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Problem 317

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Statement. Is there some constant c>0c>0 such that for every n≥1n\geq 1 there exists some δk∈{−1,0,1}\delta_k\in \{-1,0,1\} for 1≤k≤n1\leq k\leq n with

0<∣∑1≤k≤nδkk∣<c2n?0< \left\lvert \sum_{1\leq k\leq n}\frac{\delta_k}{k}\right\rvert < \frac{c}{2^n}?

Is it true that for sufficiently large nn, for any δk∈{−1,0,1}\delta_k\in \{-1,0,1\},

∣∑1≤k≤nδkk∣>1[1,…,n]\left\lvert \sum_{1\leq k\leq n}\frac{\delta_k}{k}\right\rvert > \frac{1}{[1,\ldots,n]}

whenever the left-hand side is not zero?

Formulation. The site's wording (page last edited 6 January 2026). Two questions about the signed sums ∑k≤nδk/k\sum_{k\le n}\delta_k/k with δk∈{−1,0,1}\delta_k\in\{-1,0,1\}. The first asks for an absolute cc such that for every nn some nonzero signed sum is below c/2nc/2^n in absolute value; since every such sum is a difference q−q′q-q' of two reciprocal subset sums of {1,…,n}\{1,\ldots,n\} (take A={δk=1}A=\{\delta_k=1\}, B={δk=−1}B=\{\delta_k=-1\}) and every such difference is a signed sum, this asks whether the set QnQ_n of reciprocal subset sums has two distinct members within c/2nc/2^n. The second asks whether, for all large nn, every nonzero signed sum exceeds 1/lcm(1,…,n)1/\mathrm{lcm}(1,\ldots,n) strictly; the weak inequality ≥\ge is immediate, since lcm(1,…,n)∑δk/k\mathrm{lcm}(1,\ldots,n)\sum\delta_k/k is a nonzero integer, and equality occurs for small nn (n=4n=4: 12−13−14=−112\frac12-\frac13-\frac14=-\frac1{12}). The two questions are separate; the status attaches to both.

Status. Open. First question: only a weak version is known, a nonzero signed sum of absolute value at most 2−n(log⁡log⁡log⁡n)1+o(1)/log⁡n2^{-n(\log\log\log n)^{1+o(1)}/\log n} (site commentary crediting Kovač and van Doorn; comment arguments resting on the refereed count of distinct reciprocal subset sums of Problem 320), far from c/2nc/2^n, and a heuristic in the thread suggests the weak bound may be the truth. Second question: the strict inequality fails at n=4n=4 (the monograph's example); no proof for all large nn exists, and even its special case for sums of the form 1−∑n∈A1/n1-\sum_{n\in A}1/n is described as nontrivial in the thread of Problem 311. No source beyond the monograph and the site's commentary was found in the search whose scope the Current assessment records; this is a bounded negative finding.

Source. erdosproblems.com/317, accessed 2026-09-18: the problem page (labeled OPEN, with the site's standard note that no finite computation can settle it; source key [ErGr80, p. 42]; last edited 6 January 2026; the formalized-statement field marked yes), its discussion thread (nine visible comments and one deleted post; the page's counter says eight) and its empty proof-claim tab. The site thanks Zachary Chase, Vjekoslav Kovač and Wouter van Doorn. Cite as: T. F. Bloom, Erdős Problem #317, https://www.erdosproblems.com/317, accessed 2026-09-18.

References.

  • [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. 42. Library home: erdos_1980_old_new_problems_results_combinatorial_number_theory.
  • [BlEr75] Bleicher, M. N. and Erdős, P., The number of distinct subsums of ∑i=1N1/i\sum_{i=1}^N1/i. Math. Comp. 29 (1975), 29--42. The refereed count behind the weak bound; used here only through the site's comment argument and the monograph's p. 43 statement of it. Library home: bleicher_1975_number_distinct_subsums_sum_n_1.

Formalization. Statement only here. The file ErdosProblems/317.lean of formal-conjectures at the pinned commit (main) declares erdos_317 : answer(sorry) ↔ ∃ c > 0, ∀ n ≥ 1, ∃ δ : Fin n → ℚ, range δ ⊆ {-1, 0, 1} ∧ 0 < |∑ δ k / (k+1)| ∧ |∑ δ k / (k+1)| < c / 2^n (the first question) and erdos_317.variants.claim2 (the second: for all sufficiently large nn and all such δ\delta, a nonzero sum exceeds 1 / (Icc 1 n).lcm id), both under category research open with proof sorry; claim2_inequality, the weak inequality, is left as sorry under category textbook; and erdos_317.variants.counterexample proves that strictness fails at n=4n=4 with the coefficients (0,1,−1,−1)(0,1,-1,-1). The community database (teorth/erdosproblems, data/problems.yaml as of 2026-09-18) records status open (31 August 2025), a formalized statement (18 November 2025), formal status unformalized and no OEIS entry. No external Lean artifact is linked from the thread; a prize-program pull request of 17 September 2026 offering a Lean proof of the weak inequality is recorded below as a lead.

