cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 22 results. Next

A052334 Record primes reached in A052333.

Original entry on oeis.org

3, 5, 7, 19, 31, 43, 103, 223, 367, 463, 5503, 738197503
Offset: 1

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Author

Christian G. Bower, Dec 19 1999

Keywords

Comments

Start with n=1, take 2n+1, if composite take 2n+1 again, keep going until you reach a prime. Repeat for n=2, 3... If prime is a record, add to sequence. If never reach a prime, skip that value of n.
a(13) is a 771-digit prime reached after 2552 iterations starting from 73. - Warut Roonguthai. This was proved to be a prime by Paul Jobling (Paul.Jobling(AT)WhiteCross.com) using PrimeForm and by Ignacio Larrosa Cañestro using Titanix (http://www.znz.freesurf.fr/pages/titanix.html). [Oct 30 2000]

Crossrefs

Programs

  • Mathematica
    record = 2; Reap[For[n = 1, n <= 100, n++, k = n; While[ !PrimeQ[k = 2*k + 1]]; If[k > record, Print[k]; Sow[k]; record = k]]][[2, 1]] (* Jean-François Alcover, Jul 19 2013 *)

Formula

a(n) = A052333(A051914(n)). - Amiram Eldar, Jul 25 2019

A076336 (Provable) Sierpiński numbers: odd numbers n such that for all k >= 1 the numbers n*2^k + 1 are composite.

Original entry on oeis.org

78557, 271129, 271577, 322523, 327739, 482719, 575041, 603713, 903983, 934909, 965431, 1259779, 1290677, 1518781, 1624097, 1639459, 1777613, 2131043, 2131099, 2191531, 2510177, 2541601, 2576089, 2931767, 2931991, 3083723, 3098059, 3555593, 3608251
Offset: 1

Views

Author

N. J. A. Sloane, Nov 07 2002

Keywords

Comments

"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)

References

  • R. K. Guy, Unsolved Problems in Number Theory, Section B21.
  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 420.
  • Paulo Ribenboim, The Book of Prime Number Records, 2nd. ed., 1989, p. 282.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 237-238.

Crossrefs

A101036 Riesel numbers n (n*2^k-1 is composite for all k>0, n odd) that have a covering set.

Original entry on oeis.org

509203, 762701, 777149, 790841, 992077, 1106681, 1247173, 1254341, 1330207, 1330319, 1715053, 1730653, 1730681, 1744117, 1830187, 1976473, 2136283, 2251349, 2313487, 2344211, 2554843, 2924861, 3079469, 3177553, 3292241, 3419789, 3423373, 3580901
Offset: 1

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Author

David W. Wilson, Jan 17 2005

Keywords

Comments

Conjecture: there are infinitely many Riesel numbers that do not arise from a covering system. See page 16 of the Filaseta et al. reference. - Arkadiusz Wesolowski, Nov 17 2014
a(1) = 509203 is also the smallest odd n for which either n^p*2^k - 1 or abs(n^p - 2^k) is composite for every k > 0 and every prime p > 3. - Arkadiusz Wesolowski, Oct 12 2015
Theorem 11 of Filaseta et al. gives a Riesel number which is thought to violate the assumption of a periodic sequence of prime divisors mentioned in the title of this sequence. - Jeppe Stig Nielsen, Mar 16 2019
If the Riesel number mentioned in the previous comment does in fact not have a covering set, then this sequence is different from A076337, because then that number, 3896845303873881175159314620808887046066972469809^2, is a term of A076337, but not of this sequence. - Felix Fröhlich, Sep 09 2019
Named after the Swedish mathematician Hans Ivar Riesel (1929-2014). - Amiram Eldar, Jun 20 2021
Conjecture: if R is a Riesel number (that has a covering set), then there exists a prime P such that R^p is also a Riesel number for every prime p > P. - Thomas Ordowski, Jul 12 2022
Problem: are there numbers K such that K + 2^m is a Riesel number for every m > 0? If so, then (K + 2^m)*2^n - 1 is composite for every pair of positive integers m,n. Also, by the dual Riesel conjecture, |K + 2^m - 2^n| are always composite. Note that, by the dual Riesel conjecture, if p is an odd prime and n is a positive integer, 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

References

  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 238.

