The Hardy-Littlewood circle method (with Vinogradov's improvement) states that given a set $A \subset \mathbb{N}\cup \left \{ 0 \right \} $ and given a natural number $n$, if we consider the sum:
$$f(z)=\sum_{r\in A,\; r\leq n} z^r$$
then
$$ \text # \left \{ (r_1,r_2,..,r_k)\in A^k: \ r_1+r_2+...+r_k=n \right \}=\int_{0}^{1} f^{k}(e(\alpha )) e(-n\alpha )\text d\alpha$$ $$e(\alpha)=\text{exp}(2\pi i\alpha)$$
However, this representation counts permutations, for example if we consider $A=\mathbb{Z}^{+}$ then
$$S(n,k):=\int_{0}^{1} f^{k}(e(\alpha )) e(-n\alpha )\text d\alpha$$
It gives me the "partitions" of $n$ into $k$ parts, however we know that the partitions of 6 into three parts are 3:
$$(2,2,2), (1,2,3), (1,1,4)$$
But $S(6,3)=10$, since the function counts the permutations of the three representations. (2,2,2) does not have different permutations, so add 1, (1,2,3) gives me six different permutations adding another 6 and finally (1,1,4) has three different permutations adding 3, which in total are the 10 that I get from $S(6,3)$.
My question is this: Is there a way to modify the Circle Method so that the function obtained tells me the number of representations but without counting permutations?
