6. (1988 Sichuan Province High School Mathematics Competition) 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 $a_{1}, a_{2}, a_{3}$.
6. Using the recurrence relation, we get
$$
\begin{aligned}
a_{6} & =a_{5}\left(a_{1}+2 a_{3}\right)=a_{1}\left(a_{3}+2 a_{2}\right)\left(a_{1}+2 a_{3}\right) \\
& =a_{3}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right)\left[a_{3}\left(a_{2}+2 a_{1}\right)+2 a_{3}\right] \\
& =a_{3}^{2}\left(a_{3}+2 a_{2}\right)\left(a_{2}+2 a_{1}\right)\left(a_{2}+2 a_{1}+2\right) .
\end{aligned}
$$

Since $a_{6}=2288=2^{1} \times 11 \times 13$, 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
$$
a_{2}+2 a_{1}=11, a_{2}+2 a_{1}+2=13 \text {. }
$$

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, $a_{3}=2, a_{2}=1$, and from $a_{2}+2 a_{1}=11$, we get $a_{1}=5$.