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Problem 1205

../

claims/: The 1 claim page of Problem 1205, one per claimant's result; the problem's standing derives from them.


Statement. Let F(x)F(x) be maximal such that there exists, for all n≤xn\leq x, a congruence class an(modn)a_n\pmod{n} such that every m≤xm\leq x satisfies at least F(x)F(x) of these congruences.

Estimate F(x)F(x).

Formulation. The site's wording (page last edited 8 April 2026). One residue class is chosen for every modulus n≤xn\le x, and F(x)F(x) is the largest multiplicity that some choice guarantees for every m≤xm\le x; the class modulo 11 is satisfied by every mm, so F(x)≥1F(x)\ge1. Erdős's 1980 survey (printed p. 108) defines the function right after the prime-modulus version that is Problem 689: "Denote by F(x)F(x) the largest integer for which there is a system of congruences an(modn)a_n\pmod n (8) (where nn runs through the integers not exceeding xx) so that every integer t≤xt\le x satisfies at least F(x)F(x) of the congruences (8). Now it is a simple exercise to prove that F(x)→∞F(x)\to\infty as x→∞x\to\infty. The only problem is to determine or estimate how fast F(x)F(x) tends to infinity." The site's statement is this question with mm for tt.

Status. SOLVED, the site's label, which the page glosses as a resolution by some means other than a proof or disproof: the site's own commentary gives an argument that F(x)∼log⁡xF(x)\sim\log x, with the two-sided bound log⁡x−O(log⁡xlog⁡log⁡x)≤F(x)≤log⁡x+O(1)\log x-O(\sqrt{\log x\log\log x})\le F(x)\le\log x+O(1). The status-defining source is that commentary (last edited 8 April 2026), which this compilation admits as a source for status. No paper, preprint or independent review of the argument exists, and the label is the claimant's own (the site's curator, T. F. Bloom, wrote the argument, so the curator's label is no independent review), so the argument is a pending claim (the site's argument, April 2026) and the derived standing is claimed, not solved. The argument is presented below as the site's account with its steps named and is not checked step by step here.

Source. erdosproblems.com/1205, accessed 2026-09-18 at 10:27 UTC: the problem page (SOLVED, with the site's gloss that the resolution is by some means other than a proof or disproof; last edited 8 April 2026; source key [Er80, p. 108]; an OEIS indicator reading possible), its empty discussion thread and its empty proof-claim tab. Cite as: T. F. Bloom, Erdős Problem #1205, https://www.erdosproblems.com/1205, accessed 2026-09-18.

References.

Formalization. None found: there is no ErdosProblems/1205.lean in google-deepmind/formal-conjectures and the problem page shows no formalized statement. The community database (teorth/erdosproblems,) records the problem solved (record last updated 4 April 2026), the statement not formalized, formal_status unformalized and no formal-proof URL.

Current assessment

The question (site formulation). The statement above; SOLVED, last edited 8 April 2026. The commentary, which is the status-defining source, recalls that Erdős called F(x)→∞F(x)\to\infty a "simple exercise", points to Problem 689 as the prime-modulus analogue, and then gives an argument, which it says was suggested by the comments on that problem, for F(x)∼log⁡xF(x)\sim\log x: a pigeonhole upper bound F(x)≤∑n≤x1/n=log⁡x+O(1)F(x)\le\sum_{n\le x}1/n=\log x+O(1), a lower bound from uniformly random classes for the moduli n≤x/2n\le x/2 with a Chernoff-type concentration step leaving O(x/(log⁡x)2)O(x/(\log x)^2) exceptional integers, and a greedy choice of the classes for x/2<n≤xx/2<n\le x covering those exceptions ≫(log⁡x)2\gg(\log x)^2 times; its conclusion is the displayed two-sided bound. The thread and the proof-claim tab are empty. The claim page [[problems/integer_sequences/E1205/claims/2026_04_08_bloom|the site's argument, April 2026]] records it as a pending full claim.

The origin. Item 6 of Section 6, "Some problems on sieve methods", of the survey, printed p. 108, quoted under Formulation. It follows the prime-modulus function of Problem 689 ("Let f(x)f(x) be the largest integer for which there is a system of congruences ap(modp)a_p\pmod p ... so that every integer n<xn<x should satisfy at least f(x)f(x) of the congruences ... I can not even prove that f(x)≥2f(x)\ge2 for x>x0x>x_0"), and Erdős asserts F(x)→∞F(x)\to\infty as an exercise without an argument or a rate.

The site's argument, as the site's. Three steps, named here and not re-proved. An averaging upper bound: each modulus nn puts about x/nx/n of the integers m≤xm\le x into its class, so the average number of congruences an integer m≤xm\le x satisfies is ∑n≤x1/n+O(1)=log⁡x+O(1)\sum_{n\le x}1/n+O(1)=\log x+O(1), and some mm is at or below the average. A random lower bound for the moduli n≤x/2n\le x/2: each mm lies in the class of nn with probability 1/n1/n, the expected count is log⁡x+O(1)\log x+O(1), and a Chernoff-type concentration bound leaves at most O(x/(log⁡x)2)O(x/(\log x)^2) integers m≤xm\le x covered fewer than log⁡x−O(log⁡xlog⁡log⁡x)\log x-O(\sqrt{\log x\log\log x}) times. A greedy repair with the moduli x/2<n≤xx/2<n\le x: each of these classes can be aimed at one of the exceptional integers, and there are far more such moduli than exceptional integers, so the exceptional ones end up covered ≫(log⁡x)2\gg(\log x)^2 times. Together these give F(x)=log⁡x+O(log⁡xlog⁡log⁡x)F(x)=\log x+O(\sqrt{\log x\log\log x}), hence F(x)∼log⁡xF(x)\sim\log x. The compilation records this as the site's account: the concentration step and the greedy count are not written out on the site, no published version exists, and no step is checked here; an independent whole-argument review, an elementary one, is the evidence that would accept it. The site's remark that the argument was inspired by the comments on Problem 689 is recorded; that page carries the prime-modulus question, which remains open.

Context, not consumed. Sun's 1999 theorem on the covering multiplicity of a finite system of residue classes (card) concerns systems covering every integer at least mm times and relates mm to sums of the reciprocal moduli; the present question restricts the covered integers to m≤xm\le x and chooses the classes, so the theorem does not apply to it as stated and is not cited as a result here.

Search scope. None of the routes below found a published proof of F(x)∼log⁡xF(x)\sim\log x, a sharper estimate of the second-order term, or a dispute of the site's argument.

  • The site: problem page, discussion thread and proof-claim tab; the formal-conjectures directory listing and tree (no file); the community database.
  • arXiv: the API query abs:"congruence" AND abs:"covering" AND abs:"multiplicity" (twenty records, none on this function); the API searches titles and abstracts only, so this zero is weak.
  • The primary source: [Er80] p. 108.

Not searched: MathSciNet, zbMATH, Google Scholar, X.

Remaining gaps. (1) The label rests on the site's own argument, which has no publication and no independent review, so its claim page is claimed and the problem's standing is claimed; a written proof, a review or a refereed source is the evidence that would accept it. (2) The second-order term is open: the argument gives log⁡x−O(log⁡xlog⁡log⁡x)≤F(x)≤log⁡x+O(1)\log x-O(\sqrt{\log x\log\log x})\le F(x)\le\log x+O(1) and nothing narrower was found. (3) No formalization exists.

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