$8 \cdot 4$ Let the sequence of positive integers $\left\{a_{n}\right\}$ satisfy
$$
a_{n+3}=a_{n+2}\left(a_{n+1}+2 a_{n}\right), n=1,2, \cdots
$$

and $a_{6}=2288$. Find $a_{1}, a_{2}, a_{3}$.
［Solution］From the recursive relation, we get
$$
\begin{aligned}
a_{6} & =a_{5}\left(a_{4}+2 a_{3}\right)=a_{4}\left(a_{3}+2 a_{2}\right)\left(a_{4}+2 a_{3}\right) \\
& =a_{3}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right)\left[a_{3}\left(a_{2}+2 a_{1}\right)+2 a_{3}\right] \\
& =a_{3}^{2}\left(a_{3}+2 a_{2}\right)\left(a_{2}+2 a_{1}\right)\left(a_{2}+2 a_{1}+2\right) .
\end{aligned}
$$

Since \( a_{6}=2288=2^{4} \times 11 \times 13 \), and \( a_{2}+2 a_{1}+2 \) differs from \( a_{2}+2 a_{1} \) by 2, and both are factors of 2288, it follows that
$$
a_{2}+2 a_{1}=11, \quad a_{2}+2 a_{1}+2=13 .
$$

Thus, \( a_{3}^{2}\left(a_{3}+2 a_{2}\right)=2^{4} \),
which implies that \( a_{3} \) can only be 1 or 2. If \( a_{3}=1 \), then \( a_{2}=\frac{12}{5} \) is not a natural number, leading to a contradiction! Therefore, \( a_{3}=2, a_{2}=1 \). From \( a_{2}+2 a_{1}=11 \), we get \( a_{1}=5 \).