We study quantitative density of orbits of the free semigroup Γ generated by two maps of (0,∞). For the Möbius pair x↦1/x, x↦x+1 we show that every orbit is O(n−1)-dense in [0,∞), that every irrational number u is approached at a rate governed by the continued fraction of u (at least n−2, exponential when u has bounded partial quotients, and exp(−cn/logn) for almost every u), and that for rational u and rational starting points the rate n−1 is optimal. For the affine pair T0(x)=x/a, T1(x)=bx+1 with a,b>1, which was studied by Bergelson, Misiurewicz and Senti, we prove that for every x>0 and u≥0 there are infinitely many γ∈Γ with ∣γ(x)−u∣<Cλ−∣γ∣,λ=exp(loga−1a2blogaloga−1a)>1, where ∣γ∣ is the word length of γ and C is independent of γ. This exponential rate holds whether or not logab is rational, and the exponential order cannot be improved for almost every u. We also give a corrected version of the Kronecker-type estimate ∣γ(x)−u∣<(u+1)(t1/∣γ∣−1). Finally, we discuss the approximation of real numbers by ratios of primes and by ratios of logarithms of primes, updated in light of recent results on gaps between primes.
Introduction
The Dirichlet approximation theorem states that, given a real number u and a positive integer N, there exist integers p,q with 0<q≤N such that
qp−u<qN1.(1)
In particular, if u is irrational then ∣p/q−u∣<q−2 for infinitely many reduced fractions p/q. We give a formulation of this theorem in terms of semigroup actions which leads to natural generalizations.
Let Γ be the free semigroup generated by two letters σ and τ, with identity λ0 (the empty word). Given two maps T0,T1:(0,∞)→(0,∞), we obtain an action of Γ on (0,∞) by letting σ act as T0 and τ act as T1; words act as compositions, so that (w1w2)(x)=w1(w2(x)) (the rightmost letter acts first). We write γ∈Γ(T0,T1) when we want to emphasize the choice of T0 and T1. The length ∣⋅∣:Γ→Z≥0 is defined by ∣λ0∣=0, ∣σ∣=∣τ∣=1 and ∣w1w2∣=∣w1∣+∣w2∣. For x,y>0 we say that y is reachable from x if y=γ(x) for some γ∈Γ, and we define the reachability function
R(x,y)=min{∣γ∣:γ∈Γ,γ(x)=y},(2)
with R(x,y)=∞ if y is not reachable from $x.
Definition 1.1. For x>0 let Γx denote the orbit of x under Γ(T0,T1). Let f:N→(0,∞) be a non-increasing function with f(n)→0. We say that Γx is O(f)-dense in [0,∞) if for every u≥0 there exist a constant C=C(T0,T1,x,u)>0 and infinitely many γ∈Γ such that
∣γ(x)−u∣<Cf(∣γ∣).
Since only finitely many words have a given length, “infinitely many γ” forces ∣γ∣→∞ along the approximating words.
The Möbius pair
In Section 2 we take
T0(x)=x1,T1(x)=x+1.(3)
Here γ(1) is always a positive rational number, and Lemma 2.2 shows R(1,p/q)≤p+q; combined with (1) this shows that every irrational u is approached at rate O(∣γ∣−2). We prove Theorem 2.4 that every orbit is O(n−1)-dense in [0,∞), with the rate n−2 at irrational targets. We then refine this in two directions. First (Theorem 2.5 and Corollary 2.6), the true rate at an irrational u is governed by the continued fraction expansion of $u: it is exponential in ∣γ∣ when u has bounded partial quotients, and it is at least exp(−c∣γ∣/log∣γ∣) for Lebesgue almost every u, for every c<π2/6. Second (Proposition 2.9), for rational targets and rational starting points the exponent 1 is optimal: by an argument with the Stern–Brocot tree, ∣γ(x)−k/l∣≥1/(l2(s(x)+∣γ∣)) whenever γ(x)=k/l, where s(x) is the sum of the partial quotients of x.
