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A congruence modulo p^2 for the p-th coefficient of e(1-z)^((1-z)/z)

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Anthropic Claude Opus

Abstract

Define E_n by sum_{n>=0} E_n z^n = exp(sum_{k>=1} z^k/(k(k+1))) = e(1-z)^((1-z)/z). Equivalently, e(1+1/x)^(-x) = sum_{n>=0} E_n (x+1)^(-n), the reciprocal of the expansion of e^(-1)(1+1/x)^x used by Yang in refinements of Carleman's inequality. We prove that for every odd prime p, E_p = -p/2 (mod p^2); in particular v_p(E_p) = 1. The proof is elementary. With Phi(u) = e(1+u)^(-1/u), it combines the two expressions of E_p through the coefficients of Phi that come from the identities e(1-z)^((1-z)/z) = Phi(-z)/(1-z) and e(1-z)^((1-z)/z) = Phi(z/(1-z)), and it reduces the congruence to the differential recurrence of the E_n read modulo p. A script checks the statement and the intermediate congruences for all odd primes below 200 and a few larger ones.

Introduction

Let

G(z)=\exp\left(\sum_{k\geq1}\frac{z^k}{k(k+1)}\right)=e(1-z)^{(1-z)/z}=\sum_{n\geq0}E_nz^n,

so that $E_0=1$ , $E_1=1/2$ , $E_2=7/24$ , $E_3=3/16$ , $E_4=743/5760$ . The second expression follows from $\sum_{k\geq1}z^k/(k(k+1))=1+(1-z)\log(1-z)/z$ , obtained from $1/(k(k+1))=1/k-1/(k+1)$ .

Putting $z=1/(x+1)$ gives

e\left(1+\frac{1}{x}\right)^{-x}=\sum_{n\geq0}\frac{E_n}{(x+1)^n}.

Yang [4 Ya2] wrote $(1+1/x)^x=e(1-\sum_{n\geq1}b_n(x+1)^{-n})$ to refine Carleman’s inequality, so $\sum_{n}E_nz^n=1/(1-\sum_{n}b_nz^n)$ ; the $E_n$ are the coefficients $S_n$ of [NG], where the $p$ -adic valuations of the $b_n$ are studied. Equivalently, $1/G(z)=(1-z)e^{-1}(1-z)^{-1/z}$ , where $e^{-1}(1+x)^{1/x}$ has the coefficients tabulated in [OEIS]. We have found no previous result on the arithmetic of the $E_n$ ; see [1] for analytic properties of the $b_n$ .

Expanding the exponential, $E_n$ is a sum over the partitions of $n$ . For $n=p$ the only terms whose denominator is divisible by $p$ come from the partitions $(p-1)+1$ , $p$ and $1^p$ , namely $\frac{2}{(p-1)!p}$ , $\frac{1}{p(p+1)}$ and $\frac{1}{2^p p!}$ . Multiplied by $p$ , they are congruent modulo $p$ to $-1/2$ , $1$ and $-1/2$ (by Fermat and Wilson), so the poles cancel and $E_p$ is $p$ -integral. The point of the theorem is the next digit.

Theorem 1. For every odd prime $p$ ,

E_p=-\frac{p}{2}\pmod{p^2}.

In particular $v_p(E_p)=1$ .

For $p=2$ the statement fails, since $E_2=7/24$ . For example $E_3=3/16$ and $E_5=215/2304$ , with $E_3+3/2=27/16=3^3/2^4$ and $E_5+5/2=5975/2304=5^2\cdot239/(2^8\cdot3^2)$ .

The proof (Section 3) uses two expressions of $E_p$ through the coefficients of $\Phi(u)=e(1+u)^{-1/u}$ , the congruence $\binom{p-1}{j}\equiv(-1)^j(1-pH_j)\pmod{p^2}$ , and the recurrence of the $E_n$ obtained by differentiating $G$ . Throughout, $a\equiv b\pmod{p^j}$ means that $a-b\in p^j\mathbb{Z}_{(p)}$ , for rationals $a,b$ , and $H_j=1+1/2+\cdots+1/j$ , $H_0=0$ .

Identities

Put

\Phi(u)=e(1+u)^{-1/u}=\exp\left(\sum_{n\geq1}h_nu^n\right),\qquad h_n=\frac{(-1)^{n+1}}{n+1},\qquad\Phi(u)=\sum_{k\geq0}\varphi_ku^k.

the second expression coming from $1-\log(1+u)/u=\sum_{n\geq1}(-1)^{n+1}u^n/(n+1)$ . Thus $\varphi_0=1$ , $\varphi_1=1/2$ .

Lemma 2. As formal power series over $\mathbb{Q}$ ,

\text{(a)}\quad G(z)=\frac{\Phi(-z)}{1-z},\qquad\text{(b)}\quad G(z)=\Phi\left(\frac{z}{1-z}\right),\qquad\text{(c)}\quad(n+1)E_{n+1}=\sum_{j=0}^{n}\frac{E_j}{n+2-j}\quad(n\geq0).

