<s>[INST] Disjoint sets in a combinatoral sum (continued) Let $f_S(m)$ be defined as in my previous question: 

Let $S = \{1/n^2 : n \in \mathbb{N} \}$. Let $f_S(m)$ be the sum of the products of all $m$-tuples chosen from $S$. That is 
  $$f_S(m) = \sum_{X \in {S \choose m}} \prod_{x \in X} x $$

I have used inclusion-exclusion to derive a formula for $f_S(m)$. 
For example: Let $a,b,c$ be the $3$ multiplicands when expanding 
$$\zeta(2)^3 = \left( 1 + \frac 1 4 + \frac 1 9 + \cdots \right)^3$$
Then $f_S(3)$ can be solved by excluding all products that satisfy at least one of these rules: $a=b, \ b=c, \ a=c$. 


*

*The values that satisfy $a=b$ let $c$ be any value, so the sum of their products is $\left(1 + \left(\frac 1 4\right)^2 + \left(\frac 1 9\right)^2 + \cdots \right) \left( 1 + \frac 1 4 + \frac 1 9 + \cdots \right)
=\zeta(4)\zeta(2)$. The same applies for $b=c$ and $a=c$. 

*The values that satisfy $a=b \wedge b=c$ have $a=b=c$, so the sum of the products is $\zeta(6)$. The same applies for $a=b \wedge a=c$ and $a=c \wedge b=c$. 

*For $a=b \wedge b=c \wedge a=c$ the result is again $\zeta(6)$. 
By inclusion-exclusion and symmetry ordering $a,b,c$ 
$$f_S(3) = \frac{1}{3!}(\zeta(2)^3 - 3\zeta(4)\zeta(2) + 3\zeta(6) - \zeta(6)) 
= \frac 1 6 (\zeta(2)^3 - 3\zeta(4)\zeta(2) + 2\zeta(6)) $$
In fact, for larger values of $m$ satisfying rules like $a=b$ and $c=d$ turns out to be the disjoint union problem. If the rules we satisfy are $a=b$ and $c=d$, then we have the disjoint sets $(a,b), (c,d)$ and the result will be $\zeta(4)^2$. If the rules are like $a=b$ and $b=c$ then the disjoint sets are $(a,b,c), (d)$ and the result is $\zeta(6)\zeta(2)$. 
I wrote a Mathematica program to calculate $f_S(m)$. Unfortunately it's quite slow, from evaluating all the inclusion-exclusion subsets.
altTotal[x_] := Total@x[[;; ;; 2]] - Total@x[[2 ;; ;; 2]];

inex[n_, zeta_] :=
 Module[{productPairs, summands},
  productPairs = Subsets[Range[n], {2}];
  summands = Table[
    conComp = ConnectedComponents /@ 
      Map[Join[#, Table[{i, i}, {i, 1, n}]] &, 
       Subsets[productPairs, {depth}]];
    Total[Times @@@ Map[zeta[2 Length[#]] &, conComp, {2}]],
    {depth, 1, Length[productPairs]}];
  Return[(zeta[2]^n - altTotal[summands])/n!]
  ]

I have determined the following:
\begin{align}
f_S(3) &=  \frac 1 6 (\zeta(2)^3 - 3\zeta(4)\zeta(2) + 2\zeta(6)) \\
f_S(4) &= \frac{1}{24} \left(\zeta(2)^4-6 \zeta(4) \zeta(2)^2+8 \zeta(6) \zeta(2)+3 \zeta(4)^2-6 \zeta(8)\right) \\
f_S(5) &= \frac{1}{120} \left(\zeta (2)^5-10 \zeta (4) \zeta (2)^3+20 \zeta (6) \zeta (2)^2+15 \zeta (4)^2 \zeta (2)-30 \zeta (8) \zeta (2)-20 \zeta (4) \zeta (6)+24 \zeta (10)\right) \\
f_S(6) &= \frac{1}{720} \left(\zeta (2)^6-15 \zeta (4) \zeta (2)^4+40 \zeta (6) \zeta (2)^3+45 \zeta (4)^2 \zeta (2)^2-90 \zeta (8) \zeta (2)^2-120 \zeta (4) \zeta (6) \zeta (2)+144 \zeta (10) \zeta (2)-15 \zeta (4)^3+40 \zeta (6)^2+90 \zeta (4) \zeta (8)-120 \zeta (12)\right)
\end{align}
My question is how can I derive these expressions? (my linked answer has the closed-form already) The coefficients match A181897. Wikiversity: Permutations by cycle type looks insightful; unfortunately I don't know any group theory. 
What is the connection between the cycle index of the symmetric group and this inclusion-exclusion process for $f$? 
(Program that calculates much more efficiently)
f[m_, zeta_] := ((-1)^m) CycleIndexPolynomial[SymmetricGroup[m], -zeta /@ Range[2, 2 m, 2]]

