Suppose I want to find the number of solutions in the natural numbers to solve $ax + by + cz = N$ where $a, b, c, N \in \mathbb{Z}$ (not necessarily all positive). How would I set this problem up using generating functions?
Number of integer solutions
1
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combinatorics
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0If (like me) you don't count zero as a natural number then you have to use, for example, $q^a+q^{2a}+\cdots$ where @anon has $1+q^a+q^{2a}+\cdots$. – 2012-12-13