On sums involving products of three binomial coefficients

Zhi-Wei Sun

paper

On sums involving products of three binomial coefficients

arXiv:1012.3141

Abstract

In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p > 3$ be a prime. We prove that $k = 0 \sum p - 1 \frac{( k 2 k ) ^{2} ( k + d 2 k )}{6 4 ^{k}} \equiv 0 (mod p^{2})$ for all $d \in {0, \dots, p - 1}$ with $d \equiv (p + 1) /2 (mod 2)$ . If $p \equiv 1 (mod 4)$ and $p = x^{2} + y^{2}$ with $x \equiv 1 (mod 4)$ and $y \equiv 0 (mod 2)$ , then we show $$\sum_{k=0}^{p-1}\frac{\binom{2k}k^2\binom{2k}{k+1}}{(-8)^k}\equiv 2p-2x^2\pmod{p^2}\ \ \mbox{and}\ \ \sum_{k=0}^{p-1}\frac{\binom{2k}k\binom{2k}{k+1}^2}{(-8)^k}\equiv-2p\pmod{p^2}$$ by means of determining $x$ mod $p^{2}$ via $(- 1)^{(p - 1) /4} x \equiv k = 0 \sum (p - 1) /2 \frac{k + 1}{8 ^{k}} (k 2 k)^{2} \equiv k = 0 \sum (p - 1) /2 \frac{2 k + 1}{( - 16 ) ^{k}} (k 2 k)^{2} (mod p^{2}) .$ We also solve the remaining open cases of Rodriguez-Villegas' conjectural congruences on $k = 0 \sum p - 1 \frac{( k 2 k ) ^{2} ( k 3 k )}{10 8 ^{k}}, k = 0 \sum p - 1 \frac{( k 2 k ) ^{2} ( 2 k 4 k )}{25 6 ^{k}}, k = 0 \sum p - 1 \frac{( k 2 k ) ( k 3 k ) ( 3 k 6 k )}{1 2 ^{3 k}}$ modulo $p^{2}$ .

21 pages, final published version

On sums involving products of three binomial coefficients

Abstract

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