
Re: Finding a closed form for a generating function
Posted:
Oct 10, 2012 2:58 PM


On October 10, 2012, 8:38AM, (unknown) wrote: > I have the generating function: > > C(z) = [1/(1  z^10)] * [1/(1  z^25)] > > > > And I'm trying to find a closed form for this. > > I can't use a substitution for z^10 that would simplify > > my denominator because it doesn't divide 25. > > > > Any ideas on how I could simplify C(z) and further > > reduce it to a closed form for finding the coefficient > > of any z^i term?
First, let's set u = z^5 in your equation, and express it using geometric series expansion. It becomes:
(1  u^2)^(1) * (1  u^5)^(1) = Sum[u^(2 * i + 5 * j), {i, 0, Infinity}, {j, 0, Infinity}]
If I expand this sum, it leads to:
(1  u^2)^(1) * (1  u^5)^(1) = Sum[s(n) * u^n, {n, 0, Infinity}]
where s(n) is the number of (i, j) pairs, i , j >= 0 and integer, satisfying the diophantine equation 2 * i + 5 * j = n. For example, s(0) = 1, s(1) = 0, and s(12)=2, because only (6,0) and (1,2) satisfy the equation.
The next question : is there a formula that gives this number of pairs?

