A049096 Numbers k such that 2^k + 1 is divisible by a square > 1.

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

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Comments

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

That sequence is A072936. - Robert Israel, Nov 20 2015

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

From Robert Israel, Nov 20 2015: (Start)

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

Examples

			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.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Robert Israel, When p^2 divides 2^n + 1

Crossrefs

Cf. A001220, A049093, A049094, A049095, A072936, A282269, A282270.

Cf. A086982, which is just the same with base b = 10 instead of b = 2.

Programs

Magma

[n: n in [3..220] | not IsSquarefree(2^n+1)]; // Vincenzo Librandi, Mar 08 2018

Maple

remove(n -> numtheory:-issqrfree(2^n+1), [$1..250]); # Robert Israel, Jul 07 2014

Mathematica

Select[Range[243], !SquareFreeQ[2^# + 1] &] (* Vladimir Joseph Stephan Orlovsky, Mar 18 2011*)

PARI

is(n)=!issquarefree(2^n+1) \\ Altug Alkan, Nov 20 2015

Formula

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

Extensions

More terms from James Sellers, Dec 16 1999

More terms from Vladeta Jovovic, Apr 12 2002

Missing term 182 added by Rainer Rosenthal, Nov 01 2005