Alternating Recurrence relation $a_n = b_{n-1} + 5$ and $b_n = na_{n-1}$ I am racking my brain on solving the relation where:
$$a_n = b_{n-1} + 5$$
$$b_n = na_{n-1}$$
where $a_0$ = $b_0$ = 1
I am trying to find the closed form for $a_n$.  I have tried to shifting $b_n = na_{n-1}$ to $b_{n-1} = (n-1)a_{n-2}$ by substituting n with n-1.  Then I plug $b_{n-1}$ in to $a_n = b_{n-1} + 5$ to get $a_n = (n-1)a_{n-2} + 5$ 
I am trying to use the method of generating functions to do this.  I have that $A(x) = \sum_{n=0}^{\infty} a_{n}x^{n}$.  Now I move to the n = 3 term of the summand and get $1+6x+6x^2 + \sum_{n=3}^{\infty} a_{n}x^{n}$ as $a_0 = 1, a_1 = a_2 = 6$ Then we substitute in the relation to get $1+6x+6x^2 + x^2\sum_{n=3}^{\infty} ((n-1)a_{n-2}+5)x^{n-2}$ = $1+6x+6x^2 + x^2\sum_{n=3}^{\infty} (n-1)a_{n-2}x^{n-2}+\sum_{n=3}^{\infty}5x^{n-2}$
I know what to do with the latter term as $\sum_{n=3}^{\infty}5x^{n-2} = 5x/(1-x)$.  I however am not sure what to do with the $\sum_{n=3}^{\infty} (n-1)a_{n-2}x^{n-2}$ term as there is an (n-1) in it.  I do however see that $(n-1)a_{n-2}x^{n-2}$ looks like it could be integrated with respect to x to become $a_{n-2}x^{n-1}$.  If I then pull out a x I could end up with:
$1+6x+6x^2 + x^3/(1-x)+ 5x/(1-x)$
But I am first shaky on how I came to this, but am furthermore wondering how this would be a closed form i.e. we don't even have x's in the sequence.
Thoughts would be very appreciated.
Thanks,
Brian

Start with $a_n = (n − 1) a_{n − 2} + 5$, and note that this is a first order linear recurrence in $b_n = a_{2 n + 1}$:
$$
b_{n + 1} = 2 n b_n + 5
$$
Summing factor is $\prod_{0 \le k \le n} 2 n = 2^n n!$:
$$
\frac{b_{n + 1}}{2^n n!} - \frac{b_n}{2^{n - 1} (n - 1)!}
  = \frac{5}{2^n n!}
$$
Sum for $1 \le k \le n$ to get:
$$
\frac{b_{n + 1}}{2^n n!} - b_1 = 5 \sum_{1 \le k \le n} \frac{2^{-n}}{n!}
$$
The right hand side is (almost) $5 \left(e^{1/2} - 1 \right)$, so $a_{2 n + 1} = b_n \approx a_1 + 5 \cdot 2^{n - 1} (n - 1)! \left(e^{1/2} - 1\right)$
