A Fibonacci analogue of Stirling numbers

Abstract: 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 numbers are the connections coefficients between the rising factorial basis ${(x){\uparrow_n}:n \geq 0}$ and the falling factorial basis ${(x){\downarrow_n}:n \geq 0}$ in the polynomial ring $\mathbb{Q}[x]$. The goal of this paper is to find Fibonacci analogues for the Stirling numbers of the first and second kind and the Lah numbers. Our idea is to replace the falling factorial basis and the rising factorial basis by the Fibo-falling factorial basis ${(x){\downarrow{F,n}}:n \geq 0}$ and the Fibo-rising factorial basis ${(x){\uparrow{F,n}}:n \geq 0}$ where $(x){\downarrow{F,0}} = (x){\uparrow{F,0}} = 1$ and for $k \geq 1$, $(x){\downarrow{F,k}} = x(x-F_1) \cdots (x-F_{k-1})$ and $(x){\uparrow{F,k}} = x(x+F_1) \cdots (x+F_{k-1})$. Then we study the combinatorics of the connection coefficients betweenthe usual power basis, the Fibo-falling factorial basis, and the Fibo-rising factorial basis. In each case, we can give a rook theory model for the connections coefficients and show how this rook theory model can give combinatorial explanations for many of the properties of these coefficients.