Introduction
For k ≥ 1 k \ge1 k ≥ 1
f k = ∑ Ω ( n ) = k 1 n log n . f_k=\sum_{\Omega(n)=k}\frac{1}{n\log n}. f k = Ω ( n ) = k ∑ n log n 1 . Erdős [4 ] proved that the corresponding sum is bounded uniformly over primitive sets. In particular each f k f_k f k [10 ] proved f k < f 1 f_k<f_1 f k < f 1 k > 1 k>1 k > 1 [11 ] proved f 2 > f 3 f_2>f_3 f 2 > f 3 [1 ] conjectured that
f 1 > f 2 > f 3 > ⋯ . (1) f_1>f_2>f_3>\cdots. \tag*{(1)} f 1 > f 2 > f 3 > ⋯ . (1) Lichtman [2 ] disproved (1) , showing that f k → 1 f_k\to1 f k → 1 k = 6 k=6 k = 6 [3 ] then established
f k = 1 − 2 − k ( a 1 k 2 + O ( k log ( k + 1 ) ) ) , a 1 = d log 2 4 > 0 , (2) f_k=1-2^{-k}(a_1k^2+O(k\log(k+1))),\qquad a_1=\frac{d\log2}{4}>0, \tag*{(2)} f k = 1 − 2 − k ( a 1 k 2 + O ( k log ( k + 1 ))) , a 1 = 4 d log 2 > 0 , (2) where
d = 1 4 ∏ p ≥ 3 ( 1 − 2 p ) − 1 ( 1 − 1 p ) 2 . (3) d=\frac{1}{4}\prod_{p\ge3}\left(1-\frac{2}{p}\right)^{-1}\left(1-\frac{1}{p}\right)^2. \tag*{(3)} d = 4 1 p ≥ 3 ∏ ( 1 − p 2 ) − 1 ( 1 − p 1 ) 2 . (3) Neither of these yields monotonicity beyond the minimum. A global minimum at k = 6 k=6 k = 6 f k < f k + 1 f_k<f_{k+1} f k < f k + 1
Theorem 1.1. For every k ≥ 6 k \ge6 k ≥ 6 f k < f k + 1 f_k<f_{k+1} f k < f k + 1
2020 Mathematics Subject Classification. Primary 11N25; Secondary 11Y60.
Key words and phrases. Erdős sums, almost primes, primitive sets, ball arithmetic.
Let
g k = ∑ n o d d Ω ( n ) = k 1 n log n . g_k=\sum_{\substack{n\ \mathrm{odd}\\ \Omega(n)=k}}\frac{1}{n\log n}. g k = n odd Ω ( n ) = k ∑ n log n 1 . The second theorem shows that deleting the prime 2 2 2
Theorem 1.2. For every k ≥ 1 k\ge1 k ≥ 1 we have g k > g k + 1 g_k>g_{k+1} g k > g k + 1 and g k → 1 2 g_k\to\frac{1}{2} g k → 2 1
The proof of Theorem 1.2 gives the more precise expansion
g k = 1 2 + β 2 − k + R k , ∣ R k ∣ ≤ 4900 ( k + 3 ) 3 3 − k ( k ≥ 59 ) , (4) g_k=\frac{1}{2}+\beta2^{-k}+R_k,\qquad|R_k|\le4900(k+3)^3 3^{-k}\qquad(k\ge59), \tag*{(4)} g k = 2 1 + β 2 − k + R k , ∣ R k ∣ ≤ 4900 ( k + 3 ) 3 3 − k ( k ≥ 59 ) , (4) where γ = 0.5772156649 … \gamma=0.5772156649\ldots γ = 0.5772156649 …
β = d ( γ + log 2 − ∑ p ≥ 3 log p ( p − 1 ) ( p − 2 ) ) > 0. (5) \beta=d\left(\gamma+\log2-\sum_{p\ge3}\frac{\log p}{(p-1)(p-2)}\right)>0. \tag*{(5)} β = d ( γ + log 2 − p ≥ 3 ∑ ( p − 1 ) ( p − 2 ) log p ) > 0. (5) The 2 − k 2^{-k} 2 − k z = 2 z=2 z = 2 z = 3 z=3 z = 3 z = 2 z=2 z = 2 k 2 k^2 k 2 (2) .
This contrast is not an accident of the two cases. In Section 7 we make it a general statement. Let P \mathcal{P} P q ≥ 3 q\ge3 q ≥ 3 Q \mathcal{Q} Q ∑ p ∈ Q ( log p ) / p < ∞ \sum_{p\in\mathcal{Q}}(\log p)/p<\infty ∑ p ∈ Q ( log p ) / p < ∞
f k P = ∏ p ∈ Q ( 1 − 1 p ) + c P 2 − k + o ( 2 − k ) , c P = H P ( 1 , 2 ) ( γ + 1 2 ∂ s log H P ( 1 , 2 ) ) . f_k^{\mathcal{P}}=\prod_{p\in\mathcal{Q}}\left(1-\frac{1}{p}\right)+c_{\mathcal{P}}2^{-k}+o(2^{-k}),\qquad c_{\mathcal{P}}=H_{\mathcal{P}}(1,2)\left(\gamma+\frac{1}{2}\partial_s\log H_{\mathcal{P}}(1,2)\right). f k P = p ∈ Q ∏ ( 1 − p 1 ) + c P 2 − k + o ( 2 − k ) , c P = H P ( 1 , 2 ) ( γ + 2 1 ∂ s log H P ( 1 , 2 ) ) . and Corollary 7.3 gives c P > 0 c_{\mathcal{P}}>0 c P > 0 Q = { 2 } \mathcal{Q}=\{2\} Q = { 2 } z = 2 z=2 z = 2 q ≥ 3 q\ge3 q ≥ 3 P = { p ≡ 1 ( m o d 4 ) } \mathcal{P}=\{p\equiv1\pmod4\} P = { p ≡ 1 ( mod 4 )} 0 0 0
The remaining case is 2 ∈ P 2\in\mathcal{P} 2 ∈ P
The proof has two kinds of input. Identities, coefficient reductions and inequalities are proved in the text. Bounds produced by interval arithmetic are isolated as certified propositions, so that a phrase of the form “the computation gives” never stands in for an analytic step: the exact inequality the computation supplies is stated before it is used. What the interval arithmetic certifies is the computation itself. Transcribing an enclosure into a numeral printed here, and combining several such numerals, is ordinary writing, and we check it by script rather than by ball arithmetic.
Generating functions and the analytic boundary
For s > 1 s>1 s > 1
P k ( s ) = ∑ Ω ( n ) = k n − s , P k o d d ( s ) = ∑ n o d d Ω ( n ) = k n − s . P_k(s)=\sum_{\Omega(n)=k}n^{-s},\qquad P_k^{\mathrm{odd}}(s)=\sum_{\substack{n\ \mathrm{odd}\\ \Omega(n)=k}}n^{-s}. P k ( s ) = Ω ( n ) = k ∑ n − s , P k odd ( s ) = n odd Ω ( n ) = k ∑ n − s . Lemma 2.1. For every k ≥ 1 k \ge1 k ≥ 1
f k = ∫ 1 ∞ P k ( s ) d s , g k = ∫ 1 ∞ P k o d d ( s ) d s . (6) f_k = \int_{1}^{\infty} P_k(s)\,\mathrm{d}s,\qquad g_k = \int_{1}^{\infty} P_k^{\mathrm{odd}}(s)\,\mathrm{d}s. \tag*{(6)} f k = ∫ 1 ∞ P k ( s ) d s , g k = ∫ 1 ∞ P k odd ( s ) d s . (6) Proof. The first identity is [2 (3.1)] : one has ∫ 1 ∞ n − s d s = 1 / ( n log n ) \int_{1}^{\infty} n^{-s}\,\mathrm{d}s = 1/(n\log n) ∫ 1 ∞ n − s d s = 1/ ( n log n ) n ≥ 2 n \ge2 n ≥ 2 n n n
The generating functions are, by [2 Prop. 3.1] ,
∑ k ≥ 0 P k ( s ) z k = ∏ p ( 1 − z p − s ) − 1 , ∑ k ≥ 0 P k o d d ( s ) z k = ∏ p ≥ 3 ( 1 − z p − s ) − 1 . (7) \sum_{k\ge0} P_k(s)z^k = \prod_p(1-zp^{-s})^{-1},\qquad\sum_{k\ge0} P_k^{\mathrm{odd}}(s)z^k = \prod_{p\ge3}(1-zp^{-s})^{-1}. \tag*{(7)} k ≥ 0 ∑ P k ( s ) z k = p ∏ ( 1 − z p − s ) − 1 , k ≥ 0 ∑ P k odd ( s ) z k = p ≥ 3 ∏ ( 1 − z p − s ) − 1 . (7) Write these products as ζ ( s ) z H ( s , z ) \zeta(s)^zH(s,z) ζ ( s ) z H ( s , z ) ζ ( s ) z H o ( s , z ) \zeta(s)^zH_o(s,z) ζ ( s ) z H o ( s , z )
H ( s , z ) = ∏ p ( 1 − z p − s ) − 1 ( 1 − p − s ) z , (8) H(s,z)=\prod_p(1-zp^{-s})^{-1}(1-p^{-s})^z, \tag*{(8)} H ( s , z ) = p ∏ ( 1 − z p − s ) − 1 ( 1 − p − s ) z , (8) H o ( s , z ) = ( 1 − 2 − s ) z ∏ p ≥ 3 ( 1 − z p − s ) − 1 ( 1 − p − s ) z . (9) H_o(s,z)=(1-2^{-s})^z\prod_{p\ge3}(1-zp^{-s})^{-1}(1-p^{-s})^z. \tag*{(9)} H o ( s , z ) = ( 1 − 2 − s ) z p ≥ 3 ∏ ( 1 − z p − s ) − 1 ( 1 − p − s ) z . (9) Proposition 2.2. Let σ = Re s \sigma=\operatorname{Re}s σ = Re s H H H ( s , z ) (s,z) ( s , z ) { σ > 1 / 2 , ∣ z ∣ < 2 σ } \{\sigma>1/2,\ |z|<2^\sigma\} { σ > 1/2 , ∣ z ∣ < 2 σ } H o H_o H o { σ > 1 / 2 , ∣ z ∣ < 3 σ } \{\sigma>1/2,\ |z|<3^\sigma\} { σ > 1/2 , ∣ z ∣ < 3 σ } s > 1 / 2 s>1/2 s > 1/2 H ( s , ⋅ ) H(s,\cdot) H ( s , ⋅ ) ∣ z ∣ < 2 s |z|<2^s ∣ z ∣ < 2 s H o ( s , ⋅ ) H_o(s,\cdot) H o ( s , ⋅ ) ∣ z ∣ < 3 s |z|<3^s ∣ z ∣ < 3 s ∂ s r H o ( s , ⋅ ) \partial_s^rH_o(s,\cdot) ∂ s r H o ( s , ⋅ ) ∣ z ∣ < 3 s |z|<3^s ∣ z ∣ < 3 s
d r d s r [ z j ] H o ( s , z ) = [ z j ] ∂ s r H o ( s , z ) ; (10) \frac{\mathrm{d}^r}{\mathrm{d}s^r}[z^j]H_o(s,z)=[z^j]\partial_s^rH_o(s,z); \tag*{(10)} d s r d r [ z j ] H o ( s , z ) = [ z j ] ∂ s r H o ( s , z ) ; (10) the same holds for H H H ∣ z ∣ < 2 s |z|<2^s ∣ z ∣ < 2 s
Proof. It is essential that s s s z z z
Let x = p − s x=p^{-s} x = p − s ∣ x ∣ = p − σ < 2 − 1 / 2 |x|=p^{-\sigma}<2^{-1/2} ∣ x ∣ = p − σ < 2 − 1/2 ( 1 − x ) z (1-x)^z ( 1 − x ) z 1 − x 1-x 1 − x 2 − 1 / 2 < 1 2^{-1/2}<1 2 − 1/2 < 1
Fix a compact subset of the domain, so that σ ≥ s 0 > 1 / 2 \sigma\ge s_0>1/2 σ ≥ s 0 > 1/2 ∣ z ∣ ≤ R < 2 s 0 |z|\le R<2^{s_0} ∣ z ∣ ≤ R < 2 s 0 R < 3 s 0 R<3^{s_0} R < 3 s 0 p ≥ 3 p\ge3 p ≥ 3 ∣ z x ∣ < 1 |zx|<1 ∣ z x ∣ < 1 p p p p p p
− log ( 1 − z x ) + z log ( 1 − x ) = ∑ m ≥ 1 z m − z m x m = ∑ m ≥ 2 z m − z m x m = O ( p − 2 σ ) , (11) -\log(1-zx)+z\log(1-x)=\sum_{m\ge1}\frac{z^m-z}{m}x^m=\sum_{m\ge2}\frac{z^m-z}{m}x^m=O(p^{-2\sigma}), \tag*{(11)} − log ( 1 − z x ) + z log ( 1 − x ) = m ≥ 1 ∑ m z m − z x m = m ≥ 2 ∑ m z m − z x m = O ( p − 2 σ ) , (11) the m = 1 m=1 m = 1 s s s σ \sigma σ p − s = e − s log p p^{-s}=e^{-s\log p} p − s = e − s l o g p ( s , z ) (s,z) ( s , z ) ∣ z ∣ < 2 σ ≤ p σ = ∣ p s ∣ |z|<2^\sigma\le p^\sigma=|p^s| ∣ z ∣ < 2 σ ≤ p σ = ∣ p s ∣ p p p ∣ z ∣ < 3 σ ≤ ∣ p s ∣ |z|<3^\sigma\le|p^s| ∣ z ∣ < 3 σ ≤ ∣ p s ∣ p ≥ 3 p\ge3 p ≥ 3
As ∑ p p − 2 s 0 < ∞ \sum_p p^{-2s_0}<\infty ∑ p p − 2 s 0 < ∞ C 2 \mathbb{C}^2 C 2 z = p s z=p^s z = p s p σ p^\sigma p σ 2 σ 2^\sigma 2 σ 3 σ 3^\sigma 3 σ [ z j ] [z^j] [ z j ] d / d s \mathrm{d}/\mathrm{d}s d / d s
At s = 1 s=1 s = 1 z = 2 z=2 z = 2 H H H H O H_O H O
Lemma 2.3. Let u = log ζ ( s ) u=\log\zeta(s) u = log ζ ( s ) s > 1 s>1 s > 1 s = s ( u ) s=s(u) s = s ( u ) ϵ ( u ) = s ( u ) − 1 \epsilon(u)=s(u)-1 ϵ ( u ) = s ( u ) − 1 J ( u ) = − s ′ ( u ) J(u)=-s'(u) J ( u ) = − s ′ ( u ) u → ∞ u\to\infty u → ∞
ϵ ( u ) = e − u + γ e − 2 u + O ( e − 3 u ) , J ( u ) = e − u + 2 γ e − 2 u + O ( e − 3 u ) . (12) \epsilon(u)=e^{-u}+\gamma e^{-2u}+O(e^{-3u}), \qquad J(u)=e^{-u}+2\gamma e^{-2u}+O(e^{-3u}). \tag*{(12)} ϵ ( u ) = e − u + γ e − 2 u + O ( e − 3 u ) , J ( u ) = e − u + 2 γ e − 2 u + O ( e − 3 u ) . (12) Proof. The point of the argument is that the expansions below are convergent power series and not merely asymptotic, which is what allows the second of them to be differentiated term by term.
By the Laurent expansion of ζ \zeta ζ [8 Ch. 1] , the function ϵ ↦ ϵ ζ ( 1 + ϵ ) \epsilon\mapsto\epsilon\zeta(1+\epsilon) ϵ ↦ ϵ ζ ( 1 + ϵ ) ϵ = 0 \epsilon=0 ϵ = 0 1 + γ ϵ + O ( ϵ 2 ) 1+\gamma\epsilon+O(\epsilon^2) 1 + γ ϵ + O ( ϵ 2 )
x = 1 ζ ( 1 + ϵ ) = ϵ ϵ ζ ( 1 + ϵ ) x=\frac{1}{\zeta(1+\epsilon)}=\frac{\epsilon}{\epsilon\zeta(1+\epsilon)} x = ζ ( 1 + ϵ ) 1 = ϵ ζ ( 1 + ϵ ) ϵ is analytic near ϵ = 0 \epsilon=0 ϵ = 0 x ( 0 ) = 0 x(0)=0 x ( 0 ) = 0 x = ϵ − γ ϵ 2 + O ( ϵ 3 ) x=\epsilon-\gamma\epsilon^2+O(\epsilon^3) x = ϵ − γ ϵ 2 + O ( ϵ 3 ) d x / d ϵ = 1 ≠ 0 dx/d\epsilon=1\ne0 d x / d ϵ = 1 = 0 ϵ \epsilon ϵ x x x x = 0 x=0 x = 0
ϵ = ψ ( x ) = x + γ x 2 + O ( x 3 ) , \epsilon=\psi(x)=x+\gamma x^2+O(x^3), ϵ = ψ ( x ) = x + γ x 2 + O ( x 3 ) , which is the first formula in (12) once x = e − u x=e^{-u} x = e − u ψ \psi ψ ψ ′ ( x ) = 1 + 2 γ x + O ( x 2 ) \psi'(x)=1+2\gamma x+O(x^2) ψ ′ ( x ) = 1 + 2 γ x + O ( x 2 ) x = e − u x=e^{-u} x = e − u d x / d u = − x dx/du=-x d x / d u = − x
ϵ ′ ( u ) = ψ ′ ( x ) d x d u = − x ( 1 + 2 γ x + O ( x 2 ) ) = − x − 2 γ x 2 + O ( x 3 ) , \epsilon'(u)=\psi'(x)\frac{dx}{du}=-x(1+2\gamma x+O(x^2))=-x-2\gamma x^2+O(x^3), ϵ ′ ( u ) = ψ ′ ( x ) d u d x = − x ( 1 + 2 γ x + O ( x 2 )) = − x − 2 γ x 2 + O ( x 3 ) , and J ( u ) = − ϵ ′ ( u ) = x + 2 γ x 2 + O ( x 3 ) J(u)=-\epsilon'(u)=x+2\gamma x^2+O(x^3) J ( u ) = − ϵ ′ ( u ) = x + 2 γ x 2 + O ( x 3 )
The odd sums
Throughout this section write
h j ( s ) = [ z j ] H O ( s , z ) , b j = h j ′ ( 1 ) + 2 γ h j ( 1 ) . h_j(s)=[z^j]H_O(s,z), \qquad b_j=h_j'(1)+2\gamma h_j(1). h j ( s ) = [ z j ] H O ( s , z ) , b j = h j ′ ( 1 ) + 2 γ h j ( 1 ) . Since ζ ( s ) z = ∑ m ≥ 0 ( log ζ ( s ) ) m z m / m ! \zeta(s)^z=\sum_{m\geq0}(\log\zeta(s))^m z^m/m! ζ ( s ) z = ∑ m ≥ 0 ( log ζ ( s ) ) m z m / m !
P k o d d ( s ) = ∑ j = 0 k h j ( s ) ( log ζ ( s ) ) k − j ( k − j ) ! . (13) P_k^{\mathrm{odd}}(s)=\sum_{j=0}^{k}h_j(s)\frac{(\log\zeta(s))^{k-j}}{(k-j)!}. \tag*{(13)} P k odd ( s ) = j = 0 ∑ k h j ( s ) ( k − j )! ( log ζ ( s ) ) k − j . (13) Using u = log ζ ( s ) u=\log\zeta(s) u = log ζ ( s ) s s s ∞ \infty ∞ 1 1 1 u u u 0 0 0 ∞ \infty ∞
g k = ∑ j = 0 k ∫ 0 ∞ u k − j ( k − j ) ! h j ( s ( u ) ) J ( u ) d u . (14) g_k=\sum_{j=0}^{k}\int_{0}^{\infty}\frac{u^{k-j}}{(k-j)!}\,h_j(s(u))J(u)\,du. \tag*{(14)} g k = j = 0 ∑ k ∫ 0 ∞ ( k − j )! u k − j h j ( s ( u )) J ( u ) d u . (14) The first two terms
Proposition 3.1. With h j h_j h j b j b_j b j E j E_j E j
R k = ∑ j = 0 k ∫ 0 ∞ u k − j ( k − j ) ! E j ( u ) d u − ∑ j > k h j ( 1 ) − 2 − k − 1 ∑ j > k b j 2 j . (15) R_k=\sum_{j=0}^{k}\int_{0}^{\infty}\frac{u^{k-j}}{(k-j)!}E_j(u)\,\mathrm{d}u-\sum_{j>k}h_j(1)-2^{-k-1}\sum_{j>k}b_j2^j. \tag*{(15)} R k = j = 0 ∑ k ∫ 0 ∞ ( k − j )! u k − j E j ( u ) d u − j > k ∑ h j ( 1 ) − 2 − k − 1 j > k ∑ b j 2 j . (15) Then
g k = H o ( 1 , 1 ) + 2 − k − 1 ( ∂ s H o ( 1 , 2 ) + 2 γ H o ( 1 , 2 ) ) + R k . (16) g_k=H_o(1,1)+2^{-k-1}\left(\partial_sH_o(1,2)+2\gamma H_o(1,2)\right)+R_k. \tag*{(16)} g k = H o ( 1 , 1 ) + 2 − k − 1 ( ∂ s H o ( 1 , 2 ) + 2 γ H o ( 1 , 2 ) ) + R k . (16) Proof. The idea is to write the integrand of (14) as its first two terms in e − u e^{-u} e − u J ( u ) d u J(u)\,\mathrm{d}u J ( u ) d u h j ( s ( u ) ) h_j(s(u)) h j ( s ( u )) s = 1 s=1 s = 1 (17) .
