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Claim. Let S(N)S(N) be the number of distinct values of ∑n∈A1/n\sum_{n\in A}1/n over A⊆{1,…,N}A\subseteq\{1,\ldots,N\} and let log⁡j\log_j be the jj-fold iterated logarithm. Then

log⁡S(N) ≍ Nlog⁡N∏j=3klog⁡jN,\log S(N)\ \asymp\ \frac{N}{\log N}\prod_{j=3}^{k}\log_jN,

where k=k(N)k=k(N) is an index with 1≤log⁡kN=O(1)1\le\log_kN=O(1), such as the last index with log⁡kN≥1\log_kN\ge1; every such choice gives the same order. The claim's summary writes κ(N)\kappa(N) without defining it; the audit note in the claimants' Lean repository at the pinned commit takes log⁡κ(N)N>2\log_{\kappa(N)}N>2 and says any fixed threshold above 11 gives the same order. The new half is the upper bound: the integers up to NN are sorted by a large prime factor, the sorting bounds S(N)S(N) by values of SS at smaller arguments, and that bound is iterated with explicit estimates for the counting function of the primes; this is the strategy of Bleicher and Erdős's 1976 Theorem 3, whose iteration lost a factor log⁡rN\log_rN. The lower bound is the refereed Theorem 1 of Bettin, Grenié, Molteni and Sanna. The claim determines the order of magnitude of log⁡S(N)\log S(N) and, as its own notes say, no asymptotic formula; the site reads this as the resolution of "Estimate S(N)S(N)", the question of Problem 320, and this page follows that reading. The same manuscript's consequence for the companion problem has its own page, the order of magnitude of R(N).

Submission note. Posted to erdosproblems.com as a proof claim by RayYoung, Keheng Zhu, Yanping Luo (account RayYoung) on 15 July 2026, giving "GPT 5.6 Sol Pro" as the AI used, which the site marks as accepted as correct:

We prove that

log⁡S(N)≍Nlog⁡>N∏j=3κ(N)log⁡jN.\log S(N)\asymp \frac{N}{\log > N}\prod_{j=3}^{\kappa(N)}\log_jN.

The upper bound comes from decomposing the

denominators according to their large prime factors, deriving a recursive inequality for S(N)S(N), and iterating it using uniform prime-counting estimates. The lower bound follows from the result of Bettin, Grenié, Molteni, and Sanna. Notes: This result was obtained with the assistance of generative AI, particularly during the exploratory stage of the argument. We subsequently reorganized and rewrote the original AI-assisted proof to improve its readability, logical structure, attribution, and mathematical transparency. The present result determines the order of magnitude of (\log S(N)), but does not establish an exact asymptotic formula. It remains possible that the argument can be sharpened to identify a leading asymptotic constant, or more generally to determine the corresponding lower and upper limiting constants under an appropriate normalization. We warmly welcome comments, corrections, and further discussion from the community.

Depends on. Bettin, Grenié, Molteni and Sanna's refereed lower bound supplies the lower half of the order of magnitude.

Provenance. The claim was submitted to the site's proof-claim tab on 15 July 2026 by the account RayYoung for RayYoung, Keheng Zhu and Yanping Luo. The tab names the AI system GPT 5.6 Sol Pro, and the notes say that the result was obtained with generative AI, particularly during the exploratory stage, and that the authors reorganized and rewrote the proof for readability, structure, attribution and transparency. The manuscript sits behind an Overleaf read link, which served no document to a request on 2026-09-18. The site maintainer's exposition of 1 September 2026 on the problem page is the readable account: with s(x)=log⁡S(x)s(x)=\log S(x), the decomposition gives s(x)≪x/log⁡x+∑x/log⁡x<p≤xs(x/p)s(x)\ll x/\log x+\sum_{x/\log x<p\le x}s(x/p), the prime number theorem bounds the sum by xlog⁡x∑m≪log⁡xs(m)/m2\frac{x}{\log x}\sum_{m\ll\log x}s(m)/m^2 up to a constant, and with s∗(x)=max⁡n≤xlog⁡nns(n)s^*(x)=\max_{n\le x}\frac{\log n}{n}s(n) this becomes s∗(x)≪1+(log⁡3x) s∗(log⁡x)s^*(x)\ll1+(\log_3x)\,s^*(\log x), which induction turns into s∗(x)≪∏j=3tlog⁡jxs^*(x)\ll\prod_{j=3}^{t}\log_jx with log⁡tx\log_tx bounded. The exposition describes this as the 1976 strategy carried out with more care in the quantitative analysis, the 1976 paper having applied its induction only to part of the sum. The argument has not been checked by this corpus.

Formalization link. The Lean repository at the pinned commit of 11 July 2026 holds one file, HarmonicSubsetSums.lean, which proves finite statements only: a family's subset sums number at most 2∣A∣2^{|A|}, with equality exactly for a dissociated family; 2∣A∣≤S2^{|A|}\le S for a dissociated AA inside the index set; the product bound for a disjoint union of blocks; invariance of the count under scaling by a nonzero rational; and that the set U(N)\mathcal U(N) of Bettin, Grenié, Molteni and Sanna is dissociated. Its README says the quantitative prime number theorem and the lower-bound estimate remain cited inputs. The file contains no theorem about the order of log⁡S(N)\log S(N) and has not been built by this corpus, so it is a link on the claimants' page and no formalized evidence.

Acceptance. The site's curator, Thomas Bloom, marked the claim on the tab as accepted by the site as correct, marked the problem resolved on 16 July 2026 with a thread comment saying that the order of magnitude of log⁡S(N)\log S(N) is now known and that finer questions such as an asymptotic remain open, and wrote the exposition of 1 September 2026 on the problem page. The curator is independent of the claimants, and that acceptance is the reviewed evidence. No refereed publication and no independent review of the claim were found on 2026-09-18. Among the eight comments under the claim (2026-10-07), the curator's comment of 16 July 2026 says that the proof is correct and sketches the argument. A later full claim on the same problem, Kominers and Neu's asymptotic, asserts a stronger statement and reports that its authors have not verified this one.