Let $f:\mathbb{R}\to \mathbb{R}$ be a sufficiently well behaved function and $g:\mathbb{N}\to \mathbb{N}$. Suppose that $a:\mathbb{N}\to \mathbb{R}$ is a sequence defined by the following equation:
$$\sum_{n=1}^{\infty} \frac{a(n)}{n!}x^nf(x)^{g(n)} = x.$$
Question: Is there any good way to get information about the asymptotics of $a(n)$? If not, is there any good way to simply determine a single value of the sequence, e.g., $a(100)$?
The specific example that I am looking at is $f(x)=e^{-x}$ and $g(n) = \frac{n(n+1)}{2}$, i.e.,
$$\sum_{n=1}^{\infty} \frac{a(n)}{n!}x^ne^{-x\frac{n(n+1)}{2}} = x.$$
The sequence $a(n)$ is entry A195737 in the Online Encyclopedia of Integer Sequences (OEIS) and the first 200 terms of it can be found there.
In my example, it seems to me that $a(n)$ should be roughly $\frac{n^nn!}{e^{o(n)}}$ but I would like a rigorous argument for this (and, if possible, better aymptotics).
By the way, for any function $f$ (perhaps with the condition that $f$ is injective on $(0,\varepsilon)$, or something like that) if we had $g(n)=n$, then $\sum_{n=1}^\infty \frac{a(n)}{n!}x^n$ would simply be the inverse of $xf(x)$ (this is easy to see), and this could help us get some information about $a(n)$. The main difficulty that I am having is that, in my example, $g(n)\neq n$.
Edit: In the answers below, I give a simple way of obtaining a recursive expression for $a(n)$, provided that $f(0)\neq 0$. However, this doesn't address the issue of the asymptotics of $a(n)$. I would be happy to get an answer to the following question:
Question: Can something like the saddle-point method/Cauchy's Integral Formula be applied here? Normally, it seems that this method is used to estimate the entries of a sequence $a(1),\dots, a(n)$ by integrating its generating function on a circle around the origin in the complex plane. The situation here seems somewhat different, but can the method be somehow adapted?