The number of rooted, ordered, incomplete, unlabeled $k$-ary trees with $n$ vertices is given by
$$C^{(k)}_n=\frac1{(k-1)n+1}\binom{kn}n\;.$$
These are sometimes called Fuss-Catalan numbers; see Concrete Mathematics (p. 347) and MathWorld (which gives two references). Their generating function $C^{(k)}(x)=\sum_0^\infty C^{(k)}_nx^n$ satisfies $C^{(k)}(x)=1+xC^{(k)}(x)^k$. The numbers of rooted, ordered, incomplete, unlabeled ternary ($k=3$), quartic ($k=4$), qunitic ($k=5$), sextic ($k=6$), heptic ($k=7$) and octic ($k=8$) trees form OEIS sequences A001764, A002293, A002294, A002295, A002296 and A007556, respectively. To get the number of labeled trees, just multiply by $n!$.
For Catalan trees let us recall the combinatorial class equation
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{C} = \mathcal{Z} \times \textsc{SEQ}(\mathcal{C}).$$
which gives the functional equation
$$C(z) = z \frac{1}{1-C(z)}.$$
Marking the root degree we get
$$\mathcal{Z} \times
\mathcal{U}^0 \textsc{SEQ}_{=0}(\mathcal{C})
+ \mathcal{Z} \times
\mathcal{U}^1 \textsc{SEQ}_{=1}(\mathcal{C})
+ \mathcal{Z} \times
\mathcal{U}^2 \textsc{SEQ}_{=2}(\mathcal{C})
+ \cdots$$
This gives the generating function
$$R(z, u) = z \sum_{k\ge 0} u^k C(z)^k
= z \frac{1}{1-u C(z)}.$$
As a sanity check put $u=1$ to get $z/(1-C(z))$ which is correct.
Differentiate and put $u=1$ to obtain
$$\left. \frac{\partial}{\partial u} R(z, u) \right|_{u=1}
= \left. \frac{z}{(1-uC(z))^2} C(z) \right|_{u=1}
= \frac{1}{z} C(z)^2 C(z) = \frac{1}{z} C(z)^3.$$
Now using Lagrange inversion by residues (Egorychev method) we seek
$$[z^n] \frac{1}{z} C(z)^3 = [z^{n+1}] C(z)^3
= \frac{1}{n+1} [z^n] 3 C(z)^2 C'(z)
\\ = \frac{1}{n+1} \;\underset{z}{\mathrm{res}}\;
\frac{1}{z^{n+1}} 3 C(z)^2 C'(z).$$
Next put $C(z)=w$ to get with $z= w (1-w)$
$$\frac{1}{n+1} \;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{n+1} (1-w)^{n+1}} 3w^2
= \frac{3}{n+1} [w^{n-2}] \frac{1}{(1-w)^{n+1}}
= \frac{3}{n+1} {n+n-2\choose n}
\\ = \frac{3}{n+1} {2n-2\choose n}.$$
It follows that the average degree of the root is
$$\frac{3}{n+1} {2n-2\choose n} n {2n-2\choose n-1}^{-1}
= \frac{3}{n+1} \frac{(n-1)!^2}{(n-1)! (n-2)!}$$
or
$$\bbox[5px,border:2px solid #00A000]{
3\frac{n-1}{n+1}.}$$
This result was verified with the combstruct package, see below.
with(combstruct);
with(combinat);
trees :=
proc(n)
option remember;
local spec, k, res;
res := 0;
for k to n-1 do
spec := { T1= Prod(Z, Sequence(T1)),
Z=Atom,
T2 = Prod(Z, Sequence(T1, card=k))};
res := res +
k*count([T2, spec, unlabeled], size=n);
od;
res/(1/n*binomial(2*n-2,n-1));
end;
Best Answer
Here is an answer using Egorychev method by way of enrichment. Note that the group acting on the cycle is the dihedral group $D_q$ and not just the cyclic group $C_q.$ We assume that two graphs where we get the second by turning over the first are considered the same. We thus have the combinatorial class equation
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{U} = \textsc{DHD}_{\ge 3}(\mathcal{T}).$$
This gives the EGF
$$U(z) = \sum_{q\ge 3} \frac{T(z)^q}{2q} = \frac{1}{2} \log\frac{1}{1-T(z)} - \frac{T(z)}{2} - \frac{T(z)^2}{4}.$$
The quantity we seek is $$n! [z^n] U(z) = (n-1)! [z^{n-1}] U'(z).$$ This is given by
$$(n-1)! \;\underset{z}{\mathrm{res}}\; \frac{1}{z^n} U'(z) = (n-1)! \;\underset{z}{\mathrm{res}}\; \frac{1}{z^n} \left[\frac{1}{2}\frac{1}{1-T(z)} - \frac{1}{2} - \frac{1}{2} T(z) \right] T'(z).$$
Now put $T(z) = w.$ We have from the combinatorial class specification that
$$\mathcal{T} = \mathcal{Z} \times \textsc{SET}(\mathcal{T})$$
so that
$$T(z) = z \exp T(z) \quad\text{and}\quad z = T(z) \exp(-T(z)).$$
We thus obtain
$$(n-1)! \;\underset{w}{\mathrm{res}}\; \frac{\exp(nw)}{w^n} \left[\frac{1}{2}\frac{1}{1-w} - \frac{1}{2} - \frac{1}{2} w \right] \\ = (n-1)! [w^{n-1}] \exp(nw) \left[\frac{1}{2}\frac{1}{1-w} - \frac{1}{2} - \frac{1}{2} w \right].$$
Extract the coefficient to obtain
$$\frac{1}{2} (n-1)! \times \left[ \sum_{q=0}^{n-1} \frac{n^q}{q!} - \frac{n^{n-1}}{(n-1)!} - \frac{n^{n-2}}{(n-2)!} \right].$$
We thus have the sum formula
$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{2} (n-1)! \sum_{q=0}^{n-3} \frac{n^q}{q!}.}$$
This gives the sequence OEIS A057500
$$0, 0, 1, 15, 222, 3660, 68295, 1436568, 33779340, 880107840, \ldots$$
where we see we have the right data. The traditional Lagrange inversion probably refers to the functional equation of the tree function $T(z).$
Note that the OEIS says we are counting connected labeled graphs with $n$ edges and $n$ nodes. This is the same statistic, however. Supposing such a graph had more than one cycle then there is a cycle from which we may remove one edge to obtain a graph with at least one other cycle remaining. The graph is still connected as the rest of the cycle can replace the removed edge. But if it is connected and has $n-1$ edges it must be a tree, a contradiction as there were cycles left over that we didn't touch by removing the edge. So these graphs contain at most one cycle. And this is indeed what happens since otherwise we would have a tree, again a contradiction by the number of nodes.