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arxiv: 0909.5648 · v12 · pith:YRPSY7GTnew · submitted 2009-09-30 · 🧮 math.NT · math.CO

Binomial coefficients, Catalan numbers and Lucas quotients

classification 🧮 math.NT math.CO
keywords fracbinompmodcatalandetermineequivleftlucas
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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 $\sum_{0<k<p^a,\, k\equiv r\pmod{p-1}}C_k$ modulo $p^2$ for any integer $r$, where $C_k$ denotes the Catalan number $\binom{2k}k/(k+1)$. We also pose some related conjectures.

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