 0 , … , 2 k − 1 } \{0,\ldots,2k-1\} { 0 , … , 2 k − 1 } H ν ( y ) H_\nu(y) H ν ( y ) Z / 2 k \mathbb{Z}/2k Z /2 k ν y \nu y ν y ν \nu ν ν k ≡ k ( m o d 2 k ) \nu k \equiv k \pmod{2k} ν k ≡ k ( mod 2 k ) z z z y y y
H ( z + k ) = 1 − H ( z ) , H ν ( y + k ) = 1 − H ν ( y ) . (45) H(z+k)=1-H(z),\qquad H_\nu(y+k)=1-H_\nu(y). \tag*{(45)} H ( z + k ) = 1 − H ( z ) , H ν ( y + k ) = 1 − H ν ( y ) . (45) If also y y y ν y \nu y ν y ν y ≢ 0 , k ( m o d 2 k ) \nu y\not\equiv0,k\pmod{2k} ν y ≡ 0 , k ( mod 2 k )
H ν ( − y ) = 1 − H ν ( y ) , H ν ( k − y ) = H ν ( y ) . (46) H_\nu(-y)=1-H_\nu(y),\qquad H_\nu(k-y)=H_\nu(y). \tag*{(46)} H ν ( − y ) = 1 − H ν ( y ) , H ν ( k − y ) = H ν ( y ) . (46) the second because ν ( k − y ) ≡ k − ν y \nu(k-y)\equiv k-\nu y ν ( k − y ) ≡ k − ν y z ↦ k − z z\mapsto k-z z ↦ k − z U = { 1 , … , k − 1 } U=\{1,\ldots,k-1\} U = { 1 , … , k − 1 } V = { k + 1 , … , 2 k − 1 } V=\{k+1,\ldots,2k-1\} V = { k + 1 , … , 2 k − 1 } Z / 2 k \mathbb{Z}/2k Z /2 k
Definition 8.5 (half-arc carrier). Let ν \nu ν t ∈ { 0 , k } t\in\{0,k\} t ∈ { 0 , k } y y y t y = 0 t_y=0 t y = 0 y y y t y = t t_y=t t y = t y y y half-arc carrier Q ( ν ; t ) Q(\nu;t) Q ( ν ; t ) S S S a a a y ∈ { 1 , … , k − 1 } ∖ S y\in\{1,\ldots,k-1\}\setminus S y ∈ { 1 , … , k − 1 } ∖ S H ( ν y + t y ) = 1 H(\nu y+t_y)=1 H ( ν y + t y ) = 1
So the even free positions follow the half-arc of ν \nu ν t t t Q ∩ ( { 0 } ∪ S ) = { a } Q\cap(\{0\}\cup S)=\{a\} Q ∩ ({ 0 } ∪ S ) = { a } σ = a \sigma=a σ = a
Lemma 8.6 (free reads). Let Q = Q ( ν ; t ) Q=Q(\nu;t) Q = Q ( ν ; t ) ν \nu ν 0 < p < 2 k 0<p<2k 0 < p < 2 k p ∈ P p\in P p ∈ P H ( ν p + t p ) = 1 H(\nu p+t_p)=1 H ( ν p + t p ) = 1
Proof. If p < k p<k p < k p ≠ a p\ne a p = a p > k p>k p > k p ′ = p − k p'=p-k p ′ = p − k p p p S ∪ { 0 } S\cup\{0\} S ∪ { 0 } p ∈ P p\in P p ∈ P p ′ ∉ Q p'\notin Q p ′ ∈ / Q H ( ν p ′ + t p ) = 0 H(\nu p'+t_p)=0 H ( ν p ′ + t p ) = 0 H ( ν p ′ + t p + k ) = 1 H(\nu p'+t_p+k)=1 H ( ν p ′ + t p + k ) = 1 (45) , and ν p ′ + k ≡ ν p ( m o d 2 k ) \nu p'+k\equiv\nu p\pmod{2k} ν p ′ + k ≡ ν p ( mod 2 k )
The clauses at odd indices read only even positions, and all of these are free except the deleted position k k k S S S Z / 2 k \mathbb{Z}/2k Z /2 k
Lemma 8.7 (odd indices). Let S S S ν \nu ν t ∈ { 0 , k } t\in\{0,k\} t ∈ { 0 , k } H ν ( u ) ≠ H ν ( v ) H_\nu(u)\ne H_\nu(v) H ν ( u ) = H ν ( v ) u , v ∈ L u,v\in L u , v ∈ L σ = a \sigma=a σ = a Q = Q ( ν ; t ) Q=Q(\nu;t) Q = Q ( ν ; t )
Proof. Let i i i § 8.1 , the only deleted tap is k − i k-i k − i S S S j j j p = i + j p=i+j p = i + j H ( ν p ) = H ( ν i + ν j ) H(\nu p)=H(\nu i+\nu j) H ( ν p ) = H ( ν i + ν j ) Lemma 8.6 . So the clause fails exactly when the phases ν j \nu j ν j 2 k 2k 2 k { − ν i , … , − ν i + k − 1 } \{-\nu i,\ldots,-\nu i+k-1\} { − ν i , … , − ν i + k − 1 } { − ν i + k , … , − ν i + 2 k − 1 } \{-\nu i+k,\ldots,-\nu i+2k-1\} { − ν i + k , … , − ν i + 2 k − 1 }
First I locate the phases. They are odd, so each lies in U U U V V V (46) the tap k − j k-j k − j k − ν j k-\nu j k − ν j ν j \nu j ν j H ν ( u ) ≠ H ν ( v ) H_\nu(u)\ne H_\nu(v) H ν ( u ) = H ν ( v ) x − x_- x − U U U u 0 u_0 u 0 y − y_- y − V V V v 0 v_0 v 0 k − u 0 k-u_0 k − u 0 k − v 0 k-v_0 k − v 0 k − x − ∈ U k-x_-\in U k − x − ∈ U 3 k − y − ∈ V 3k-y_-\in V 3 k − y − ∈ V 2 x − ≤ k 2x_-\le k 2 x − ≤ k 2 y − ≤ 3 k 2y_-\le3k 2 y − ≤ 3 k x − ≤ k − x − < y − ≤ 3 k − y − x_-\le k-x_-<y_-\le3k-y_- x − ≤ k − x − < y − ≤ 3 k − y − k − 2 x − k-2x_- k − 2 x − x − + y − − k x_-+y_--k x − + y − − k 3 k − 2 y − 3k-2y_- 3 k − 2 y − x − + y − − k x_-+y_--k x − + y − − k k k k k k k x − = k / 2 x_-=k/2 x − = k /2 y − = 3 k / 2 y_-=3k/2 y − = 3 k /2 k − 1 k-1 k − 1 k − 1 k-1 k − 1 m m m 2 k − m 2k-m 2 k − m 2 k − m ≤ k − 1 2k-m\le k-1 2 k − m ≤ k − 1
(a) no k k k
(b) if k + 1 k+1 k + 1 x − = k / 2 x_-=k/2 x − = k /2 y − = 3 k / 2 y_-=3k/2 y − = 3 k /2 k / 2 k/2 k /2 3 k / 2 3k/2 3 k /2 z ∈ U z\in U z ∈ U z ≥ x − z\ge x_- z ≥ x − k − z ≥ x − k-z\ge x_- k − z ≥ x − V V V
If k − i ∉ S k-i \notin S k − i ∈ / S s 0 = k − i ∈ S s_0=k-i\in S s 0 = k − i ∈ S ν i ≡ k − ν s 0 \nu_i\equiv k-\nu_{s_0} ν i ≡ k − ν s 0 (45) the tap j ≠ s 0 j\ne s_0 j = s 0 1 − H ( ν j − ν s 0 ) 1-H(\nu_j-\nu_{s_0}) 1 − H ( ν j − ν s 0 ) k k k { ν s 0 , … , ν s 0 + k − 1 } \{\nu_{s_0},\ldots,\nu_{s_0}+k-1\} { ν s 0 , … , ν s 0 + k − 1 } ν s 0 \nu_{s_0} ν s 0 { ν s 0 + k , … , ν s 0 + 2 k − 1 } \{\nu_{s_0}+k,\ldots,\nu_{s_0}+2k-1\} { ν s 0 + k , … , ν s 0 + 2 k − 1 } ν s 0 \nu_{s_0} ν s 0 k + 1 k+1 k + 1 { ν s 0 + k , … , ν s 0 + 2 k } \{\nu_{s_0}+k,\ldots,\nu_{s_0}+2k\} { ν s 0 + k , … , ν s 0 + 2 k } { k / 2 , 3 k / 2 } \{k/2,3k/2\} { k /2 , 3 k /2 } j ≠ s 0 j\ne s_0 j = s 0 ν j ≡ ν s 0 \nu_j\equiv\nu_{s_0} ν j ≡ ν s 0 u u u k − u k-u k − u 2 u < k 2u<k 2 u < k v v v k − v k-v k − v s 0 s_0 s 0 s 0 s_0 s 0
Supports with one element below k / 2 k/2 k /2
If 1 ∉ S 1\notin S 1 ∈ / S L = { a } L=\{a\} L = { a } ∣ S ∣ ≥ 3 |S|\ge3 ∣ S ∣ ≥ 3 S = { a , k / 2 , k − a } S=\{a,k/2,k-a\} S = { a , k /2 , k − a } k / 2 k/2 k /2 Lemma 8.7 does not apply, since L L L
Theorem 8.8 (three-element supports). Let k = 2 h k=2h k = 2 h h h h S = { a , h , k − a } S=\{a,h,k-a\} S = { a , h , k − a } 3 ≤ a < h 3\le a<h 3 ≤ a < h gcd ( a , h ) = 1 \gcd(a,h)=1 g cd( a , h ) = 1 β ∈ { 2 , 4 } \beta\in\{2,4\} β ∈ { 2 , 4 } β ≡ a + h ( m o d 4 ) \beta\equiv a+h\pmod{4} β ≡ a + h ( mod 4 ) 3 β < h 3\beta<h 3 β < h h ≥ 7 h\ge7 h ≥ 7 β = 2 \beta=2 β = 2 h ≥ 13 h\ge13 h ≥ 13 β = 4 \beta=4 β = 4 ν ≡ a − 1 ( 3 h + β ) ( m o d 2 k ) \nu\equiv a^{-1}(3h+\beta)\pmod{2k} ν ≡ a − 1 ( 3 h + β ) ( mod 2 k ) σ = a \sigma=a σ = a Q = Q ( ν ; 0 ) Q=Q(\nu;0) Q = Q ( ν ; 0 ) σ \sigma σ S S S
Proof. All residues in this proof are modulo 2 k = 4 h 2k=4h 2 k = 4 h a a a h h h ν \nu ν ν a ≡ 3 h + β \nu a\equiv3h+\beta ν a ≡ 3 h + β a 2 ≡ 1 ( m o d 4 ) a^2\equiv1\pmod{4} a 2 ≡ 1 ( mod 4 ) 3 h + β ≡ 3 h + a + h ≡ a ( m o d 4 ) 3h+\beta\equiv3h+a+h\equiv a\pmod{4} 3 h + β ≡ 3 h + a + h ≡ a ( mod 4 ) ν ≡ a ( 3 h + β ) ≡ 1 ( m o d 4 ) \nu\equiv a(3h+\beta)\equiv1\pmod{4} ν ≡ a ( 3 h + β ) ≡ 1 ( mod 4 ) ν \nu ν ν h ≡ h \nu h\equiv h ν h ≡ h
ν a ≡ 3 h + β , ν h ≡ h , ν ( k − a ) ≡ k − ν a ≡ 3 h − β . \nu a\equiv3h+\beta,\qquad\nu h\equiv h,\qquad\nu(k-a)\equiv k-\nu a\equiv3h-\beta. ν a ≡ 3 h + β , ν h ≡ h , ν ( k − a ) ≡ k − ν a ≡ 3 h − β . (B1) holds by construction. By Lemma 8.6 , a tap reading a free position p p p H ( ν p ) = H ( ν i + ν j ) H(\nu p)=H(\nu i+\nu j) H ( ν p ) = H ( ν i + ν j ) j j j
Deletions and pins. The deleted positions are k k k a a a k + a k+a k + a h h h k − a k-a k − a k + h k+h k + h 2 k − a 2k-a 2 k − a a a a k + a k+a k + a i + j = p i+j=p i + j = p j ∈ S j\in S j ∈ S p p p 0 < i < k 0<i<k 0 < i < k
a , h , k − a , h − a , k − 2 a , h + a , 2 a . a,\qquad h,\qquad k-a,\qquad h-a,\qquad k-2a,\qquad h+a,\qquad2a. a , h , k − a , h − a , k − 2 a , h + a , 2 a . Generic indices. At any other index all three taps read free positions, with phases ν i + { 3 h + β , h , 3 h − β } \nu i+\{3h+\beta,h,3h-\beta\} ν i + { 3 h + β , h , 3 h − β } h < 3 h − β < 3 h + β h<3h-\beta<3h+\beta h < 3 h − β < 3 h + β 2 h − β 2h-\beta 2 h − β 2 β 2\beta 2 β 2 h − β 2h-\beta 2 h − β 2 h 2h 2 h { c , … , c + 2 h − 1 } \{c,\ldots,c+2h-1\} { c , … , c + 2 h − 1 }
The special indices. At each of the seven special indices one names a tap reading 0 and a tap reading 1, using the rule that the phase of a position is the sum of the phases of the index and the tap, and that the position reads 1 exactly when its phase lies in { 0 , … , 2 h − 1 } \{0,\ldots,2h-1\} { 0 , … , 2 h − 1 }
The even indices with a unit multiplier
Now let ν \nu ν 2 k 2k 2 k ν \nu ν Z / 2 k \mathbb{Z}/2k Z /2 k
Φ = { ν s m o d 2 k : s ∈ S } , α = ν a m o d 2 k . \Phi=\{\nu s\bmod2k:s\in S\},\qquad\alpha=\nu a\bmod2k. Φ = { ν s mod 2 k : s ∈ S } , α = ν a mod 2 k . The elements of Φ \Phi Φ Φ = k − Φ \Phi=k-\Phi Φ = k − Φ (46) . Moreover H ν H_\nu H ν S S S Φ \Phi Φ U U U V V V
Let t ∈ { 0 , k } t \in\{0,k\} t ∈ { 0 , k } Q = Q ( ν ; t ) Q=Q(\nu;t) Q = Q ( ν ; t ) 0 < i < k 0<i<k 0 < i < k ϕ = ν i m o d 2 k \phi=\nu i \bmod2k ϕ = ν i mod 2 k ϕ \phi ϕ ϕ ∉ { 0 , k } \phi\notin\{0,k\} ϕ ∈ / { 0 , k } ν \nu ν i ≢ 0 , k ( m o d 2 k ) i\not\equiv0,k\pmod{2k} i ≡ 0 , k ( mod 2 k ) ϕ + k \phi+k ϕ + k j j j p = i + j p=i+j p = i + j ν p = ϕ + ν j \nu p=\phi+\nu j ν p = ϕ + ν j p > j ≥ a p>j\geq a p > j ≥ a p = a + k p=a+k p = a + k p p p α + k \alpha+k α + k p ↦ ν p p\mapsto\nu p p ↦ ν p p p p ν p ∈ Φ \nu p\in\Phi ν p ∈ Φ ν p ∈ Φ + k \nu p\in\Phi+k ν p ∈ Φ + k z z z
0 if z ∈ Φ ; 1 if z ∈ Φ + k ; otherwise H ( z + t ) = { 1 [ z ∈ U ] , t = 0 , 1 [ z ∈ V ] , t = k . (47) \begin{aligned}
&0 &&\text{if } z\in\Phi; \qquad1 &&\text{if } z\in\Phi+k; \qquad\text{otherwise } H(z+t)=
\begin{cases}
\mathbf{1}[z\in U], & t=0,\\
\mathbf{1}[z\in V], & t=k.
