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\hypersetup{pdftitle={A congruence modulo p squared for the p-th coefficient of e(1-z) to the power (1-z)/z},
pdfauthor={David Niedbala Giraudin}}
\title[A congruence for the $p$-th coefficient of $e(1-z)^{(1-z)/z}$]{A congruence modulo $p^2$
for the $p$-th coefficient of $e\,(1-z)^{(1-z)/z}$}
\author{David Niedbala Giraudin}
\address{ORCID 0009-0009-1526-1178}
\date{4 November 2026}

\begin{document}

\begin{abstract}
Define $E_n$ by $\sum_{n\ge0}E_nz^n=\exp\bigl(\sum_{k\ge1}z^k/(k(k+1))\bigr)=e\,(1-z)^{(1-z)/z}$.
Equivalently,
$e\,(1+1/x)^{-x}=\sum_{n\ge0}E_n(x+1)^{-n}$, 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 $p$,
$E_p\equiv-p/2\pmod{p^2}$; 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.
\end{abstract}

\maketitle

\section{Introduction}

Let
\[
G(z)=\exp\Bigl(\sum_{k\ge1}\frac{z^k}{k(k+1)}\Bigr)=e\,(1-z)^{(1-z)/z}=\sum_{n\ge0}E_nz^n,
\]
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 $\sum_{k\ge1}z^k/(k(k+1))=1+(1-z)\log(1-z)/z$, obtained from $1/(k(k+1))=1/k-1/(k+1)$.
Putting $z=1/(x+1)$ gives
\[
e\Bigl(1+\frac1x\Bigr)^{-x}=\sum_{n\ge0}\frac{E_n}{(x+1)^n}.
\]
Yang \cite{Ya1,Ya2} wrote $(1+1/x)^x=e\bigl(1-\sum_{n\ge1}b_n(x+1)^{-n}\bigr)$ to refine
Carleman's inequality, so $\sum_nE_nz^n=1/(1-\sum_nb_nz^n)$; the $E_n$ are the coefficients $S_n$
of \cite{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
\cite{OEIS}. We have found no previous result on the arithmetic of the $E_n$; see \cite{HM} 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
$\frac{1}{2(p-1)p}$, $\frac1{p(p+1)}$ and $\frac{1}{2^pp!}$. 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.

\begin{theorem}\label{thm:main}
For every odd prime $p$,
\[
E_p\equiv-\frac p2\pmod{p^2}.
\]
In particular $\vp(E_p)=1$.
\end{theorem}

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~\ref{sec:proof}) 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 $H_j=1+1/2+\dots+1/j$, $H_0=0$.

\section{Identities}

Put
\[
\Phi(u)=e\,(1+u)^{-1/u}=\exp\Bigl(\sum_{n\ge1}h_nu^n\Bigr),\qquad h_n=\frac{(-1)^{n+1}}{n+1},
\qquad \Phi(u)=\sum_{k\ge0}\varphi_ku^k,
\]
the second expression coming from $1-\log(1+u)/u=\sum_{n\ge1}(-1)^{n+1}u^n/(n+1)$. Thus
$\varphi_0=1$, $\varphi_1=1/2$.

\begin{lemma}\label{lem:id}
As formal power series over $\mathbb{Q}$,
\[
\text{(a)}\ \ G(z)=\frac{\Phi(-z)}{1-z},\qquad \text{(b)}\ \ G(z)=\Phi\Bigl(\frac z{1-z}\Bigr),\qquad
\text{(c)}\ \ (n+1)E_{n+1}=\sum_{j=0}^n\frac{E_j}{n+2-j}\quad(n\ge0).
\]
Consequently, for $n,k\ge0$ and every integer $p\ge1$,
\begin{equation}\label{eq:cons}
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.
\end{equation}
\end{lemma}

\begin{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
$\log G=\sum_kz^k/(k(k+1))$: $G'=G\sum_{m\ge0}z^m/(m+2)$, and compare the coefficients of $z^n$.

The first two identities in \eqref{eq:cons} follow from (a), since $1/(1-z)=\sum_jz^j$. For the
third, $[z^p](z/(1-z))^k=[z^{p-k}](1-z)^{-k}=\binom{p-1}{k-1}$ for $1\le k\le p$, and $0$ for
$k=0$ or $k>p$; apply (b).
\end{proof}

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

\begin{lemma}\label{lem:int}
\begin{enumerate}
\item[(i)] $h_n$ is $p$-integral for $n\le p-2$, and $\psi_k$ is $p$-integral for $k\le p-1$.
\item[(ii)] $\varphi_k=\psi_k$ for $k\le p-2$; in particular $\varphi_k$ and $E_k$ are $p$-integral
for $k\le p-2$.
\item[(iii)] $\varphi_{p-1}=\psi_{p-1}-1/p$, and $\varphi_p+1/p$ is $p$-integral.
\item[(iv)] $E_p=E_{p-2}+\varphi_{p-1}-\varphi_p$; hence $E_p$ is $p$-integral.
\end{enumerate}
\end{lemma}

