Exponential Generating function for inhomogenous recurrence relation

Question

I want to determine all solutions to the recurrence relation

$$ a_n−2na_{n−1}+n(n−1)a_{n−2} = 2n·n!, \ \ \ n≥2, a_0 = a_1 = 1. $$

by using exponential generating functions.

My idea was something like that: $$ \sum_{n\geq2} a_n \frac{z^n}{n!}−\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}+\sum_{n\geq2}n(n−1)a_{n−2} \frac{z^n}{n!} - \sum_{n\geq2}2n\frac{z^n}{n!} = 0 \\ $$

Denoting

$$Â(z) =\sum_{n\geq0} a_n \frac{z^n}{n!}$$

which is the exponential generating function, I can derive that $\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}$ is $z^2 Â'(z)$ and for $\sum_{n\geq2}2n\frac{z^n}{n!}$, that $zÂ(2z)$ (which I am also not 100% sure of). However, I get stuck for the third summation term. How would I continue, am I even on the right path?

Answer 1

For the first term $$\sum_{n\geq2} a_n \frac{z^n}{n!}$$ I get $Â(z)-1-z$ by taking care of the first two initial conditions. For the second term $$\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}$$ I get $2(zÂ(z)-z)$, notice $\frac{n}{n!}=\frac{1}{(n-1)!}$ and that taking derivatives shift the index of $a_n$ up, multiplying by z shifts the index of $a_n$ down. The -z is obtained by accounting for initial conditions.

For the third term $$\sum_{n\geq2}n(n−1)a_{n−2} \frac{z^n}{n!}$$ I get $z^2Â(z)$. Notice that $\frac{n(n-1)}{n!}=\frac{1}{(n-2)!}$.

For the fourth term is actually $$ \sum_{n\geq 2}2nz^n$$ I get $\frac{2(z-2)z^2}{(z-1)^2}$. Because $\sum_{n\geq 1}nz^n=\frac{z}{(1-z)^2}$ and taking out the first term we get $\sum_{n\geq 2}nz^n=\frac{z}{(1-z)^2}-z=\frac{z(1-(1-z)^2)}{(1-z)^2}=\frac{z^2(2-z)}{(1-z)^2}$.