{"paper":{"title":"Supercongruences motivated by e","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":"2010-11-15T19:59:35Z","abstract_excerpt":"In this paper we establish some new supercongruences motivated by the well-known fact $\\lim_{n\\to\\infty}(1+1/n)^n=e$. Let $p>3$ be a prime. We prove that $$\\sum_{k=0}^{p-1}\\binom{-1/(p+1)}k^{p+1}\\equiv 0\\ \\pmod{p^5}\\ \\ \\ \\mbox{and}\\ \\ \\ \\sum_{k=0}^{p-1}\\binom{1/(p-1)}k^{p-1}\\equiv \\frac{2}{3}p^4B_{p-3}\\ \\pmod{p^5},$$ where $B_0,B_1,B_2,\\ldots$ are Bernoulli numbers. We also show that for any $a\\in\\mathbb Z$ with $p\\nmid a$ we have $$\\sum_{k=1}^{p-1}\\frac1k\\left(1+\\frac ak\\right)^k\\equiv -1\\pmod{p}\\ \\ \\ \\mbox{and}\\ \\ \\ \\sum_{k=1}^{p-1}\\frac1{k^2}\\left(1+\\frac ak\\right)^k\\equiv 1+\\frac 1{2a}\\pmo"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1011.3487","kind":"arxiv","version":8},"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"}