\end{cases} \tag*{(47)}
\end{aligned} 0 if z ∈ Φ ; 1 if z ∈ Φ + k ; otherwise H ( z + t ) = { 1 [ z ∈ U ] , 1 [ z ∈ V ] , t = 0 , t = k . (47) by Lemma 8.6; here z z z U U U V V V i i i read set ϕ + Φ \phi+\Phi ϕ + Φ α + k \alpha+k α + k τ \tau τ
C U ( τ ) : ( τ + Φ ) ∩ U ⊆ Φ , C V ( τ ) : ( τ + Φ ) ∩ V ⊆ Φ . C_U(\tau):\quad(\tau+\Phi)\cap U\subseteq\Phi,\qquad C_V(\tau):\quad(\tau+\Phi)\cap V\subseteq\Phi. C U ( τ ) : ( τ + Φ ) ∩ U ⊆ Φ , C V ( τ ) : ( τ + Φ ) ∩ V ⊆ Φ. Lemma 8.9 (failing read sets). With the notation above, suppose that the clause at the even index i i i
(i) If every surviving tap reads 1 1 1 C U ( ϕ + k ) C_U(\phi+k) C U ( ϕ + k ) t = 0 t=0 t = 0 C V ( ϕ + k ) C_V(\phi+k) C V ( ϕ + k ) t = k t=k t = k
(ii) If every surviving tap reads 0 0 0 C U ( ϕ ) C_U(\phi) C U ( ϕ ) t = 0 t=0 t = 0 C V ( ϕ ) C_V(\phi) C V ( ϕ ) t = k t=k t = k i i i α + k \alpha+k α + k U U U t = 0 t=0 t = 0 V V V t = k t=k t = k
(iii) If some tap is deleted at i i i i ≠ 2 a i\ne2a i = 2 a 1 1 1
Proof. (i) Take t = 0 t=0 t = 0 x ∈ Φ x\in\Phi x ∈ Φ z = ϕ + x ∈ V z=\phi+x\in V z = ϕ + x ∈ V z z z z = α + k ∈ Φ + k z=\alpha+k\in\Phi+k z = α + k ∈ Φ + k z z z (47) it is not in Φ \Phi Φ z ∉ U z\notin U z ∈ / U Φ + k \Phi+k Φ + k ( ϕ + Φ ) ∩ V ⊆ Φ + k (\phi+\Phi)\cap V\subseteq\Phi+k ( ϕ + Φ ) ∩ V ⊆ Φ + k k k k C U ( ϕ + k ) C_U(\phi+k) C U ( ϕ + k ) t = k t=k t = k U U U V V V
(ii) Take t = 0 t=0 t = 0 x ∈ Φ x\in\Phi x ∈ Φ z = ϕ + x ∈ U z=\phi+x\in U z = ϕ + x ∈ U z z z α + k ∈ U \alpha+k\in U α + k ∈ U z z z z ∉ Φ + k z\notin\Phi+k z ∈ / Φ + k z ∈ U z\in U z ∈ U (47) forces z ∈ Φ z\in\Phi z ∈ Φ t = k t=k t = k
(iii) The deleted tap is j 0 = a + k − i j_0=a+k-i j 0 = a + k − i k − a ∈ S k-a\in S k − a ∈ S i + k − a = k + ( k − j 0 ) i+k-a=k+(k-j_0) i + k − a = k + ( k − j 0 ) k − j 0 ∈ S k-j_0\in S k − j 0 ∈ S k − j 0 = a k-j_0=a k − j 0 = a i = 2 a i=2a i = 2 a i ≠ 2 a i\ne2a i = 2 a
Two failures of opposite type, one in U U U V V V Φ \Phi Φ ν S \nu S ν S
Lemma 8.10 (no coset). Let ν \nu ν 2 k 2k 2 k S ⊆ { 1 , … , k − 1 } S\subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } Φ = ν S \Phi=\nu S Φ = ν S c + g Z / 2 k Z c+g\mathbb{Z}/2k\mathbb{Z} c + g Z /2 k Z g ∣ 2 k g\mid2k g ∣ 2 k 0 < g < 2 k 0<g<2k 0 < g < 2 k
Proof. Multiplication by ν − 1 \nu^{-1} ν − 1 Z / 2 k \mathbb{Z}/2k Z /2 k s 0 + g Z / 2 k Z s_0+g\mathbb{Z}/2k\mathbb{Z} s 0 + g Z /2 k Z S S S 2 k 2k 2 k g ∣ 2 k g\mid2k g ∣ 2 k g < 2 k g<2k g < 2 k g ≤ k g\leq k g ≤ k s 0 , s 0 + g , s 0 + 2 g , … s_0,s_0+g,s_0+2g,\ldots s 0 , s 0 + g , s 0 + 2 g , … k k k k k k { k , … , 2 k − 1 } \{k,\ldots,2k-1\} { k , … , 2 k − 1 } S S S
Lemma 8.11 (two failures force a coset). Let k k k Φ \Phi Φ 2 k 2k 2 k Φ = k − Φ \Phi=k-\Phi Φ = k − Φ U U U V V V ϕ , ϕ ′ \phi,\phi' ϕ , ϕ ′ { 0 , k } \{0,k\} { 0 , k } C U ( ϕ ) C_U(\phi) C U ( ϕ ) C V ( ϕ ′ ) C_V(\phi') C V ( ϕ ′ ) Φ \Phi Φ c + g Z / 2 k Z c+g\mathbb{Z}/2k\mathbb{Z} c + g Z /2 k Z g ∣ 2 k g\mid2k g ∣ 2 k 0 < g < k 0<g<k 0 < g < k
Proof. Reductions. C U ( ϕ ) C_U(\phi) C U ( ϕ ) C U ( − ϕ ) C_U(-\phi) C U ( − ϕ ) x ∈ Φ x \in\Phi x ∈ Φ z = x − ϕ ∈ U z=x-\phi\in U z = x − ϕ ∈ U k − x ∈ Φ k-x \in\Phi k − x ∈ Φ ϕ + ( k − x ) = k − z ∈ U \phi+(k-x)=k-z \in U ϕ + ( k − x ) = k − z ∈ U k − z ∈ Φ k-z \in\Phi k − z ∈ Φ z ∈ Φ z \in\Phi z ∈ Φ C V ( ϕ ′ ) C_V(\phi') C V ( ϕ ′ ) C V ( − ϕ ′ ) C_V(-\phi') C V ( − ϕ ′ ) z ↦ k − z z \mapsto k-z z ↦ k − z V V V 0 < ϕ , ϕ ′ < k 0<\phi,\phi'<k 0 < ϕ , ϕ ′ < k Φ + k \Phi+k Φ + k Φ + k = k − ( Φ + k ) \Phi+k=k-(\Phi+k) Φ + k = k − ( Φ + k ) U U U V V V C U ( τ ) C_U(\tau) C U ( τ ) Φ \Phi Φ C V ( τ ) C_V(\tau) C V ( τ ) Φ + k \Phi+k Φ + k Φ \Phi Φ Φ + k \Phi+k Φ + k C U C_U C U C V C_V C V C U ( ϕ ) C_U(\phi) C U ( ϕ ) C V ( ϕ ′ ) C_V(\phi') C V ( ϕ ′ ) Φ \Phi Φ C U ( ϕ ′ ) C_U(\phi') C U ( ϕ ′ ) C V ( ϕ ) C_V(\phi) C V ( ϕ ) Φ + k \Phi+k Φ + k Φ + k \Phi+k Φ + k ϕ \phi ϕ ϕ ′ \phi' ϕ ′ Φ + k \Phi+k Φ + k Φ \Phi Φ k k k ϕ > ϕ ′ \phi>\phi' ϕ > ϕ ′ Φ \Phi Φ Φ + k \Phi+k Φ + k ϕ ≤ ϕ ′ \phi\leq\phi' ϕ ≤ ϕ ′
Translation into ( 0 , k ) (0,k) ( 0 , k ) Put Y + = Φ ∩ U Y_+=\Phi\cap U Y + = Φ ∩ U Y − = { z : z + k ∈ Φ ∩ V } Y_-=\{z:z+k\in\Phi\cap V\} Y − = { z : z + k ∈ Φ ∩ V } ( 0 , k ) (0,k) ( 0 , k ) Φ = k − Φ \Phi=k-\Phi Φ = k − Φ z ↦ k − z z\mapsto k-z z ↦ k − z Y − Y_- Y − k − ( m + k ) ≡ ( k − m ) + k k-(m+k)\equiv(k-m)+k k − ( m + k ) ≡ ( k − m ) + k
(G1) if p ∈ Y + p\in Y_+ p ∈ Y + p + ϕ < k p+\phi<k p + ϕ < k p + ϕ ∈ Y + p+\phi\in Y_+ p + ϕ ∈ Y + C U ( ϕ ) C_U(\phi) C U ( ϕ )
(G2) if m ∈ Y − m\in Y_- m ∈ Y − m < ϕ m<\phi m < ϕ ϕ − m ∈ Y + \phi-m\in Y_+ ϕ − m ∈ Y + 2 k − m ≡ k − ( m + k ) 2k-m\equiv k-(m+k) 2 k − m ≡ k − ( m + k ) Φ \Phi Φ ϕ − m ∈ U \phi-m\in U ϕ − m ∈ U
(G3) if m ∈ Y − m\in Y_- m ∈ Y − m + ϕ ′ < k m+\phi'<k m + ϕ ′ < k m + ϕ ′ ∈ Y − m+\phi'\in Y_- m + ϕ ′ ∈ Y − C V ( ϕ ′ ) C_V(\phi') C V ( ϕ ′ )
(G4) if p ∈ Y + p\in Y_+ p ∈ Y + p < ϕ ′ p<\phi' p < ϕ ′ ϕ ′ − p ∈ Y − \phi'-p\in Y_- ϕ ′ − p ∈ Y − k − p k-p k − p Φ \Phi Φ k + ϕ ′ − p ∈ V k+\phi'-p\in V k + ϕ ′ − p ∈ V
Call a set Π \Pi Π ( 0 , k ) (0,k) ( 0 , k ) n n n p ∈ Π p\in\Pi p ∈ Π 0 < z < k 0<z<k 0 < z < k z ≡ p ( m o d n ) z\equiv p\pmod n z ≡ p ( mod n ) z ∈ Π z\in\Pi z ∈ Π
The engine. Let Π \Pi Π n n n n > 0 n>0 n > 0 p ∈ Π p\in\Pi p ∈ Π p ′ ∈ Π p'\in\Pi p ′ ∈ Π p ′ ≡ p + f ( m o d n ) p'\equiv p+f\pmod n p ′ ≡ p + f ( mod n ) Π \Pi Π gcd ( n , f ) \gcd(n,f) g cd( n , f ) g = gcd ( n , f ) g=\gcd(n,f) g = g cd( n , f ) y ≥ 0 y\geq0 y ≥ 0 y f ≡ g ( m o d n ) yf\equiv g\pmod n y f ≡ g ( mod n ) p ∈ Π p\in\Pi p ∈ Π 0 < z < k 0<z<k 0 < z < k z ≡ p ( m o d g ) z\equiv p\pmod g z ≡ p ( mod g ) z ≡ p + j g ( m o d n ) z\equiv p+jg\pmod n z ≡ p + j g ( mod n ) j ≥ 0 j\geq0 j ≥ 0 j y jy j y p ′ ′ ∈ Π p''\in\Pi p ′′ ∈ Π p ′ ′ ≡ p + j y f ≡ z ( m o d n ) p''\equiv p+jyf\equiv z\pmod n p ′′ ≡ p + j y f ≡ z ( mod n ) z ∈ Π z\in\Pi z ∈ Π
Step 1: Y + Y_+ Y + γ \gamma γ γ = gcd ( ϕ , ϕ ′ ) \gamma=\gcd(\phi,\phi') γ = g cd( ϕ , ϕ ′ ) By (G1) and the symmetry of Y + Y_+ Y + p − ϕ = k − ( ( k − p ) + ϕ ) p-\phi=k-((k-p)+\phi) p − ϕ = k − (( k − p ) + ϕ ) Y + Y_+ Y + ± ϕ \pm\phi ± ϕ ( 0 , k ) (0,k) ( 0 , k ) ϕ \phi ϕ p ∈ Y + p\in Y_+ p ∈ Y + p 0 = p m o d ϕ p_0=p\bmod\phi p 0 = p mod ϕ 0 < p 0 < ϕ 0<p_0<\phi 0 < p 0 < ϕ p 0 ∈ Y + p_0\in Y_+ p 0 ∈ Y + r ≥ 0 r\geq0 r ≥ 0 p 0 + r ϕ < ϕ ′ p_0+r\phi<\phi' p 0 + r ϕ < ϕ ′ p 0 + r ϕ ∈ Y + p_0+r\phi\in Y_+ p 0 + r ϕ ∈ Y + m = ϕ ′ − p 0 − r ϕ ∈ Y − m=\phi'-p_0-r\phi\in Y_- m = ϕ ′ − p 0 − r ϕ ∈ Y − 0 < m ≤ ϕ 0<m\leq\phi 0 < m ≤ ϕ m m m m < ϕ m<\phi m < ϕ ϕ − m ∈ Y + \phi-m\in Y_+ ϕ − m ∈ Y + ( ϕ − m ) + ϕ ′ = p 0 + ( r + 1 ) ϕ ≡ p ( m o d ϕ ) (\phi-m)+\phi'=p_0+(r+1)\phi\equiv p\pmod\phi ( ϕ − m ) + ϕ ′ = p 0 + ( r + 1 ) ϕ ≡ p ( mod ϕ ) p ∈ Y + p\in Y_+ p ∈ Y + p ′ ∈ Y + p'\in Y_+ p ′ ∈ Y + p ′ + ϕ ′ ≡ p ( m o d ϕ ) p'+\phi'\equiv p\pmod\phi p ′ + ϕ ′ ≡ p ( mod ϕ ) k − p k-p k − p p ∈ Y + p\in Y_+ p ∈ Y + p ′ ′ = k − p ′ ∈ Y + p''=k-p'\in Y_+ p ′′ = k − p ′ ∈ Y + p ′ ′ ≡ p + ϕ ′ ( m o d ϕ ) p''\equiv p+\phi'\pmod\phi p ′′ ≡ p + ϕ ′ ( mod ϕ ) n = ϕ n=\phi n = ϕ f = ϕ ′ f=\phi' f = ϕ ′
Step 2: if 0 < z < k 0<z<k 0 < z < k z + p ≡ 0 ( m o d γ ) z+p\equiv0\pmod\gamma z + p ≡ 0 ( mod γ ) p ∈ Y + p\in Y_+ p ∈ Y + z ∈ Y − z\in Y_- z ∈ Y − z 0 = z m o d ϕ ′ z_0=z\bmod\phi' z 0 = z mod ϕ ′ 0 < z 0 < ϕ ′ 0<z_0<\phi' 0 < z 0 < ϕ ′ ϕ ′ − z 0 ≡ − z ≡ p ( m o d γ ) \phi'-z_0\equiv-z\equiv p\pmod\gamma ϕ ′ − z 0 ≡ − z ≡ p ( mod γ ) ϕ ′ − z 0 ∈ Y + \phi'-z_0\in Y_+ ϕ ′ − z 0 ∈ Y + z 0 ∈ Y − z_0\in Y_- z 0 ∈ Y − z = z 0 + q ϕ ′ ∈ Y − z=z_0+q\phi'\in Y_- z = z 0 + q ϕ ′ ∈ Y −
Step 3: Y + Y_+ Y + g g g g = gcd ( γ , 2 k ) g=\gcd(\gamma,2k) g = g cd( γ , 2 k ) p ∈ Y + p\in Y_+ p ∈ Y + z = ( k + p ) m o d γ z=(k+p)\bmod\gamma z = ( k + p ) mod γ 0 < z < γ ≤ ϕ 0<z<\gamma\leq\phi 0 < z < γ ≤ ϕ ( k − z ) + p ≡ 0 ( m o d γ ) (k-z)+p\equiv0\pmod\gamma ( k − z ) + p ≡ 0 ( mod γ ) k − z ∈ Y − k-z\in Y_- k − z ∈ Y − z ∈ Y − z\in Y_- z ∈ Y − ϕ − z ∈ Y + \phi-z\in Y_+ ϕ − z ∈ Y + k − ϕ + z ∈ Y + k-\phi+z\in Y_+ k − ϕ + z ∈ Y + γ ∣ ϕ \gamma\mid\phi γ ∣ ϕ k − ϕ + z ≡ k + z ≡ p + 2 k ( m o d γ ) k-\phi+z\equiv k+z\equiv p+2k\pmod\gamma k − ϕ + z ≡ k + z ≡ p + 2 k ( mod γ ) n = γ n=\gamma n = γ f = 2 k f=2k f = 2 k
Step 4: coset. The number g g g 2 k 2k 2 k 0 < g ≤ ϕ < k 0<g\leq\phi<k 0 < g ≤ ϕ < k c ∈ Y + c\in Y_+ c ∈ Y + w w w ≡ c ( m o d g ) \equiv c\pmod g ≡ c ( mod g ) [ 0 , 2 k ) [0,2k) [ 0 , 2 k ) w < k w<k w < k w ∈ Y + ⊆ Φ w\in Y_+\subseteq\Phi w ∈ Y + ⊆ Φ w > k w>k w > k z = w − k z=w-k z = w − k − z ≡ k − c ( m o d g ) -z\equiv k-c\pmod g − z ≡ k − c ( mod g ) z ′ = ( − z ) m o d γ z'=(-z)\bmod\gamma z ′ = ( − z ) mod γ 0 < z ′ < γ < k 0<z'<\gamma<k 0 < z ′ < γ < k z ′ ≡ k − c ( m o d g ) z'\equiv k-c\pmod g z ′ ≡ k − c ( mod g ) k − c ∈ Y + k-c\in Y_+ k − c ∈ Y + z ′ ∈ Y + z'\in Y_+ z ′ ∈ Y + z + z ′ ≡ 0 ( m o d γ ) z+z'\equiv0\pmod\gamma z + z ′ ≡ 0 ( mod γ ) z ∈ Y − z\in Y_- z ∈ Y − w ∈ Φ w\in\Phi w ∈ Φ c + g Z / 2 k Z ⊆ Φ c+g\mathbb{Z}/2k\mathbb{Z}\subseteq\Phi c + g Z /2 k Z ⊆ Φ
The coset lemma disposes of every pair of failures except one: the all-0 failure of a read set that has lost its deleted point. By Lemma 8.9 (iii) this can only happen at the index 2 a 2a 2 a ν \nu ν L L L 2 a 2a 2 a a a a
Definition 8.12 (the three-point condition). Let 1 ∉ S 1 \notin S 1 ∈ / S a = min S a = \min S a = min S ν \nu ν 2 k 2k 2 k satisfies the three-point condition (T) if
(Ta) some x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } H ν ( x ) ≠ H ν ( a ) H_{\nu}(x) \ne H_{\nu}(a) H ν ( x ) = H ν ( a )
(Tb) some j ∈ S ∖ { k − a } j \in S \setminus\{k-a\} j ∈ S ∖ { k − a } ( 2 a + j ) m o d k ∉ S (2a+j) \bmod k \notin S ( 2 a + j ) mod k ∈ / S H ν ( 2 a + j ) ≠ H ν ( a ) H_{\nu}(2a+j) \ne H_{\nu}(a) H ν ( 2 a + j ) = H ν ( a )
Proposition 8.13 (the half-arc carrier under (T)). Let S S S all-odd with 1 ∉ S 1 \notin S 1 ∈ / S ν \nu ν unit satisfying (T) . Then for t = 0 t=0 t = 0 t = k t=k t = k σ = a \sigma=a σ = a Q ( ν ; t ) Q(\nu;t) Q ( ν ; t ) σ \sigma σ S S S
Proof. (B1) holds by construction, and the clauses at odd indices hold for both values of t t t Lemma 8.7 , applied with u = x u=x u = x v = a v=a v = a t t t
By (Ta), 2 a < 2 x < k 2a<2x<k 2 a < 2 x < k 2 a 2a 2 a Φ \Phi Φ U U U V V V t t t i 0 i_0 i 0 t = 0 t=0 t = 0 i 1 i_1 i 1 t = k t=k t = k
I claim first that the failure at i 0 i_0 i 0 C U ( τ ) C_U(\tau) C U ( τ ) τ ∉ { 0 , k } \tau\notin\{0,k\} τ ∈ / { 0 , k } Lemma 8.9 (i) and (ii) this holds with τ ∈ { ν i 0 , ν i 0 + k } \tau\in\{\nu i_0,\nu i_0+k\} τ ∈ { ν i 0 , ν i 0 + k } i 0 i_0 i 0 0 0 0 α + k ∈ U \alpha+k \in U α + k ∈ U i 0 = 2 a i_0=2a i 0 = 2 a Lemma 8.9 (iii). Moreover α ∈ V \alpha\in V α ∈ V H ν ( a ) = 0 H_{\nu}(a)=0 H ν ( a ) = 0 j j j j j j 2 a 2a 2 a 2 a + j ≠ a + k 2a+j \ne a+k 2 a + j = a + k Lemma 8.6 its read is H ( ν ( 2 a + j ) ) = H ν ( 2 a + j ) = 1 H(\nu(2a+j))=H_{\nu}(2a+j)=1 H ( ν ( 2 a + j )) = H ν ( 2 a + j ) = 1 i 1 i_1 i 1 C V ( τ ′ ) C_V(\tau') C V ( τ ′ ) τ ′ ∉ { 0 , k } \tau' \notin\{0,k\} τ ′ ∈ / { 0 , k } α + k ∈ V \alpha+k \in V α + k ∈ V H ν ( a ) = 1 H_{\nu}(a)=1 H ν ( a ) = 1 j j j H ( ν ( 2 a + j ) + k ) = 1 − H ν ( 2 a + j ) = 1 H(\nu(2a+j)+k)=1-H_{\nu}(2a+j)=1 H ( ν ( 2 a + j ) + k ) = 1 − H ν ( 2 a + j ) = 1 Lemma 8.11 now puts a coset c + g Z / 2 k Z c+g\mathbb{Z}/2k\mathbb{Z} c + g Z /2 k Z 0 < g < k 0<g<k 0 < g < k Φ \Phi Φ Lemma 8.10 says that there is none.
The three-point condition
It remains to find a unit satisfying (T). For k > 90 k>90 k > 90 x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } 2 a − x 2a-x 2 a − x a a a x x x x < 2 a x<2a x < 2 a x − 2 a ∉ S x-2a \notin S x − 2 a ∈ / S ) ; or $L$ is the chain $a,3a,\ldots,(2m+1)a$, and {xref}Theorem 9.1, applied to the rescaled triple $(k;1,2m+3)$, supplies the unit. Triples with $\gcd(a,x,k)>1$ may have no witness, but these are rescaled copies of the 59 failures at orders up to 90 listed in {xref}Lemma 8.16 (see {xref}Lemma 8.17), and {xref}Lemma 8.18` shows that they cannot block both routes, except for a = 15 a=15 a = 15 k = 120 k=120 k = 120 k ≤ 90 k \le90 k ≤ 90
Definition 8.14 (witnesses and admissible triples). Let k k k a , x a,x a , x ν \nu ν 2 k 2k 2 k witnesses the triple ( k ; a , x ) (k;a,x) ( k ; a , x )
H ν ( 2 a − x ) = H ν ( a ) ≠ H ν ( x ) , (48) H_{\nu}(2a-x)=H_{\nu}(a)\ne H_{\nu}(x), \tag*{(48)} H ν ( 2 a − x ) = H ν ( a ) = H ν ( x ) , (48) and ( k ; a , x ) (k;a,x) ( k ; a , x ) a < x a<x a < x 2 x < k 2x<k 2 x < k x ≠ 3 a x\ne3a x = 3 a gcd ( a , x , k ) = 1 \gcd(a,x,k)=1 g cd( a , x , k ) = 1 x = 3 a x=3a x = 3 a 2 a − x = − a 2a-x=-a 2 a − x = − a H ν ( − a ) = 1 − H ν ( a ) H_{\nu}(-a)=1-H_{\nu}(a) H ν ( − a ) = 1 − H ν ( a ) ν \nu ν (46) .)
Note that 2 a − x 2a-x 2 a − x a a a x x x Theorem 9.1 , proved in § 9 : every admissible triple ( k ; a , x ) (k;a,x) ( k ; a , x ) k > 90 k>90 k > 90
Lemma 8.15 (the progression route). Let S S S primitive and all-odd with 1 ∉ S 1 \notin S 1 ∈ / S
(i) Let x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } x < 2 a x<2a x < 2 a x − 2 a ∉ S x-2a \notin S x − 2 a ∈ / S ν \nu ν ( k ; a , x ) (k;a,x) ( k ; a , x )
(ii) Suppose that ∣ L ∣ ≥ 2 |L|\ge2 ∣ L ∣ ≥ 2 x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } x > 2 a x>2a x > 2 a x − 2 a ∈ S x-2a \in S x − 2 a ∈ S gcd ( a , k ) = 1 \gcd(a,k)=1 g cd( a , k ) = 1 m ≥ 1 m\ge1 m ≥ 1 2 ( 2 m + 1 ) a < k 2(2m+1)a<k 2 ( 2 m + 1 ) a < k ( 2 i + 1 ) a ∈ S (2i+1)a \in S ( 2 i + 1 ) a ∈ S 0 ≤ i ≤ m 0\le i\le m 0 ≤ i ≤ m ( 2 m + 3 ) a ∉ S (2m+3)a \notin S ( 2 m + 3 ) a ∈ / S
Proof. (i) (Ta) holds with this x x x H ν ( x ) ≠ H ν ( a ) H_\nu(x) \ne H_\nu(a) H ν ( x ) = H ν ( a ) j = k − x ∈ S j=k-x \in S j = k − x ∈ S k − a k-a k − a x < 2 a x<2a x < 2 a ( 2 a + j ) mod k = 2 a − x ∈ ( 0 , a ) (2a+j)\mathbin{\operatorname{mod}} k=2a-x\in(0,a) ( 2 a + j ) mod k = 2 a − x ∈ ( 0 , a ) S S S a a a x > 2 a x>2a x > 2 a 2 a + j = k − ( x − 2 a ) < k 2a+j=k-(x-2a)<k 2 a + j = k − ( x − 2 a ) < k S S S x − 2 a x-2a x − 2 a (45) , H ν ( 2 a + j ) = H ν ( 2 a − x + k ) = 1 − H ν ( 2 a − x ) = 1 − H ν ( a ) H_\nu(2a+j)=H_\nu(2a-x+k)=1-H_\nu(2a-x)=1-H_\nu(a) H ν ( 2 a + j ) = H ν ( 2 a − x + k ) = 1 − H ν ( 2 a − x ) = 1 − H ν ( a )
(ii) Each x ∈ L ∖ { a } x\in L\setminus\{a\} x ∈ L ∖ { a } x − 2 a ∈ L x-2a\in L x − 2 a ∈ L x x x 2 a 2a 2 a L L L a a a L L L ≡ a ( m o d 2 a ) \equiv a\pmod{2a} ≡ a ( mod 2 a ) a a a f = gcd ( a , k ) f=\gcd(a,k) f = g cd( a , k ) L L L k − x k-x k − x x ∈ L x\in L x ∈ L k / 2 k/2 k /2 f f f k k k f f f gcd ( S ∪ { k } ) = 1 \gcd(S\cup\{k\})=1 g cd( S ∪ { k }) = 1 ( 2 m + 1 ) a (2m+1)a ( 2 m + 1 ) a L L L m ≥ 1 m\ge1 m ≥ 1 ∣ L ∣ ≥ 2 |L|\ge2 ∣ L ∣ ≥ 2 ( 2 i + 1 ) a ∈ S (2i+1)a\in S ( 2 i + 1 ) a ∈ S 0 ≤ i ≤ m 0\le i\le m 0 ≤ i ≤ m y = ( 2 m + 3 ) a ∈ S y=(2m+3)a\in S y = ( 2 m + 3 ) a ∈ S 2 y < k 2y<k 2 y < k 2 y = k 2y=k 2 y = k a ∣ k a\mid k a ∣ k 2 y > k 2y>k 2 y > k k − y k-y k − y L L L a ∣ k − y a\mid k-y a ∣ k − y a ∣ k a\mid k a ∣ k a ≥ 3 a\ge3 a ≥ 3 gcd ( a , k ) = 1 \gcd(a,k)=1 g cd( a , k ) = 1
Up to order 90 some admissible triples have no witness, and these failures propagate upwards by scaling.
