cp's OEIS Frontend

From 2002 to 2014 this was given as the sequence of Brier numbers. These /are/ Brier numbers, but there are many smaller numbers - see A076335.

"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)

509203 has been proved to be a member of the sequence, and is conjectured to be the smallest member. However, as of 2009, there are still several smaller numbers which are candidates and have not yet been ruled out (see links).

Riesel numbers are proved by exhibiting a periodic sequence p of prime divisors with p(k) | n*2^k-1 and disproved by finding prime n*2^k-1. It is conjectured that numbers that cannot be proved Riesel in this way are non-Riesel. However, some numbers resist both proof and disproof.

Others conjecture the opposite: that there are infinitely many Riesel numbers that do not arise from a covering system, see A101036. The word "odd" is needed in the definition because otherwise for any term n, all numbers n*2^m, m >= 1, would also be Riesel numbers, but we don't want them in this sequence (as is manifest from A101036). Since 1 and 3 obviously are not in this sequence, for any n in this sequence n-1 is an even number > 2 and therefore composite, so one could replace "k >= 1" equivalently by "k >= 0". - M. F. Hasler, Aug 20 2020

Named after the Swedish mathematician Hans Ivar Riesel (1929-2014). - Amiram Eldar, Apr 02 2022

WARNING: These are just the smallest examples known - there may be smaller ones. Even the first term is uncertain. - N. J. A. Sloane, Jun 20 2017

There are no prime Brier numbers below 10^10. - Arkadiusz Wesolowski, Jan 12 2011

It is a conjecture that every such number has more than 11 digits. In 2011 I have calculated that for any prime p < 10^11 there is a k such that either p*2^k + 1 or p*2^k - 1 has all its prime factors greater than 1321. - Arkadiusz Wesolowski, Feb 03 2016

The first term was found by Dan Ismailescu and Peter Seho Park and the next two by Christophe Clavier (see below). See also A076335. - N. J. A. Sloane, Jan 03 2014

a(4)-a(9) computed in 2017 by the author.

Fred Cohen and J. L. Selfridge showed that a(n) = -1 infinitely often.

a(n) = 0 iff n is in A045718.

Indices of records are given by A217892.

Primes arising from A194591 (or 0 if no such prime exists).

Many of these terms are in A093868.

A194607 gives the record values.

Primes arising from A194606 (or 0 if no such prime exists).

Integers for which the smallest k in A194591 such that prime(n)*2^k - 1 or prime(n)*2^k + 1 is prime (A194608) increases.

a(19) > 10^10.

A194607 gives the record values of A194606.