{"paper":{"title":"p-adic congruences motivated by series","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":"2011-11-21T19:58:22Z","abstract_excerpt":"Let $p>5$ be a prime. Motivated by the known formulae $\\sum_{k=1}^\\infty(-1)^k/(k^3\\binom{2k}{k})=-2\\zeta(3)/5$ and $\\sum_{k=0}^\\infty \\binom{2k}{k}^2/((2k+1)16^k)=4G/\\pi$$ (where $G=\\sum_{k=0}^\\infty(-1)^k/(2k+1)^2$ is the Catalan constant), we show that $$\\sum_{k=1}^{(p-1)/2}\\frac{(-1)^k}{k^3\\binom{2k}{k}}\\equiv-2B_{p-3}\\pmod{p},$$ $$\\sum_{k=(p+1)/2}^{p-1}\\frac{\\binom{2k}{k}^2}{(2k+1)16^k}\\equiv-\\frac 7{4}p^2B_{p-3}\\pmod{p^3}$$, and $$\\sum_{k=0}^{(p-3)/2}\\frac{\\binom{2k}{k}^2}{(2k+1)16^k} \\equiv-2q_p(2)-pq_p(2)^2+\\frac{5}{12}p^2B_{p-3}\\pmod{p^3},$$ where $B_0,B_1,\\ldots$ are Bernoulli number"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1111.4988","kind":"arxiv","version":4},"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"}