Crossrefs

Main sequences for Riesel problem: A038699, A040081, A046069, A050412, A052333, A076337, A101036, A108129.
See A076337 for references and additional information. Cf. A076336.

Extensions

Up to 3292241, checked by Don Reble, Jan 17 2005, who comments that up to this point each n*2^k-1 has a prime factor <= 241.
New name from Felix Fröhlich, Sep 09 2019

A076337 Riesel numbers: odd numbers n such that for all k >= 1 the numbers n*2^k - 1 are composite.

Original entry on oeis.org

509203
Offset: 1

Views

Author

N. J. A. Sloane, Nov 07 2002

Keywords

Comments

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

References

  • R. K. Guy, Unsolved Problems in Number Theory, Section B21.
  • Paulo Ribenboim, The Book of Prime Number Records, 2nd ed., 1989, p. 282.

Crossrefs

Main sequences for Riesel problem: A040081, A046069, A050412, A052333, A076337, A101036, A108129.

Extensions

Normally we require at least four terms but we will make an exception for this sequence in view of its importance. - N. J. A. Sloane, Nov 07 2002. See A101036 for the most likely extension.
Edited by N. J. A. Sloane, Nov 13 2009
Definition corrected ("odd" added) by M. F. Hasler, Aug 23 2020

A050412 Riesel problem: start with n; repeatedly double and add 1 until reaching a prime. Sequence gives number of steps to reach a prime or 0 if no prime is ever reached.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 4, 1, 1, 2, 2, 1, 2, 1, 1, 4, 1, 3, 2, 1, 3, 4, 1, 1, 2, 2, 1, 2, 1, 1, 2, 3, 1, 2, 1, 7, 24, 1, 3, 4, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 12, 2, 3, 4, 2, 1, 4, 1, 5, 2, 1, 1, 2, 4, 7, 2552, 1, 1, 2, 2, 1, 4, 3, 1, 2, 1, 5, 6, 1, 23, 4, 1, 1, 2, 3, 3, 2, 1, 1, 4, 1, 1
Offset: 1

Views

Author

Robert G. Wilson v, Dec 22 1999

Keywords

Comments

a(n) is the smallest m >= 1 such that (n+1)*2^m - 1 is prime (or 0 if no such prime exists).
It is conjectured that n = 509203 is the smallest Riesel number, i.e., n*2^k - 1 is composite for every k>0. - Robert G. Wilson v, Mar 01 2015. [This would imply that a(509203) is the first zero term in this sequence. - N. J. A. Sloane, Jul 31 2024]
Comment from N. J. A. Sloane, Aug 01 2024 (Start)
Both the Ballinger-Keller and Prime Wiki links assert that 104917*2^340181-1 is prime, but leave open the possibility that there is an m < 340181 which makes 104917*2^m - 1 a prime.
This question was finally settled by Lucas A. Brown on Jul 31 2024, who showed that m = 340181 is the smallest value that gives a prime. This implies that a(104917) = 340181.
Brown used a Python program (see below), with BPSW for the primality testing and gmpy2 to handle the arithmetic. The program was started on Jul 30 2024 and finished on Jul 31 2024.
He reports that it took about 15 hours in wall-clock time, and used 24 threads running in parallel. (End)

Examples

			For n=4; the smallest m>=1 such that (4+1)*2^m-1 is prime is m=2: 5*2^2-1=19 (prime). - _Jaroslav Krizek_, Feb 13 2011
		

Crossrefs

Main sequences for Riesel problem: A038699, A040081, A046069, A050412, A052333, A076337, A101036, A108129.