Lemma 8.16 (small orders). Let ( n ; a , x ) (n;a,x) ( n ; a , x ) n ≤ 90 n\le90 n ≤ 90 ( n ; a , x ) (n;a,x) ( n ; a , x )
Table 8.16.
Denote this list by F \mathcal{F} F
Proof. This is a finite check: for every admissible triple with n ≤ 90 n\le90 n ≤ 90 F \mathcal{F} F ν \nu ν 2 n 2n 2 n
Lemma 8.17 (lifting). If ( n ; a , x ) (n;a,x) ( n ; a , x ) ( g n ; g a , g x ) (gn;ga,gx) ( g n ; g a , g x ) g ≥ 1 g\ge1 g ≥ 1 k > 90 k>90 k > 90 ( k ; a , x ) (k;a,x) ( k ; a , x ) gcd ( a , x , k ) = 1 \gcd(a,x,k)=1 g cd( a , x , k ) = 1 ( k ; a , x ) (k;a,x) ( k ; a , x ) g = gcd ( a , x , k ) g=\gcd(a,x,k) g = g cd( a , x , k ) ( k / g ; a / g , x / g ) (k/g;a/g,x/g) ( k / g ; a / g , x / g ) F \mathcal{F} F
Proof. Let ν \nu ν 2 n 2n 2 n ( n ; a , x ) (n;a,x) ( n ; a , x ) Z / 2 g n → Z / 2 n \mathbb{Z}/2gn\to\mathbb{Z}/2n Z /2 g n → Z /2 n ν ′ \nu' ν ′ 2 g n 2gn 2 g n ν ′ ≡ ν ( m o d 2 n ) \nu'\equiv\nu\pmod{2n} ν ′ ≡ ν ( mod 2 n ) y y y ν ′ g y mod 2 g n = g ( ν y mod 2 n ) \nu'gy\mathbin{\operatorname{mod}}2gn=g(\nu y\mathbin{\operatorname{mod}}2n) ν ′ g y mod 2 g n = g ( ν y mod 2 n ) g n gn g n ν y mod 2 n < n \nu y\mathbin{\operatorname{mod}}2n<n ν y mod 2 n < n H ν ′ ( g y ) H_{\nu'}(gy) H ν ′ ( g y ) g n gn g n H ν ( y ) H_\nu(y) H ν ( y ) n n n ν ′ \nu' ν ′ ( g n ; g a , g x ) (gn;ga,gx) ( g n ; g a , g x )
For the second part, g g g a a a k / g k/g k / g ( k / g ; a / g , x / g ) (k/g;a/g,x/g) ( k / g ; a / g , x / g ) Theorem 9.1 , so k / g ≤ 90 k/g\le90 k / g ≤ 90 F \mathcal{F} F
Above order 90 the lifted failures can still occur, but only in a way that the three-point condition can absorb.
Lemma 8.18 (failures above order 90). Let k > 90 k > 90 k > 90 S S S 1 ∉ S 1 \notin S 1 ∈ / S ( k , a ) ≠ ( 120 , 15 ) (k,a) \ne(120,15) ( k , a ) = ( 120 , 15 ) x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } ( k ; a , x ) (k;a,x) ( k ; a , x ) x > 2 a x > 2a x > 2 a x − 2 a ∈ S x - 2a \in S x − 2 a ∈ S x x x ( k ; a , x ) (k;a,x) ( k ; a , x ) x ∈ L ∖ { a , 3 a } x \in L \setminus\{a,3a\} x ∈ L ∖ { a , 3 a }
Proof. Suppose that ( k ; a , x 0 ) (k;a,x_0) ( k ; a , x 0 ) x 0 ∈ L ∖ { a , 3 a } x_0 \in L \setminus\{a,3a\} x 0 ∈ L ∖ { a , 3 a } M \mathcal{M} M n n n ( n ; a ′ , x ′ ) ∈ F (n;a',x') \in\mathcal{F} ( n ; a ′ , x ′ ) ∈ F a ′ , x ′ a',x' a ′ , x ′ n ∣ k n \mid k n ∣ k a = ( k / n ) a ′ a = (k/n)a' a = ( k / n ) a ′ ℓ = lcm M \ell= \operatorname{lcm}\mathcal{M} ℓ = lcm M y ∈ L ∖ { a , 3 a } y \in L \setminus\{a,3a\} y ∈ L ∖ { a , 3 a } g y = gcd ( a , y , k ) g_y = \gcd(a,y,k) g y = g cd( a , y , k ) ( k ; a , y ) (k;a,y) ( k ; a , y ) ( k / g y ; a / g y , y / g y ) ∈ F (k/g_y;a/g_y,y/g_y) \in\mathcal{F} ( k / g y ; a / g y , y / g y ) ∈ F k / g y ∈ M k/g_y \in\mathcal{M} k / g y ∈ M M \mathcal{M} M y = x 0 y = x_0 y = x 0 M \mathcal{M} M k k k ℓ ∣ k \ell\mid k ℓ ∣ k k = ℓ f k = \ell f k = ℓ f n ∈ M n \in\mathcal{M} n ∈ M n ∣ ℓ n \mid\ell n ∣ ℓ f f f k / n k/n k / n f ∣ g y f \mid g_y f ∣ g y ( k ; a , y ) (k;a,y) ( k ; a , y ) f ∣ g x 0 ∣ a f \mid g_{x_0} \mid a f ∣ g x 0 ∣ a f f f
I claim that f f f y ∈ L y \in L y ∈ L y y y y = a y = a y = a y ≠ a y \ne a y = a ( k ; a , y ) (k;a,y) ( k ; a , y ) y − 2 a ∈ L y - 2a \in L y − 2 a ∈ L f ∣ y − 2 a f \mid y - 2a f ∣ y − 2 a f ∣ y f \mid y f ∣ y y = 3 a y = 3a y = 3 a f ∣ a ∣ y f \mid a \mid y f ∣ a ∣ y y ≠ 3 a y \ne3a y = 3 a f ∣ g y ∣ y f \mid g_y \mid y f ∣ g y ∣ y f f f k k k k / 2 k/2 k /2 k − y k-y k − y f f f gcd ( S ∪ { k } ) = 1 \gcd(S \cup\{k\}) = 1 g cd( S ∪ { k }) = 1 ℓ = k \ell= k ℓ = k
It remains to see that ℓ = k \ell= k ℓ = k n ∈ M n \in\mathcal{M} n ∈ M ( n ; a ′ , x ′ ) ∈ F (n;a',x') \in\mathcal{F} ( n ; a ′ , x ′ ) ∈ F a ′ / n = a / k a'/n = a/k a ′ / n = a / k M \mathcal{M} M a ′ / n a'/n a ′ / n F \mathcal{F} F 3 / 12 = 7 / 28 3/12 = 7/28 3/12 = 7/28 3 / 24 = 5 / 40 3/24 = 5/40 3/24 = 5/40 1 / 16 = 3 / 48 1/16 = 3/48 1/16 = 3/48 M \mathcal{M} M ℓ ≤ 90 \ell\le90 ℓ ≤ 90 ℓ \ell ℓ lcm ( 12 , 28 ) = 84 \operatorname{lcm}(12,28) = 84 lcm ( 12 , 28 ) = 84 lcm ( 24 , 40 ) = 120 \operatorname{lcm}(24,40) = 120 lcm ( 24 , 40 ) = 120 lcm ( 16 , 48 ) = 48 \operatorname{lcm}(16,48) = 48 lcm ( 16 , 48 ) = 48 ℓ = k > 90 \ell= k > 90 ℓ = k > 90 k = 120 k = 120 k = 120 a / k = 1 / 8 a/k = 1/8 a / k = 1/8 a = 15 a = 15 a = 15
Theorem 8.19 (the three-point condition). Let S S S 1 ∉ S 1 \notin S 1 ∈ / S ∣ L ∣ ≥ 2 |L| \ge2 ∣ L ∣ ≥ 2 ν \nu ν 2 k 2k 2 k ( T ) (T) ( T )
Proof. Orders k > 90 k > 90 k > 90 x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } ( k ; a , x ) (k;a,x) ( k ; a , x ) x < 2 a x < 2a x < 2 a x − 2 a ∉ S x - 2a \notin S x − 2 a ∈ / S ( T ) (T) ( T ) ( k , a ) ≠ ( 120 , 15 ) (k,a) \ne(120,15) ( k , a ) = ( 120 , 15 ) x ∈ L ∖ { a , 3 a } x \in L \setminus\{a,3a\} x ∈ L ∖ { a , 3 a } x > 2 a x > 2a x > 2 a x − 2 a ∈ S x - 2a \in S x − 2 a ∈ S x = 3 a x = 3a x = 3 a a ∈ S a \in S a ∈ S gcd ( a , k ) = 1 \gcd(a,k) = 1 g cd( a , k ) = 1 m ≥ 1 m \ge1 m ≥ 1 ( 2 i + 1 ) a ∈ S (2i + 1)a \in S ( 2 i + 1 ) a ∈ S i ≤ m i \le m i ≤ m 2 ( 2 m + 1 ) a < k 2(2m + 1)a < k 2 ( 2 m + 1 ) a < k ( 2 m + 3 ) a ∉ S (2m + 3)a \notin S ( 2 m + 3 ) a ∈ / S a ≥ 3 a \ge3 a ≥ 3 2 ( 2 m + 3 ) < 6 ( 2 m + 1 ) ≤ 2 ( 2 m + 1 ) a < k 2(2m + 3) < 6(2m + 1) \le2(2m + 1)a < k 2 ( 2 m + 3 ) < 6 ( 2 m + 1 ) ≤ 2 ( 2 m + 1 ) a < k ( k ; 1 , 2 m + 3 ) (k;1,2m + 3) ( k ; 1 , 2 m + 3 ) ν 1 \nu_1 ν 1 H ν 1 ( − ( 2 m + 1 ) ) = H ν 1 ( 1 ) ≠ H ν 1 ( 2 m + 3 ) H_{\nu_1}(-(2m + 1)) = H_{\nu_1}(1) \ne H_{\nu_1}(2m + 3) H ν 1 ( − ( 2 m + 1 )) = H ν 1 ( 1 ) = H ν 1 ( 2 m + 3 ) H ν 1 ( 2 m + 1 ) = 1 − H ν 1 ( − ( 2 m + 1 ) ) ≠ H ν 1 ( 1 ) H_{\nu_1}(2m + 1) = 1 - H_{\nu_1}(-(2m + 1)) \ne H_{\nu_1}(1) H ν 1 ( 2 m + 1 ) = 1 − H ν 1 ( − ( 2 m + 1 )) = H ν 1 ( 1 ) a a a k k k a ˉ \bar{a} a ˉ 2 k 2k 2 k ν = ν 1 a ˉ \nu= \nu_1\bar{a} ν = ν 1 a ˉ ν y a = ν 1 y \nu ya = \nu_1 y ν y a = ν 1 y 2 k 2k 2 k H ν ( y a ) = H ν 1 ( y ) H_\nu(ya) = H_{\nu_1}(y) H ν ( y a ) = H ν 1 ( y ) y y y j = ( 2 m + 1 ) a j = (2m + 1)a j = ( 2 m + 1 ) a L ∖ { a } L \setminus\{a\} L ∖ { a } ( T a ) (T_a) ( T a ) x = j x = j x = j H ν ( j ) = H ν 1 ( 2 m + 1 ) ≠ H ν 1 ( 1 ) = H ν ( a ) H_\nu(j) = H_{\nu_1}(2m + 1) \ne H_{\nu_1}(1) = H_\nu(a) H ν ( j ) = H ν 1 ( 2 m + 1 ) = H ν 1 ( 1 ) = H ν ( a ) ( T b ) (T_b) ( T b ) j ≠ k − a j \ne k-a j = k − a 2 a + j = ( 2 m + 3 ) a < 2 ( 2 m + 1 ) a < k 2a+j = (2m+3)a < 2(2m+1)a < k 2 a + j = ( 2 m + 3 ) a < 2 ( 2 m + 1 ) a < k ( 2 a + j ) (2a+j) ( 2 a + j ) k = ( 2 m + 3 ) a ∉ S k = (2m+3)a \notin S k = ( 2 m + 3 ) a ∈ / S H ν ( 2 a + j ) = H ν 1 ( 2 m + 3 ) ≠ H ν 1 ( 1 ) = H ν ( a ) H_\nu(2a+j) = H_{\nu_1}(2m+3) \ne H_{\nu_1}(1) = H_\nu(a) H ν ( 2 a + j ) = H ν 1 ( 2 m + 3 ) = H ν 1 ( 1 ) = H ν ( a )
The order k = 120 k = 120 k = 120 a = 15 a = 15 a = 15 x x x x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } x > 30 x > 30 x > 30 x − 30 ∈ S x - 30 \in S x − 30 ∈ S x ∈ L ∖ { 15 , 45 } x \in L \setminus\{15,45\} x ∈ L ∖ { 15 , 45 } ( 120 / g ; 15 / g , x / g ) ∈ F (120/g;15/g,x/g) \in\mathcal{F} ( 120/ g ; 15/ g , x / g ) ∈ F g = gcd ( 15 , x , 120 ) g = \gcd(15,x,120) g = g cd( 15 , x , 120 ) g ≠ 1 g \ne1 g = 1 F \mathcal{F} F F \mathcal{F} F ( 40 ; 5 , ⋅ ) (40;5,\cdot) ( 40 ; 5 , ⋅ ) ( 24 ; 3 , ⋅ ) (24;3,\cdot) ( 24 ; 3 , ⋅ ) ( 8 ; 1 , ⋅ ) (8;1,\cdot) ( 8 ; 1 , ⋅ ) ( 40 ; 5 , 17 ) (40;5,17) ( 40 ; 5 , 17 ) ( 24 ; 3 , 7 ) (24;3,7) ( 24 ; 3 , 7 ) x ∈ { 35 , 51 } x \in\{35,51\} x ∈ { 35 , 51 } L L L { 15 , 35 , 45 , 51 } \{15,35,45,51\} { 15 , 35 , 45 , 51 } x ∈ L x \in L x ∈ L x − 30 ∈ L x - 30 \in L x − 30 ∈ L x − 30 ∈ { 15 , 35 , 45 , 51 } x - 30 \in\{15,35,45,51\} x − 30 ∈ { 15 , 35 , 45 , 51 } x < 60 x < 60 x < 60 x = 45 x = 45 x = 45 35 ∉ L 35 \notin L 35 ∈ / L 3 3 3 L L L S S S k / 2 = 60 k/2 = 60 k /2 = 60 S = L ∪ ( k − L ) S = L \cup(k-L) S = L ∪ ( k − L ) 35 ∈ L 35 \in L 35 ∈ L 5 5 5 3 3 3 51 ∈ L 51 \in L 51 ∈ L ν = 11 \nu= 11 ν = 11 ( T ) (T) ( T ) H 11 ( 15 ) = 0 H_{11}(15) = 0 H 11 ( 15 ) = 0 11 ⋅ 15 = 165 11 \cdot15 = 165 11 ⋅ 15 = 165 ( T a ) (T_a) ( T a ) x = 51 x = 51 x = 51 11 ⋅ 51 = 81 11 \cdot51 = 81 11 ⋅ 51 = 81 ( T b ) (T_b) ( T b ) 45 ∉ S 45 \notin S 45 ∈ / S j = 15 j = 15 j = 15 2 a + j = 45 ∉ S 2a+j = 45 \notin S 2 a + j = 45 ∈ / S 11 ⋅ 45 ≡ 15 11 \cdot45 \equiv15 11 ⋅ 45 ≡ 15 45 ∈ S 45 \in S 45 ∈ S j = k − 35 = 85 j = k - 35 = 85 j = k − 35 = 85 2 a + j = 115 ∉ S 2a+j = 115 \notin S 2 a + j = 115 ∈ / S 5 < a 5 < a 5 < a 11 ⋅ 115 ≡ 65 11 \cdot115 \equiv65 11 ⋅ 115 ≡ 65
Orders k ≤ 90 k \le90 k ≤ 90 1 ∉ S 1 \notin S 1 ∈ / S ∣ L ∣ ≥ 2 |L| \ge2 ∣ L ∣ ≥ 2 k ≥ 12 k \ge12 k ≥ 12 k k k
Table 4. The supports of Lemma 8.20 , with the parameters ( ν , t ) (\nu,t) ( ν , t ) Q ( ν ; t ) Q(\nu;t) Q ( ν ; t ) σ = min S \sigma=\min S σ = min S σ \sigma σ ν \nu ν 2 k 2k 2 k { a , k / 2 , k − a } \{a,k/2,k-a\} { a , k /2 , k − a } 3 β ≥ k / 2 3\beta\ge k/2 3 β ≥ k /2 Theorem 8.8 , and the other twelve have ∣ L ∣ ≥ 2 |L| \ge2 ∣ L ∣ ≥ 2
Let F k F_k F k ( a , 3 a ) (a,3a) ( a , 3 a ) 6 a < k 6a < k 6 a < k (46) , together with the pairs ( g a ′ , g x ′ ) (ga',gx') ( g a ′ , g x ′ ) g g g ( k / g ; a ′ , x ′ ) ∈ F (k/g;a',x') \in\mathcal{F} ( k / g ; a ′ , x ′ ) ∈ F F 44 = { ( 1 , 3 ) , ( 3 , 9 ) , ( 5 , 15 ) , ( 7 , 21 ) } F_{44}=\{(1,3),(3,9),(5,15),(7,21)\} F 44 = {( 1 , 3 ) , ( 3 , 9 ) , ( 5 , 15 ) , ( 7 , 21 )} a < x < k / 2 a < x < k/2 a < x < k /2 F k F_k F k ( k ; a , x ) (k;a,x) ( k ; a , x ) ν \nu ν 2 k 2k 2 k
By Lemma 8.15 (i), S S S x ∈ L ∖ { a } x \in L \setminus\{a\} x ∈ L ∖ { a } ( a , x ) ∈ F k (a,x) \in F_k ( a , x ) ∈ F k x − 2 a ∈ L x-2a \in L x − 2 a ∈ L L L L a ≥ 3 a \ge3 a ≥ 3 1 ∉ S 1 \notin S 1 ∈ / S 2 a < k 2a < k 2 a < k L = { a } L=\{a\} L = { a } x x x k / 2 k/2 k /2 k / 2 k/2 k /2 ∣ L ∣ < 2 |L| < 2 ∣ L ∣ < 2 2
For each of these 411 supports except the twelve with ∣ L ∣ ≥ 2 |L| \ge2 ∣ L ∣ ≥ 2 ν \nu ν
Proof of Theorem 8.1
The seventeen supports that escape the arguments above are listed in Table 4, each with a half-arc carrier. The multiplier ν \nu ν 2 k 2k 2 k { 3 , 9 , 13 , 19 } \{3,9,13,19\} { 3 , 9 , 13 , 19 } { 5 , 7 , 15 , 17 } \{5,7,15,17\} { 5 , 7 , 15 , 17 } k = 22 k=22 k = 22 { 7 , 13 , 17 , 23 } \{7,13,17,23\} { 7 , 13 , 17 , 23 } k = 30 k=30 k = 30 ν \nu ν 2 k 2k 2 k t ∈ { 0 , k } t \in\{0,k\} t ∈ { 0 , k } ( min S , Q ( ν ; t ) ) (\min S,Q(\nu;t)) ( min S , Q ( ν ; t )) σ \sigma σ ν = 11 \nu=11 ν = 11
Lemma 8.20 (seventeen small supports). For each of the seventeen supports S S S ν \nu ν t t t σ = min S \sigma=\min S σ = min S Q ( ν ; t ) Q(\nu;t) Q ( ν ; t ) σ \sigma σ
Proof. This is a finite check of (B1) and (B2) for seventeen explicit pairs ( σ , Q ) (\sigma,Q) ( σ , Q )
Proof of Theorem 8.1 . If 1 ∈ S 1 \in S 1 ∈ S Proposition 8.4 applies. Assume 1 ∉ S 1 \notin S 1 ∈ / S a = min S ≥ 3 a=\min S \ge3 a = min S ≥ 3 a ∈ L a \in L a ∈ L S = { k / 2 } S=\{k/2\} S = { k /2 } S S S Lemma 8.20 applies. Otherwise there are two cases.
