Congruences for the Apéry numbers modulo $p^3$

Zhi-Hong Sun
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引用次数: 0

Abstract

Let $\{A'_n\}$ be the Ap\'ery numbers given by $A'_n=\sum_{k=0}^n\binom nk^2\binom{n+k}k.$ For any prime $p\equiv 3\pmod 4$ we show that $A'_{\frac{p-1}2}\equiv \frac{p^2}3\binom{\frac{p-3}2}{\frac{p-3}4}^{-2}\pmod {p^3}$. Let $\{t_n\}$ be given by $$t_0=1,\ t_1=5\quad\hbox{and}\quad t_{n+1}=(8n^2+12n+5)t_n-4n^2(2n+1)^2t_{n-1}\ (n\ge 1).$$ We also obtain the congruences for $t_p\pmod {p^3},\ t_{p-1}\pmod {p^2}$ and $t_{\frac{p-1}2}\pmod {p^2}$, where $p$ is an odd prime.
阿佩里数以 $p^3$ 为模数的同余式
让 ${A'_n\}$ 是由 $A'_n=\sum_{k=0}^n\binomnk^2\binom{n+k}k 给出的阿普瑞数。$ 对于任意素数 $p (3/pmod 4),我们可以证明 $A'_{frac{p-1}2} (3/binom/frac{p-3}2}{/frac{p-3}4}^{-2}/pmod{p^3}$。让 ${t_n\}$ 由 $$t_0=1,t_1=5\quad\hbox{ and}\quadt_{n+1}=(8n^2+12n+5)t_n-4n^2(2n+1)^2t_{n-1}\ (n\ge 1) 给出。$$ 我们还得到了 $t_p\pmod {p^3},t_{p-1}\pmod {p^2}$ 和 $t_{frac{p-1}2}/pod{p^2}$的共轭关系,其中 $p$ 是奇素数。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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