Solving a recurrence using substitutions I have to solve this recurrence using substitutions:
$(n+1)(n-2)a_n=n(n^2-n-1)a_{n-1}-(n-1)^3a_{n-2}$ with $a_2=a_3=1$.
The only useful substitution that I see is $b_n=(n+1)a_n$, but I don't know how to go on, could you help me please?
 A: it seems that I solved. $$(n-2)b_n=(n^2-n-1)b_{n-1}-(n-1)^2b_{n-2}$$ 
So divide it by $n-1$ then we get $$\begin{align*}
\left(1-\frac{1}{n-1}\right)b_n&=\left(n-\frac{1}{n-1}\right)b_{n-1}-(n-1)b_{n-2}\\
&=\left(n-1+1-\frac{1}{n-1}\right)b_{n-1}-(n-1)b_{n-2}\\
&=(n-1)(b_{n-1}-b_{n-2})+b_{n-1}\left(1-\frac{1}{n-1}\right)
\end{align*}$$ 
then $$\left(1-\frac{1}{n-1}\right)(b_n-b_{n-1})=(n-1)(b_{n-1}-b_{n-2})$$ 
then we make subtitution $p_n=b_n-b_{n-1}$ as $b_2=3,b_3=4$ we get that $p_3=1$. 
we have that $$\begin{align*}
&\left(1-\frac{1}{n-1}\right)p_n =(n-1)p_{n-1} \implies\\ &p_n=\frac{(n-1)^2}{n-2}p_{n-1}\implies\\
&p_n=\prod\limits_{k=4}^n{\frac{(k-1)^2}{k-2}}=\frac{n-1}{2}\frac{(n-1)!}{2} \implies\\
&b_n-b_{n-1}=\frac{n!-(n-1)!}{4} \implies\\
&b_n=4+\sum\limits_{k=4}^n{\frac{k!-(k-1)!}{4}}=4+\frac{1}{4}(n!-6)=\frac{10+n!}{4} \implies\\
&a_n=\frac{10+n!}{4(n+1)}
\end{align*}$$
A: So if $b_n=(n+1)a_n$, then $b_{n-1}=na_{n-1}$, and $b_{n-2}=(n-1)a_{n-2}$, and you equation becomes $$(n-2)b_n=(n^2-n-1)b_{n-1}-(n-1)^2b_{n-2}$$ which is a little simpler than what you started with, though I must admit I can't see offhand any easy way to get to a solution from there. Are you working from a text or some notes that have other examples of solving via substitution? Maybe there's a hint in some other example as to how best to proceed. 