∣ L ∣ ≥ 2 |L| \ge2 ∣ L ∣ ≥ 2 Theorem 8.19 gives a unit ν \nu ν Proposition 8.13 gives the σ \sigma σ ( a , Q ( ν ; t ) ) (a,Q(\nu;t)) ( a , Q ( ν ; t )) t ∈ { 0 , k } t \in\{0,k\} t ∈ { 0 , k } ∣ L ∣ = 1 |L| = 1 ∣ L ∣ = 1 L = { a } L = \{a\} L = { a } ∣ S ∣ ≥ 3 |S| \ge3 ∣ S ∣ ≥ 3 S = { a , k / 2 , k − a } S = \{a, k/2, k-a\} S = { a , k /2 , k − a } h = k / 2 h = k/2 h = k /2 a a a h h h a a a h h h k − a k-a k − a k k k gcd ( a , h ) = 1 \gcd(a,h) = 1 g cd( a , h ) = 1 3 ≤ a < h 3 \le a < h 3 ≤ a < h β \beta β Theorem 8.8 . If 3 β ≥ h 3\beta\ge h 3 β ≥ h h ≤ 11 h \le11 h ≤ 11 ( h , a ) (h,a) ( h , a ) ( 5 , 3 ) (5,3) ( 5 , 3 ) ( 7 , 5 ) (7,5) ( 7 , 5 ) ( 9 , 7 ) (9,7) ( 9 , 7 ) ( 11 , 5 ) (11,5) ( 11 , 5 ) ( 11 , 9 ) (11,9) ( 11 , 9 ) β = 4 \beta= 4 β = 4 β = 2 \beta= 2 β = 2 h = 5 h = 5 h = 5 a a a S S S 3 β < h 3\beta< h 3 β < h Theorem 8.8 gives the σ \sigma σ
Three-term progressions under unit multipliers
The all-odd case of § 8 needs one purely arithmetic fact, about how multiplication by a unit moves a short arithmetic progression around the cycle Z / 2 k \mathbb{Z}/2k Z /2 k Z / 2 k \mathbb{Z}/2k Z /2 k { 0 , … , k − 1 } \{0,\ldots,k-1\} { 0 , … , k − 1 } { k , … , 2 k − 1 } \{k,\ldots,2k-1\} { k , … , 2 k − 1 } a < x a < x a < x x < k / 2 x < k/2 x < k /2 { 0 , … , k / 2 − 1 } \{0,\ldots,k/2-1\} { 0 , … , k /2 − 1 } ν \nu ν a a a 2 a − x 2a-x 2 a − x x x x 2 a − x 2a-x 2 a − x a a a x x x e = x − a e = x-a e = x − a gcd ( a , x , k ) = 1 \gcd(a,x,k) = 1 g cd( a , x , k ) = 1 k > 90 k > 90 k > 90
The proof is by the geometry of numbers. The pairs ( ν a , ν e ) (\nu a,\nu e) ( ν a , ν e ) 2 k 2k 2 k 2 k 2k 2 k ν \nu ν [ 0 , 2 k ) 2 [0,2k)^2 [ 0 , 2 k ) 2 2 k 2k 2 k k / 2 k/2 k /2 J ( n ) J(n) J ( n ) J ( n ) J(n) J ( n ) § 9.6 ). At orders 92 ≤ k ≤ 2730 92 \le k \le2730 92 ≤ k ≤ 2730 § 9.7 ), and at the other orders up to 2730 a finite certificate does (Appendix § 12 ). On a first reading one can skip the proofs of the lemmas and Appendix § 12 .
Statement
The functions H H H H ν H_\nu H ν (44) , and witnesses and admissible triples are as in Definition 8.14 . For an integer y y y ⟨ y ⟩ ∈ { 0 , … , 2 k − 1 } \langle y\rangle\in\{0,\ldots,2k-1\} ⟨ y ⟩ ∈ { 0 , … , 2 k − 1 } 2 k 2k 2 k H ( y ) = 1 H(y)=1 H ( y ) = 1 ⟨ y ⟩ < k \langle y\rangle< k ⟨ y ⟩ < k H ( y + k ) = 1 − H ( y ) H(y+k)=1-H(y) H ( y + k ) = 1 − H ( y ) (45) .
Theorem 9.1 (three-term progressions). Every admissible triple ( k ; a , x ) (k;a,x) ( k ; a , x ) k > 90 k > 90 k > 90
Both restrictions are genuine. If x = 3 a x=3a x = 3 a 2 a − x = − a 2a-x=-a 2 a − x = − a ν a \nu a ν a 0 0 0 k k k 2 k 2k 2 k H ν ( − a ) = 1 − H ν ( a ) H_\nu(-a)=1-H_\nu(a) H ν ( − a ) = 1 − H ν ( a ) ν \nu ν k k k ( 90 ; 1 , 33 ) (90;1,33) ( 90 ; 1 , 33 )
It is convenient to work with the pair ( a , e ) (a,e) ( a , e ) e = x − a e=x-a e = x − a 2 a − x = a − e 2a-x=a-e 2 a − x = a − e x = a + e x=a+e x = a + e ν \nu ν ( k ; A , A + E ) (k;A,A+E) ( k ; A , A + E )
H ( ν ( A − E ) ) = H ( ν A ) ≠ H ( ν ( A + E ) ) , (49) H\bigl(\nu(A-E)\bigr)=H(\nu A)\ne H\bigl(\nu(A+E)\bigr), \tag*{(49)} H ( ν ( A − E ) ) = H ( ν A ) = H ( ν ( A + E ) ) , (49) and I shall say that a pair ( A , E ) (A,E) ( A , E ) witness if some unit ν \nu ν (49) . Since H H H 2 k 2k 2 k A A A E E E 2 k 2k 2 k Theorem 9.1 states that ( a , x − a ) (a,x-a) ( a , x − a )
Throughout the rest of the section ( k ; a , x ) (k; a, x) ( k ; a , x ) k > 90 k > 90 k > 90 e = x − a e = x - a e = x − a
a odd , e even , 0 < e , a + e < k / 2 , e ≠ 2 a , gcd ( a , e , 2 k ) = 1. (50) a\ \text{odd},\quad e\ \text{even},\quad0 < e,\quad a + e < k/2,\quad e \ne2a,\quad\gcd(a,e,2k)=1. \tag*{(50)} a odd , e even , 0 < e , a + e < k /2 , e = 2 a , g cd( a , e , 2 k ) = 1. (50) The last condition holds because a prime dividing a a a e e e 2 k 2k 2 k a a a k k k a a a x = a + e x = a + e x = a + e
The target region
The condition (49) only involves the residues ⟨ ν A ⟩ \langle\nu A\rangle ⟨ ν A ⟩ ⟨ ν E ⟩ \langle\nu E\rangle ⟨ ν E ⟩ [ 0 , 2 k ) 2 [0,2k)^2 [ 0 , 2 k ) 2
Δ 1 = { ( X , Y ) ∈ R 2 : k / 2 < X < k , ∣ Y − k / 2 ∣ < X − k / 2 } , Δ 2 = Δ 1 + ( k , 0 ) , \Delta_1 = \{(X,Y) \in\mathbb{R}^2 : k/2 < X < k,\ \lvert Y-k/2\rvert< X-k/2\},\qquad\Delta_2 = \Delta_1 + (k,0), Δ 1 = {( X , Y ) ∈ R 2 : k /2 < X < k , ∣ Y − k /2 ∣ < X − k /2 } , Δ 2 = Δ 1 + ( k , 0 ) , Δ 3 = { ( X , Y ) ∈ R 2 : 0 < X < k / 2 , ∣ Y − 3 k / 2 ∣ < k / 2 − X } , Δ 4 = Δ 3 + ( k , 0 ) , \Delta_3 = \{(X,Y) \in\mathbb{R}^2 : 0 < X < k/2,\ \lvert Y-3k/2\rvert< k/2-X\},\qquad\Delta_4 = \Delta_3 + (k,0), Δ 3 = {( X , Y ) ∈ R 2 : 0 < X < k /2 , ∣ Y − 3 k /2 ∣ < k /2 − X } , Δ 4 = Δ 3 + ( k , 0 ) , and R = Δ 1 ∪ Δ 2 ∪ Δ 3 ∪ Δ 4 ⊂ ( 0 , 2 k ) 2 R = \Delta_1 \cup\Delta_2 \cup\Delta_3 \cup\Delta_4 \subset(0,2k)^2 R = Δ 1 ∪ Δ 2 ∪ Δ 3 ∪ Δ 4 ⊂ ( 0 , 2 k ) 2 Δ 1 \Delta_1 Δ 1 ( k / 2 , k / 2 ) (k/2,k/2) ( k /2 , k /2 ) ( k , 0 ) (k,0) ( k , 0 ) ( k , k ) (k,k) ( k , k ) k / 2 k/\sqrt{2} k / 2 k k k
ρ = 1 2 ( k / 2 + k / 2 − k ) = 1 2 ( 2 − 1 ) k , (51) \rho= \frac{1}{2}(k/\sqrt{2} + k/\sqrt{2} - k) = \frac{1}{2}(\sqrt{2}-1)k, \tag*{(51)} ρ = 2 1 ( k / 2 + k / 2 − k ) = 2 1 ( 2 − 1 ) k , (51) and its incentre is O = ( k − ρ , k / 2 ) O = (k-\rho,k/2) O = ( k − ρ , k /2 ) Figure 2 shows the four triangles inside the square, together with the lattice of the next subsection.
Figure 2. The geometry of Lemma 9.2 and Proposition 9.8 , drawn to scale for k = 128 k = 128 k = 128 ( a , e ) = ( 9 , 20 ) (a,e) = (9,20) ( a , e ) = ( 9 , 20 ) [ 0 , 2 k ] 2 [0,2k]^2 [ 0 , 2 k ] 2 Δ 1 , … , Δ 4 \Delta_1,\ldots,\Delta_4 Δ 1 , … , Δ 4 2 k 2k 2 k ( ⟨ ν a ⟩ , ⟨ ν e ⟩ ) (\langle\nu a\rangle,\langle\nu e\rangle) (⟨ ν a ⟩ , ⟨ ν e ⟩) Λ \Lambda Λ ν \nu ν v = ( 11 , − 4 ) v = (11,-4) v = ( 11 , − 4 ) l = 137 l = \sqrt{137} l = 137 w = ( a , e ) w = (a,e) w = ( a , e ) v × w = 2 k v \times w = 2k v × w = 2 k L 0 , … , L 14 L_0,\ldots,L_{14} L 0 , … , L 14 2 k / l ≈ 21.9 2k/l \approx21.9 2 k / l ≈ 21.9 Δ 1 \Delta_1 Δ 1 O O O ρ \rho ρ L 4 L_4 L 4 0.06 k 0.06k 0.06 k O O O ν = 208 , 67 , 182 , 41 \nu= 208, 67, 182, 41 ν = 208 , 67 , 182 , 41 ( 9 , 20 ) (9,20) ( 9 , 20 ) g = 1 g = 1 g = 1 n = 2 n = 2 n = 2 Proposition 9.8 hold.
Lemma 9.2 (the target region). If gcd ( ν , 2 k ) = 1 \gcd(\nu,2k)=1 g cd( ν , 2 k ) = 1 and ( ⟨ ν A ⟩ , ⟨ ν E ⟩ ) ∈ R (\langle\nu A\rangle,\langle\nu E\rangle) \in R (⟨ ν A ⟩ , ⟨ ν E ⟩) ∈ R then ν \nu ν (49) , so ( A , E ) (A,E) ( A , E ) has a witness .
Proof. Write X = ⟨ ν A ⟩ X = \langle\nu A\rangle X = ⟨ ν A ⟩ Y = ⟨ ν E ⟩ Y = \langle\nu E\rangle Y = ⟨ ν E ⟩ H ( ν A ) = H ( X ) H(\nu A) = H(X) H ( ν A ) = H ( X ) H ( ν ( A ± E ) ) = H ( X ± Y ) H(\nu(A \pm E)) = H(X \pm Y) H ( ν ( A ± E )) = H ( X ± Y ) ( X , Y ) ∈ Δ 1 (X,Y) \in\Delta_1 ( X , Y ) ∈ Δ 1 k − X < Y < X k-X < Y < X k − X < Y < X 0 < X − Y < 2 X − k < k 0 < X-Y < 2X-k < k 0 < X − Y < 2 X − k < k k < X + Y < 2 X < 2 k k < X+Y < 2X < 2k k < X + Y < 2 X < 2 k H ( X − Y ) = 1 = H ( X ) H(X-Y) = 1 = H(X) H ( X − Y ) = 1 = H ( X ) H ( X + Y ) = 0 H(X+Y) = 0 H ( X + Y ) = 0 ( X , Y ) ∈ Δ 3 (X,Y) \in\Delta_3 ( X , Y ) ∈ Δ 3 k + X < Y < 2 k − X k+X < Y < 2k-X k + X < Y < 2 k − X X − Y X-Y X − Y ( 2 X − 2 k , − k ) (2X-2k,-k) ( 2 X − 2 k , − k ) ( 2 X , k ) (2X,k) ( 2 X , k ) 2 k 2k 2 k X + Y X+Y X + Y ( k + 2 X , 2 k ) (k+2X,2k) ( k + 2 X , 2 k ) H ( X − Y ) = 1 = H ( X ) H(X-Y) = 1 = H(X) H ( X − Y ) = 1 = H ( X ) H ( X + Y ) = 0 H(X+Y) = 0 H ( X + Y ) = 0 X X X X + k X+k X + k k k k X X X X ± Y X \pm Y X ± Y H H H H ( X − Y ) = H ( X ) ≠ H ( X + Y ) H(X-Y) = H(X) \ne H(X+Y) H ( X − Y ) = H ( X ) = H ( X + Y ) Δ 2 \Delta_2 Δ 2 Δ 4 \Delta_4 Δ 4
The lattice
The pairs ( ν a , ν e ) (\nu a,\nu e) ( ν a , ν e ) 2 k 2k 2 k
Λ = { w = ( w 1 , w 2 ) ∈ Z 2 : 2 k ∣ w 1 e − w 2 a } . \Lambda= \{w=(w_1,w_2) \in\mathbb{Z}^2 : 2k \mid w_1e-w_2a\}. Λ = { w = ( w 1 , w 2 ) ∈ Z 2 : 2 k ∣ w 1 e − w 2 a } . Since gcd ( a , e , 2 k ) = 1 \gcd(a,e,2k)=1 g cd( a , e , 2 k ) = 1 ξ 1 , ξ 2 , ξ 3 \xi_1,\xi_2,\xi_3 ξ 1 , ξ 2 , ξ 3 ξ 1 a + ξ 2 e + 2 k ξ 3 = 1 \xi_1a+\xi_2e+2k\xi_3=1 ξ 1 a + ξ 2 e + 2 k ξ 3 = 1 label of w ∈ Z 2 w \in\mathbb{Z}^2 w ∈ Z 2 λ ( w ) = ξ 1 w 1 + ξ 2 w 2 \lambda(w)=\xi_1w_1+\xi_2w_2 λ ( w ) = ξ 1 w 1 + ξ 2 w 2 w , w ′ ∈ R 2 w,w' \in\mathbb{R}^2 w , w ′ ∈ R 2 w × w ′ = w 1 w 2 ′ − w 2 w 1 ′ w \times w'=w_1w'_2-w_2w'_1 w × w ′ = w 1 w 2 ′ − w 2 w 1 ′ w ⋅ w ′ w\cdot w' w ⋅ w ′
Lemma 9.3 (labels).
(a) Λ \Lambda Λ Z 2 \mathbb{Z}^2 Z 2 2 k Z 2 2k\mathbb{Z}^2 2 k Z 2 ( ν a , ν e ) (\nu a,\nu e) ( ν a , ν e ) ν ∈ Z \nu\in\mathbb{Z} ν ∈ Z w ∈ Λ w \in\Lambda w ∈ Λ w ≡ λ ( w ) ( a , e ) ( m o d 2 k ) w \equiv\lambda(w)(a,e) \pmod{2k} w ≡ λ ( w ) ( a , e ) ( mod 2 k ) Λ = Z ( a , e ) + 2 k Z 2 \Lambda= \mathbb{Z}(a,e)+2k\mathbb{Z}^2 Λ = Z ( a , e ) + 2 k Z 2 λ \lambda λ
(b) If w , w ′ ∈ Λ w,w' \in\Lambda w , w ′ ∈ Λ 2 k ∣ w × w ′ 2k \mid w \times w' 2 k ∣ w × w ′
(c) If w ∈ Λ w \in\Lambda w ∈ Λ gcd ( λ ( w ) , 2 k ) = 1 \gcd(\lambda(w),2k)=1 g cd( λ ( w ) , 2 k ) = 1 ( ⟨ w 1 ⟩ , ⟨ w 2 ⟩ ) ∈ R (\langle w_1\rangle,\langle w_2\rangle) \in R (⟨ w 1 ⟩ , ⟨ w 2 ⟩) ∈ R ( a , e ) (a,e) ( a , e )
Proof. (a) The first claims are clear. If w 1 e − w 2 a = 2 k c w_1e-w_2a=2kc w 1 e − w 2 a = 2 k c
w 1 − λ ( w ) a = w 1 ( ξ 2 e + 2 k ξ 3 ) − ξ 2 a w 2 = ξ 2 ⋅ 2 k c + 2 k ξ 3 w 1 , w_1-\lambda(w)a=w_1(\xi_2e+2k\xi_3)-\xi_2aw_2=\xi_2\cdot2kc+2k\xi_3w_1, w 1 − λ ( w ) a = w 1 ( ξ 2 e + 2 k ξ 3 ) − ξ 2 a w 2 = ξ 2 ⋅ 2 k c + 2 k ξ 3 w 1 , and similarly w 2 − λ ( w ) e = − ξ 1 ⋅ 2 k c + 2 k ξ 3 w 2 w_2-\lambda(w)e=-\xi_1\cdot2kc+2k\xi_3w_2 w 2 − λ ( w ) e = − ξ 1 ⋅ 2 k c + 2 k ξ 3 w 2 w × w ′ = λ ( w ) λ ( w ′ ) ( a , e ) × ( a , e ) = 0 w\times w'=\lambda(w)\lambda(w')(a,e)\times(a,e)=0 w × w ′ = λ ( w ) λ ( w ′ ) ( a , e ) × ( a , e ) = 0 2 k 2k 2 k ⟨ λ ( w ) a ⟩ = ⟨ w 1 ⟩ \langle\lambda(w)a\rangle=\langle w_1\rangle ⟨ λ ( w ) a ⟩ = ⟨ w 1 ⟩ ⟨ λ ( w ) e ⟩ = ⟨ w 2 ⟩ \langle\lambda(w)e\rangle=\langle w_2\rangle ⟨ λ ( w ) e ⟩ = ⟨ w 2 ⟩ Lemma 9.2 applies with ν = λ ( w ) \nu=\lambda(w) ν = λ ( w )
Call w ∈ Λ w \in\Lambda w ∈ Λ unit point if gcd ( λ ( w ) , 2 k ) = 1 \gcd(\lambda(w),2k)=1 g cd( λ ( w ) , 2 k ) = 1 Lemma 9.3 (c), the task is to find a unit point of Λ \Lambda Λ R + 2 k Z 2 R+2k\mathbb{Z}^{2} R + 2 k Z 2 Λ \Lambda Λ
Lemma 9.4 (a shortest vector). Let v v v nonzero vector of Λ \Lambda Λ l = ∣ v ∣ l=|v| l = ∣ v ∣ g = gcd ( v 1 , v 2 ) g=\gcd(v_1,v_2) g = g cd( v 1 , v 2 ) u = ( p , q ) = v / g u=(p,q)=v/g u = ( p , q ) = v / g
(a) there is w ∈ Λ w\in\Lambda w ∈ Λ v × w = 2 k v\times w=2k v × w = 2 k
(b) 3 l 4 ≤ 16 k 2 3l^4\le16k^2 3 l 4 ≤ 16 k 2
(c) g ∣ 2 k g\mid2k g ∣ 2 k g q gq g q m u ∉ Λ mu\notin\Lambda m u ∈ / Λ 0 < m < g 0<m<g 0 < m < g
(d) g ∣ λ ( v ) g\mid\lambda(v) g ∣ λ ( v ) λ ( v ) \lambda(v) λ ( v ) 2 k 2k 2 k g g g
(e) gcd ( λ ( w ) , g ) = 1 \gcd(\lambda(w),g)=1 g cd( λ ( w ) , g ) = 1 w w w
Proof. (a) Write v 1 e − v 2 a = 2 k φ v_1e-v_2a=2k\varphi v 1 e − v 2 a = 2 k φ c = gcd ( v 1 , v 2 , φ ) c=\gcd(v_1,v_2,\varphi) c = g cd( v 1 , v 2 , φ ) v / c ∈ Λ v/c\in\Lambda v / c ∈ Λ c = 1 c=1 c = 1 μ 1 , μ 2 , μ 3 \mu_1,\mu_2,\mu_3 μ 1 , μ 2 , μ 3 μ 1 v 1 + μ 2 v 2 + μ 3 φ = 1 \mu_1v_1+\mu_2v_2+\mu_3\varphi=1 μ 1 v 1 + μ 2 v 2 + μ 3 φ = 1 w = ( − 2 k μ 2 + μ 3 a , 2 k μ 1 + μ 3 e ) w=(-2k\mu_2+\mu_3a,2k\mu_1+\mu_3e) w = ( − 2 k μ 2 + μ 3 a , 2 k μ 1 + μ 3 e ) w ∈ Λ w\in\Lambda w ∈ Λ μ 3 ( a , e ) \mu_3(a,e) μ 3 ( a , e ) 2 k Z 2 2k\mathbb{Z}^2 2 k Z 2 v × w = 2 k ( μ 1 v 1 + μ 2 v 2 ) + μ 3 ( v 1 e − v 2 a ) = 2 k v\times w=2k(\mu_1v_1+\mu_2v_2)+\mu_3(v_1e-v_2a)=2k v × w = 2 k ( μ 1 v 1 + μ 2 v 2 ) + μ 3 ( v 1 e − v 2 a ) = 2 k
(b) Replacing w w w w + t v w+tv w + t v t ∈ Z t\in\mathbb{Z} t ∈ Z v × w v\times w v × w ∣ v ⋅ w ∣ ≤ l 2 / 2 |v\cdot w|\le l^2/2 ∣ v ⋅ w ∣ ≤ l 2 /2 w ≠ 0 w\ne0 w = 0 ∣ w ∣ ≥ l |w|\ge l ∣ w ∣ ≥ l ( v × w ) 2 + ( v ⋅ w ) 2 = ∣ v ∣ 2 ∣ w ∣ 2 (v\times w)^2+(v\cdot w)^2=|v|^2|w|^2 ( v × w ) 2 + ( v ⋅ w ) 2 = ∣ v ∣ 2 ∣ w ∣ 2 4 k 2 ≥ l 4 − l 4 / 4 4k^2\ge l^4-l^4/4 4 k 2 ≥ l 4 − l 4 /4
(c) Let g ′ = gcd ( 2 k , g ) = ζ 1 2 k + ζ 2 g g'=\gcd(2k,g)=\zeta_1 2k+\zeta_2g g ′ = g cd( 2 k , g ) = ζ 1 2 k + ζ 2 g g ′ u = ζ 1 ( 2 k u ) + ζ 2 v ∈ Λ g'u=\zeta_1(2ku)+\zeta_2v\in\Lambda g ′ u = ζ 1 ( 2 k u ) + ζ 2 v ∈ Λ g ′ ∣ u ∣ ≥ l = g ∣ u ∣ g'|u|\ge l=g|u| g ′ ∣ u ∣ ≥ l = g ∣ u ∣ g ′ = g g'=g g ′ = g g ∣ 2 k g\mid2k g ∣ 2 k 0 < m < g 0<m<g 0 < m < g m u mu m u v v v m u ∉ Λ mu\notin\Lambda m u ∈ / Λ v 2 a ≡ v 1 e ( m o d 2 k ) v_2a\equiv v_1e\pmod{2k} v 2 a ≡ v 1 e ( mod 2 k ) e e e 2 k 2k 2 k v 2 a v_2a v 2 a a a a v 2 = g q v_2=gq v 2 = g q
(d) λ ( v ) = g ( ζ 1 p + ζ 2 q ) \lambda(v)=g(\zeta_1p+\zeta_2q) λ ( v ) = g ( ζ 1 p + ζ 2 q ) c c c λ ( v ) \lambda(v) λ ( v ) 2 k 2k 2 k Lemma 9.3 (a) it divides v 1 v_1 v 1 v 2 v_2 v 2 g g g
(e) For every z ∈ R 2 z\in\mathbb{R}^2 z ∈ R 2 ( v × w ) z = ( z × w ) v + ( v × z ) w (v\times w)z=(z\times w)v+(v\times z)w ( v × w ) z = ( z × w ) v + ( v × z ) w z z z z = v z=v z = v z = w z=w z = w z = ( a , e ) ∈ Λ z=(a,e)\in\Lambda z = ( a , e ) ∈ Λ Lemma 9.3 (b) both z × w z\times w z × w v × z v\times z v × z 2 k = v × w 2k=v\times w 2 k = v × w ( a , e ) = c 1 v + c 2 w (a,e)=c_1v+c_2w ( a , e ) = c 1 v + c 2 w c 1 , c 2 ∈ Z c_1,c_2\in\mathbb{Z} c 1 , c 2 ∈ Z λ \lambda λ 1 − 2 k ζ 3 = ζ 1 a + ζ 2 e = c 1 λ ( v ) + c 2 λ ( w ) 1-2k\zeta_3=\zeta_1a+\zeta_2e=c_1\lambda(v)+c_2\lambda(w) 1 − 2 k ζ 3 = ζ 1 a + ζ 2 e = c 1 λ ( v ) + c 2 λ ( w ) g g g 2 k 2k 2 k λ ( v ) \lambda(v) λ ( v ) λ ( w ) \lambda(w) λ ( w )
Replacing ( v , w ) (v,w) ( v , w ) ( − v , − w ) (-v,-w) ( − v , − w ) u u u normalised : p > 0 p>0 p > 0 p = 0 < q p=0<q p = 0 < q
Lemma 9.5 (eight excluded directions). With v = g u v=gu v = g u Lemma 9.4 and u = ( p , q ) u=(p,q) u = ( p , q ) ( g ; p , q ) (g;p,q) ( g ; p , q )
( 1 ; 1 , 0 ) , ( 1 ; 1 , 2 ) , ( 2 ; 0 , 1 ) , ( 2 ; 1 , 0 ) , ( 2 ; 1 , 1 ) , ( 2 ; 1 , − 1 ) , ( 2 ; 1 , − 2 ) , ( 4 ; 0 , 1 ) . (1;1,0),\quad(1;1,2),\quad(2;0,1),\quad(2;1,0),\quad(2;1,1),\quad(2;1,-1),\quad(2;1,-2),\quad(4;0,1). ( 1 ; 1 , 0 ) , ( 1 ; 1 , 2 ) , ( 2 ; 0 , 1 ) , ( 2 ; 1 , 0 ) , ( 2 ; 1 , 1 ) , ( 2 ; 1 , − 1 ) , ( 2 ; 1 , − 2 ) , ( 4 ; 0 , 1 ) . Proof. Each case says that 2 k 2k 2 k g ( p e − q a ) g(pe-qa) g ( p e − q a ) (50) . For ( 1 ; 1 , 0 ) (1;1,0) ( 1 ; 1 , 0 ) 2 k ∣ e 2k\mid e 2 k ∣ e 0 < e < k 0<e<k 0 < e < k ( 1 ; 1 , 2 ) (1;1,2) ( 1 ; 1 , 2 ) 2 k ∣ e − 2 a 2k\mid e-2a 2 k ∣ e − 2 a ∣ e − 2 a ∣ < k |e-2a|<k ∣ e − 2 a ∣ < k e = 2 a e=2a e = 2 a ( 2 ; 0 , 1 ) (2;0,1) ( 2 ; 0 , 1 ) k ∣ a k\mid a k ∣ a 0 < a < k 0<a<k 0 < a < k ( 2 ; 1 , 0 ) (2;1,0) ( 2 ; 1 , 0 ) k ∣ e k\mid e k ∣ e ( 2 ; 1 , ± 1 ) (2;1,\pm1) ( 2 ; 1 , ± 1 ) k ∣ e ∓ a k\mid e\mp a k ∣ e ∓ a k k k ( 2 ; 1 , − 2 ) (2;1,-2) ( 2 ; 1 , − 2 ) k ∣ e + 2 a k\mid e+2a k ∣ e + 2 a 0 < e + 2 a = x + a < k 0<e+2a=x+a<k 0 < e + 2 a = x + a < k ( 4 ; 0 , 1 ) (4;0,1) ( 4 ; 0 , 1 ) k / 2 ∣ a k/2\mid a k /2 ∣ a 0 < a < k / 2 0<a<k/2 0 < a < k /2
Lattice lines and Jacobsthal's function
For n ≥ 1 n\ge1 n ≥ 1 Jacobsthal’s function J ( n ) J(n) J ( n ) m m m m m m n n n [19 b,c] . It depends only on the set of primes dividing n n n J ( n ′ ) ≤ J ( n ) J(n')\le J(n) J ( n ′ ) ≤ J ( n ) n ′ n' n ′ n n n J ( n ) ≪ ω ( n ) 2 log 2 ω ( n ) J(n)\ll\omega(n)^2\log^2\omega(n) J ( n ) ≪ ω ( n ) 2 log 2 ω ( n ) ω ( n ) \omega(n) ω ( n ) n n n [18 ] , but the elementary bounds proved below suffice here.
