Supercongruences involving products of two binomial coefficients

preprint2013arXivOpen access

Abstract

In this paper we deduce some new supercongruences modulo powers of a prime p>3p>3. Let d∈{0,1,…,(p−1)/2}d\in\{0,1,\ldots,(p-1)/2\}. We show that $$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{8^k}\equiv 0\ (\mbox{mod}\ p)\ \ \ \mbox{if}\ d\equiv \frac{p+1}2\ (\mbox{mod}\ 2),$$ and ∑k=0(p−1)/2(2kk)(2kk+d)16k≡(−1p)+p2(−1)d4Ep−3(d+12)(modp3),\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{16^k} \equiv\left(\frac{-1}p\right)+p^2\frac{(-1)^d}4E_{p-3}\left(d+\frac12\right)\pmod{p^3}, where Ep−3(x)E_{p-3}(x) denotes the Euler polynomial of degree p−3p-3, and (−)(-) stands for the Legendre symbol. The paper also contains some other results such as ∑k=0p−1k(1+(−1p))/2(6k3k)(3kk)864k≡0(modp2).\sum_{k=0}^{p-1}k^{(1+(\frac{-1}p))/2}\frac{\binom{6k}{3k}\binom{3k}k}{864^k}\equiv0\pmod{p^2}.

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