$1 \cdot 34$ Let $a_{1}=1, a_{2}=3$, and for all positive integers $n$,
$$
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n} .
$$

Find all values of $n$ for which $a_{n}$ is divisible by 11.
(7th Balkan Mathematical Olympiad, 1990)
［Solution］From the given information,
$$
\begin{array}{c}
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}, \\
a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right) .
\end{array}
$$

Let $b_{n+1}=a_{n+1}-a_{n}$, then we have
$$
b_{n+1}=(n+1) b_{n} \text {. }
$$

From this, it is easy to see that
$$
b_{n}=n!\text {. }
$$

Thus,
$$
n!=a_{n}-a_{n-1},
$$
$$
\begin{aligned}
(n-1)!= & a_{n-1}-a_{n-2}, \\
& \cdots \cdots \cdots \cdots \\
2!= & a_{2}-a_{1} .
\end{aligned}
$$

Adding these equations and using $a_{1}=1$, we get
$$
a_{n}=1!+2!+\cdots+n!\text {. }
$$

By calculation, we find
$$
a_{4}=1+2!+3!+4!=33 .
$$

So
$11 \mid a_{4}$.
$$
a_{8}=1+2!+3!+\cdots+8!=46233 \text {. }
$$

So $11 \mid a_{8}$.

For $n \geqslant 10$,
$$
\begin{aligned}
a_{n} & =a_{8}+(9!+10!)+11!+12!+\cdots+n! \\
& =a_{8}+9!\cdot 11+11!+12!+\cdots+n!
\end{aligned}
$$

So
$$
11 \mid a_{n} \quad(n \geqslant 10) \text {. }
$$

Therefore, when $n=4, n=8, n \geqslant 10$, $11 \mid a_{n}$.