Fix v = g u v=gu v = g u w w w Lemma 9.4 , and write λ v = λ ( v ) \lambda_v=\lambda(v) λ v = λ ( v ) λ w = λ ( w ) \lambda_w=\lambda(w) λ w = λ ( w ) j ∈ Z j\in\mathbb{Z} j ∈ Z
L j = { j w + t v : t ∈ R } L_j=\{jw+tv:t\in\mathbb{R}\} L j = { j w + t v : t ∈ R } (see Figure 2 ). Since v × ( j w + t v ) = 2 k j v\times(jw+tv)=2kj v × ( j w + t v ) = 2 k j L j L_j L j 2 k j / l 2kj/l 2 k j / l L 0 L_0 L 0 2 k / l 2k/l 2 k / l j w + i v jw+iv j w + i v i ∈ Z i\in\mathbb{Z} i ∈ Z Λ \Lambda Λ l l l j λ w + i λ v j\lambda_w+i\lambda_v j λ w + i λ v
Lemma 9.6 (unit points on a line). Let n ≥ 1 n\geq1 n ≥ 1 2 k 2k 2 k gcd ( n , g ) = 1 \gcd(n,g)=1 g cd( n , g ) = 1 2 k 2k 2 k n n n g g g n n n 2 k 2k 2 k g g g gcd ( j , g ) = 1 \gcd(j,g)=1 g cd( j , g ) = 1 J ( n ) J(n) J ( n ) j w + i v jw+iv j w + i v L j L_j L j
Proof. A common divisor of λ v \lambda_v λ v n n n 2 k 2k 2 k g g g Lemma 9.4 (d), and it is 1 since gcd ( n , g ) = 1 \gcd(n,g)=1 g cd( n , g ) = 1 ι \iota ι ι λ v ≡ 1 ( m o d n ) \iota\lambda_v\equiv1\pmod n ι λ v ≡ 1 ( mod n ) j λ w + i λ v ≡ λ v ( i + ι j λ w ) ( m o d n ) j\lambda_w+i\lambda_v\equiv\lambda_v(i+\iota j\lambda_w)\pmod n j λ w + i λ v ≡ λ v ( i + ι j λ w ) ( mod n ) i i i J ( n ) J(n) J ( n ) i + ι j λ w i+\iota j\lambda_w i + ι j λ w j λ w + i λ v j\lambda_w+i\lambda_v j λ w + i λ v n n n ℓ \ell ℓ g g g λ v \lambda_v λ v Lemma 9.4 (d)) but neither j j j λ w \lambda_w λ w Lemma 9.4 (e)), so it does not divide j λ w + i λ v j\lambda_w+i\lambda_v j λ w + i λ v 2 k 2k 2 k n n n g g g 2 k 2k 2 k
Lemma 9.7 (chords). Let n n n Lemma 9.6 , gcd ( j , g ) = 1 \gcd(j,g)=1 g cd( j , g ) = 1 Δ \Delta Δ Δ 1 , … , Δ 4 \Delta_1,\ldots,\Delta_4 Δ 1 , … , Δ 4 z ∈ Z 2 z\in\mathbb{Z}^2 z ∈ Z 2 t 0 < t 1 t_0<t_1 t 0 < t 1 t 1 − t 0 > J ( n ) t_1-t_0>J(n) t 1 − t 0 > J ( n ) j w + t v ∈ Δ + 2 k z jw+tv\in\Delta+2kz j w + t v ∈ Δ + 2 k z t ∈ ( t 0 , t 1 ) t\in(t_0,t_1) t ∈ ( t 0 , t 1 ) ( a , e ) (a,e) ( a , e )
Proof. The integers i 0 , … , i 0 + J ( n ) − 1 i_0,\ldots,i_0+J(n)-1 i 0 , … , i 0 + J ( n ) − 1 i 0 = ⌊ t 0 ⌋ + 1 i_0=\lfloor t_0\rfloor+1 i 0 = ⌊ t 0 ⌋ + 1 ( t 0 , t 0 + J ( n ) ] ⊂ ( t 0 , t 1 ) (t_0,t_0+J(n)]\subset(t_0,t_1) ( t 0 , t 0 + J ( n )] ⊂ ( t 0 , t 1 ) Lemma 9.6 one of the points j w + i v jw+iv j w + i v i i i Δ + 2 k z \Delta+2kz Δ + 2 k z Δ ⊂ ( 0 , 2 k ) 2 \Delta\subset(0,2k)^2 Δ ⊂ ( 0 , 2 k ) 2 2 k 2k 2 k Δ ⊂ R \Delta\subset R Δ ⊂ R Lemma 9.3 (c).
Two ways to find a chord
The first way uses a band of lines through the incircle of Δ 1 \Delta_1 Δ 1
Proposition 9.8 (the band). With v = g u v=gu v = g u Lemma 9.4 and n n n Lemma 9.6 , suppose that
J ( g ) < 0.102 l and J ( n ) l < 0.3604 k . J(g)<0.102l\qquad\text{and}\qquad J(n)l<0.3604k. J ( g ) < 0.102 l and J ( n ) l < 0.3604 k . Then ( a , e ) (a,e) ( a , e )
Proof. Let D D D ρ \rho ρ O O O Δ 1 \Delta_1 Δ 1 Δ 1 \Delta_1 Δ 1 τ = ( v × O ) / 2 k \tau=(v\times O)/2k τ = ( v × O ) /2 k O O O 2 k τ / l 2k\tau/l 2 k τ / l L 0 L_0 L 0 J ( g ) J(g) J ( g ) ⌊ τ − J ( g ) / 2 ⌋ + 1 , … , ⌊ τ − J ( g ) / 2 ⌋ + J ( g ) \lfloor\tau-J(g)/2\rfloor+1,\ldots,\lfloor\tau-J(g)/2\rfloor+J(g) ⌊ τ − J ( g ) /2 ⌋ + 1 , … , ⌊ τ − J ( g ) /2 ⌋ + J ( g ) j j j g g g ∣ j − τ ∣ ≤ J ( g ) / 2 |j-\tau|\leq J(g)/2 ∣ j − τ ∣ ≤ J ( g ) /2 O O O L j L_j L j ( 2 k / l ) J ( g ) / 2 = J ( g ) k / l < 0.102 k (2k/l)J(g)/2=J(g)k/l<0.102k ( 2 k / l ) J ( g ) /2 = J ( g ) k / l < 0.102 k L j L_j L j D D D 2 ρ 2 − ( 0.102 k ) 2 2\sqrt{\rho^2-(0.102k)^2} 2 ρ 2 − ( 0.102 k ) 2 (51) , ρ 2 = 1 4 ( 3 − 2 2 ) k 2 > 0.04289 k 2 \rho^2=\frac{1}{4}(3-2\sqrt{2})k^2>0.04289k^2 ρ 2 = 4 1 ( 3 − 2 2 ) k 2 > 0.04289 k 2 ( 0.102 k ) 2 = 0.010404 k 2 (0.102k)^2=0.010404k^2 ( 0.102 k ) 2 = 0.010404 k 2 2 0.03248 k > 0.3604 k > J ( n ) l 2\sqrt{0.03248}k>0.3604k>J(n)l 2 0.03248 k > 0.3604 k > J ( n ) l t t t L j L_j L j J ( n ) J(n) J ( n ) Δ 1 \Delta_1 Δ 1 Lemma 9.7 completes the proof.
The second way handles the short directions. Its key feature is that the geometry does not depend on k k k a a a e e e 2 k 2k 2 k Δ ^ i = Δ i / 2 k \widehat{\Delta}_i=\Delta_i/2k Δ i = Δ i /2 k
Δ ^ 1 = { ( X , Y ) : 1 4 < X < 1 2 , ∣ Y − 1 4 ∣ < X − 1 4 } , Δ ^ 2 = Δ ^ 1 + ( 1 2 , 0 ) , \widehat{\Delta}_1=\{(X,Y):\tfrac{1}{4}<X<\tfrac{1}{2},\ |Y-\tfrac{1}{4}|<X-\tfrac{1}{4}\},\qquad
\widehat{\Delta}_2=\widehat{\Delta}_1+(\tfrac{1}{2},0), Δ 1 = {( X , Y ) : 4 1 < X < 2 1 , ∣ Y − 4 1 ∣ < X − 4 1 } , Δ 2 = Δ 1 + ( 2 1 , 0 ) , Δ ^ 3 = { ( X , Y ) : 0 < X < 1 4 , ∣ Y − 3 4 ∣ < 1 4 − X } , Δ ^ 4 = Δ ^ 3 + ( 1 2 , 0 ) . \widehat{\Delta}_3=\{(X,Y):0<X<\tfrac{1}{4},\ |Y-\tfrac{3}{4}|<\tfrac{1}{4}-X\},\qquad
\widehat{\Delta}_4=\widehat{\Delta}_3+(\tfrac{1}{2},0). Δ 3 = {( X , Y ) : 0 < X < 4 1 , ∣ Y − 4 3 ∣ < 4 1 − X } , Δ 4 = Δ 3 + ( 2 1 , 0 ) . For a primitive u = ( p , q ) u=(p,q) u = ( p , q ) g ≥ 1 g\ge1 g ≥ 1 j ∈ Z j\in\mathbb{Z} j ∈ Z
L ^ j ( g , u ) = { j g ∣ u ∣ 2 ( − q , p ) + s u : s ∈ R } . \widehat{\mathcal{L}}_j(g,u)=\left\{\frac{j}{g\lvert u\rvert^2}(-q,p)+su:s\in\mathbb{R}\right\}. L j ( g , u ) = { g ∣ u ∣ 2 j ( − q , p ) + s u : s ∈ R } . If v = g u v=gu v = g u w w w Lemma 9.4 , then ( − q , p ) ⋅ w = u × w = 2 k / g (-q,p)\cdot w=u\times w=2k/g ( − q , p ) ⋅ w = u × w = 2 k / g w w w u u u ( − q , p ) (-q,p) ( − q , p )
j w + t v 2 k = j g ∣ u ∣ 2 ( − q , p ) + ( s ∗ + g 2 k t ) u , s ∗ = j ( u ⋅ w ) 2 k ∣ u ∣ 2 . (52) \frac{jw+tv}{2k}=\frac{j}{g\lvert u\rvert^2}(-q,p)+\left(s_*+\frac{g}{2k}t\right)u,\qquad s_*=\frac{j(u\cdot w)}{2k\lvert u\rvert^2}. \tag*{(52)} 2 k j w + t v = g ∣ u ∣ 2 j ( − q , p ) + ( s ∗ + 2 k g t ) u , s ∗ = 2 k ∣ u ∣ 2 j ( u ⋅ w ) . (52) So L j / 2 k = L ^ j ( g , u ) \mathcal{L}_j/2k=\widehat{\mathcal{L}}_j(g,u) L j /2 k = L j ( g , u ) s s s ψ \psi ψ t t t 2 k ψ / g 2k\psi/g 2 k ψ / g
Definition 9.9. A chord witness for ( g ; p , q ) (g;p,q) ( g ; p , q ) j j j gcd ( j , g ) = 1 \gcd(j,g)=1 g cd( j , g ) = 1 i ∈ { 1 , 2 , 3 , 4 } i\in\{1,2,3,4\} i ∈ { 1 , 2 , 3 , 4 } z ∈ Z 2 z\in\mathbb{Z}^2 z ∈ Z 2 s 0 < s 1 s_0<s_1 s 0 < s 1 L ^ j ( g , u ) \widehat{\mathcal{L}}_j(g,u) L j ( g , u ) s s s Δ ^ i + z \widehat{\Delta}_i+z Δ i + z s ∈ ( s 0 , s 1 ) s\in(s_0,s_1) s ∈ ( s 0 , s 1 ) ψ = s 1 − s 0 \psi=s_1-s_0 ψ = s 1 − s 0
A chord witness is checked by evaluating finitely many signs. Each Δ ^ i + z \widehat{\Delta}_i+z Δ i + z { f > 0 } \{f>0\} { f > 0 } f f f f ≥ 0 f\ge0 f ≥ 0 s 0 s_0 s 0 s 1 s_1 s 1 f > 0 f>0 f > 0 f > 0 f>0 f > 0 f f f
Proposition 9.10 (a tabulated chord). With v = g u v=gu v = g u Lemma 9.4 and n n n Lemma 9.6 , suppose ( g ; p , q ) (g;p,q) ( g ; p , q ) ψ \psi ψ 2 k ψ > J ( n ) g 2k\psi>J(n)g 2 k ψ > J ( n ) g ( a , e ) (a,e) ( a , e )
Proof. By (52) , j w + t v ∈ Δ i + 2 k z jw+tv\in\Delta_i+2kz j w + t v ∈ Δ i + 2 k z t t t 2 k ψ / g > J ( n ) 2k\psi/g>J(n) 2 k ψ / g > J ( n ) Lemma 9.7 applies.
