New congruences for central binomial coefficients

Abstract

Let p be a prime and let a be a positive integer. In this paper we determine ${\sum }_{k=0}^{p^a-1}\left(\genfrac{}{}{0ex}{}{2k}{k+d}\right)/m^k$ and ${\sum }_{k=1}^{p-1}\left(\genfrac{}{}{0ex}{}{2k}{k+d}\right)/(k{m}^{k-1})$ modulo $p$ for all d=0,...,p^a, where m is any integer not divisible by p. For example, we show that if $p\ne 2,5$ then $$\sum _{k=1}^{p-1}(-1{)}^k\frac{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}{k}=-5\frac{F_{p-(\frac{p}{5})}}{p}(modp),$$ where F_n is the n-th Fibonacci number and (-) is the Jacobi symbol. We also prove that if p>3 then $$\sum _{k=1}^{p-1}\frac{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}{k}=8/9p^2{B}_{p-3}(mod{p}^3),$$ where B_n denotes the n-th Bernoulli number.

References in corpus (1)

• On some new congruences for binomial coefficients

Cited by in corpus (4)

• Congruences involving the reciprocals of central binomial coefficients
• Asymptotics of partial sums of central binomial coefficients and Catalan numbers
• Congruences involving alternating multiple harmonic sum
• q-Analogs of some congruences involving Catalan numbers