OFFSET
1,1
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
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Robert Israel, When p^2 divides 2^n + 1
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
EXAMPLE
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.
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*)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms from James A. Sellers, Dec 16 1999
More terms from Vladeta Jovovic, Apr 12 2002
Missing term 182 added by Rainer Rosenthal, Nov 01 2005
STATUS
approved