Doorn 2026 length interval distinct multiples

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theorem_1: Van Doorn's proof of the Erdős–Pomerance conjecture that some interval needs much more room than the interval just above n to hold distinct multiples of 1 through n; the second question of Problem 711.

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W. van Doorn, On the length of an interval that contains distinct multiples of the first $n$ positive integers. Integers 26 (2026), #A7.

The retained folder-name PDF is the journal's file, headed "#A7 INTEGERS 26 (2026)" with the dates "Received: 2/19/25, Revised: 8/19/25, Accepted: 11/27/25, Published: 1/5/26" and the DOI 10.5281/zenodo.18154085 (three pages; pdfTeX). The volume page https://math.colgate.edu/~integers/vol26.html lists it as A7, and the same paper is arXiv:2601.16972v1 (23 January 2026), whose arXiv record carries the journal reference and DOI. Integers is a refereed journal. Read status: claims checked for Theorem 1, Lemma 2 and Lemma 3 (text layer; p. 1 also on the page image); the half-page proof was read through and not independently reviewed. The deposit DOI printed on p. 1 leads to the Zenodo record of the journal's deposit, which names the license "Creative Commons Attribution 4.0 International" (https://zenodo.org/records/18154085, read 2026-10-02), and the journal's site states "All works of this journal are licensed under a Creative Commons Attribution 4.0 International License" (https://math.colgate.edu/~integers/, read 2026-10-02): the Creative Commons Attribution 4.0 license.

Let f(n,m) be the least length such that the interval (m, m + f(n,m)] contains n distinct integers a_1, ..., a_n with i dividing a_i for every i. Erdos and Pomerance conjectured that max_m f(n,m) - f(n,n) tends to infinity with n, a question Erdos backed with a prize. Theorem 1 proves the much stronger quantitative statement max_m f(n,m) - f(n,n) > 0.36 n log n / (log log n) for all sufficiently large n, so in particular some interval of that length contains no distinct multiples of 1, 2, ..., n. The proof is short: Lemma 2 gives the scaling inequality kn + f(kn, kn) <= k^2 n + f(n, k^2 n) by explicitly choosing multiples a_i = ki for i in (n, kn], and combining it with the Erdos-Pomerance bounds of Lemma 3, namely (2/sqrt(e) + o(1)) n sqrt(log n / log log n) < f(n,n) < (2 + o(1)) n sqrt(log n), with the choice k about 0.6 sqrt(log n / log log n) yields the theorem. The paper states that this settles the conjecture, listed as part of Problem 711, and since its content is exactly the growth of f(n,m) and f(n,n) it also bears on the closely related Problem 709, on the multiplier f(n) for which any f(n) max(A) consecutive integers hold distinct multiples of the members of every n-element set A of integers at least 2.

Source: http://math.colgate.edu/~integers/vol26.html.

Bears on. #709: the site's commentary deduces the lower bound f(n) >> log n / log log n for that problem from Theorem 1, and a comment of 15 March 2026 in the problem's discussion thread obtains it by restricting to the set {2, ..., n+1}; the deduction is the site's; #711: Theorem 1 (p. 1) answers the problem's second question, max_m (f(n,m) - f(n,n)) -> infinity, in quantitative form; the site's f(n,m) uses the open interval (m, m+f(n,m)), one more than the paper's half-open convention, and the difference is the same in both.

Results to transcribe.

• Theorem 1 (p. 1): max_m f(n,m) - f(n,n) > 0.36 n log n / log log n for all sufficiently large n; hence an interval of that length exists containing no distinct multiples of 1, ..., n.
• Lemma 2 (p. 1, display (1)): For all positive integers k and n, kn + f(kn, kn) <= k^2 n + f(n, k^2 n).
• Lemma 3 (p. 2; Erdos-Pomerance, quoted): (2/sqrt(e) + o(1)) n sqrt(log n / log log n) < f(n,n) < (2 + o(1)) n sqrt(log n) for all large n.