{"record_type":"pith_number_record","schema_url":"https://pith.science/schemas/pith-number/v1.json","pith_number":"pith:2009:YRPSY7GT3L67JFBFJ3SPORTZQ7","short_pith_number":"pith:YRPSY7GT","schema_version":"1.0","canonical_sha256":"c45f2c7cd3dafdf494254ee4f7467987d49bfd646eccfb9a693f94b27e828ede","source":{"kind":"arxiv","id":"0909.5648","version":12},"attestation_state":"computed","paper":{"title":"Binomial coefficients, Catalan numbers and Lucas quotients","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.CO"],"primary_cat":"math.NT","authors_text":"Zhi-Wei Sun","submitted_at":"2009-09-30T19:57:32Z","abstract_excerpt":"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 $m \\not\\equiv0\\pmod p$. 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