{"paper":{"title":"New congruences involving products of two binomial coefficients","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.CO"],"primary_cat":"math.NT","authors_text":"Guo-Shuai Mao, Zhi-Wei Sun","submitted_at":"2016-01-19T03:18:05Z","abstract_excerpt":"Let $p>3$ be a prime and let $a$ be a positive integer. We show that if $p\\equiv1\\pmod 4$ or $a>1$ then $$\\sum_{k=0}^{\\lfloor\\frac34p^a\\rfloor}\\frac{\\binom{2k}k^2}{16^k}\\equiv\\l(\\frac{-1}{p^a}\\r)\\pmod{p^3}$$ with $(-)$ the Jacobi symbol, which confirms a conjecture of Z.-W. Sun. We also establish the following new congruences: \\begin{align*}\\sum_{k=0}^{(p-1)/2}\\frac{\\binom{2k}k\\binom{3k}k}{27^k}\\equiv&\\l(\\frac p3\\r)\\frac{2^p+1}3\\pmod{p^2}, \\\\\\sum_{k=0}^{(p-1)/2}\\frac{\\binom{6k}{3k}\\binom{3k}k}{(2k+1)432^k}\\equiv&\\l(\\frac p3\\r)\\frac{3^p+1}4\\pmod{p^2}, \\\\\\sum_{k=0}^{(p-1)/2}\\frac{\\binom{4k}{2k}\\"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1601.04782","kind":"arxiv","version":1},"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"}