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Problem 321

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claims/: The 4 claim pages of Problem 321, one per claimant's result; the problem's standing derives from them.


Statement. What is the size of the largest A⊆{1,…,N}A\subseteq \{1,\ldots,N\} such that all sums ∑n∈S1n\sum_{n\in S}\frac{1}{n} are distinct for S⊆AS\subseteq A?

Formulation. The site's wording, accessed 2026-09-18 (page last edited 16 July 2026; source key [ErGr80, p.43]). Write R(N)R(N) for the largest size of a set A⊆{1,…,N}A\subseteq\{1,\ldots,N\} whose 2∣A∣2^{|A|} subset sums ∑n∈S1/n\sum_{n\in S}1/n, S⊆AS\subseteq A, are pairwise distinct; such a set is dissociated in the reciprocal sense, and equivalently no two disjoint non-empty subsets of AA have equal reciprocal sums. Since the subset sums of such an AA are 2R(N)2^{R(N)} distinct values among those counted by S(N)S(N) in Problem 320, 2R(N)≤S(N)2^{R(N)}\le S(N) (the inequality the site calls trivial). "What is the size" is read, as the site's resolution comment of 16 July 2026 reads it, as asking for the order of magnitude of R(N)R(N); that comment leaves finer questions, such as an asymptotic, open, and no exact formula for R(N)R(N) is known. The monograph writes r(n)r(n). The exact values are OEIS A391592: R(N)=1,2,3,4,5,5,6,7,8,9,10,10,11,12,12,13,14,14,15,15,15,…R(N)=1,2,3,4,5,5,6,7,8,9,10,10,11,12,12,13,14,14,15,15,15,\ldots, known for N≤54N\le54; the first sixteen were recomputed here by exhaustive search. The site's displayed asymptotic contains a slip, ∏j=3klog⁡kN\prod_{j=3}^k\log_kN for ∏j=3klog⁡jN\prod_{j=3}^k\log_jN.

Status. Solved, in the site's label, which the site glosses as a resolution other than a proof or disproof, for the order of magnitude

R(N) ≍ log⁡S(N) ≍ Nlog⁡N∏j=3klog⁡jN,log⁡kN=O(1).R(N)\ \asymp\ \log S(N)\ \asymp\ \frac{N}{\log N}\prod_{j=3}^{k}\log_jN,\qquad \log_kN=O(1).

The lower bound is refereed and is written out below: Bleicher and Erdős's set of products of rapidly growing primes (1975) has distinct subset reciprocal sums and gives the site's displayed lower bound, and the set U(N)\mathcal U(N) in Bettin, Grenié, Molteni and Sanna's proof (Math. Comp., online 22 January 2026) has the same property and gives R(N)≥2(1−3/2log⁡kN)Nlog⁡N∏j=3klog⁡jNR(N)\ge2(1-\frac{3/2}{\log_kN})\frac{N}{\log N}\prod_{j=3}^k\log_jN, a bound the paper does not state and the site calls implicit in it. The refereed upper bound, R(N)≤log⁡S(N)/log⁡2R(N)\le\log S(N)/\log2 with Bleicher and Erdős's 1976 Theorem 3, carries the extra factor log⁡rN\log_rN; the upper bound of the same order as the lower is the bound for log⁡S(N)\log S(N) obtained with the AI system GPT 5.6 Sol Pro and accepted by the site on Problem 320 (claim of 15 July 2026, unrefereed), restated for this problem in the accepted claim on this page's tab. So the matching of the two bounds behind the label rests on that accepted claim, whose page for this problem is the order of magnitude of R(N). The three refereed bounds have their own partial claim pages, linked under Progress and known results, and the standing derives from the four pages.

