Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
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Statement
is "the maximal length of a sequence of consecutive integers each divisible by one of arbitrarily chosen primes" (p. 225), and "the maximal length of the sequence of consecutive integers each divisible by one of the first primes" (p. 226).
Corollary (p. 226, quoted). "We have ."
It follows the Theorem on the same page: for an absolute and arbitrary primes , , each interval of length contains at least integers coprime to . The paper prints no step between the two; the step is that decreases in , so the product is at most by Mertens's theorem, and an interval of length then contains an integer coprime to , so no run of consecutive integers each divisible by one of the is that long (a filing remark, not the paper's text). The page also records the context: "Then the results of [4] imply while in [3] it is proved (1) . Jacobsthal asked whether and whether . The aim of this paper is to prove (1) for ." The introduction (p. 225) has the weaker from the Jurkat--Richert sieving limit and remarks that "by the sieve method the exponent 2 cannot be reduced". The note added in proof (p. 230) records Vaughan's , derived from [3].
In the problems' notation. Three one-line steps made here and named as such; the paper states none of them.
- Problem 970's is the least such that, for each with at most distinct prime factors, any consecutive integers contain one coprime to , that is with Jacobsthal's function. A run of consecutive integers each sharing a factor with is a run each divisible by one of the primes of , so , and is nondecreasing (a run for primes is a run for those primes and one more), so . Erdős's 1965 lecture writes the same over . The Corollary is therefore , the site's statement.
- Problem 687's , the longest initial interval covered by one residue class per prime , is , the longest run of consecutive integers each divisible by a prime ([FGKMT18] display (1.3)). Since , , using and . This is the ", which comes from Iwaniec's work [26]" of [FGKMT18] p. 4; the paper itself credits the primorial bound (1) to its [3] and proves the general bound here.
- Problem 929's is the least with . If and then , so ; this is Erdős's "Iwaniec's result " of 1979, and it is stronger than the site's from Rosser's sieve.
Source. H. Iwaniec, On the problem of Jacobsthal, Demonstratio Math. 11 (1978), no. 1, 225--231; the Corollary, the Theorem, the definition of , display (1) and Jacobsthal's questions on printed p. 226 (PDF p. 2 of the publisher's scan), the definition of and the Jurkat--Richert bound on p. 225 (PDF p. 1), the note added in proof on p. 230 (PDF p. 6), read on the page images (the text layer garbles the displays). The edition read is identified in the source digest.
Read depth. Claims checked: the Corollary, the Theorem, both definitions, display (1) and the questions were read clause by clause on the page images on 2026-09-22. The proof of the Theorem (pp. 228--230) was followed at the level of its displays and not checked; the step from the Theorem to the Corollary is the filing remark above. Nothing here is independently reviewed.
Proof pointer
Page 226: the Corollary is stated directly after the Theorem, with no printed argument; the Mertens step above supplies it. The Theorem is proved in § 3 (pp. 228--230) by the shifted sieve of § 2 with the linear-sieve weights and the two estimates (5) and (6) quoted from the author's 1971 paper; see the theorem page.
Dependencies
The Theorem (p. 226) and Mertens's theorem for the product over the first primes. The translation to uses Chebyshev's bound and the identity of Lemma 1.1 with (1.3).
Bears on
- Problem 970: the upper bound that the site attributes to the paper, now read at its source; the displayed question is Jacobsthal's second question as p. 226 reports it, and the paper leaves it open.
- Problem 687: the upper bound , through ; the paper does not state the bound in this form, and it settles neither displayed question there.
- Problem 929: the lower bound by inversion of , the strongest lower bound on that page.