labelled rooted tree with odd degree Let $F(z)$ be the exponential generating function for labelled rooted trees with every vertex having even out-degree and let $H(z)$ be the exponential generating function for labelled rooted trees in which every vertex has odd degree.
How can I use the compositional formula to show that $$F(z) = \frac{z}{2}\left(e^{F(z)}+ e^{-F(z)}\right)$$ and then express $H(z)$ in terms of $F(z)$. Please, any help would be appreciated.
 A: For the first one we have from first principles using the notation from
Analytic Combinatorics the combinatorial class equation
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{F}
= \mathcal{Z} \times \textsc{SET}_{\text{even}}(\mathcal{F}).$$
We attach  a set of  even-outdegree trees  of even cardinality  at the
root and  obtain even  out-degree at all  nodes, recursively.  For the
second one observe that the  elements of $\mathcal{F}$ have odd degree
at all nodes  except the root, hence  we must attach an  odd number of
these to the root, getting
$$\mathcal{H}
= \mathcal{Z} \times \textsc{SET}_{\text{odd}}(\mathcal{F}).$$
Note that
$$\textsc{SET}_{\text{even}}(\mathcal{Z})
= \textsc{SET}_{\text{=0}}(\mathcal{Z})
+ \textsc{SET}_{\text{=2}}(\mathcal{Z})
+ \textsc{SET}_{\text{=4}}(\mathcal{Z}) + \cdots$$
which gives the generating function
$$\frac{z^0}{0!}+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots
= \frac{1}{2} (\exp(z)+\exp(-z)).$$
Similarly,
$$\textsc{SET}_{\text{odd}}(\mathcal{Z})
= \textsc{SET}_{\text{=1}}(\mathcal{Z})
+ \textsc{SET}_{\text{=3}}(\mathcal{Z})
+ \textsc{SET}_{\text{=5}}(\mathcal{Z}) + \cdots$$
this time with generating function
$$\frac{z^1}{1!}+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots
= \frac{1}{2} (\exp(z)-\exp(-z)).$$
The conclusion is that
$$F(z) = z \frac{1}{2} (\exp(F(z)) + \exp(-F(z))
= z \cosh F(z)$$
and
$$H(z) = z \frac{1}{2} (\exp(F(z)) - \exp(-F(z))
= z \sinh F(z).$$
Solving  these  we  get OEIS  A036778  and
OEIS A060279.
Remark. If we want to extract coefficients from $F(z)$
use the Cauchy Coeffcient Formula and write
$$F(z) = \sum_{n\ge 1} Q_n \frac{z^n}{n!}$$
to get
$$\frac{Q_n}{(n-1)!} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}} F'(z) \; dz$$
We have
$$z = \frac{2F(z)}{\exp(F(z)) + \exp(-F(z))}$$
and put $F(z) = w$ so that $F'(z) \; dz = dw$ to obtain
$$\frac{Q_{n}}{(n-1)!} =
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{2^{n} w^{n}} (\exp(w)+\exp(-w))^{n} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{2^{n} w^{n}}
\sum_{p=0}^n {n\choose p} \exp(pw) \exp(-(n-p)w) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{2^{n} w^{n}}
\sum_{p=0}^n {n\choose p} \exp((2p-n)w) \; dw.$$
This is
$$Q_n = \frac{(n-1)!}{2^n}
\sum_{p=0}^n {n\choose p} \frac{(2p-n)^{n-1}}{(n-1)!}$$
or
$$\bbox[5px,border:2px solid #00A000]{
Q_n = \frac{1}{2^n}
\sum_{p=0}^n {n\choose p} (2p-n)^{n-1}.}$$
The sequence is
$$1, 0, 3, 0, 65, 0, 3787, 0, 427905, 0, 79549811, 0,
\\ 22036379521, 0, 8513206310715, 0, 4374455745966593, \ldots$$
