Example 9 Let $a_{1}=1, a_{2}=3$, for all positive integers $n$ we have $a_{n+2}=(n+3) a_{n+1}-(n+$ 2) $a_{n}$, find the values of all numbers divisible by 11.
Given the recurrence relation, transform the problem through substitution, and finally find the general formula for $a_{n}$, thereby completing this problem.

Solution: From the given conditions, we have $\left(a_{n+2}-a_{n+1}\right)=(n+2)\left(a_{n+1}-a_{n}\right)$. Let $b_{n+1}=a_{n+1}-a_{n}(n \geqslant 1)$, then from the conditions, we have $b_{n+1}=(n+1)\left(a_{n}-a_{n-1}\right)=(n+1) b_{n}(n \geqslant 2)$, hence $b_{n}=n b_{n-1}=n(n-1) b_{n-2}=n(n-1)(n-2) \cdots 3 b_{2}=n!(n \geqslant 2)$. Therefore, $a_{n}=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right)+a_{1}=b_{n}+b_{n-1}+\cdots+b_{2}+1=\sum_{k=1}^{n} k!$. From this, we can calculate $a_{4}=\sum_{k=1}^{n} k!=33=11 \times 3, a_{8}=\sum_{k=1}^{8} k!=46233=11 \times 4203$, $a_{10}=\sum_{k=1}^{10} k!=4037913=11 \times 367083$. When $n \geqslant 11$, note that $\sum_{k=11}^{n} k!$ is divisible by 11, thus $a_{n}=\sum_{k=1}^{10} k!+\sum_{k=11}^{n} k!$ is also divisible by 11. Therefore, when $n=4, n=8$ or when $n \geqslant 10$, $a_{n}$ is divisible by 11.