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Supercongruences involving products of two binomial coefficients (1011.6676v6)

Published 30 Nov 2010 in math.NT and math.CO

Abstract: In this paper we deduce some new supercongruences modulo powers of a prime $p>3$. Let $d\in{0,1,\ldots,(p-1)/2}$. We show that $$\sum_{k=0}{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{8k}\equiv 0\ (\mbox{mod}\ p)\ \ \ \mbox{if}\ d\equiv \frac{p+1}2\ (\mbox{mod}\ 2),$$ and $$\sum_{k=0}{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{16k} \equiv\left(\frac{-1}p\right)+p2\frac{(-1)d}4E_{p-3}\left(d+\frac12\right)\pmod{p3},$$ where $E_{p-3}(x)$ denotes the Euler polynomial of degree $p-3$, and $(-)$ stands for the Legendre symbol. The paper also contains some other results such as $$\sum_{k=0}{p-1}k{(1+(\frac{-1}p))/2}\frac{\binom{6k}{3k}\binom{3k}k}{864k}\equiv0\pmod{p2}.$$

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