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 14 results. Next

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

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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

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

A052333 Riesel problem: start with n; repeatedly double and add 1 until reach a prime. Sequence gives a(n) = prime reached, or 0 if no prime is ever reached.

Original entry on oeis.org

3, 5, 7, 19, 11, 13, 31, 17, 19, 43, 23, 103, 223, 29, 31, 67, 71, 37, 79, 41, 43, 367, 47, 199, 103, 53, 223, 463, 59, 61, 127, 131, 67, 139, 71, 73, 151, 311, 79, 163, 83, 5503, 738197503, 89, 367, 751, 191, 97, 199, 101, 103, 211, 107, 109, 223, 113, 463
Offset: 1

Views

Author

Christian G. Bower, Dec 19 1999

Keywords

Comments

Equivalently, a(n) = smallest prime of form (n+1)*2^k-1 for k >= 1, or 0 if no such prime exists.
a(509202) = 0 (i.e. never reaches a prime) - Chris Nash (chris_nash(AT)hotmail.com). (Of course this is related to the entry 509203 of A076337.)
a(73) is a 771-digit prime reached after 2552 iterations - 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]

Examples

			a(4)=19 because 4 -> 9 (composite) -> 19 (prime).
		

Crossrefs

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

Programs

  • Mathematica
    Table[NestWhile[2#+1&,2n+1,!PrimeQ[#]&,1,1000],{n,60}] (* Harvey P. Dale, May 08 2011 *)
  • PARI
    a(n)=while(!isprime(n=2*n+1),);n \\ oo loop when a(n) = 0. - Charles R Greathouse IV, May 08 2011

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

A108129 Riesel problem: let k=2n-1; then a(n)=smallest m >= 1 such that k*2^m-1 is prime, or -1 if no such prime exists.

Original entry on oeis.org

2, 1, 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, 1, 1, 2
Offset: 1

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Author

Jorge Coveiro, Jun 04 2005

Keywords

Comments

It is conjectured that the integer k = 509203 is the smallest Riesel number, that is, the first n such that a(n) = -1 is 254602.
Browkin & Schinzel, having proved that 509203*2^k - 1 is composite for all k > 0, ask for the first such number with this property, noting that the question is implicit in Aigner 1961. - Charles R Greathouse IV, Jan 12 2018
Record values begin a(1) = 2, a(7) = 3, a(12) = 4, a(22) = 7, a(30) = 12, a(64) = 25, a(96) = 226, a(330) = 800516; the next record appears to be a(1147), unless a(1147) = -1. (The value for a(330), i.e., for k = 659, is from the Ballinger & Keller link, which also lists k = 2293, i.e., n = (k+1)/2 = (2293+1)/2 = 1147, as the smallest of 50 values of k < 509203 for which no prime of the form k*2^m-1 had yet been found.) - Jon E. Schoenfield, Jan 13 2018
Same as A046069 except for a(2) = 1. - Georg Fischer, Nov 03 2018

References

  • Hans Riesel, Några stora primtal, Elementa 39 (1956), pp. 258-260.

Crossrefs

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

Programs

  • Mathematica
    Array[Function[k, SelectFirst[Range@300, PrimeQ[k 2^# - 1] &]][2 # - 1] &, 102] (* Michael De Vlieger, Jan 12 2018 *)
    smk[n_]:=Module[{m=1,k=2n-1},While[!PrimeQ[k 2^m-1],m++];m]; Array[smk,120] (* Harvey P. Dale, Dec 26 2023 *)
  • PARI
    forstep(k=1,301,2,n=1;while(!isprime(k*2^n-1),n++);print1(n,","))

Extensions

Edited by Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 25 2006
Name corrected by T. D. Noe, Feb 13 2011

A033809 Smallest m>0 such that (2n-1)2^m+1 is prime, or -1 if no such value exists.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 3, 6, 1, 1, 2, 2, 1, 8, 1, 1, 2, 1, 1, 2, 2, 583, 2, 1, 1, 4, 2, 5, 4, 1, 1, 2, 1, 3, 2, 1, 3, 2, 1, 1, 4, 2, 1, 8, 2, 1, 2, 1, 3, 16, 1, 3, 6, 1, 1, 2, 3, 1, 8, 6, 1, 2, 3, 1, 4, 1, 3, 2, 1, 53, 6, 8, 3, 4, 1, 1, 8, 6, 3, 2, 1, 7, 2, 8, 1, 2, 2, 1, 4, 1, 3, 6, 1, 1, 2, 4, 15, 2
Offset: 1

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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

Cf. A046067 (except for initial term a(1) identical to this sequence), A046068, A046069, A046070.

A090733 a(n) = 25*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 25.

Original entry on oeis.org

2, 25, 623, 15550, 388127, 9687625, 241802498, 6035374825, 150642568127, 3760028828350, 93850078140623, 2342491924687225, 58468448039040002, 1459368709051312825, 36425749278243780623, 909184363247043202750
Offset: 0

Views

Author

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 18 2004

Keywords

Comments

A Chebyshev T-sequence with Diophantine property.
a(n) gives the general (nonnegative integer) solution of the Pell equation a^2 - 69*(3*b)^2 =+4 together with the companion sequence b(n)=A097780(n-1), n>=0.

Examples

			(x,y) =(2,0), (25;1), (623;25), (15550;624), ... give the nonnegative integer solutions to x^2 - 69*(3*y)^2 =+4.
		

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Crossrefs

a(n)=sqrt(4 + 69*(3*A097780(n-1))^2), n>=1.
Cf. A077428, A078355 (Pell +4 equations).
Cf. A097779 for 2*T(n, 23/2).

Programs

  • Mathematica
    a[0] = 2; a[1] = 25; a[n_] := 25a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
  • Sage
    [lucas_number2(n,25,1) for n in range(0,20)] # Zerinvary Lajos, Jun 26 2008

Formula

a(n) = S(n, 25) - S(n-2, 25) = 2*T(n, 25/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 25)=A097780(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (25+3*sqrt(69))/2 and am := (25-3*sqrt(69))/2.
G.f.: (2-25*x)/(1-25*x+x^2).

Extensions

Extension, Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A046070 Second smallest m such that (2n-1)2^m-1 is prime, or -1 if no such value exists.

Original entry on oeis.org

3, 1, 4, 5, 3, 26, 7, 2, 4, 3, 2, 6, 9, 2, 16, 5, 3, 6, 2553, 24, 10, 31, 2, 14, 5, 9, 6, 3, 2, 16, 5, 3, 6, 9, 4, 14, 11, 3, 4, 3, 5, 4, 11, 2, 8, 3, 4, 6, 9, 4, 18, 7, 3, 12, 149, 3, 14, 3, 2, 16, 3, 3, 4, 113, 3, 14, 11, 9, 18, 5, 2, 4, 13, 2, 16, 221, 4, 8, 5, 4, 6, 31, 3, 6, 5, 3, 4, 3
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

Programs

  • Mathematica
    max = 10000 (* this maximum value of m is sufficient up to n=168 *); a[n_] := Reap[ For[m = 0; cnt = 0, m <= max && cnt < 2, m++, If[m == max, Sow[-1], If[PrimeQ[(2*n - 1)*2^m - 1], cnt++; Sow[m]]]]][[2, 1]]; Table[a[n][[2]], {n, 1, 88}] (* Jean-François Alcover, Feb 28 2013 *)
Showing 1-10 of 14 results. Next