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\title[$p$-adic valuations of Yang's coefficients]{The $p$-adic valuations of the coefficients
in Yang's expansion of \texorpdfstring{$(1+1/x)^x$}{(1+1/x)\^{}x}}
\author{David Niedbala Giraudin}
\address{ORCID 0009-0009-1526-1178}
\date{3 November 2026}

\begin{document}

\begin{abstract}
Yang's coefficients $b_n$, defined by $(1+1/x)^x=e\bigl(1-\sum_{n\ge1}b_n(x+1)^{-n}\bigr)$, appear
in refinements of Carleman's inequality, and the literature on them concerns their analytic
properties. We study their $p$-adic valuations for an odd prime $p$. From the identity
$1-\sum_n b_nu^n=\exp\bigl(-\sum_{k\ge1}u^k/(k(k+1))\bigr)$ one gets $v_p(b_n)\ge-M(n)$, where
$M(n)$ is the optimum of a knapsack problem on the compositions of $n$. We compute $M(n)$ in
closed form and show that, for $n\ge2$, the optimum is attained by a unique composition exactly when
$n=(p-1)m$ with $m$ in an explicit set $U_p$ defined by a digit condition. For these $n$ the
bound is an equality: $v_p(b_n)=-(m+v_p(m!))$. The set $U_p$ has density $1/(p-1)$, so the
valuation of $b_n$ is determined on a set $N_p$ of indices of density $1/(p-1)^2$. The set $N_p$
contains the indices $p^K-1$, at which $b_{p^K-1}$ is the unique coefficient of minimal valuation among
$b_1,\dots,b_{p^K-1}$. The same statements hold for the coefficients of $e(1-u)^{1/u}$, and the
minimality statement for those of $e(1-u)^{1/u-1}$. A script checks all statements numerically
in finite ranges.
\end{abstract}

\maketitle

\section{Introduction}

Yang \cite{Ya} introduced the rational numbers $b_n$ defined by
\begin{equation}\label{eq:def}
\Bigl(1+\frac1x\Bigr)^x=e\Bigl(1-\sum_{n\ge1}\frac{b_n}{(x+1)^n}\Bigr),\qquad x>0,
\end{equation}
in a strengthened form of Carleman's inequality \cite{Ca}; the first values are $b_1=1/2$,
$b_2=1/24$, $b_3=1/48$, $b_4=73/5760$, $b_5=11/1280$, $b_6=3625/580608$ (the value of $b_6$
agrees with Finch \cite{Fi}); numerators and denominators are tabulated in OEIS A249276 and
A249277 \cite{OEIS}. The work on this sequence concerns its analytic properties; see Hu and
Mortici \cite{HM} and the references there. We study its $p$-adic valuations.

Put $u=1/(x+1)$, so that $1+1/x=1/(1-u)$ and $x=(1-u)/u$. Since $\log(1-u)=-\sum_{k\ge1}u^k/k$,
\[
x\log\Bigl(1+\frac1x\Bigr)=\frac{1-u}{u}\sum_{k\ge1}\frac{u^k}{k}=1-\sum_{k\ge1}\frac{u^k}{k(k+1)},
\]
so that, with $B(u)=\sum_{n\ge1}b_nu^n$,
\begin{equation}\label{eq:B}
1-B(u)=\exp\Bigl(-\sum_{k\ge1}\frac{u^k}{k(k+1)}\Bigr).
\end{equation}
We also consider
\[
T(u)=\exp\Bigl(-\sum_{k\ge1}\frac{u^k}{k+1}\Bigr)=e\,(1-u)^{1/u},\qquad
S(u)=\frac{T(u)}{1-u}=\frac1{1-B(u)}=e\,(1-u)^{1/u-1},
\]
with coefficients $T_n$ and $S_n$; thus $S_n=T_0+\dots+T_n$. The series $1/T(u)=(1-u)^{-1/u}/e$
has the coefficients $a_n=A055505(n)/A055535(n)$ of \cite{OEIS} (see also \cite[p.~293]{Co}), and
$b_n=a_{n-1}-a_n$. Differentiating \eqref{eq:B} gives the recurrence $b_1=1/2$,
$nb_n=1/(n+1)-\sum_{j=1}^{n-1}b_j/(n-j+1)$.

