English

Congruences for sums involving products of three binomial coefficients

Number Theory 2022-02-15 v2 Combinatorics

Abstract

Let p>3p>3 be a prime, and let aa be a rational pp-adic integer, using WZ method we establish the congruences modulo p3p^3 for k=0p1(ak)(1ak)(2kk)w(k)4k,\sum_{k=0}^{p-1} \binom ak\binom{-1-a}k\binom{2k}k\frac {w(k)}{4^k}, where w(k)=1,1k+1,1(k+1)2,1(k+1)3,12k1,1k+2,1k+3,k,k2,k3,1a+k,1a+k1.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=12,13,14,16a=-\frac 12,-\frac 13,-\frac 14,-\frac 16 we deduce many congruences modulo p3p^3 and so solve some conjectures posed by the author earlier.

Cite

@article{arxiv.2202.05077,
  title  = {Congruences for sums involving products of three binomial coefficients},
  author = {Zhi-Hong Sun},
  journal= {arXiv preprint arXiv:2202.05077},
  year   = {2022}
}

Comments

add Section 11 and more conjectures