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

(ii), (iii) Write $\Phi=\Psi\cdot R$ with $R=\exp\bigl(\sum_{n\ge p-1}h_nu^n\bigr)$. Since
$2(p-1)\ge p+1$, $R\equiv1+h_{p-1}u^{p-1}+h_pu^p\pmod{u^{p+1}}$, where $h_{p-1}=(-1)^p/p=-1/p$ and
$h_p=1/(p+1)$ is $p$-integral. Hence $\varphi_k=\psi_k$ for $k\le p-2$,
$\varphi_{p-1}=\psi_{p-1}-1/p$ 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 $\le p-2$ 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 \eqref{eq:cons}, (i) and the first part of (ii), $E_k=\sum_{i\le k}(-1)^i\varphi_i$ is
$p$-integral for $k\le p-2$.

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

\section{Proof of Theorem~\ref{thm:main}}\label{sec:proof}

\emph{Step A: reduction to a congruence modulo $p$.} For $0\le j\le p-1$,
$\binom{p-1}j=\prod_{i=1}^j(p-i)/i=(-1)^j\prod_{i=1}^j(1-p/i)\equiv(-1)^j(1-pH_j)\pmod{p^2}$. In the
last expression of \eqref{eq:cons} the terms $k\le p-2$ 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~\ref{lem:int}(iv),
\[
2E_p\equiv1+p\varphi_{p-1}-pH^*=p\,(\psi_{p-1}-H^*)\pmod{p^2},
\]
by Lemma~\ref{lem:int}(iii). Since $\psi_{p-1}$ and $H^*$ are $p$-integral, this already gives
$E_p\equiv0\pmod p$, and the theorem is equivalent to
\begin{equation}\label{eq:A}
\psi_{p-1}-H^*\equiv-1\pmod p.
\end{equation}

\emph{Step B: the residue $\psi_{p-1}$.} From $\Psi'=\Psi\sum_{j=1}^{p-2}jh_ju^{j-1}$, 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~\ref{lem:int}(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
\[
\psi_{p-1}\equiv\sum_{k=1}^{p-2}(-1)^k\Bigl(1+\frac1k\Bigr)\varphi_k\pmod p.
\]
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 \eqref{eq:A} is equivalent to
\begin{equation}\label{eq:C}
\sum_{k=1}^{p-2}(-1)^kH_k\varphi_k\equiv-E_{p-2}\pmod p.
\end{equation}

\emph{Step C: the recurrence.} By \eqref{eq:cons} and summation by parts,
\begin{align*}
\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{align*}
Since $H_{p-1}\equiv0\pmod p$ (pair $i$ with $p-i$), $H_{p-2}=H_{p-1}-1/(p-1)\equiv1$. So
\eqref{eq:C} 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~\ref{lem:id}(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 \eqref{eq:C}, hence \eqref{eq:A} and the theorem.\qed

\begin{remark}
After Step~A only $E_0,\dots,E_{p-2}$ and $\varphi_1,\dots,\varphi_{p-2}$ 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 $E_3=3/16\equiv-3/2\pmod 9$.
\end{remark}

\section{Numerical verification and a remark}

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

The next $p$-adic digit $\bigl((E_p+p/2)/p^2\bmod p\bigr)$ is $0,1,3,9,0,12,14,3$ for
$p=3,5,\dots,23$ 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 $p$.

\subsection*{Acknowledgement}
Technical assistance: Claude (Anthropic).

\begin{thebibliography}{OEIS}
\bibitem[HM]{HM} Y.~Hu, C.~Mortici, On the coefficients of an expansion of $(1+1/x)^x$ related to
Carleman's inequality, arXiv:1401.2236 (2014).
\bibitem[NG]{NG} D.~Niedbala Giraudin, The $p$-adic valuations of the coefficients in Yang's
expansion of $(1+1/x)^x$, preprint hexagon:2611.00074, \url{https://hexagonmath.org} (2026).
\bibitem[OEIS]{OEIS} The On-Line Encyclopedia of Integer Sequences, sequences A055505 and A055535,
\url{https://oeis.org}.
\bibitem[Ya1]{Ya1} X.~Yang, On Carleman's inequality, \emph{J. Math. Anal. Appl.} \textbf{253}
(2001), 691--694.
\bibitem[Ya2]{Ya2} X.~Yang, Approximations for constant $e$ and their applications, \emph{J. Math.
Anal. Appl.} \textbf{262} (2001), 651--659.
\end{thebibliography}

\end{document}