Throughout, $p$ is an odd prime, $c=p-2\ge1$, and $s_p(m)$ is the sum of the base-$p$ digits of
$m$, so that $\vp(m!)=(m-s_p(m))/(p-1)$ (Legendre). Let
\[
U_p=\bigl\{m\ge1:\ i+s_p(m+ic)>s_p(m)\ \text{for every } i\ge1\bigr\},\qquad
N_p=(p-1)\,U_p .
\]
It suffices to test $1\le i<s_p(m)$, since for $i\ge s_p(m)$ the inequality holds because
$s_p(m+ic)\ge1$.

\begin{theorem}\label{thm:exact}
Let $n=(p-1)m$ with $m\in U_p$. Then
\[
\vp(b_n)=\vp(T_n)=-\bigl(m+\vp(m!)\bigr)=-\frac{pm-s_p(m)}{p-1}.
\]
\end{theorem}

\begin{theorem}\label{thm:density}
For $X\ge p-2$,
\[
-\frac{p-2}{2}\ \le\ \#\bigl(U_p\cap[1,X]\bigr)-\frac{X}{p-1}\ \le\ (p-2)\bigl(1+\log_p(X+p-2)\bigr).
\]
In particular $U_p$ has density $1/(p-1)$, and $N_p$ has density $1/(p-1)^2$ in the positive
integers.
\end{theorem}

\begin{theorem}\label{thm:pK}
Let $K\ge1$, $R=p^K-1$, $R_K=R/(p-1)=1+p+\dots+p^{K-1}$ and
\[
D_K=R_K+\vp(R_K!)=\frac{p^{K+1}-p-K(p-1)}{(p-1)^2}=\sum_{j=0}^{K-1}(K-j)p^j.
\]
Then $R_K\in U_p$, so $\vp(b_R)=\vp(T_R)=-D_K$, while $\vp(b_n)>-D_K$ and $\vp(T_n)>-D_K$ for
$1\le n<R$. Moreover $\vp(S_R)=-D_K$ and $\vp(S_n)>-D_K$ for $1\le n<R$. Thus each of $b_R$,
$T_R$, $S_R$ is the unique coefficient of minimal $p$-adic valuation in its sequence on
$[1,R]$.
\end{theorem}

We know of no previous result on the $p$-adic valuations of the $b_n$. The proofs rest on a knapsack problem (Section~\ref{sec:knap}), which we solve in closed form
(Proposition~\ref{prop:reduction}), with a criterion for the uniqueness of the optimum
(Theorem~\ref{thm:unique}). A related phenomenon, for the powers of $\log(1+x)/x$, was found by
Davis \cite{Da}, where the coefficients of index $(p-1)m$ also behave like those of a series in
$x^{p-1}/p$. Krattenthaler and Rivoal \cite{KR} give general upper bounds for the denominators of
the Taylor coefficients of differentially algebraic power series; the statements here are exact
valuations for specific series.

\section{The knapsack bound}\label{sec:knap}

For $k\ge1$ put $\alpha(k)=\vp(k)+\vp(k+1)$ and $\alpha'(k)=\vp(k+1)$, so $\alpha'(k)\le\alpha(k)$. For a
composition $\kappa=(k_1,\dots,k_m)$ of $n$ (an ordered sequence of positive integers with sum
$n$) define the scores
\[
\sigma(\kappa)=\sum_{i=1}^m\alpha(k_i)+\vp(m!),\qquad \sigma'(\kappa)=\sum_{i=1}^m\alpha'(k_i)+\vp(m!),
\]
and let $M(n)$, $M'(n)$ be their maxima over the compositions of $n$.

