I am looking to find the number of planted plane trees (PPT) on $mn+1$ nodes, such that the number of upward branches on each node is a multiple of $m$, for integer $m$ $\ge$ 2.
Here is what I have so far:
Let $P(x)$ be the generating function with $x$ being all nodes. Then the following $P(x)$ enumerates the number of PPT on $n$ nodes, for which the number of upward branches on each node is exactly equal to $m$.
\begin{align} P(x) & = x(m+mP+mP^2+mP^3+...) \\\\ & = x \cdot m(1+P+P^2+P^3+...) \\\\ & = x \cdot m(1-P)^{-1} \end{align}
Then to find the exact number of PPT on this many nodes, we would have to extract the coefficient $[x^nt^m]$, by using the Lagrange Inversion Formula where $\phi(t) = (1-t)^{-1}$.
However, I am looking for the number of branches on each node to be a multiple of m, not necessarily m itself (or am I correct in my original interpretation?). Also, I am looking for the total number of nodes to be $mn+1$, not $n$.
Any thoughts on how I can rework the formula?
It looks like this is asking about the combinatorial class $\mathcal{T}$ where
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{T} = \mathcal{Z} + \mathcal{Z} \times \textsc{SEQ}_{=m, =2m, =3m, \ldots}(\mathcal{T}).$$
These are rooted plane trees with outdegree a multiple of $m.$ This yields the functional equation
$$T(z) = z + z \times ( T(z)^m + T(z)^{2m} + T(z)^{3m} + \cdots) \\ = z \times (1 + T(z)^m + T(z)^{2m} + T(z)^{3m} + \cdots) \\ = z \frac{1}{1-T(z)^m}$$
which also produces
$$z = (1-T(z)^m) T(z).$$
To extract coefficients on this consult the entry for the Lagrange Inversion Formula at Wikipedia and take $\phi(w)=1/(1-w^m)$ to obtain for $q$ nodes
$$\frac{1}{q} [w^{q-1}] \frac{1}{(1-w^m)^q}.$$
Here we see that $q-1$ must be a multiple of $m$ so we put $q=mn+1$ to get
$$\frac{1}{mn+1} [w^{mn}] \frac{1}{(1-w^m)^{mn+1}} = \frac{1}{mn+1} [w^n] \frac{1}{(1-w)^{mn+1}}.$$
We have the closed form
$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{mn+1} {mn+n\choose n}.}$$