This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A343101 #63 Sep 03 2025 01:36:48 %S A343101 2,8,6,48,14,224,30,960,75,1215,62,3968,126,16128,254,65024,510, %T A343101 261120,1022,1046528,2046,4190208,4094,16769024,8190,67092480,16382, %U A343101 268402688,32766,1073676288,65534,4294836224 %N A343101 Pairs of integers (k, m) ordered by m with 1 < k < m such that k has the same prime divisors as m, and, k+1 has the same prime divisors as m+1. %C A343101 This sequence was the subject of the 1st problem of the 3rd Benelux Mathematical Olympiad in 2011, where a pair (k, m) is called a 'Benelux pair' (see links). %C A343101 Every pair (2^q-2, 2^q*(2^q-2)) for q >= 2 is a solution, the next such pairs are (4094, 16769024), (8190, 67092480), (16382, 268402688), (32766, 1073676288), ... hence there exist infinitely many Benelux pairs. %C A343101 Only one pair is known to be not of this form (75, 1215) (see examples). %H A343101 Thomas Bloom, <a href="https://www.erdosproblems.com/850">Problem 850</a>, Erdős Problems. %H A343101 BxMO 2011, <a href="http://www.bxmo.org/problems/bxmo-problems-2011-fr.pdf">Third Benelux Mathematical Olympiad, Problème 1</a>. %H A343101 Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/3029-2014-11-29-09-23-00">A1853. Deux miniatures bénéluxiennes</a> (in French). %H A343101 Christian Hercher, <a href="https://arxiv.org/abs/2506.01099">On one of Erdős' Problems - An Efficient Search for Benelux Pairs</a>, arXiv:2506.01099 [math.NT], 2025. See p. 13. %H A343101 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>. %e A343101 First pairs are (2, 8), (6, 48), (14, 224), (30, 960), (75, 1215), (62, 3968), (126, 16128), ... %e A343101 Examples corresponding to solutions (2^q-2, 2^q*(2^q-2)): %e A343101 -> For q = 2, a(1) = 2 = 2^1 and a(2) = 8 = 2^3 while 3 = 3^1 and 9 = 3^2. %e A343101 -> For q = 3, a(3) = 6 = 2 * 3 and a(4) = 48 = 2^4 * 3 while 7 = 7^1 and 49 = 7^2. %e A343101 The only known solution not of that form: a(9) = 75 = 3 * 5^2 and a(10) = 1215 = 5 * 3^5 while 76 = 2^2 * 19 and 1216 = 2^6 * 19. %Y A343101 Cf. A000918 (2^n-2), A087914 (2nd column of the array, the m's). %K A343101 nonn,tabf,hard,more,changed %O A343101 1,1 %A A343101 _Bernard Schott_, Apr 05 2021 %E A343101 Confirmed a(23)-a(30) and extended with a(31)-a(32) by _Martin Ehrenstein_, Apr 18 2021