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Or, primes p which do not divide 2^n+1 for any n.

The possibility n=0 in the above rules out A072936(1)=2; apart from this, a(n)=A072936(n+1). - M. F. Hasler, Dec 08 2007

The order of 2 mod p is odd iff 2^k=1 mod p, where p-1=2^s*k, k odd. - M. F. Hasler, Dec 08 2007

Has density 7/24 (Hasse).

From Jianing Song, Jun 27 2025: (Start)

The multiplicative order of 2 modulo a(n) is A139686(n).

Contained in primes congruent to 1 or 7 modulo 8 (primes p such that 2 is a quadratic residue modulo p, A001132), and contains primes congruent to 7 modulo 8 (A007522). (End)

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

All terms are primes. Largest currently known prime of the form (2^n + 1)/257 is (256^23029 + 1)/257 found by Donovan Johnson 03/2005. The only currently known prime of the form (2^n + 1)/65537 is (65536^239 + 1)/65537.

From Giuseppe Coppoletta, May 18 2017: (Start)

In general, for any j > 1, if (2^(n*2^j) + 1)/Fj is a prime (where Fj = 2^2^j + 1 is the corresponding Fermat number), then n needs to be prime, as for any odd proper factor q of n, 2^(q*2^j) + 1 is another factor of the numerator. The same for j = 0, apart for the particular value n = 3^2.

For the case j = 4, I checked it again, and (65536^p + 1)/65537 indeed is not a prime at least for 239 < p < 12500, i.e. (2^n + 1)/65537 is not a prime at least up to n = 200000. Any higher upper bound available?

One can also remark that 65536 = 2^16 and 239 = 2^8 - 2^4 - 1. Is there any special reason (see Brennen's link) for that?

I checked also that (2^(p*2^j) + 1)/Fj is never a proper power (in particular it is not a prime power) for j = 0..4 and for any prime p, at least for any exponent p*2^j < 200000.

We can even conjecture that ((Fj-1)^p + 1)/Fj is always squarefree for any odd prime p and for any Fermat number Fj with j >= 0. Note that this is not true if p is not restricted to be a prime, even if p and Fj are coprime, as shown by the following counterexample relative to the case j = 1, f1 = 5: 4^91 + 1 == 0 mod 1093^2. Remark that any such counterexample has to be a Wieferich prime (A001220), but not every Wieferich prime gives a counterexample, as shown by the second known Wieferich prime (3511), which cannot match here because it belongs to A072936.

(End)

The least k, if it exists, such that prime(n) divides 2^k + 1.