What is the name of the transform which finds the number of ways to make partitions of the given sizes?

Question

I'm looking for the name of a transform which takes a sequence giving the number of 'prime' elements of a given size to the number of ways to make a number out of a sum of 'prime' elements, up to order.

For example, if there were two 'primes' of size 1 and one 'prime' of size two there are two ways to make 1 (either of the primes), four ways to make 2 (two of the first prime of size 1, two of the other, one of each, or one prime of size 2), and so forth, so that the transform of

2, 1, 0, 0, 0, ...

is

2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 90, 100, ....

I'm actually more interested in transforms of sequences which aren't zero after some point, but this made a simple example.

Also useful would be information on how to efficiently compute such transform (but if you have the name alone that would do).

Answer 1

I don't think there is a name for such a transform, although I could be wrong. It's pretty easy to compute using generating functions. If $a_n$ is your original sequence, let $$f(x) = \prod_{i=1}^\infty (1-x^i)^{-a_n}.$$ Then the coefficients of $f(x)$ are the transform of your sequence. In your example, we we have $a_1 = 2, a_2 = 1$, and then $$f(x) = (1-x)^{-2} (1-x^2)^{-1} = 1 + 2x + 4x^2 + 6x^3 + 9x^4 + 12x^5 + \ldots.$$

You can see this by expanding out:

$$f(x) = (1 + (x^1)^1 + (x^1)^2 + \ldots)(1 + (x^1)^1 + (x^1)^2 + \ldots)(1 + x^2 + (x^2)^2 + \ldots).$$

The first two factors represent your two "primes" of size one, and the third is the one ``prime'' of size $2$. To find the coefficient of $x^n$, you choose a term from each factor which represents how many times you include that prime and then sum over all possible ways of doing this.