\begin{lemma}\label{lem:bound}
For $n\ge1$, $\vp(b_n)\ge-M(n)$ and $\vp(T_n)\ge-M'(n)$. If $M(n)$ is attained by a unique
composition, then $\vp(b_n)=-M(n)$; if $M'(n)$ is attained by a unique composition, then
$\vp(T_n)=-M'(n)$.
\end{lemma}

\begin{proof}
Expanding the exponential in \eqref{eq:B},
\[
-b_n=[u^n]\exp\Bigl(-\sum_{k\ge1}\frac{u^k}{k(k+1)}\Bigr)
=\sum_{\kappa}\frac{(-1)^{m}}{m!}\prod_{i=1}^{m}\frac1{k_i(k_i+1)},
\]
a finite sum over all compositions $\kappa=(k_1,\dots,k_m)$ of $n$. The term of $\kappa$ has
valuation exactly $-\sigma(\kappa)$. The valuation of a finite sum is at least the minimum of the
valuations of its terms, with equality if this minimum is attained by a single term. The same
argument applies to $T_n$, with $1/(k_i+1)$ in place of $1/(k_i(k_i+1))$.
\end{proof}

\begin{lemma}\label{lem:part}
For $k\ge1$, $\alpha(k)\le k/(p-1)$, with equality if and only if $k=p-1$.
\end{lemma}

\begin{proof}
If $\alpha(k)=0$ the inequality is strict. If $a=\alpha(k)\ge1$, then, since $p$ divides at most one
of the consecutive integers $k$ and $k+1$, $p^a$ divides $k$ or $k+1$, so
$k\ge p^a-1=(p-1)(1+p+\dots+p^{a-1})\ge(p-1)a$. Both inequalities are equalities if and only if
$a=1$ and $k=p-1$.
\end{proof}

\section{The optimum}\label{sec:opt}

\begin{proposition}\label{prop:reduction}
For $n\ge1$,
\[
M(n)=M'(n)=\max_{0\le j\le n/(p-1)}f_n(j),\qquad f_n(j)=j+\vp\bigl((n-jc)!\bigr).
\]
More precisely, every composition of $n$ can be transformed, by a finite sequence of the moves
below, into a composition all of whose parts are equal to $1$ or $p-1$, and no move decreases
$\sigma$ or $\sigma'$. A composition with $j$ parts equal to $p-1$ and $i=n-j(p-1)$ parts equal to $1$
has $m=n-jc$ parts and $\sigma=\sigma'=f_n(j)$.
\end{proposition}

The moves replace one part $k\notin\{1,p-1\}$ of a composition, as follows.
\begin{itemize}
\item[(A)] If $(p-1)\mid k$ and $k>p-1$: by $q=k/(p-1)\ge2$ parts equal to $p-1$.
\item[(B)] If $k=q(p-1)+r$ with $q\ge1$ and $1\le r\le p-2$: by $q$ parts equal to $p-1$ and one part
equal to $r$.
\item[(C)] If $2\le k\le p-2$: by $k$ parts equal to $1$.
\end{itemize}
Every $k\notin\{1,p-1\}$ falls under exactly one move. Each move increases the number of parts,
so any sequence of moves stops, and it stops exactly at a composition with parts in
$\{1,p-1\}$.

\begin{proof}
The number of parts $m$ never decreases, so $\vp(m!)$ never decreases. The parts $1,\dots,p-2$
have $\alpha=\alpha'=0$, since $k$ and $k+1$ are then smaller than $p$, and the part $p-1$ has
$\alpha=\alpha'=1$. In move (A) the weight $q$ of the new parts exceeds $\alpha(k)\ge\alpha'(k)$, by
Lemma~\ref{lem:part}; so the move increases $\sigma$ and $\sigma'$ by at least $1$. In move (B),
Lemma~\ref{lem:part} gives $\alpha'(k)\le\alpha(k)\le k/(p-1)<q+1$, hence $\alpha'(k)\le\alpha(k)\le q$,
and the new parts have total weight $q$. In move (C) the total weight is $0$ before and after.
The last assertion follows from $\alpha(1)=\alpha'(1)=0$ and $\alpha(p-1)=\alpha'(p-1)=1$.
\end{proof}

\begin{theorem}\label{thm:unique}
Let $n\ge2$. The following are equivalent:
\begin{enumerate}
\item[(i)] $M(n)$ is attained by a unique composition of $n$;
\item[(ii)] $M'(n)$ is attained by a unique composition of $n$;
\item[(iii)] $n=(p-1)m$ with $m\in U_p$.
\end{enumerate}
In that case the optimal composition consists of $m$ parts equal to $p-1$, and
$M(n)=M'(n)=m+\vp(m!)$.
\end{theorem}

\begin{proof}
We write the proof for $\sigma$; for $\sigma'$ it is the same, since Proposition~\ref{prop:reduction}
holds for both scores. By Legendre's formula,
\begin{equation}\label{eq:f}
(p-1)f_n(j)=n+j-s_p(n-jc).
\end{equation}