The affine pair
In Section 3 we consider
T0(x)=ax,T1(x)=bx+1,a,b>1.(4)
Bergelson, Misiurewicz and Senti [4] showed that every orbit of this action is dense in [0,∞). Our main result (Theorem 3.8) is that every orbit is in fact exponentially dense: for every x>0 and u≥0 there are infinitely many γ with ∣γ(x)−u∣<Cλ−∣γ∣ where λ=λ(a,b)>1 is explicit. The proof rests on an explicit normal form for γ(x) (Lemma 3.1) and a greedy algorithm, and it does not distinguish between rational and irrational values of logab. A counting argument (Proposition 2.11) shows that for almost every u no rate faster than μ−∣γ∣ with μ>2 is possible, so the exponential order is optimal.
We also give corrected versions of two earlier estimates: a Kronecker-type bound ∣γ(x)−u∣<(u+1)(t1/∣γ∣−1) with t=a12b4 when logab is irrational (Theorem 3.2), and an exponential bound with exponent independent of u when logab is rational (Theorem 3.6). Since t1/n−1∼(logt)/n, the Kronecker-type bound is an O(n−1) statement; it is superseded by Theorem 3.8, but its proof is short and the words it produces are very simple.
The contrast between the two sections is worth noting: for the Möbius pair rational targets are approached only at rate n−1 (at least from rational starting points), whereas for the affine pair every target is approached exponentially fast.
Relation to the 3x+1 problem
The 3x+1 (Collatz) problem asks whether iterating
f(m)={m/2,3m+1,meven,modd,
eventually reaches 1 from every positive integer; see [12] for a survey. In our notation, with a=2 and b=3, a Collatz trajectory is a particular orbit segment of Γ(T0,T1), namely one in which T0 is applied exactly to even integers and T1 exactly to odd integers. Thus the Collatz conjecture implies that 1 is reachable from every positive integer via Γ(T0,T1); the converse is not clear, since the semigroup allows T1 to be applied to even integers as well.
Moreover, any sequence x1,…,xm with x1∈N, xm=1 and xi+1∈{T0(xi),T1(xi)} consists of integers. Indeed, write v2 for the 2-adic valuation on Q×. If v2(xi)<0 then v2(T0xi)=v2(xi)−1<0 and v2(T1xi)=v2(3xi+1)=v2(xi)<0, so once a non-integer appears, all later terms are non-integers; since xm=1 is an integer, no xi can be a non-integer.
Theorem 3.8 shows that, starting from any x>0, there are words γ of arbitrarily large length n with ∣γ(x)−1∣<Cλ−n, where λ=λ(2,3)≈1.2133.
Primes
In § 4 we discuss approximating real numbers by ratios of primes and by ratios of logarithms of primes, updating the discussion in light of the work of Goldston, Pintz and Yıldırım [8, 7], Zhang [18], Maynard [13] and Baker, Harman and Pintz [3].
The Möbius pair
Throughout this section T0(x)=1/x and T1(x)=x+1.
Matrices and continued fractions
Associate to T0 and T1 the matrices
Mσ=(0110),Mτ=(1011).
and to a word γ=w1⋯wn the product Mγ=Mw1⋯Mwn. If Mγ=(ACBD) then A,B,C,D are non-negative integers, AD−BC=±1, C+D≥1, and
γ(z)=Cz+DAz+B(z>0).
Every positive rational number r has a unique canonical continued fraction expansion r=[a0;a1,…,ak] with a0≥0, a1,…,ak≥1, and either k=0 (and then a0≥1) or k≥1 and $a_k\ge2. We put
s(r)=a0+a1+⋯+ak,ℓ(r)=k.
Lemma 2.1. For every positive rational r we have s(T0r)=s(r) and s(T1r)=s(r)+1. Consequently, for every positive rational x and every γ∈Γ,
s(γ(x))=s(x)+(number of occurrences of τ in γ)≤s(x)+∣γ∣.
