Inverse of composition - series with factorial

Question

If $$2y+1=\sum_{n=0}^{+\infty}\left(\frac{1}{1+\sqrt 5}\right)^n\frac{x^n}{n!},$$

find a sequence $\{a_n\}$ such that $x=\sum_{n=1}^{+\infty}a_ny^n$.

How can we derive a generating function for series that contain factorials?

We know that $\sum_{n=0}^{+\infty}\frac{x^n}{n!}=e^x$.

How to derive the generating function for $\sum_{n=0}^{+\infty}\left(\frac{1}{1+\sqrt 5}\right)^n\frac{x^n}{n!}$?

Answer 1

If $$ 2y+1=\exp\left(\frac{x}{1+\sqrt{5}}\right) \tag{1}$$ then $$ x = (1+\sqrt{5}) \log(1+2y) = (1+\sqrt{5})\sum_{n\geq 1}\frac{2^n (-1)^{n+1}}{n} y^n. \tag{5}$$ For more complex task, you may use Lagrange's inversion theorem.

Answer 2

Hints:

What you have is simply

$$2y+1=e^{\frac x{1+\sqrt5}}\implies x=(1+\sqrt5)\,\log(2y+1)$$

and now use some power series for the logarithm...