Source. erdosproblems.com/321, accessed 2026-09-18: the problem page (SOLVED; source key [ErGr80, p.43]; last edited 16 July 2026; OEIS A384927 and A391592 linked), its ten-comment discussion thread (24 November 2025 to 16 July 2026) and its proof-claim tab with one full-proof claim (15 July 2026), marked accepted by the site. The site cites [BlEr75], [BlEr76b] and [BGMS25] in its commentary and thanks Boris Alexeev, Zachary Hunter, and Dustin Mixon. Cite as: T. F. Bloom, Erdős Problem #321, https://www.erdosproblems.com/321, accessed 2026-09-18.

References.

  • [BlEr75] Bleicher, M. N. and Erdős, P., The number of distinct subsums of ∑i=1N1/i\sum_{i=1}^N 1/i. Math. Comp. 29 (1975), no. 129, 29--42, DOI 10.1090/S0025-5718-1975-0366795-4. The theorem on Qk(N)Q_k(N), p. 30; the theorem S(N)≥2Q(N)S(N)\ge2^{Q(N)} and its Lemma, pp. 39--42. Library home: bleicher_1975_number_distinct_subsums_sum_n_1.
  • [BlEr76b] Bleicher, M. N. and Erdős, P., Denominators of Egyptian fractions. II. Illinois J. Math. 20 (1976), 598--613. Theorem 3, p. 610.
  • [BGMS25] Bettin, S., Grenié, L., Molteni, G. and Sanna, C., A lower bound for the number of Egyptian fractions. arXiv:2509.10030v1 (12 September 2025); Math. Comp., DOI 10.1090/mcom/4190, published online 22 January 2026. Theorem 1 and Section 2 (the set U\mathcal U). Library home: bettin_2025_lower_bound_number_egyptian_fractions.
  • [ErGr80] Erdős, P. and Graham, R. L., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique 28, Université de Genève (1980), printed p. 43. Library home: erdos_1980_old_new_problems_results_combinatorial_number_theory.
  • [OEIS-a] Lu, C., Sequence A391592, The On-Line Encyclopedia of Integer Sequences (10 January 2026; created 18 January 2026): R(N)R(N) for N≤54N\le54; per the entry, terms 11--2020 by the account epistemologist, 2121--3636 by Stijn Cambie, 3737--5454 by Cong Lu; accessed 2026-09-18.
  • [OEIS-b] Xie, Y., Sequence A384927 (7 September 2025; last modified 21 January 2026): the largest S⊆{1,…,n}S\subseteq\{1,\ldots,n\} with t+u∤tut+u\nmid tu for distinct t,u∈St,u\in S, the sequence of Problem 327, which agrees with R(N)R(N) for N≤20N\le20 and differs at N=21N=21 (thread); accessed 2026-09-18.
  • [YZL26] Young, R., Zhu, K. and Luo, Y., the manuscript of the accepted claim, the same one as on Problem 320 (an Overleaf read link, which served no document to a request on 2026-09-18); Lean repository Zarathustra23/erdos-320-harmonic-subset-sums at its commit of 11 July 2026, pinned on the claim page.

Formalization. Statement only. The file ErdosProblems/321.lean of formal-conjectures at the linked commit (main) defines R (N : ℕ) : ℕ := sSup { #A | (A) (_ : A ⊆ Finset.Icc 1 N) (_ : Set.InjOn (fun (S : Finset ℕ) ↦ ∑ n ∈ S, (1 : ℚ) / n) A.powerset) } and declares erdos_321 (N : ℕ) : R N = answer(sorry) with proof sorry under category research open, three asymptotic variants (isTheta, isBigO, isLittleO) with answer(sorry) placeholders, and two Bleicher--Erdős bounds under category research solved, both sorry: variants.lower, Nlog⁡N∏i=3klog⁡iN≤R(N)\frac{N}{\log N}\prod_{i=3}^k\log_iN\le R(N) for k≥4k\ge4 with log⁡kN≥k\log_kN\ge k, and variants.upper, R(N)≤1log⁡2log⁡rNNlog⁡N∏i=3rlog⁡iNR(N)\le\frac1{\log2}\log_rN\frac{N}{\log N}\prod_{i=3}^r\log_iN for r≥1r\ge1 with log⁡iN≥1\log_iN\ge1 for all i≤2ri\le2r. The file's categories were not updated for the site's July 2026 resolution, and no declaration carries a formal_proof attribute. The community database records the statement as formalized, with 6 October 2025 as that entry's last-update date, formal status unformalized, status solved as of the status entry's last update on 31 August 2025, OEIS A384927 and A391592. Nothing in the file has been built by this corpus, and the site's label carries no Lean suffix.