Proof. Let r=[a0;a1,…,ak] be canonical. Then T1r=[a0+1;a1,…,ak] is canonical. If a0=0 then k≥1 and T0r=[a1;a2,…,ak] is canonical. If a0≥1 and k≥1, then T0r=[0;a0,a1,…,ak] is canonical; if a0≥2 and k=0 then T0r=[0;a0] is canonical; and if r=[1] then T0r=r. In all cases the sum of the partial quotients is preserved by T0 and increased by one by T1. □
Lemma 2.2. Let r=[a0;a1,…,ak] be a positive rational number in canonical form. Then
s(r)−1≤R(1,r)≤s(r)+ℓ(r)−1.(5)
In particular, if p,q are positive integers then R(1,p/q)≤p+q.
Proof. The lower bound follows from Lemma 2.1, since s(1)=1 and each word reaching r from 1 contains at least s(r)−1 letters τ. For the upper bound, put zk=ak and zi=ai+1/zi+1 for i<k, so that z0=r. Then zk=T1ak−1(1) and zi=T1aiT0(zi+1), so r is reached from 1 by a word of length (ak−1)+∑i<k(ai+1)=s(r)+k−1.
For the last statement, apply the bound on p+q (one may also deduce it from (5)). If p=q then R(1,p/q)=0. If p>q then p/q=T1((p−q)/q), so R(1,p/q)≤1+R(1,(p−q)/q)≤1+p. If 1<p<q then p/q=T0T1((q−p)/p), so R(1,p/q)≤2+R(1,(q−p)/p)≤q+2≤p+q. Finally, if p=1 then 1/q=T0T1q−1(1) and R(1,1/q)≤q. □
Lemma 2.3. Let γ∈Γ and write γ(1)=p/q in lowest terms. Then for every ε≥0,
∣γ(1+ε)−γ(1)∣=q(q+Cε)ε≤q2ε,
where C is the lower left entry of Mγ.
Proof. With Mγ=(ACBD) we have γ(1)=(A+B)/(C+D), and this fraction is in lowest terms because AD−BC=±1; thus p=A+B and q=C+D. Then
Define Q:[0,∞)→{1,2} by Q(u)=1 if u is rational and Q(u)=2 if u is irrational.
Theorem 2.4. For every x>0 the orbit Γx is O(n−1)-dense in [0,∞). More precisely, for every x>0 and u≥0 there exist C=C(x,u)>0 and infinitely many γ∈Γ such that
∣γ(x)−u∣<C∣γ∣−Q(u).(6)
Proof.Reduction. If x<1, replace x by T1(x)=x+1; a word γ for x+1 gives the word γτ for x, of length ∣γ∣+1≤2∣γ∣. So we may assume x=1+ε with ε≥0.
The case u=0. For m≥1 let γ=στm. Then γ(x)=1/(x+m)<1/m<2/∣γ∣.
Rational u=k/l>0. For m≥1 put rm=(km+1)/(lm), so rm−u=1/(lm). By Lemma 2.2 there is γm with γm(1)=rm and ∣γm∣≤(k+l)m+1≤(k+l+1)m. Since gcd(km+1,m)=1, the reduced denominator qm of rm is at least lm/gcd(km+1,l)≥m. By Lemma 2.3,
∣γm(x)−u∣≤qm2ε+lm1≤mε+1≤∣γm∣x(k+l+1).
The rm are distinct, so the γm are distinct. (In the original version of this argument one took rm=(klm+1)/lm+1; this fails when u is an integer, since then l=1 and rm does not depend on m.)
Irrational u. This follows from Corollary 2.6(a) below. □
Irrational targets and continued fractions
Let u=[a0;a1,a2,…] be irrational, with convergents pk/qk, and put Sk=a0+a1+⋯+ak.
Theorem 2.5. Let u>0 be irrational and x>0. For every k≥1 there is γk∈Γ with
∣γk∣≤Sk+kand∣γk(x)−u∣<qk2x+1.
Proof. The canonical expansion of pk/qk is either [a0;…,ak] (if ak≥2) or [a0;…,ak−1+1] (if ak=1); in both cases its partial quotients sum to Sk and it has at most k terms after a0. By Lemma 2.2 there is δk with δk(1)=pk/qk and ∣δk∣≤Sk+k−1. Recall ∣pk/qk−u∣<1/(qkqk+1)≤1/qk2.
