{"paper":{"title":"On sums involving products of three binomial coefficients","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":"2010-12-14T19:26:16Z","abstract_excerpt":"In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p>3$ be a prime. We prove that $$\\sum_{k=0}^{p-1}\\frac{\\binom{2k}k^2\\binom{2k}{k+d}}{64^k}\\equiv 0\\pmod{p^2}$$ for all $d\\in\\{0,\\ldots,p-1\\}$ with $d\\equiv (p+1)/2\\pmod2$. If $p\\equiv 1\\pmod4$ and $p=x^2+y^2$ with $x\\equiv 1\\pmod4$ and $y\\equiv 0\\pmod2$, then we show $$\\sum_{k=0}^{p-1}\\frac{\\binom{2k}k^2\\binom{2k}{k+1}}{(-8)^k}\\equiv 2p-2x^2\\pmod{p^2}\\ \\ \\mbox{and}\\ \\ \\sum_{k=0}^{p-1}\\frac{\\binom{2k}k\\binom{2k}{k+1}^2}{(-8)^k}\\equiv-2p\\pmod{p^2}"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1012.3141","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"}