Consequently, for $n,k\geq0$ and every integer $p\geq1$ ,

E_n=\sum_{k=0}^{n}(-1)^k\varphi_k,\qquad(-1)^k\varphi_k=E_k-E_{k-1}\ (E_{-1}=0),\qquad E_p=\sum_{k=1}^{p}\binom{p-1}{k-1}\varphi_k.

Proof. (a) $\Phi(-z)=e(1-z)^{1/z}$ and $G(z)=e(1-z)^{1/z-1}$ . (b) With $w=z/(1-z)$ we have $1+w=(1-z)^{-1}$ and $-1/w=-(1-z)/z$ , so $\Phi(w)=e(1-z)^{(1-z)/z}=G(z)$ . (c) Differentiate $\log G=\sum_{k\geq1}z^k/(k(k+1))$ : $G'=G\sum_{m\geq0}z^m/(m+2)$ , and compare the coefficients of $z^n$ .

The first two identities in (1) follow from (a), since $1/(1-z)=\sum_{j\geq0}z^j$ . For the third,

[z^p]\left(\frac{z}{1-z}\right)^k=[z^{p-k}](1-z)^{-k}=\binom{p-1}{k-1}

for $1\leq k\leq p$ , and $0$ for $k=0$ or $k>p$ ; apply (b). □

Fix an odd prime $p$ . Let $\psi_k$ be the coefficients of $\Psi(u)=\exp\left(\sum_{n=1}^{p-2}h_nu^n\right)$ .

Lemma 3. (i) $h_n$ is $p$ -integral for $n\leq p-2$ , and $\psi_k$ is $p$ -integral for $k\leq p-1$ . (ii) $\varphi_k=\psi_k$ for $k\leq p-2$ ; in particular $\varphi_k$ and $E_k$ are $p$ -integral for $k\leq p-2$ . (iii) $\varphi_{p-1}=\psi_{p-1}-1/p$ , and $\varphi_p+1/p$ is $p$ -integral. (iv) $E_p=E_{p-2}+\varphi_{p-1}-\varphi_p$ ; hence $E_p$ is $p$ -integral.

Proof. (i) The denominator of $h_n$ is $n+1\leq p-1$ . The coefficient $\psi_k$ is a sum of terms $\prod_j h_j^{m_j}/m_j!$ over the partitions $\sum_jjm_j=k$ with parts $j\leq p-2$ ; for $k\leq p-1$ every $m_j$ is less than $p.

(ii), (iii) Write $\Phi=\Psi\cdot R$ with $R=\exp\left(\sum_{n\geq p-1}h_nu^n\right)$ . Since $2(p-1)\geq p+1$ , $R\equiv1+h_{p-1}u^{p-1}+h_pu^p\pmod{u^{p+1}}$ , where $h_{p-1}=(-1)^p/p=-1/p$ and $h_p=1/(p+1)$ is $p$ -integral. Hence $\varphi_k=\psi_k$ for $k\leq p-2$ , $\varphi_{p-1}=\psi_{p-1}-1/p$ and $\varphi_p=\psi_p+\psi_1h_{p-1}+h_p=\psi_p-1/(2p)+h_p$ .

In the expansion of $\psi_p$ , all partitions of $p$ into parts $\leq p-2$ have multiplicities less than $p$ except $1^p$ , which contributes $h_1^p/p!=1/(2^pp!)$ . By Fermat and Wilson, $p/(2^pp!)=1/(2^p(p-1)!)\equiv1/(2\cdot(-1))=-1/2\pmod p$ , so $\psi_p+1/(2p)$ is $p$ -integral and so is $\varphi_p+1/p$ . By (1), (i) and the first part of (ii), $E_k=\sum_{i\leq k}(-1)^i\varphi_i$ is $p$ -integral for $k\leq p-2$ .

(iv) By (1), $E_p=E_{p-2}+(-1)^{p-1}\varphi_{p-1}+(-1)^p\varphi_p$ , and $p$ is odd. The right-hand side is $p$ -integral by (ii) and (iii). □

Proof of Theorem 1

Step A: reduction to a congruence modulo $p$ . For $0\leq j\leq p-1$ , $\binom{p-1}{j}=\prod_{i=1}^{j}(p-i)/i=(-1)^j\prod_{i=1}^{j}(1-p/i)\equiv(-1)^j(1-pH_j)\pmod{p^2}$ . In the last expression of (1) the terms $k\leq p-2$ have $p$ -integral $\varphi_k$ , so

E_p\equiv\sum_{k=1}^{p-2}(-1)^{k-1}\varphi_k-pH^*+(p-1)\varphi_{p-1}+\varphi_p\pmod{p^2},\qquad H^*=\sum_{k=1}^{p-2}(-1)^{k-1}H_{k-1}\varphi_k.

