9. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1} a_{n-1}=a_{n} a_{n-1}+a_{n}^{2}\left(n \in \mathbf{N}^{*}, n \geqslant 2\right), \frac{a_{n+1}}{a_{n}}=k n+1$. Let $f(n)=\left\{\begin{array}{l}a_{n}+5 a_{n-1}\left(n=2 l-1, l \in \mathbf{N}^{*}\right) \\ 3 a_{n}+2 a_{n-1}\left(n=2 l, l \in \mathbf{N}^{*}\right)\end{array}\right.$, find the positive integer $m$, such that $f(m+15)=5 A_{m+14}^{15}$ $f(m)$ holds.
$$
\begin{array}{l}
\text { 9. Solution: } a_{n+1} a_{n-1}=a_{n} a_{n-1}+a_{n}^{2}, \\
\therefore \frac{a_{n+1}}{a_{n}} \times \frac{a_{n-1}}{a_{n}}=\frac{a_{n-1}}{a_{n}}+1 . \\
\frac{k n+1}{k(n-1)+1}=\frac{1}{k(n-1)+1}+1, \\
k n+1=k(n-1)+1+1, \therefore k=1 . \\
\therefore a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_{2}}{a_{1}} \cdot a_{1}=n \cdot(n-1) \cdots 2 \cdot 1=n!. \\
\therefore f(n)=\left\{\begin{array}{l}
n!+5(n-1)!\left(n=2 l-1, l \in \mathbf{N}^{*}\right) . \\
3 n!+2(n-1)!\quad\left(n=2 l, l \in \mathbf{N}^{*}\right)
\end{array}\right.
\end{array}
$$
(1) When $m$ is odd, $m+15$ is even.

From $f(m+15)=5 A_{m+14}^{15} f(m)$ we get
$$
\begin{array}{c}
{[3(m+15)!+2(m+14)!]=5 A_{m+14}^{15}[m!+5(m-1)!],} \\
(m+14)!(3 m+47)=5 A_{m+14}^{15}(m-1)!(m+5),
\end{array}
$$

So $3 m+47=5 m+25, m=11$.
(2) When $m$ is even, $m+15$ is odd.

From $f(m+15)=5 A_{m+14}^{15} f(m)$ we get
$$
\begin{array}{c}
{[(m+15)!+5(m+14)!]=5 A_{m+14}^{15}[3 m!+2(m-1)!],} \\
(m+14)!(m+20)=5 A_{m+14}^{15}(m-1)!(3 m+2)
\end{array}
$$

So $m+20=15 m+10, m=\frac{5}{7} \notin \mathbf{N}^{*}$ (discard).
In summary, there exists a unique positive integer $m=11$, such that $f(m+15)=5 A_{m+14}^{15} f(m)$ holds.
$$
$$