Let x = e − u x=e^{-u} x = e − u J ( u ) = x + 2 γ x 2 + O ( x 3 ) J(u)=x+2\gamma x^2+O(x^3) J ( u ) = x + 2 γ x 2 + O ( x 3 ) ε ( u ) J ( u ) = x 2 + O ( x 3 ) \varepsilon(u)J(u)=x^2+O(x^3) ε ( u ) J ( u ) = x 2 + O ( x 3 ) s = 1 s=1 s = 1
h j ( 1 + ε ) = h j ( 1 ) + ε h j ′ ( 1 ) + ε 2 2 h j ′ ′ ( 1 + θ j , u ε ) h_j(1+\varepsilon)=h_j(1)+\varepsilon h_j'(1)+\frac{\varepsilon^2}{2}h_j''(1+\theta_{j,u}\varepsilon) h j ( 1 + ε ) = h j ( 1 ) + ε h j ′ ( 1 ) + 2 ε 2 h j ′′ ( 1 + θ j , u ε ) for some 0 < θ j , u < 1 0<\theta_{j,u}<1 0 < θ j , u < 1
r 0 ( u ) = J ( u ) − e − u − 2 γ e − 2 u , r 1 ( u ) = ε ( u ) J ( u ) − e − 2 u . r_0(u)=J(u)-e^{-u}-2\gamma e^{-2u},\qquad r_1(u)=\varepsilon(u)J(u)-e^{-2u}. r 0 ( u ) = J ( u ) − e − u − 2 γ e − 2 u , r 1 ( u ) = ε ( u ) J ( u ) − e − 2 u . Multiplication by J ( u ) J(u) J ( u )
h j ( s ( u ) ) J ( u ) = h j ( 1 ) e − u + b j e − 2 u + E j ( u ) , h_j(s(u))J(u)=h_j(1)e^{-u}+b_je^{-2u}+E_j(u), h j ( s ( u )) J ( u ) = h j ( 1 ) e − u + b j e − 2 u + E j ( u ) , E j ( u ) = h j ( 1 ) r 0 ( u ) + h j ′ ( 1 ) r 1 ( u ) + ε ( u ) 2 J ( u ) 2 h j ′ ′ ( 1 + θ j , u ε ( u ) ) . (17) E_j(u)=h_j(1)r_0(u)+h_j'(1)r_1(u)+\frac{\varepsilon(u)^2J(u)}{2}h_j''(1+\theta_{j,u}\varepsilon(u)). \tag*{(17)} E j ( u ) = h j ( 1 ) r 0 ( u ) + h j ′ ( 1 ) r 1 ( u ) + 2 ε ( u ) 2 J ( u ) h j ′′ ( 1 + θ j , u ε ( u )) . (17) No error is suppressed in the notation E j E_j E j
Insert (17) into (14) . Since
∫ 0 ∞ u m m ! e − r u d u = r − m − 1 ( r > 0 ) , \int_0^\infty\frac{u^m}{m!}e^{-ru}\,\mathrm{d}u=r^{-m-1}\qquad(r>0), ∫ 0 ∞ m ! u m e − r u d u = r − m − 1 ( r > 0 ) , the first two terms become
∑ j = 0 k h j ( 1 ) = H o ( 1 , 1 ) − ∑ j > k h j ( 1 ) , (18) \sum_{j=0}^{k}h_j(1)=H_o(1,1)-\sum_{j>k}h_j(1), \tag*{(18)} j = 0 ∑ k h j ( 1 ) = H o ( 1 , 1 ) − j > k ∑ h j ( 1 ) , (18) ∑ j = 0 k b j 2 − ( k − j + 1 ) = 2 − k − 1 ( ∂ s H o ( 1 , 2 ) + 2 γ H o ( 1 , 2 ) ) − 2 − k − 1 ∑ j > k b j 2 j . (19) \sum_{j=0}^{k}b_j2^{-(k-j+1)}
=2^{-k-1}\left(\partial_sH_o(1,2)+2\gamma H_o(1,2)\right)-2^{-k-1}\sum_{j>k}b_j2^j. \tag*{(19)} j = 0 ∑ k b j 2 − ( k − j + 1 ) = 2 − k − 1 ( ∂ s H o ( 1 , 2 ) + 2 γ H o ( 1 , 2 ) ) − 2 − k − 1 j > k ∑ b j 2 j . (19) Substituting (18) and (19) , and collecting the E j E_j E j (15) , and this gives (16) . Note that (15) is an identity, so every later estimate is an estimate of those three explicit quantities. □
Lemma 3.2. One has
H o ( 1 , 1 ) = 1 2 , H o ( 1 , 2 ) = d , (20) H_o(1,1)=\frac{1}{2},\qquad H_o(1,2)=d, \tag*{(20)} H o ( 1 , 1 ) = 2 1 , H o ( 1 , 2 ) = d , (20) and
∂ s log H o ( 1 , 2 ) = 2 log 2 − 2 ∑ p ≥ 3 log p ( p − 1 ) ( p − 2 ) . (21) \partial_s\log H_o(1,2)=2\log2-2\sum_{p\geq3}\frac{\log p}{(p-1)(p-2)}. \tag*{(21)} ∂ s log H o ( 1 , 2 ) = 2 log 2 − 2 p ≥ 3 ∑ ( p − 1 ) ( p − 2 ) log p . (21) Proof. The first two values are immediate from (9): at s = z = 1 s=z=1 s = z = 1 p ≥ 3 p\ge3 p ≥ 3 H o ( 1 , 1 ) = 1 − 2 − 1 = 1 2 H_o(1,1)=1-2^{-1}=\frac{1}{2} H o ( 1 , 1 ) = 1 − 2 − 1 = 2 1 ( s , z ) = ( 1 , 2 ) (s,z)=(1,2) ( s , z ) = ( 1 , 2 ) H o ( 1 , 2 ) = d H_o(1,2)=d H o ( 1 , 2 ) = d
For the derivative, take logarithms:
log H o ( s , 2 ) = 2 log ( 1 − 2 − s ) + ∑ p ≥ 3 [ − log ( 1 − 2 p − s ) + 2 log ( 1 − p − s ) ] \log H_o(s,2)=2\log(1-2^{-s})+\sum_{p\ge3}\left[-\log(1-2p^{-s})+2\log(1-p^{-s})\right] log H o ( s , 2 ) = 2 log ( 1 − 2 − s ) + p ≥ 3 ∑ [ − log ( 1 − 2 p − s ) + 2 log ( 1 − p − s ) ] The differentiated p p p O ( ( log p ) p − 2 s ) O((\log p)p^{-2s}) O (( log p ) p − 2 s ) s = 1 s=1 s = 1 s = 1 s=1 s = 1 2 log 2 2\log2 2 log 2 p p p
− 2 log p p − 2 + 2 log p p − 1 = − 2 log p ( p − 1 ) ( p − 2 ) . -\frac{2\log p}{p-2}+\frac{2\log p}{p-1}=-\frac{2\log p}{(p-1)(p-2)}. − p − 2 2 log p + p − 1 2 log p = − ( p − 1 ) ( p − 2 ) 2 log p . Summing over p ≥ 3 p\ge3 p ≥ 3
Combining Proposition 3.1 and Lemma 3.2 gives
g k = 1 2 + β 2 − k + R k , β = d ( γ + log 2 − ∑ p ≥ 3 log p ( p − 1 ) ( p − 2 ) ) . (22) g_k=\frac{1}{2}+\beta2^{-k}+R_k,\qquad\beta=d\left(\gamma+\log2-\sum_{p\ge3}\frac{\log p}{(p-1)(p-2)}\right). \tag*{(22)} g k = 2 1 + β 2 − k + R k , β = d ( γ + log 2 − p ≥ 3 ∑ ( p − 1 ) ( p − 2 ) log p ) . (22) The remainder
Three weighted ℓ 1 \ell^1 ℓ 1 s ( 3 ) = s ( u ) ∣ u = 3 = 1.0512 … s(3)=s(u)|_{u=3}=1.0512\ldots s ( 3 ) = s ( u ) ∣ u = 3 = 1.0512 … s s s log ζ ( s ) = 3 \log\zeta(s)=3 log ζ ( s ) = 3
Σ 0 = ∑ m ≥ 0 3 m ∣ [ z m ] ( 1 − z / 3 ) H o ( 1 , z ) ∣ , \Sigma_0=\sum_{m\ge0}3^m\left|[z^m](1-z/3)H_o(1,z)\right|, Σ 0 = m ≥ 0 ∑ 3 m ∣ [ z m ] ( 1 − z /3 ) H o ( 1 , z ) ∣ , Σ 1 = ∑ m ≥ 0 3 m ∣ [ z m ] ( 1 − z / 3 ) 2 ∂ s H o ( 1 , z ) ∣ , \Sigma_1=\sum_{m\ge0}3^m\left|[z^m](1-z/3)^2\partial_sH_o(1,z)\right|, Σ 1 = m ≥ 0 ∑ 3 m [ z m ] ( 1 − z /3 ) 2 ∂ s H o ( 1 , z ) , Σ 2 = sup 1 ≤ s ≤ s ( 3 ) ( ∑ m ≥ 0 3 m ∣ [ z m ] ( 1 − z / 3 ) 3 ∂ s 2 H o ( s , z ) ∣ ) . (23) \Sigma_2=\sup_{1\le s\le s(3)}\left(\sum_{m\ge0}3^m\left|[z^m](1-z/3)^3\partial_s^2H_o(s,z)\right|\right). \tag*{(23)} Σ 2 = 1 ≤ s ≤ s ( 3 ) sup ( m ≥ 0 ∑ 3 m [ z m ] ( 1 − z /3 ) 3 ∂ s 2 H o ( s , z ) ) . (23) Finiteness of Σ 0 \Sigma_0 Σ 0 Σ 1 \Sigma_1 Σ 1 r = 1 , 2 r=1,2 r = 1 , 2 ( 1 − z / 3 ) r (1-z/3)^r ( 1 − z /3 ) r H o ( 1 , ⋅ ) H_o(1,\cdot) H o ( 1 , ⋅ ) ∣ z ∣ < 3 |z|<3 ∣ z ∣ < 3 3 m 3^m 3 m s = 1 s=1 s = 1 z z z z = 3 z=3 z = 3 ( 1 − z / 3 ) − 1 (1-z/3)^{-1} ( 1 − z /3 ) − 1 r − 1 r-1 r − 1 s s s r r r ( 1 − z / 3 ) r (1-z/3)^r ( 1 − z /3 ) r ∣ z ∣ < 5 |z|<5 ∣ z ∣ < 5 z = 5 z=5 z = 5 5 − m 5^{-m} 5 − m
For Σ 2 \Sigma_2 Σ 2 s > 1 s>1 s > 1 z = 3 s z=3^s z = 3 s ( 1 − z / 3 ) 3 (1-z/3)^3 ( 1 − z /3 ) 3 3 s > 3 3^s>3 3 s > 3 s → 1 + s\to1^+ s → 1 + s s s s = 1 s=1 s = 1 3 / 3 s < 1 3/3^s<1 3/ 3 s < 1 s > 1 s>1 s > 1 [ 1 , s ( 3 ) ] [1,s(3)] [ 1 , s ( 3 )] Σ 0 \Sigma_0 Σ 0 Σ 1 \Sigma_1 Σ 1
Next, split the remainder of Proposition 3.1 at u = 3 u=3 u = 3 E j E_j E j
R k , f a r = ∑ j = 0 k ∫ 3 ∞ u k − j ( k − j ) ! E j ( u ) d u , R k , n e a r = ∑ j = 0 k ∫ 0 3 u k − j ( k − j ) ! E j ( u ) d u , (24) R_{k,\mathrm{far}}=\sum_{j=0}^{k}\int_{3}^{\infty}\frac{u^{k-j}}{(k-j)!}E_j(u)\,du,\qquad R_{k,\mathrm{near}}=\sum_{j=0}^{k}\int_{0}^{3}\frac{u^{k-j}}{(k-j)!}E_j(u)\,du, \tag*{(24)} R k , far = j = 0 ∑ k ∫ 3 ∞ ( k − j )! u k − j E j ( u ) d u , R k , near = j = 0 ∑ k ∫ 0 3 ( k − j )! u k − j E j ( u ) d u , (24) so that R k R_k R k R k , f a r + R k , n e a r R_{k,\mathrm{far}}+R_{k,\mathrm{near}} R k , far + R k , near (18) –(19) .
The far range needs one further constant. Write ε = s − 1 \varepsilon=s-1 ε = s − 1 J = ζ ( s ) / ( − ζ ′ ( s ) ) J=\zeta(s)/(-\zeta'(s)) J = ζ ( s ) / ( − ζ ′ ( s )) J J J Lemma 2.3 expressed in s s s u u u d u / d s = ζ ′ ( s ) / ζ ( s ) \mathrm{d}u/\mathrm{d}s=\zeta'(s)/\zeta(s) d u / d s = ζ ′ ( s ) / ζ ( s ) − s ′ ( u ) -s'(u) − s ′ ( u ) ζ ( s ) / ( − ζ ′ ( s ) ) \zeta(s)/(-\zeta'(s)) ζ ( s ) / ( − ζ ′ ( s )) ζ − 1 = e − u \zeta^{-1}=e^{-u} ζ − 1 = e − u
C b = sup ∣ ε ζ ∣ , C f = sup ∣ ζ 2 ( J − ζ − 1 ) ∣ , C J = sup ∣ ζ 3 ( J − ζ − 1 − 2 γ ζ − 2 ) ∣ , C P = sup ∣ ζ 3 ( ε J − ζ − 2 ) ∣ , C E = C J + C P + C b 2 ( C f + 1 ) , (25) \begin{aligned}
C_b&=\sup|\varepsilon\zeta|,\qquad C_f=\sup|\zeta^2(J-\zeta^{-1})|,\\
C_J&=\sup|\zeta^3(J-\zeta^{-1}-2\gamma\zeta^{-2})|,\qquad C_P=\sup|\zeta^3(\varepsilon J-\zeta^{-2})|,\\
C_E&=C_J+C_P+C_b^2(C_f+1), \tag*{(25)}
\end{aligned} C b C J C E = sup ∣ εζ ∣ , C f = sup ∣ ζ 2 ( J − ζ − 1 ) ∣ , = sup ∣ ζ 3 ( J − ζ − 1 − 2 γ ζ − 2 ) ∣ , C P = sup ∣ ζ 3 ( ε J − ζ − 2 ) ∣ , = C J + C P + C b 2 ( C f + 1 ) , (25) every supremum being over 1 < s ≤ s ( 3 ) 1<s\le s(3) 1 < s ≤ s ( 3 ) ζ \zeta ζ s = 1 s=1 s = 1 § 8 : evaluating them on boxes of s s s
Proposition 3.3 (Certified odd input). With Σ i \Sigma_i Σ i (23) and C E C_E C E (25) , the following are certified by interval arithmetic.
(i) The coefficient norms satisfy
Σ 0 ≤ 44.5987 , Σ 1 ≤ 486.1946 , Σ 2 ≤ 14414 , C E ≤ 5.48910. (26) \Sigma_0\le44.5987,\qquad\Sigma_1\le486.1946,\qquad\Sigma_2\le14414,\qquad C_E\le5.48910. \tag*{(26)} Σ 0 ≤ 44.5987 , Σ 1 ≤ 486.1946 , Σ 2 ≤ 14414 , C E ≤ 5.48910. (26) (ii) On the far range,
∑ j = 0 k ∫ 3 ∞ u k − j ( k − j ) ! ∣ E j ( u ) ∣ d u ≤ C E Σ 2 18 ( k + 3 ) 3 3 − k . (27) \sum_{j=0}^{k}\int_{3}^{\infty}\frac{u^{k-j}}{(k-j)!}|E_j(u)|\,\mathrm{d}u\le\frac{C_E\Sigma_2}{18}(k+3)^3 3^{-k}. \tag*{(27)} j = 0 ∑ k ∫ 3 ∞ ( k − j )! u k − j ∣ E j ( u ) ∣ d u ≤ 18 C E Σ 2 ( k + 3 ) 3 3 − k . (27) (iii) On the near range the two groups of terms in (17) are bounded separately:
∑ j = 0 k ∫ 0 3 u k − j ( k − j ) ! ∣ h j ( s ( u ) ) ∣ ∣ J ( u ) ∣ d u ≤ C N 3 − k ( 59 ≤ k ≤ 200 ) , C N ≤ 367.8436 , (28) \sum_{j=0}^{k}\int_{0}^{3}\frac{u^{k-j}}{(k-j)!}|h_j(s(u))||J(u)|\,\mathrm{d}u\le C_N3^{-k}\qquad(59\le k\le200),\quad C_N\le367.8436, \tag*{(28)} j = 0 ∑ k ∫ 0 3 ( k − j )! u k − j ∣ h j ( s ( u )) ∣∣ J ( u ) ∣ d u ≤ C N 3 − k ( 59 ≤ k ≤ 200 ) , C N ≤ 367.8436 , (28) ∑ j = 0 k ∫ 0 3 u k − j ( k − j ) ! ( ∣ h j ( 1 ) ∣ e − u + ∣ b j ∣ e − 2 u ) d u ≤ C U ( k + 3 ) 3 3 − k , C U ≤ 26.0962. (29) \sum_{j=0}^{k}\int_{0}^{3}\frac{u^{k-j}}{(k-j)!}\left(|h_j(1)|e^{-u}+|b_j|e^{-2u}\right)\,\mathrm{d}u\le C_U(k+3)^3 3^{-k},\qquad C_U\le26.0962. \tag*{(29)} j = 0 ∑ k ∫ 0 3 ( k − j )! u k − j ( ∣ h j ( 1 ) ∣ e − u + ∣ b j ∣ e − 2 u ) d u ≤ C U ( k + 3 ) 3 3 − k , C U ≤ 26.0962. (29) (iv) β > 0 \beta>0 β > 0
4900 ( 62 ) 3 3 − 59 ≤ β 3 2 − 59 . (30) 4900(62)^3 3^{-59}\le\frac{\beta}{3}2^{-59}. \tag*{(30)} 4900 ( 62 ) 3 3 − 59 ≤ 3 β 2 − 59 . (30) (v) Direct interval evaluation encloses g k g_k g k 1 ≤ k ≤ 75 1\le k\le75 1 ≤ k ≤ 75 g k > g k + 1 g_k>g_{k+1} g k > g k + 1 1 ≤ k ≤ 74 1\le k\le74 1 ≤ k ≤ 74 g 74 g_{74} g 74 g 75 g_{75} g 75 g 76 g_{76} g 76
Part (v) is a sweep, not a coefficient bound. The range in (28) is two-sided, and its ends are discharged differently: the lower one is harmless, monotonicity below the analytic threshold coming from (v) and never appealing to (32) , while the upper one is removed in the proof of Proposition 3.5, where a Cauchy estimate extends the bound past k = 200 k=200 k = 200
The lower restriction is a feature of the method, not of the inequality. The h j h_j h j a ( s ) = 2 ∣ h 1 ( s ) ∣ a(s)=2|h_1(s)| a ( s ) = 2∣ h 1 ( s ) ∣ s s s k k k u k − j / ( k − j ) ! u^{k-j}/(k-j)! u k − j / ( k − j )! C N 3 − k C_N3^{-k} C N 3 − k 9 ≤ k ≤ 25 9\le k\le25 9 ≤ k ≤ 25 699.87 ⋅ 3 − k 699.87\cdot3^{-k} 699.87 ⋅ 3 − k k = 14 k=14 k = 14 34.14 ⋅ 3 − k 34.14\cdot3^{-k} 34.14 ⋅ 3 − k k = 59 k=59 k = 59 k ≥ 26 k\ge26 k ≥ 26 k ≥ 59 k\ge59 k ≥ 59 (32) is applied. The numeral C N ≤ 367.8436 C_N\le367.8436 C N ≤ 367.8436 34.14 34.14 34.14 k ≥ 59 k\ge59 k ≥ 59 sup k ≥ 1 3 k P k o d d ( s ( 3 ) ) \sup_{k\ge1} 3^kP_k^{\mathrm{odd}}(s(3)) sup k ≥ 1 3 k P k odd ( s ( 3 )) 367.84350 … 367.84350\ldots 367.84350 … k = 11 k=11 k = 11
Lemma 3.4. Let A ( z ) = ∑ j ≥ 0 a j z j A(z)=\sum_{j\ge0}a_jz^j A ( z ) = ∑ j ≥ 0 a j z j ∑ m 3 m ∣ [ z m ] ( 1 − z / 3 ) A ( z ) ∣ ≤ S \sum_m 3^m\lvert[z^m](1-z/3)A(z)\rvert\le S ∑ m 3 m ∣[ z m ] ( 1 − z /3 ) A ( z )∣ ≤ S ∣ a j ∣ ≤ S 3 − j \lvert a_j\rvert\le S3^{-j} ∣ a j ∣ ≤ S 3 − j ∑ m 3 m ∣ [ z m ] ( 1 − z / 3 ) 2 A ( z ) ∣ ≤ S \sum_m 3^m\lvert[z^m](1-z/3)^2A(z)\rvert\le S ∑ m 3 m ∣[ z m ] ( 1 − z /3 ) 2 A ( z )∣ ≤ S ∣ a j ∣ ≤ S ( j + 1 ) 3 − j \lvert a_j\rvert\le S(j+1)3^{-j} ∣ a j ∣ ≤ S ( j + 1 ) 3 − j ∑ m 3 m ∣ [ z m ] ( 1 − z / 3 ) 3 A ( z ) ∣ ≤ S \sum_m 3^m\lvert[z^m](1-z/3)^3A(z)\rvert\le S ∑ m 3 m ∣[ z m ] ( 1 − z /3 ) 3 A ( z )∣ ≤ S
∑ j = 0 k 3 j ∣ a j ∣ ≤ ( k + 3 3 ) S . (31) \sum_{j=0}^{k}3^j\lvert a_j\rvert\le\binom{k+3}{3}S. \tag*{(31)} j = 0 ∑ k 3 j ∣ a j ∣ ≤ ( 3 k + 3 ) S . (31) Sketch. Write ( 1 − z / 3 ) r A ( z ) = ∑ m w m z m (1-z/3)^rA(z)=\sum_m w_mz^m ( 1 − z /3 ) r A ( z ) = ∑ m w m z m ( 1 − z / 3 ) − r (1-z/3)^{-r} ( 1 − z /3 ) − r A A A ∑ q ( q + r − 1 r − 1 ) 3 − q z q \sum_q\binom{q+r-1}{r-1}3^{-q}z^q ∑ q ( r − 1 q + r − 1 ) 3 − q z q ∑ m w m z m \sum_m w_mz^m ∑ m w m z m j ≤ k j\le k j ≤ k ∑ q ≤ k − m ( q + 2 2 ) = ( k − m + 3 3 ) ≤ ( k + 3 3 ) \sum_{q\le k-m}\binom{q+2}{2}=\binom{k-m+3}{3}\le\binom{k+3}{3} ∑ q ≤ k − m ( 2 q + 2 ) = ( 3 k − m + 3 ) ≤ ( 3 k + 3 )
Proposition 3.5. For every k ≥ 59 k\ge59 k ≥ 59
∣ R k ∣ ≤ 4900 ( k + 3 ) 3 3 − k . (32) \lvert R_k\rvert\le4900(k+3)^3 3^{-k}. \tag*{(32)} ∣ R k ∣ ≤ 4900 ( k + 3 ) 3 3 − k . (32) The threshold is not an artefact of the method but of the near range: the coefficients h j h_j h j (36) , and the resulting bound exceeds (32) for small k k k k ≤ 75 k\le75 k ≤ 75 (32) ; see the proof of Theorem 1.2 .
Proof. By (15) the remainder has four parts: the error integrals ∫ E j \int E_j ∫ E j u ≥ 3 u\ge3 u ≥ 3 0 < u ≤ 3 0<u\le3 0 < u ≤ 3 (18) –(19) . Each has to be brought to the common shape ( k + 3 ) 3 3 − k (k+3)^33^{-k} ( k + 3 ) 3 3 − k
The two tails are the easiest: they are geometric series in the h j h_j h j b j b_j b j (27) controls it pointwise, and the only real task is to see how the three summands of (17) depend on k k k k ≥ 59 k\ge59 k ≥ 59 h j h_j h j evaluate the sum instead bounds it from the wrong side, and one is pushed onto a majorant that is loosest precisely where the weights put most of their mass. We take the four in that order.