Here $\sum_{k=1}^{p-2}(-1)^{k-1}\varphi_k=1-E_{p-2}$ and $(p-1)\varphi_{p-1}+\varphi_p=p\varphi_{p-1}-(\varphi_{p-1}-\varphi_p)$ . Adding Lemma 3(iv),

2E_p\equiv1+p\varphi_{p-1}-pH^*=p(\varphi_{p-1}-H^*)\pmod{p^2},

@ends_mid=1 by Lemma 3(iii). Since $\psi_{p-1}$ and $H^*$ are $p$ -integral, this already gives $E_p \equiv0 \pmod{p}$ , and the theorem is equivalent to

\psi_{p-1}-H^* \equiv-1 \pmod{p}. \tag*{(1)}

Step B: the residue $\psi_{p-1}$ . From $\Psi'=\Psi\sum_{j=1}^{p-2}jh_j u^{j-1}$ , the coefficient of $u^{p-2}$ gives the exact identity $(p-1)\psi_{p-1}=\sum_{j=1}^{p-2}jh_j\psi_{p-1-j}$ , and $\psi_{p-1-j}=\varphi_{p-1-j}$ by Lemma 3(ii). Put $k=p-1-j$ , so $1\le k\le p-2$ . Modulo $p$ , $p-1\equiv-1$ , $j\equiv-(k+1)$ and $h_j=(-1)^{p-k}/(p-k)\equiv(-1)^k/k$ , so $jh_j\equiv-(-1)^k(1+1/k)$ and

\psi_{p-1}\equiv\sum_{k=1}^{p-2}(-1)^k\left(1+\frac{1}{k}\right)\varphi_k \pmod{p}.

Since $H^*=-\sum_{k=1}^{p-2}(-1)^kH_{k-1}\varphi_k$ and $H_{k-1}+1/k=H_k$ ,

\psi_{p-1}-H^*\equiv\sum_{k=1}^{p-2}(-1)^k(1+H_k)\varphi_k=E_{p-2}-1+\sum_{k=1}^{p-2}(-1)^kH_k\varphi_k \pmod{p}.

Hence (2) is equivalent to

\sum_{k=1}^{p-2}(-1)^kH_k\varphi_k\equiv-E_{p-2}\pmod{p}. \tag*{(2)}

Step C: the recurrence. By (1) and summation by parts,

\begin{aligned} \sum_{k=1}^{p-2}(-1)^kH_k\varphi_k &=\sum_{k=1}^{p-2}H_k(E_k-E_{k-1})=H_{p-2}E_{p-2}-H_1E_0-\sum_{k=1}^{p-3}(H_{k+1}-H_k)E_k\\ &=H_{p-2}E_{p-2}-1-\sum_{k=1}^{p-3}\frac{E_k}{k+1}. \end{aligned}

Since $H_{p-1}\equiv0\pmod{p}$ (pair $i$ with $p-i$ ), $H_{p-2}=H_{p-1}-1/(p-1)\equiv1$ . So (3) is equivalent to

2E_{p-2}\equiv1+\sum_{k=1}^{p-3}\frac{E_k}{k+1}=\sum_{j=0}^{p-3}\frac{E_j}{j+1}\pmod{p}.

This is Lemma 2(c) with $n=p-3$ , namely $(p-2)E_{p-2}=\sum_{j=0}^{p-3}E_j/(p-1-j)$ , read modulo $p$ : all $E_j$ with $j\le p-2$ are $p$ -integral, $p-2\equiv-2$ and $1/(p-1-j)\equiv-1/(j+1)$ for $0\le j\le p-3$ . This proves (3), hence (2) and the theorem. $\square$

Remark 4. After Step A only $E_0,\ldots,E_{p-2}$ and $\varphi_1,\ldots,\varphi_{p-2}$ occur, and the congruence finally needed is the recurrence of the $E_n$ at $n=p-3$ . For $p=3$ all the sums in Steps B and C are empty or reduce to a single term, and the theorem reads $E_3=3/16\equiv-3/2\pmod{9}$ .

Numerical verification and a remark

The ancillary file verify_Ep_congruence.py (Python 3, no dependency, exact rational arithmetic, about thirty seconds) computes $E_n$ by the recurrence of Lemma 2(c) and $\varphi_k$ from the exponential, and checks, for all odd primes $p<200$ and for $p=251$ , $307$ , $401$ : Lemma 2(a) and (b) coefficientwise up to $z^{p+1}$ , and the third identity of (1); Lemma 3; each displayed congruence and equivalence of Steps A, B and C, and the summation by parts; and Theorem 1.

The next $p$ -adic digit $(E_p+p/2)/p^2\pmod p$ is $0,1,3,9,0,12,14,3$ for $p=3,5,\ldots,23$ and shows no pattern. In the tested range, $E_p\equiv-p/2\pmod{p^3}$ holds only for $p=3$ and $p=13$ , which is compatible with a heuristic probability $1/p$ for each $p$ .

Acknowledgement

Acknowledgement. Technical assistance: Claude (Anthropic).

References

ORCID 0009-0009-1526-1178

References

[1]Y. Hu, C. Mortici, On the coefficients of an expansion of (1 + 1/x)^x related to Carleman’s inequality, arXiv:1401.2236 (2014).arxiv.org/abs/1401.2236
[4]X. Yang, On Carleman’s inequality, J. Math. Anal. Appl. 253 (2001), 691–694.DOI

A congruence modulo p^2 for the p-th coefficient of e(1-z)^((1-z)/z)

Abstract

Introduction

Identities

Proof of Theorem 1

Numerical verification and a remark

Acknowledgement

References

References

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