Sum of products of $m$-tuples chosen from the set of squared reciprocals

Question

Let $S = \{1/n^2 : n \in \mathbb{N} \}$. We know $\sum S = \zeta(2) = \pi^2/ 6$.

Let $f(S, m)$ be the sum of the products of all $m$-tuples chosen from $S$. That is $$f(S,m) = \sum_{X \in {S \choose m}} \prod_{x \in X} x $$

$f(S,2)$ is the sum of the product of all pairs of distinct values from $S$. I believe \begin{align}f(S,2) &= \frac 1 2 \left( \left( 1 + \frac 1 4 + \frac 1 9 + \cdots\right)^2 - \left( 1^2 + \left(\frac 1 4\right)^2 + \left(\frac 1 9\right)^2 + \cdots\right) \right) \\ &= \frac 1 2 (\zeta(2)^2 - \zeta(4)) = \frac{\pi^4}{120}\end{align}

If we visualize multiplication as a table where each row and column is an entry of $S$, this is just removing the diagonal and taking one triangular half.

How can I calculate $f(S,3), f(S,4), \dots$? This sort of triangular region of unique tuples comes up a lot in my combinatorics calculations.

Edit: I was thinking about Jack D'Aurizio's deleted comment if I wanted to solve $f(S,n)$ without generating functions. I think it's possible to generalize my inclusion-exclusion approach I used for $f(S, 2)$. I've asked a new question: Disjoint sets in a combinatoral sum (continued)

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I also theorize that if we let $S_p = \{1/p^2 : p \ \text{prime}\}$ then by inclusion-exclusion the proportion of square-free integers approaches $$1 - f(S_p,1) + f(S_p,2) - f(S_p,3) + \cdots \to 6 /\pi^2$$

(I'm aware there's a much simpler estimation method)

Answer 1

In general we have

$$ f(S,n) = [x^n]\prod_{m\geq 1}\left(1+\frac{x}{m^2}\right)=[x^{2n}]\prod_{m\geq 1}\left(1+\frac{x^2}{m^2}\right)=[x^{2n}]\frac{\sinh(\pi x)}{\pi x}=\color{red}{\frac{\pi^{2n}}{(2n+1)!}} $$ and in a similar way $$ f(S_p,n) = [x^n]\prod_{p}\left(1+\frac{x}{p^2}\right), $$ such that $$ 1- f(S_p,1)+f(S_p,2)-f(S_p,3)+\ldots = \prod_{p}\left(1-\frac{1}{p^2}\right)=\frac{1}{\zeta(2)} = \frac{6}{\pi^2} $$ by Euler's product.