11. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a_{2}=a_{3}=1$, for any positive integer $n$, we have
$$
a_{n+3}=\frac{a_{n+1} a_{n+2}+1}{a_{n}} .
$$

Find the general term formula for the sequence $\left\{a_{n}\right\}$.
For any positive integer $n$, we have
$$
\begin{array}{l}
a_{n} a_{n+3}=a_{n+1} a_{n+2}+1 \\
\Rightarrow a_{n+1} a_{n+4}=a_{n+2} a_{n+3}+1 .
\end{array}
$$

Subtracting and rearranging the two equations, we get
$$
\frac{a_{n+2}+a_{n+4}}{a_{n+3}}=\frac{a_{n}+a_{n+2}}{a_{n+1}} .
$$

Since $a_{4}=\frac{a_{2} a_{3}+1}{a_{1}}=2$, we have
$$
\begin{array}{l}
\frac{a_{2 k-1}+a_{2 k+1}}{a_{2 k}}=\frac{a_{1}+a_{3}}{a_{2}}=2, \\
\frac{a_{2 k}+a_{2 k+2}}{a_{2 k+1}}=\frac{a_{2}+a_{4}}{a_{3}}=3 .
\end{array}
$$

Thus, $a_{2 k-1}+a_{2 k+1}=2 a_{2 k}$,
$$
a_{2 k}+a_{2 k+2}=3 a_{2 k+1} \text {. }
$$

From equation (1), we get
$$
\begin{array}{l}
a_{2 k}=\frac{1}{2}\left(a_{2 k-1}+a_{2 k+1}\right) \\
\Rightarrow a_{2 k+2}=\frac{1}{2}\left(a_{2 k+1}+a_{2 k+3}\right) .
\end{array}
$$

Substituting into equation (2), we get
$$
a_{2 k+3}-4 a_{2 k+1}+a_{2 k-1}=0 \text {. }
$$

Similarly, $a_{2 k+4}-4 a_{2 k+2}+a_{2 k}=0$.
Let $a_{2 k-1}=x_{k}, a_{2 k}=y_{k}$.
Then $x_{k+2}-4 x_{k+1}+x_{k}=0\left(x_{1}=x_{2}=1\right)$,
$$
y_{k+2}-4 y_{k+1}+y_{k}=0\left(y_{1}=1, y_{2}=2\right) \text {. }
$$

By the method of characteristic roots, we get
$$
\begin{array}{l}
a_{2 k-1}=x_{k} \\
=\frac{9-5 \sqrt{3}}{6}(2+\sqrt{3})^{k}+\frac{9+5 \sqrt{3}}{6}(2-\sqrt{3})^{k}, \\
a_{2 k}=y_{k} \\
=\frac{2-\sqrt{3}}{2}(2+\sqrt{3})^{k}+\frac{2+\sqrt{3}}{2}(2-\sqrt{3})^{k} .
\end{array}
$$