When neither proposition applies directly, the lattice can be pinned down completely. Since g u ∈ Λ gu\in\Lambda g u ∈ Λ ( a , e ) (a,e) ( a , e ) g 2 g^2 g 2
Lemma 9.11 (lattice classes). Let v = g u v=gu v = g u Lemma 9.4 and n 0 = 2 k / g n_0=2k/g n 0 = 2 k / g η \eta η 2 k 2k 2 k 0 ≤ s , t < g 0\le s,t<g 0 ≤ s , t < g ( A , E ) = ( p + n 0 s , q + n 0 t ) (A,E)=(p+n_0s,q+n_0t) ( A , E ) = ( p + n 0 s , q + n 0 t )
η ( a , e ) ≡ ( A , E ) ( m o d 2 k ) . \eta(a,e)\equiv(A,E)\pmod{2k}. η ( a , e ) ≡ ( A , E ) ( mod 2 k ) . Moreover gcd ( A , E , 2 k ) = 1 \gcd(A,E,2k)=1 g cd( A , E , 2 k ) = 1 2 k ∤ m ( p E − q A ) 2k\nmid m(pE-qA) 2 k ∤ m ( pE − q A ) 0 < m < g 0<m<g 0 < m < g ( A , E ) (A,E) ( A , E ) ( a , e ) (a,e) ( a , e )
Proof. By Lemma 9.4 (d), λ v = λ ( v ) = g λ ∗ \lambda_v=\lambda(v)=g\lambda^* λ v = λ ( v ) = g λ ∗ gcd ( λ ∗ , n 0 ) = 1 \gcd(\lambda^*,n_0)=1 g cd( λ ∗ , n 0 ) = 1 c c c g c gc g c 2 k 2k 2 k λ v \lambda_v λ v g g g c = 1 c=1 c = 1 Lemma 9.3 (a), g ( p , q ) = v ≡ g λ ∗ ( a , e ) ( m o d 2 k ) g(p,q)=v\equiv g\lambda^*(a,e)\pmod{2k} g ( p , q ) = v ≡ g λ ∗ ( a , e ) ( mod 2 k ) λ ∗ ( a , e ) ≡ ( p , q ) ( m o d n 0 ) \lambda^*(a,e)\equiv(p,q)\pmod{n_0} λ ∗ ( a , e ) ≡ ( p , q ) ( mod n 0 ) n 0 n_0 n 0 2 k 2k 2 k η = λ ∗ + n 0 r \eta=\lambda^*+n_0r η = λ ∗ + n 0 r r r r 2 k 2k 2 k n 0 n_0 n 0 λ ∗ \lambda^* λ ∗ ℓ \ell ℓ 2 k 2k 2 k ℓ ∣ n 0 \ell\mid n_0 ℓ ∣ n 0 η ≡ λ ∗ ≢ 0 ( m o d ℓ ) \eta\equiv\lambda^*\not\equiv0\pmod{\ell} η ≡ λ ∗ ≡ 0 ( mod ℓ ) gcd ( λ ∗ , n 0 ) = 1 \gcd(\lambda^*,n_0)=1 g cd( λ ∗ , n 0 ) = 1 ℓ ∤ n 0 \ell\nmid n_0 ℓ ∤ n 0 ℓ ∣ λ ∗ \ell\mid\lambda^* ℓ ∣ λ ∗ ℓ ∤ r \ell\nmid r ℓ ∤ r η ≡ n 0 r ≢ 0 \eta\equiv n_0r\not\equiv0 η ≡ n 0 r ≡ 0 ℓ \ell ℓ ℓ ∣ r \ell\mid r ℓ ∣ r η ≡ λ ∗ ≢ 0 \eta\equiv\lambda^*\not\equiv0 η ≡ λ ∗ ≡ 0 η a = p + n 0 K 1 \eta a=p+n_0K_1 η a = p + n 0 K 1 η e = q + n 0 K 2 \eta e=q+n_0K_2 η e = q + n 0 K 2 K 1 , K 2 K_1,K_2 K 1 , K 2 s , t s,t s , t K 1 , K 2 K_1,K_2 K 1 , K 2 g g g η ( a , e ) ≡ ( A , E ) ( m o d n 0 g ) \eta(a,e)\equiv(A,E)\pmod{n_0g} η ( a , e ) ≡ ( A , E ) ( mod n 0 g )
If c c c A A A E E E 2 k 2k 2 k η a \eta a η a η e \eta e η e η ( ξ 1 a + ξ 2 e + 2 k ξ 3 ) = η \eta(\xi_1a+\xi_2e+2k\xi_3)=\eta η ( ξ 1 a + ξ 2 e + 2 k ξ 3 ) = η c = 1 c=1 c = 1 2 k ∣ m ( p E − q A ) 2k\mid m(pE-qA) 2 k ∣ m ( pE − q A ) 2 k ∣ m η ( p e − q a ) 2k\mid m\eta(pe-qa) 2 k ∣ m η ( p e − q a ) 2 k ∣ m ( p e − q a ) 2k\mid m(pe-qa) 2 k ∣ m ( p e − q a ) m u ∈ Λ mu\in\Lambda m u ∈ Λ 0 < m < g 0<m<g 0 < m < g Lemma 9.4 (c). For the last statement, if a unit ν \nu ν (49) for ( A , E ) (A,E) ( A , E ) ν η \nu\eta ν η (49) for ( a , e ) (a,e) ( a , e ) ν η ( a , e ) ≡ ν ( A , E ) ( m o d 2 k ) \nu\eta(a,e)\equiv\nu(A,E)\pmod{2k} ν η ( a , e ) ≡ ν ( A , E ) ( mod 2 k )
The two conditions on ( A , E ) (A,E) ( A , E ) Lemma 9.11 are used in reverse: a class ( s , t ) (s,t) ( s , t ) gcd ( A , E , 2 k ) > 1 \gcd(A,E,2k)>1 g cd( A , E , 2 k ) > 1 2 k ∣ m ( p E − q A ) 2k\mid m(pE-qA) 2 k ∣ m ( pE − q A ) 0 < m < g 0<m<g 0 < m < g ( a , e ) (a,e) ( a , e ) g u gu g u
The chord table and orders above 2730
The two propositions of the last subsection settle every large order. The band Proposition 9.8 needs a long shortest vector, and the short ones are covered once and for all by a table of chords.
Let V \mathcal{V} V ( g ; p , q ) (g;p,q) ( g ; p , q ) g ≥ 1 g \ge1 g ≥ 1 ( p , q ) (p,q) ( p , q ) g q gq g q g 2 ( p 2 + q 2 ) ≤ 6400 g^{2}(p^{2}+q^{2}) \le6400 g 2 ( p 2 + q 2 ) ≤ 6400 ( g ; p , q ) (g;p,q) ( g ; p , q ) Lemma 9.5 . By Lemma 9.4 (c) and Lemma 9.5 , a normalised shortest vector v = g u v=gu v = g u Λ \Lambda Λ l ≤ 80 l \le80 l ≤ 80 ( g ; p , q ) ∈ V (g;p,q) \in\mathcal{V} ( g ; p , q ) ∈ V g ∣ 2 k g \mid2k g ∣ 2 k Theorem 9.1 ; the other two parts, (C2) and (C3), are in Appendix § 12 .
(C1) The chord table. For each of the 5012 triples of V \mathcal{V} V V \mathcal{V} V g ≤ 80 g \le80 g ≤ 80 ( p , q ) (p,q) ( p , q ) 0 ≤ p ≤ 80 / g 0 \le p \le80/g 0 ≤ p ≤ 80/ g ∣ q ∣ ≤ 80 / g |q| \le80/g ∣ q ∣ ≤ 80/ g
g ≤ 339.8 ψ , (53) g \le339.8\psi, \tag*{(53)} g ≤ 339.8 ψ , (53) where ψ \psi ψ ( 76 ; 1 , 0 ) (76;1,0) ( 76 ; 1 , 0 ) ψ = 17 / 76 \psi=17/76 ψ = 17/76 g / ψ = 339.76 … g/\psi=339.76\ldots g / ψ = 339.76 …
Say that an even order k k k tail condition with bound m m m
17.8 m 2 ≤ k and 169.9 m < k , (54) 17.8m^{2} \le k \qquad\text{and}\qquad169.9m < k, \tag*{(54)} 17.8 m 2 ≤ k and 169.9 m < k , (54) that is, 178 m 2 ≤ 10 k 178m^{2} \le10k 178 m 2 ≤ 10 k 1699 m < 10 k 1699m < 10k 1699 m < 10 k Proposition 9.8 ) needs, J ( n ) l < 0.3604 k J(n)l < 0.3604k J ( n ) l < 0.3604 k l 2 ≤ 4 k / 3 l^{2} \le4k/\sqrt{3} l 2 ≤ 4 k / 3 Proposition 9.10 ) needs, 2 k ψ > J ( n ) g 2k\psi> J(n)g 2 k ψ > J ( n ) g Corollary A.5 ).
Proposition 9.12. Let k k k tail condition (53) with bound m = J ( 2 k ) m=J(2k) m = J ( 2 k ) Theorem 9.1 holds at the order k k k m = J ( 2 k ) m=J(2k) m = J ( 2 k ) k > 2730 k>2730 k > 2730
Proof. For the last statement, write J = J ( 2 k ) J=J(2k) J = J ( 2 k ) Corollary A.5 (a), 17.8 J 2 ≤ k 17.8J^{2}\le k 17.8 J 2 ≤ k J ≥ 10 J\ge10 J ≥ 10 169.9 J < 178 J ≤ 17.8 J 2 ≤ k 169.9J<178J\le17.8J^{2}\le k 169.9 J < 178 J ≤ 17.8 J 2 ≤ k J ≤ 9 J\le9 J ≤ 9 169.9 J < 1530 < k 169.9J<1530<k 169.9 J < 1530 < k
Now let v = g u v=gu v = g u Λ \Lambda Λ n n n 2 k 2k 2 k g g g n n n Lemma 9.6 and J ( n ) ≤ J J(n)\le J J ( n ) ≤ J 17.8 J 2 ≤ k 17.8J^{2}\le k 17.8 J 2 ≤ k k > 169.9 J k>169.9J k > 169.9 J
Short vectors, l 2 ≤ 6400 l^{2}\le6400 l 2 ≤ 6400 Then ( g ; p , q ) ∈ V (g;p,q)\in\mathcal{V} ( g ; p , q ) ∈ V Lemma 9.4 (c) and Lemma 9.5 , and (52) gives 2 k ψ ≥ k g / 169.9 > J g ≥ J ( n ) g 2k\psi\ge kg/169.9>Jg\ge J(n)g 2 k ψ ≥ k g /169.9 > J g ≥ J ( n ) g Proposition 9.10 applies.
Long vectors, l > 80 l>80 l > 80 By Corollary A.5 (b), either J ( g ) ≤ 8 < 0.102 ⋅ 80 < 0.102 l J(g)\le8<0.102\cdot80<0.102l J ( g ) ≤ 8 < 0.102 ⋅ 80 < 0.102 l J ( g ) < 0.102 g ≤ 0.102 l J(g)<0.102g\le0.102l J ( g ) < 0.102 g ≤ 0.102 l Lemma 9.4 (b), l 2 ≤ 4 k / 3 l^{2}\le4k/\sqrt{3} l 2 ≤ 4 k / 3
J ( n ) 2 l 2 ≤ k 17.8 ⋅ 4 k 3 < 0.12975 k 2 , J(n)^{2}l^{2}\le\frac{k}{17.8}\cdot\frac{4k}{\sqrt{3}}<0.12975k^{2}, J ( n ) 2 l 2 ≤ 17.8 k ⋅ 3 4 k < 0.12975 k 2 , and J ( n ) l < 0.36021 k J(n)l<0.36021k J ( n ) l < 0.36021 k Proposition 9.8 applies.
Orders with at most two odd primes
When k k k Theorem 9.1 has a direct proof, with no lattice. It covers 1193 of the 1320 orders 92 ≤ k ≤ 2730 92\le k\le2730 92 ≤ k ≤ 2730 c c c ν \nu ν ( c A , c E ) (cA,cE) ( c A , c E ) ν c \nu c ν c ( A , E ) (A,E) ( A , E ) ( A , E ) (A,E) ( A , E ) ( c A , c E ) (cA,cE) ( c A , c E )
Lemma 9.13 (two normal forms). Let k k k
(a) If m m m 0 < m < k 0 < m < k 0 < m < k k + m < ⟨ m D ⟩ < 2 k − m k+m < \langle mD\rangle< 2k-m k + m < ⟨ m D ⟩ < 2 k − m m m m ( 1 , D ) (1,D) ( 1 , D )
(b) If ν \nu ν 0 < ν < k 0 < \nu< k 0 < ν < k ⟨ ν A ⟩ = r < k \langle\nu A\rangle= r < k ⟨ ν A ⟩ = r < k ⟨ 2 r − ν ⟩ ≥ k \langle2r-\nu\rangle\ge k ⟨ 2 r − ν ⟩ ≥ k ν \nu ν ( A , A − 1 ) (A,A-1) ( A , A − 1 )
Proof. (a) Write Y = ⟨ m D ⟩ Y=\langle mD\rangle Y = ⟨ m D ⟩ H ( m ) = 1 H(m)=1 H ( m ) = 1 m ( 1 + D ) ≡ m + Y m(1+D)\equiv m+Y m ( 1 + D ) ≡ m + Y k + 2 m < m + Y < 2 k k+2m<m+Y<2k k + 2 m < m + Y < 2 k H ( m ( 1 + D ) ) = 0 H(m(1+D))=0 H ( m ( 1 + D )) = 0 m ( 1 − D ) ≡ m − Y + 2 k m(1-D)\equiv m-Y+2k m ( 1 − D ) ≡ m − Y + 2 k 2 m < m − Y + 2 k < k 2m<m-Y+2k<k 2 m < m − Y + 2 k < k H ( m ( 1 − D ) ) = 1 H(m(1-D))=1 H ( m ( 1 − D )) = 1 H ( ν ) = 1 = H ( ν A ) H(\nu)=1=H(\nu A) H ( ν ) = 1 = H ( ν A ) ν ( 2 A − 1 ) ≡ 2 r − ν \nu(2A-1)\equiv2r-\nu ν ( 2 A − 1 ) ≡ 2 r − ν H ( ν ( 2 A − 1 ) ) = 0 H(\nu(2A-1))=0 H ( ν ( 2 A − 1 )) = 0
The multipliers for a unit middle term come in chains . A chain is a list ( m 1 , i 1 ) , … , ( m N , i N ) (m_1,i_1),\ldots,(m_N,i_N) ( m 1 , i 1 ) , … , ( m N , i N ) m j m_j m j i j ≥ 0 i_j\ge0 i j ≥ 0 K K K
2 i 1 + 1 m 1 + 1 K ≤ 1 8 , 2 i j + 1 + 1 m j + 1 + 1 K < 2 i j + 2 m j − 1 K , 2 i N + 2 m N − 1 K > 1 2 , (55) \frac{2i_1+1}{m_1}+\frac{1}{K}\le\frac{1}{8},\qquad
\frac{2i_{j+1}+1}{m_{j+1}}+\frac{1}{K}<\frac{2i_j+2}{m_j}-\frac{1}{K},\qquad
\frac{2i_N+2}{m_N}-\frac{1}{K}>\frac{1}{2}, \tag*{(55)} m 1 2 i 1 + 1 + K 1 ≤ 8 1 , m j + 1 2 i j + 1 + 1 + K 1 < m j 2 i j + 2 − K 1 , m N 2 i N + 2 − K 1 > 2 1 , (55) the middle condition for j = 1 , … , N − 1 j=1,\ldots,N-1 j = 1 , … , N − 1 ( m j , i j ) (m_j,i_j) ( m j , i j ) D / k D/k D / k ( ( 2 i j + 1 ) / m j , ( 2 i j + 2 ) / m j ) ((2i_j+1)/m_j,(2i_j+2)/m_j) (( 2 i j + 1 ) / m j , ( 2 i j + 2 ) / m j ) m j D m_jD m j D ( ( 2 i j + 1 ) k , ( 2 i j + 2 ) k ) ((2i_j+1)k,(2i_j+2)k) (( 2 i j + 1 ) k , ( 2 i j + 2 ) k ) 1 / K 1/K 1/ K [ 1 / 8 , 1 / 2 ] [1/8,1/2] [ 1/8 , 1/2 ] Table 5 lists ten chains, one for each pattern of small primes dividing k k k
Table 5. The chains used in the proof of Proposition 9.14 , for even k ≥ 92 k \ge92 k ≥ 92 m i j m_i^j m i j ( m , i ) (m,i) ( m , i ) K K K k ≥ K k \ge K k ≥ K
Proposition 9.14 (two odd primes). Let k ≥ 92 k\ge92 k ≥ 92 ( k ; a , x ) (k;a,x) ( k ; a , x )
Proof. The units modulo 2 k 2k 2 k k k k k k k e = x − a e=x-a e = x − a k k k a − e a-e a − e a a a a + e a+e a + e a a a e e e
The case where a a a Let a ˉ \bar a a ˉ 2 k 2k 2 k D = ⟨ a e ˉ ⟩ D=\langle a e\bar{\ }\rangle D = ⟨ a e ˉ ⟩ ( 1 , D ) (1,D) ( 1 , D ) a D ≡ e aD\equiv e a D ≡ e D = 0 , 2 , k − 2 , k , k + 2 , 2 k − 2 D=0,2,k-2,k,k+2,2k-2 D = 0 , 2 , k − 2 , k , k + 2 , 2 k − 2 e ≡ 0 , 2 a , k − 2 a , k , k + 2 a , − 2 a ( m o d 2 k ) e\equiv0,2a,k-2a,k,k+2a,-2a\pmod{2k} e ≡ 0 , 2 a , k − 2 a , k , k + 2 a , − 2 a ( mod 2 k ) 0 < e 0<e 0 < e e ≠ 2 a e\ne2a e = 2 a e + 2 a < k e+2a<k e + 2 a < k Lemma 9.13 (a) it suffices to find a unit m m m 0 < m < k 0<m<k 0 < m < k
( 2 i + 1 ) k + m < m D < ( 2 i + 2 ) k − m for some integer i . (56) (2i+1)k+m<mD<(2i+2)k-m\qquad\text{for some integer }i. \tag*{(56)} ( 2 i + 1 ) k + m < m D < ( 2 i + 2 ) k − m for some integer i . (56) If k < D < 2 k k<D<2k k < D < 2 k m = 1 m=1 m = 1 i = 0 i=0 i = 0 m m m m m m k − D k-D k − D i i i D D D i i i ( m − 3 ) / 2 − i (m-3)/2-i ( m − 3 ) /2 − i m D = m k − m ( k − D ) mD=mk-m(k-D) m D = mk − m ( k − D ) 8 D ≤ k 8D \le k 8 D ≤ k m m m I = ( k / ( D − 1 ) , 2 k / ( D + 1 ) ) I = (k/(D - 1), 2k/(D + 1)) I = ( k / ( D − 1 ) , 2 k / ( D + 1 )) m ( D − 1 ) > k m(D - 1) > k m ( D − 1 ) > k m ( D + 1 ) < 2 k m(D + 1) < 2k m ( D + 1 ) < 2 k (56) with i = 0 i = 0 i = 0 m < 2 k / 5 m < 2k/5 m < 2 k /5 I I I k ( D − 3 ) / ( D 2 − 1 ) k(D - 3)/(D^2 - 1) k ( D − 3 ) / ( D 2 − 1 ) D ≤ 11 D \le11 D ≤ 11 k ≥ 92 k \ge92 k ≥ 92 D ≥ 12 D \ge12 D ≥ 12 k ≥ 8 D k \ge8D k ≥ 8 D 8 D ( D − 3 ) − 6 ( D 2 − 1 ) = 2 D ( D − 12 ) + 6 > 0 8D(D - 3) - 6(D^2 - 1) = 2D(D - 12) + 6 > 0 8 D ( D − 3 ) − 6 ( D 2 − 1 ) = 2 D ( D − 12 ) + 6 > 0 I I I
If 8 D > k 8D > k 8 D > k Table 5 for the case of k k k k k k (56) , as m 1 D > m 1 k / 8 ≥ ( 2 i 1 + 1 ) k + m 1 k / K ≥ ( 2 i 1 + 1 ) k + m 1 m_1D > m_1k/8 \ge(2i_1 + 1)k + m_1k/K \ge(2i_1 + 1)k + m_1 m 1 D > m 1 k /8 ≥ ( 2 i 1 + 1 ) k + m 1 k / K ≥ ( 2 i 1 + 1 ) k + m 1 k ≥ K k \ge K k ≥ K m j m_j m j D / k ≥ ( 2 i j + 2 ) / m j − 1 / k D/k \ge(2i_j + 2)/m_j - 1/k D / k ≥ ( 2 i j + 2 ) / m j − 1/ k (55) gives the left inequality for m j + 1 m_{j+1} m j + 1 m N m_N m N D / k ≤ 1 / 2 D/k \le1/2 D / k ≤ 1/2 m j m_j m j (56) .