\emph{Step 1: the composition with all parts equal to $1$ is never the unique optimum.} If
$n\le p-2$, all parts of all compositions of $n$ are at most $p-2$ and $m\le n<p$, so every
composition of $n$ has score $0$, and $n\ge2$ has at least two compositions. If $n\ge p-1$, let
$J=\lfloor n/(p-1)\rfloor\ge1$ and $n=(p-1)J+\rho$ with $0\le\rho\le p-2$. Then $n-Jc=J+\rho$ and
$s_p(J+\rho)\le J+\rho$, so \eqref{eq:f} gives $(p-1)(f_n(J)-f_n(0))\ge s_p(n)-\rho$. Since
$s_p(n)\equiv n\equiv\rho\pmod{p-1}$ and $s_p(n)\ge1$, we have $s_p(n)\ge\rho$. Hence
$f_n(J)\ge f_n(0)$, and the composition with $J$ parts $p-1$ followed by $\rho$ parts $1$ scores at
least as much as the composition with all parts $1$.

\emph{Step 2: (i) implies (iii).} Let $\kappa$ be the unique optimal composition. Applying a move to
$\kappa$ would give a different composition with at least the same score, so no move applies, and
$\kappa$ has $j$ parts $p-1$ and $i$ parts $1$. If $i\ge1$ and $j\ge1$, reordering the parts gives a
different composition with the same score. By Step~1, $j=0$ is excluded. Hence $i=0$, $n=(p-1)j$,
and $m:=j$. If $f_n(j')\ge f_n(m)$ for some $j'<m$, a composition with $j'$ parts $p-1$ and
$n-j'(p-1)\ge1$ parts $1$ would also be optimal. So $f_n(m)>f_n(m-i)$ for $1\le i\le m$. Since
$n-(m-i)c=m+ic$, \eqref{eq:f} turns this into $i+s_p(m+ic)>s_p(m)$ for $1\le i\le m$, which is
the condition $m\in U_p$ (the values $i>m\ge s_p(m)$ need no test).

\emph{Step 3: (iii) implies (i).} Let $n=(p-1)m$ with $m\in U_p$. By \eqref{eq:f} as in Step~2,
$f_n(m)>f_n(j)$ for $0\le j<m$, so by Proposition~\ref{prop:reduction} $M(n)=f_n(m)=m+\vp(m!)$.
Let $\kappa$ be any optimal composition, and transform it by moves into a composition $\kappa'$ with
parts in $\{1,p-1\}$. Then $\kappa'$ is optimal, so it has $m$ parts $p-1$ and no part $1$. A move of
type (A) increases the score, so none occurs. A move of type (B) or (C) creates a part smaller
than $p-1$, and no move decreases the number of parts smaller than $p-1$ (move (C) replaces such
a part by parts $1$), so none occurs either. Thus $\kappa=\kappa'$, which is the
composition with $m$ parts equal to $p-1$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:exact}]
Combine Lemma~\ref{lem:bound} and Theorem~\ref{thm:unique}; the last expression follows from
Legendre's formula.
\end{proof}

\begin{remark}
For $n\ge2$ outside $N_p$ the optimum is not unique, and $\vp(b_n)>-M(n)$ does occur: for $p=3$
this happens at $23$ of the indices $n\le80$, the smallest being $n=5$. The script checks this
count and the inequality of Lemma~\ref{lem:bound} for $p\le13$ and $n\le100$.
\end{remark}

