{"paper":{"title":"A refinement of a congruence result by van Hamme and Mortenson","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":"2010-11-08T20:55:31Z","abstract_excerpt":"Let $p$ be an odd prime. In 2008 E. Mortenson proved van Hamme's following conjecture: $$\\sum_{k=0}^{(p-1)/2}(4k+1)\\binom{-1/2}k^3\\equiv (-1)^{(p-1)/2}p\\pmod{p^3}.$$ In this paper we show further that \\begin{align*}\\sum_{k=0}^{p-1}(4k+1)\\binom{-1/2}k^3\\equiv &\\sum_{k=0}^{(p-1)/2}(4k+1)\\binom{-1/2}k^3 \\\\\\equiv & (-1)^{(p-1)/2}p+p^3E_{p-3} \\pmod{p^4},\\end{align*}where $E_0,E_1,E_2,\\ldots$ are Euler numbers. We also prove that if $p>3$ then $$\\sum_{k=0}^{(p-1)/2}\\frac{20k+3}{(-2^{10})^k}\\binom{4k}{k,k,k,k}\\equiv(-1)^{(p-1)/2}p(2^{p-1}+2-(2^{p-1}-1)^2)\\pmod{p^4}.$$"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1011.1902","kind":"arxiv","version":6},"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"}