The case where a − e a - e a − e Let y = 2 a − x = a − e y = 2a - x = a - e y = 2 a − x = a − e y ˉ \bar{y} y ˉ A = ⟨ y a ˉ ⟩ A = \langle y\bar{a}\rangle A = ⟨ y a ˉ ⟩ y ˉ e ≡ A − 1 \bar{y}e \equiv A^{-1} y ˉ e ≡ A − 1 ( A , A − 1 ) (A, A^{-1}) ( A , A − 1 ) a a a f = gcd ( A , 2 k ) = gcd ( a , k ) f = \gcd(A, 2k) = \gcd(a, k) f = g cd( A , 2 k ) = g cd( a , k ) 1 < f ≤ a < k / 2 1 < f \le a < k/2 1 < f ≤ a < k /2 k / f k/f k / f A = f A 0 A = fA_0 A = f A 0 M = 2 k / f M = 2k/f M = 2 k / f A 0 A_0 A 0 M M M ν 0 ∈ ( 0 , M ) \nu_0 \in(0, M) ν 0 ∈ ( 0 , M ) M M M ν ≡ c ν 0 ( m o d M ) \nu\equiv c\nu_0 \pmod{M} ν ≡ c ν 0 ( mod M ) ν A ≡ f c ( m o d 2 k ) \nu A \equiv fc \pmod{2k} ν A ≡ f c ( mod 2 k ) lifts ν ≡ ν 0 ( m o d M ) \nu\equiv\nu_0 \pmod{M} ν ≡ ν 0 ( mod M ) M M M r = f r = f r = f ( 2 f , k ) (2f, k) ( 2 f , k ) ⟨ 2 f − ν ⟩ = 2 k + 2 f − ν ≥ k \langle2f - \nu\rangle= 2k + 2f - \nu\ge k ⟨ 2 f − ν ⟩ = 2 k + 2 f − ν ≥ k k k k M M M f f f
If f ≥ 15 f \ge15 f ≥ 15 2 f + 3 M ≤ k 2f + 3M \le k 2 f + 3 M ≤ k ( 2 f , k ) (2f, k) ( 2 f , k ) ν \nu ν ν + M \nu+ M ν + M ν + 2 M \nu+ 2M ν + 2 M 2 M 2M 2 M ν \nu ν 5 ≤ f ≤ 13 5 \le f \le13 5 ≤ f ≤ 13 f f f ℓ \ell ℓ 2 f + 2 M ≤ k 2f + 2M \le k 2 f + 2 M ≤ k k ≥ 92 k \ge92 k ≥ 92 f ∈ { 5 , 7 } f \in\{5, 7\} f ∈ { 5 , 7 } ( 2 f , k ) (2f, k) ( 2 f , k ) ν \nu ν ν + M \nu+ M ν + M ℓ \ell ℓ ℓ \ell ℓ ν \nu ν M M M f = 3 f = 3 f = 3 9 ∣ k 9 \mid k 9 ∣ k 3 ∣ M 3 \mid M 3 ∣ M 2 f + M ≤ k 2f + M \le k 2 f + M ≤ k ( 6 , k ) (6, k) ( 6 , k )
There remains f = 3 f = 3 f = 3 3 ∤ M 3 \nmid M 3 ∤ M k = 3 q k = 3q k = 3 q q q q 3 ∤ q 3 \nmid q 3 ∤ q q ≥ 32 q \ge32 q ≥ 32 M = 2 q M = 2q M = 2 q M M M ν 0 \nu_0 ν 0 6 < ν 0 < q 6 < \nu_0 < q 6 < ν 0 < q ν 0 \nu_0 ν 0 ν 0 + M \nu_0 + M ν 0 + M ( 6 , k ) (6, k) ( 6 , k ) ν 0 > q \nu_0 > q ν 0 > q 3 ∤ ν 0 3 \nmid\nu_0 3 ∤ ν 0 ν = ν 0 \nu= \nu_0 ν = ν 0 ν 0 > q \nu_0 > q ν 0 > q 3 ∣ ν 0 3 \mid\nu_0 3 ∣ ν 0 c = q − 3 c = q - 3 c = q − 3 ν \nu ν c ν 0 c\nu_0 c ν 0 M M M ν A ≡ 3 c ≡ k − 9 \nu A \equiv3c \equiv k - 9 ν A ≡ 3 c ≡ k − 9 ν \nu ν M M M q − 3 q - 3 q − 3 q q q ν 0 \nu_0 ν 0 q ν 0 ≡ q ( m o d M ) q\nu_0 \equiv q \pmod{M} q ν 0 ≡ q ( mod M ) ν ≡ q − 3 ν 0 \nu\equiv q - 3\nu_0 ν ≡ q − 3 ν 0 q < ν 0 < 2 q q < \nu_0 < 2q q < ν 0 < 2 q ν \nu ν 5 q − 3 ν 0 5q - 3\nu_0 5 q − 3 ν 0 7 q − 3 ν 0 7q - 3\nu_0 7 q − 3 ν 0 ⟨ 2 ( k − 9 ) − ν ⟩ = 2 k − 18 − ν ≥ k \langle2(k - 9) - \nu\rangle= 2k - 18 - \nu\ge k ⟨ 2 ( k − 9 ) − ν ⟩ = 2 k − 18 − ν ≥ k ν < M ≤ k − 18 \nu< M \le k - 18 ν < M ≤ k − 18 ν 0 < 6 \nu_0 < 6 ν 0 < 6 ν 0 + M \nu_0 + M ν 0 + M ( 6 , k ) (6, k) ( 6 , k ) ν 0 ∈ { 1 , 5 } \nu_0 \in\{1,5\} ν 0 ∈ { 1 , 5 } q ≡ ν 0 ( m o d 3 ) q \equiv\nu_0 \pmod3 q ≡ ν 0 ( mod 3 ) s = 5 s = 5 s = 5 5 ∤ k 5 \nmid k 5 ∤ k 5 ∣ k 5 \mid k 5 ∣ k 5 ∣ M 5 \mid M 5 ∣ M ν 0 = 1 \nu_0 = 1 ν 0 = 1 s = 11 s = 11 s = 11 k k k c = q − s c = q - s c = q − s ν = q − s ν 0 ≡ c ν 0 ( m o d M ) \nu= q - s\nu_0 \equiv c\nu_0 \pmod M ν = q − s ν 0 ≡ c ν 0 ( mod M ) ν A ≡ 3 c ≡ k − 3 s \nu A \equiv3c \equiv k - 3s ν A ≡ 3 c ≡ k − 3 s ν \nu ν q q q s ν 0 s\nu_0 s ν 0 ν ≡ ν 0 ( 1 − s ) ≠ 0 ( m o d 3 ) \nu\equiv\nu_0(1 - s) \ne0 \pmod3 ν ≡ ν 0 ( 1 − s ) = 0 ( mod 3 ) 2 k − 6 s − ν ≥ k 2k - 6s - \nu\ge k 2 k − 6 s − ν ≥ k 2 q ≥ 64 ≥ s ( 6 − ν 0 ) 2q \ge64 \ge s(6 - \nu_0) 2 q ≥ 64 ≥ s ( 6 − ν 0 )
The case where a + e a + e a + e Suppose finally that neither a a a a − e a - e a − e y ′ = a + e = x y' = a + e = x y ′ = a + e = x y ˉ ′ \bar{y}' y ˉ ′ A ′ = ⟨ y ˉ ′ a ⟩ A' = \langle\bar{y}'a\rangle A ′ = ⟨ y ˉ ′ a ⟩ y ˉ ′ e ≡ 1 − A ′ \bar{y}'e \equiv1 - A' y ˉ ′ e ≡ 1 − A ′ ( A ′ , 1 − A ′ ) (A', 1 - A') ( A ′ , 1 − A ′ ) 2 A ′ − 1 , A ′ , 1 2A' - 1, A', 1 2 A ′ − 1 , A ′ , 1 A = A ′ + k A = A' + k A = A ′ + k ν \nu ν H ( ν A ) = 1 − H ( ν A ′ ) H(\nu A) = 1 - H(\nu A') H ( ν A ) = 1 − H ( ν A ′ ) ν ( 2 A − 1 ) ≡ ν ( 2 A ′ − 1 ) \nu(2A - 1) \equiv\nu(2A' - 1) ν ( 2 A − 1 ) ≡ ν ( 2 A ′ − 1 ) ν \nu ν (49) for ( A , A − 1 ) (A, A - 1) ( A , A − 1 ) ( A ′ , 1 − A ′ ) (A', 1 - A') ( A ′ , 1 − A ′ ) A ′ A' A ′ k k k gcd ( A , 2 k ) = gcd ( A ′ , 2 k ) = gcd ( a , k ) \gcd(A, 2k) = \gcd(A', 2k) = \gcd(a, k) g cd( A , 2 k ) = g cd( A ′ , 2 k ) = g cd( a , k ) A A A
Proof of Theorem 9.1. If k k k k ≥ 2730 k \ge2730 k ≥ 2730 k ≤ 2730 k \le2730 k ≤ 2730 k k k § 12 .
For 92 ≤ k ≤ 2730 92 \le k \le2730 92 ≤ k ≤ 2730 ν \nu ν 2 k 2k 2 k 100 435 222 100\,435\,222 100 435 222 ( k ; a , x ) (k; a, x) ( k ; a , x )
Proof of the main theorem
By Corollary 2.9 it is enough to show the following: if an equation (8) is globally periodic, its support S S S c j = c k − j c_j=c_{k-j} c j = c k − j Definition 2.6 ). So let such an equation be given. Since c j = c k − j c_j=c_{k-j} c j = c k − j
∣ S ∣ ≤ 2 |S| \le2 ∣ S ∣ ≤ 2 § 4 , either S = ∅ S=\varnothing S = ∅ k = 2 k=2 k = 2 S = { 1 } S=\{1\} S = { 1 } Theorem 4.1 gives case (K2); or k = 3 k=3 k = 3 S = { 1 , 2 } S=\{1,2\} S = { 1 , 2 } Theorem 4.2 gives case (K3); or Theorem 4.3 shows that the equation is not globally periodic, a contradiction.
∣ S ∣ ≥ 3 |S| \ge3 ∣ S ∣ ≥ 3 S S S { 1 , … , k − 1 } \{1,\ldots,k-1\} { 1 , … , k − 1 } k ≥ 4 k\ge4 k ≥ 4 Theorem 5.1 applies. So suppose S S S k k k Theorem 6.1 applies. If k = 6 k=6 k = 6 S = { 1 , 3 , 5 } S=\{1,3,5\} S = { 1 , 3 , 5 } Proposition 3.10 applies. Otherwise k k k ( k , S ) ≠ ( 6 , { 1 , 3 , 5 } ) (k,S)\ne(6,\{1,3,5\}) ( k , S ) = ( 6 , { 1 , 3 , 5 }) S S S σ \sigma σ Theorem 7.1 if S S S Theorem 8.1 if every element of S S S Theorem 3.14 shows that no equation with support S S S
So every globally periodic equation (8) with primitive support and palindromic coefficients has normalised data of known type. By Corollary 2.9, a globally periodic equation (3) is therefore the shift or has the coefficients listed in Lemma 2.7 , and these are exactly the equations listed in Theorem 1.1 . This completes the proof of Theorem 1.1 . □
The argument for supports with three or more elements never uses the coefficients, and it therefore also proves a statement about the min-plus recursion alone, extending the result of Csörnyei and Laczkovich [ [10 ] for the full support (§ 5 ).
Corollary 10.1 (the min-plus map is not periodic). Let S ⊆ { 1 , … , k − 1 } S\subseteq\{1,\ldots,k-1\} S ⊆ { 1 , … , k − 1 } ∣ S ∣ ≥ 3 |S|\ge3 ∣ S ∣ ≥ 3 θ ∈ { 0 , 1 } \theta\in\{0,1\} θ ∈ { 0 , 1 } T T T p ≥ 1 p\ge1 p ≥ 1 w ∈ Z k w\in\mathbb{Z}^k w ∈ Z k T p ( w ) ≠ w T^p(w)\ne w T p ( w ) = w
Proof. Suppose first that S S S Theorem 1.1 above provides an integer solution e e e (15) for ( k , S , θ ) (k,S,\theta) ( k , S , θ ) d d d Proposition 6.4 , with the solutions of the reduced recursion given in the proof of Theorem 6.11 , or by a σ \sigma σ { 1 , 3 , 5 } \{1,3,5\} { 1 , 3 , 5 } 6 6 6 σ \sigma σ Lemma 3.13 . None of these constructions involves the coefficients c j c_j c j Theorem 3.6 (ii) applies.
If S S S ℓ = gcd ( S ∪ { k } ) \ell=\gcd(S\cup\{k\}) ℓ = g cd( S ∪ { k }) S ′ S' S ′ k ′ k' k ′ Lemma 2.8 ; then S ′ S' S ′ ∣ S ′ ∣ = ∣ S ∣ ≥ 3 |S'|=|S|\ge3 ∣ S ′ ∣ = ∣ S ∣ ≥ 3 Lemma 2.8 (ii), if z z z ( k ′ , S ′ , θ ) (k',S',\theta) ( k ′ , S ′ , θ ) x n = z ⌊ n / ℓ ⌋ x_n=z_{\lfloor n/\ell\rfloor} x n = z ⌊ n / ℓ ⌋ ( k , S , θ ) (k,S,\theta) ( k , S , θ ) T p = id T^p=\operatorname{id} T p = id x x x p p p z m + p = x m ℓ + p ℓ = x m ℓ = z m z_{m+p}=x_{m\ell+p\ell}=x_{m\ell}=z_m z m + p = x m ℓ + p ℓ = x m ℓ = z m ( k ′ , S ′ , θ ) (k',S',\theta) ( k ′ , S ′ , θ ) p p p
Since Z k ⊂ R k \mathbb{Z}^k\subset\mathbb{R}^k Z k ⊂ R k R k \mathbb{R}^k R k T T T w w w p p p ∣ S ∣ ≥ 3 |S|\ge3 ∣ S ∣ ≥ 3 θ = 1 \theta=1 θ = 1 S = { 1 } S=\{1\} S = { 1 } 2 2 2 S = { 1 , 2 } S=\{1,2\} S = { 1 , 2 } 3 3 3 5 5 5 8 8 8 § 4 ).
The corollary leaves the two-term supports { j , k − j } \{j,k-j\} { j , k − j } Theorem 1.1 settles them with the linearisation. For k = 4 k=4 k = 4 S = { 1 , 3 } S=\{1,3\} S = { 1 , 3 } E = ( − 1 , − 1 , − 1 , − 1 ) E=(-1,-1,-1,-1) E = ( − 1 , − 1 , − 1 , − 1 )
θ = 1 : q = 11 , V = ( − 1 , − 1 , 0 , − 1 ) , W = ( 0 , 0 , 1 , 1 ) , ( w t ) = 1 , 1 , 1 , 3 , 0 , 1 , 3 , 3 , 3 , 3 , 1 , θ = 0 : q = 66 , V = ( − 1 , − 1 , 0 , 0 ) , W = ( 4 , 0 , 0 , 0 ) , ( w t ) = 1 6 3 5 1 6 3 5 1 6 3 6 1 5 3 6 1 5 3 6 1 5 3 5 , \begin{aligned}
\theta=1 &: \quad q=11,\ V=(-1,-1,0,-1),\ W=(0,0,1,1),\ (w_t)=1,1,1,3,0,1,3,3,3,3,1,\\
\theta=0 &: \quad q=66,\ V=(-1,-1,0,0),\ W=(4,0,0,0),\ (w_t)=1^{6}3^{5}1^{6}3^{5}1^{6}3^{6}1^{5}3^{6}1^{5}3^{6}1^{5}3^{5},
\end{aligned} θ = 1 θ = 0 : q = 11 , V = ( − 1 , − 1 , 0 , − 1 ) , W = ( 0 , 0 , 1 , 1 ) , ( w t ) = 1 , 1 , 1 , 3 , 0 , 1 , 3 , 3 , 3 , 3 , 1 , : q = 66 , V = ( − 1 , − 1 , 0 , 0 ) , W = ( 4 , 0 , 0 , 0 ) , ( w t ) = 1 6 3 5 1 6 3 5 1 6 3 6 1 5 3 6 1 5 3 6 1 5 3 5 , (1 6 1^6 1 6 E E E ( 4 , { 1 , 3 } , θ ) (4,\{1,3\},\theta) ( 4 , { 1 , 3 } , θ ) θ \theta θ { m , 2 m } \{m,2m\} { m , 2 m } 3 m 3m 3 m θ = 1 \theta=1 θ = 1
Acknowledgements
This proof was AI-generated with fairly limited input from the human author. The models used were Claude Opus 5.5 and Fable 5.1.
The finite range of the progression theorem
This appendix completes the proof of Theorem 9.1 . It gives the certificate for the 127 orders 92 ≤ k ≤ 2730 92 \le k \le2730 92 ≤ k ≤ 2730 Proposition 9.14 or Proposition 9.12 , and proves the two bounds on Jacobsthal’s function used in § 9.6 for the orders above 2730. The notation is that of § 9 .
The orders up to 2730 with three or more odd primes
By Lemma 9.4 (b), (c) and Lemma 9.5 , the normalised shortest vector of Λ \Lambda Λ k ≤ 2730 k \le2730 k ≤ 2730 V \mathcal{V} V g ∣ 2 k g \mid2k g ∣ 2 k 3 l 4 ≤ 16 k 2 3l^4 \le16k^2 3 l 4 ≤ 16 k 2 l 4 ≤ 16 ⋅ 2730 2 / 3 l^4 \le16 \cdot2730^2/3 l 4 ≤ 16 ⋅ 273 0 2 /3 l 2 ≤ 6304 < 6400 l^2 \le6304 < 6400 l 2 ≤ 6304 < 6400 l ≤ 80 l \le80 l ≤ 80 § 9.6 , the finite certificate has two more parts.
(C2) Jacobsthal values. A list of 224 pairs ( n , m ) (n,m) ( n , m ) J ( n ) ≤ m J(n) \le m J ( n ) ≤ m y y y n n n 0 ≤ i < m 0 \le i < m 0 ≤ i < m gcd ( y + i , n ) = 1 \gcd(y+i,n)=1 g cd( y + i , n ) = 1
(C3) The per-order certificate. For each of the 127 even k k k 92 ≤ k ≤ 2730 92 \le k \le2730 92 ≤ k ≤ 2730 ( n , m ) (n,m) ( n , m ) n n n 2 k 2k 2 k k k k (54) with bound m m m tail alternative ), or, for each ( g ; p , q ) ∈ V (g;p,q) \in\mathcal{V} ( g ; p , q ) ∈ V g ∣ 2 k g \mid2k g ∣ 2 k 3 ( g 2 ( p 2 + q 2 ) ) 2 ≤ 16 k 2 3\left(g^2(p^2+q^2)\right)^2 \le16k^2 3 ( g 2 ( p 2 + q 2 ) ) 2 ≤ 16 k 2
(i) a pair ( n , m ) (n,m) ( n , m ) n ∣ 2 k n \mid2k n ∣ 2 k gcd ( n , g ) = 1 \gcd(n,g)=1 g cd( n , g ) = 1 2 k 2k 2 k n g ng n g 2 k ψ > m g 2k\psi> mg 2 k ψ > m g ψ \psi ψ
(ii) for each of the g 2 g^2 g 2 0 ≤ s , t < g 0 \le s,t < g 0 ≤ s , t < g ( A , E ) = ( p + 2 k g s , q + 2 k g t ) (A,E)=\left(p+\frac{2k}{g}s,q+\frac{2k}{g}t\right) ( A , E ) = ( p + g 2 k s , q + g 2 k t ) ν < 100 \nu< 100 ν < 100 (49) , or gcd ( A , E , 2 k ) > 1 \gcd(A,E,2k)>1 g cd( A , E , 2 k ) > 1 m m m 0 < m < g 0<m<g 0 < m < g 2 k ∣ m ( p E − q A ) 2k \mid m(pE-qA) 2 k ∣ m ( pE − q A )
The tail alternative of (C3) holds at 65 of these 127 orders, and alternative (ii) is needed only at 18 orders, all at most 924.
Proposition A.1. Theorem 9.1 holds at the 127 even orders 92 ≤ k ≤ 2730 92 \le k \le2730 92 ≤ k ≤ 2730
Proof. If the tail alternative of (C3) holds, then J ( 2 k ) = J ( n ) ≤ m J(2k)=J(n) \le m J ( 2 k ) = J ( n ) ≤ m k k k m m m J ( 2 k ) J(2k) J ( 2 k ) Proposition 9.12 applies. Otherwise, let ν = q v \nu=qv ν = q v Λ \Lambda Λ ( d ; p , q ) ∈ V (d;p,q) \in\mathcal{V} ( d ; p , q ) ∈ V g ∣ 2 k g \mid2k g ∣ 2 k 3 l 4 ≤ 16 k 2 3l^4 \le16k^2 3 l 4 ≤ 16 k 2 J ( n ) ≤ m J(n) \le m J ( n ) ≤ m 2 k ψ > J ( n ) g 2k\psi> J(n)g 2 k ψ > J ( n ) g Proposition 9.10 shows that ( a , e ) (a,e) ( a , e ) n n n Lemma 9.6 . In case (ii), let ( s , t ) (s,t) ( s , t ) Lemma 9.11 . Its two exclusions are ruled out by that lemma, so the certificate gives a unit ν \nu ν ( A , E ) (A,E) ( A , E ) ( a , e ) (a,e) ( a , e ) Lemma 9.11 . □
Elementary bounds on Jacobsthal's function
This subsection proves Corollary A.5 , the two inequalities for Jacobsthal's function used in Proposition 9.12 , by a union bound and a progression trick.