If x=1+ε≥1, let γk=δk; by Lemma 2.3, ∣γk(x)−u∣<(ε+1)/qk2=x/qk2. If x<1, let γk=δkτ, so that γk(x)=δk(1+x) and ∣γk(x)−u∣<(1+x)/qk2. □
Corollary 2.6. Let x>0 and let u=[a0;a1,…] be irrational.
(a) There is C=C(x,u) such that ∣γ(x)−u∣<C∣γ∣−2 for infinitely many γ.
(b) If ai≤A for all i≥1, then ∣γ(x)−u∣<Cφ−2∣γ∣/(A+1) for infinitely many γ, where φ=(1+5)/2 and C=C(x,u).
(c) For Lebesgue almost every u>0 and every c<π2/6, there are infinitely many γ with ∣γ(x)−u∣<exp(−c∣γ∣/log∣γ∣).
Proof. (a) We have q1=a1 and, for k≥2, qk=akqk−1+qk−2≥(ak+qk−1−1)+qk−2≥ak+qk−1, using (ak−1)(qk−1−1)≥0 and qk−2≥1. Hence qk≥a1+⋯+ak=Sk−a0≥k, and
∣γk∣≤Sk+k≤a0+2qk.
By Theorem 2.5, ∣γk(x)−u∣<(x+1)qk−2≤4(x+1)(∣γk∣−a0)−2, which is at most C∣γk∣−2 for k large. The words γk are pairwise distinct, because the values δk(1)=pk/qk are distinct.
(b) Now qk≥φk−1 and ∣γk∣≤a0+(A+1)k, so
qk−2≤φ2φ−2k≤φ2+2a0/(A+1)φ−2∣γk∣/(A+1).
(c) By Lévy’s theorem, k1logqk→β:=12log2π2 for almost every u. By Khinchin’s weak law for the sums of partial quotients, Sk/(klogk)→1/log2 in measure; hence there is a fixed sequence kj→∞ along which Skj/(kjlogkj)→1/log2 almost everywhere (see [10, 11]).
Fix η>0 and such a u. For j large, nj:=∣γkj∣≤(1+η)kjlogkj/log2, hence kj≥1+ηlog2lognjnj (as kj≤nj), and
Choosing η small in terms of c<π2/6 gives the claim. □
Remark 2.7. Part (a) is the estimate ∣γ(x)−u∣≤C∣γ∣−2 asserted for irrational u in Theorem 2.4. Parts (b) and (c) show that it is far from optimal for most u; for instance, for the golden ratio the orbit of 1 approaches φ exponentially fast. On the other hand, by Proposition 2.11 below, for almost every u no rate μ−∣γ∣ with μ>2 is possible.
Optimality for rational targets
We use the Stern–Brocot tree (see [9 §4.5]). Its nodes are the positive rationals, each occurring exactly once; the root is 1/1, and the node r=[a0;a1,…,ak] (canonical) has depth s(r)−1. Every node u=k/l has a nearest left ancestor kL/lL<u and a nearest right ancestor kR/lR>u (with the conventions 0/1 and 1/0 at the boundary) such that k=kL+kR, l=lL+lR and klL−lkL=kRl−lRk=1; moreover, the rationals in the open interval (kL/lL,kR/lR) are exactly the nodes of the subtree rooted at u.
Lemma 2.8. Let u=k/l≥0 be rational in lowest terms and let r>0 be rational with r=u. Then
∣r−u∣≥l2s(r)1.
Proof. If u=0 (so l=1): if r≥1 then r≥1≥1/s(r); otherwise r=[0;a1,…,am] with s(r)=a1+⋯+am, and r=1/a1≥1/s(r) if m=1, while r>1/(a1+1)≥1/s(r) if m≥2 (as then $a_m\ge2).