\section{The indices \texorpdfstring{$p^K-1$}{p\^{}K-1}}\label{sec:pK}

\begin{lemma}\label{lem:carries}
Let $A$ be a nonnegative integer whose base-$p$ digits are all $0$ or $1$. Let $w\ge1$
and $B=wc$, and let $\kappa$ be the number of carries in the base-$p$ addition $A+B$. If $\kappa\ge1$, then
$s_p(B)\ge\kappa c+1$ and $w\ge\kappa$. In all cases, $\kappa\le w$ and $s_p(A+B)\ge s_p(A)+1-\kappa\ge s_p(A)+1-w$.
\end{lemma}

\begin{proof}
A carry out of a position where $A$ has digit $1$ requires a digit of $B$ at least $p-2=c$ if a
carry comes in, and at least $p-1=c+1$ if none comes in. A carry out of a position where $A$ has
digit $0$ requires the digit $p-1$ of $B$ and an incoming carry. The lowest carry position
receives no carry. Hence the digits of $B$ at the $\kappa$ carry positions are at least $c$, and the
one at the lowest carry position is at least $c+1$; this gives $s_p(B)\ge\kappa c+1$, and also
$B\ge c\sum_{\text{carry positions } t}p^t\ge c\kappa$, that is $w\ge\kappa$. Kummer's theorem gives
$s_p(A+B)=s_p(A)+s_p(B)-(p-1)\kappa\ge s_p(A)+1-\kappa$, and if $\kappa=0$ then $s_p(A+B)=s_p(A)+s_p(B)\ge
s_p(A)+1$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:pK}]
The two expressions of $D_K$ follow from $p^K-1=(p-1)R_K$, $s_p(R_K)=K$ and Legendre's formula.

\emph{$R_K\in U_p$.} Apply Lemma~\ref{lem:carries} with $A=R_K$ and $B=ic$: for $i\ge1$,
$i+s_p(R_K+ic)\ge i+K+1-\kappa\ge K+1>K=s_p(R_K)$, since $i\ge\kappa$. By Theorem~\ref{thm:exact},
$\vp(b_R)=\vp(T_R)=-(R_K+\vp(R_K!))=-D_K$.

\emph{$M(p^K-2)\le D_K-1$.} Let $n=p^K-2$. Then $\lfloor n/(p-1)\rfloor=R_K-1$, and for
$0\le j\le R_K-1$ write $w=R_K-j\ge1$. Since $R_Kc=p^K-1-R_K$, we get $n-jc=(R_K-1)+wc$. The
integer $R_K-1=p+\dots+p^{K-1}$ has digit $0$ in position $0$ and $s_p(R_K-1)=K-1$, so
Lemma~\ref{lem:carries} gives $s_p(n-jc)\ge K-w$. By \eqref{eq:f},
\[
(p-1)f_n(j)=n+j-s_p(n-jc)\le p^K-2+R_K-K=(p-1)D_K-1,
\]
using $(p-1)D_K=(p-1)R_K+R_K-K=p^K-1+R_K-K$. Since $f_n(j)$ is an integer, $f_n(j)\le D_K-1$, and
Proposition~\ref{prop:reduction} gives $M(p^K-2)=M'(p^K-2)\le D_K-1$ (equality holds in all the
cases computed by the script). (For $K=1$, $n=p-2$, and the
statement is $M(n)=0$.)

\emph{Minimality.} Appending a part $1$ to a composition does not decrease $\sigma$ or $\sigma'$, so $M$
and $M'$ are nondecreasing. For $1\le n<R$, Lemma~\ref{lem:bound} gives
$\vp(b_n)\ge-M(n)\ge-M(R-1)\ge-(D_K-1)$, and the same for $T_n$. Finally $S_n=T_0+\dots+T_n$ with
$T_0=1$; for $n<R$ every term has valuation $>-D_K$, and for $n=R$ the term $T_R$ is the only one
of valuation $-D_K$. Hence $\vp(S_n)>-D_K$ for $n<R$ and $\vp(S_R)=-D_K$.
\end{proof}

