How can find this sequence $ a_{n+1}=a_{n}+na_{n-1},$ let $a_{n+1}=a_{n}+na_{n-1},a_{1}=1，a_{2}=2$.
find the $a_{n}=?$
my ideas: $\dfrac{a_{n+1}}{(n+1)!}=\dfrac{1}{n+1}\dfrac{a_{n}}{n!}+\dfrac{1}{n+1}\dfrac{a_{n-1}}{(n-1)!},$
and let $b_{n}=\dfrac{a_{n}}{n!}$,then we have 
$(n+1)b_{n+1}=b_{n}+b_{n-1},b_{1}=1,b_{2}=1$,but I failure,
so let $f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n$,and assume that $b_{0}=0$
$f'(x)=\displaystyle\sum_{n=0}^{\infty}nb_{n}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(n+1)b_{n+1}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(b_{n}+b_{n-1})x^{n-1}=1+\dfrac{f(x)}{x}+f(x)$
then $f'(x)-(1+1/x)f(x)=1,f'(0)=1$
so
$f(x)=xe^x(\displaystyle\int \dfrac{e^{-x}}{x}dx+c)$
so $c=0$
and $f(x)=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$
and  my problem：How can prove that:
if:$f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$
then we have 
$b_{n}=\displaystyle\sum_{k=0}^{[n/2]}\dfrac{1}{（n-2k)!2^kk!}?$

If $f(x)=\sum_{n\ge 1}a_nx^{n+1}$ then 
$$
x^3f'(x)+(x-1)f(x)+x^3+x^2=0
$$