Current assessment

The question (site formulation, accessed 2026-09-18). The statement above; SOLVED; last edited 16 July 2026. The site's commentary writes R(N)R(N) for the maximal size and records the bounds that [BlEr75] and [BlEr76b] imply, Nlog⁡N∏i=3klog⁡iN≤R(N)≤1log⁡2log⁡rN(Nlog⁡N∏i=3rlog⁡iN)\frac{N}{\log N}\prod_{i=3}^k\log_iN\le R(N)\le\frac1{\log2}\log_rN\bigl(\frac{N}{\log N}\prod_{i=3}^r\log_iN\bigr) for k≥4k\ge4 with log⁡kN≥k\log_kN\ge k and r≥1r\ge1 with log⁡2rN≥1\log_{2r}N\ge1; notes the trivial inequality 2R(N)≤S(N)2^{R(N)}\le S(N) with the S(N)S(N) of Problem 320; states that R(N)≍log⁡S(N)≍Nlog⁡N∏j=3klog⁡jNR(N)\asymp\log S(N)\asymp\frac{N}{\log N}\prod_{j=3}^k\log_jN with log⁡kN=O(1)\log_kN=O(1) is now known (its display carries the slip noted above); describes the lower bound as implicit in [BGMS25]; and attributes the upper bound to the AI system GPT 5.6 Sol prompted by Young, Zhu and Luo, referring to Problem 320 for the details. The thread (ten comments): on 24 November 2025 a commenter reports that the first values of R(N)R(N) agree with OEIS A384927 and asks how the two conditions are related, with a notebook in a public gist; Woett notes that 1a+1b\frac1a+\frac1b is a unit fraction exactly when a+b∣aba+b\mid ab; Tao posts a call for help. On 25 November Cambie shows the coincidence breaks at n=21n=21 (R(21)=15R(21)=15 while A384927 has 1616), explains why, and computes R(n)R(n) to n=39n=39 by linear programming over the forbidden pairs of disjoint subsets with equal reciprocal sums (with the observation that primes p>n/3p>n/3, their doubles and 16,27,25,3216,27,25,32 never occur in such a pair); Tao replies. On 30 December 2025 Cong Lu reports code and values to N=54N=54 on issue 161 of the community database's repository. On 16 July 2026 the maintainer marks this problem and Problem 320 resolved because the order of magnitude of R(N)R(N) is now known, noting that finer questions such as an asymptotic stay open and guessing that the order of magnitude would have been good enough for Erdős. The community database record: solved as of its last update on 31 August 2025, statement formalized, OEIS A384927 and A391592.

