13. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{2}=5, a_{n+1}=\frac{a_{n} a_{n-1}}{\sqrt{a_{n}^{2}+a_{n-1}^{2}+1}}$, find the general term formula of the sequence $\left\{a_{n}\right\}$.
13. From the known recurrence relation, we have $a_{n+1}^{2}=\frac{a_{n}^{2} a_{n-1}^{2}}{a_{n}^{2}+a_{n-1}^{2}+1}$.

Thus, $\frac{1}{a_{n+1}^{2}}=\frac{1}{a_{n}^{2}}+\frac{1}{a_{n-1}^{2}}+\frac{1}{a_{n}^{2} a_{n-1}^{2}}$, which means $1+\frac{1}{a_{n+1}^{2}}=\left(1+\frac{1}{a_{n}^{2}}\right)\left(1+\frac{1}{a_{n-1}^{2}}\right) \cdots$ (1)
We define the sequence $\left\{F_{n}\right\}$ as follows: $F_{1}=1, F_{2}=0, F_{n}=F_{n-1}+F_{n-2}(n=3,4, \cdots)$.

Proof:
When $n=1,2$, it is easy to verify that (2) holds;
Assume that (2) holds for $n=k-1, k$, i.e.,

then $1+\frac{1}{a_{k+1}^{2}}=\left(1+\frac{1}{a_{k}^{2}}\right)\left(1+\frac{1}{a_{k-1}^{2}}\right)$
$$
\begin{array}{l}
=\left(1+\frac{1}{a_{1}^{2}}\right)^{F_{k}}\left(1+\frac{1}{a_{2}^{2}}\right)^{F_{k+1}} \cdot\left(1+\frac{1}{a_{1}^{2}}\right)^{F_{k-1}}\left(1+\frac{1}{a_{2}^{2}}\right)^{F_{k}} \\
=\left(1+\frac{1}{a_{1}^{2}}\right)^{F_{k}+F_{k-1}}\left(1+\frac{1}{a_{2}^{2}}\right)^{F_{k+1}+F_{k}}=\left(1+\frac{1}{a_{1}^{2}}\right)^{F_{k+1}}\left(1+\frac{1}{a_{2}^{2}}\right)^{F_{k+2}} .
\end{array}
$$

Therefore, (2) holds for $n=k+1$.
By the principle of mathematical induction, (2) holds for all positive integers $n$.
From (2), we get $1+\frac{1}{a_{n}^{2}}=2^{F_{n+2}} \cdot 13^{F_{n+1}} \cdot 5^{-2 F_{n+1}}$.
Hence, $a_{n}=\left(2^{F_{n+2}} \cdot 13^{F_{n+1}} \cdot 5^{-2 F_{n+1}}-1\right)^{\frac{1}{2}}\left(n \in \mathbf{N}^{*}\right)$, where $F_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n-2}-\right.$
$$
\left.\left(\frac{1-\sqrt{5}}{2}\right)^{n-2}\right]\left(n \in \mathbf{N}^{*}\right)
$$