English

Various congruences involving binomial coefficients and higher-order Catalan numbers

Number Theory 2009-09-28 v2 Combinatorics

Abstract

Let pp be a prime and let aa be a positive integer. In this paper we investigate k=0pa1([()h+1)k,k+d]/mk\sum_{k=0}^{p^a-1}\binom[(h+1)k,k+d]/m^k modulo a prime pp, where dd and mm are integers with h<d<=pa-h<d<=p^a and m0(modp)m\not=0 (mod p). We also study congruences involving higher-order Catalan numbers Ck(h)=([()h+1)k,k]/(hk+1)C_k^{(h)}=\binom[(h+1)k,k]/(hk+1) and Cˉk(h)=\binomal[(h+1)k,k]h/(k+1)\bar C_k^{(h)}=\binomal[(h+1)k,k]*h/(k+1). Our tools include linear recurrences and the theory of cubic residues. Here are some typical results in the paper. (i) If pa=1(mod6)p^a=1 (mod 6) then k=1pa1([3)k,k]/6k=2(pa1)/31(modp).\sum_{k=1}^{p^a-1}\binom[3k,k]/6^k=2^{(p^a-1)/3}-1 (mod p). Also, \sum_{k=0}^{p^a-1}\binom[3k,k]/7^k=\cases-2&if p^a=\pm2 (mod 7), \\1&otherwise. (ii) We have \sum_{k=1}^{p^a-1}\binom[4k,k]/5^k=\cases1 (mod p) if p\not=11 and p^a=1 (mod 5), \1/11 (mod p)&if p^a=2,3 (mod 5), \9/11 (mod p) if p^a=4 (mod 5). Also, \sum_{k=0}^{p^a-1}C_k^{(3)}/5^k=\cases1 (mod p) if p^a=1,3 (mod 5), \2 (mod p) if p^a=2 (mod 5), \\0 (mod p)& p^a=4 (mod 5).

Keywords

Cite

@article{arxiv.0909.3808,
  title  = {Various congruences involving binomial coefficients and higher-order Catalan numbers},
  author = {Zhi-Wei Sun},
  journal= {arXiv preprint arXiv:0909.3808},
  year   = {2009}
}

Comments

33 pages. Extended version