Origin. Printed p. 43 of the 1980 monograph, after the question of Problem 320: "A related question is the following. How many integers a1<a2<⋯<ar(n)≤na_1<a_2<\cdots<a_{r(n)}\le n can we have so that all the sums ∑i=1r(n)εiai\sum_{i=1}^{r(n)}\frac{\varepsilon_i}{a_i} are distinct? Estimates of Bleicher and Erdös [Bl-Er (75)] imply that nlog⁡n∏i=3slog⁡in<r(n)<nlog⁡snlog⁡n∏i=3slog⁡in\frac{n}{\log n}\prod_{i=3}^s\log_in<r(n)<\frac{n\log_sn}{\log n}\prod_{i=3}^s\log_in for any fixed ss. Is it true that log⁡t(n)r(n)→∞\frac{\log t(n)}{r(n)}\to\infty with nn? Here one can also ask for a maximal such set of aia_i's having as few elements as possible. It is easy to see that the number of elements in such a set has a greater order of magnitude than π(n)\pi(n). We don't know whether this is the case if instead we require all products ∏i=1r(n)1aiεi\prod_{i=1}^{r(n)}\frac1{a_i^{\varepsilon_i}} to be distinct." Here t(n)t(n) is the S(N)S(N) of Problem 320. The rider "log⁡t(n)/r(n)→∞\log t(n)/r(n)\to\infty?" is answered in the negative by R(N)≍log⁡S(N)R(N)\asymp\log S(N), if the accepted upper bound stands; the refereed bounds alone leave it open (below).

Lower bounds (refereed sources; the deductions written here). Two explicit families of integers with distinct subset reciprocal sums are in the refereed literature, though neither paper states the consequence for R(N)R(N).

  1. Bleicher and Erdős (1975). Let Q(N)\mathcal Q(N) be the set of n≤Nn\le N that are products p1⋯pkp_1\cdots p_k of primes with pi>e3pi−1/2p_i>e^{3p_{i-1}/2}, over all kk. The Lemma of p. 40 of [BlEr75] (theorem page) says that two sequences of distinct elements of Q(N)\mathcal Q(N) have equal reciprocal sums only if they coincide up to order, that is, Q(N)\mathcal Q(N) has distinct subset reciprocal sums; the paper uses it to prove S(N)≥2Q(N)S(N)\ge2^{Q(N)} with Q(N)=∣Q(N)∣Q(N)=|\mathcal Q(N)|. Hence R(N)≥Q(N)≥Qk(N)≥Nlog⁡N∏j=3k+1log⁡jNR(N)\ge Q(N)\ge Q_k(N)\ge\frac{N}{\log N}\prod_{j=3}^{k+1}\log_jN for log⁡k+1N≥k+1\log_{k+1}N\ge k+1 (the theorem of p. 30), which is the site's displayed lower bound with k+1k+1 renamed kk and the monograph's attribution. Read depth: claims checked for both theorems and the Lemma; proofs not verified.
  2. Bettin, Grenié, Molteni and Sanna (2025; Math. Comp. 2026). Their proof of Theorem 1 works with the set U\mathcal U of integers uu such that 1/u1/u is not a {−1,0,1}\{-1,0,1\}-combination of 1/1,…,1/(u−1)1/1,\ldots,1/(u-1) (Section 2; u∈Uu\in\mathcal U exactly when S(u)=2S(u−1)S(u)=2S(u-1), Lemma 1), proves the lower bound for ∣U(N)∣=∣U∩[1,N]∣|\mathcal U(N)|=|\mathcal U\cap[1,N]| and concludes with Lemma 2, S(N)≥2∣U(N)∣S(N)\ge2^{|\mathcal U(N)|}. The set U(N)\mathcal U(N) has distinct subset reciprocal sums: if two distinct subsets had equal sums, dropping their common elements and taking the largest remaining element uu would write 1/u1/u as a {−1,0,1}\{-1,0,1\}-combination of smaller reciprocals, contradicting u∈Uu\in\mathcal U (the three-line argument on the theorem page, written for this problem). Hence, under the theorem's conditions k≥4k\ge4 and ln⁡kN≥3/2\ln_kN\ge3/2,
R(N) ≥ ∣U(N)∣ ≥ 2(1−3/2ln⁡kN)Nln⁡N∏j=3kln⁡jN,R(N)\ \ge\ |\mathcal U(N)|\ \ge\ 2\Bigl(1-\frac{3/2}{\ln_kN}\Bigr)\frac{N}{\ln N}\prod_{j=3}^{k}\ln_jN,

which is the lower bound the site calls implicit in that work and the lower bound of the accepted claim, whose summary takes it from the same dissociated set; the accepted claim's Lean file proves the finite half of this (BGMSU_dissociated, pow_card_BGMSU_le_harmonic_subsetSums). The theorem is refereed (Mathematics of Computation, online 22 January 2026, per the Crossref record); the statements here follow arXiv v1, which was not compared with the published text, and the proof is checked for structure only.