Two remarks before starting. The far term and the near-range main term are cubic in k k k 3 − k 3^{-k} 3 − k ( k + 3 ) 3 ≥ 1 (k+3)^3\ge1 ( k + 3 ) 3 ≥ 1
Far range $u\ge 3. The exact error here is the sum of the three terms of (17) , so the certified Taylor model of (27) applies to the integrand as it stands. What that computation cannot supply is the dependence on k k k (17) call on three different parts of it, since they contain h j ( 1 ) h_j(1) h j ( 1 ) h j ′ ( 1 ) h_j'(1) h j ′ ( 1 ) ∂ s 2 h j \partial_s^2h_j ∂ s 2 h j u ≥ 3 u\ge3 u ≥ 3 O ( e − 3 u ) O(e^{-3u}) O ( e − 3 u ) ζ \zeta ζ (25) encode, and ∫ 0 ∞ u m e − 3 u d u / m ! = 3 − m − 1 \int_0^\infty u^me^{-3u}\,\mathrm{d}u/m!=3^{-m-1} ∫ 0 ∞ u m e − 3 u d u / m ! = 3 − m − 1
∑ j ≤ k 3 j ∣ h j ( 1 ) ∣ ≤ ( k + 1 ) Σ 0 , ∑ j ≤ k 3 j ∣ h j ′ ( 1 ) ∣ ≤ ( k + 2 2 ) Σ 1 , ∑ j ≤ k 3 j ∣ ∂ s 2 h j ∣ ≤ ( k + 3 3 ) Σ 2 , \sum_{j\le k}3^j|h_j(1)|\le(k+1)\Sigma_0,\qquad\sum_{j\le k}3^j|h'_j(1)|\le\binom{k+2}{2}\Sigma_1,\qquad\sum_{j\le k}3^j|\partial_s^2h_j|\le\binom{k+3}{3}\Sigma_2, j ≤ k ∑ 3 j ∣ h j ( 1 ) ∣ ≤ ( k + 1 ) Σ 0 , j ≤ k ∑ 3 j ∣ h j ′ ( 1 ) ∣ ≤ ( 2 k + 2 ) Σ 1 , j ≤ k ∑ 3 j ∣ ∂ s 2 h j ∣ ≤ ( 3 k + 3 ) Σ 2 , so with C J C_J C J C P C_P C P C b 2 ( C f + 1 ) C_b^2(C_f+1) C b 2 ( C f + 1 ) (25) , the last of them halved by the Taylor remainder’s 1 2 ε 2 \frac{1}{2}\varepsilon^2 2 1 ε 2
3 − k 3 ( C J ( k + 1 ) Σ 0 + C P ( k + 2 2 ) Σ 1 + 1 2 C b 2 ( C f + 1 ) ( k + 3 3 ) Σ 2 ) . \frac{3^{-k}}{3}\left(C_J(k+1)\Sigma_0+C_P\binom{k+2}{2}\Sigma_1+\frac{1}{2}C_b^2(C_f+1)\binom{k+3}{3}\Sigma_2\right). 3 3 − k ( C J ( k + 1 ) Σ 0 + C P ( 2 k + 2 ) Σ 1 + 2 1 C b 2 ( C f + 1 ) ( 3 k + 3 ) Σ 2 ) . Only the third term is cubic in k k k ( C J + C P + 1 2 C b 2 ( C f + 1 ) ) ( k + 3 3 ) Σ 2 \left(C_J+C_P+\frac{1}{2}C_b^2(C_f+1)\right)\binom{k+3}{3}\Sigma_2 ( C J + C P + 2 1 C b 2 ( C f + 1 ) ) ( 3 k + 3 ) Σ 2 C E ( k + 3 3 ) Σ 2 C_E\binom{k+3}{3}\Sigma_2 C E ( 3 k + 3 ) Σ 2 1 2 C b 2 ( C f + 1 ) < C b 2 ( C f + 1 ) \frac{1}{2}C_b^2(C_f+1)<C_b^2(C_f+1) 2 1 C b 2 ( C f + 1 ) < C b 2 ( C f + 1 )
∣ R k , f a r ∣ ≤ C E Σ 2 18 ( k + 3 ) 3 3 − k , (33) |R_{k,\mathrm{far}}|\le\frac{C_E\Sigma_2}{18}(k+3)^3 3^{-k}, \tag*{(33)} ∣ R k , far ∣ ≤ 18 C E Σ 2 ( k + 3 ) 3 3 − k , (33) the 18 = 6 ⋅ 3 18=6\cdot3 18 = 6 ⋅ 3 ( k + 3 3 ) ≤ ( k + 3 ) 3 / 6 \binom{k+3}{3}\le(k+3)^3/6 ( 3 k + 3 ) ≤ ( k + 3 ) 3 /6
Collapsing the three constants into C E C_E C E 7.1 ⋅ 10 4 7.1\cdot10^4 7.1 ⋅ 1 0 4 k = 1 k=1 k = 1 8.4 ⋅ 10 5 8.4\cdot10^5 8.4 ⋅ 1 0 5 (33) , a factor of 12; the factor falls to 4.9 by k = 59 k=59 k = 59 C E / ( 1 2 C b 2 ( C f + 1 ) ) = 4.67 C_E/(\frac{1}{2}C_b^2(C_f+1))=4.67 C E / ( 2 1 C b 2 ( C f + 1 )) = 4.67 Proposition 3.5 closes comfortably with it, and because one constant in the statement is easier to check than three; the slack is recorded here rather than spent.
First tail. By Lemma 3.4 (i),
∣ ∑ j > k h j ( 1 ) ∣ ≤ Σ 0 ∑ j > k 3 − j = Σ 0 2 3 − k . \left|\sum_{j>k}h_j(1)\right|\le\Sigma_0\sum_{j>k}3^{-j}=\frac{\Sigma_0}{2}3^{-k}. j > k ∑ h j ( 1 ) ≤ Σ 0 j > k ∑ 3 − j = 2 Σ 0 3 − k . Second tail. Put B ( z ) = ∂ s H o ( 1 , z ) + 2 γ H o ( 1 , z ) B(z)=\partial_sH_o(1,z)+2\gamma H_o(1,z) B ( z ) = ∂ s H o ( 1 , z ) + 2 γ H o ( 1 , z ) ( 1 − z / 3 ) 2 (1-z/3)^2 ( 1 − z /3 ) 2 Σ 1 \Sigma_1 Σ 1 Proposition 3.3 . For the second, write ( 1 − z / 3 ) 2 H o = ( 1 − z / 3 ) ⋅ [ ( 1 − z / 3 ) H o ] (1-z/3)^2H_o=(1-z/3)\cdot[(1-z/3)H_o] ( 1 − z /3 ) 2 H o = ( 1 − z /3 ) ⋅ [( 1 − z /3 ) H o ] ℓ 1 \ell^1 ℓ 1 ∑ m 3 m ∣ [ z m ] ⋅ ∣ \sum_m3^m|[z^m]\cdot| ∑ m 3 m ∣ [ z m ] ⋅ ∣ ( 1 − z / 3 ) (1-z/3) ( 1 − z /3 ) 2 γ ⋅ 2 Σ 0 = 4 γ Σ 0 2\gamma\cdot2\Sigma_0=4\gamma\Sigma_0 2 γ ⋅ 2 Σ 0 = 4 γ Σ 0 γ < 0.578 \gamma<0.578 γ < 0.578
∑ m ≥ 0 3 m ∣ [ z m ] ( 1 − z / 3 ) 2 B ( z ) ∣ ≤ Σ 1 + 4 γ Σ 0 < 590 , \sum_{m\ge0}3^m|[z^m](1-z/3)^2B(z)|\le\Sigma_1+4\gamma\Sigma_0<590, m ≥ 0 ∑ 3 m ∣ [ z m ] ( 1 − z /3 ) 2 B ( z ) ∣ ≤ Σ 1 + 4 γ Σ 0 < 590 , so ∣ b j ∣ ≤ 590 ( j + 1 ) 3 − j |b_j|\le590(j+1)3^{-j} ∣ b j ∣ ≤ 590 ( j + 1 ) 3 − j Lemma 3.4 (ii). Summing the resulting geometric series,
2 − k − 1 ∑ j > k ∣ b j ∣ 2 j ≤ 590 ⋅ 2 − k − 1 ∑ j > k ( j + 1 ) ( 2 / 3 ) j = 590 ( k + 4 ) 3 − k , 2^{-k-1}\sum_{j>k}|b_j|2^j\le590\cdot2^{-k-1}\sum_{j>k}(j+1)(2/3)^j=590(k+4)3^{-k}, 2 − k − 1 j > k ∑ ∣ b j ∣ 2 j ≤ 590 ⋅ 2 − k − 1 j > k ∑ ( j + 1 ) ( 2/3 ) j = 590 ( k + 4 ) 3 − k , and ( k + 4 ) ≤ 5 64 ( k + 3 ) 3 (k+4)\le\frac{5}{64}(k+3)^3 ( k + 4 ) ≤ 64 5 ( k + 3 ) 3 k ≥ 1 k\ge1 k ≥ 1 k = 1 k=1 k = 1
Near range 0 < u ≤ 3 0<u\le3 0 < u ≤ 3 This is the delicate part. Here ∣ E j ∣ |E_j| ∣ E j ∣ N U ( k + 3 ) 3 3 − k N_U(k+3)^3 3^{-k} N U ( k + 3 ) 3 3 − k (29) . For the former, (13) and the substitution s = s ( u ) s=s(u) s = s ( u ) u = log ζ ( s ) u=\log\zeta(s) u = log ζ ( s ) J ( u ) d u = d s J(u)\,\mathrm{d}u=\mathrm{d}s J ( u ) d u = d s
∑ j = 0 k ∫ 0 3 u k − j ( k − j ) ! ∣ h j ( s ( u ) ) ∣ J ( u ) d u = ∫ s ( 3 ) ∞ A k ( s ) d s , A k ( s ) = ∑ j = 0 k ( log ζ ( s ) ) k − j ( k − j ) ! ∣ h j ( s ) ∣ . (34) \sum_{j=0}^{k}\int_0^3\frac{u^{k-j}}{(k-j)!}|h_j(s(u))|J(u)\,\mathrm{d}u=\int_{s(3)}^\infty A_k(s)\,\mathrm{d}s,\qquad A_k(s)=\sum_{j=0}^{k}\frac{(\log\zeta(s))^{k-j}}{(k-j)!}|h_j(s)|. \tag*{(34)} j = 0 ∑ k ∫ 0 3 ( k − j )! u k − j ∣ h j ( s ( u )) ∣ J ( u ) d u = ∫ s ( 3 ) ∞ A k ( s ) d s , A k ( s ) = j = 0 ∑ k ( k − j )! ( log ζ ( s ) ) k − j ∣ h j ( s ) ∣. (34) It is A k A_k A k Π ( t ) = ∑ p ≥ 3 p − t \Pi(t)=\sum_{p\ge3}p^{-t} Π ( t ) = ∑ p ≥ 3 p − t z z z log H o \log H_o log H o (9) is
h 1 ( s ) = Π ( s ) − log ζ ( s ) = log ( 1 − 2 − s ) − ∑ m ≥ 2 Π ( m s ) m < 0 , (35) h_1(s)=\Pi(s)-\log\zeta(s)=\log(1-2^{-s})-\sum_{m\ge2}\frac{\Pi(ms)}{m}<0, \tag*{(35)} h 1 ( s ) = Π ( s ) − log ζ ( s ) = log ( 1 − 2 − s ) − m ≥ 2 ∑ m Π ( m s ) < 0 , (35) both terms being negative. So the h j h_j h j ∑ j u k − j h j ( s ) / ( k − j ) ! = P k o d d ( s ) \sum_j u^{k-j}h_j(s)/(k-j)! = P_k^{\mathrm{odd}}(s) ∑ j u k − j h j ( s ) / ( k − j )! = P k odd ( s ) A k A_k A k (32) .
We majorize instead. What makes this cheap is that, of the coefficients of log H o \log H_o log H o (35) every coefficient of log H o ( s , z ) = h 1 ( s ) z + ∑ m ≥ 2 Π ( m s ) z m / m \log H_o(s,z)=h_1(s)z+\sum_{m\ge2}\Pi(ms)z^m/m log H o ( s , z ) = h 1 ( s ) z + ∑ m ≥ 2 Π ( m s ) z m / m h 1 h_1 h 1 ∣ h 1 ∣ \lvert h_1\rvert ∣ h 1 ∣
H o + ( s , z ) = H o ( s , z ) e − 2 h 1 ( s ) z , H_o^{+}(s,z)=H_o(s,z)e^{-2h_1(s)z}, H o + ( s , z ) = H o ( s , z ) e − 2 h 1 ( s ) z , a power series with non-negative coefficients dominating ∣ h j ∣ \lvert h_j\rvert ∣ h j ∣ a ( s ) = 2 ∣ h 1 ( s ) ∣ a(s)=2\lvert h_1(s)\rvert a ( s ) = 2 ∣ h 1 ( s )∣
A k ( s ) ≤ [ z k ] ( ζ ( s ) H o + ( s , z ) ) = ∑ i = 0 k P k − i o d d ( s ) a ( s ) i i ! = : P k + ( s ) . (36) A_k(s)\le[z^k]\left(\zeta(s)^{}H_o^{+}(s,z)\right)=\sum_{i=0}^{k}P_{k-i}^{\mathrm{odd}}(s)\frac{a(s)^i}{i!}=:P_k^{+}(s). \tag*{(36)} A k ( s ) ≤ [ z k ] ( ζ ( s ) H o + ( s , z ) ) = i = 0 ∑ k P k − i odd ( s ) i ! a ( s ) i =: P k + ( s ) . (36) It remains to integrate P k + P_k^{+} P k + [ s ( 3 ) , ∞ ) [s(3),\infty) [ s ( 3 ) , ∞ ) s 0 > s ( 3 ) s_0>s(3) s 0 > s ( 3 )
On the compact piece [ s ( 3 ) , s 0 ] [s(3),s_0] [ s ( 3 ) , s 0 ] P j o d d P_j^{\mathrm{odd}} P j odd (35) , a a a P k + P_k^{+} P k + s s s [ s ( 3 ) , s 0 ] [s(3),s_0] [ s ( 3 ) , s 0 ] ∫ s ( 3 ) s 0 P k + \int_{s(3)}^{s_0}P_k^{+} ∫ s ( 3 ) s 0 P k +
On the tail [ s 0 , ∞ ) [s_0,\infty) [ s 0 , ∞ ) n n n Ω ( n ) = k − i \Omega(n)=k-i Ω ( n ) = k − i n ≥ 3 k − i n\ge3^{k-i} n ≥ 3 k − i s ≥ s 0 s\ge s_0 s ≥ s 0 t = s − s 0 t=s-s_0 t = s − s 0 P k − i o d d ( s ) ≤ 3 − ( k − i ) t P k − i o d d ( s 0 ) P_{k-i}^{\mathrm{odd}}(s)\le3^{-(k-i)t}P_{k-i}^{\mathrm{odd}}(s_0) P k − i odd ( s ) ≤ 3 − ( k − i ) t P k − i odd ( s 0 ) a a a n ≥ 2 n\ge2 n ≥ 2 a ( s ) ≤ 2 − t a ( s 0 ) a(s)\le2^{-t}a(s_0) a ( s ) ≤ 2 − t a ( s 0 ) (35) . Since 3 − x ≤ 2 − x 3^{-x}\le2^{-x} 3 − x ≤ 2 − x x ≥ 0 x\ge0 x ≥ 0 i i i P k + P_k^{+} P k + 3 − ( k − i ) t 2 − i t ≤ 2 − k t 3^{-(k-i)t}2^{-it}\le2^{-kt} 3 − ( k − i ) t 2 − i t ≤ 2 − k t ∫ 0 ∞ 2 − k t d t = 1 / ( k log 2 ) \int_0^\infty2^{-kt}\,dt=1/(k\log2) ∫ 0 ∞ 2 − k t d t = 1/ ( k log 2 )
∫ s 0 ∞ P k + ( s ) d s ≤ P k + ( s 0 ) k log 2 . (37) \int_{s_0}^{\infty}P_k^{+}(s)\,ds\le\frac{P_k^{+}(s_0)}{k\log2}. \tag*{(37)} ∫ s 0 ∞ P k + ( s ) d s ≤ k log 2 P k + ( s 0 ) . (37) Adding the Riemann sum over [ s ( 3 ) , s 0 ] [s(3),s_0] [ s ( 3 ) , s 0 ] (37) bounds ∫ s ( 3 ) ∞ P k + \int_{s(3)}^\infty P_k^{+} ∫ s ( 3 ) ∞ P k + ∫ s ( 3 ) ∞ A k \int_{s(3)}^\infty A_k ∫ s ( 3 ) ∞ A k (36) . For one k k k (28) is its outcome for each of the 142 indices 59 ≤ k ≤ 200 59\le k\le200 59 ≤ k ≤ 200
That leaves the dependence on k k k H o + ( s , ⋅ ) H_o^{+}(s,\cdot) H o + ( s , ⋅ ) ∣ z ∣ = R \lvert z\rvert=R ∣ z ∣ = R z = R z=R z = R R < 3 s ( 3 ) R<3^{s(3)} R < 3 s ( 3 ) s ≥ s ( 3 ) s\ge s(3) s ≥ s ( 3 )
A k ( s ) ≤ R − k ζ ( s ) R H o ( s , R ) e a ( s ) R = R − k ∏ p ≥ 3 ( 1 − R p − s ) − 1 e a ( s ) R , (38) A_k(s)\le R^{-k}\zeta(s)^RH_o(s,R)e^{a(s)R}=R^{-k}\prod_{p\ge3}(1-Rp^{-s})^{-1}e^{a(s)R}, \tag*{(38)} A k ( s ) ≤ R − k ζ ( s ) R H o ( s , R ) e a ( s ) R = R − k p ≥ 3 ∏ ( 1 − R p − s ) − 1 e a ( s ) R , (38) a bound whose only k k k R − k R^{-k} R − k [ s ( 3 ) , ∞ ) [s(3),\infty) [ s ( 3 ) , ∞ ) 1 1 1 0 0 0 s → ∞ s\to\infty s → ∞ [ s ( 3 ) , s 1 ] [s(3),s_1] [ s ( 3 ) , s 1 ] s 1 s_1 s 1 s 1 s_1 s 1 [ s ( 3 ) , s 1 ] [s(3),s_1] [ s ( 3 ) , s 1 ] (38) gives ∫ s ( 3 ) s 1 A k ≤ R − k Θ 1 \int_{s(3)}^{s_1}A_k\le R^{-k}\Theta_1 ∫ s ( 3 ) s 1 A k ≤ R − k Θ 1 Θ 1 = ∫ s ( 3 ) s 1 ∏ p ≥ 3 ( 1 − R p − s ) − 1 e a ( s ) R d s \Theta_1=\int_{s(3)}^{s_1}\prod_{p\ge3}(1-Rp^{-s})^{-1}e^{a(s)R}\,ds Θ 1 = ∫ s ( 3 ) s 1 ∏ p ≥ 3 ( 1 − R p − s ) − 1 e a ( s ) R d s k k k s 1 s_1 s 1 (37) applies with s 0 = s 1 s_0=s_1 s 0 = s 1 P k + ( s 1 ) P_k^{+}(s_1) P k + ( s 1 ) (38) in turn gives ∫ s 1 ∞ A k ≤ R − k Θ 2 / ( k log 2 ) \int_{s_1}^{\infty}A_k\le R^{-k}\Theta_2/(k\log2) ∫ s 1 ∞ A k ≤ R − k Θ 2 / ( k log 2 ) Θ 2 = ∏ p ≥ 3 ( 1 − R p − s 1 ) − 1 e a ( s 1 ) R \Theta_2=\prod_{p\ge3}(1-Rp^{-s_1})^{-1}e^{a(s_1)R} Θ 2 = ∏ p ≥ 3 ( 1 − R p − s 1 ) − 1 e a ( s 1 ) R k k k
Hence
3 k ∫ s ( 3 ) ∞ A k ≤ ( 3 R ) k ( Θ 1 + Θ 2 k log 2 ) 3^k\int_{s(3)}^\infty A_k \le\left(\frac{3}{R}\right)^k\left(\Theta_1+\frac{\Theta_2}{k\log2}\right) 3 k ∫ s ( 3 ) ∞ A k ≤ ( R 3 ) k ( Θ 1 + k log 2 Θ 2 ) Take R = 3.15 R=3.15 R = 3.15 3 s ( 3 ) = 3.1738 … 3^{s(3)}=3.1738\ldots 3 s ( 3 ) = 3.1738 … s 1 = 8 s_1=8 s 1 = 8 Θ 1 + Θ 2 / ( 200 log 2 ) ≤ 175346 \Theta_1+\Theta_2/(200\log2)\le175346 Θ 1 + Θ 2 / ( 200 log 2 ) ≤ 175346 ( 3 / 3.15 ) 201 = 5.5075 ⋅ 10 − 5 (3/3.15)^{201}=5.5075\cdot10^{-5} ( 3/3.15 ) 201 = 5.5075 ⋅ 1 0 − 5 k > 200 k>200 k > 200
∫ s ( 3 ) ∞ A k ( s ) d s ≤ C N 3 − k for all k ≥ 59 , (39) \int_{s(3)}^\infty A_k(s)\,\mathrm{d}s\le C_N3^{-k}\qquad\text{for all }k\ge59, \tag*{(39)} ∫ s ( 3 ) ∞ A k ( s ) d s ≤ C N 3 − k for all k ≥ 59 , (39) which is the range in which (32) is used. Neither branch is anywhere near binding: over 59 ≤ k ≤ 200 59\le k\le200 59 ≤ k ≤ 200 34.14 ⋅ 3 − k 34.14\cdot3^{-k} 34.14 ⋅ 3 − k k = 59 k=59 k = 59 9.66 ⋅ 3 − k 9.66\cdot3^{-k} 9.66 ⋅ 3 − k C N = 367.8436 C_N=367.8436 C N = 367.8436
The companion bound (29) , for the two subtracted main terms, reduces the same way and needs no new idea. Here ∣ h j ( 1 ) ∣ ≤ Σ 0 3 − j |h_j(1)|\le\Sigma_0 3^{-j} ∣ h j ( 1 ) ∣ ≤ Σ 0 3 − j Lemma 3.4 (i), while b j = h j ′ ( 1 ) + 2 γ h j ( 1 ) b_j=h_j'(1)+2\gamma h_j(1) b j = h j ′ ( 1 ) + 2 γ h j ( 1 ) ∣ h j ′ ( 1 ) ∣ ≤ ( j + 1 ) Σ 1 3 − j |h_j'(1)|\le(j+1)\Sigma_1 3^{-j} ∣ h j ′ ( 1 ) ∣ ≤ ( j + 1 ) Σ 1 3 − j 2 γ h j ( 1 ) 2\gamma h_j(1) 2 γ h j ( 1 ) ∣ b j ∣ ≤ ( ( j + 1 ) Σ 1 + 2 γ Σ 0 ) 3 − j |b_j|\le((j+1)\Sigma_1+2\gamma\Sigma_0)3^{-j} ∣ b j ∣ ≤ (( j + 1 ) Σ 1 + 2 γ Σ 0 ) 3 − j P ( m , x ) ≤ x m / m ! P(m,x)\le x^m/m! P ( m , x ) ≤ x m / m ! U = 3 U=3 U = 3 Σ 0 e 3 3 − k \Sigma_0 e^3 3^{-k} Σ 0 e 3 3 − k 1 2 ( ( k + 1 ) Σ 1 + 2 γ Σ 0 ) e 3 3 − k \frac{1}{2}((k+1)\Sigma_1+2\gamma\Sigma_0)e^3 3^{-k} 2 1 (( k + 1 ) Σ 1 + 2 γ Σ 0 ) e 3 3 − k k k k ( k + 3 ) 3 3 − k (k+3)3^{3-k} ( k + 3 ) 3 3 − k k k k k = 400 k=400 k = 400 N U N_U N U 1 ≤ k ≤ 400 1\le k\le400 1 ≤ k ≤ 400 k = 7 k=7 k = 7 (32) is used.
