Find generating function for sequence I am suppose to find generating function for sequence $(e_n)_0^\infty$ where $e_n$ is number of ways how to write a number $n$ as a sum of four natural odd numbers ($e_n$ is basically a number of ordered fours $(\alpha, \beta, \gamma, \delta)$ odd natural numbers that $\alpha + \beta + \gamma + \delta = n$)
I don't know how should I start or "imagine" this problem. 

Well, I guess I have figured that out:
\begin{multline*}
\sum_{(\alpha, \beta, \gamma, \delta) \in L^4} x^{\alpha + \beta + \gamma + \delta} = \sum_{\alpha \in L} \sum_{\beta \in L} \sum_{\gamma \in L} \sum_{\delta \in L} x^{\alpha + \beta + \gamma + \delta} = \\ 
 = \sum_{(\alpha, \beta, \gamma) \in L^3} \left(x^{\alpha + \beta + \gamma} \cdot \sum_{\delta \in L} x^\delta\right) = \sum_{(\alpha, \beta, \gamma) \in L^3} \left(x^{\alpha + \beta + \gamma} \frac{x}{1 - x^2}\right) = \\
 = \left (\frac{1}{1 - x^2} \right) \sum_{(\alpha, \beta) \in L^2}x^{\alpha + \beta} \cdot \sum_{\gamma \in L} x^\gamma = \left(\frac{x}{1 - x^2}\right)^2 \sum_{\alpha \in L}\left(x^\alpha \sum_{\beta \in L} x^\beta\right) = \\
 = \left(\frac{x}{1 - x^2}\right)^3 \sum_{\alpha \in L}x^\alpha = \left(\frac{x}{1 - x^2}\right)^4 = A(x)
\end{multline*}
Where $L$ is a set of odd numbers. There will $x^n$ exactly so many times how many are there odd four numbers that $\alpha + \beta + \gamma + \delta = n$ and that is $e_n$-times.
It is basically what @Brian M. Scott was saying.