Let u>0, let d0 be the depth of u, and let D=s(r)−1 be the depth of r. If r∈/(kL/lL,kR/lR) then ∣r−u∣≥min{1/(llL),1/(llR)}≥1/l2 since lL,lR≤l. Otherwise r lies in the subtree of u; suppose r<u (the case r>u is symmetric). The nodes cj=(kL+jk)/(lL+jl), j≥1, form the right spine of the left subtree of u; cj has depth d0+j, and the rationals in (cj,u) are exactly the nodes of the subtree rooted at cj+1, all of depth at least d0+j+1. Here D≥d0+1; taking j=D−d0≥1, the node r is not in (cj,u), so r≤cj
Proposition 2.9. Let x>0 be rational and u=k/l≥0 rational in lowest terms. Then for every γ∈Γ with γ(x)=u,
∣γ(x)−u∣≥l2(s(x)+∣γ∣)1.
In particular, the exponent Q(u)=1 in Theorem 2.4 cannot be improved when x and u are rational.
Proof. Combine Lemma 2.8 with Lemma 2.1.
Remark. 2.10. We expect the same lower bound, with a constant depending on x, for irrational starting points x, but we have not proved it.
A general upper limit on the rate
The following observation applies to any pair of maps.
Proposition 2.11. Let T0,T1:(0,∞)→(0,∞) be arbitrary, let x>0 and let μ>2. Then for Lebesgue almost every u≥0 there are only finitely many γ∈Γ(T0,T1) with ∣γ(x)−u∣<μ−∣γ∣.
Proof. Let Iγ=(γ(x)−μ−∣γ∣,γ(x)+μ−∣γ∣). There are 2n words of length n, so ∑γ∣Iγ∣=∑n2n⋅2μ−n<∞, and the Borel–Cantelli lemma shows that almost every u lies in only finitely many Iγ.
The affine pair
Throughout this section T0(x)=x/a and T1(x)=bx+1 with a,b>1, and σ,τ act as T0,T1. We put α=logab>0.
A normal form
Lemma 3.1. Let γ∈Γ contain N letters τ and M letters σ. For 0≤j<N let ℓj be the number of letters σ to the left of the (j+1)-st letter τ, counted from the left. Then 0≤ℓ0≤ℓ1≤⋯≤ℓN−1≤M and
γ(x)=aMbNx+j=0∑N−1aℓjbj(x∈R).(7)
Conversely, for all integers N≥0 and 0≤ℓ0≤⋯≤ℓN−1≤M, the word γ=σℓ0τσℓ1−ℓ0τ⋯σℓN−1−ℓN−2τσM−ℓN−1 has length N+M and satisfies (7).
Proof. Write γ=w1⋯wL. Each Ti is affine, T0′=1/a and T1′=b, so γ′=bNa−M. The constant term of γ is the sum, over the positions i with wi=τ, of the constant 1 of T1 multiplied by the derivative of w1⋯wi−1. If wi is the (j+1)-st τ from the left, this derivative is bja−ℓj. The converse is immediate.
The Kronecker-type estimate
We use the following inhomogeneous form of Kronecker’s theorem ([5, Ch. III, Thm. II], [16, §2.4]): if α is irrational and β∈R, then there are infinitely many pairs of integers (m,n) with m>0 and
∣mα−n−β∣<m3.(8)
Theorem 3.2. Suppose α=logab is irrational, and put t=a12b4. Then for every x>0 and u≥0 there are infinitely many γ∈Γ with
∣γ(x)−u∣<(u+1)(t1/∣γ∣−1).(9)
Proof. If u=0, then γ=σm gives γ(x)=xa−m<t1/m−1 for m large. Let u>0 and β=loga((b−1)u). By (8) there are infinitely many pairs of positive integers (m,n) with
∣mα−n−β∣<m3,(10)
and since ∣loga(bm−1)−mα∣=∣loga(1−b−m)∣<1/m for m large,
∣loga(bm−1)−n−β∣<m4.(11)
Let η=σnτm−1. Since τm−1(1)=(bm−1)/(b−1), we have η(1)=(bm−1)/((b−1)an), and (11) says ∣logaη(1)−logau∣<4/m. Hence
∣η(1)−u∣<u(a4/m−1).
