Does anybody out there know of a general way to expand the product of two infinite series that are not identical (if they were identical I would just use multinomial expansion and be done with it).

This is in relation to generating functions in combinatorics.

Does anybody out there know of a general way to expand the product of two infinite series that are not identical (if they were identical I would just use multinomial expansion and be done with it).

This is in relation to generating functions in combinatorics.

Sure. Your generating functions are two power series, ∑a(n)*x^n and ∑b(n)*x^n where n goes from 0 to infinity. The product is again an infinite series, say ∑c(n)*x^n. Then you know c(n)=∑a(i)*b(n-i) where i ranges from 0 to n here. This is really as general as you can get. If a(n) and b(n) are nice enough, it might be possible to get a simple formula for the c(n)'s. At any rate this will let you compute any coefficient in the product you like.

If your initial generating functions have nice closed forms, like ∑x^n=1/(1-x), then you can multiply these formulae to get a nice formula for the product, which you can then convert to a power series and get the c(n)'s this way.

My generating functions are x^3n and x^2n, unfortunately. If it were two x^n series I'd just put them into a more convenient form.

My generating functions are x^3n and x^2n, unfortunately. If it were two x^n series I'd just put them into a more convenient form.

So your product is just (1+x^3+x^6+x^9+...) *(1+x^2+x^4+x^6+...) correct?

Let ∑c(n)*x^n be the product of your two series. Then c(n) is just the number of non-negative integer solutions to n=3*p+2*q. The p corresponds to chosing x^3p from (1+x^3+x^6+...) and the q is from chosing x^2q in (1+x^2+x^4+...) when you multiply the products out.

If n is even, c(n)=[n/6]+1, where [.] is the floor function. If n is odd, c(n)=[(n-3)/6]+1. Suitable for your purposes?

You can see this easily enough from the formula c(n)=∑a(i)*b(n-i). Here a(n) is zero unless 3 divides n evenly, then it's 1. Likewise, b(n) is 0 unless 2 divides n evenly, then it's 1. Suppose n is even. Then b(n-i) is 1 when 2 divides i and 0 otherwise. a(i) is 0 when 3 divides i and 0 otherwise. Thus a(i)*b(n-i) is 1 when 6 divides i and 0 otherwise. So we just need to count multiples of 6 that are no greater than n. There are a total of [n/6]+1 multiples of 6 no greater than n (0 counts as a multiple of 6 here). This needs only slight modification to give c(n) for n odd.

Ah, many thanks. Alas, too late, but will be useful in the future.