* Let $f(n)$ be an arbitrary function such that $f(n)\in\mathbb{Z}$.
* Let
$$
F(x)=\sum\limits\_{m\geqslant 0}f(m)x^m
$$
* Define the operator $\operatorname{SR}$, which is associated with the [series reversion](https://mathworld.wolfram.com/SeriesReversion.html).
* Let $a(n)$ be an integer sequence with generating function $A(x)$ where
$$
A(x)=\frac{1}{x}\operatorname{SR}(xF(x))
$$
* Let $b(n)$ be an integer sequence with generating function $\frac{1}{G(0)}$ where
$$
G(j)=F\left(\frac{x}{G(j+1)}\right)
$$
Here we have
$$
G(0)=F\left(\frac{x}{G(1)}\right)=F\left(\frac{x}{F\left(\frac{x}{G(2)}\right)}\right)=F\left(\frac{x}{F\left(\frac{x}{F\left(\frac{x}{G(3)}\right)}\right)}\right)
$$
and so on.


I conjecture that $$a(n)=b(n).$$


Is there a way to prove it?



Series reversion using something like continued fraction
We assume $F(0) \neq 0$, since otherwise we don't satisfy the assumptions for the series reversion. Let $G = G(0)$ be the fixpoint of the recurrence given:


$$G(x) = F\left(\frac{x}{G(x)}\right)$$


Multiply both sides by $\frac{x}{G(x)}$:


$$x = \frac{x}{G(x)} F\left(\frac{x}{G(x)}\right)$$


By inspection of the structure we have $\frac{x}{G(x)} = \operatorname{SR}(xF(x))$, or $$\frac{1}{G(x)} = \frac{1}{x} \operatorname{SR}(xF(x))$$ as desired.


