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]]
I think I finally have some intuition on the relationship between my inclusion-exlcusion "rules" and cycles in permutations.
Suppose I were at the stage in my inclusion-exclusion calculations where I needed to consider $(a = b) \wedge (b=c) \wedge (d=e)$. This corresponds to $a=b=c$ and $d=e$. If we consider $a=b$ as $1 \mapsto 2$, $b=c$ as $2 \mapsto 3$, and $d=e$ as $4 \mapsto 5$, there is a bijection between these rules and a permutation consisting of a 3-cycle and 2-cycle: $(1 \ 2 \ 3)(4 \ 5)$. (Here $a=1, b=2,\dots$)
Similarly if my rules were $(a=c) \wedge (b=c)$ this corresponds to $1 \mapsto 3 \mapsto 2$ or the cycle $(1 \ 3 \ 2)$. Since every cycle can be represented uniquely by having its least element written first, I believe every combination of rules corresponds exactly to every permutation in $S_n$, organized by which cycles it contains.