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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.

Douglas S. Stones
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StationaryTraveller
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3 Answers3

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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.

Marc van Leeuwen
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    Can you be a bit more elaborate on you what you mean by finding the integral solutions to the equations $4a = n$ etc. If I were to solve this I would probably do this $$ (x^0 + x^4 + x^8 + \cdots) (x^0 + x^2 + x^4 + \cdots) (x^0 + x^1 + x^2 + \cdots)(x^0 + x^3 + \cdots) $$ – Vigneshwaren May 04 '13 at 18:03
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    @Vigneshwaren: Yes the coefficient of $x^n$ in $(1+x^4+x^8+\cdots)$ is the number of non-negative integral solutions $a$ to $4a=n$. Actually the question says "positive", so that would mean drop the term $1$ from each factor. – Marc van Leeuwen May 04 '13 at 18:10
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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

Vigneshwaren
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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.

user669065
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  • Try out $\frac{1}{1-x}$, $x^{4}$. – dantopa Apr 27 '19 at 22:15