Current assessment

The question (site formulation, accessed 2026-09-18). The statement above; OPEN, last edited 6 January 2026; source [ErGr80, p. 42]. The commentary says that the weak inequality of the second question is obvious and the strict one is the problem, failing for small nn with the example 12−13−14=−112\frac12-\frac13-\frac14=-\frac1{12}; that comment arguments of Kovač and van Doorn prove a weak form of the first question, a nonzero signed sum of absolute value at most 2−n(log⁡log⁡log⁡n)1+o(1)/log⁡n2^{-n(\log\log\log n)^{1+o(1)}/\log n}; and that a heuristic of van Doorn suggests this bound may be the true order of magnitude. The thread (none of it verified by the site): 25 August 2025, a commenter proposes the prime-denominator variant min⁡∣∑δi/pi∣\min|\sum\delta_i/p_i| and reports computed values of $F(n)=\mathrm{lcm}(1,\ldots,n)\min_{\ne0}|\sum_{i\le n} \delta_i/p_i|$ (F(19)=874F(19)=874, F(29)=F(30)=13340F(29)=F(30)=13340, and others), with a random-model heuristic and the count 6354534463545344 of distinct subset sums of reciprocals for n=32n=32; Kovač replies the same day that both questions very likely have affirmative answers, as the small cases suggest, that disproving the first is much harder than proving the second, that lcm(1,…,n)=en+o(n)\mathrm{lcm}(1,\ldots,n)=e^{n+o(n)} is much larger than 2n2^n, and that sums of reciprocals of distinct primes up to nn, being distinct and well concentrated, give two sums within 2−(1+o(1))n/log⁡n2^{-(1+o(1))n/\log n}, somewhat weaker than the bound asked for; 4 January 2026, van Doorn reformulates the first question as min⁡q≠q′∈Qn∣q−q′∣<c/2n\min_{q\ne q'\in Q_n}|q-q'|<c/2^n, notes ∣Qn∣=exp⁡(nf(n)/log⁡n)|Q_n|=\exp(nf(n)/\log n) with f(n)→∞f(n)\to\infty at rate (log⁡log⁡log⁡n)1+o(1)(\log\log\log n)^{1+o(1)} by the results of Problem 320, and since Qn⊆[0,log⁡n+1]Q_n\subseteq[0,\log n+1] obtains by the pigeonhole principle two members within $(\log n+1)/|Q_n|=2^{-n(\log\log\log n)^{1+o(1)}/\log n}$; he adds that ∣Qn∣|Q_n| uniformly random points in [0,log⁡n+1][0,\log n+1] would have minimum gap of order ∣Qn∣−2−o(1)|Q_n|^{-2-o(1)}, the shape of the upper bound, which he reads as some doubt about the first question. The proof-claim tab is empty. The community database says open.

Origin. Printed p. 42 of the 1980 monograph, after the discussion of r(n)=min⁡εi∣α−∑i≤nεi/i∣r(n)=\min_{\varepsilon_i}|\alpha-\sum_{i\le n}\varepsilon_i/i|: "These questions lead to the consideration of the distribution of the sums ∑k=0nδkk\sum_{k=0}^n\frac{\delta_k}{k} [sic] where δk=0\delta_k=0 or ±1\pm1. It should be easy to see that there is a c>0c>0 so that the inequality min⁡δk{∣∑k=1nδkk∣}<c2n\min_{\delta_k}\{|\sum_{k=1}^n\frac{\delta_k}{k}|\}<\frac{c}{2^n}, δk=0,±1\delta_k=0,\pm1 holds for all nn where the value 00 is not allowed. Unfortunately, we do not see how to prove this at present. It seems quite likely that there is a c>0c>0 independent of nn so that lim⁡n(2+c)nmin⁡δk∑k=1nδkk=0\lim_n(2+c)^n\min_{\delta_k}\sum_{k=1}^n\frac{\delta_k}{k}=0. Of course, ∣∑k=1nδkk∣≥1Ln|\sum_{k=1}^n\frac{\delta_k}{k}|\ge\frac1{L_n} where Ln=lcm{2,3,…,n}L_n=\mathrm{lcm}\{2,3,\ldots,n\}. For large nn we no doubt must have inequality but this we cannot prove. Examples of equality exist for small nn, e.g., 12−13−14=−112\frac12-\frac13-\frac14=-\frac1{12}." The lower limit k=0k=0 of the first sum is the print's misprint for k=1k=1, as the displayed inequalities show. The book thus expects a yes to both questions and even decay faster than (2+c)−n(2+c)^{-n}; the site's two questions are the book's first sentence and its strict-inequality remark.

