TITLE: How find the $a_{n}$ close form
QUESTION [6 upvotes]: Question:

let $a_{1}=1,a_{2}=3$, and $a_{n+2}=(n+1)a_{n+1}+a_{n}$

find the close form $a_{n}$
my try:

let
  $$\dfrac{a_{n+2}}{(n+1)!}=\dfrac{a_{n+1}}{n!}+\dfrac{a_{n}}{(n+1)!}$$
  so
  $$\dfrac{a_{n+2}}{(n+1)!}-\dfrac{a_{2}}{1!}=\sum_{i=1}^{n}\dfrac{a_{i}}{(i+1)!}$$

But we can't find the $a_{n}$ Thank you

REPLY [3 votes]: Hint: 
Use the following recurrence equations:
$$I_n(z)=\frac{2(n+1)}{z}I_{n+1}(z)+I_{n+2}(z)$$
and:
$$K_n(z)=\frac{2(n+1)}{z}K_{n+1}(z)-K_{n+2}(z)$$
If you put $z=2$ and $a_n=I_n(-2)+K_n(2)$ you get:
your recurrence equation in $a_n$
You can find what you need here