Given $\{a$.$\}$ is a sequence of positive integers,
$$
\begin{aligned}
& a_{n+3}=a_{n+2}\left(a_{n+1}+2 a_{0}\right), \quad(a \in N) . \\
a_{0}= & 2238 .
\end{aligned}
$$

Find $a_{1}, a_{2}, a_{3}$.
Solution: $\because a_{4}=a_{3}\left(a_{2}+2 a_{1}\right)$,
$a_{5}=a_{3}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{3}\right)$
$\therefore a_{6}=a_{3}^{2}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right)$
- $\left(a_{2}+2 a_{1}+2\right)$.
Also, $a_{0}=2288=2^{4} \times 11 \times 13$.
$\because\left\{a_{n}\right\}$ is a sequence of positive integers.
$\therefore a_{3}+2 a_{2} \geqslant 3, a_{2}+2 a_{1} \geqslant 3$.

Since $\left(a_{2}+2 a_{1}+2\right)$ is 2 more than $\left(a_{2}+2 a_{1}\right)$,
and in $a_{0}=2^{4} \times 11 \times 13$ only 13 and 11 are factors not less than 3
and their difference is 2, it must be that
$$
\left\{\begin{array}{l}
a_{2}+2 a_{1}+2=13, \\
a_{2}+2 a_{1}=11 .
\end{array}\right.
$$

Therefore, $a_{3}^{2}\left(a_{3}+2 a_{2}\right)=2^{4}$.
Thus, $a_{3}=1$ or $a_{3}=2$.
If $a_{3}=1$, then $1+2 a_{2}=2^{4}$.
$\therefore a_{2}$ is not an integer, which is impossible, so $a_{3}$ can only be 2.
Substituting $a_{3}=2$ into $a_{3}^{2}\left(a_{3}+2 a_{2}\right)=2^{4}$,

we get $a_{2}=1$.
Thus, $a_{1}=5, a_{2}=1, a_{3}=2$.