Let $A$ be a finite set of integers. The generating function for the number of ways of writing a given integer $n$ as the sum of $k$ elements from $A$ not necessarily distinct is given by:
$$\left(\sum_{a \in A}{x^a}\right)^k=\sum_n{r(n,k)x^n}$$
Is there a generating function for the number of ways of writing an integer $n$ as a sum of $k$ distinct elements of $A$?
Using the Polya Enumeration Theorem (PET) the closed form is given by
$$[z^n] Z(P_k)\left(\sum_{a\in A} z^a \right)$$
where $Z(P_k) = Z(A_k)-Z(S_k)$ is the difference between the cycle index of the alternating group and the cycle index of the symmetric group. This cycle index is known in species theory as the set operator $\mathfrak{P}_{=k}$ (unlabeled) and the species equation here is
$$\mathfrak{P}_{=k}\left(\sum_{a\in A} \mathcal{Z}^a\right).$$
Recall the recurrence by Lovasz for the cycle index $Z(P_k)$ of the set operator $\mathfrak{P}_{=k}$ on $k$ slots, which is $$Z(P_k) = \frac{1}{k} \sum_{l=1}^k (-1)^{l-1} a_l Z(P_{k-l}) \quad\text{where}\quad Z(P_0) = 1.$$
This recurrence lets us calculate the cycle index $Z(P_n)$ very easily. For example when $n=3$ the cycle index is
$$Z(P_3) = 1/6\,{a_{{1}}}^{3}-1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}.$$
These cycle indices are also given by the exponential formula, which says that
$$Z(P_k) = [w^k] \exp\left(a_1 w - a_2 \frac{w^2}{2} + a_3 \frac{w^3}{3} - a_4 \frac{w^4}{4} + \cdots\right).$$
For example suppose $A$ consists of powers of two. By inspection we should get from
$$\sum_{k\ge 0} [w^k] \exp\left(\sum_{l\ge 1} (-1)^{l-1} a_l \frac{w^l}{l}\right)$$
evaluated at $$a_l = \sum_{q\ge 0} z^{l2^q} \quad\text{the value}\quad \frac{1}{1-z}.$$
And indeed we get for the sum term
$$\sum_{l\ge 1} (-1)^{l-1} \frac{w^l}{l} \sum_{q\ge 0} z^{l2^q} = \sum_{q\ge 0} \sum_{l\ge 1} (-1)^{l-1} z^{l2^q} \frac{w^l}{l} = \sum_{q\ge 0} \log (1+wz^{2^q}).$$
We obtain
$$\sum_{k\ge 0} [w^k] \prod_{q\ge 0} (1+wz^{2^q}) = \left. \prod_{q\ge 0} (1+wz^{2^q}) \right|_{w=1} \\ = \prod_{q\ge 0} (1+z^{2^q}) = \frac{1}{1-z}.$$
The reader is invited to verify that the conversion from the exponential formula to the product representation of the set operator does in fact always carry through independent of the choice of $A$ so that we get
$$[w^k] \prod_{a\in A} (1+wz^a).$$
Here is some Maple code to explore these cycle indices. We would always prefer the recurrence in a practical setting.
pet_cycleind_set :=
proc(k)
option remember;
if k=0 then return 1; fi;
expand(1/k*add((-1)^(l-1)*a[l]*
pet_cycleind_set(k-l), l=1..k));
end;
pet_cycleind_set2 :=
proc(k)
option remember;
local gf;
gf := exp(add((-1)^(l+1)*a[l]*w^l/l, l=1..k));
coeftayl(gf, w=0, k);
end;
Remark. We can derive the recurrence from the exponential formula. Introducing
$$G(w) = \exp\left(\sum_{l\ge 1} (-1)^{l-1} a_l \frac{w^l}{l}\right)$$
we differentiate to obtain
$$G'(w) = G(w) \left(\sum_{l\ge 1} (-1)^{l-1} a_l w^{l-1}\right) = G(w) \left( \sum_{l\ge 0} (-1)^{l} a_{l+1} w^{l}\right).$$
Extracting coefficients we get
$$[w^k] G'(w) = (k+1) [w^{k+1}] G(w) = \sum_{q=0}^k (-1)^q a_{q+1} [w^{k-q}] G(w) \\ = \sum_{q=1}^{k+1} (-1)^{q-1} a_q [w^{k+1-q}] G(w).$$
This is our recurrence precisely.
Remark, II. We may ask why the exponential formula is the OGF of the set operator $\mathfrak{P}_{=k}$ and the multiset operator $\mathfrak{M}_{=k}.$ This is obtained from the labeled species of permutations being factored into disjoint cycles. Marking cycle sizes with the variable $\mathcal{A}_q$ we obtain the species
$$\mathfrak{P}(\mathcal{A}_1 \mathfrak{C}_{=1}(\mathcal{W}) + \mathcal{A}_2 \mathfrak{C}_{=2}(\mathcal{W}) + \mathcal{A}_3 \mathfrak{C}_{=3}(\mathcal{W}) + \mathcal{A}_4 \mathfrak{C}_{=4}(\mathcal{W}) + \cdots).$$
The exponential formula then follows. (This is an EGF which means the coefficients include an inverse factorial, which produces an OGF for the cycle indices as these are averaged over all $k!$ permutations.)