It remains only to add up. Substituting the certified constants, the four contributions are
∣ R k ∣ ≤ ( 5.48910 ⋅ 14414 18 + 44.5987 2 + 5 ⋅ 590 64 + C N + N U ) ( k + 3 ) 3 − k < 4900 ( k + 3 ) 3 − k , |R_k|\le\left(\frac{5.48910\cdot14414}{18}+\frac{44.5987}{2}+\frac{5\cdot590}{64}+C_N+N_U\right)(k+3)3^{-k}<4900(k+3)3^{-k}, ∣ R k ∣ ≤ ( 18 5.48910 ⋅ 14414 + 2 44.5987 + 64 5 ⋅ 590 + C N + N U ) ( k + 3 ) 3 − k < 4900 ( k + 3 ) 3 − k , with C N = 367.8436 C_N=367.8436 C N = 367.8436 N U = 26.0962 N_U=26.0962 N U = 26.0962
Proof of Theorem 1.2. The limit follows at once from (22) and Proposition 3.5 . For monotonicity,
g k − g k + 1 = β 2 − k − 1 + R k − R k + 1 . (40) g_k-g_{k+1}=\beta2^{-k-1}+R_k-R_{k+1}. \tag*{(40)} g k − g k + 1 = β 2 − k − 1 + R k − R k + 1 . (40) The main term β 2 − k − 1 \beta2^{-k-1} β 2 − k − 1 k k k k + 1 k+1 k + 1 ∣ R m ∣ < β 2 − m / 3 |R_m|<\beta2^{-m}/3 ∣ R m ∣ < β 2 − m /3 m = k , k + 1 m=k,k+1 m = k , k + 1
∣ R k ∣ + ∣ R k + 1 ∣ < 2 β 3 2 − k − 1 + β 3 2 − k − 1 = β 2 − k − 1 . |R_k|+|R_{k+1}|<\frac{2\beta}{3}2^{-k-1}+\frac{\beta}{3}2^{-k-1}=\beta2^{-k-1}. ∣ R k ∣ + ∣ R k + 1 ∣ < 3 2 β 2 − k − 1 + 3 β 2 − k − 1 = β 2 − k − 1 . Define
q ( k ) = 3 ⋅ 4900 β ( k + 3 ) 3 ( 2 / 3 ) k . q(k)=\frac{3\cdot4900}{\beta}(k+3)^3(2/3)^k. q ( k ) = β 3 ⋅ 4900 ( k + 3 ) 3 ( 2/3 ) k . The certified inequality (30) says q ( 59 ) < 1 q(59)<1 q ( 59 ) < 1
q ( k + 1 ) q ( k ) = 2 3 ( k + 4 k + 3 ) 3 < 1 ( k ≥ 4 ) . \frac{q(k+1)}{q(k)}=\frac{2}{3}\left(\frac{k+4}{k+3}\right)^3<1\qquad(k\ge4). q ( k ) q ( k + 1 ) = 3 2 ( k + 3 k + 4 ) 3 < 1 ( k ≥ 4 ) . Thus q ( k ) < 1 q(k)<1 q ( k ) < 1 k ≥ 59 k\ge59 k ≥ 59 Proposition 3.5 gives ∣ R k ∣ < β 2 − k / 3 |R_k|<\beta2^{-k}/3 ∣ R k ∣ < β 2 − k /3 (40) proves monotonicity for k ≥ 59 k\ge59 k ≥ 59 1 ≤ k ≤ 74 1\le k\le74 1 ≤ k ≤ 74 k ≥ 1 k\ge1 k ≥ 1 59 ≤ k ≤ 74 59\le k\le74 59 ≤ k ≤ 74
An exact criterion for the unrestricted sums
The asymptotic (2) already says that f k f_k f k O ( k log ( k + 1 ) ) O(k\log(k+1)) O ( k log ( k + 1 )) E ( k ) E(k) E ( k )
f k = 1 − 2 − k ( a 1 k 2 + E ( k ) ) , a 1 = d log 2 4 , (41) f_k=1-2^{-k}\left(a_1k^2+E(k)\right),\qquad a_1=\frac{d\log2}{4}, \tag*{(41)} f k = 1 − 2 − k ( a 1 k 2 + E ( k ) ) , a 1 = 4 d log 2 , (41) so that E ( k ) E(k) E ( k )
f k + 1 − f k = 2 − k − 1 ( a 1 ( k 2 − 2 k − 1 ) + 2 E ( k ) − E ( k + 1 ) ) . (42) f_{k+1}-f_k=2^{-k-1}\left(a_1(k^2-2k-1)+2E(k)-E(k+1)\right). \tag*{(42)} f k + 1 − f k = 2 − k − 1 ( a 1 ( k 2 − 2 k − 1 ) + 2 E ( k ) − E ( k + 1 ) ) . (42) Thus f k < f k + 1 f_k<f_{k+1} f k < f k + 1
a 1 ( k 2 − 2 k − 1 ) + 2 E ( k ) − E ( k + 1 ) > 0. (43) a_1(k^2-2k-1)+2E(k)-E(k+1)>0. \tag*{(43)} a 1 ( k 2 − 2 k − 1 ) + 2 E ( k ) − E ( k + 1 ) > 0. (43) We now derive the decomposition of E ( k ) E(k) E ( k )
F ( s , z ) = ∏ p ( 1 − z p − s ) − 1 , G 2 ( s , z ) = ( s − 1 ) z ( 1 − z 2 − s ) F ( s , z ) , F(s,z)=\prod_p(1-zp^{-s})^{-1},\qquad G_2(s,z)=(s-1)z(1-z2^{-s})F(s,z), F ( s , z ) = p ∏ ( 1 − z p − s ) − 1 , G 2 ( s , z ) = ( s − 1 ) z ( 1 − z 2 − s ) F ( s , z ) , and write c i = [ z i ] G 2 ( 1 , z ) c_i=[z^i]G_2(1,z) c i = [ z i ] G 2 ( 1 , z ) G 2 ( s , z ) = ( ( s − 1 ) ζ ( s ) ) z H O ( s , z ) G_2(s,z)=((s-1)\zeta(s))^zH_O(s,z) G 2 ( s , z ) = (( s − 1 ) ζ ( s ) ) z H O ( s , z ) s > 1 s>1 s > 1 ( ( s − 1 ) ζ ( s ) ) z → 1 ((s-1)\zeta(s))^z\to1 (( s − 1 ) ζ ( s ) ) z → 1 s → 1 + s\to1^+ s → 1 + ( s − 1 ) ζ ( s ) (s-1)\zeta(s) ( s − 1 ) ζ ( s ) 1 1 1 1 1 1 s = 1 s=1 s = 1 0 ⋅ ∞ 0\cdot\infty 0 ⋅ ∞ G 2 ( 1 , z ) = H O ( 1 , z ) G_2(1,z)=H_O(1,z) G 2 ( 1 , z ) = H O ( 1 , z ) 3 3 3
∑ i ≥ 0 c i = 1 2 , ∑ i ≥ 0 2 i c i = d . (44) \sum_{i\geq0}c_i=\frac{1}{2},\qquad\sum_{i\geq0}2^ic_i=d. \tag*{(44)} i ≥ 0 ∑ c i = 2 1 , i ≥ 0 ∑ 2 i c i = d . (44) The coefficients are those of Section 3:
c i = h i ( 1 ) . (45) c_i=h_i(1). \tag*{(45)} c i = h i ( 1 ) . (45) Define the absolutely convergent signed moments
U = ∑ i ≥ 0 i 2 i c i , V = ∑ i ≥ 0 i 2 2 i c i . (46) U=\sum_{i\geq0}i2^ic_i,\qquad V=\sum_{i\geq0}i^22^ic_i. \tag*{(46)} U = i ≥ 0 ∑ i 2 i c i , V = i ≥ 0 ∑ i 2 2 i c i . (46) Following [3 §3] , put
I k = ∫ 1 2 [ z k ] F ( s , z ) d s , (47) I_k=\int_1^2[z^k]F(s,z)\,ds, \tag*{(47)} I k = ∫ 1 2 [ z k ] F ( s , z ) d s , (47) I k ′ = ∑ i + j + ℓ = k c i ∫ 1 2 ( − log ( s − 1 ) ) ℓ ℓ ! 2 − j s d s . (48) I'_k=\sum_{i+j+\ell=k}c_i\int_1^2\frac{(-\log(s-1))^\ell}{\ell!}2^{-js}\,ds. \tag*{(48)} I k ′ = i + j + ℓ = k ∑ c i ∫ 1 2 ℓ ! ( − log ( s − 1 ) ) ℓ 2 − j s d s . (48) For m ≥ 0 m\geq0 m ≥ 0
A m = 2 m ∑ j + ℓ = m ∫ 1 2 ( − log ( s − 1 ) ) ℓ ℓ ! 2 − j s d s (49) A_m=2^m\sum_{j+\ell=m}\int_1^2\frac{(-\log(s-1))^\ell}{\ell!}2^{-js}\,ds \tag*{(49)} A m = 2 m j + ℓ = m ∑ ∫ 1 2 ℓ ! ( − log ( s − 1 ) ) ℓ 2 − j s d s (49) and
δ m = A m − 2 m + 1 + log 2 4 m 2 , Δ ( k ) = ∑ i = 0 k 2 i c i δ k − i . (50) \delta_m=A_m-2^{m+1}+\frac{\log2}{4}m^2,\qquad\Delta(k)=\sum_{i=0}^k2^ic_i\delta_{k-i}. \tag*{(50)} δ m = A m − 2 m + 1 + 4 log 2 m 2 , Δ ( k ) = i = 0 ∑ k 2 i c i δ k − i . (50) Proposition 4.1. Let
E S 1 ( k ) = − 2 k ( I k − I k ′ ) , (51) E_{\mathrm{S1}}(k)=-2^k(I_k-I'_k), \tag*{(51)} E S1 ( k ) = − 2 k ( I k − I k ′ ) , (51) L k = 2 ∑ i > k c i , Q k = ∑ i > k 2 i c i ( k − i ) 2 , (52) L_k=2\sum_{i>k}c_i,\qquad Q_k=\sum_{i>k}2^i c_i(k-i)^2, \tag*{(52)} L k = 2 i > k ∑ c i , Q k = i > k ∑ 2 i c i ( k − i ) 2 , (52) T k = f k − I k , (53) T_k=f_k-I_k, \tag*{(53)} T k = f k − I k , (53) E exp ( k ) = 2 k L k − log 2 4 Q k − 2 k T k . (54) E_{\exp}(k)=2^kL_k-\frac{\log2}{4}Q_k-2^kT_k. \tag*{(54)} E e x p ( k ) = 2 k L k − 4 log 2 Q k − 2 k T k . (54) Then the identity
E ( k ) = E S 1 ( k ) + log 2 4 ( V − 2 k U ) − Δ ( k ) + E exp ( k ) (55) E(k)=E_{\mathrm{S1}}(k)+\frac{\log2}{4}(V-2kU)-\Delta(k)+E_{\exp}(k) \tag*{(55)} E ( k ) = E S1 ( k ) + 4 log 2 ( V − 2 k U ) − Δ ( k ) + E e x p ( k ) (55) holds for every k ≥ 1 k\ge1 k ≥ 1
Proof. From (48) and (49) ,
I k ′ = 2 − k ∑ i = 0 k 2 i c i A k − i . I'_k=2^{-k}\sum_{i=0}^{k}2^i c_i A_{k-i}. I k ′ = 2 − k i = 0 ∑ k 2 i c i A k − i . Insert the definition of δ k − i \delta_{k-i} δ k − i
I k ′ = 2 ∑ i = 0 k c i − log 2 4 2 − k ∑ i = 0 k 2 i c i ( k − i ) 2 + 2 − k Δ ( k ) . (56) I'_k=2\sum_{i=0}^{k}c_i-\frac{\log2}{4}2^{-k}\sum_{i=0}^{k}2^i c_i(k-i)^2+2^{-k}\Delta(k). \tag*{(56)} I k ′ = 2 i = 0 ∑ k c i − 4 log 2 2 − k i = 0 ∑ k 2 i c i ( k − i ) 2 + 2 − k Δ ( k ) . (56) By (44) and (46) ,
2 ∑ i = 0 k c i = 1 − L k , 2\sum_{i=0}^{k}c_i=1-L_k, 2 i = 0 ∑ k c i = 1 − L k , ∑ i = 0 k 2 i c i ( k − i ) 2 = ∑ i ≥ 0 2 i c i ( k − i ) 2 − Q k \sum_{i=0}^{k}2^i c_i(k-i)^2=\sum_{i\ge0}2^i c_i(k-i)^2-Q_k i = 0 ∑ k 2 i c i ( k − i ) 2 = i ≥ 0 ∑ 2 i c i ( k − i ) 2 − Q k = d k 2 − 2 k U + V − Q k . =dk^2-2kU+V-Q_k. = d k 2 − 2 k U + V − Q k . Since a 1 = log 2 4 d a_1=\frac{\log2}{4}d a 1 = 4 l o g 2 d (56) gives
I k ′ = 1 − 2 − k ( a 1 k 2 + log 2 4 ( V − 2 k U ) − Δ ( k ) + 2 k L k − log 2 4 Q k ) . I'_k=1-2^{-k}\left(a_1k^2+\frac{\log2}{4}(V-2kU)-\Delta(k)+2^kL_k-\frac{\log2}{4}Q_k\right). I k ′ = 1 − 2 − k ( a 1 k 2 + 4 log 2 ( V − 2 k U ) − Δ ( k ) + 2 k L k − 4 log 2 Q k ) . Finally f k = I k ′ + ( I k − I k ′ ) + T k f_k=I'_k+(I_k-I'_k)+T_k f k = I k ′ + ( I k − I k ′ ) + T k (41) yields (55) . □
Bounds for the four source terms
By (55) the criterion (43) needs a lower bound for 2 E ( k ) − E ( k + 1 ) 2E(k)-E(k+1) 2 E ( k ) − E ( k + 1 ) E E E S 1 S_1 S 1 k k k Δ \Delta Δ k log k k\log k k log k k k k k k k
One point of method recurs. Each bound is taken on the combination 2 E ∙ ( k ) − E ∙ ( k + 1 ) 2E_\bullet(k)-E_\bullet(k+1) 2 E ∙ ( k ) − E ∙ ( k + 1 ) E ∙ ( k ) E_\bullet(k) E ∙ ( k ) E ∙ ( k + 1 ) E_\bullet(k+1) E ∙ ( k + 1 )
The S 1 S_1 S 1
Set
α i = sup 1 ≤ t ≤ 2 ∣ [ z i ] ∂ s G 2 ( t , z ) ∣ , α + = 1 2 ∑ i ≥ 0 α i 2 i . \alpha_i=\sup_{1\leq t\leq2}\left|\left[z^i\right]\partial_sG_2(t,z)\right|,\qquad\alpha_+=\frac{1}{2}\sum_{i\geq0}\alpha_i2^i. α i = 1 ≤ t ≤ 2 sup [ z i ] ∂ s G 2 ( t , z ) , α + = 2 1 i ≥ 0 ∑ α i 2 i . The series converges, and a pole location at one value of t t t t t t
First, G 2 G_2 G 2 s s s [ 1 , 2 ] [1,2] [ 1 , 2 ] ( s − 1 ) z (s-1)^z ( s − 1 ) z s = 1 s=1 s = 1 F ( s , z ) = ζ ( s ) z H ( s , z ) F(s,z)=\zeta(s)^zH(s,z) F ( s , z ) = ζ ( s ) z H ( s , z ) (8) , so
G 2 ( s , z ) = ( ( s − 1 ) ζ ( s ) ) z ( 1 − z 2 − s ) H ( s , z ) , G_2(s,z)=((s-1)\zeta(s))^z(1-z2^{-s})H(s,z), G 2 ( s , z ) = (( s − 1 ) ζ ( s ) ) z ( 1 − z 2 − s ) H ( s , z ) , and ( s − 1 ) ζ ( s ) (s-1)\zeta(s) ( s − 1 ) ζ ( s ) s = 1 s=1 s = 1 1 1 1 ζ \zeta ζ 1 1 1 ( ( s − 1 ) ζ ( s ) ) z = exp ( z log ( ( s − 1 ) ζ ( s ) ) ) ((s-1)\zeta(s))^z=\exp(z\log((s-1)\zeta(s))) (( s − 1 ) ζ ( s ) ) z = exp ( z log (( s − 1 ) ζ ( s ))) H H H s = 1 s=1 s = 1 G 2 ( 1 , z ) = H 0 ( 1 , z ) G_2(1,z)=H_0(1,z) G 2 ( 1 , z ) = H 0 ( 1 , z )
The first z z z ∂ s G 2 ( t , ⋅ ) \partial_sG_2(t,\cdot) ∂ s G 2 ( t , ⋅ ) z = 3 t z=3^t z = 3 t t ∈ [ 1 , 2 ] t\in[1,2] t ∈ [ 1 , 2 ] ∣ z ∣ < 3 |z|<3 ∣ z ∣ < 3 r r r 2 < r < 3 2<r<3 2 < r < 3 M = sup 1 ≤ t ≤ 2 sup ∣ z ∣ = r ∣ ∂ s G 2 ( t , z ) ∣ M=\sup_{1\leq t\leq2}\sup_{|z|=r}|\partial_sG_2(t,z)| M = sup 1 ≤ t ≤ 2 sup ∣ z ∣ = r ∣ ∂ s G 2 ( t , z ) ∣ [ 1 , 2 ] × { ∣ z ∣ = r } [1,2]\times\{|z|=r\} [ 1 , 2 ] × { ∣ z ∣ = r } ∂ s G 2 \partial_sG_2 ∂ s G 2 ∣ z ∣ = r |z|=r ∣ z ∣ = r α i ≤ M r − i \alpha_i\leq Mr^{-i} α i ≤ M r − i t t t ∑ i ≥ 0 α i 2 i ≤ M ( 1 − 2 / r ) − 1 < ∞ \sum_{i\geq0}\alpha_i2^i\leq M(1-2/r)^{-1}<\infty ∑ i ≥ 0 α i 2 i ≤ M ( 1 − 2/ r ) − 1 < ∞
Lemma 5.1. For every k ≥ 1 k\geq1 k ≥ 1
∣ E S 1 ( k ) ∣ ≤ α + ( k + 1 ) , (57) |E_{S_1}(k)|\leq\alpha_+(k+1), \tag*{(57)} ∣ E S 1 ( k ) ∣ ≤ α + ( k + 1 ) , (57) and hence
∣ 2 E S 1 ( k ) − E S 1 ( k + 1 ) ∣ ≤ α + ( 3 k + 4 ) . (58) |2E_{S_1}(k)-E_{S_1}(k+1)|\leq\alpha_+(3k+4). \tag*{(58)} ∣2 E S 1 ( k ) − E S 1 ( k + 1 ) ∣ ≤ α + ( 3 k + 4 ) . (58) Proof. Since [ z i ] G 2 ( ⋅ , z ) \left[z^i\right]G_2(\cdot,z) [ z i ] G 2 ( ⋅ , z ) (10) and the analyticity of G 2 G_2 G 2 s s s
[ z i ] G 2 ( s , z ) − c i = ∫ 1 s [ z i ] ∂ t G 2 ( t , z ) d t , \left[z^i\right]G_2(s,z)-c_i=\int_1^s\left[z^i\right]\partial_tG_2(t,z)\,dt, [ z i ] G 2 ( s , z ) − c i = ∫ 1 s [ z i ] ∂ t G 2 ( t , z ) d t , so ∣ [ z i ] G 2 ( s , z ) − c i ∣ ≤ ( s − 1 ) α i \left|\left[z^i\right]G_2(s,z)-c_i\right|\leq(s-1)\alpha_i [ z i ] G 2 ( s , z ) − c i ≤ ( s − 1 ) α i 1 ≤ s ≤ 2 1\leq s\leq2 1 ≤ s ≤ 2 α i \alpha_i α i (48) from the coefficient expansion for I k I_k I k s = 1 + e − t s=1+e^{-t} s = 1 + e − t
∣ I k − I k ′ ∣ ≤ ∑ i + j + ℓ = k α i ∫ 0 ∞ t ℓ ℓ ! 2 − j ( 1 + e − t ) e − 2 t d t ≤ ∑ i + j + ℓ = k α i 2 − j ∫ 0 ∞ t ℓ ℓ ! e − 2 t d t = ∑ i + j + ℓ = k α i 2 − j − ℓ − 1 . \begin{aligned}
\left|I_k-I'_k\right|\leq\sum_{i+j+\ell=k}\alpha_i\int_0^\infty\frac{t^\ell}{\ell!}2^{-j(1+e^{-t})}e^{-2t}\,dt \\
&\leq\sum_{i+j+\ell=k}\alpha_i2^{-j}\int_0^\infty\frac{t^\ell}{\ell!}e^{-2t}\,dt \\
&=\sum_{i+j+\ell=k}\alpha_i2^{-j-\ell-1}.
