Given $f(x) = \sum_{n \geq 0} a_nx^n$, can we find a function $g(x) = \sum_{n \geq 0} a_na_{n+1}x^n$?
When we solve the recurrences of the form $a_na_{n+1} = h(n)$, it is normal to define $b_n = (-1)^n \ln (a_n)$ as it turns our equation in something like $b_{n+1} - b_n = H(n)$, so I wonder what would be the generating function of $g(x) = \ln (f(x))$... I'm not even sure we can center it at origin tho, any ideas?
Here is one try:
$$f(x) = \sum_{n \geq 0} a_nx^n, g(x) = \sum_{n \geq 0} b_nx^n \implies$$
$$f(x)\cdot g(x) = \sum_{n \geq 0} c_n x^n$$ with $$c_n = \sum_{k=0}^n b_ka_{n-k}$$
So, if we can find $b_n$ such that
$$a_na_{n+1} = \sum_{k=0}^n b_k a_{n-k}$$
we can solve the problem with simply Cauchy's product. But how do we invert such a sum? The difference operator doesn't apply directly. The Cauchy product may be the way to find the series of $h(x) = \exp (f(x))$ but that takes a lot of calculations.
EDIT: Oleksandr Kulkov gave us a nice insight.
If $$f(x) = \sum_{n \geq 0} a_nx^n = a_0 + x \sum_{n \geq 0} a_{n+1}x^n$$
So $$\frac{f(x)-a_0}x = g(x) = \sum_{n \geq 0} a_{n+1}x^n$$
I want this:
$$\frac1{2\pi} \int_0^{2\pi}f(\sqrt x e^{it})g(\sqrt x e^{-it}) dt$$