cp's OEIS Frontend

It appears this sequence gives the positive integers m such that the sum of the first m Fibonacci numbers divides their product. For example, if n=2 and m=a(2)=6, we have the sum 1+1+2+3+5+8=20 which clearly divides the corresponding product 480. See A175553 for the analogous sequence when using the triangular numbers. Sum_{k=1..n} Fibonacci(k) divides Product_{k=1..n} Fibonacci(k). - John W. Layman, Jul 10 2010

Keller proved that the occurrence of 2 consecutive Woodall numbers that are divisible by the same prime is restricted to primes p with even h(p), the order of 2 mod p, and that there are an infinity of such pairs.

Right side of the binomial sum Sum_{i = 0..n} (n-2*i)^2 * binomial(n, i) = n*2^n. - Yong Kong (ykong(AT)curagen.com), Dec 28 2000

Let W be a binary relation on the power set P(A) of a set A having n = |A| elements such that for all elements x, y of P(A), xRy if x is a proper subset of y and there are no z in P(A) such that x is a proper subset of z and z is a proper subset of y, or y is a proper subset of x and there are no z in P(A) such that y is a proper subset of z and z is a proper subset of x. Then a(n) = |W|. - Ross La Haye, Sep 26 2007

Partial sums give A036799. - Vladimir Joseph Stephan Orlovsky, Jul 09 2011

a(n) = n with the bits shifted to the left by n places (new bits on the right hand side are zeros). - Indranil Ghosh, Jan 05 2017

Satisfies Benford's law [Theodore P. Hill, Personal communication, Feb 06, 2017]. - N. J. A. Sloane, Feb 08 2017

Also the circumference of the n-cube connected cycle graph. - Eric W. Weisstein, Sep 03 2017

a(n) is also the number of derangements in S_{n+3} with a descent set of {i, i+1} such that i ranges from 1 to n-2. - Isabella Huang, Mar 17 2018

a(n-1) is also the number of multiplications required to compute the permanent of general n X n matrices using Glynn's formula (see Theorem 2.1 in Glynn). - Stefano Spezia, Oct 27 2021

Binomial transform is A084859. Inverse binomial transform is A004277. - Paul Barry, Jun 12 2003

Let A be the Hessenberg matrix of order n defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1] =-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)= (-1)^(n-1)*coeff(charpoly(A,x),x). - Milan Janjic, Jan 26 2010

Indices of primes are listed in A005849. - M. F. Hasler, Jan 18 2015

Add the list of fractions beginning with 1/2 + 3/4 + 7/8 + ... + (2^n - 1)/2^n and take the sums pairwise from left to right. For 1/2 + 3/4 = 5/4, 5 + 4 = 9 = a(2); for 5/4 + 7/8 = 17/8, 17 + 8 = 25 = a(3); for 17/8 + 15/16 = 49/16, 49 + 16 = 65 = a(4); for 49/16 + 31/32 = 129/32, 129 + 32 = 161 = a(5). For each pairwise sum a/b, a + b = n*2^(n+1). - J. M. Bergot, May 06 2015

Number of divisors of (2^n)^(2^n). - Gus Wiseman, May 03 2021

Named after the Irish Jesuit priest James Cullen (1867-1933), who checked the primality of the terms up to n=100. - Amiram Eldar, Jun 05 2021

"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)

a(34) = 17016602 is tentative until the range 16838832..17016601 is fully searched. - Eric W. Weisstein, Mar 22 2018

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

a(1), a(4), and a(6)-a(8) computed by Christophe Clavier, Dec 31 2013 (see link below). 10439679896374780276373 had been found earlier in 2013 by Dan Ismailescu and Peter Seho Park (see reference below). a(3), a(5), and a(9) computed in 2014 by Emmanuel Vantieghem.

These are just the smallest examples known - there may be smaller ones.

There are no Brier numbers below 10^9. - Arkadiusz Wesolowski, Aug 03 2009

Other Brier numbers are 143665583045350793098657, 1547374756499590486317191, 3127894363368981760543181, 3780564951798029783879299, but these may not be the /next/ Brier numbers after those shown. From 2002 to 2013 these four numbers were given here as the smallest known Brier numbers, so the new entry A234594 has been created to preserve that fact. - N. J. A. Sloane, Jan 03 2014

143665583045350793098657 computed in 2007 by Michael Filaseta, Carrie Finch, and Mark Kozek.

It is a conjecture that every such number has more than 10 digits. In 2011 I have calculated that for any n < 10^10 there is a k such that either n*2^k + 1 or n*2^k - 1 has all its prime factors greater than 1321. - Arkadiusz Wesolowski, Feb 03 2016 [Editor's note: The comment below states that the conjecture is now proved. - M. F. Hasler, Oct 06 2021]

There are no Brier numbers below 10^10. For each n < 10^10, there exists at least one prime of the form n*2^k-1 or n*2^k+1 with k <= 356981. The largest necessary prime is 1355477231*2^356981+1. - Kellen Shenton, Oct 25 2020