\end{aligned} ∣ I k − I k ′ ∣ ≤ i + j + ℓ = k ∑ α i ∫ 0 ∞ ℓ ! t ℓ 2 − j ( 1 + e − t ) e − 2 t d t ≤ i + j + ℓ = k ∑ α i 2 − j ∫ 0 ∞ ℓ ! t ℓ e − 2 t d t = i + j + ℓ = k ∑ α i 2 − j − ℓ − 1 . Multiplying by 2 k 2^k 2 k k = i + j + ℓ k=i+j+\ell k = i + j + ℓ
∣ E S 1 ( k ) ∣ ≤ 1 2 ∑ i = 0 k α i 2 i ( k − i + 1 ) ≤ α + ( k + 1 ) . |E_{S_1}(k)|\leq\frac{1}{2}\sum_{i=0}^k\alpha_i2^i(k-i+1)\leq\alpha_+(k+1). ∣ E S 1 ( k ) ∣ ≤ 2 1 i = 0 ∑ k α i 2 i ( k − i + 1 ) ≤ α + ( k + 1 ) . The triangle inequality now gives
∣ 2 E S 1 ( k ) − E S 1 ( k + 1 ) ∣ ≤ 2 α + ( k + 1 ) + α + ( k + 2 ) = α + ( 3 k + 4 ) . |2E_{S_1}(k)-E_{S_1}(k+1)|\leq2\alpha_+(k+1)+\alpha_+(k+2)=\alpha_+(3k+4). ∣2 E S 1 ( k ) − E S 1 ( k + 1 ) ∣ ≤ 2 α + ( k + 1 ) + α + ( k + 2 ) = α + ( 3 k + 4 ) . The signed moment term
The combination of signed moments that has to be bounded is not chosen; it is what the difference of two consecutive source terms produces:
2 ( V − 2 k U ) − ( V − 2 ( k + 1 ) U ) = V + ( 2 − 2 k ) U . (59) 2(V - 2kU) - (V - 2(k+1)U) = V + (2 - 2k)U. \tag*{(59)} 2 ( V − 2 k U ) − ( V − 2 ( k + 1 ) U ) = V + ( 2 − 2 k ) U . (59) Since this is an identity, bounding the right-hand side is an exact optimization over the rectangle of admissible ( U , V ) (U,V) ( U , V ) ∑ i 2 i ∣ c i ∣ \sum_i 2^i|c_i| ∑ i 2 i ∣ c i ∣ k k k
Lemma 5.2. If ∣ U ∣ ≤ U max |U| \le U_{\max} ∣ U ∣ ≤ U m a x ∣ V ∣ ≤ V max |V| \le V_{\max} ∣ V ∣ ≤ V m a x k ≥ 2 k \ge2 k ≥ 2
log 2 4 ( V + ( 2 − 2 k ) U ) ≥ − log 2 4 ( V max + ( 2 k − 2 ) U max ) . (60) \frac{\log2}{4}\bigl(V + (2 - 2k)U\bigr) \ge-\frac{\log2}{4}\bigl(V_{\max} + (2k - 2)U_{\max}\bigr). \tag*{(60)} 4 log 2 ( V + ( 2 − 2 k ) U ) ≥ − 4 log 2 ( V m a x + ( 2 k − 2 ) U m a x ) . (60) Proof. The coefficient of V V V 2 − 2 k 2-2k 2 − 2 k U U U [ − U max , U max ] × [ − V max , V max ] [-U_{\max},U_{\max}]\times[-V_{\max},V_{\max}] [ − U m a x , U m a x ] × [ − V m a x , V m a x ] U = U max U=U_{\max} U = U m a x V = − V max V=-V_{\max} V = − V m a x
The term
The δ \delta δ Δ ( k ) \Delta(k) Δ ( k )
Lemma 5.3. For j ≥ 0 j \ge0 j ≥ 0 u ≥ 0 u \ge0 u ≥ 0
B j ( u ) = 2 − j e − u − 1 + j log 2 e − u B_j(u)=2^{-j e^{-u}}-1+j\log2\,e^{-u} B j ( u ) = 2 − j e − u − 1 + j log 2 e − u and
η i ( j ) = 1 i ! ∫ 0 ∞ u i e − u B j ( u ) d u . \eta_i(j)=\frac{1}{i!}\int_0^\infty u^i e^{-u}B_j(u)\,\mathrm{d}u. η i ( j ) = i ! 1 ∫ 0 ∞ u i e − u B j ( u ) d u . Then B j ( u ) ≥ 0 B_j(u)\ge0 B j ( u ) ≥ 0 η i ( j ) ≥ 0 \eta_i(j)\ge0 η i ( j ) ≥ 0
δ m = − 1 − log 2 4 m + ∑ i = 0 m 2 i η i ( m − i ) . (61) \delta_m=-1-\frac{\log2}{4}m+\sum_{i=0}^{m}2^i\eta_i(m-i). \tag*{(61)} δ m = − 1 − 4 log 2 m + i = 0 ∑ m 2 i η i ( m − i ) . (61) Proof. With s = 1 + e − u s=1+e^{-u} s = 1 + e − u
A m = ∑ i = 0 m 2 i i ! ∫ 0 ∞ u i e − u 2 − ( m − i ) e − u d u . A_m=\sum_{i=0}^{m}\frac{2^i}{i!}\int_0^\infty u^i e^{-u}2^{-(m-i)e^{-u}}\,\mathrm{d}u. A m = i = 0 ∑ m i ! 2 i ∫ 0 ∞ u i e − u 2 − ( m − i ) e − u d u . Use 2 − j e − u = 1 − j log 2 e − u + B j ( u ) 2^{-j e^{-u}}=1-j\log2\,e^{-u}+B_j(u) 2 − j e − u = 1 − j log 2 e − u + B j ( u )
∑ i = 0 m 2 i = 2 m + 1 − 1. \sum_{i=0}^{m}2^i=2^{m+1}-1. i = 0 ∑ m 2 i = 2 m + 1 − 1. The linear term contributes
− log 2 ∑ i = 0 m 2 i ( m − i ) i ! ∫ 0 ∞ u i e − 2 u d u = − log 2 2 ∑ i = 0 m ( m − i ) -\log2\sum_{i=0}^{m}\frac{2^i(m-i)}{i!}\int_0^\infty u^i e^{-2u}\,\mathrm{d}u
=-\frac{\log2}{2}\sum_{i=0}^{m}(m-i) − log 2 i = 0 ∑ m i ! 2 i ( m − i ) ∫ 0 ∞ u i e − 2 u d u = − 2 log 2 i = 0 ∑ m ( m − i ) = − log 2 4 m ( m + 1 ) . =-\frac{\log2}{4}m(m+1). = − 4 log 2 m ( m + 1 ) . The remaining term is ∑ i = 0 m 2 i η i ( m − i ) \sum_{i=0}^{m}2^i\eta_i(m-i) ∑ i = 0 m 2 i η i ( m − i ) 2 m + 1 2^{m+1} 2 m + 1 ( log 2 ) m 2 / 4 (\log2)m^2/4 ( log 2 ) m 2 /4 x ≥ 0 x\ge0 x ≥ 0 e − x ≥ 1 − x e^{-x}\ge1-x e − x ≥ 1 − x x = j log 2 e − u x=j\log2\,e^{-u} x = j log 2 e − u B j ( u ) ≥ 0 B_j(u)\ge0 B j ( u ) ≥ 0
Lemma 5.4. For every m ≥ 0 m\ge0 m ≥ 0
∣ δ m ∣ ≤ D ( m ) , D ( m ) = ( log 2 ) m log ( m + 1 ) + 1. (62) |\delta_m|\le D(m),\qquad D(m)=(\log2)m\log(m+1)+1. \tag*{(62)} ∣ δ m ∣ ≤ D ( m ) , D ( m ) = ( log 2 ) m log ( m + 1 ) + 1. (62) Proof. For the lower bound, (61) gives δ m ≥ − 1 − log 2 4 m ≥ − D ( m ) \delta_m \ge-1-\frac{\log2}{4}m \ge-D(m) δ m ≥ − 1 − 4 l o g 2 m ≥ − D ( m ) m ≥ 1 m \ge1 m ≥ 1 log ( m + 1 ) ≥ log 2 > 1 / 4 \log(m+1) \ge\log2 > 1/4 log ( m + 1 ) ≥ log 2 > 1/4 δ 0 = − 1 = − D ( 0 ) \delta_0=-1=-D(0) δ 0 = − 1 = − D ( 0 )
The upper bound is where the log ( m + 1 ) \log(m+1) log ( m + 1 ) Λ m = ∑ i = 0 m 2 i η i ( m − i ) \Lambda_m=\sum_{i=0}^{m}2^i\eta_i(m-i) Λ m = ∑ i = 0 m 2 i η i ( m − i ) e − x − 1 + x ≤ min ( x , x 2 / 2 ) e^{-x}-1+x\le\min(x,x^2/2) e − x − 1 + x ≤ min ( x , x 2 /2 ) x = ( m − i ) log 2 e − u x=(m-i)\log2e^{-u} x = ( m − i ) log 2 e − u u u u u u u u = 0 u=0 u = 0 m m m x = ( m − i ) log 2 e − u x=(m-i)\log2e^{-u} x = ( m − i ) log 2 e − u
Λ m ≤ ∫ 0 ∞ e − u ∑ i = 0 m ( 2 u ) i i ! min ( ( m − i ) log 2 e − u , ( m − i ) 2 ( log 2 ) 2 e − 2 u 2 ) d u . (63) \Lambda_m\le\int_0^\infty e^{-u}\sum_{i=0}^{m}\frac{(2u)^i}{i!}\min\left((m-i)\log2e^{-u},\frac{(m-i)^2(\log2)^2e^{-2u}}{2}\right)\,du. \tag*{(63)} Λ m ≤ ∫ 0 ∞ e − u i = 0 ∑ m i ! ( 2 u ) i min ( ( m − i ) log 2 e − u , 2 ( m − i ) 2 ( log 2 ) 2 e − 2 u ) d u . (63) The crossover is at v m = max ( 0 , log m + 1 3 ) v_m=\max(0,\log\frac{m+1}{3}) v m = max ( 0 , log 3 m + 1 ) 0 ≤ u ≤ v m 0\le u\le v_m 0 ≤ u ≤ v m ∑ i = 0 m ( m − i ) ( 2 u ) i / i ! ≤ m e 2 u \sum_{i=0}^{m}(m-i)(2u)^i/i!\le me^{2u} ∑ i = 0 m ( m − i ) ( 2 u ) i / i ! ≤ m e 2 u u ≥ v m u\ge v_m u ≥ v m ∑ i = 0 m ( m − i ) 2 ( 2 u ) i / i ! ≤ m 2 e 2 u \sum_{i=0}^{m}(m-i)^2(2u)^i/i!\le m^2e^{2u} ∑ i = 0 m ( m − i ) 2 ( 2 u ) i / i ! ≤ m 2 e 2 u
Λ m ≤ m log 2 v m + m 2 ( log 2 ) 2 2 e − v m . (64) \Lambda_m\le m\log2\,v_m+\frac{m^2(\log2)^2}{2}e^{-v_m}. \tag*{(64)} Λ m ≤ m log 2 v m + 2 m 2 ( log 2 ) 2 e − v m . (64) For m ≥ 2 m\ge2 m ≥ 2 v m = log m + 1 3 v_m=\log\frac{m+1}{3} v m = log 3 m + 1 e − v m = 3 m + 1 e^{-v_m}=\frac{3}{m+1} e − v m = m + 1 3 3 m 2 2 ( m + 1 ) ≤ 3 2 m \frac{3m^2}{2(m+1)}\le\frac{3}{2}m 2 ( m + 1 ) 3 m 2 ≤ 2 3 m
Λ m ≤ m log 2 log m + 1 3 + 3 m 2 ( log 2 ) 2 2 ( m + 1 ) ≤ m log 2 ( log ( m + 1 ) − log 3 + 3 2 log 2 ) ≤ m log 2 log ( m + 1 ) , \begin{aligned}
\Lambda_m\le m\log2\log\frac{m+1}{3}+\frac{3m^2(\log2)^2}{2(m+1)} \\
&\le m\log2\left(\log(m+1)-\log3+\frac{3}{2}\log2\right)\le m\log2\log(m+1),
\end{aligned} Λ m ≤ m log 2 log 3 m + 1 + 2 ( m + 1 ) 3 m 2 ( log 2 ) 2 ≤ m log 2 ( log ( m + 1 ) − log 3 + 2 3 log 2 ) ≤ m log 2 log ( m + 1 ) , the last step because 3 > 2 3 / 2 3>2^{3/2} 3 > 2 3/2 m ≤ 1 m\le1 m ≤ 1 v m = 0 v_m=0 v m = 0 Λ m ≤ m 2 ( log 2 ) 2 / 2 \Lambda_m\le m^2(\log2)^2/2 Λ m ≤ m 2 ( log 2 ) 2 /2 m log 2 log ( m + 1 ) m\log2\log(m+1) m log 2 log ( m + 1 ) m log 2 2 ≤ log ( m + 1 ) \frac{m\log2}{2}\le\log(m+1) 2 m l o g 2 ≤ log ( m + 1 )
δ m = − 1 − log 2 4 m + Λ m ≤ Λ m ≤ ( log 2 ) m log ( m + 1 ) < D ( m ) , \delta_m=-1-\frac{\log2}{4}m+\Lambda_m\le\Lambda_m\le(\log2)m\log(m+1)<D(m), δ m = − 1 − 4 log 2 m + Λ m ≤ Λ m ≤ ( log 2 ) m log ( m + 1 ) < D ( m ) , which with the lower bound proves (62). □
Lemma 5.5. If S ∗ = ∑ i ≥ 0 2 i ∣ c i ∣ S^*=\sum_{i\ge0}2^i|c_i| S ∗ = ∑ i ≥ 0 2 i ∣ c i ∣ D D D
∣ 2 Δ ( k ) − Δ ( k + 1 ) ∣ ≤ S ∗ ( 2 D ( k ) + D ( k + 1 ) ) . (65) |2\Delta(k)-\Delta(k+1)|\le S^*(2D(k)+D(k+1)). \tag*{(65)} ∣2Δ ( k ) − Δ ( k + 1 ) ∣ ≤ S ∗ ( 2 D ( k ) + D ( k + 1 )) . (65) Sketch. Since D D D ∣ Δ ( k ) ∣ ≤ ∑ i ≤ k 2 i ∣ c i ∣ D ( k − i ) ≤ S ∗ D ( k ) |\Delta(k)|\le\sum_{i\le k}2^i|c_i|D(k-i)\le S^*D(k) ∣Δ ( k ) ∣ ≤ ∑ i ≤ k 2 i ∣ c i ∣ D ( k − i ) ≤ S ∗ D ( k ) 2 Δ ( k ) − Δ ( k + 1 ) 2\Delta(k)-\Delta(k+1) 2Δ ( k ) − Δ ( k + 1 )
The exponentially small term
The three terms in (54) are respectively a coefficient tail at z = 1 z=1 z = 1 z = 2 z=2 z = 2 s ≥ 2 s\ge2 s ≥ 2
Proposition 5.6 (Certified unrestricted input). The interval computations give
d ≥ 0.3786949 , S ∗ ≤ 6.9764 , α + ≤ 15.425661 , d\ge0.3786949,\qquad S^*\le6.9764,\qquad\alpha_+\le15.425661, d ≥ 0.3786949 , S ∗ ≤ 6.9764 , α + ≤ 15.425661 , ∣ U ∣ ≤ U max = 0.08436 , ∣ V ∣ ≤ V max = 1.907462. (66) |U|\le U_{\max}=0.08436,\qquad|V|\le V_{\max}=1.907462. \tag*{(66)} ∣ U ∣ ≤ U m a x = 0.08436 , ∣ V ∣ ≤ V m a x = 1.907462. (66) For every k ≥ 900 k\ge900 k ≥ 900
2 k ∣ T k ∣ ≤ 1.45 k , (67) 2^k|T_k|\le\frac{1.45}{k}, \tag*{(67)} 2 k ∣ T k ∣ ≤ k 1.45 , (67) and a contour bound for the pole-free part of the generating function of the c i c_i c i ρ \rho ρ G 2 ( 1 , ⋅ ) G_2(1,\cdot) G 2 ( 1 , ⋅ ) z = 3 z=3 z = 3 ∣ ρ ∣ ≤ 0.1749324 |\rho|\le0.1749324 ∣ ρ ∣ ≤ 0.1749324
sup ∣ z ∣ = 4.2 ∣ G 2 ( 1 , z ) − ρ 3 − z ∣ ≤ 101.245. (68) \sup_{|z|=4.2}\left|G_2(1,z)-\frac{\rho}{3-z}\right|\le101.245. \tag*{(68)} ∣ z ∣ = 4.2 sup G 2 ( 1 , z ) − 3 − z ρ ≤ 101.245. (68) Here the subtraction matters. For the regularized product G ~ 2 ( z ) = ( 1 − z / 3 ) G 2 ( 1 , z ) \widetilde{G}_2(z)=(1-z/3)G_2(1,z) G 2 ( z ) = ( 1 − z /3 ) G 2 ( 1 , z ) 228.94 ≤ sup ∣ z ∣ = 4.2 ∣ G ~ 2 ∣ ≤ 228.95 228.94\le\sup_{|z|=4.2}|\widetilde{G}_2|\le228.95 228.94 ≤ sup ∣ z ∣ = 4.2 ∣ G 2 ∣ ≤ 228.95 G ~ 2 \widetilde{G}_2 G 2 (68) , with about 5.7% to spare. Both suprema are certified in numerics/contour-certified.py (log contour.log), which also recovers ρ \rho ρ
∣ E exp ( k ) ∣ ≤ 1.45 k + 10 − 3 ( k ≥ 900 ) , (69) |E_{\exp}(k)|\le\frac{1.45}{k}+10^{-3}\qquad(k\ge900), \tag*{(69)} ∣ E e x p ( k ) ∣ ≤ k 1.45 + 1 0 − 3 ( k ≥ 900 ) , (69) as derived in §5.4. Of the three terms of E exp E_{\exp} E e x p k k k (68) ; only T k T_k T k k k k (67) . Both are recorded here rather than inside that derivation so that every certified input for the unrestricted case appears in one place. Finally, f k < f k + 1 f_k<f_{k+1} f k < f k + 1
6 ≤ k ≤ 199 , 78 ≤ k ≤ 899 , 900 ≤ k ≤ 2439 , (70) 6\le k\le199,\qquad78\le k\le899,\qquad900\le k\le2439, \tag*{(70)} 6 ≤ k ≤ 199 , 78 ≤ k ≤ 899 , 900 ≤ k ≤ 2439 , (70) whose union is the single range $6\le k\le 2439$. These three ranges are not established the same way, and the distinction matters to anyone checking the certificates. On the first, f k f_k f k k k k f k f_k f k 2 ≤ k ≤ 500 2\le k\le500 2 ≤ k ≤ 500 f k + 1 − f k f_{k+1}-f_k f k + 1 − f k verified: false from k = 298 k=298 k = 298 6 ≤ k ≤ 199 6\le k\le199 6 ≤ k ≤ 199 (43) that is evaluated in ball arithmetic at each k k k φ \varphi φ § 6 , evaluated per k k k 78 ≤ k ≤ 734 78\le k\le734 78 ≤ k ≤ 734 900 ≤ k ≤ 2439 900\le k\le2439 900 ≤ k ≤ 2439 E 1 E_1 E 1 β i \beta_i β i α + \alpha_+ α + k = 728 k=728 k = 728 735 ≤ k ≤ 899 735\le k\le899 735 ≤ k ≤ 899 (70) robust to a change in either sweep.
The coefficient tails in (69) are obtained only after the pole at z = 3 z=3 z = 3
H ( z ) = G 2 ( 1 , z ) − ρ 3 − z = ∑ m ≥ 0 w m z m , H(z)=G_2(1,z)-\frac{\rho}{3-z}=\sum_{m\ge0}w_mz^m, H ( z ) = G 2 ( 1 , z ) − 3 − z ρ = m ≥ 0 ∑ w m z m , which is holomorphic on ∣ z ∣ < 5 |z|<5 ∣ z ∣ < 5 G 2 ( 1 , ⋅ ) G_2(1,\cdot) G 2 ( 1 , ⋅ ) z = 5 z=5 z = 5 ∣ z ∣ = 4.2 |z|=4.2 ∣ z ∣ = 4.2 (68) gives ∣ w m ∣ ≤ 101.245 4.2 − m |w_m|\le101.245\,4.2^{-m} ∣ w m ∣ ≤ 101.245 4. 2 − m [ z i ] ρ / ( 3 − z ) = ρ 3 − i − 1 [z^i]\rho/(3-z)=\rho\,3^{-i-1} [ z i ] ρ / ( 3 − z ) = ρ 3 − i − 1
∣ c i ∣ ≤ ∣ ρ ∣ 3 i + 1 + 101.245 ( 3 / 4.2 ) i 3 − i ≤ 101.313 − i < 354.363 − i . (71) |c_i|\le\frac{|\rho|}{3^{i+1}}+101.245\,(3/4.2)^i3^{-i}\le101.313^{-i}<354.363^{-i}. \tag*{(71)} ∣ c i ∣ ≤ 3 i + 1 ∣ ρ ∣ + 101.245 ( 3/4.2 ) i 3 − i ≤ 101.31 3 − i < 354.36 3 − i . (71) The weaker constant 354.36 is the one carried below; the sharper 101.31 is recorded only so that the chain can be checked. By (45) the c i c_i c i h i ( 1 ) h_i(1) h i ( 1 ) (71) to ∣ c i ∣ ≤ ∑ 3 − i |c_i| \le\sum3^{-i} ∣ c i ∣ ≤ ∑ 3 − i C = 354.36 C = 354.36 C = 354.36
2 k ∣ L k ∣ ≤ C ( 2 / 3 ) k , 2^k |L_k| \le C(2/3)^k, 2 k ∣ L k ∣ ≤ C ( 2/3 ) k , log 2 4 ∣ Q k ∣ ≤ log 2 4 C ( 2 / 3 ) k ∑ r ≥ 1 r 2 ( 2 / 3 ) r = 30 C log 2 4 ( 2 / 3 ) k . \frac{\log2}{4}|Q_k| \le\frac{\log2}{4}C(2/3)^k \sum_{r \ge1} r^2(2/3)^r = \frac{30C\log2}{4}(2/3)^k. 4 log 2 ∣ Q k ∣ ≤ 4 log 2 C ( 2/3 ) k r ≥ 1 ∑ r 2 ( 2/3 ) r = 4 30 C log 2 ( 2/3 ) k . Both terms are less than 10 − 3 10^{-3} 1 0 − 3 k ≥ 900 k \ge900 k ≥ 900 k k k
The third term admits an elementary bound, which we give because (67) is otherwise the one input of Proposition 5.6 not visibly tied to a computation. Since [ z k ] F ( s , z ) = ∑ Ω ( n ) = k n − s [z^k]F(s,z) = \sum_{\Omega(n)=k} n^{-s} [ z k ] F ( s , z ) = ∑ Ω ( n ) = k n − s (53) and (47) give
T k = ∫ 2 ∞ ∑ Ω ( n ) = k n − s d s = ∑ Ω ( n ) = k n − 2 log n . T_k = \int_2^\infty\sum_{\Omega(n)=k} n^{-s}\,ds = \sum_{\Omega(n)=k} \frac{n^{-2}}{\log n}. T k = ∫ 2 ∞ Ω ( n ) = k ∑ n − s d s = Ω ( n ) = k ∑ log n n − 2 . a sum of positive terms, so T k > 0 T_k > 0 T k > 0 n n n 2 k 2^k 2 k log n ≥ k log 2 \log n \ge k\log2 log n ≥ k log 2
2 k ∣ T k ∣ ≤ 2 k P k ( 2 ) k log 2 , P k ( 2 ) = [ z k ] F ( 2 , z ) . 2^k|T_k| \le\frac{2^kP_k(2)}{k\log2}, \qquad P_k(2) = [z^k]F(2,z). 2 k ∣ T k ∣ ≤ k log 2 2 k P k ( 2 ) , P k ( 2 ) = [ z k ] F ( 2 , z ) . As F ( 2 , ⋅ ) F(2,\cdot) F ( 2 , ⋅ ) ∣ z ∣ < 4 |z| < 4 ∣ z ∣ < 4 ∣ z ∣ = 3 |z| = 3 ∣ z ∣ = 3 P k ( 2 ) ≤ M 3 − k P_k(2) \le M3^{-k} P k ( 2 ) ≤ M 3 − k M = ∏ p ( 1 − 3 p − 2 ) − 1 ≤ 7.96896 M = \prod_p(1-3p^{-2})^{-1} \le7.96896 M = ∏ p ( 1 − 3 p − 2 ) − 1 ≤ 7.96896
2 k ∣ T k ∣ ≤ M k log 2 ( 2 3 ) k , (72) 2^k|T_k| \le\frac{M}{k\log2}\left(\frac{2}{3}\right)^k, \tag*{(72)} 2 k ∣ T k ∣ ≤ k log 2 M ( 3 2 ) k , (72) which is below 1.45 / k 1.45/k 1.45/ k k ≥ 6 k \ge6 k ≥ 6 k = 900 k = 900 k = 900 10 158 10^{158} 1 0 158 (67) is true with room of that order; it is stated in that form because 1.45 / k 1.45/k 1.45/ k (70) rather than forced here. (The first two terms of (54) are below 10 − 3 10^{-3} 1 0 − 3 k ≥ 37 k \ge37 k ≥ 37 E e x p E_{\rm exp} E exp k ≥ 900 k \ge900 k ≥ 900 M M M numerics/tk_certified.py, log tk.log.
