The p-adic valuations of the coefficients in Yang's expansion of (1+1/x)^x

Introduction

Yang [Ya] introduced the rational numbers $b_n$ defined by

in a strengthened form of Carleman’s inequality [1]; 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 [4]); numerators and denominators are tabulated in OEIS A249276 and A249277 [7]. The work on this sequence concerns its analytic properties; see Hu and Mortici [5] 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 $\log(1-u)=-\sum_{k\geq1}u^k/k$,

so that, with $B(u)=\sum_{n\geq1}b_nu^n$,

We also consider

with coefficients $T_n$ and $S_n$; thus $S_n=T_0+\cdots+T_n$. The series $1/T(u)=(1-u)^{-1/u}/e$ has the coefficients $a_n=A055505(n)/A055535(n)$ of [7] (see also [2 p. 293]), and $b_n=a_{n-1}-a_n$.

Differentiating (2) 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, $c=p-2\geq1$, and $s_p(m)$ is the sum of the base-$p$ digits of $m$, so that $v_p(m!)=(m-s_p(m))/(p-1)$ (Legendre). Let

It suffices to test $1 \le i < s_p(m)$, since for $i \ge s_p(m)$ the inequality holds because $s_p(m+ic) \ge1$.

We know of no previous result on the $p$-adic valuations of the $b_n$. The proofs rest on a knapsack problem (§ 2), which we solve in closed form (Proposition 6), with a criterion for the uniqueness of the optimum (Theorem 7). A related phenomenon, for the powers of $\log(1+x)/x$, was found by Davis [3], where the coefficients of index $(p-1)m$ also behave like those of a series in $x^{p-1/p}$. Krattenthaler and Rivoal [6] 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.

The knapsack bound

For $k\ge1$ put $\alpha(k)=v_p(k)+v_p(k+1)$ and $\alpha'(k)=v_p(k+1)$, so $\alpha'(k)\le\alpha(k)$. For a composition $\kappa=(k_1,\ldots,k_m)$ of $n$ (an ordered sequence of positive integers with sum $n$) define the scores

and let $M(n)$, $M'(n)$ be their maxima over the compositions of $n$.

Proof. Expanding the exponential in (2),

a finite sum over all compositions $\kappa=(k_1,\ldots,k_m)$ 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 $1/(k_i(k_i+1))$. □

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 $k\ge p^a-1=(p-1)(1+p+\cdots+p^{a-1})\ge(p-1)a$. Both inequalities are equalities if and only if $a=1$ and $k=p-1$. □

The optimum

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 $p-1$.

Proof of Theorem 1. Combine Lemma 4 and Theorem 7; the last expression follows from Legendre’s formula.

Remark 8. For $n\ge2$ outside $N_p$ the optimum is not unique, and $v_p(b_n)>-M(n)$ 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 4 for $p\le13$ and $n\le100$.

The indices $p^K-1$

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 $s_p(A+B)=s_p(A)+s_p(B)\ge s_p(A)+1$.

Proof of Theorem 3. The two expressions of $D_K$ follow from $p^K-1=(p-1)R_K$, $s_p(R_K)=K$ and Legendre’s formula.

$R_K\in U_p$. Apply Lemma 9 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 1, $v_p(b_R)=v_p(T_R)=-(R_K+v_p(R_K!))=-D_K$.

$M(p^K-2)\le D_K-1$. 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 $R_K-1=p+\cdots+p^{K-1}$ has digit 0 in position 0 and $s_p(R_K-1)=K-1$, so Lemma 9 gives $s_p(n-jc)\ge K-w$. By (3),

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 6 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 $M(n)=0$.)

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 4 gives $v_p(b_n)\ge-M(n)\ge-M(R-1)\ge-(D_K-1)$, and the same for $T_n$. Finally $S_n=T_0+\cdots+T_n$ 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 $v_p(S_n)>-D_K$ for $n<R$ and $v_p(S_R)=-D_K$.

The density of $U_p$

Put $\Lambda(y)=y+cs_p(y)$ for $y\ge1$. For $m\ge1$ and $i\ge1$, $\Lambda(m+ic)-\Lambda(m)=c(i+s_p(m+ic)-s_p(m))$, so

Proof. Let $\Delta=\min_{i\ge1}(i+s_p(m+ic)-s_p(m))$; 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 $\Delta\le1+c=p-1$.

Proof of Theorem 2. Fix a residue class modulo $c$, with elements $x_0 < x_1 < \cdots$, 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 (4), 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 10, $L(x_{k+1}) - L(x_k) = c(p-1)$. Let $x_\ell\le X < x_{\ell+1}$. Summing over $0 \le k \le\ell$,

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 $X+1,\ldots,X+c$ and the numbers $x_0$ are $1,\ldots,c$, so the differences $x_{\ell+1} - x_0$ add up to $cX$. Since $s_p(x_0) = x_0$ and $s_p(y) \le(p-1)(1+\log_p y)$, 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 $(p-1) \mid n$.

Verification

The ancillary file 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:

• (2) against the recurrence of the introduction; the coefficients of $1/T$ against the sequences A055505 and A055535 for $n \le17$; $b_n = a_{n-1} - a_n$; $S = 1/(1-B)$;

• for $p \le13$ and $n \le100$: $M(n)$ and $M'(n)$ computed by dynamic programming over all compositions against Proposition 6; the uniqueness of the optimum against Theorem 7; the inequalities of Lemma 4;

• for $p \le31$: Theorem 1 on all $n \in N_p$ with $n \le250$; Theorem 3 for all $p^K - 1 \le250$; Lemma 10 and the bounds of Theorem 2 for $X = 60000$.

These computations check the statements; the proofs do not depend on them.

Acknowledgement. Technical assistance: Claude (Anthropic).

ORCID 0009-0009-1526-1178