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arXiv 1503.03154math.NT
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关于模$p^αq^β$交错调和和的一个同余

A congruence involving alternating harmonic sums modulo $p^αq^β$

Zhongyan Shen, Tianxin Cai

摘要:

在2014年,Wang和Cai建立了以下调和同余式:对于任何奇素数$p$和正整数$r$,\begin{equation*} \sum\limits_{i+j+k=p^{r}\atop{i,j,k\in \mathcal{P}_{p}}}\frac{1}{ijk}\equiv-2p^{r-1}B_{p-3} ~(\bmod ~ p^{r}), \end{equation*},其中$\mathcal{P}_{n}$表示与$n$互素的正整数的集合。本文得到了不同奇素数$p,~q$与正整数$α,~β$、\begin{equation*} \sum\limits_{i+j+k=p^αq^β\atop{i,j,k\in\mathcal{P}_{pq}\atop{i\equiv j\equiv k\equiv 1\pmod{2}}}}\frac{1}{ijk}\equiv\frac{7}{8}(2-q)(1-\frac{1}{q^{3}})p^{α-1}q^{β-1}B_{p-3}\pmod{p^α} \end{equation*}和\begin{equation*} \sum\limits_{i+j+k=p^αq^β\atop{i,j,k\in \mathcal{P}_{pq}}}\frac{(-1)^{i}}{ijk} \equiv \frac{1}{2}(q-2)(1-\frac{1}{q^{3}})p^{α-1}q^{β-1}B_{p-3}\pmod{p^α}. \end{equation*}的同余关系。最后,我们提出了一个猜想:对于$n>1$和奇素数幂$p^α||n$、$α\geq1$、\begin{eqnarray} \nonumber \sum\limits_{i+j+k=n\atop{i,j,k\in\mathcal{P}_{n}}}\frac{(-1)^{i}}{ijk} \equiv \prod\limits_{q|n\atop{q\neq p}}(1-\frac{2}{q})(1-\frac{1}{q^{3}})\frac{n}{2p}B_{p-3}\pmod{p^α} \end{eqnarray}和\begin{eqnarray} \nonumber \sum\limits_{i+j+k=n\atop{i,j,k\in\mathcal{P}_{n}\atop{i\equiv j\equiv k\equiv 1\pmod{2}}}}\frac{1}{ijk} \equiv \prod\limits_{q|n\atop{q\neq p}}(1-\frac{2}{q})(1-\frac{1}{q^{3}})(-\frac{7n}{8p})B_{p-3}\pmod{p^α}. \end{eqnarray}

英文摘要:

In 2014, Wang and Cai established the following harmonic congruence for any odd prime $p$ and positive integer $r$, \begin{equation*} \sum\limits_{i+j+k=p^{r}\atop{i,j,k\in \mathcal{P}_{p}}}\frac{1}{ijk}\equiv-2p^{r-1}B_{p-3} ~(\bmod ~ p^{r}), \end{equation*} where $\mathcal{P}_{n}$ denote the set of positive integers which are prime to $n$. In this note, we obtain the congruences for distinct odd primes $p,~q$ and positive integers $α,~β$, \begin{equation*} \sum\limits_{i+j+k=p^αq^β\atop{i,j,k\in\mathcal{P}_{pq}\atop{i\equiv j\equiv k\equiv 1\pmod{2}}}}\frac{1}{ijk}\equiv\frac{7}{8}(2-q)(1-\frac{1}{q^{3}})p^{α-1}q^{β-1}B_{p-3}\pmod{p^α} \end{equation*} and \begin{equation*} \sum\limits_{i+j+k=p^αq^β\atop{i,j,k\in \mathcal{P}_{pq}}}\frac{(-1)^{i}}{ijk} \equiv \frac{1}{2}(q-2)(1-\frac{1}{q^{3}})p^{α-1}q^{β-1}B_{p-3}\pmod{p^α}. \end{equation*} Finally, we raise a conjecture that for $n>1$ and odd prime power $p^α||n$, $α\geq1$, \begin{eqnarray} \nonumber \sum\limits_{i+j+k=n\atop{i,j,k\in\mathcal{P}_{n}}}\frac{(-1)^{i}}{ijk} \equiv \prod\limits_{q|n\atop{q\neq p}}(1-\frac{2}{q})(1-\frac{1}{q^{3}})\frac{n}{2p}B_{p-3}\pmod{p^α} \end{eqnarray} and \begin{eqnarray} \nonumber \sum\limits_{i+j+k=n\atop{i,j,k\in\mathcal{P}_{n}\atop{i\equiv j\equiv k\equiv 1\pmod{2}}}}\frac{1}{ijk} \equiv \prod\limits_{q|n\atop{q\neq p}}(1-\frac{2}{q})(1-\frac{1}{q^{3}})(-\frac{7n}{8p})B_{p-3}\pmod{p^α}. \end{eqnarray}

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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