Generating Functions for n+1 Integer Set: Closed Form

[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.

 

[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 x₁^m₁x₂^m₂/(1-x₁^a₂x₂^a₁)
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 x₂^β₁f(x₁,x₂)x₃^α₃, 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.