Consider the recurrence $$\mu_1=1, \mu_2=2, \mu_3=4, \mu_4=8, \mu_5=16, \mu_6=32 $$ and $$\mu_{n+6} = \mu_n + \mu_{n+1} + \mu_{n+2} + \mu_{n+3} + \mu_{n+4} + \mu_{n+5}, n\geqslant 1. $$ The generating function is $$P(s) = \sum_{n=1}^\infty \mu_ns^n.$$ By looking at the first few terms of the sequence, it appears that $$P(s) = \sum_{n=0}^\infty(s+s^2+s^3+s^4+s^5+s^6)^n. $$ If I can write $$\sum_{n=0}^\infty(s+s^2+s^3+s^4+s^5+s^6)^n = \sum_{n=1}^\infty a_nx^n $$ then I could show that $a_n$ satisfies the recurrence by induction. It seems that there is a closed form for $a_n$: $$a_n = \sum_{i=1}^6\frac{x_i^n}{-x_i^5+x_i^3+2x_i^2+10x_1-1}$$ where $x_i$ are the roots of $x^6-x^5-x^4-x^3-x^2-x-1$. That isn't very feasible, so is there another way to show that this generating function satisfies the recurrence?
For context, $\mu_n$ is the number of ways to throw die until the sum of the faces is $n$. I found the recurrence by inspection, but am not sure how to find the generating function from the recurrence.