English

Congruences concerning binomial coefficients and binary quadratic forms

Number Theory 2022-11-28 v4 Combinatorics

Abstract

Let p>3p>3 be a prime. In this paper, we obtain the congruences for k=0p1w(k)(2kk)3(8)k, k=0p1w(k)(2kk)2(3kk)(192)k, k=0p1w(k)(2kk)2(4k2k)(144)k and k=0p1w(k)(2kk)2(4k2k)648k\sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^3}{(-8)^k},\ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{3k}k}{(-192)^k},\ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{4k}{2k}}{(-144)^k}\ \text{and} \ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{4k}{2k}}{648^k} modulo p2p^2, and partial results for k=0(p1)/2(2kk)3w(k)mk\sum_{k=0}^{(p-1)/2} \binom{2k}k^3\frac{w(k)}{m^k} modulo p2p^2, where m{1,16,64,256,512,4096}m\in\{1,16,-64,256,-512,4096\} and w(k){k2,k3,1k+1,1(k+1)2,1(k+1)3,12k1,1k+2}w(k)\in\{k^2,k^3,\frac 1{k+1},\frac 1{(k+1)^2},\frac 1{(k+1)^3}, \frac 1{2k-1},\frac 1{k+2}\}.

Keywords

Cite

@article{arxiv.2210.17255,
  title  = {Congruences concerning binomial coefficients and binary quadratic forms},
  author = {Zhi-Hong Sun},
  journal= {arXiv preprint arXiv:2210.17255},
  year   = {2022}
}

Comments

correct some typos

R2 v1 2026-06-28T04:50:30.946Z