The first question: the weak version. The only proved statements are the comment arguments. Kovač's: the 2π(n)2^{\pi(n)} sums of reciprocals of distinct primes up to nn are pairwise distinct (by unique factorization of their denominators) and concentrated: a uniformly random subset sum has mean 12∑p≤n1/p\frac12\sum_{p\le n}1/p and variance 14∑p≤n1/p2<1/8\frac14\sum_{p\le n}1/p^2<1/8, so by Chebyshev's inequality at least half of the sums lie within 1/21/2 of the mean, and the pigeonhole principle gives two of them differing by at most 2−π(n)+O(1)=2−(1+o(1))n/log⁡n2^{-\pi(n)+O(1)}=2^{-(1+o(1))n/\log n}. (The sums range over an interval of length ∑p≤n1/p=log⁡log⁡n+O(1)\sum_{p\le n}1/p=\log\log n+O(1), which is unbounded, so the concentration step is needed; the full range alone gives two sums within (log⁡log⁡n+O(1))2−π(n)(\log\log n+O(1))2^{-\pi(n)}, which is still 2−(1+o(1))n/log⁡n2^{-(1+o(1))n/\log n}.) Their difference is a signed harmonic sum. Van Doorn's improvement uses the count t(n)=∣Qn∣t(n)=|Q_n| of distinct reciprocal subset sums: the monograph (printed p. 43) quotes the refereed bounds of Bleicher and Erdős, log⁡t(n)/log⁡2≥nlog⁡n∏i=3klog⁡in\log t(n)/\log2\ge\frac{n}{\log n}\prod_{i=3}^k\log_in for k≥4k\ge4 and log⁡kn≥k\log_kn\ge k, so ∣Qn∣≥2n(log⁡log⁡log⁡n)1+o(1)/log⁡n|Q_n|\ge2^{n(\log\log\log n)^{1+o(1)}/\log n}, and the pigeonhole step gives two subset sums, hence a nonzero signed sum, of absolute value at most 2−n(log⁡log⁡log⁡n)1+o(1)/log⁡n2^{-n(\log\log\log n)^{1+o(1)}/\log n}. The count is refereed ([BlEr75], on the card of Problem 320); the two-line deductions are the site's commentary and are reproduced above. Neither approaches c/2nc/2^n, which would require the exponent nn in place of $n(\log\log\log n)^{1+o(1)}/\log n$; the heuristic of the same comment suggests that the pigeonhole bound may be sharp, in which case the answer to the first question would be no, against the monograph's expectation.

The second question. The weak inequality is trivial and the strict one open; the monograph's example at n=4n=4 gives equality min⁡≠0∣∑δk/k∣=1/lcm(1,…,n)\min_{\ne0}|\sum\delta_k/k|=1/\mathrm{lcm}(1,\ldots,n). Sawin's comment of 1 July 2026 in the thread of Problem 311 observes that even the special case δ(N)>1/lcm(1,…,N)\delta(N)>1/\mathrm{lcm}(1,\ldots,N) for the sums 1−∑n∈A1/n1-\sum_{n\in A}1/n seems nontrivial: equality would force $\mathrm{lcm}(1,\ldots,N)/p\equiv\pm1 \pmod p$ with one common sign for every prime p∈(N/2,N]p\in(N/2,N], which he regards as very unlikely but possibly hard to exclude. The exact zero sums that the second question excludes are the subject of Problem 319.

Leads (not status). The prime-denominator variant and its computed values (thread, 25 August 2025) concern a different minimum and are recorded only as a variant. A Lean formalization of the weak inequality ∣∑δk/k∣≥1/lcm(1,…,n)|\sum\delta_k/k|\ge1/\mathrm{lcm}(1,\ldots,n) for all nn, together with the n=4n=4 equality, was submitted on 17 September 2026 as a pull request to a prize program's repository (TheJustinSunPrize/awards, PR 493; closed), its description stating that neither question is claimed; the corpus has not built or checked it.

Search scope (2026-09-18 UTC). The problem, discussion and proof-claim pages (and the thread of Problem 311 for the special case); the community database record; the formal-conjectures file at the pinned commit; arXiv API searches for abstracts naming reciprocals with signs and harmonic or unit fractions (one unrelated record), Egyptian fractions with subset sums (one unrelated record) and "Erdos problem" with the problem number (none); the monograph's pp. 42 and 43; the GitHub API record of the pull request above; one general web search in three rounds, whose results were papers on other unit-fraction problems (the count of subsets with reciprocal sum one, Problem 297; one of them, arXiv:2403.17041, concerns Problem 297, not this problem, although the search engine's summary suggested otherwise). Not searched: MathSciNet, zbMATH, Google Scholar full text, X. Nothing found proves or refutes either question; this is a bounded negative finding.

Remaining gaps. (1) Both questions are open; the weak version of the first rests on a refereed count plus comment arguments, and no source states it as a theorem. (2) The heuristic against the first question is unverified. (3) There is nothing to compile: no source proves or disproves either statement.

Progress and known results

  • Erdős and Graham (1980, printed p. 42): both questions, the expectation of a yes to both, the stronger guess lim⁡n(2+c)nmin⁡=0\lim_n(2+c)^n\min=0, and the equality example at n=4n=4.
  • First question, weak version: a nonzero signed sum below 2−(1+o(1))n/log⁡n2^{-(1+o(1))n/\log n} (distinct prime reciprocals) and below 2−n(log⁡log⁡log⁡n)1+o(1)/log⁡n2^{-n(\log\log\log n)^{1+o(1)}/\log n} (pigeonhole on the count of Problem 320); site commentary, 2025 and 2026.
  • Second question: equality at n=4n=4 (the monograph's example); open in general, with the special case of Problem 311 also open.

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.