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A049096
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Numbers k such that 2^k + 1 is divisible by a square > 1.
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13
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3, 9, 10, 15, 21, 27, 30, 33, 39, 45, 50, 51, 55, 57, 63, 68, 69, 70, 75, 78, 81, 87, 90, 93, 99, 105, 110, 111, 117, 123, 129, 130, 135, 141, 147, 150, 153, 159, 165, 170, 171, 177, 182, 183, 189, 190, 195, 201, 204, 207, 210, 213, 219, 225, 230, 231, 234, 237, 243
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OFFSET
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1,1
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COMMENTS
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Conjecture: lim n -> infinity a(n)/n = C exists and 4 < C < 9/2. There seems to be a sequence of primes p such that p^2 never divides numbers of the form 2^x + 1: the first few are 2, 7, 23, 31. - Benoit Cloitre, Aug 20 2002
The first case where 2^n + 1 is divisible by a square that is coprime to n is n = 182 (where 2^182 + 1 is divisible by 1093^2). - Robert Israel, Jul 07 2014
Numbers n such that gcd(n, 2^n + 1) > 1 or n = k m where k is odd and 2 m is the order of 2 modulo a Wieferich prime. See link "When p^2 divides 2^n + 1".
If n is in the sequence, then so is k*n for any odd k. (End)
The sequence consists of all odd multiples of { 3, 10, 55, 68, 78, 182, 301, 406, 666, ... }. - M. F. Hasler, Mar 06 2018
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LINKS
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FORMULA
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For any a(n+1) - a(n) <= 6 since numbers of form 3^a*(2k+1) a > 0, k >= 0, are in the sequence (2^(3*(2k+1) + 1 is divisible by 9). So are numbers of the form 20k + 10 since 2^(20k+10) + 1 is divisible by 25, 110k + 55 since 2^(110k+55) + 1 is divisible by 11^2, 78 + 156k since 2^(156k+78) + 1 is divisible by 13^2 ... - Benoit Cloitre, Aug 20 2002
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EXAMPLE
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9 is here because 2^9 + 1 = 513 is divisible by 9.
99 is here because 2^99 + 1 = 3^3*19*67*683*5347*20857*242099935645987 is divisible by 9, i.e. is not squarefree.
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MAPLE
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remove(n -> numtheory:-issqrfree(2^n+1), [$1..250]); # Robert Israel, Jul 07 2014
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MATHEMATICA
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PROG
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(PARI) is(n)=!issquarefree(2^n+1) \\ Altug Alkan, Nov 20 2015
(Magma) [n: n in [3..220] | not IsSquarefree(2^n+1)]; // Vincenzo Librandi, Mar 08 2018
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CROSSREFS
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Cf. A086982, which is just the same with base b = 10 instead of b = 2.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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