14. (7th Balkan Mathematical Olympiad, 1990) Let $a_{1}=1, a_{2}=3$, and for all positive integers $n$,
$$
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n} \text {. }
$$

Find all values of $n$ for which $a_{n}$ is divisible by 11.
14. From the known
math
number
theory
problem
$$
\begin{array}{l}
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}, \\
a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right) . \\
\text { Let } b_{n+1}=a_{n+1}-a_{n} \text {, then we have } \\
b_{n+1}=(n+1) b_{n} .
\end{array}
$$

Let $b_{n+1}=a_{n+1}-a_{n}$, then we have

It is easy to see that $b_{n}=n!$.
Thus, $n!=a_{n}-a_{n-1}$,
$(n-1)!=a_{n-1}-a_{n-2}$,
$\qquad$
$2!=a_{2}-a_{1}$.
Adding these equations, and using $a_{1}=1$, we get
$$
a_{n}=1!+2!+\cdots+n!\text {. }
$$

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

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

so $11 \mid a_{8}$.
For $n \geqslant 10$,
$$
a_{n}=a_{8}+(9!+10!)+11!+12!+\cdots+n!=a_{8}+9!\cdot 11+11!+12!+\cdots+n!\text {, }
$$

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