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arXiv 2202.05077math.NTmath.CO
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三个二项式系数乘积和的同余式

Congruences for sums involving products of three binomial coefficients

Zhi-Hong Sun

摘要:

设$p>3$是素数,$a$是有理$p$整数,利用WZ方法建立了$$\sum_{k=0}^{p-1} \binom ak\binom{-1-a}k\binom{2k}k\frac {w(k)}{4^k},$$的模$p^3$同余式,其中$$w(k)=1,\frac 1{k+1},\frac 1{(k+1)^2},\frac 1{(k+1)^3},\frac 1{2k-1},\frac 1{k+2}, \frac 1{k+3}, k,k^2,k^3,\frac 1{a+k},\frac 1{a+k-1}.$$作为推论,利用$a=-\frac 12,-\frac 13,-\frac 14,-\frac 16$导出了许多模$p^3$的同余式,从而解决了作者以前提出的一些问题.

英文摘要:

Let $p>3$ be a prime, and let $a$ be a rational $p$-adic integer, using WZ method we establish the congruences modulo $p^3$ for $$\sum_{k=0}^{p-1} \binom ak\binom{-1-a}k\binom{2k}k\frac {w(k)}{4^k},$$ where $$w(k)=1,\frac 1{k+1},\frac 1{(k+1)^2},\frac 1{(k+1)^3},\frac 1{2k-1},\frac 1{k+2}, \frac 1{k+3}, k,k^2,k^3,\frac 1{a+k},\frac 1{a+k-1}.$$ As consequences, taking $a=-\frac 12,-\frac 13,-\frac 14,-\frac 16$ we deduce many congruences modulo $p^3$ and so solve some conjectures posed by the author earlier.

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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