TITLE: How to Solve Advanced Recurrence Relations
QUESTION [5 upvotes]: For some research I’m doing, I had to solve recurrences of the form: $a(n)=n \cdot a(n-1)-(n-3) \cdot a(n-2)$. 
WolframAlpha gave the solution:
$$ a(n) = 4c_2 \Gamma (n-1) ( \Gamma (n-1) F(1,n-2;n,n;1) -(c_1-4c_2(e-3))(n-2) )$$
Where F is the regularized generalized hypergeometric function.
I was taught a little bit about creating closed forms from recurrence relations, but have no clue to derive something such as this. It looks really similar to examples where you use exponential generating functions, but I can’t see exactly how it would work in this case. So my question is, how would a human get this solution, and where could I go to learn this and similar advanced techniques.

REPLY [3 votes]: Here are two references which might be useful when learning recurrence relations.

Some interesting techniques are presented in Mathematics for the Analysis of Algorithms
  by D. E. Knuth and D. H. Greene. One of them is the so-called repertoire method. A real life application is presented in this post .
Recurrences are also a prominent theme in Concrete Mathematics
  by R.L. Graham, D.E. Knuth and O. Patashnik.


Hint: Sometimes we are lucky and can find a solution also by elementary means. Assuming initial values $a_0, a_1$ are given, we derive from
\begin{align*}
a_n=na_{n-1}-(n-3)a_{n-2}\qquad n\geq 2
\end{align*}
successively
\begin{align*}
\color{blue}{1}a_2&=2a_1+a_0\\
a_3&=\color{blue}{3}a_2\\
a_4&=4a_3-a_2=\color{blue}{11}a_2\\
a_5&=5a_4-2a_3=\color{blue}{49}a_2\\
a_6&=6a_5-3a_4=\color{blue}{261}a_2\\
&\cdots\\
\end{align*}
Looking for the values $1,3,11,49,261$  in OEIS we find the sequence A001339 with representation
\begin{align*}
a_{n+2}=\sum_{k=0}^n\binom{n}{k}(k+1)!\qquad\qquad n\geq 0
\end{align*}