English

New congruences for central binomial coefficients

Number Theory 2010-04-02 v4 Combinatorics

Abstract

Let p be a prime and let a be a positive integer. In this paper we determine k=0pa1(2kk+d)/mk\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k and k=1p1(2kk+d)/(kmk1)\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1}) modulo pp for all d=0,...,p^a, where m is any integer not divisible by p. For example, we show that if p2,5p\not=2,5 then k=1p1(1)k(2kk)k=5Fp(p5)p(modp),\sum_{k=1}^{p-1}(-1)^k\frac{\binom{2k}k}k=-5\frac{F_{p-(\frac p5)}}p (mod p), where F_n is the n-th Fibonacci number and (-) is the Jacobi symbol. We also prove that if p>3 then k=1p1(2kk)k=8/9p2Bp3(modp3),\sum_{k=1}^{p-1}\frac{\binom{2k}k}k={8/9} p^2B_{p-3} (mod p^3), where B_n denotes the n-th Bernoulli number.

Keywords

Cite

@article{arxiv.0805.0563,
  title  = {New congruences for central binomial coefficients},
  author = {Zhi-Wei Sun and Roberto Tauraso},
  journal= {arXiv preprint arXiv:0805.0563},
  year   = {2010}
}
R2 v1 2026-06-21T10:37:29.774Z