New congruences for central binomial coefficients

preprint2010arXivOpen access

Abstract

Let p be a prime and let a be a positive integer. In this paper we determine ∑k=0pa−1(2kk+d)/mk\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k and ∑k=1p−1(2kk+d)/(kmk−1)\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 p≠2,5p\not=2,5 then ∑k=1p−1(−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=1p−1(2kk)k=8/9p2Bp−3(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.

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