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Consider a finite set S of positive integers, and define $q(S) = \sum_{s \in S}{1/s^2}$. Letting $\rho = \pi^2/6$, we have $q(S)$ in the ranges $[0, \rho - 1), [1, \rho)$. I conjecture that for every rational $r$ within those ranges, $\exists S: q(S) = r$. Can anyone prove or disprove this?

Note that for plain unit fractions $u(S) = \sum_{s \in S}{1/s}$ it is straightforward to show by a greedy algorithm that $\exists S: u(S) = r \forall r \in \mathbb{Q}^+$; conversely for prime unit fractions $p(S) = \sum_{s \in S}{1/p_s}$ almost all rationals are unreachable, since for each $S$, $p(S)$ has a unique (squarefree) denominator. Given that the primes are denser than the squares, it would thus be a somewhat interesting result if the conjecture were to be proven.

Computationally I have established existence of $S$ for all rationals with denominators up to 50 within the range; see results on github.

Update: added "finite" to clarify first sentence; corrected $\mathbb{R}$ to $\mathbb{Q}$.

Hugo van der Sanden
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    I think that is answered here: https://math.stackexchange.com/a/3482498/42969 – Martin R Feb 13 '24 at 15:25
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    I don't know exactly how to interpret your results on Github, but it looks like the examples you have there mostly consist of finite subsets. Are you happy for $S$ to be infinite? – Izaak van Dongen Feb 13 '24 at 15:49
  • @MartinR thanks, it does look like it; will have to study the proof there. – Hugo van der Sanden Feb 13 '24 at 16:44
  • @IzaakvanDongen apologies, my intention was the S should be finite. I'll edit the question to clarify this. – Hugo van der Sanden Feb 13 '24 at 16:45
  • @MartinR oh, maybe not: I'm not sure what the "" operator means, but I think the referenced answer deals only with infinite sets; as (just) clarified, I'm looking for rationals that can be generated by finite sets. – Hugo van der Sanden Feb 13 '24 at 16:56
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    No worries! Indeed Martin's link is about infinite sums. ($A \setminus B$ is the "set difference" of $A$ and $B$ - it's just ${x \in A : x \notin B}$, aka "the set of things in $A$ but not in $B$". When $B \subseteq A$ as in the link, that's just "the complement of $B$ in $A$"). – Izaak van Dongen Feb 13 '24 at 17:16

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You have rediscovered a celebrated theorem of R. L. Graham, who proved this exact statement (as a special case of something much more general) in 1964.

Greg Martin
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