Now let γ=ητσm=σnτmσm, so that γ(x)=η(bxa−m+1). Since η is affine with derivative bm−1a−n,
∣γ(x)−η(1)∣=am+nbmx≤(b−1)uxa3/m−m,
using bma−n=amα−n<(b−1)ua3/m from (10). The right side is exponentially small in m, so for m large
∣γ(x)−u∣≤(b−1)uxa3/m−m+u(a4/m−1)<(u+1)(a4/m−1).
Finally, (10) gives n<(α+1)m for m large, hence ∣γ∣=2m+n<(α+3)m, and
a4/m<a4(α+3)/∣γ∣=(a12b4)1/∣γ∣.
Remark 3.3. Since t1/n−1∼(logt)/n, Theorem 3.2 says that every orbit is O(n−1)-dense, with the explicit constant ∣γ(x)−u∣≲4(u+1)(3loga+logb)/∣γ∣.
Rational logab
Lemma 3.4. Let a>1 and u>0. There are non-negative integers d0,d1,… with d0<u and di≤⌈a⌉−1<a for i≥1, such that
0<u−i=0∑kdia−i≤a−kfor all k≥0,(12)
and dk>0 for infinitely many $k.
Proof. Let d0=⌈u−1⌉ and ρ0=u−d0∈(0,1]. Given ρk∈(0,a−k], let dk+1=⌈ρkak+1−1⌉ and ρk+1=ρk−dk+1a−k−1. Since 0<ρkak+1≤a, we get 0≤dk+1<ρkak+1≤a and ρk+1=a−k−1(ρkak+1−dk+1)∈(0,a−k−1]. If dk=0 for all k≥K, then ρk=ρK>0 for k≥K, contradicting ρk≤a−k→0. □
Lemma 3.5. Suppose a=b>1 and let d0,…,dk be non-negative integers with dk>0. Then there is η∈Γ with η(1)=∑i=0kdia−i, ∣η∣≤k+2∑i=0kdi, and η′=a−k.
Proof. Note τσ(z)=z+1. For k=0 take η=(τσ)d0−1. For the inductive step write P=∑i=0k+1dia−i=d0+Q/a with Q=∑i=0kdi+1a−i; if μ(1)=Q with ∣μ∣≤k+2∑i≥1di and μ′=a−k, then η=(τσ)d0σμ satisfies η(1)=P, ∣η∣≤k+1+2∑i≥0di and η′=a−k−1.
Theorem 3.6. Suppose logab=n0/m0 with m0,n0 positive integers, and put A=am0=bn0 and
ρ=A1/((2A+1)max{m0,n0})>1.
Then for every x>0 and u≥0 there are C=C(a,b,x,u) and infinitely many γ∈Γ with
∣γ(x)−u∣<Cρ−∣γ∣.
In particular, if a=b one may take ρ=a1/(2a+1).
Proof. The case a=b. If u=0, use γ=σM. Let u>0 and let di be as in Lemma 3.4. For each k with dk>0, Lemma 3.5 gives γk with γk(1)=∑i≤kdia−i, γk′=a−k and
The values γk(1) are strictly increasing in k, so the γk are distinct.
The general case. Let c=∑i=0n0−1bi and θ(z)=cz. Then
S0:=θ−1∘T0m0∘θ:z↦z/A,S1:=θ−1∘T1n0∘θ:z↦Az+1,
because T1n0(y)=bn0y+c. Applying the first case to (S0,S1), with base A, to v=x/c and w=u/c, we get infinitely many words μ in S0,S1 with ∣μ(v)−w∣<C′A−∣μ′∣/(2A+1), where ∣μ′∣ is the length in S0,S1. Each such μ equals θ−1γθ for a word γ in T0,T1 with ∣γ∣≤max{m0,n0}∣μ′∣, and ∣γ(x)−u∣=c∣μ(v)−w∣.