Upper bounds. Refereed: R(N)≤log⁡S(N)/log⁡2R(N)\le\log S(N)/\log2 and Theorem 3 of [BlEr76b] (p. 610) give R(N)≤1log⁡2Nlog⁡rNlog⁡N∏j=3rlog⁡jNR(N)\le\frac1{\log2}\frac{N\log_rN}{\log N}\prod_{j=3}^r\log_jN for r≥1r\ge1 and log⁡2rN≥1\log_{2r}N\ge1, the site's displayed upper bound. Its ratio to the refereed lower bounds is unbounded in NN (the factor log⁡rN\log_rN is a tower above the iterated logarithms that separate the two products, as the Problem 320 page explains), so the refereed results determine R(N)R(N) only up to such a factor. The matching upper bound is the site-accepted bound, obtained with an AI system, log⁡S(N)≪Nlog⁡N∏j=3klog⁡jN\log S(N)\ll\frac{N}{\log N}\prod_{j=3}^k\log_jN (log⁡kN=O(1)\log_kN=O(1)) of Problem 320, whose account and provenance are on that page; through R(N)≤log⁡S(N)/log⁡2R(N)\le\log S(N)/\log2 it gives R(N)≍log⁡S(N)R(N)\asymp\log S(N) and answers the monograph's rider question in the negative.

The accepted claim (provenance recorded, not judged). The proof-claim tab lists a full-proof claim submitted 2026-07-15 23:49:54 by the account RayYoung for RayYoung, Keheng Zhu and Yanping Luo, marked on the tab as accepted by the site as correct. Its summary asserts that R(N)R(N) has order Nlog⁡N∏j=3κ(N)log⁡jN\frac{N}{\log N}\prod_{j=3}^{\kappa(N)}\log_jN: the upper bound comes from 2R(N)≤S(N)2^{R(N)}\le S(N) and the claimants' bound for log⁡S(N)\log S(N), a route the summary presents as growing out of Erdős's ideas, and the lower bound from the dissociated set in the proof of Bettin, Grenié, Molteni and Sanna. Its notes say that the manuscript submitted for Problem 320 also contains a proposed order-of-magnitude resolution of this problem and no exact asymptotic formula yet. The claim's tab names the AI system GPT 5.6 Sol Pro; the claim links the same Overleaf manuscript and the same Lean repository as the Problem 320 claim; its one comment, the curator's of 16 July 2026, says that the proof is correct, that the lower bound is from the earlier work the claim cites and that the upper bound follows at once from the resolution of Problem 320. The claim page is the order of magnitude of R(N). The Overleaf read link served no document to a request; the Lean file's content is described on Problem 320 and contains no theorem about the order of R(N)R(N) beyond the finite dissociation bridge. No refereed publication or independent review was found.

Data leads, not status. OEIS A391592 gives R(N)R(N) for N≤54N\le54; the first sixteen values (1,2,3,4,5,5,6,7,8,9,10,10,11,12,12,131,2,3,4,5,5,6,7,8,9,10,10,11,12,12,13) were recomputed here by exhaustive search over all subsets and agree. The thread records that R(n+1)=R(n)R(n+1)=R(n) tends to happen when n+1n+1 is smooth, with exceptions such as 3636, and that the first twenty values coincide with A384927 by an accident that ends at 2121 (Cambie's explanation: at n=21n=21 several equalities of longer reciprocal sums involve 2121, such as 1+121=12+13+17+1141+\frac1{21}=\frac12+\frac13+\frac17+\frac1{14}, so no 15-element dissociated subset of [1,20][1,20] extends by 2121). The public gist and the repository of Cong Lu (head of 31 December 2025) hold the code; it was not run by this corpus.

