{"paper":{"title":"A simple discharging method for forbidden subposet problems","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CO","authors_text":"Abhishek Methuku, Andrew Uzzell, Ryan R. Martin, Shanise Walker","submitted_at":"2017-10-13T19:27:02Z","abstract_excerpt":"The poset $Y_{k+1, 2}$ consists of $k+2$ distinct elements $x_1$, $x_2$, \\dots, $x_{k}$, $y_1$,$y_2$, such that $x_1 \\le x_2 \\le \\dots \\le x_{k} \\le y_1$,~$y_2$. The poset $Y'_{k+1, 2}$ is the dual of $Y_{k+1, 2}$ Let $\\rm{La}^{\\sharp}(n,\\{Y_{k+1, 2}, Y'_{k+1, 2}\\})$ be the size of the largest family $\\mathcal{F} \\subset 2^{[n]}$ that contains neither $Y_{k+1,2}$ nor $Y'_{k+1,2}$ as an induced subposet. Methuku and Tompkins proved that $\\rm{La}^{\\sharp}(n, \\{Y_{3,2}, Y'_{3,2}\\}) = \\Sigma(n,2)$ for $n \\ge 3$ and they conjectured the generalization that if $k \\ge 2$ is an integer and $n \\ge k+1$"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1710.05057","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":""},"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"}