$4 \cdot 31$ Given $\left\{a_{n}\right\}$ is a sequence of positive integers
$$
a_{n+3}=a_{n+2}\left(a_{n+1}+2 a_{n}\right) \quad(n \in N), \quad a_{6}=2288 .
$$

Find $a_{1}, a_{2}, a_{3}$.
(China Sichuan Province High School Mathematics Competition, 1988)
［Solution］From the recursive formula,
$$
\begin{array}{l}
a_{4}=a_{3}\left(a_{2}+2 a_{1}\right), \\
a_{5}=a_{3}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right), \\
a_{6}=a_{3}^{2}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right)\left(a_{2}+2 a_{1}+2\right) .
\end{array}
$$

Also, $a_{6}=2288=2^{4} \cdot 11 \cdot 13$.
Since the sequence $\left\{a_{n}\right\}$ consists of positive integers, we have
$$
a_{3}+2 a_{2} \geqslant 3, \quad a_{2}+2 a_{1} \geqslant 3 .
$$

Furthermore, $\quad\left(a_{2}+2 a_{1}+2\right)-\left(a_{2}+2 a_{1}\right)=2$, and in $a_{6}=2^{4} \cdot 11 \cdot 13$ only 13 and 11 are factors that are not less than 3 and have a difference of 2. Therefore, it must be that
$$
\left\{\begin{array}{l}
a_{2}+2 a_{1}+2=13, \\
a_{2}+2 a_{1}=11 .
\end{array}\right.
$$

Thus,
$$
a_{3}^{2}\left(a_{3}+2 a_{2}\right)=2^{4} \text {. }
$$

If $a_{3}$ is odd, then the left side of the equation is odd, while the right side is even, which is impossible.
Therefore, $a_{3}$ must be a power of 2. Given $a_{3}+2 a_{2} \geqslant 3$, we have $a_{3}=2$.
Substituting $a_{3}=2$ into (1) gives
$$
\begin{array}{c}
4\left(2+2 a_{2}\right)=16 . \\
a_{2}=1 .
\end{array}
$$

From $a_{2}+2 a_{1}=11$, we get $a_{1}=5$.
Thus, $\quad a_{1}=5, a_{2}=1, a_{3}=2$.