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A343101
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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.
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1
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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
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn,tabf,hard,more
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AUTHOR
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EXTENSIONS
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Confirmed a(23)-a(30) and extended with a(31)-a(32) by Martin Ehrenstein, Apr 18 2021
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STATUS
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approved
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