cp's OEIS Frontend

"Provable" in the definition means provable by any method (whether using a covering set or not). - N. J. A. Sloane, Aug 03 2024

It is only a conjecture that this sequence is complete up to 3000000 - there may be missing terms.

It is conjectured that 78557 is the smallest Sierpiński number. - T. D. Noe, Oct 31 2003

Sierpiński numbers may be proved by exhibiting a periodic sequence p of prime divisors with p(k) | n*2^k+1 and disproved by finding a prime n*2^k+1. It is conjectured by some people that numbers that cannot be proved to be Sierpiński by this method are non-Sierpiński. However, some numbers resist both proof and disproof. - David W. Wilson, Jan 17 2005 [Edited by N. J. A. Sloane, Aug 03 2024]

Sierpiński showed that this sequence is infinite.

There are four related sequences that arise in this context:

S1: Numbers n such that n*2^k + 1 is composite for all k (this sequence)

S2: Odd numbers n such that 2^k + n is composite for all k (apparently it is conjectured that S1 and S2 are the same sequence)

S3: Numbers n such that n*2^k + 1 is prime for all k (empty)

S4: Numbers n such that 2^k + n is prime for all k (empty)

The following argument, due to Michael Reid, attempts to show that S3 and S4 are empty: If p is a prime divisor of n + 1, then for k = p - 1, the term (either n*2^k + 1 or 2^k + n) is a multiple of p (and also > p, so not prime). [However, David McAfferty points that for the case S3, this argument fails if p is of the form 2^m-1. So it may only be a conjecture that the set S3 is empty. - N. J. A. Sloane, Jun 27 2021]

a(1) = 78557 is also the smallest odd n for which either n^p*2^k + 1 or n^p + 2^k is composite for every k > 0 and every prime p greater than 3. - Arkadiusz Wesolowski, Oct 12 2015

n = 4008735125781478102999926000625 = (A213353(1))^4 is in this sequence but is thought not to satisfy the conjecture mentioned by David W. Wilson above. For this multiplier, all n*2^(4m + 2) + 1 are composite by an Aurifeuillean factorization. Only the remaining cases, n*2^k + 1 where k is not 2 modulo 4, are covered by a finite set of primes (namely {3, 17, 97, 241, 257, 673}). See Izotov link for details (although with another prime set). - Jeppe Stig Nielsen, Apr 14 2018

Conjecture: if S is a (provable) Sierpiński number, then there exists a prime P such that S^p is also a Sierpiński number for every prime p > P. - Thomas Ordowski, Jul 12 2022

Problem: are there odd numbers K such that K - 2^m is a Sierpiński number for every 1 < 2^m < K? If so, then all positive values of (K - 2^m)*2^n + 1 are composite. Also, by the dual Sierpiński conjecture, K - 2^m + 2^n is composite for every 1 < 2^m < K and for every n > 0. Note that, by the dual Sierpiński conjecture, if p is an odd prime and 1 < 2^m < p, then there exists n such that (p - 2^m)*2^n + 1 is prime. So if such a number K exists, it must be composite. - Thomas Ordowski, Jul 20 2022

From M. F. Hasler, Jul 23 2022: (Start)

1) The above Conjecture is true for Sierpiński numbers provable by a "covering set", with P equal to the largest prime factor of the elements of that set*, according to the explanation from Michael Filaseta posted Jul 12 2022 on the SeqFan mailing list, cf. links. (*More generally: for S^p with any p coprime to all elements of the covering set, but not necessarily prime.)

2) Wilson's comment from 2005 (also the first part, not only the conjecture) is misleading if not wrong because there are provable Sierpiński numbers for which a covering set is not known (maybe even believed not to exist), as explained by Nielsen in his above comment from 2018. (End)

There are k such that k*2^m + 1 is not prime for any m (then k is called a Sierpiński number). Erdős once conjectured that for such a k, the smallest prime factor of k*2^m + 1 would be bounded as m tends to infinitiy. The proven Sierpiński number k=44745755^4 is thought to be the first counterexample to this conjecture.

This sequence is either unbounded (in which case 44745755^4 is in fact a counterexample) or periodic.

a(229) <= 3034663491871541. - Chai Wah Wu, Aug 09 2020

k = 44745755^4 has the property that k + 2^m is composite for all m. However, it is conjectured that this sequence is unbounded. This is the case if and only if A336347 is unbounded; because a full covering set for k*2^m + 1 would also be a full covering for k + 2^m, and vice versa.