Search scope. The problem, discussion and proof-claim pages; the community database record; the formal-conjectures file at the pinned commit; OEIS A391592 and A384927 (JSON records); issue 161 of the community database's repository and the gist (GitHub API); the Crossref and Semantic Scholar records of [BGMS25]'s DOI (no citing paper indexed) and the arXiv abstract page of 2509.10030 (v1 only); one request to the Overleaf read link; the GitHub API for the accepted claim's Lean repository and its file at the head commit; arXiv API searches for abstracts naming dissociated sets with reciprocals or unit fractions (nine records, none relevant) and the listing of the seventy-six most recent abstracts mentioning Egyptian or unit fractions (to 7 September 2026; none on this problem); printed p. 43 of [ErGr80] and pp. 30, 39, 40 and 42 of [BlEr75]. Not searched: MathSciNet, zbMATH, Google Scholar, X. No refereed proof of the matching upper bound was found.

Remaining gaps. (1) The matching upper bound is the site-accepted claim of Problem 320, obtained with an AI system; without it the refereed bounds determine R(N)R(N) only up to an unbounded iterated-logarithm factor. (2) No source gives an asymptotic for R(N)R(N) or its leading constant; the Kominers--Neu claim on Problem 320 concerns log⁡S(N)\log S(N), not R(N)R(N). (3) The monograph's further questions, maximal dissociated sets with as few elements as possible and the multiplicative variant (distinct products), were not addressed by any source found. (4) The refereed statements are compiled as statements; no proof was checked, and the two dissociation deductions above are author-recorded.

Progress and known results

  • Bleicher and Erdős (1975): the set Q(N)\mathcal Q(N) of products of rapidly growing primes has distinct subset reciprocal sums (Lemma and theorem, pp. 39--42); its size (p. 30) gives R(N)≥Nlog⁡N∏j=3k+1log⁡jNR(N)\ge\frac{N}{\log N}\prod_{j=3}^{k+1}\log_jN for log⁡k+1N≥k+1\log_{k+1}N\ge k+1. Claim page: the products of rapidly growing primes.
  • Bleicher and Erdős (1976): Theorem 3 with 2R(N)≤S(N)2^{R(N)}\le S(N) gives R(N)≤1log⁡2Nlog⁡rNlog⁡N∏j=3rlog⁡jNR(N)\le\frac1{\log2}\frac{N\log_rN}{\log N}\prod_{j=3}^r\log_jN for log⁡2rN≥1\log_{2r}N\ge1. Claim page: the 1976 upper bound.
  • Bettin, Grenié, Molteni and Sanna (2025; Math. Comp. 2026): Theorem 1 and its set U(N)\mathcal U(N) give R(N)≥2(1−3/2log⁡kN)Nlog⁡N∏j=3klog⁡jNR(N)\ge2(1-\frac{3/2}{\log_kN})\frac{N}{\log N}\prod_{j=3}^k\log_jN for k≥4k\ge4, log⁡kN≥3/2\log_kN\ge3/2 (deduction above). Claim page: the dissociated set of the right size.
  • Young, Zhu and Luo (2026; obtained with GPT 5.6 Sol Pro; site-accepted, unrefereed): R(N)≍log⁡S(N)≍Nlog⁡N∏j=3klog⁡jNR(N)\asymp\log S(N)\asymp\frac{N}{\log N}\prod_{j=3}^k\log_jN with log⁡kN=O(1)\log_kN=O(1), through the upper bound of Problem 320.
  • Exact values: R(N)R(N) for N≤54N\le54 (OEIS A391592; thread computations of 2025); the first sixteen checked here.

Linked library material

These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.