{"paper":{"title":"Pentagonal number recurrence relations for $p(n)$","license":"http://creativecommons.org/licenses/by/4.0/","headline":"","cross_cats":["math.CO"],"primary_cat":"math.NT","authors_text":"Ajit Singh, Hasan Saad, Ken Ono, Kevin Gomez","submitted_at":"2024-11-25T22:37:34Z","abstract_excerpt":"We revisit Euler's partition function recurrence, which asserts, for integers $n\\geq 1,$ that $$ p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\\dots = \\sum_{k\\in \\mathbb{Z}\\setminus \\{0\\}} (-1)^{k+1} p(n-\\omega(k)), $$ where $\\omega(m):=(3m^2+m)/2$ is the $m$th pentagonal number. We prove that this classical result is the $\\nu=0$ case of an infinite family of ``pentagonal number'' recurrences. For each $\\nu\\geq 0,$ we prove for positive $n$ that\n  $$ p(n)=\\frac{1}{g_{\\nu}(n,0)}\\left(\\alpha_{\\nu}\\cdot \\sigma_{2\\nu-1}(n)+ \\mathrm{Tr}_{2\\nu}(n) +\\sum_{k\\in \\mathbb{Z}\\setminus \\{0\\}} (-1)^{k+1} g_{\\nu}(n,k)\\cd"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"2411.16968","kind":"arxiv","version":2},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"integrity":{"clean":true,"summary":{"advisory":0,"critical":0,"by_detector":{},"informational":0},"endpoint":"/pith/2411.16968/integrity.json","findings":[],"available":true,"detectors_run":[],"snapshot_sha256":"c28c3603d3b5d939e8dc4c7e95fa8dfce3d595e45f758748cecf8e644a296938"},"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"}