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.

2, 8, 6, 48, 14, 224, 30, 960, 75, 1215, 62, 3968, 126, 16128, 254, 65024, 510, 261120, 1022, 1046528, 2046, 4190208, 4094, 16769024, 8190, 67092480, 16382, 268402688, 32766, 1073676288, 65534, 4294836224

Offset

1,1

Comments

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).

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.

Only one pair is known to be not of this form (75, 1215) (see examples).

Links

Table of n, a(n) for n=1..32.

BxMO 2011, Third Benelux Mathematical Olympiad, Problème 1.

Diophante, A1853. Deux miniatures bénéluxiennes (in French).

Index to sequences related to Olympiads.

Example

First pairs are (2, 8), (6, 48), (14, 224), (30, 960), (75, 1215), (62, 3968), (126, 16128), ...
Examples corresponding to solutions (2^q-2, 2^q*(2^q-2)):
-> For q = 2, a(1) = 2 = 2^1 and a(2) = 8 = 2^3 while 3 = 3^1 and 9 = 3^2.
-> For q = 3, a(3) = 6 = 2 * 3 and a(4) = 48 = 2^4 * 3 while 7 = 7^1 and 49 = 7^2.
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.

Crossrefs

Cf. A000918 (2^n-2), A087914 (2nd column of the array, the m's).

Sequence in context: A085590 A079538 A098221 * A334519 A009214 A021781

Adjacent sequences: A343098 A343099 A343100 * A343102 A343103 A343104

Keyword

nonn,tabf,hard,more

Author

Bernard Schott, Apr 05 2021

Extensions

Confirmed a(23)-a(30) and extended with a(31)-a(32) by Martin Ehrenstein, Apr 18 2021

Status

approved