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](https://mathoverflow.net/a/378165) to [your previous question](https://mathoverflow.net/questions/378103/calculating-factorial-sequence-of-a-rational-function), 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](https://oeis.org/A265165), and so they do represent the same sequence (up to signs and shift of indices).