15. Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ is always divisible by 9, find the minimum value of $n$.

When $m \geqslant n$, $a_{m}$'s value can be divided by 9, find the smallest $n$.
15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1)$ $\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots \cdot 4 \cdot 3 \cdot\left(a_{2}-a_{1}\right)=(n+2)!$,
it follows that $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=1+2!+3!+\cdots+n!\quad(n \geqslant 1)$.
Given $a_{1}=1, a_{2}=3, a_{3}=9, a_{4}=33, a_{5}=153$, at this point 153 is divisible by 9.
When $m \geqslant 5$, $a_{n}=a_{5}+\sum_{k=6}^{m} k!$ and for $k \geqslant 6$, $k!$ is divisible by 9, so for $m \geqslant 5$, $a_{n}$ is divisible by 9. Therefore, the smallest value of $n$ is 5.