Substituting (67) into (54) proves (69) , and no asymptotic O O O
Assembly of the unrestricted theorem
The four bounds of Section 5 now combine into a single explicit function φ \varphi φ φ \varphi φ
For k ≥ 900 k \ge900 k ≥ 900
R ( k ) = log 2 4 ( ( 2 k − 2 ) U max + 2 V max + 4 U max ) + S ∗ ( 2 D ( k ) + D ( k + 1 ) ) + α + ( 3 k + 4 ) + 3 ( 1.45 k + 10 − 3 ) , (73) R(k) = \frac{\log2}{4}\left((2k-2)U_{\max}+2V_{\max}+4U_{\max}\right) + S^*(2D(k)+D(k+1))+\alpha_+(3k+4)+3\left(\frac{1.45}{k}+10^{-3}\right), \tag*{(73)} R ( k ) = 4 log 2 ( ( 2 k − 2 ) U m a x + 2 V m a x + 4 U m a x ) + S ∗ ( 2 D ( k ) + D ( k + 1 )) + α + ( 3 k + 4 ) + 3 ( k 1.45 + 1 0 − 3 ) , (73) and
φ ( k ) = a 1 ( k 2 − 2 k − 1 ) − R ( k ) . (74) \varphi(k) = a_1(k^2-2k-1)-R(k). \tag*{(74)} φ ( k ) = a 1 ( k 2 − 2 k − 1 ) − R ( k ) . (74) Proposition 6.1. For k ≥ 900 k \ge900 k ≥ 900
2 E ( k ) − E ( k + 1 ) ≥ − R ( k ) . (75) 2E(k)-E(k+1) \ge-R(k). \tag*{(75)} 2 E ( k ) − E ( k + 1 ) ≥ − R ( k ) . (75) Consequently φ ( k ) > 0 \varphi(k)>0 φ ( k ) > 0 f k < f k + 1 f_k<f_{k+1} f k < f k + 1
Proof. Apply (55) at k k k k + 1 k+1 k + 1 S 1 S_1 S 1
log 2 4 ( V + ( 2 − 2 k ) U ) , \frac{\log2}{4}\left(V+(2-2k)U\right), 4 log 2 ( V + ( 2 − 2 k ) U ) , which is bounded below by (60). The first line of (73) is slightly larger than the resulting quantity, since
V max + ( 2 k − 2 ) U max ≤ ( 2 k − 2 ) U max + 2 V max + 4 U max . V_{\max}+(2k-2)U_{\max}\le(2k-2)U_{\max}+2V_{\max}+4U_{\max}. V m a x + ( 2 k − 2 ) U m a x ≤ ( 2 k − 2 ) U m a x + 2 V m a x + 4 U m a x . The Δ \Delta Δ
∣ 2 E exp ( k ) − E exp ( k + 1 ) ∣ ≤ 2 ( 1.45 k + 10 − 3 ) + ( 1.45 k + 1 + 10 − 3 ) ≤ 3 ( 1.45 k + 10 − 3 ) . \left|2E_{\exp}(k)-E_{\exp}(k+1)\right|
\le2\left(\frac{1.45}{k}+10^{-3}\right)
+\left(\frac{1.45}{k+1}+10^{-3}\right)
\le3\left(\frac{1.45}{k}+10^{-3}\right). ∣ 2 E e x p ( k ) − E e x p ( k + 1 ) ∣ ≤ 2 ( k 1.45 + 1 0 − 3 ) + ( k + 1 1.45 + 1 0 − 3 ) ≤ 3 ( k 1.45 + 1 0 − 3 ) . Adding the four lower bounds proves (75). Equation (42) then gives
f k + 1 − f k ≥ 2 − k − 1 φ ( k ) , f_{k+1}-f_k\ge2^{-k-1}\varphi(k), f k + 1 − f k ≥ 2 − k − 1 φ ( k ) , which proves the last assertion.
Lemma 6.2. With the certified constants in Proposition 5.6, φ ( 2440 ) > 1155 \varphi(2440)>1155 φ ( 2440 ) > 1155
Proof. Substitution in (74) gives the following certified one-sided bounds:
a 1 ( 2440 2 − 2 ⋅ 2440 − 1 ) > 390371.6 , a_1(2440^2-2\cdot2440-1)>390371.6, a 1 ( 244 0 2 − 2 ⋅ 2440 − 1 ) > 390371.6 , log 2 4 ( 4878 U max + 2 V max + 4 U max ) < 72.1 , \frac{\log2}{4}(4878U_{\max}+2V_{\max}+4U_{\max})<72.1, 4 log 2 ( 4878 U m a x + 2 V m a x + 4 U m a x ) < 72.1 , S ∗ ( 2 D ( 2440 ) + D ( 2441 ) ) < 276166.8 , S^*(2D(2440)+D(2441))<276166.8, S ∗ ( 2 D ( 2440 ) + D ( 2441 )) < 276166.8 , α + ( 3 ⋅ 2440 + 4 ) < 112977.6 , \alpha_+(3\cdot2440+4)<112977.6, α + ( 3 ⋅ 2440 + 4 ) < 112977.6 , 3 ( 1.45 2440 + 10 − 3 ) < 0.005. 3\left(\frac{1.45}{2440}+10^{-3}\right)<0.005. 3 ( 2440 1.45 + 1 0 − 3 ) < 0.005. Therefore
φ ( 2440 ) > 390371.6 − 72.1 − 276166.8 − 112977.6 − 0.005 > 1155. \varphi(2440)>390371.6-72.1-276166.8-112977.6-0.005>1155. φ ( 2440 ) > 390371.6 − 72.1 − 276166.8 − 112977.6 − 0.005 > 1155. The same substitution at k = 2433 k=2433 k = 2433 φ ( 2433 ) > 135 \varphi(2433)>135 φ ( 2433 ) > 135 φ ( 2432 ) < 0 < φ ( 2433 ) \varphi(2432)<0<\varphi(2433) φ ( 2432 ) < 0 < φ ( 2433 ) φ ( 2432 ) < 0 \varphi(2432)<0 φ ( 2432 ) < 0 lower bound on R ( 2432 ) R(2432) R ( 2432 ) 2440 2440 2440
Lemma 6.3. The function k ↦ φ ( k ) k\mapsto\varphi(k) k ↦ φ ( k ) k ≥ 2440 k\ge2440 k ≥ 2440
Proof. First,
a 1 ( ( k + 1 ) 2 − 2 ( k + 1 ) − 1 − ( k 2 − 2 k − 1 ) ) = a 1 ( 2 k − 1 ) . a_1\left((k+1)^2-2(k+1)-1-(k^2-2k-1)\right)=a_1(2k-1). a 1 ( ( k + 1 ) 2 − 2 ( k + 1 ) − 1 − ( k 2 − 2 k − 1 ) ) = a 1 ( 2 k − 1 ) . Moreover
D ( m + 1 ) − D ( m ) = log 2 ( log ( m + 2 ) + m log ( 1 + 1 m + 1 ) ) ≤ log 2 ( log ( m + 2 ) + 1 ) . D(m+1)-D(m)=\log2\left(\log(m+2)+m\log\left(1+\frac{1}{m+1}\right)\right)
\le\log2(\log(m+2)+1). D ( m + 1 ) − D ( m ) = log 2 ( log ( m + 2 ) + m log ( 1 + m + 1 1 ) ) ≤ log 2 ( log ( m + 2 ) + 1 ) . The last term in (73) decreases with k k k
R ( k + 1 ) − R ( k ) ≤ log 2 2 U max + 3 α + + S ∗ log 2 ( 2 log ( k + 2 ) + log ( k + 3 ) + 3 ) . (76) R(k+1)-R(k)\leq\frac{\log2}{2}U_{\max}+3\alpha_{+}+S^*\log2\bigl(2\log(k+2)+\log(k+3)+3\bigr). \tag*{(76)} R ( k + 1 ) − R ( k ) ≤ 2 log 2 U m a x + 3 α + + S ∗ log 2 ( 2 log ( k + 2 ) + log ( k + 3 ) + 3 ) . (76) Define, for real x ≥ 2440 x\geq2440 x ≥ 2440
M ( x ) = a 1 ( 2 x − 1 ) − log 2 2 U max − 3 α + − S ∗ log 2 ( 2 log ( x + 2 ) + log ( x + 3 ) + 3 ) . M(x)=a_1(2x-1)-\frac{\log2}{2}U_{\max}-3\alpha_{+}-S^*\log2\bigl(2\log(x+2)+\log(x+3)+3\bigr). M ( x ) = a 1 ( 2 x − 1 ) − 2 log 2 U m a x − 3 α + − S ∗ log 2 ( 2 log ( x + 2 ) + log ( x + 3 ) + 3 ) . Then
M ′ ( x ) = 2 a 1 − S ∗ log 2 ( 2 x + 2 + 1 x + 3 ) > 0 , M'(x)=2a_1-S^*\log2\left(\frac{2}{x+2}+\frac{1}{x+3}\right)>0, M ′ ( x ) = 2 a 1 − S ∗ log 2 ( x + 2 2 + x + 3 1 ) > 0 , because the certified lower bound for d d d 2 a 1 > 0.1312 2a_1>0.1312 2 a 1 > 0.1312 < 0.006 <0.006 < 0.006 x ≥ 2440 x\geq2440 x ≥ 2440 M ( 2440 ) > 146 M(2440)>146 M ( 2440 ) > 146 M ( k ) > 0 M(k)>0 M ( k ) > 0 k ≥ 2440 k\geq2440 k ≥ 2440 (76) , this yields φ ( k + 1 ) − φ ( k ) > 0 \varphi(k+1)-\varphi(k)>0 φ ( k + 1 ) − φ ( k ) > 0
Proof of Theorem 1.1. By Lemmas 6.2 and 6.3, φ ( k ) > 0 \varphi(k)>0 φ ( k ) > 0 k ≥ 2440 k\geq2440 k ≥ 2440 f k < f k + 1 f_k<f_{k+1} f k < f k + 1 (70) cover every integer from 6 through 2439. Together the analytic and certified ranges cover all k ≥ 6 k\geq6 k ≥ 6
A general principle
Conjecture 7.1. Let P \mathcal{P} P q q q
∑ p ∉ P log p p < ∞ . (77) \sum_{p\notin\mathcal{P}}\frac{\log p}{p}<\infty. \tag*{(77)} p ∈ / P ∑ p log p < ∞. (77) and let f k P f_k^{\mathcal{P}} f k P 1 / ( n log n ) 1/(n\log n) 1/ ( n log n ) n n n Ω ( n ) = k \Omega(n)=k Ω ( n ) = k P \mathcal{P} P f k P f_k^{\mathcal{P}} f k P q = 2 q=2 q = 2
Some hypothesis is needed. For P = { 2 } \mathcal{P}=\{2\} P = { 2 } f k P = 1 / ( 2 k k log 2 ) f_k^{\mathcal{P}}=1/(2^k k\log2) f k P = 1/ ( 2 k k log 2 ) P = { 2 , 3 } \mathcal{P}=\{2,3\} P = { 2 , 3 } q = 2 q=2 q = 2 P \mathcal{P} P ∑ p ∈ P p − s \sum_{p\in\mathcal{P}}p^{-s} ∑ p ∈ P p − s s → 1 + s\to1^+ s → 1 + ζ ( s ) z \zeta(s)z ζ ( s ) z f k P f_k^{\mathcal{P}} f k P q q q
Those examples, however, are already excluded by the weaker condition ∑ p ∉ P 1 / p < ∞ \sum_{p\notin\mathcal{P}}1/p<\infty ∑ p ∈ / P 1/ p < ∞ ∏ p ∉ P ( 1 − 1 / p ) \prod_{p\notin\mathcal{P}}(1-1/p) ∏ p ∈ / P ( 1 − 1/ p ) H P ( 1 , 1 ) H_{\mathcal{P}}(1,1) H P ( 1 , 1 ) (79) below, and it is a convergent product with positive value precisely when ∑ p ∉ P 1 / p < ∞ \sum_{p\notin\mathcal{P}}1/p<\infty ∑ p ∈ / P 1/ p < ∞ q ≥ 3 q\geq3 q ≥ 3 P \mathcal{P} P f k → 1 f_k\to1 f k → 1 [2 ] ; and for the two sparse examples the product is 0, as are their limits. So the unweighted condition separates a sequence with a positive limit, whose monotonicity is then decided by a correction term, from one that decays to 0. The sparse examples are the degenerate case of a single product, and not a separate phenomenon.
Why then does (77) carry the weight log p \log p log p (79) it is the second, over the complement, that needs a hypothesis at all. The first converges locally uniformly on σ > 1 / 2 \sigma>1/2 σ > 1/2 log L p = O ( p − 2 s ) \log L_p=O(p^{-2s}) log L p = O ( p − 2 s ) ≍ − z p − s \asymp-zp^{-s} ≍ − z p − s ∑ p ∈ Q p − σ < ∞ \sum_{p\in Q}p^{-\sigma}<\infty ∑ p ∈ Q p − σ < ∞ s s s s = 1 s=1 s = 1 2 − k 2^{-k} 2 − k ∑ p ∉ P log p p − 1 \sum_{p\notin\mathcal{P}}\frac{\log p}{p-1} ∑ p ∈ / P p − 1 l o g p (77) , and the weaker condition does not give it. Take the complement to be blocks of primes in [ X j , X j 1 + η j ] [X_j, X_j^{1+\eta_j}] [ X j , X j 1 + η j ] η j → 0 \eta_j \to0 η j → 0 log X j \log X_j log X j ∑ p ∉ P 1 / p \sum_{p\notin\mathcal{P}}1/p ∑ p ∈ / P 1/ p ∑ p ∉ P ( log p ) / p \sum_{p\notin\mathcal{P}}(\log p)/p ∑ p ∈ / P ( log p ) / p P \mathcal{P} P
The singularity structure is the one the two theorems above exploit. On ∣ z ∣ < q σ |z|<q^\sigma ∣ z ∣ < q σ ∏ p ∈ P ( 1 − z p − s ) − 1 \prod_{p\in\mathcal{P}}(1-zp^{-s})^{-1} ∏ p ∈ P ( 1 − z p − s ) − 1 z z z z = q σ z=q^\sigma z = q σ
∏ p ∈ P ( 1 − z p − s ) − 1 = ∏ p ( 1 − z p − s ) − 1 ∏ p ∉ P ( 1 − z p − s ) \prod_{p\in\mathcal{P}}(1-zp^{-s})^{-1}=\prod_p(1-zp^{-s})^{-1}\prod_{p\notin\mathcal{P}}(1-zp^{-s}) p ∈ P ∏ ( 1 − z p − s ) − 1 = p ∏ ( 1 − z p − s ) − 1 p ∈ / P ∏ ( 1 − z p − s ) then relates it to ζ ( s ) z H P ( s , z ) \zeta(s)^zH_{\mathcal{P}}(s,z) ζ ( s ) z H P ( s , z ) ∣ z ∣ < 2 σ |z|<2^\sigma ∣ z ∣ < 2 σ s = 1 s=1 s = 1 z = 2 z=2 z = 2 q = 2 q=2 q = 2 z = 2 z=2 z = 2 γ e − 2 ν \gamma e^{-2\nu} γ e − 2 ν J J J q ≥ 3 q\geq3 q ≥ 3 z = 2 z=2 z = 2 2 − k 2^{-k} 2 − k q = 2 q=2 q = 2 k 2 k^2 k 2 [3 ] for P = { all primes } \mathcal{P}=\{\text{all primes}\} P = { all primes }
In the interior case q ≥ 3 q\geq3 q ≥ 3
L p ( s , z ) = ( 1 − p − s ) z ( 1 − z p − s ) − 1 , log L p ( s , z ) = ∑ m ≥ 2 z m − z m p − m s . (78) L_p(s,z)=(1-p^{-s})^z(1-zp^{-s})^{-1},\qquad\log L_p(s,z)=\sum_{m\geq2}\frac{z^m-z}{m}p^{-ms}. \tag*{(78)} L p ( s , z ) = ( 1 − p − s ) z ( 1 − z p − s ) − 1 , log L p ( s , z ) = m ≥ 2 ∑ m z m − z p − m s . (78) the second identity being (11). Define H P H_{\mathcal{P}} H P H H H H o H_o H o
H P ( s , z ) = ∏ p ∉ P ( 1 − p − s ) z ∏ p ∈ P ( 1 − z p − s ) − 1 ( 1 − p − s ) z = ∏ p ∉ P ( 1 − p − s ) z ∏ p ∈ P L p ( s , z ) . (79) H_{\mathcal{P}}(s,z)=\prod_{p\notin\mathcal{P}}(1-p^{-s})^z\prod_{p\in\mathcal{P}}(1-zp^{-s})^{-1}(1-p^{-s})^z=\prod_{p\notin\mathcal{P}}(1-p^{-s})^z\prod_{p\in\mathcal{P}}L_p(s,z). \tag*{(79)} H P ( s , z ) = p ∈ / P ∏ ( 1 − p − s ) z p ∈ P ∏ ( 1 − z p − s ) − 1 ( 1 − p − s ) z = p ∈ / P ∏ ( 1 − p − s ) z p ∈ P ∏ L p ( s , z ) . (79) which is (8) when P \mathcal{P} P P = { p : p ≥ 3 } \mathcal{P}=\{p:p\geq3\} P = { p : p ≥ 3 } s > 1 s>1 s > 1 ζ ( s ) z \zeta(s)^z ζ ( s ) z P \mathcal{P} P
∑ k ≥ 0 P k P ( s ) z k = ∏ p ∈ P ( 1 − z p − s ) − 1 = ζ ( s ) z H P ( s , z ) , P k P ( s ) = ∑ Ω ( n ) = k p ∣ n ⇒ p ∈ P n − s . (80) \sum_{k\geq0}P_k^{\mathcal{P}}(s)z^k=\prod_{p\in\mathcal{P}}(1-zp^{-s})^{-1}=\zeta(s)^zH_{\mathcal{P}}(s,z),\qquad P_k^{\mathcal{P}}(s)=\sum_{\substack{\Omega(n)=k\\ p\mid n\Rightarrow p\in\mathcal{P}}}n^{-s}. \tag*{(80)} k ≥ 0 ∑ P k P ( s ) z k = p ∈ P ∏ ( 1 − z p − s ) − 1 = ζ ( s ) z H P ( s , z ) , P k P ( s ) = Ω ( n ) = k p ∣ n ⇒ p ∈ P ∑ n − s . (80) by the same computation, since ζ ( s ) − z = ∏ p ( 1 − p − s ) z \zeta(s)^{-z}=\prod_p(1-p^{-s})^z ζ ( s ) − z = ∏ p ( 1 − p − s ) z ζ ( s ) − z ∏ p ( 1 − z p − s ) − 1 \zeta(s)^{-z}\prod_p(1-zp^{-s})^{-1} ζ ( s ) − z ∏ p ( 1 − z p − s ) − 1 s = 1 s=1 s = 1 h j P ( s ) = [ z j ] H P ( s , z ) h_j^{\mathcal{P}}(s)=[z^j]H_{\mathcal{P}}(s,z) h j P ( s ) = [ z j ] H P ( s , z )
Proposition 7.2. Let P \mathcal{P} P q ≥ 3 q\geq3 q ≥ 3 Q \mathcal{Q} Q R ∈ ( 2 , 3 ) R\in(2,3) R ∈ ( 2 , 3 ) R < q R<q R < q q ≥ 3 q\geq3 q ≥ 3
(i) for every real s ≥ 1 s\geq1 s ≥ 1 ∣ z ∣ < q s |z|<q^s ∣ z ∣ < q s H P ( s , ⋅ ) H_{\mathcal{P}}(s,\cdot) H P ( s , ⋅ )
(ii) M : = sup s ≥ 1 sup ∣ z ∣ = R ∣ H P ( s , z ) ∣ < ∞ M:=\sup_{s\geq1}\sup_{|z|=R}|H_{\mathcal{P}}(s,z)|<\infty M := sup s ≥ 1 sup ∣ z ∣ = R ∣ H P ( s , z ) ∣ < ∞ ∣ h j P ( s ) ∣ ≤ M R − j |h_j^{\mathcal{P}}(s)|\leq MR^{-j} ∣ h j P ( s ) ∣ ≤ M R − j j ≥ 0 j\geq0 j ≥ 0 s ≥ 1 s\geq1 s ≥ 1
(iii) each h j P h_j^{\mathcal{P}} h j P C 1 C^1 C 1 [ 1 , ∞ ) [1,\infty) [ 1 , ∞ ) sup s ≥ 1 ∣ ( h j P ) ′ ( s ) ∣ ≤ M ′ R − j \sup_{s\geq1}|(h_j^{\mathcal{P}})'(s)|\leq M'R^{-j} sup s ≥ 1 ∣ ( h j P ) ′ ( s ) ∣ ≤ M ′ R − j M ′ < ∞ M'<\infty M ′ < ∞
1 2 ∂ s log H P ( 1 , 2 ) = − ∑ p ∈ P log p ( p − 1 ) ( p − 2 ) + ∑ p ∈ Q log p p − 1 . (81) \frac{1}{2}\partial_s\log H_{\mathcal{P}}(1,2)=-\sum_{p\in\mathcal{P}}\frac{\log p}{(p-1)(p-2)}+\sum_{p\in\mathcal{Q}}\frac{\log p}{p-1}. \tag*{(81)} 2 1 ∂ s log H P ( 1 , 2 ) = − p ∈ P ∑ ( p − 1 ) ( p − 2 ) log p + p ∈ Q ∑ p − 1 log p . (81) both sums finite, the second precisely by (77);
(iv) consequently
f k P = ∏ p ∈ Q ( 1 − 1 p ) + c P 2 − k + o ( 2 − k ) , c P = H P ( 1 , 2 ) ( γ + 1 2 ∂ s log H P ( 1 , 2 ) ) , (82) f_k^{\mathcal{P}}=\prod_{p\in\mathcal{Q}}\left(1-\frac{1}{p}\right)+c_{\mathcal{P}}2^{-k}+o(2^{-k}),\qquad c_{\mathcal{P}}=H_{\mathcal{P}}(1,2)\left(\gamma+\frac{1}{2}\partial_s\log H_{\mathcal{P}}(1,2)\right), \tag*{(82)} f k P = p ∈ Q ∏ ( 1 − p 1 ) + c P 2 − k + o ( 2 − k ) , c P = H P ( 1 , 2 ) ( γ + 2 1 ∂ s log H P ( 1 , 2 ) ) , (82) with H P ( 1 , 2 ) = ∏ p ∈ P ( 1 − p − 1 ) 2 ( 1 − 2 p − 1 ) − 1 ∏ p ∈ Q ( 1 − p − 1 ) 2 ∈ ( 0 , ∞ ) H_{\mathcal{P}}(1,2)=\prod_{p\in\mathcal{P}}(1-p^{-1})^{2}(1-2p^{-1})^{-1}\prod_{p\in\mathcal{Q}}(1-p^{-1})^{2}\in(0,\infty) H P ( 1 , 2 ) = ∏ p ∈ P ( 1 − p − 1 ) 2 ( 1 − 2 p − 1 ) − 1 ∏ p ∈ Q ( 1 − p − 1 ) 2 ∈ ( 0 , ∞ )
Proof. Parts (i)–(iii) are the analytic groundwork: that H P H_{\mathcal{P}} H P s s s ∣ z ∣ = R |z|=R ∣ z ∣ = R R ∈ ( 2 , 3 ) R\in(2,3) R ∈ ( 2 , 3 ) h j P h_j^{\mathcal{P}} h j P R − j R^{-j} R − j P \mathcal{P} P R > 2 R>2 R > 2 o ( 2 − k ) o(2^{-k}) o ( 2 − k ) R < 3 R<3 R < 3
(i) Fix s ≥ 1 s\geq1 s ≥ 1 K ⊂ { ∣ z ∣ < q s } K\subset\{|z|<q^s\} K ⊂ { ∣ z ∣ < q s } ∣ z ∣ ≤ R ′ < q s |z|\leq R'<q^s ∣ z ∣ ≤ R ′ < q s P \mathcal{P} P (78) gives log L p ( s , z ) = O R ′ ( p − 2 ) \log L_p(s,z)=O_{R'}(p^{-2}) log L p ( s , z ) = O R ′ ( p − 2 ) K K K (11) , whose hypothesis ∣ z p − s ∣ < 1 |zp^{-s}|<1 ∣ z p − s ∣ < 1 ∣ z ∣ ≤ R ′ < q s ≤ p s |z|\leq R'<q^s\leq p^s ∣ z ∣ ≤ R ′ < q s ≤ p s p ∈ P p\in\mathcal{P} p ∈ P ∑ p p − 2 s < ∞ \sum_p p^{-2s}<\infty ∑ p p − 2 s < ∞ s ≥ 1 s\geq1 s ≥ 1 Q \mathcal{Q} Q z log ( 1 − p − s ) = O R ′ ( p − s ) z\log(1-p^{-s})=O_{R'}(p^{-s}) z log ( 1 − p − s ) = O R ′ ( p − s )
