{"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 $\\sum_{0<k<p^a,\\, k\\equiv r\\pmod{p-1}}C_k$ modulo $p^2$ for"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"0909.5648","kind":"arxiv","version":12},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}