{"paper":{"title":"A Fibonacci analogue of Stirling numbers","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CO","authors_text":"Jeffrey B. Remmel, Quang T. Bach, Roshil Paudyal","submitted_at":"2015-10-14T20:54:16Z","abstract_excerpt":"Consider the Fibonacci numbers defined by setting $F_1=1=F_2$ and $F_n =F_{n-1}+F_{n-2}$ for $n \\geq 3$. We let $n_F! = F_1 \\cdots F_n$ and $\\binom{n}{k}_F = \\frac{n_F!}{k_F!(n-k)_F!}$. Let $(x)_{\\downarrow_0} = (x)_{\\uparrow_0} = 1$ and for $k \\geq 1$, $(x)_{\\downarrow_k} = x(x-1) \\cdots (x-k+1)$ and $(x)_{\\uparrow_k} = x(x+1) \\cdots (x+k-1)$. Then the Stirling numbers of the first and second kind are the connections coefficients between the usual power basis $\\{x^n:n \\geq 0\\}$ and the falling factorial basis $\\{(x)_{\\downarrow_n}:n \\geq 0\\}$ in the polynomial ring $\\mathbb{Q}[x]$ and the Lah"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1510.04310","kind":"arxiv","version":3},"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"}