∑ p ∈ Q p − s ≤ ∑ p ∈ Q 1 p ≤ 1 2 + 1 log 3 ∑ p ∈ Q log p p < ∞ ( s ≥ 1 ) , \sum_{p\in\mathcal{Q}}p^{-s}\leq\sum_{p\in\mathcal{Q}}\frac{1}{p}\leq\frac{1}{2}+\frac{1}{\log3}\sum_{p\in\mathcal{Q}}\frac{\log p}{p}<\infty\qquad(s\geq1), p ∈ Q ∑ p − s ≤ p ∈ Q ∑ p 1 ≤ 2 1 + log 3 1 p ∈ Q ∑ p log p < ∞ ( s ≥ 1 ) , the comparison 1 / p ≤ ( log p ) / ( p log 3 ) 1/p\leq(\log p)/(p\log3) 1/ p ≤ ( log p ) / ( p log 3 ) p ≥ 3 p\geq3 p ≥ 3 p = 2 p=2 p = 2 Q \mathcal{Q} Q P \mathcal{P} P q ≥ 3 q\geq3 q ≥ 3 log 2 < 1 \log2<1 log 2 < 1 1 / 2 1/2 1/2 (77) gives ∑ p ∈ Q 1 / p < ∞ \sum_{p\in\mathcal{Q}}1/p<\infty ∑ p ∈ Q 1/ p < ∞ z z z ∣ z ∣ < q s |z|<q^s ∣ z ∣ < q s
(ii) The two estimates in (i), applied with R ′ = R < 3 ≤ q ≤ q s R'=R<3\leq q\leq q^s R ′ = R < 3 ≤ q ≤ q s s s s p − s ≤ p − 1 p^{-s}\leq p^{-1} p − s ≤ p − 1
sup s ≥ 1 sup ∣ z ∣ = R ∣ log H P ( s , z ) ∣ ≤ C R ∑ p p − 2 + R ∑ p ∈ Q log 1 1 − 1 / p < ∞ , \sup_{s\geq1}\sup_{|z|=R}|\log H_{\mathcal{P}}(s,z)|\leq C_R\sum_p p^{-2}+R\sum_{p\in\mathcal{Q}}\log\frac{1}{1-1/p}<\infty, s ≥ 1 sup ∣ z ∣ = R sup ∣ log H P ( s , z ) ∣ ≤ C R p ∑ p − 2 + R p ∈ Q ∑ log 1 − 1/ p 1 < ∞ , which bounds M M M h j P h_j^{\mathcal{P}} h j P ∣ z ∣ = R |z|=R ∣ z ∣ = R
(iii) Differentiating (79) logarithmically term by term,
∂ s log H P ( s , z ) = − ∑ p ∈ P log p ∑ m ≥ 2 ( z m − z ) p − m s + z ∑ p ∈ Q log p p s − 1 . (83) \partial_s\log H_{\mathcal{P}}(s,z)=-\sum_{p\in\mathcal{P}}\log p\sum_{m\geq2}(z^m-z)p^{-ms}+z\sum_{p\in\mathcal{Q}}\frac{\log p}{p^s-1}. \tag*{(83)} ∂ s log H P ( s , z ) = − p ∈ P ∑ log p m ≥ 2 ∑ ( z m − z ) p − m s + z p ∈ Q ∑ p s − 1 log p . (83) For s ≥ 1 s\geq1 s ≥ 1 ∣ z ∣ ≤ R |z|\leq R ∣ z ∣ ≤ R C R ( log p ) p − 2 C_R(\log p)p^{-2} C R ( log p ) p − 2 R ( log p ) / ( p − 1 ) R(\log p)/(p-1) R ( log p ) / ( p − 1 ) Q \mathcal{Q} Q (77) . Both bounds are independent of s s s M M M [ 1 , ∞ ) × { ∣ z ∣ ≤ R } [1,\infty)\times\{|z|\leq R\} [ 1 , ∞ ) × { ∣ z ∣ ≤ R } ∂ s H P = H P ∂ s log H P \partial_sH_{\mathcal{P}}=H_{\mathcal{P}}\partial_s\log H_{\mathcal{P}} ∂ s H P = H P ∂ s log H P [ z j ] [z^j] [ z j ] ∣ z ∣ = R |z|=R ∣ z ∣ = R s s s ( h j P ) ′ ( s ) = [ z j ] ∂ s H P ( s , z ) (h_j^{\mathcal{P}})'(s)=[z^j]\partial_sH_{\mathcal{P}}(s,z) ( h j P ) ′ ( s ) = [ z j ] ∂ s H P ( s , z ) (10) , and Cauchy’s estimate then bounds it by M ′ R − j M'R^{-j} M ′ R − j z = 2 z=2 z = 2 (83) is
∑ m ≥ 2 ( 2 m − 2 ) p − m = 4 p ( p − 2 ) − 2 p ( p − 1 ) = 2 ( p − 1 ) ( p − 2 ) , \sum_{m\geq2}(2^m-2)p^{-m}=\frac{4}{p(p-2)}-\frac{2}{p(p-1)}=\frac{2}{(p-1)(p-2)}, m ≥ 2 ∑ ( 2 m − 2 ) p − m = p ( p − 2 ) 4 − p ( p − 1 ) 2 = ( p − 1 ) ( p − 2 ) 2 , which gives (81) after halving.
(iv) The plan is the one of Section 3, carried out with P \mathcal{P} P u = log ζ ( s ) u=\log\zeta(s) u = log ζ ( s ) J ( u ) d u J(u)\,du J ( u ) d u h j P h_j^{\mathcal{P}} h j P s = 1 s=1 s = 1 2 − k 2^{-k} 2 − k o ( 2 − k ) o(2^{-k}) o ( 2 − k )
Write P k P ( s ) = ∑ n − s P_k^{\mathcal{P}}(s)=\sum n^{-s} P k P ( s ) = ∑ n − s n n n f k P f_k^{\mathcal{P}} f k P f k P = ∫ 1 ∞ P k P ( s ) d s f_k^{\mathcal{P}}=\int_1^\infty P_k^{\mathcal{P}}(s)\,ds f k P = ∫ 1 ∞ P k P ( s ) d s ζ ( s ) z = e u z \zeta(s)^z=e^{uz} ζ ( s ) z = e u z u = log ζ ( s ) u=\log\zeta(s) u = log ζ ( s ) H P ( s , z ) = ∑ j h j P ( s ) z j H_{\mathcal{P}}(s,z)=\sum_j h_j^{\mathcal{P}}(s)z^j H P ( s , z ) = ∑ j h j P ( s ) z j P k P ( s ) = ∑ j = 0 k h j P ( s ) u k − j / ( k − j ) ! P_k^{\mathcal{P}}(s)=\sum_{j=0}^k h_j^{\mathcal{P}}(s)u^{k-j}/(k-j)! P k P ( s ) = ∑ j = 0 k h j P ( s ) u k − j / ( k − j )! P \mathcal{P} P u u u
f k P = ∫ 0 ∞ ∑ j = 0 k h j P ( s ( u ) ) u k − j ( k − j ) ! J ( u ) d u . f_k^{\mathcal{P}}=\int_0^\infty\sum_{j=0}^k h_j^{\mathcal{P}}(s(u))\frac{u^{k-j}}{(k-j)!}J(u)\,du. f k P = ∫ 0 ∞ j = 0 ∑ k h j P ( s ( u )) ( k − j )! u k − j J ( u ) d u . Split the integral at u = 1 u=1 u = 1 s 1 = s ( 1 ) s_1=s(1) s 1 = s ( 1 ) s 1 > 1 s_1>1 s 1 > 1 u u u u u u u ∈ ( 0 , 1 ] u\in(0,1] u ∈ ( 0 , 1 ] s ≥ s 1 s\ge s_1 s ≥ s 1 ∫ s 1 ∞ P k P ( s ) d s \int_{s_1}^{\infty}P_k^{\mathcal{P}}(s)\,ds ∫ s 1 ∞ P k P ( s ) d s n n n P k P P_k^{\mathcal{P}} P k P n ≥ q k ≥ 3 k n\ge q^k\ge3^k n ≥ q k ≥ 3 k P k P ( s ) ≤ 3 − k ( s − s 1 ) P k P ( s 1 ) P_k^{\mathcal{P}}(s)\le3^{-k(s-s_1)}P_k^{\mathcal{P}}(s_1) P k P ( s ) ≤ 3 − k ( s − s 1 ) P k P ( s 1 ) s ≥ s 1 s\ge s_1 s ≥ s 1
∫ s 1 ∞ P k P ( s ) d s ≤ P k P ( s 1 ) k log 3 ≤ 2 − k k log 3 ∏ p ∈ P ( 1 − 2 p − s 1 ) − 1 = O ( 2 − k k ) = o ( 2 − k ) , \int_{s_1}^{\infty}P_k^{\mathcal{P}}(s)\,ds\le\frac{P_k^{\mathcal{P}}(s_1)}{k\log3}\le\frac{2^{-k}}{k\log3}\prod_{p\in\mathcal{P}}(1-2p^{-s_1})^{-1}=O\left(\frac{2^{-k}}{k}\right)=o(2^{-k}), ∫ s 1 ∞ P k P ( s ) d s ≤ k log 3 P k P ( s 1 ) ≤ k log 3 2 − k p ∈ P ∏ ( 1 − 2 p − s 1 ) − 1 = O ( k 2 − k ) = o ( 2 − k ) , the middle step because the generating function (7) for P \mathcal{P} P P k P ( s 1 ) ≤ 2 − k ∑ j P j P ( s 1 ) 2 j P_k^{\mathcal{P}}(s_1)\le2^{-k}\sum_jP_j^{\mathcal{P}}(s_1)2^j P k P ( s 1 ) ≤ 2 − k ∑ j P j P ( s 1 ) 2 j 2 < 3 s 1 ≤ q s 1 2<3^{s_1}\le q^{s_1} 2 < 3 s 1 ≤ q s 1 [ 1 , ∞ ) [1,\infty) [ 1 , ∞ ) O ( e − 3 u ) O(e^{-3u}) O ( e − 3 u ) e 3 u e^{3u} e 3 u [ 1 , ∞ ) [1,\infty) [ 1 , ∞ ) u → ∞ u\to\infty u → ∞ [ 1 , ∞ ) [1,\infty) [ 1 , ∞ ) ∫ 0 ∞ u m / m ! e − a u d u = a − ( m + 1 ) \int_0^\infty u^m/m!e^{-au}\,du=a^{-(m+1)} ∫ 0 ∞ u m / m ! e − a u d u = a − ( m + 1 )
∫ 1 ∞ u m m ! J d u = 1 + γ 2 − m + O ( 3 − m ) , ∫ 1 ∞ u m m ! ε J d u = 1 2 2 − m + O ( 3 − m ) . (84) \int_1^\infty\frac{u^m}{m!}J\,du=1+\gamma2^{-m}+O(3^{-m}),\qquad\int_1^\infty\frac{u^m}{m!}\varepsilon J\,du=\frac{1}{2}2^{-m}+O(3^{-m}). \tag*{(84)} ∫ 1 ∞ m ! u m J d u = 1 + γ 2 − m + O ( 3 − m ) , ∫ 1 ∞ m ! u m ε J d u = 2 1 2 − m + O ( 3 − m ) . (84) the restriction to [ 1 , ∞ ) [1,\infty) [ 1 , ∞ ) ∫ 0 1 u m / m ! d u ≤ 1 / ( m + 1 ) ! \int_0^1u^m/m!\,du\le1/(m+1)! ∫ 0 1 u m / m ! d u ≤ 1/ ( m + 1 )!
h j P ( s ( u ) ) = h j P ( 1 ) + ε ( u ) ( ( h j P ) ′ ( 1 ) + r j ( u ) ) , ∣ r j ( u ) ∣ ≤ ε ( u ) Ω ( ε ( u ) ) R − j , h_j^{\mathcal{P}}(s(u))=h_j^{\mathcal{P}}(1)+\varepsilon(u)\left((h_j^{\mathcal{P}})'(1)+r_j(u)\right),\qquad|r_j(u)|\le\varepsilon(u)\Omega(\varepsilon(u))R^{-j}, h j P ( s ( u )) = h j P ( 1 ) + ε ( u ) ( ( h j P ) ′ ( 1 ) + r j ( u ) ) , ∣ r j ( u ) ∣ ≤ ε ( u ) Ω ( ε ( u )) R − j , where
Ω ( δ ) = sup ∣ z ∣ = R sup 1 ≤ s ≤ 1 + δ ∣ ∂ s H P ( s , z ) − ∂ s H P ( 1 , z ) ∣ . \Omega(\delta)=\sup_{|z|=R}\sup_{1\le s\le1+\delta}\left|\partial_sH_{\mathcal{P}}(s,z)-\partial_sH_{\mathcal{P}}(1,z)\right|. Ω ( δ ) = ∣ z ∣ = R sup 1 ≤ s ≤ 1 + δ sup ∣ ∂ s H P ( s , z ) − ∂ s H P ( 1 , z ) ∣ . Here r j ( u ) = ∫ 1 s ( u ) ( ( h j P ) ′ ( t ) − ( h j P ) ′ ( 1 ) ) d t r_j(u)=\int_1^{s(u)}((h_j^{\mathcal{P}})'(t)-(h_j^{\mathcal{P}})'(1))\,dt r j ( u ) = ∫ 1 s ( u ) (( h j P ) ′ ( t ) − ( h j P ) ′ ( 1 )) d t ∣ z ∣ = R |z|=R ∣ z ∣ = R ∂ s H P ( t , ⋅ ) − ∂ s H P ( 1 , ⋅ ) \partial_sH_{\mathcal{P}}(t,\cdot)-\partial_sH_{\mathcal{P}}(1,\cdot) ∂ s H P ( t , ⋅ ) − ∂ s H P ( 1 , ⋅ ) Ω ( t − 1 ) R − j \Omega(t-1)R^{-j} Ω ( t − 1 ) R − j r j r_j r j Ω \Omega Ω s ( u ) − 1 = ε ( u ) s(u)-1=\varepsilon(u) s ( u ) − 1 = ε ( u ) Ω ( δ ) → 0 \Omega(\delta)\to0 Ω ( δ ) → 0 δ → 0 \delta\to0 δ → 0 ∂ s H P \partial_sH_{\mathcal{P}} ∂ s H P [ 1 , ∞ ) × { ∣ z ∣ ≤ R } [1,\infty)\times\{|z|\le R\} [ 1 , ∞ ) × { ∣ z ∣ ≤ R } [ 1 , 2 ] × { ∣ z ∣ = R } [1,2]\times\{|z|=R\} [ 1 , 2 ] × { ∣ z ∣ = R } δ ≤ 1 \delta\le1 δ ≤ 1 R R R ∑ j > k \sum_{j>k} ∑ j > k R > 2 R>2 R > 2 O ( 3 − m ) O(3^{-m}) O ( 3 − m ) R < 3 R<3 R < 3
First group , the values h j P ( 1 ) h_j^{\mathcal{P}}(1) h j P ( 1 ) ∑ j > k \sum_{j>k} ∑ j > k ∑ j h j P ( 1 ) \sum_jh_j^{\mathcal{P}}(1) ∑ j h j P ( 1 ) 2 − k ∑ j h j P ( 1 ) 2 j 2^{-k}\sum_jh_j^{\mathcal{P}}(1)2^j 2 − k ∑ j h j P ( 1 ) 2 j O ( R − k ) O(R^{-k}) O ( R − k ) O ( 2 − k ( 2 / R ) k ) O(2^{-k}(2/R)^k) O ( 2 − k ( 2/ R ) k ) o ( 2 − k ) o(2^{-k}) o ( 2 − k ) R > 2 R>2 R > 2 O ( 3 − m ) O(3^{-m}) O ( 3 − m ) O ( 3 − k ∑ j ≤ k M R − j 3 j ) = O ( R − k ) = o ( 2 − k ) O\left(3^{-k}\sum_{j\le k}MR^{-j}3^j\right)=O(R^{-k})=o(2^{-k}) O ( 3 − k ∑ j ≤ k M R − j 3 j ) = O ( R − k ) = o ( 2 − k ) R < 3 R<3 R < 3 H P ( 1 , 1 ) + γ H P ( 1 , 2 ) 2 − k + o ( 2 − k ) H_{\mathcal P}(1,1)+\gamma H_{\mathcal P}(1,2)2^{-k}+o(2^{-k}) H P ( 1 , 1 ) + γ H P ( 1 , 2 ) 2 − k + o ( 2 − k )
Second group, the derivatives ( h j P ) ′ ( 1 ) (h_j^{\mathcal P})'(1) ( h j P ) ′ ( 1 ) The same argument with (iii) in place of (ii). The tail of 2 − k ∑ j ( h j P ) ′ ( 1 ) 2 j 2^{-k}\sum_j(h_j^{\mathcal P})'(1)2^j 2 − k ∑ j ( h j P ) ′ ( 1 ) 2 j O ( 2 − k ( 2 / R ) k ) O(2^{-k}(2/R)^k) O ( 2 − k ( 2/ R ) k ) O ( 3 − m ) O(3^{-m}) O ( 3 − m ) O ( R − k ) O(R^{-k}) O ( R − k ) 1 2 ∂ s H P ( 1 , 2 ) 2 − k + o ( 2 − k ) \frac{1}{2}\partial_sH_{\mathcal P}(1,2)2^{-k}+o(2^{-k}) 2 1 ∂ s H P ( 1 , 2 ) 2 − k + o ( 2 − k ) ∑ j ( h j P ) ′ ( 1 ) 2 j \sum_j(h_j^{\mathcal P})'(1)2^j ∑ j ( h j P ) ′ ( 1 ) 2 j ∂ s H P ( 1 , 2 ) \partial_sH_{\mathcal P}(1,2) ∂ s H P ( 1 , 2 ) R > 2 R>2 R > 2
Third group, the remainders r j r_j r j This is the only place where Ω \Omega Ω ε \varepsilon ε δ > 0 \delta>0 δ > 0 U U U Ω ( ε ( U ) ) < δ \Omega(\varepsilon(U))<\delta Ω ( ε ( U )) < δ ε ( U ) → 0 \varepsilon(U)\to0 ε ( U ) → 0 [ U , ∞ ) [U,\infty) [ U , ∞ ) δ R − j \delta R^{-j} δ R − j O ( δ 2 − k ) O(\delta2^{-k}) O ( δ 2 − k ) [ 1 , U ] [1,U] [ 1 , U ] O U ( 1 ) O_U(1) O U ( 1 ) u k − j / ( k − j ) ! u^{k-j}/(k-j)! u k − j / ( k − j )! ∫ 1 U u m / m ! d u ≤ U m + 1 / ( m + 1 ) ! \int_1^U u^m/m!\,du\le U^{m+1}/(m+1)! ∫ 1 U u m / m ! d u ≤ U m + 1 / ( m + 1 )! O U ( ∑ j ≤ k R − j U k − j + 1 / ( k − j + 1 ) ! ) = O U ( R − k ) = o ( 2 − k ) O_U\left(\sum_{j\le k}R^{-j}U^{k-j+1}/(k-j+1)!\right)=O_U(R^{-k})=o(2^{-k}) O U ( ∑ j ≤ k R − j U k − j + 1 / ( k − j + 1 )! ) = O U ( R − k ) = o ( 2 − k ) δ \delta δ o ( 2 − k ) o(2^{-k}) o ( 2 − k )
Collecting the two surviving coefficients gives c P = γ H P ( 1 , 2 ) + 1 2 ∂ s H P ( 1 , 2 ) c_{\mathcal P}=\gamma H_{\mathcal P}(1,2)+\frac{1}{2}\partial_sH_{\mathcal P}(1,2) c P = γ H P ( 1 , 2 ) + 2 1 ∂ s H P ( 1 , 2 ) L p ( s , 1 ) = 1 L_p(s,1)=1 L p ( s , 1 ) = 1 H P ( s , 1 ) = ∏ p ∈ Q ( 1 − p − s ) H_{\mathcal P}(s,1)=\prod_{p\in\mathcal Q}(1-p^{-s}) H P ( s , 1 ) = ∏ p ∈ Q ( 1 − p − s ) H P ( 1 , 2 ) H_{\mathcal P}(1,2) H P ( 1 , 2 ) ( s , z ) = ( 1 , 2 ) (s,z)=(1,2) ( s , z ) = ( 1 , 2 ) p ≥ q ≥ 3 p\ge q\ge3 p ≥ q ≥ 3 P \mathcal P P ( s , z ) = ( 1 , 2 ) (s,z)=(1,2) ( s , z ) = ( 1 , 2 ) 2 < 3 ≤ q 2<3\le q 2 < 3 ≤ q
By (82) it is positivity of c P c_{\mathcal P} c P
Corollary 7.3. c P > 0 c_{\mathcal P}>0 c P > 0 P \mathcal P P q ≥ 3 q\ge3 q ≥ 3 c P ≥ H P ( 1 , 2 ) β / d c_{\mathcal P}\ge H_{\mathcal P}(1,2)\,\beta/d c P ≥ H P ( 1 , 2 ) β / d f k P f_k^{\mathcal P} f k P
Proof. Put a p = log p ( p − 1 ) ( p − 2 ) ≥ 0 a_p=\frac{\log p}{(p-1)(p-2)}\ge0 a p = ( p − 1 ) ( p − 2 ) l o g p ≥ 0 b p = log p p − 1 ≥ 0 b_p=\frac{\log p}{p-1}\ge0 b p = p − 1 l o g p ≥ 0 p ≥ 3 p\ge3 p ≥ 3 q ≥ 3 q\ge3 q ≥ 3 2 ∈ Q 2\in\mathcal Q 2 ∈ Q P = { p ≥ 3 } ∖ ( Q ∖ { 2 } ) \mathcal P=\{p\ge3\}\setminus(\mathcal Q\setminus\{2\}) P = { p ≥ 3 } ∖ ( Q ∖ { 2 }) b 2 = log 2 b_2=\log2 b 2 = log 2 Q \mathcal Q Q
1 2 ∂ s log H P ( 1 , 2 ) = log 2 − ∑ p ≥ 3 a p + ∑ p ∈ Q ∖ { 2 } ( a p + b p ) . \frac{1}{2}\partial_s\log H_{\mathcal P}(1,2)=\log2-\sum_{p\ge3}a_p+\sum_{p\in\mathcal Q\setminus\{2\}}(a_p+b_p). 2 1 ∂ s log H P ( 1 , 2 ) = log 2 − p ≥ 3 ∑ a p + p ∈ Q ∖ { 2 } ∑ ( a p + b p ) . all three sums converging by (77). The last sum has non-negative terms, so the right side is least when it is empty, that is at Q = { 2 } \mathcal Q=\{2\} Q = { 2 } P = { p : p ≥ 3 } \mathcal P=\{p:p\ge3\} P = { p : p ≥ 3 } γ + 1 2 ∂ s log H P ( 1 , 2 ) \gamma+\frac{1}{2}\partial_s\log H_{\mathcal P}(1,2) γ + 2 1 ∂ s log H P ( 1 , 2 ) γ + log 2 − ∑ p ≥ 3 log p ( p − 1 ) ( p − 2 ) = β / d = 0.410970 … \gamma+\log2-\sum_{p\ge3}\frac{\log p}{(p-1)(p-2)}=\beta/d=0.410970\ldots γ + log 2 − ∑ p ≥ 3 ( p − 1 ) ( p − 2 ) l o g p = β / d = 0.410970 … H P ( 1 , 2 ) > 0 H_{\mathcal P}(1,2)>0 H P ( 1 , 2 ) > 0 c P > 0 c_{\mathcal P}>0 c P > 0 f k P − f k + 1 P = c P 2 − k − 1 + o ( 2 − k ) > 0 f_k^{\mathcal P}-f_{k+1}^{\mathcal P}=c_{\mathcal P}2^{-k-1}+o(2^{-k})>0 f k P − f k + 1 P = c P 2 − k − 1 + o ( 2 − k ) > 0 k k k
So the single inequality β > 0 \beta>0 β > 0 f k P f_k^{\mathcal P} f k P q ≥ 3 q\ge3 q ≥ 3 P \mathcal P P f k P f_k^{\mathcal P} f k P q = 2 q=2 q = 2 q = 2 q=2 q = 2 R > 2 R>2 R > 2 R < q R<q R < q P \mathcal P P
The constants of Proposition 7.2(iv) have been checked numerically, and the two checks should not be conflated. The constant formulas specialize correctly: at Q = { 2 } \mathcal Q=\{2\} Q = { 2 } H P ( 1 , 2 ) = 0.378695020 H_{\mathcal P}(1,2)=0.378695020 H P ( 1 , 2 ) = 0.378695020 d = 0.3786950315 d=0.3786950315 d = 0.3786950315 c P = 0.155632714 c_{\mathcal P}=0.155632714 c P = 0.155632714 β \beta β 10 − 7 10^{-7} 1 0 − 7 sequence f k P f_k^{\mathcal P} f k P q q q Q = { 2 } \mathcal Q=\{2\} Q = { 2 } O ( k 3 3 − k ) O(k^3 3^{-k}) O ( k 3 3 − k ) ( f k P − c 0 ) 2 k (f_k^{\mathcal P}-c_0)2^k ( f k P − c 0 ) 2 k c P c_{\mathcal P} c P k = 14 k=14 k = 14 Q = { 2 , 3 } \mathcal Q=\{2,3\} Q = { 2 , 3 } q = 5 q=5 q = 5 q ≥ 5 q\ge5 q ≥ 5 Q = { 2 } Q=\{2\} Q = { 2 } genP_check.py; alone among the computations here it uses ordinary floating-point rather than ball arithmetic, carries no certificate, and nothing else in the paper depends on it.
