Why does this "factorial sequence" appear in the OEIS?

Question

For a reciprocal of a polynomial, $f = \frac{1}{p}$, we (presumably) may construct a sequence $(c_n)_{n=0}^\infty$ such that for all $N\ge 0$ $$f(k)k! = \sum_{n=0}^{N-1} c_n(k-n)! + O((k-N)!). $$ I ask about constructing such sequences in Calculating "factorial sequence" of a rational function.

Through some messy calculations, I get $$c_0,c_1,\dotsc = 0,0,0,0,1,-2,7,-32,179,-1182,8993,-77440,744425,-7901410,\dotsc$$ when $f = \frac{1}{(k-1)(k(k-1)(k-2)-k)}$. It just so happens that $\lvert c_{i+3}\rvert = a(i)$ for the OEIS A265165. I have tried looking at Banderier, Baril, and Dos Santos - Right-jumps & pattern avoiding permutations which the OEIS page links, but I couldn't see a reason why these should yield the same sequence. Can you find one? (Funnily enough, I came across this sequence $c_n$ by considering permutations, but in a totally different setting than the paper.)

Also, are "factorial sequences" $(c_n)$ a concept which has been studied before? Are there any other interesting OEIS sequences that happen to have a factorial sequence?

Answer 1

We have $$\frac{1}{x}f(\frac1x) = \sum_{n\geq 1} (F_{2n-3}-1)x^n,$$ where $F_{2n+1}$ are Fibonacci numbers. Per answers to your previous question, it follows that $$c_{n+1} = \sum_{i=0}^n s(n,i) (F_{2i-3}-1).$$

For $n\geq 2$, "$-1$" can be dropped, reducing the formula to $$c_{n+1} = \sum_{i=0}^n s(n,i) F_{2i-3}.$$

Using the generating function for Stirling numbers and Binet's formula for Fibonacci numbers, we obtain the following exponential generating function for $c_{n+1}$: $$\sum_{n\geq 0} c_{n+1} \frac{z^n}{n!} = \frac{(1+z)^{1+\phi}(2\phi-3)+(1+z)^{2-\phi}(2\phi+1)}{\sqrt{5}}-1,$$ where $\phi\mathrel{:=}\frac{1+\sqrt{5}}2$. Under substitution of $z=-x$, this matches the e.g.f. given in the sequence OEIS A265165, and so they do represent the same sequence (up to signs and shift of indices).