11. The sequence $\left\{a_{n}\right\}$ satisfies -
$$
\begin{array}{l}
a_{1}=8, a_{2}=26, \\
a_{n}=a_{n-1}+a_{n-2}+a_{n-1} a_{n-2} .
\end{array}
$$

Then $a_{10}=$
11. $3^{144}-1$

According to the condition, we have $a_{n}+1=\left(a_{n-1}+1\right)\left(a_{n-2}+1\right)$ and $\left\{a_{n}\right\}$ is a strictly increasing sequence of positive integers.
$$
\begin{array}{l}
\text { and } \log _{3}\left(a_{n}+1\right) \\
=\log _{3}\left(a_{n-1}+1\right)+\log _{3}\left(a_{n-2}+1\right) . \\
\text { Let } b_{n}=\log _{3}\left(a_{n}+1\right) \text {. Then } \\
b_{1}=2, b_{2}=3, b_{n}=b_{n-1}+b_{n-2}(n \geqslant 3) .
\end{array}
$$

Thus, $b_{10}=144$. Therefore, $a_{10}=3^{b_{10}}-1=3^{144}-1$.