Remark 3.7. The original version of this theorem gave the bound exp(−C∣γ∣/(u+1)), whose exponent deteriorates as u grows; the estimate ∣γk∣≤(2a+1)k+2u shows that u only affects the constant. (Also, in the case u=0 the original choice of constant C=1/2 fails when a<e1/2.)
Then for every x>0 and u≥0 there are C=C(a,b,x,u) and infinitely many γ∈Γ such that
∣γ(x)−u∣<Cλ−∣γ∣.(13)
In particular, every orbit is O(λ−n)-dense in [0,∞).
Proof. The case u=0. We have logλ≤loga, because loga−1a≤loga−1a2b. So γ=σM gives
γ(x)=xa−M≤xλ−M.
A greedy expansion. Let u>0. Set ℓ−1=0 and r−1=u. For j=0,1,2,…, given an integer ℓj−1 and rj−1>0, let ℓj∗ be the least integer ℓ with bja−ℓ<rj−1, and put
ℓj=max{ℓj−1,ℓj∗},tj=bja−ℓj,rj=rj−1−tj.
Since ℓj≥ℓj∗ we have tj<rj−1, so rj>0 and the recursion continues; also 0=ℓ−1≤ℓ0≤ℓ1≤⋯ and rj=u−∑i≤jti. Call step j free if ℓj=ℓj∗.
(i) If step j is free, thentj≥rj−1/a; hence rj≤(1−1/a)rj−1 and rj≤(a−1)tj. Indeed, by minimality of ℓj∗, atj=bja−(ℓj∗−1)≥rj−1.
(ii) If step j is free, so is step j+1. We must show ℓj+1∗≥ℓj, i.e. that ℓ=ℓj−1 violates bj+1a−ℓ<rj. But bj+1a−(ℓj−1)=abtj>(a−1)tj≥rj by (i).
(iii) Some step is free. If steps 0,…,j are not free, then ℓ0=⋯=ℓj=0 and ti=bi, so 1+b+⋯+bj<u. Hence the first free step J satisfies bJ≤1+(b−1)u.
Put ϑ=κ−α=logaa−1a>0, so that a−ϑ=1−1/a and λ=aϑ/(1+κ).
Case MN=ℓN−1. By (14), ∣γN(x)−u∣≤(x+ba−1)aNα−MN. Since MN≥κN and ϑ+α=κ, a direct computation gives
(1+κ)(MN−Nα)−ϑ(N+MN)=(1+α)(MN−κN)≥0,
that is, MN−Nα≥ϑL/(1+κ). Hence ∣γN(x)−u∣≤(x+ba−1)λ−L.
Case MN=⌈κN⌉>ℓN−1. Then bNa−MN≤a−ϑN=(1−1/a)N, and by (14),
∣γN(x)−u∣≤(x+u(a−1a)J)(1−a1)N.
As L≤(1+κ)N+1, we have (1−1/a)N=λ−(1+κ)N≤λ1−L.
In both cases ∣γN(x)−u∣≤Cλ−∣γN∣ with C=λ(x+u(a−1a)J)+ba−1, and the words γN, N>J, are distinct since they contain different numbers of letters τ. Replacing C by 2C gives strict inequality. □
Remark 3.9. (1) Theorem 3.8 makes no use of the arithmetic of logab, and it supersedes Theorem 3.2 and, for small a, Theorem 3.6. For instance, λ(2,3)≈1.2133 and λ(2,2)=21/3, whereas Theorem 3.6 gives ρ=21/5 when a=b=2; for a=b=10, on the other hand, Theorem 3.6 gives the larger base 101/21≈1.116 against λ≈1.053.
(2) By Proposition 2.11, for almost every u the base λ in (13) can never exceed 2; thus exponential density is the optimal order, although we have not tried to optimize λ. Numerical experiments suggest that the greedy words of the proof perform considerably better than the bound λ−∣γ∣.
(3) The constant C depends on u only through u(a−1a)J≤u(1+(b−1)u)logba−1a, a polynomial in u.
