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arXiv 1702.08401math.NT
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多重调和和与Bernoulli数的超同余

Super Congruences Involving Multiple Harmonic Sums and Bernoulli Numbers

Kevin Chen, Jianqiang Zhao

备注：
19 pages. We updated the proof of Lemma 5.1 and corrected some misprints
发表信息：
Journal of Integer Sequences, volume 20 (2017), Article 17.6.8
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摘要：

令$m$、$r$和$n$为正整数。我们用${\bf k}\vdash n$表示任何奇正整数元组${\bf k}=(k_1,\dots,k_t)$，使得对于所有的$j$有$k_1+\dots+k_t=n$和$k_j\ge 3$。本文证明了：对每个充分大的素数$p$ $$ \sum_{\substack{l_1+l_2+\cdots+l_n=mp^r p\nmid l_1 l_2 \cdots l_n }} \frac1{l_1l_2\cdots l_n} \equiv p^{r-1} \sum_{{\bf k}\vdash n} C_{m,{\bf k}} B_{p-{\bf k}} \pmod{p^r} $$，其中$B_{p-{\bf k}}=B_{p-k_1}B_{p-k_2}\cdots B_{p-k_t}$是Bernoulli数的乘积，系数$C_{m,{\bf k}}$是$m$的多项式，与$p$和$r$无关.这推广了许多不同作者以前的结果，并证实了作者及其合作者的猜想。

英文摘要：

Let $m$, $r$ and $n$ be positive integers. We denote by ${\bf k}\vdash n$ any tuple of odd positive integers ${\bf k}=(k_1,\dots,k_t)$ such that $k_1+\dots+k_t=n$ and $k_j\ge 3$ for all $j$. In this paper we prove that for every sufficiently large prime $p$ $$ \sum_{\substack{l_1+l_2+\cdots+l_n=mp^r p\nmid l_1 l_2 \cdots l_n }} \frac1{l_1l_2\cdots l_n} \equiv p^{r-1} \sum_{{\bf k}\vdash n} C_{m,{\bf k}} B_{p-{\bf k}} \pmod{p^r} $$ where $B_{p-{\bf k}}=B_{p-k_1}B_{p-k_2}\cdots B_{p-k_t}$ are products of Bernoulli numbers and the coefficients $C_{m,{\bf k}}$ are polynomials of $m$ independent of $p$ and $r$. This generalizes previous results by many different authors and confirms a conjecture by the authors and their collaborators.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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