Lemma A.2. (a) Suppose every m m m c c c N N N T \mathcal{T} T ∑ p ∈ T ⌈ m / p ⌉ < c \sum_{p\in\mathcal{T}} \lceil m/p\rceil< c ∑ p ∈ T ⌈ m / p ⌉ < c J ( n ) ≤ m J(n) \le m J ( n ) ≤ m n ≥ 1 n \ge1 n ≥ 1 N N N T \mathcal{T} T
(b) If gcd ( f , N ) = 1 \gcd(f,N)=1 g cd( f , N ) = 1 J ( f N ) ≤ f J ( N ) J(fN) \le fJ(N) J ( f N ) ≤ f J ( N )
Proof. (a) Among m m m ⌈ m / p ⌉ \lceil m/p\rceil ⌈ m / p ⌉ p p p c c c T \mathcal{T} T c c c N N N n n n
(b) Given a run y , y + 1 , … , y + f J ( N ) − 1 y,y+1,\ldots,y+fJ(N)-1 y , y + 1 , … , y + f J ( N ) − 1 0 ≤ r < f 0\le r<f 0 ≤ r < f y + r ≡ 1 ( m o d f ) y+r\equiv1\pmod f y + r ≡ 1 ( mod f ) y i = y + r + f i y_i=y+r+fi y i = y + r + f i 0 ≤ i < J ( N ) 0\le i<J(N) 0 ≤ i < J ( N ) y i y_i y i f f f f ′ f' f ′ f f f N N N f ′ y i ≡ f ′ ( y + r ) + i ( m o d N ) f'y_i\equiv f'(y+r)+i\pmod N f ′ y i ≡ f ′ ( y + r ) + i ( mod N ) J ( N ) J(N) J ( N ) y i y_i y i N N N f N fN f N
Let ϖ 1 < ϖ 2 < ⋯ < ϖ 8 \varpi_1<\varpi_2<\cdots<\varpi_8 ϖ 1 < ϖ 2 < ⋯ < ϖ 8 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 17,19,23,29,31,37,41,43 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 Π r = ϖ 1 ⋯ ϖ r \Pi_r=\varpi_1\cdots\varpi_r Π r = ϖ 1 ⋯ ϖ r Π 0 = 1 \Pi_0=1 Π 0 = 1 ω ( n ) \omega(n) ω ( n ) n n n
Lemma A.3 (at most eight prime factors). Let n ≥ 1 n\ge1 n ≥ 1 ω ( n ) ≤ 8 \omega(n)\le8 ω ( n ) ≤ 8 B B B p ≤ 13 p\le13 p ≤ 13 n n n r r r p > 13 p>13 p > 13 n n n B Π r ≤ n B\Pi_r\le n B Π r ≤ n m m m J ( n ) ≤ m J(n)\le m J ( n ) ≤ m
(i) if B B B 178 m 2 ≤ 10 max ( B Π r , B ( ⌊ 2730 / B ⌋ + 1 ) ) 178m^2\le10\max(B\Pi_r,B(\lfloor2730/B\rfloor+1)) 178 m 2 ≤ 10 max ( B Π r , B (⌊ 2730/ B ⌋ + 1 ))
(ii) m ≤ 8 m\le8 m ≤ 8 1000 m < 102 B Π r 1000m<102B\Pi_r 1000 m < 102 B Π r
The maximum in (i) is the better of two lower bounds for an order k k k k ≥ B Π r k\ge B\Pi_r k ≥ B Π r k k k B B B 2730 2730 2730 k ≥ B ( ⌊ 2730 / B ⌋ + 1 ) k\ge B(\lfloor2730/B\rfloor+1) k ≥ B (⌊ 2730/ B ⌋ + 1 ) J ( g ) ≤ 8 J(g)\le8 J ( g ) ≤ 8 J ( g ) < 0.102 g J(g)<0.102g J ( g ) < 0.102 g
Proof. The i i i ϖ i \varpi_i ϖ i p 1 < ⋯ < p r p_1<\cdots<p_r p 1 < ⋯ < p r n n n p i ≥ ϖ i p_i\ge\varpi_i p i ≥ ϖ i n ≥ B p 1 ⋯ p r ≥ B Π r n\ge Bp_1\cdots p_r\ge B\Pi_r n ≥ B p 1 ⋯ p r ≥ B Π r ∑ i ⌈ m / p i ⌉ ≤ ∑ i ⌈ m / ϖ i ⌉ \sum_i\lceil m/p_i\rceil\le\sum_i\lceil m/\varpi_i\rceil ∑ i ⌈ m / p i ⌉ ≤ ∑ i ⌈ m / ϖ i ⌉ m m m
The rest is a finite table. For each of the 64 sets P \mathcal{P} P ≤ 13 \le13 ≤ 13 B B B r r r 0 ≤ r ≤ 8 − ∣ P ∣ 0\le r\le8-|\mathcal{P}| 0 ≤ r ≤ 8 − ∣ P ∣ ( B , r ) (B,r) ( B , r ) m m m c c c
(P1) every m m m c c c B B B
(P2) ∑ i = 1 r ⌈ m / ϖ i ⌉ < c \sum_{i=1}^{r}\lceil m/\varpi_i\rceil<c ∑ i = 1 r ⌈ m / ϖ i ⌉ < c
(P3) (i) and (ii) hold for this m m m
Condition (P1) is checked by sliding a window of length m m m { 0 , … , B − 1 } \{0,\ldots,B-1\} { 0 , … , B − 1 } m m m Lemma A.2 (a), with N = B N=B N = B T \mathcal{T} T n n n J ( n ) ≤ m J(n)\le m J ( n ) ≤ m
Lemma A.4 (at least nine prime factors). Let n n n ω = ω ( n ) ≥ 9 \omega=\omega(n)\ge9 ω = ω ( n ) ≥ 9 B B B p ≤ ω p\le\omega p ≤ ω n n n R R R r r r n n n B R ≤ n BR\le n B R ≤ n J ( n ) ≤ B ( r + 1 ) J(n)\le B(r+1) J ( n ) ≤ B ( r + 1 )
178 B ( r + 1 ) 2 ≤ 10 R , 1000 ( r + 1 ) < 102 R . 178B(r+1)^2\le10R,\qquad1000(r+1)<102R. 178 B ( r + 1 ) 2 ≤ 10 R , 1000 ( r + 1 ) < 102 R . Proof. Every prime of R R R ω ≥ r \omega\ge r ω ≥ r r + 1 r+1 r + 1 Lemma A.2 (a) (with N = 1 N=1 N = 1 c = m = r + 1 c=m=r+1 c = m = r + 1 J ( R ) ≤ r + 1 J(R) \le r+1 J ( R ) ≤ r + 1 gcd ( B , R ) = 1 \gcd(B,R)=1 g cd( B , R ) = 1 Lemma A.2 (b) gives J ( B R ) ≤ B ( r + 1 ) J(BR) \le B(r+1) J ( B R ) ≤ B ( r + 1 ) J ( n ) = J ( B R ) J(n)=J(BR) J ( n ) = J ( B R )
For the inequalities, the r r r R R R ω \omega ω R ≥ ( ω + 1 ) ( ω + 2 ) ⋯ ( ω + r ) R \ge(\omega+1)(\omega+2)\cdots(\omega+r) R ≥ ( ω + 1 ) ( ω + 2 ) ⋯ ( ω + r ) ω − r \omega-r ω − r ω \omega ω B B B ω # = ∏ p ≤ ω p \omega\#=\prod_{p\le\omega}p ω # = ∏ p ≤ ω p r ≥ ω − π ( ω ) r\ge\omega-\pi(\omega) r ≥ ω − π ( ω ) ω ≤ 63 \omega\le63 ω ≤ 63 178 ( ω # ) ( r + 1 ) 2 ≤ 10 ∏ i = 1 r ( ω + i ) 178(\omega\#)(r+1)^2\le10\prod_{i=1}^{r}(\omega+i) 178 ( ω # ) ( r + 1 ) 2 ≤ 10 ∏ i = 1 r ( ω + i ) 1000 ( r + 1 ) < 10 2 ∏ i = 1 r ( ω + i ) 1000(r+1)<10^2\prod_{i=1}^{r}(\omega+i) 1000 ( r + 1 ) < 1 0 2 ∏ i = 1 r ( ω + i ) 9 ≤ ω ≤ 63 9\le\omega\le63 9 ≤ ω ≤ 63 ω − π ( ω ) ≤ r ≤ ω \omega-\pi(\omega)\le r\le\omega ω − π ( ω ) ≤ r ≤ ω ω ≥ 64 \omega\ge64 ω ≥ 64 ω # ≤ 4 ω \omega\#\le4^\omega ω # ≤ 4 ω π ( ω ) ≤ ω / 2 + 1 \pi(\omega)\le\omega/2+1 π ( ω ) ≤ ω /2 + 1 ω \omega ω [ 3 , ω ] [3,\omega] [ 3 , ω ] ω # \omega\# ω # ω ≤ 2 \omega\le2 ω ≤ 2 ω > 2 \omega>2 ω > 2 ω # = ( ω − 1 ) # \omega\#=(\omega-1)\# ω # = ( ω − 1 ) # ω = 2 m + 1 \omega=2m+1 ω = 2 m + 1 ( m + 1 , 2 m + 1 ] (m+1,2m+1] ( m + 1 , 2 m + 1 ] ( 2 m + 1 m ) \binom{2m+1}{m} ( m 2 m + 1 ) 4 m 4^m 4 m ( 2 m + 1 m + 1 ) \binom{2m+1}{m+1} ( m + 1 2 m + 1 ) 2 2 m + 1 2^{2m+1} 2 2 m + 1 ( 2 m + 1 ) # ≤ ( m + 1 ) # 4 m ≤ 4 m + 1 4 m (2m+1)\#\le(m+1)\#4^m\le4^{m+1}4^m ( 2 m + 1 ) # ≤ ( m + 1 ) # 4 m ≤ 4 m + 1 4 m r ≥ ω / 2 − 1 ≥ 31 r\ge\omega/2-1\ge31 r ≥ ω /2 − 1 ≥ 31 ω ≤ 2 r + 2 \omega\le2r+2 ω ≤ 2 r + 2 ∏ i = 1 r ( ω + i ) ≥ 64 r \prod_{i=1}^{r}(\omega+i)\ge64^r ∏ i = 1 r ( ω + i ) ≥ 6 4 r B ≤ 4 ω ≤ 16 ⋅ 16 r B\le4^\omega\le16\cdot16^r B ≤ 4 ω ≤ 16 ⋅ 1 6 r 178 B ( r + 1 ) 2 ≤ 16 r ⋅ 2848 ( r + 1 ) 2 ≤ 16 r ⋅ 10 ⋅ 4 r = 10 ⋅ 64 r 178B(r+1)^2\le16^r\cdot2848(r+1)^2\le16^r\cdot10\cdot4^r=10\cdot64^r 178 B ( r + 1 ) 2 ≤ 1 6 r ⋅ 2848 ( r + 1 ) 2 ≤ 1 6 r ⋅ 10 ⋅ 4 r = 10 ⋅ 6 4 r 2848 ( r + 1 ) 2 ≤ 10 ⋅ 4 r 2848(r+1)^2\le10\cdot4^r 2848 ( r + 1 ) 2 ≤ 10 ⋅ 4 r r ≥ 8 r\ge8 r ≥ 8 r = 8 r=8 r = 8
Corollary A.5. (a) Every even k > 2730 k>2730 k > 2730 178 J ( 2 k ) 2 ≤ 10 k 178J(2k)^2\le10k 178 J ( 2 k ) 2 ≤ 10 k 17.8 J ( 2 k ) 2 ≤ k 17.8J(2k)^2\le k 17.8 J ( 2 k ) 2 ≤ k
(b) Every g ≥ 1 g\ge1 g ≥ 1 J ( g ) ≤ 8 J(g)\le8 J ( g ) ≤ 8 1000 J ( g ) < 10 2 g 1000J(g)<10^2g 1000 J ( g ) < 1 0 2 g
Proof. (a) Note J ( 2 k ) = J ( k ) J(2k)=J(k) J ( 2 k ) = J ( k ) ω ( k ) ≤ 8 \omega(k)\le8 ω ( k ) ≤ 8 Lemma A.3 to n = k n=k n = k B B B k ≥ B Π r k\ge B\Pi_r k ≥ B Π r k k k B B B 2730 2730 2730 k ≥ B ( ⌊ 2730 / B ⌋ + 1 ) k\ge B(\lfloor2730/B\rfloor+1) k ≥ B (⌊ 2730/ B ⌋ + 1 ) 178 J ( k ) 2 ≤ 178 m 2 ≤ 10 k 178J(k)^2\le178m^2\le10k 178 J ( k ) 2 ≤ 178 m 2 ≤ 10 k ω ( k ) ≥ 9 \omega(k)\ge9 ω ( k ) ≥ 9 Lemma A.4 gives 178 J ( k ) 2 ≤ 178 B 2 ( r + 1 ) 2 ≤ 10 B R ≤ 10 k 178J(k)^2\le178B^2(r+1)^2\le10BR\le10k 178 J ( k ) 2 ≤ 178 B 2 ( r + 1 ) 2 ≤ 10 B R ≤ 10 k ω ( g ) ≤ 8 \omega(g)\le8 ω ( g ) ≤ 8 Lemma A.3 (ii) gives J ( g ) ≤ m ≤ 8 J(g)\le m\le8 J ( g ) ≤ m ≤ 8 1000 J ( g ) ≤ 1000 m < 10 2 B Π r ≤ 10 2 g 1000J(g)\le1000m<10^2B\Pi_r\le10^2g 1000 J ( g ) ≤ 1000 m < 1 0 2 B Π r ≤ 1 0 2 g ω ( g ) ≥ 9 \omega(g)\ge9 ω ( g ) ≥ 9 Lemma A.4 gives 1000 J ( g ) ≤ 1000 B ( r + 1 ) < 10 2 B R ≤ 10 2 g 1000J(g)\le1000B(r+1)<10^2BR\le10^2g 1000 J ( g ) ≤ 1000 B ( r + 1 ) < 1 0 2 B R ≤ 1 0 2 g
Detailed case checks
This appendix contains two elementary case checks that complete proofs in the main text.
The case M = 1 M=1 M = 1
This completes the proof of Lemma 7.13, in the notation of that proof and with the description of Q Q Q
At each index i i i a a a b b b 2 a 2a 2 a a + b a+b a + b § 3.5 ), a star marking a wrapped tap; reads follow from (22) and (23) .
Table 7.
Positions not described as odd or as elements of S S S ( 0 , b ) ∪ ( 2 a , k ) (0,b)\cup(2a,k) ( 0 , b ) ∪ ( 2 a , k ) Q Q Q S S S a a a 3 a ∈ Q 3a\in Q 3 a ∈ Q ( i − a , i − a − b , 2 b − a , i − 2 a ) (i-a,i-a-b,2b-a,i-2a) ( i − a , i − a − b , 2 b − a , i − 2 a ) { a , 3 a , b − a } \{a,3a,b-a\} { a , 3 a , b − a } ( i + a , i + 2 a ) (i+a,i+2a) ( i + a , i + 2 a ) { a , 3 a , b − a } \{a,3a,b-a\} { a , 3 a , b − a } b − a ∈ Q b-a\in Q b − a ∈ Q
These memberships are checked with a < b < 2 a a<b<2a a < b < 2 a k = 2 a + b k=2a+b k = 2 a + b i + a ∈ ( a , a + b ) i+a\in(a,a+b) i + a ∈ ( a , a + b ) 3 a 3a 3 a i < b < 2 a i<b<2a i < b < 2 a b − a b-a b − a i − a ∈ ( b − a , b ) i-a\in(b-a,b) i − a ∈ ( b − a , b ) b − a b-a b − a 3 a 3a 3 a a a a i ≠ 2 a i\ne2a i = 2 a i + a ∈ ( a + b , k ) i+a\in(a+b,k) i + a ∈ ( a + b , k ) 3 a 3a 3 a i ≠ 2 a i\ne2a i = 2 a i − a − b ∈ ( 0 , a ) i-a-b\in(0,a) i − a − b ∈ ( 0 , a ) b − a b-a b − a i ≠ 2 b i\ne2b i = 2 b 2 b − a 2b-a 2 b − a a a a 3 a 3a 3 a b − a b-a b − a b ≠ a , 2 a , 0 b\ne a,2a,0 b = a , 2 a , 0 i + 2 a ∈ ( 2 a , 3 a ) i+2a\in(2a,3a) i + 2 a ∈ ( 2 a , 3 a ) i − 2 a ∈ ( b − a , b ) i-2a\in(b-a,b) i − 2 a ∈ ( b − a , b ) a a a i ≠ 3 a i\ne3a i = 3 a 3 a 3a 3 a 2 a < a + b < 3 a < k 2a<a+b<3a<k 2 a < a + b < 3 a < k a + b < 2 b < k a+b<2b<k a + b < 2 b < k σ = a \sigma=a σ = a
This proves (B2), and with it Lemma 7.13.
The special indices of the three-element supports
This completes the proof of Theorem 8.8 , in its notation (residues modulo 4 h 4h 4 h
In each case below, every position named that is not called a pin is free, and no tap used is deleted; this is checked after the list. The inequality 3 β < h 3\beta<h 3 β < h
i = a i=a i = a 3 h + β 3h+\beta 3 h + β a a a 2 a 2a 2 a 6 h + 2 β ≡ 2 h + 2 β 6h+2\beta\equiv2h+2\beta 6 h + 2 β ≡ 2 h + 2 β h h h a + h a+h a + h 4 h + β ≡ β 4h+\beta\equiv\beta 4 h + β ≡ β
i = h i=h i = h h h h a a a a + h a+h a + h k − a k-a k − a k + h − a k+h-a k + h − a 4 h − β 4h-\beta 4 h − β
i = k − a i=k-a i = k − a 3 h − β 3h-\beta 3 h − β h h h k + h − a k+h-a k + h − a k − a k-a k − a 2 k − 2 a 2k-2a 2 k − 2 a 6 h − 2 β ≡ 2 h − 2 β 6h-2\beta\equiv2h-2\beta 6 h − 2 β ≡ 2 h − 2 β
i = h − a i=h-a i = h − a ν h − ν a ≡ 2 h − β \nu h-\nu a\equiv2h-\beta ν h − ν a ≡ 2 h − β a a a h h h k − a k-a k − a 3 h − 2 a 3h-2a 3 h − 2 a 5 h − 2 β ≡ h − 2 β 5h-2\beta\equiv h-2\beta 5 h − 2 β ≡ h − 2 β
i = k − 2 a i=k-2a i = k − 2 a 2 h − 2 ( 3 h + β ) ≡ − 2 β 2h-2(3h+\beta)\equiv-2\beta 2 h − 2 ( 3 h + β ) ≡ − 2 β a a a k − a k-a k − a h h h 3 h − 2 a 3h-2a 3 h − 2 a h − 2 β h-2\beta h − 2 β
i = h + a i=h+a i = h + a β \beta β k − a k-a k − a k + h k+h k + h a a a h + 2 a h+2a h + 2 a 3 h + 2 β 3h+2\beta 3 h + 2 β
i = 2 a i=2a i = 2 a 2 h + 2 β 2h+2\beta 2 h + 2 β a a a 3 a 3a 3 a 5 h + 3 β ≡ h + 3 β 5h+3\beta\equiv h+3\beta 5 h + 3 β ≡ h + 3 β h h h h + 2 a h+2a h + 2 a 3 h + 2 β 3h+2\beta 3 h + 2 β
The positions 2 a 2a 2 a a + h a+h a + h k + h − a k+h-a k + h − a 2 k − 2 a 2k-2a 2 k − 2 a ≠ 0 ( m o d k ) \ne0\pmod{k} = 0 ( mod k ) 3 h − 2 a 3h-2a 3 h − 2 a h + 2 a h+2a h + 2 a 3 a 3a 3 a ( 0 , 2 k ) (0,2k) ( 0 , 2 k ) k k k a a a h h h k − a k-a k − a h = a h=a h = a h = 3 a h=3a h = 3 a h = 2 a h=2a h = 2 a 3 a = 2 h 3a=2h 3 a = 2 h a < h a<h a < h gcd ( a , h ) = 1 \gcd(a,h)=1 g cd( a , h ) = 1 a ≥ 3 a\ge3 a ≥ 3 a a a h h h h h h k − a k-a k − a k + h k+h k + h 2 k − a 2k-a 2 k − a k k k a a a k + a k+a k + a
So the clause holds at each of the seven special indices, which completes the proof of Theorem 8.8 .
The formalisation
The proof of Theorem 1.1 has been formalised in the Lean 4 proof assistant, using the Mathlib library. The formal statement is, in the notation of (3) ,
if every positive solution of (3) satisfies x n + p = x n for some fixed p ≥ 1 , then (3) \text{if every positive solution of (3) satisfies } x_{n+p}=x_n \text{ for some fixed } p\geq1,\text{ then (3)} if every positive solution of (3) satisfies x n + p = x n for some fixed p ≥ 1 , then (3) is one of the equations listed in Theorem 1.1 ,
with each of the five cases written out as explicit conditions on the coefficients A i A_i A i B i B_i B i Theorem 1.1 is globally periodic, is formalised as well, so that the formal statement is an equivalence: (3) is globally periodic if and only if it is one of the listed equations. The main theorem and the equivalence are, respectively,
PeriodicDifference.Problem11.problem11_complete and
PeriodicDifference.Problem11.problem11_iff.The tropical Corollary 10.1 is formalised too, as
PeriodicDifference.Trop.trop_nonperiodic,which I state for the window map of Definition 3.1, with the constant slot 0 added to S S S θ = 1 \theta=1 θ = 1 https://github.com/johnrudge/periodic_difference .
The formalisation follows the structure of this paper result by result, including the period-sum proof of Theorem 2.5 . In a few places the formal proof takes a different, usually more elementary, route:
Linearisation (Theorem 2.2 and Theorem 2.4 ). Instead of the chain rule for F p F^p F p F ( u ∗ + ϵ w ) = u ∗ + ϵ G ϵ ( w ) F(u^*+\epsilon w)=u^*+\epsilon G_\epsilon(w) F ( u ∗ + ϵ w ) = u ∗ + ϵ G ϵ ( w ) G ϵ G_\epsilon G ϵ ϵ \epsilon ϵ G 0 = D F ( u ∗ ) G_0=DF(u^*) G 0 = D F ( u ∗ ) G ϵ p = i d G_\epsilon^p=\mathrm{id} G ϵ p = id ϵ > 0 \epsilon>0 ϵ > 0 ϵ → 0 \epsilon\to0 ϵ → 0 G 0 p = i d G_0^p=\mathrm{id} G 0 p = id
Order three with c 0 = 0 c_0=0 c 0 = 0 (Theorem 4.2 ). Instead of the algebraic-integer step, a root λ \lambda λ λ 2 − 3 2 λ + 1 \lambda^2-\frac{3}{2}\lambda+1 λ 2 − 2 3 λ + 1 λ m = α m λ + β m \lambda^m=\alpha_m\lambda+\beta_m λ m = α m λ + β m α m , β m \alpha_m,\beta_m α m , β m 2 m − 1 α m 2^{m-1}\alpha_m 2 m − 1 α m m ≥ 1 m\geq1 m ≥ 1 α m ≠ 0 \alpha_m\ne0 α m = 0 λ \lambda λ λ m ≠ 1 \lambda^m\ne1 λ m = 1
Small orders in Theorem 4.3 . Step 2 of the proof (that n 0 ≥ 9 n_0\geq9 n 0 ≥ 9 ζ 2 − ϵ ζ + 1 = 0 \zeta^2-\epsilon\zeta+1=0 ζ 2 − ϵ ζ + 1 = 0 ζ r = P r ( ϵ ) ζ + Q r ( ϵ ) \zeta^r=P_r(\epsilon)\zeta+Q_r(\epsilon) ζ r = P r ( ϵ ) ζ + Q r ( ϵ ) P r , Q r P_r,Q_r P r , Q r ζ r = 1 \zeta^r=1 ζ r = 1 P r ( ϵ ) = 0 P_r(\epsilon)=0 P r ( ϵ ) = 0 1 ≤ r ≤ 8 1\leq r\leq8 1 ≤ r ≤ 8 0 < ϵ ≤ 1 2 0<\epsilon\leq\frac{1}{2} 0 < ϵ ≤ 2 1
Odd indices (Lemma 8.7 ). Instead of the argument with the gaps between phases, the formal proof is local: if H ν ( u ) ≠ H ν ( v ) H_\nu(u)\ne H_\nu(v) H ν ( u ) = H ν ( v ) u u u u − v u-v u − v v v v k − v k-v k − v ν i \nu i ν i ν u \nu u ν u ν v \nu v ν v 2 k 2k 2 k
The band (Proposition 9.8 ). To stay in rational arithmetic, the incircle of Δ 1 \Delta_1 Δ 1 Δ 1 \Delta_1 Δ 1 0.207 k 0.207\,k 0.207 k 0.102 k 0.102\,k 0.102 k 0.36024 k 0.36024\,k 0.36024 k 0.3604 k 0.3604\,k 0.3604 k J ( n ) ≤ 0.36021 k J(n)\leq0.36021\,k J ( n ) ≤ 0.36021 k Proposition 9.12 still suffices.
Every finite computation in the paper is carried out in exact integer arithmetic inside Lean’s kernel by the decide tactic, so no step of the proof relies on compiled code or on an external program. In some the kernel finds the required objects itself by search (or none are needed); in the others they were produced outside Lean and are imported as data, which the kernel only checks:
Table 4 (PDF p. 61).
The case tables of Appendix § 13 and the chains of Table 5 hold for all values of the parameters, and they are formalised as case analyses.