 [/INST] These may be computed with the unlabeled set operator
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}$$
applied to the Riemann Zeta function
$$\zeta(s) = \sum_{n\ge 1} \frac{1}{n^s}.$$
We use  the recurrence by Lovasz  for the cycle index  $Z(P_n)$ of the
set operator $\textsc{SET}_{=n}$ on $n$ slots, which is
$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l})
\quad\text{where}\quad
Z(P_0) = 1.$$
This  recurrence  lets us  calculate  the  cycle index  $Z(P_n)$  very
easily. We are interested in
$$f(s, m) = Z(P_m; \zeta(s))$$
and hence the recurrence becomes
$$f(s,m) = \frac{1}{m} \sum_{l=1}^m (-1)^{l-1} \zeta(ls) f(s, m-l)
\quad\text{where}\quad
f(s,0) = 1.$$
This yields e.g. for $f(2,4)$
$$1/24\, \left( \zeta \left( 2 \right)  \right) ^{4}-1/4
\,\zeta \left( 4 \right)  \left( \zeta \left( 2
 \right)  \right) ^{2}+1/3\,\zeta \left( 6 \right)
\zeta \left( 2 \right) +1/8\, \left( \zeta \left( 4
 \right)  \right) ^{2}-1/4\,\zeta \left( 8 \right)$$
and for $f(2,5)$
$${\frac { \left( \zeta \left( 2 \right)  \right) ^{5}}{
120}}-1/12\,\zeta \left( 4 \right)  \left( \zeta
 \left( 2 \right)  \right) ^{3}+1/6\,\zeta \left( 6
 \right)  \left( \zeta \left( 2 \right)  \right) ^{2}+1
/8\,\zeta \left( 2 \right)  \left( \zeta \left( 4
 \right)  \right) ^{2}\\-1/4\,\zeta \left( 8 \right)
\zeta \left( 2 \right) -1/6\,\zeta \left( 4 \right)
\zeta \left( 6 \right) +1/5\,\zeta \left( 10 \right)$$
We also have for $f(3,7)$
$$-1/12\, \left( \zeta \left( 3 \right)  \right) ^{2}
\zeta \left( 6 \right) \zeta \left( 9 \right) +1/8\,
\zeta \left( 3 \right) \zeta \left( 6 \right) \zeta
 \left( 12 \right) -1/6\,\zeta \left( 18 \right) \zeta
 \left( 3 \right)\\ -{\frac {\zeta \left( 6 \right)
 \left( \zeta \left( 3 \right)  \right) ^{5}}{240}}+{
\frac {\zeta \left( 9 \right)  \left( \zeta \left( 3
 \right)  \right) ^{4}}{72}}+1/48\, \left( \zeta
 \left( 3 \right)  \right) ^{3} \left( \zeta \left( 6
 \right)  \right) ^{2}-1/24\,\zeta \left( 12 \right)
 \left( \zeta \left( 3 \right)  \right) ^{3}\\+1/10\,
\zeta \left( 15 \right)  \left( \zeta \left( 3 \right)
 \right) ^{2}-1/48\,\zeta \left( 3 \right)  \left(
\zeta \left( 6 \right)  \right) ^{3}\\+1/18\,\zeta
 \left( 3 \right)  \left( \zeta \left( 9 \right)
 \right) ^{2}+1/24\, \left( \zeta \left( 6 \right)
 \right) ^{2}\zeta \left( 9 \right) -1/10\,\zeta
 \left( 6 \right) \zeta \left( 15 \right) \\-1/12\,\zeta
 \left( 9 \right) \zeta \left( 12 \right) +1/7\,\zeta
 \left( 21 \right) +{\frac { \left( \zeta \left( 3
 \right)  \right) ^{7}}{5040}}.$$
(Sorted and formatted by Maple.)
</s>