5. A certain public figure and all of his descendants, regardless of gender, are included in the family tree, but their spouses are not included in the family tree. The family tree only indicates whether a member is male or female and their parent-child relationships. Given that the family tree of this public figure includes a total of $n$ male and female members, each member has at most two children, and if they have two children, they must be one male and one female. How many possible family trees are there for this public figure?
4. Let $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+\cdots$
$$
\begin{array}{l}
-2 x \cdot f(x)=-2 a_{0} x-2 a_{1} x^{2}-\cdots-2 a_{n-1} x^{n}-\cdots, \\
-3 x^{2} \cdot f(x)=-3 a_{0} x^{2}-\cdots-3 a_{n-2} x^{n}-\cdots, \\
-\frac{1}{1-3 x}=-1-3 x-3^{2} x^{2}-\cdots-3^{n} x^{n}-\cdots .
\end{array}
$$

Adding the above four equations and using the given conditions, we get $\left(1-2 x-3 x^{2}\right) \cdot f(x)-\frac{1}{1-3 x}=-2$. Therefore, $f(x)=\frac{6 x-1}{(1+x)(1-3 x)^{2}}=\frac{A}{1+x}+\frac{B}{(1-3 x)^{2}}+\frac{C}{1-3 x}$,
4. The family tree of a certain person can be represented by a tree diagram similar to the one on the right. This family tree can be constructed as follows: each member is represented by a black dot, and members with father-son (or daughter) and mother-son (or daughter) relationships are connected by line segments; the son of any member is drawn below and to the left, and the daughter is drawn below and to the right. Let $a_{n}$ be the number of ways to construct such a family tree, and it is clear that $a_{1}=1, a_{2}=2$, and let $a_{0}=1$. Now consider $n \geqslant 3$, and remove the certain person from the family tree, turning it into two separate family trees for the person's son and daughter. There are the following possible scenarios: the son's lineage has $k$ people, and the family tree has $a_{k}$ possible configurations, while the daughter's lineage has $n-k-1$ people, and the family tree has $a_{n-k-1}$ possible configurations, i.e., there are $a_{k} a_{n-k-1}$ possible configurations, where $k=0,1,2, \cdots, n-1$. Thus, we have the relation $a_{n}=a_{0} a_{n-1}+a_{1} a_{n-2}+\cdots+a_{n-1} a_{0}$.
Let the generating function of the sequence $\left\{a_{n}\right\}$ be $f(x)=\sum_{k=0}^{\infty} a_{k} x^{k}$, then
$$
\begin{aligned}
a_{0}=a_{1} & =1, x \cdot f^{2}(x)=x\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+\cdots\right)^{2} \\
& =a_{0}^{2} x+\left(o_{0} a_{1}+a_{2} a_{0}\right) x^{2}+\cdots+\left(a_{0} a_{n-1}+a_{1} a_{n-2}+\cdots+a_{n-1} a_{0}\right) x^{n}+\cdots \\
& =a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+\cdots=f(x)-1 .
\end{aligned}
$$

Thus, the generating function $f(x)$ satisfies the equation $x f^{2}(x)-f(x)+1=0$. Therefore, $2 x f(x)=1 \pm \sqrt{1-4 x}$, but since $f(0)=a_{0}=1$, we have $f(x)=\frac{1}{2 x}(1-\sqrt{1-4 x})$.

Applying formula 5, expand $(1-4 x)^{\frac{1}{2}}$ and substitute it into (*), we can find (the process is omitted) $f(x)=\sum_{k=0}^{\infty} \frac{1}{k+1} C_{2 k}^{k} \cdot x^{k}$. Hence, $a_{n}=\frac{1}{n+1} C_{2 n}^{n}(n \geqslant 0)$ is the solution.