Programs

  • Maple
    A050412 := proc(n)
        local twox1,k ;
        twox1 := 2*n+1 ;
        k := 1;
        while not isprime(twox1) do
            twox1 := 2*twox1+1 ;
            k := k+1 ;
        end do:
        return k;
    end proc: # R. J. Mathar, Jul 23 2015
  • Mathematica
    a[n_] := Block[{s=n, c=1}, While[ ! PrimeQ[2*s+1], s = 2*s+1; c++]; c]; Table[ a[n], {n, 1, 99} ] (* Jean-François Alcover, Feb 06 2012, after Pari *)
    a[n_] := Block[{k = 1}, While[ !PrimeQ[2^k (n + 1) - 1], k++];k]; Array[a, 100] (* Robert G. Wilson v, Feb 14 2015 *) (* Corrected by Paolo Xausa, Jul 30 2024 *)
  • PARI
    a(n)=if(n<0,0,s=n; c=1; while(isprime(2*s+1)==0,s=2*s+1; c++); c)
    (Python, designed specifically for n = 104917)
    #! /usr/bin/env python3
    from labmath import primegen, isprime, mpz, count
    from multiprocessing import Pool
    primes = list(primegen(1000000))
    def test(n):
        for p in primes:
            if (104917 * pow(2, n, p)) % p == 1:
                return (n, False)
        return (n, isprime(104917 * mpz(2)**n - 1, tb=[]))
    with Pool(24) as P:
        for (n, result) in P.imap(test, count()):
            print('\b'*80, n, end='', flush=True)
            if result:
                break # Lucas A. Brown, Aug 01 2024

Formula

If a(n) = k with k>1, then a(2n+1) = k-1. - Robert G. Wilson v, Mar 01 2015
If a(n) = 0, then a(2n+1) is also 0. Conjecture: If a(n) = 1, then a(2n+1) is not 0. - Jeppe Stig Nielsen, Feb 12 2023

Extensions

More terms from Christian G. Bower, Dec 23 1999
Second definition corrected by Jaroslav Krizek, Feb 13 2011

A076335 Brier numbers: numbers that are both Riesel and Sierpiński [Sierpinski], or odd n such that for all k >= 1 the numbers n*2^k + 1 and n*2^k - 1 are composite.

Original entry on oeis.org

3316923598096294713661, 10439679896374780276373, 11615103277955704975673, 12607110588854501953787, 17855036657007596110949, 21444598169181578466233, 28960674973436106391349, 32099522445515872473461, 32904995562220857573541
Offset: 1

Views

Author

Olivier Gérard, Nov 07 2002

Keywords

Comments

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

Crossrefs

Extensions

Many terms reported in Problem 29 from "The Prime Problems & Puzzles Connection" from Carlos Rivera, May 30 2010
Entry revised by Arkadiusz Wesolowski, May 17 2012
Entry revised by Carlos Rivera and N. J. A. Sloane, Jan 03 2014
Entry revised by Arkadiusz Wesolowski, Feb 15 2014

A040081 Riesel problem: a(n) = smallest m >= 0 such that n*2^m-1 is prime, or -1 if no such prime exists.

Original entry on oeis.org

2, 1, 0, 0, 2, 0, 1, 0, 1, 1, 2, 0, 3, 0, 1, 1, 2, 0, 1, 0, 1, 1, 4, 0, 3, 2, 1, 3, 4, 0, 1, 0, 2, 1, 2, 1, 1, 0, 3, 1, 2, 0, 7, 0, 1, 3, 4, 0, 1, 2, 1, 1, 2, 0, 1, 2, 1, 3, 12, 0, 3, 0, 2, 1, 4, 1, 5, 0, 1, 1, 2, 0, 7, 0, 1, 1, 2, 2, 1, 0, 3, 1, 2, 0, 5, 6, 1, 23, 4, 0, 1, 2, 3, 3, 2, 1, 1, 0, 1, 1, 10, 0, 3
Offset: 1

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Author

Keywords

Crossrefs

Main sequences for Riesel problem: A038699, A040081, A046069, A050412, A052333, A076337, A101036, A108129.

Programs

  • Haskell
    a040081 = length . takeWhile ((== 0) . a010051) .
                           iterate  ((+ 1) . (* 2)) . (subtract 1)
    -- Reinhard Zumkeller, Mar 05 2012
    
  • Mathematica
    Table[m = 0; While[! PrimeQ[n*2^m - 1], m++]; m, {n, 100}] (* Arkadiusz Wesolowski, Sep 04 2011 *)
  • PARI
    a(n)=for(k=0,2^16,if(ispseudoprime(n*2^k-1), return(k))) \\ Eric Chen, Jun 01 2015
    
  • Python
    from sympy import isprime
    def a(n):
      m = 0
      while not isprime(n*2**m - 1): m += 1
      return m
    print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Feb 01 2021

A038699 Riesel problem: Smallest prime of form n*2^m-1, m >= 0, or 0 if no such prime exists.