Prime approximation of real numbers
Let pn denote the n-th prime. Goldston, Pintz and Yıldırım [8] (see also [7]) proved that
n→∞liminflogpnpn+1−pn=0,(15)
equivalently ∣pn+1/pn−1∣=o(logpn/pn) for infinitely many n. Zhang [18] proved that liminf(pn+1−pn)<∞, and Maynard [13] and the Polymath project [15] improved this to liminf(pn+1−pn)≤246. Hence ∣pn+1/pn−1∣≤246/pn for infinitely many n. This suggests the following.
Conjecture 4.1 (Prime approximation). For every u≥0 and ε>0 there are infinitely many pairs of distinct primes (p,q) with
qp−u<εqlogq.(16)
The cases u=0 (take p=2) and u=1 (by the bounded gaps theorem) are known, the latter in the stronger form ∣p/q−1∣≤246/q. A naive probabilistic model suggests that even ∣p/q−u∣<c/q should have infinitely many solutions for every c>0, since for each prime q the interval (uq−c,uq+c) should contain a prime with probability about 2c/logq, and ∑q1/logq diverges.
Unconditionally, Baker, Harman and Pintz [3] proved that [x−x0.525,x] contains a prime for all sufficiently large x. Choosing a prime p in [uq−(uq)0.525,uq] shows that, for every fixed u>0 and every sufficiently large q (prime or not),
0≤u−qp≤u0.525q−0.475.(17)
Assuming the Riemann Hypothesis, Cramér [6] showed pn+1−pn=O(pnlogpn), which improves (17) to ∣p/q−u∣≤Cuq−1/2logq, still far from (16).
Ratios of logarithms of primes
We need a Mills-type lemma [14, 17]; we include the short proof.
Lemma 4.2.Suppose 0<β<1 and pn+1−pn=O(pnβ). Then for every real u>1/(1−β) there is θ>1 such that ⌊θun⌋ is prime for every n≥1.
Proof. The hypothesis gives a prime in [y,y+Kyβ] for all y≥2 and some K. For a large integer P, the interval [Pu,(P+1)u−2] has length at least uPu−1−2, which exceeds K(Pu)β for P large because u−1>uβ. So we may choose primes P1<P2<⋯, with P1 large, such that Pnu≤Pn+1≤(Pn+1)u−2. Then Pnu−n is non-decreasing and (Pn+1)u−n is strictly decreasing, so θ=limPnu−n satisfies Pn≤θun<Pn+1, i.e. ⌊θun⌋=Pn, for all n≥1. □
Theorem 4.3.For every u>40/19 there is an infinite sequence of primes q1<q2<⋯ such that, for all n,
−qn+1logqn1<logqnlogqn+1−u<qnlogqnu.(18)
Assuming the Riemann Hypothesis, the same holds for every u>2.
Proof. By [3], the hypothesis of Lemma 4.2 holds with β=0.525, and 1/(1−0.525)=40/19. (Under the Riemann Hypothesis, Cramér’s bound gives it for every β>1/2.) Let qn=⌊θun⌋.
using logqn+1<ulog(qn+1) and log(1+1/qn)<1/qn. □
The feature of Theorem 4.3 is that the approximations form a chain: each qn serves as the numerator of one approximation and the denominator of the next. For a single pair, a much stronger estimate holds for all u>0:
Proposition 4.4.Let u>0. For every sufficiently large prime q there is a prime p with
0≤u−logqlogp≤logq2q−0.475u.
Proof. Let X=qu and pick a prime p∈[X−X0.525,X][3]. Then 0≤logX−logp≤−log(1−X−0.475)≤2X−0.475; divide by logq. □
For u>40/19 we have q−0.475u=o(q−1), so Proposition 4.4 is stronger than (18) pair by pair; its proof, however, does not produce chains.
Finally, if 0<v<19/40, applying Theorem 4.3 to u=1/v gives a sequence of primes with
logqn+1logqn−v<qnlogqn+11.
This implies qn>qn+1vexp(−1/qn), and therefore, for every fixed ε>0 and n large,
logqn+1logqn−v<qn+1vlogqn+11+ε.
References
Department of Mathematics and Statistics, Siena University, Loudonville, NY 12211
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