Generating function for the number of positive integer solutions to the equation: $4a+2b+c+3d=n$ Let $a_n$ be the number of positive integer solutions to the equation: $4a+2b+c+3d=n$, What's the generating function for the series $a_n$?
I'm pretty much stuck on how to even start.
 A: Suppose you have found the generating functions for the numbers of positive integer solutions to each of the equations $4a=n$, $2b=n$, $c=n$ and $3d=n$, and you multiply those series together, then you will have a generating function for the problem in the question. Try to understand why this is so (it is a standard reasoning with generating functions), and then solve the four given sub-problems, which are similar and easy.
A: Thanks @Leeuwen
I had to work this out for my own good, so bear with me for stating the obvious
You should agree that this is the G.F. that we arrive at: $$ (x^4 + x^8 + \cdots)(x^2 + x^4 + \cdots)(x^1 + x^2 + \cdots)(x^3 + x^6 + \cdots) \tag{1}$$
This becomes, after some rearranging and GP reduction: $$ x^{10}\frac{1}{1-x}\frac{1}{1-x^2} \frac{1}{1-x^3}\frac{1}{1-x^4}$$
After which I'm officially stuck so if anyone can suggest how I can complete this proof. I do know that I got to reduce the last 4 terms to one infinite series. SO if anyone can suggest the identity or power series that achieves this would be great. THanks
A: You can write :
1/[(1-x)(1-x^2)(1-x^3)(1-x^4)]= A/(1-x) +B/(1-x^2) +C/(1-x^3) +D/(1-x^4) 
and solve for A,B,C,D by comparing co- efficients.
