Counting endofunctions with a certain recurrent condition. Problem: Let $\phi:X\rightarrow X$ be an endofunction. A vertex $v\in X$ is recurrent if there is some positive integer $k\geq 1$ such that $\phi^k(v)=v$. Let $\mathcal{Q}$ be the species of endofunctions $\phi:X\rightarrow X$ such that if $v\in X$ has $|\phi^{-1}(v)|\geq 2$, then $v$ is recurrent. Obtain a formula for the exponential generating function $$Q(x)=\sum_{n=0}^\infty |\mathcal{Q}_n|\frac{x^n}{n!},$$ where $|\mathcal{Q}_n|$ is the number of such endofunctions on a set $|X|=n$.
My approach: The species of endofunctions is naturally equivalent to the species of sets of cycles of rooted labelled-trees, which we denote by $\mathcal{F}\equiv \mathcal{E}[\mathcal{C}[\mathcal{R}]]$. In this setting its easy to observe that a vertex is recurrent if and only if it lives on a cycle. The condition that $v\in X$ has $|\phi^{-1}(v)|\geq 2$, then $v$ is recurrent means that only each tree root vertex can have degree at most 2, while all others tree vertices must be degree 1 or 0. So to find a decomposition of $\mathcal{Q}$ (and hence its e.g.f $Q(x)$) we should replace the species of rooted labelled-trees in the general endofunction decomposition with another species that decomposes like $\mathcal{X}*\left(\mathcal{L}\oplus \mathcal{L}*\mathcal{L}\right)$ i.e. a root followed by a linear order or a pair of linear orders (here I'm thinking about the linear orders like trees with no branches). The species of linear orders $\mathcal{L}$ has exponential generating function $L(x)=(1-x)^{-1}$, and so we replace the e.g.f $R(x)$ in the endofunction e.g.f $$F(x)=\exp(\log(1-R(x))^{-1}$$ with $$A(x)=x\left((1-x)^{-1}+(1-x)^{-2}\right).$$ This gives $$Q(x)=\exp(\log(1-A(x))^{-1}=\frac{(1-x)^2}{(2x^2-4x+1)}.$$ Using SAGE the coefficients $\left[\frac{x^n}{n!}\right]Q(x)=|\mathcal{Q}_n|$ gave me 1,2,7,24,82... which seems reasonable and is https://oeis.org/A003480 in the OEIS.
My question: Does my description of the species replacing the rooted labelled trees give the correct description of $\mathcal{Q}$, and if so is there a formula for these objects that is not a rational expression?
 A: I disagree with your statement "each tree root vertex can have degree at most $2$". For example, the endofunction $\phi:\{1,2,3,4\}\to \{1,2,3,4\}$ defined by
$$
\phi(1)=\phi(2)=\phi(3)=\phi(4)=1
$$
satisfies "$|\phi^{-1}(v)\ge 2|$ implies $v$ is recurrent," and $1$ is a root vertex in the composition you described, yet $|\phi^{-1}(1)|=4$. 

Let $\mathcal P$ be the species of permtutations (linear orderings), so that $\mathcal P=\mathcal E[\mathcal C]$. Your first observation can be stated as
$$
\mathcal F=\mathcal P[\mathcal R]
$$
That is, you are decomposing your endofunction as a permutation of certain rooted trees. In each tree, the root can have arbtrary in-degree, but all other vertices must have in-degree $0$ or $1$. This means that each root has some set of nonempty chains dangling off of it. Therefore, your species $\mathcal R$ is consists of a root, together with a set of nonempty linearly ordered chains, or 
$$
\mathcal R=\{\bullet\}\times \mathcal E[\mathcal P^+]
$$
where $\mathcal P^+$ is the species of  nonempty permtuations, so that $\mathcal P=\mathcal P^++1$. This should give
$$
R(x) = x\cdot \exp\left(\frac{x}{1-x}\right)
$$
Combined with $\mathcal F=\mathcal P[\mathcal R]$ and $P(x)=\frac1{1-x}$, we get
$$
Q(x)=\frac1{1-R(x)}=\frac1{1-x\cdot \exp\left(\frac{x}{1-x}\right)}
$$
