# Generating functions

1. Jul 31, 2012

### anthony2005

Hi everyone,
given a set of $n+1$ integers $a_{1},a_{2},...a_{n},\rho$, is it possible to get the closed form of the generating function

$f\left(x_{1},...x_{n}\right)=\sum_{i_{1},..i_{n}=0}^{\infty}\delta_{i_{1}a_{1}+...+i_{n}a_{n},\rho}x_{1}^{i_{1}}...x_{n}^{i_{n}}$

It seems to be always a rational function. For instance here are two specific results:

$\sum_{i,j,k,h=0}^{\infty}\delta_{i+j-h-l,0}x_{1}^{i}x_{2}^{j}x_{3}^{k}x_{4}^{h}=\frac{1-x_{1}x_{2}x_{3}x_{4}}{\left(x_{1}x_{3}-1\right)\left(x_{2}x_{3}-1\right)\left(x_{1}x_{4}-1\right)\left(x_{2}x_{4}-1\right)}$

$\sum_{i,j,k=0}^{\infty}\delta_{2i+j-h,0}x_{1}^{i}x_{2}^{j}x_{3}^{k}=\frac{1}{\left(x_{2}x_{3}-1\right)\left(x_{1}x_{3}^{2}-1\right)}$

Also references that could help me to tackle this problem would be appreciated. Thank you.

2. Aug 1, 2012

### haruspex

I think you can do this inductively on n.
Start with just x1, x2. We can normalise to (a1,a2)=1. If both positive, there is only a finite sum, so assume a2 < 0. There is a unique pair of integers m1, m2 s.t. a1 m1+a2 m2 = ρ and m1 is minimised but > 0. This also minimises m2 subject to m2 > 0.
The function is x1m1x2m2/(1-x1a2x2a1)
Now introduce x3. For now I'll assume a1 > 0, a2 < 0, a3 < 0. Let c = (a1,a2), so (c,a3)=1.
Consider a given α3. We have a1α1 + a2α2 = ρ - a3α3, and c|(ρ-ρ - a3α3). I.e. (a1/c)α1 + (a2/c)α2 = (ρ - a3α3)/c.
Again, we find the unique smallest m1, m2, but these depend on α3. I believe I'm right in saying, but have not proved, that as α3 increases, one of m1, m2 stays constant (m1 say, depends on which of a1 a2 is larger in modulus?), while m2 increases in steps of a1/c. This means the function can be rewritten as a sum of terms x2β1f(x1,x2)x3α3, where f is a constant (independent of α3) rational function of x1, x2. β and α3 satisfy a relationship similar to that of α1, α2 previously.
Sorry, in a bit of a rush and this was a bit complicated to write out. Hope it makes sense.