Original entry on oeis.org

3, 3, 2, 3, 19, 5, 13, 7, 17, 19, 43, 11, 103, 13, 29, 31, 67, 17, 37, 19, 41, 43, 367, 23, 199, 103, 53, 223, 463, 29, 61, 31, 131, 67, 139, 71, 73, 37, 311, 79, 163, 41, 5503, 43, 89, 367, 751, 47, 97, 199, 101, 103, 211, 53, 109, 223, 113, 463, 241663, 59, 487, 61
Offset: 1

Views

Author

N. J. A. Sloane, Dec 30 1999

Keywords

Crossrefs

Primes arising in A040081 (or 0 if no prime exists).
Main sequences for Riesel problem: A038699, A040081, A046069, A050412, A052333, A076337, A101036, A108129.

Programs

  • Haskell
    a038699 = until ((== 1) . a010051) ((+ 1) . (* 2)) . (subtract 1)
    -- Reinhard Zumkeller, Mar 05 2012
  • Mathematica
    getm[n_]:=Module[{m=0},While[!PrimeQ[n 2^m-1],m++];n 2^m-1]; Array[getm,80]  (* Harvey P. Dale, Apr 24 2011 *)

Extensions

More terms from Henry Bottomley, Apr 24 2001

A180247 Prime Brier numbers: primes p such that for all k >= 1 the numbers p*2^k + 1 and p*2^k - 1 are composite.

Original entry on oeis.org

10439679896374780276373, 21444598169181578466233, 105404490005793363299729, 178328409866851219182953, 239365215362656954573813, 378418904967987321998467, 422280395899865397194393, 474362792344501650476113, 490393518369132405769309
Offset: 1

Views

Author

Arkadiusz Wesolowski, Aug 19 2010

Keywords

Comments

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.

Crossrefs

Extensions

Entry revised by N. J. A. Sloane, Jan 03 2014
Entry revised by Arkadiusz Wesolowski, May 29 2017

A046069 Riesel Problem: Smallest m >= 0 such that (2n-1)2^m-1 is prime, or -1 if no such value exists.

Original entry on oeis.org

2, 0, 2, 1, 1, 2, 3, 1, 2, 1, 1, 4, 3, 1, 4, 1, 2, 2, 1, 3, 2, 7, 1, 4, 1, 1, 2, 1, 1, 12, 3, 2, 4, 5, 1, 2, 7, 1, 2, 1, 3, 2, 5, 1, 4, 1, 3, 2, 1, 1, 10, 3, 2, 10, 9, 2, 8, 1, 1, 12, 1, 2, 2, 25, 1, 2, 3, 1, 2, 1, 1, 2, 5, 1, 4, 5, 3, 2, 1, 1, 2, 3, 2, 4, 1, 2, 2, 1, 1, 8, 3, 4, 2, 1, 3, 226, 3, 1, 2
Offset: 1

Views

Author

Keywords

Comments

There exist odd integers 2k-1 such that (2k-1)2^n-1 is always composite.

References

  • Ribenboim, P., The New Book of Prime Number Records. New York: Springer-Verlag, pp. 357-359, 1996.

Crossrefs

Main sequences for Riesel problem: A038699, A040081, A046069, A050412, A052333, A076337, A101036, A108129.
Bisection of A040081.

Programs

  • Mathematica
    max = 10^6; (* this maximum value of m is sufficient up to n=1000 *) a[1] = 2; a[2] = 0; a[n_] := For[m = 1, m <= max, m++, If[PrimeQ[(2*n - 1)*2^m - 1], Return[m]]] /. Null -> -1; Reap[ Do[ Print[ "a(", n, ") = ", a[n]]; Sow[a[n]], {n, 1, 100}]][[2, 1]] (* Jean-François Alcover, Nov 15 2013 *)
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