\section{The density of \texorpdfstring{$U_p$}{U\_p}}\label{sec:density}

Put $\Lambda(y)=y+c\,s_p(y)$ for $y\ge1$. For $m\ge1$ and $i\ge1$,
$\Lambda(m+ic)-\Lambda(m)=c\bigl(i+s_p(m+ic)-s_p(m)\bigr)$, so
\begin{equation}\label{eq:tail}
m\in U_p\iff \Lambda(m)<\Lambda(y)\ \text{for every } y>m \text{ with } y\equiv m\pmod c .
\end{equation}

\begin{lemma}\label{lem:jump}
If $m\in U_p$, then $\min_{i\ge1}\bigl(\Lambda(m+ic)-\Lambda(m)\bigr)=c(p-1)$.
\end{lemma}

\begin{proof}
Let $\Delta=\min_{i\ge1}\bigl(i+s_p(m+ic)-s_p(m)\bigr)$; the minimum exists since the quantity tends to
infinity with $i$, and $\Delta\ge1$ because $m\in U_p$. Since $s_p(y)\equiv y\pmod{p-1}$, we have
$\Lambda(y)\equiv(1+c)y=(p-1)y\equiv0\pmod{p-1}$ for every $y$, so $c\Delta\equiv0\pmod{p-1}$. As $c=p-2$ and
$p-1$ are coprime, $(p-1)\mid\Delta$, hence $\Delta\ge p-1$. On the other hand, taking $i=1$ and using
$s_p(m+c)\le s_p(m)+s_p(c)=s_p(m)+c$, we get $\Delta\le1+c=p-1$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:density}]
Fix a residue class modulo $c$, with elements $x_0<x_1<\dots$, where $1\le x_0\le c$ and
$x_{k+1}=x_k+c$. Put $L(x)=\min\{\Lambda(y):y\ge x,\ y\equiv x\pmod c\}$, which exists since
$\Lambda(y)\ge y$. Then $x\le L(x)\le\Lambda(x)=x+c\,s_p(x)$. By \eqref{eq:tail}, if $x_k\notin U_p$ then
$L(x_k)=L(x_{k+1})$, and if $x_k\in U_p$ then $L(x_k)=\Lambda(x_k)$ and, by Lemma~\ref{lem:jump},
$L(x_{k+1})-L(x_k)=c(p-1)$. Let $x_\ell\le X<x_{\ell+1}$. Summing over $0\le k\le\ell$,
\[
c(p-1)\cdot\#\bigl(U_p\cap\{x_0,\dots,x_\ell\}\bigr)=L(x_{\ell+1})-L(x_0),
\]
which lies between $x_{\ell+1}-x_0-c\,s_p(x_0)$ and $x_{\ell+1}-x_0+c\,s_p(x_{\ell+1})$. Over the $c$
classes, the numbers $x_{\ell+1}$ are $X+1,\dots,X+c$ and the numbers $x_0$ are $1,\dots,c$, so the
differences $x_{\ell+1}-x_0$ add up to $cX$. Since $s_p(x_0)=x_0$ and
$s_p(y)\le(p-1)(1+\log_py)$, summing and dividing by $c(p-1)$ gives the inequalities of the
theorem. The density of $N_p$ follows, since $n\in N_p$ requires $(p-1)\mid n$.
\end{proof}

\section{Verification}

The ancillary file \texttt{verify\_carleman\_padic.py} (Python~3, no dependency, exact
rational arithmetic, about fifteen seconds) computes $b_n$, $T_n$, $S_n$ for $n\le250$ and checks:
\begin{itemize}
\item \eqref{eq:B} against the recurrence of the introduction; the coefficients of $1/T$ against
the sequences A055505 and A055535 for $n\le17$; $b_n=a_{n-1}-a_n$; $S=1/(1-B)$;
\item for $p\le13$ and $n\le100$: $M(n)$ and $M'(n)$ computed by dynamic programming over all
compositions against Proposition~\ref{prop:reduction}; the uniqueness of the optimum against
Theorem~\ref{thm:unique}; the inequalities of Lemma~\ref{lem:bound};
\item for $p\le31$: Theorem~\ref{thm:exact} on all $n\in N_p$ with $n\le250$; Theorem~\ref{thm:pK}
for all $p^K-1\le250$; Lemma~\ref{lem:jump} and the bounds of Theorem~\ref{thm:density} for
$X=60000$.
\end{itemize}
These computations check the statements; the proofs do not depend on them.

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

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\bibitem[Da]{Da} D.~M.~Davis, Coefficients in powers of the log series, arXiv:1001.3068 (2010).
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\bibitem[HM]{HM} Y.~Hu, C.~Mortici, On the coefficients of an expansion of $(1+1/x)^x$ related to
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\end{thebibliography}

\end{document}