The boundary case q = 2 q=2 q = 2 L 2 L_2 L 2 (79) gives the exact identity H P = H P ∖ { 2 } / ( 1 − z 2 − s ) H_P=H_{P\setminus\{2\}}/(1-z2^{-s}) H P = H P ∖ { 2 } / ( 1 − z 2 − s ) z z z z = 2 s z=2^s z = 2 s 3 3 3 k 2 k^2 k 2 (2) comes from the pole. Whether the resulting coefficient depends on that numerator only through its value at the critical point, which would give Conjecture 7.1, we have been unable to settle, and we will return to it separately.
Certification
The numerical propositions are Proposition 3.3 and Proposition 5.6. Their statements are finite systems of interval inequalities, each established in ball arithmetic by Arb/FLINT [13 ] through its Python bindings, so that every enclosure carries a rigorous error radius. The scripts, their logs and the certificate are in the repository cited below, each script recording its settings in its docstring.
What ball arithmetic certifies is the computation. Transcribing an enclosure into a printed numeral, and combining several into a displayed chain, is ordinary writing, and that is where the errors in this manuscript have been found. A late adversarial pass turned up three: a stated bound rounded down rather than up, so that the text asserted what its own certificate did not establish; the two-part assembly of Σ 2 \Sigma_2 Σ 2 G 2 ( 1 , z ) − ρ / ( 3 − z ) G_2(1,z)-\rho/(3-z) G 2 ( 1 , z ) − ρ / ( 3 − z ) ∣ z ∣ = 4.2 |z|=4.2 ∣ z ∣ = 4.2 check_proof_spine.py compares every printed numeral against the log that certifies it, failing if a stated bound falls below it; verify.py checks the sweep tables and the Γ \Gamma Γ β > 0 \beta>0 β > 0 S = ∑ p ≥ 3 ( log p ) / ( ( p − 1 ) ( p − 2 ) ) S=\sum_{p\ge3}(\log p)/((p-1)(p-2)) S = ∑ p ≥ 3 ( log p ) / (( p − 1 ) ( p − 2 )) odd_beta_certified.py as a partial sum over p ≤ 10 8 p\le10^8 p ≤ 1 0 8 d d d d d d
Two constants are stated above more conservatively than the computation requires: the threshold 2440 of Lemma 6.2 in place of 2433, the first index at which φ \varphi φ
Σ 2 \Sigma_2 Σ 2 j ≤ 16 j\le16 j ≤ 16 s s s j > 16 j>16 j > 16 sup ∣ z ∣ = 4.5 \sup_{|z|=4.5} sup ∣ z ∣ = 4.5 1.9672165 ⋅ 10 6 1.9672165\cdot10^6 1.9672165 ⋅ 1 0 6 ( 3 / 4.5 ) 17 ( 1 − 3 / 4.5 ) − 1 (3/4.5)^{17}(1-3/4.5)^{-1} ( 3/4.5 ) 17 ( 1 − 3/4.5 ) − 1 Σ 2 ≤ 14414 \Sigma_2\le14414 Σ 2 ≤ 14414 C E C_E C E s s s
The constant C E C_E C E ( 1 , s ( 3 ) ] (1,s(3)] ( 1 , s ( 3 )] (25) are ratios in which the pole of ζ \zeta ζ s = 1 s=1 s = 1 s = 1 s=1 s = 1 1 + E / 2 1+E/2 1 + E /2 E / 2 E/2 E /2 [ 1 , s ( 3 ) ] [1,s(3)] [ 1 , s ( 3 )] Σ 2 \Sigma_2 Σ 2 s s s s − 1 s-1 s − 1 ( 1 , s ( 3 ) ] (1,s(3)] ( 1 , s ( 3 )] ζ \zeta ζ σ = s − s 0 \sigma=s-s_0 σ = s − s 0 s 0 = 1 + E / 2 s_0=1+E/2 s 0 = 1 + E /2 E = s ( 3 ) − 1 E=s(3)-1 E = s ( 3 ) − 1 ∣ σ ∣ ≤ E / 2 |\sigma|\le E/2 ∣ σ ∣ ≤ E /2 10 − 20 10^{-20} 1 0 − 20
This is the one constant here with no headroom: the four suprema are attained at s ( 3 ) s(3) s ( 3 ) C E = 5.4890955 … C_E=5.4890955\ldots C E = 5.4890955 … 4 ⋅ 10 − 6 4\cdot10^{-6} 4 ⋅ 1 0 − 6
The constant C N C_N C N Π ( t ) = ∑ p ≥ 3 p − t \Pi(t)=\sum_{p\ge3}p^{-t} Π ( t ) = ∑ p ≥ 3 p − t t = s , 2 s , … , 200 s t=s,2s,\ldots,200s t = s , 2 s , … , 200 s s = s ( 3 ) s=s(3) s = s ( 3 ) s ≈ 1.05 s\approx1.05 s ≈ 1.05 [12 ] applied beyond 10 8 10^8 1 0 8 Π \Pi Π Π ( t ) = ∑ n ≥ 1 μ ( n ) n − 1 log ζ ( n t ) − 2 − t \Pi(t)=\sum_{n\ge1}\mu(n)n^{-1}\log\zeta(nt)-2^{-t} Π ( t ) = ∑ n ≥ 1 μ ( n ) n − 1 log ζ ( n t ) − 2 − t [7 ] , as used in [2 (3.6)] . The recursion is run at 600 bits, cancellation costing about one digit every 2.3 steps over the 200 terms used; at 1200 bits the head agrees to every printed digit.
For the odd theorem, the coefficient bounds in (26) are obtained by cancelling the pole at z = 3 z=3 z = 3 0 < u ≤ 3 0<u\le3 0 < u ≤ 3 u ≥ 3 u\ge3 u ≥ 3 Proposition 3.1 .
For the unrestricted theorem, the s s s α + \alpha_+ α + s s s [ 1 , 2 ] [1,2] [ 1 , 2 ] U U U V V V (59) .
Availability and versions
Since the contribution is in part a certification, the artefacts and the versions they were produced with are part of the claim. The scripts, their logs and the certificate are at https://github.com/jaideepsaipadhi/erdos-sums-almost-primes. The numerics use python-flint 0.9.0; the certificate’s versions block records that binding’s version but not the FLINT release under it, which should be added before submission. The certificate records the parameters of the quadrature run, namely k max k_{\max} k m a x
Appendix A. Certified quadrature
Since 1 / ( n log n ) = ∫ 1 ∞ n − s d s 1/(n\log n)=\int_1^\infty n^{-s}\,ds 1/ ( n log n ) = ∫ 1 ∞ n − s d s f k = ∫ 1 ∞ P k ( s ) d s f_k=\int_1^\infty P_k(s)\,ds f k = ∫ 1 ∞ P k ( s ) d s s = 1 + e − u s=1+e^{-u} s = 1 + e − u
f k = ∫ − ∞ ∞ F ( u ) d u , F ( u ) = P k ( 1 + e − u ) e − u , (85) f_k=\int_{-\infty}^{\infty}F(u)\,du,\qquad F(u)=P_k(1+e^{-u})e^{-u}, \tag*{(85)} f k = ∫ − ∞ ∞ F ( u ) d u , F ( u ) = P k ( 1 + e − u ) e − u , (85) the integrand being analytic on the strip ∣ Im u ∣ < π / 2 |\operatorname{Im}u|<\pi/2 ∣ Im u ∣ < π /2 [ u − , u + ] [u_-,u_+] [ u − , u + ] u − = − 5.10 u_-=-5.10 u − = − 5.10 u + = 3.5 k max u_+=3.5k_{\max} u + = 3.5 k m a x k k k k k k H = 0.8 H=0.8 H = 0.8 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] N = 52 N = 52 N = 52 k max k_{\max} k m a x
Four of the six error sources in the certificate are arithmetic: ball arithmetic throughout, the explicit truncation balls in the Möbius series [7 ] for the prime zeta function, the discretization bound derived below, and interval-Newton enclosures for the Gauss nodes and weights. The remaining two are the tails of (85) , which are bounded as follows and are not negligible by inspection.
For the left tail, u ≤ u − u \le u_{-} u ≤ u − s ≥ s − = 1 + e 5.10 = 165.02 s \ge s_{-} = 1+e^{5.10} = 165.02 s ≥ s − = 1 + e 5.10 = 165.02 P 1 ( s ) = ∑ p p − s ≤ 2.78 ⋅ 2 − s P_1(s) = \sum_p p^{-s} \le2.78 \cdot2^{-s} P 1 ( s ) = ∑ p p − s ≤ 2.78 ⋅ 2 − s P 1 ( s ) 2 s = 1 + ∑ p ≥ 3 ( 2 / p ) s P_1(s)2^s = 1+\sum_{p\ge3}(2/p)^s P 1 ( s ) 2 s = 1 + ∑ p ≥ 3 ( 2/ p ) s s s s s = 2 s = 2 s = 2 P k ( s ) ≤ P 1 ( s ) k P_k(s) \le P_1(s)^k P k ( s ) ≤ P 1 ( s ) k P 1 ( s ) k P_1(s)^k P 1 ( s ) k n n n Ω ( n ) = k \Omega(n) = k Ω ( n ) = k P k ( s ) ≤ 2.78 k 2 − k s P_k(s) \le2.78^k2^{-ks} P k ( s ) ≤ 2.7 8 k 2 − k s
∫ − ∞ u − ∣ F ∣ d u = ∫ s − ∞ P k ( s ) d s ≤ 2.78 k 2 − k s − k log 2 . (86) \int_{-\infty}^{u_{-}} |F|\,du = \int_{s_{-}}^{\infty} P_k(s)\,ds \le\frac{2.78^k2^{-ks_{-}}}{k\log2}. \tag*{(86)} ∫ − ∞ u − ∣ F ∣ d u = ∫ s − ∞ P k ( s ) d s ≤ k log 2 2.7 8 k 2 − k s − . (86) For the right tail, P k ( s ) ≤ ( log ζ ( s ) ) k / k ! P_k(s) \le(\log\zeta(s))^k/k! P k ( s ) ≤ ( log ζ ( s ) ) k / k ! s > 1 s > 1 s > 1 [2 Lemma 5.6] . Substituting u = log ζ ( s ) u = \log\zeta(s) u = log ζ ( s ) d s = J ( u ) d u ds = J(u)\,du d s = J ( u ) d u J = ε ( 1 + γ ε + ⋯ ) J = \varepsilon(1+\gamma\varepsilon+\cdots) J = ε ( 1 + γ ε + ⋯ ) ε = 1 / ζ ⋅ ( 1 + γ ε + ⋯ ) ≥ e − u \varepsilon= 1/\zeta\cdot(1+\gamma\varepsilon+\cdots)\ge e^{-u} ε = 1/ ζ ⋅ ( 1 + γ ε + ⋯ ) ≥ e − u J ≥ e − u J \ge e^{-u} J ≥ e − u ≤ \le ≤ ∫ u + ∞ u k e − u d u / k ! \int_{u_{+}}^\infty u^k e^{-u}\,du/k! ∫ u + ∞ u k e − u d u / k ! J ( u ) ≤ e − u ( 1 + C e − u ) J(u) \le e^{-u}(1+Ce^{-u}) J ( u ) ≤ e − u ( 1 + C e − u ) u ≥ 1 u \ge1 u ≥ 1 C C C u ≥ u + = 1750 u \ge u_{+} = 1750 u ≥ u + = 1750 1 + O ( e − 1750 ) 1+O(e^{-1750}) 1 + O ( e − 1750 ) 2 ∫ u + ∞ u k e − u d u / k ! 2\int_{u_{+}}^\infty u^k e^{-u}\,du/k! 2 ∫ u + ∞ u k e − u d u / k ! Γ ( k + 1 , u + ) / k ! ≤ 2 u + k e − u + / k ! \Gamma(k+1,u_{+})/k! \le2u_{+}^k e^{-u_{+}}/k! Γ ( k + 1 , u + ) / k ! ≤ 2 u + k e − u + / k ! u + ≥ 2 k + 2 u_{+} \ge2k+2 u + ≥ 2 k + 2 [9 §8.11] , so
∫ u + ∞ ∣ F ∣ d u ≤ 2 u + k e − u + k ! . (87) \int_{u_{+}}^{\infty} |F|\,du \le\frac{2u_{+}^k e^{-u_{+}}}{k!}. \tag*{(87)} ∫ u + ∞ ∣ F ∣ d u ≤ k ! 2 u + k e − u + . (87) The hypothesis u + ≥ 2 k + 2 u_{+} \ge2k+2 u + ≥ 2 k + 2 u + = 3.5 k max u_{+} = 3.5k_{\max} u + = 3.5 k m a x k ≤ k max k \le k_{\max} k ≤ k m a x f k + 1 − f k ≍ a 1 k 2 2 − k f_{k+1}-f_k \asymp a_1k^2 2^{-k} f k + 1 − f k ≍ a 1 k 2 2 − k (87) increases in k k k u + u_{+} u + u + k / k ! u_{+}^k/k! u + k / k ! k ≤ u + k \le u_{+} k ≤ u + (86) is ( 2.78 ⋅ 2 − s − ) k / ( k log 2 ) (2.78\cdot2^{-s_{-}})^k/(k\log2) ( 2.78 ⋅ 2 − s − ) k / ( k log 2 ) k max = 500 k_{\max} = 500 k m a x = 500 u + = 1750 u_{+} = 1750 u + = 1750 (87) is below 10 − 272 10^{-272} 1 0 − 272 k = 500 k = 500 k = 500 (86) below 2.5 ⋅ 10 − 99 2.5\cdot10^{-99} 2.5 ⋅ 1 0 − 99 k = 2 k = 2 k = 2 6 ≤ k ≤ 199 6 \le k \le199 6 ≤ k ≤ 199 10 − 57 10^{-57} 1 0 − 57
For f f f E ρ E_{\rho} E ρ M M M N N N
∣ ∫ − 1 1 f ( x ) d x − I ^ N ∣ ≤ 64 15 M ρ − 2 N 1 − ρ − 2 . (88) \left|\int_{-1}^{1} f(x)\,dx-\widehat{I}_N\right| \le\frac{64}{15}\frac{M\rho^{-2N}}{1-\rho^{-2}}. \tag*{(88)} ∫ − 1 1 f ( x ) d x − I N ≤ 15 64 1 − ρ − 2 M ρ − 2 N . (88) [5 Thm. 4.5] , where the rule is indexed so that I n I_n I n n + 1 n+1 n + 1 ρ B − 2 ( n + 2 ) \rho_B^{-2(n+2)} ρ B − 2 ( n + 2 ) ρ B − 2 N \rho_B^{-2N} ρ B − 2 N [6 Thm. 19.3] . Here ρ B \rho_B ρ B ρ \rho ρ (88) with the interval-scaling factor H / 2 H/2 H /2 δ = 1.45 < π / 2 \delta= 1.45 < \pi/2 δ = 1.45 < π /2 u u u ρ B ≈ 7.385 \rho_B \approx7.385 ρ B ≈ 7.385 M M M ∣ P k ( s ) ∣ ≤ P k ( Re s ) |P_k(s)| \le P_k(\operatorname{Re}s) ∣ P k ( s ) ∣ ≤ P k ( Re s ) P k P_k P k
The per-panel bound is not monotone in the panel’s position, and it would be wrong to locate its maximum by inspection: on the right-hand side F ( u ) ≈ u k e − u / k ! F(u)\approx u^k e^{-u}/k! F ( u ) ≈ u k e − u / k ! (85) , which increases until u ≈ k u\approx k u ≈ k M M M k ≤ k max k\le k_{\max} k ≤ k m a x 4.6 × 10 − 87 4.6\times10^{-87} 4.6 × 1 0 − 87 k = 2 k=2 k = 2 k = 2 k = 2 k = 2 f k f_k f k
The pinned coefficient sweeps use primes up to 10 5 10^5 1 0 5 s s s α + \alpha_+ α + uv_constants.json and gamma_constant.json respectively, not in certificate.json, whose parameters block covers the quadrature run alone.
Acknowledgements
I am grateful to Jared Duker Lichtman for his guidance throughout this work, and for many helpful discussions.
Statement on the use of AI
During the preparation of this work the author used Claude Opus 5 (Anthropic), a large language model, in order to develop and check proofs, write and run the interval-arithmetic code, search and check the literature, and draft and edit the text. The proofs were checked by additional adversarial review passes of the model, and the numerical claims were certified in ball arithmetic by scripts available in the repository cited in Section 8. After using this tool, the author reviewed and edited the content as needed and takes full responsibility for the content of the article.
Department of Computer Science, Purdue University, West Lafayette, IN 47907, USA
Email address: jpadhi@purdue.edu
