Binomial coefficients, Catalan numbers and Lucas quotients

preprint2016arXivOpen access

Abstract

Let pp be an odd prime and let a,ma,m be integers with a>0a>0 and m≢0(modp)m \not\equiv0\pmod p. In this paper we determine ∑k=0pa−1(2kk+d)/mk\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k mod p2p^2 for d=0,1d=0,1; for example, ∑k=0pa−1(2kk)mk≡(m2−4mpa)+(m2−4mpa−1)up−(m2−4mp)(modp2),\sum_{k=0}^{p^a-1}\frac{\binom{2k}k}{m^k}\equiv\left(\frac{m^2-4m}{p^a}\right)+\left(\frac{m^2-4m}{p^{a-1}}\right)u_{p-(\frac{m^2-4m}{p})}\pmod{p^2}, where (−)(-) is the Jacobi symbol, and {un}n⩾0\{u_n\}_{n\geqslant0} is the Lucas sequence given by u0=0u_0=0, u1=1u_1=1 and un+1=(m−2)un−un−1u_{n+1}=(m-2)u_n-u_{n-1} for n=1,2,3,…n=1,2,3,\ldots. As an application, we determine ∑0<k<pa, k≡r(modp−1)Ck\sum_{0<k<p^a,\, k\equiv r\pmod{p-1}}C_k modulo p2p^2 for any integer rr, where CkC_k denotes the Catalan number (2kk)/(k+1)\binom{2k}k/(k+1). We also pose some related conjectures.

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