Suppose that we have the following sequence satisfying for every $n\geq1$ and $1\leq k\leq n$
\begin{array}{l}
f(n,k)=f(n-1,k-1)-(2n-k-2)f(n-1,k),\\\\
f(n,n)=1,
\end{array}
where we formally set $f(n,0)=0$, for every $n$.
Is it possible to find a generating function? Thank you for all the help.
I obtained : $\displaystyle \;\boxed{f(n,k)= \frac{(2n-k-1)!}{(k-1)!\,(n-k)!\,(-2)^{n-k}}}\quad$ for $\,1\le k\le n\,$ and $\,0\,$ else.
Short derivation for $1<k<n$ : \begin{align} f(n,k)&=\frac{(2n-k-1)!}{(k-1)!\,(n-k)!\,(-2)^{n-k}}\\ &=\frac {2n-k-1}{k-1}\frac{(2n-k-2)!}{(k-2)!\,(n-k)!\,(-2)^{n-k}}\\ &=\frac {k-1+2(n-k)}{k-1}\frac{(2n-k-2)!}{(k-2)!\,(n-k)!\,(-2)^{n-k}}\\ &=f(n-1,k-1)+\frac{2(2n-k-2)\,(2n-k-3)!}{-2(k-1)(k-2)!\,(n-1-k)!\,(-2)^{n-1-k}}\\ &=f(n-1,k-1)-(2n-k-2)f(n-1,k)\\ \end{align}
Generating functions may be found using A001147 and others...