TITLE: Counting $k$-ary labelled trees
QUESTION [7 upvotes]: The (full) binary counting tree problems gives the number of binary trees can be formed using $N$ nodes  $T(n)= C_n$, where $C_i$ are the Catalan numbers.    The recursion form is $T_n = \sum_{i=0}^{n-1}T_iT_{n-1-i}$.
Now I want to generalize the binary counting tree by:

Label the node, so that the order matters.  This seems simple enough, the number of trees now is $T_n = n!C_n$.  The recursion form is $n\sum_{i=0}^{n-1}{{n-1\choose i}T_iT_{n-1-i}}$
$k$-ary tree: instead of binary, now it's $k$-ary (and of course with labelled nodes).  I don't know if there's a name for this problem but I can't seem to find a "nice" recursion form or closed formula for $T_n$.

The question is thus asking for the recurrence form (and closed form if possible) of the $k$-ary labelled trees problem above. 

What about a simpler version of counting ternary trees (no label) ?  The recurrence form is easy to get but what about the closed form of it ?

REPLY [2 votes]: Expanding on Marko Riedl's answer, we want to solve for the series $T(z)$ where:
$$
T(z) = 1 + z T(z)^k
$$
Change variables to $u \mapsto T(z) - 1$ so that:
$$
u = z (u + 1)^k
$$
The Bürmann-Lagrange inversion formula for $u = z \phi(u)$ gives
$$
[z^n] u(z) = \frac{1}{n} [u^{n - 1}] \phi^n (u)
$$
In our case:
\begin{align}
[z^n] u(z) 
  &= \frac{1}{n} [u^{n - 1}] (u + 1)^{k n} \\
  &= \frac{1}{n} \binom{k n}{n - 1}
\end{align}
This is valid for $n > 0$. But:
$$
\frac{1}{n} \binom{k n}{n - 1}
  = \frac{(k n)!}{n (n - 1)! (k n - n + 1)!}
  = \frac{1}{n (k - 1) + 1} \binom{k n}{n}
$$
By happy coincidence, this gives the correct value 1 for $n = 0$. And for $k = 2$ it gives the familiar Catalan numbers.