Is there any way to solve recurrence equations with variable coefficients? So far I have done some problems that are best solved using generating functions. These mostly contain variable coefficients. 
A simple one is $H(n) = (n+2)H(n-2)$. I have found solutions to these equations using mathematical induction, which requires a bit of conjecturing (by checking the result for initial values) and then proving it.
But what about bigger equations,* is there a definite way of solving them and obtain a simple formula (without resorting to generating functions)? 

edit:
*Functions like $H(n) =f_1(n)H(n-1) + f_2(n)H(n-2) + \cdots + f_k(n)H(n-k)$. Where $f_1, f_2,\dots, f_k$ are functions of $n$.
$n\to n+2$ :

$H(n+2)=(n+4)H(n)$

$n\to 2n$ :

$H(2n+2)=(2n+4)H(2n)$

$H(2(n+1))=2(n+2)H(2n)$

$H(2n)=\Theta_1(n)\prod\limits_n(2(n+2))$ , where $\Theta_1(n)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_product#Rules ,

$H(2n)=\Theta_1(n)2^n\Gamma(n+2)$ , where $\Theta_1(n)$ is an arbitrary periodic function with unit period

$H(n)=\Theta(n)2^{\frac{n}{2}}\Gamma\left(\dfrac{n}{2}+2\right)$ , where $\Theta(n)$ is an arbitrary periodic function with period $2$
$n\to n+2$ :

$H(n+2)=(n+4)H(n)$

$n\to 2n$ :

$H(2n+2)=(2n+4)H(2n)$

$H(2(n+1))=2(n+2)H(2n)$

$H(2n)=\Theta_1(n)\prod\limits_n(2(n+2))$ , where $\Theta_1(n)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_product#Rules ,

$H(2n)=\Theta_1(n)2^n\Gamma(n+2)$ , where $\Theta_1(n)$ is an arbitrary periodic function with unit period

$H(n)=\Theta(n)2^{\frac{n}{2}}\Gamma\left(\dfrac{n}{2}+2\right)$ , where $\Theta(n)$ is an arbitrary periodic function with period $2$