I have this exponential generating function for arranging $n$ books into 1 shelf: $$\hat{a}(x) = \sum_{n=1}^\infty a_n\frac{x^n}{n!} , \text{ with } a_n = n! \text{ for } n \geq 1 $$
My transformation to get the algebraic expression of $\hat{a}(x)$ is: $$\hat{a}(x) = \sum_{n=1}^\infty x^n = \left(\sum_{n=0}^\infty x^n\right) - 1 = \frac{1}{1-x} - 1=\frac{x}{1-x}$$
Now I think the generating function for $(t_n)_{n \geq 2}$ is $\ \hat{t}(x) = \hat{a}^2(x)$, but I have a few questions:
The indices in this case are a bit odd: normally I would have multiplication of generating functions written as $\hat{t}(x)_{n \geq 0} = \hat{a}(x)_{n \geq 0} \times \hat{b}(x)_{n \geq 0}$, but in this case it is $\hat{t}(x)_{n \geq 2} = \hat{a}(x)_{n \geq 1} \times \hat{a}(x)_{n \geq 1}$. Is it an issue?
Also a way to amend the situation is to set $t_0 = t_1 = 0$, and $a_0 = 0$, which seems to agree with the way the problem is worded. Then we can have $\hat{t}(x)_{n \geq 2} = \hat{t}(x)_{n \geq 0}$, and $\hat{a}(x)_{n \geq 1} = \hat{a}(x)_{n \geq 0}$. Does this work?
The standard method after getting $\hat{t}(x) = \frac{x^2}{(1-x)^2}$ (if we proceed with the multiplication above), is to find the partial fractions decomposition $\frac{x^2}{(1-x)^2} = \frac{A}{1-x} + \frac{B}{1-x}$. But this doesn't seem possible. Where did I go wrong?