TITLE: Lagrange inversion for power-series with rational powers
QUESTION [10 upvotes]: One can use Lagrange inversion to find the
power series $F(x)$, which solves $F(x) = x(1+F(x)^p)$,
where $p$ is a positive integer.
Now, what if $p$ is not an integer, but rather a positive rational number,
say $p=7/3$?
As a concrete example, we are looking for a formal solution to $F(x) = x(1+F(x)^{7/3})$,
but now, $F(x) \in \mathbb{C}[x^{1/3}]$.
The Lagrange inversion formula still seem to work in this case,
that is, the function
$$
F(x) := \sum_{r>0} x^r \left( [t^{r-1}]\frac{1}{r} (1+t^p)^r \right)
$$
is a solution to $F(x) = x(1+F(x)^p)$, but now we must have $F(x) \in \mathbb{C}[x^{1/d}]$,
where $d$ is the denominator of $p$, and the sum ranges over all positive integer multiples of $1/d$.
Is there some reference which proves this extension of Lagrange inversion?
Edit: I think I managed to prove that Lagrange inversion generalizes to this setting, i.e,
instead of having $f,g \in \mathbb{C}[x]$,
we have $f,g \in \mathbb{C}[x^{1/d}]$,
and we wish to express the coefficients of $g$,
in terms of the coefficients of $f$, where $f(g(x))=x$.

REPLY [4 votes]: I managed to write up a proof based on these lecture notes (which cites Stanley's EC2)
The original proof uses residue calculus, and properties of analytic functions. This proof does not generalize to rational exponents, but the proof linked above, extends without any issues to "power series" with rational powers.