{ "id": "0909.5648", "version": "v12", "published": "2009-09-30T19:57:32.000Z", "updated": "2016-02-15T16:20:00.000Z", "title": "Binomial coefficients, Catalan numbers and Lucas quotients", "authors": [ "Zhi-Wei Sun" ], "comment": "24 pages. Correct few typos", "journal": "Sci. China Math. 53(2010), no.9, 2473-2488", "doi": "10.1007/s11425-010-3151-3", "categories": [ "math.NT", "math.CO" ], "abstract": "Let $p$ be an odd prime and let $a,m$ be integers with $a>0$ and $m \\not\\equiv0\\pmod p$. In this paper we determine $\\sum_{k=0}^{p^a-1}\\binom{2k}{k+d}/m^k$ mod $p^2$ for $d=0,1$; for example, $$\\sum_{k=0}^{p^a-1}\\frac{\\binom{2k}k}{m^k}\\equiv\\left(\\frac{m^2-4m}{p^a}\\right)+\\left(\\frac{m^2-4m}{p^{a-1}}\\right)u_{p-(\\frac{m^2-4m}{p})}\\pmod{p^2},$$ where $(-)$ is the Jacobi symbol, and $\\{u_n\\}_{n\\geqslant0}$ is the Lucas sequence given by $u_0=0$, $u_1=1$ and $u_{n+1}=(m-2)u_n-u_{n-1}$ for $n=1,2,3,\\ldots$. As an application, we determine $\\sum_{00 and m not=0 (mod p). In this paper we determine $\\sum_{k=0}^{p^a-1}\\binom[2k,k+d]/m^k$ mod p^2 for d=0,1; for example, $$\\sum_{k=0}^{p^a-1}\\binom[2k,k]/m^k=(\\frac{m^2-4m}{p^a})+(\\frac{m^2-4m}{p^{a-1}})u_{p-(\\frac{m^2-4m}{p})} (mod p^2),$$ where (-) is the Jacobi symbol, u_0=0, u_1=1 and u_{n+1}=(m-2)u_n-u_{n-1} for n=1,2,3,.... As an application, we determine $\\sum_{0