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 1-4 of 4 results.

A186311 Least k such that the interval 100k to 100k+99 has exactly n primes.

Original entry on oeis.org

16718, 1559, 3020, 588, 314, 188, 186, 59, 48, 41, 21, 13, 11, 19, 5, 8, 2, 4, 1228537713709, 14688670051164208, 203860951641372730864, 1
Offset: 0

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Author

T. D. Noe, Feb 22 2011

Keywords

Comments

It is known that a(25)=0. Terms for n = 22 and 23 are unknown. Glaisher tabulates the number of centuries having 0, 1, 2, ... primes for numbers up to 9000000. Glaisher's 1883 book is still in print!
a(24) does not exist because the only century having 24 primes is 0 to 99 -- the same century having 25 primes. From A020497, we see that a range of 101 numbers is required to find 24 primes. Dickson's conjecture implies that a(n) exists for n=18..23. - Charles R Greathouse IV, Feb 24 2011
To see that Dickson's conjecture is applicable to the preceding statement, the appropriate general sequence to consult is A364678, which affirms that 23 primes are permissible between adjacent multiples of 100, as opposed to in an arbitrary interval of 99 integers. - Peter Munn, Sep 04 2023
a(n) for n = 18..23 is greater than 10^10. Ribenboim discusses Dickson's conjecture in two books. - T. D. Noe, Feb 24 2011
a(19) <= 1108851311300675700427. - Donovan Johnson, Feb 28 2011
a(20) <= 394338677302163715754576644. - Tim Johannes Ohrtmann, Aug 27 2015

References

  • James Glaisher, Factor Table for the Sixth Million, Taylor and Francis, London, 1883.
  • Paulo Ribenboim, The New Book of Prime Number Records, Springer-Verlag NY, 1995, p. 372.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY, 2004, p. 250.

Crossrefs

Cf. A038822 (number of primes between 100n and 100n+99).
Cf. A181098 (centuries without primes).
Cf. A186393-A186408 (centuries having 1 to 16 primes).
Cf. A186509 (centuries having 17 primes).
Cf. A361723 (centuries having 18 primes).

Programs

  • Mathematica
    t = Differences[PrimePi[100*Range[0, 20000]]]; Flatten[Table[Position[t, n, 1, 1], {n, 0, 17}] - 1]
  • PARI
    a(n)=for(k=0,9e99,if(sum(i=100*k+1,100*k+99,ispseudoprime(i))==n, return(k))) \\ Charles R Greathouse IV, Feb 24 2011

Extensions

a(18) from Donovan Johnson, Feb 28 2011
a(19) from Brian Kehrig, Apr 08 2023
a(20)-a(21) from Brian Kehrig, May 28 2024

A100418 Numbers k such that 30*k + {1,11,13,17,19,23,29} are all prime.

Original entry on oeis.org

49, 34083, 41545, 48713, 140609, 524027, 616812, 855281, 1314397, 1324750, 1636152, 2281293, 2927134, 3401412, 3605413, 4989341, 5212221, 5284979, 5406303, 5645269, 6141254, 6342728, 7231434, 7347697, 7637329, 8027068, 8161657, 8372756, 8392776, 8567216, 8986096, 9145563
Offset: 1

Views

Author

Ferenc Adorjan (fadorjan(AT)freemail.hu), Nov 19 2004

Keywords

Comments

Values are 0 mod 7.
From Peter Munn, Sep 06 2023: (Start)
In each case, the 7 primes are necessarily consecutive.
As A065706 demonstrates, many intervals of 27 integers contain 8 primes, but only A364678(30) = 7 primes can occur between adjacent positive multiples of 30. This is because there are 8 values {1,7,11,13,17,19,23,29} coprime to 30, but they cover every residue class modulo 7, which means at least one of 30*k + {1,7,11,13,17,19,23,29} is divisible by 7.
1 and 29 are in the same residue class, but if we remove any of the other coprime integers there is a class that is not represented in the set. For this sequence, we remove 7, so when k is a multiple of 7, none of 30*k + {1,11,13,17,19,23,29} is a multiple of 2, 3, 5 or 7 and the set can potentially be 7 consecutive primes.
The sequences for the other appropriate subsets of 7 coprime values are A100419-A100423.
(End)

Crossrefs

Programs

  • Magma
    [ n: n in [0..70000000 by 7] | forall{ q: q in [1, 11, 13, 17, 19, 23, 29] | IsPrime(30*n+q) } ]; // Klaus Brockhaus, Feb 24 2011
  • Mathematica
    Select[Range[803*10^4],AllTrue[30#+{1,11,13,17,19,23,29},PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Jun 11 2019 *)
  • PARI
    {pav7(mx)= local(wp=[1,11,13,17,19,23,29],v=[],i,j,m); for(k=1,mx, i=k*30;j=1;m=1;while(m&&(j<8),m=isprime(i+wp[j]);j+=1);if(m,v=concat(v,k))); return(v)}
    

Extensions

Edited by Don Reble, Nov 17 2005

A023193 a(n) gives the largest number k for which there is at least one admissible k-tuple taken from [0, 1, ..., n-1] if the tuple starts with 0. Admissibility is defined in a comment.

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 20, 20
Offset: 1

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Author

Keywords

Comments

The (wrong) old name was: Largest number of pairwise coprime numbers that can occur in an interval of length n. - Wolfdieter Lang, Oct 10 2017
Conjecturally, a(n) is the largest number of primes that occurs on an infinite number of intervals of n consecutive integers. The conjecture is apparently due to Dickson; Hardy & Littlewood's Conjecture B concerns only pairs (p, p + 2n).
According to the link at www.opertech.com, a(3159) >= 447 > 446 = pi(3159). The k-tuples conjecture then implies that, for an infinitude of n, the interval [n+1, n+3159] includes 447 primes. For these n, pi(n+3159) >= pi(n)+447 > pi(n)+446 = pi(n)+pi(3159), contradicting the conjecture that pi(x+y) <= pi(x)+pi(y). - David W. Wilson, May 23 2005
From Wolfdieter Lang, Oct 10 2017: (Start)
The following admissibility definition is adapted from the Hensley and Richards [H-R] or Richards [R] links. A k-tuple B_k = [b_1, b_2, ..., b_k] of integers with 0 <= b_1 < b_2 < ... < b_k is admissible if, for each prime p, there exists at least one congruence class modulo p which contains none of the B_k members. Because complete residue systems modulo p are equivalent under translation one can consider the length n interval [0, 1, ..., n-1] and admissible k-tuples starting with 0. The prime p = 2 allows then only even tuple numbers from I_n = [0, 2, ..., floor((n-1)/2)]. Only primes p <= k have to be tested.
a(n) is then the maximal k for which there is at least one such admissible B_k tuple from the interval I_n. This function a(n) is called rho^* in (H-R) and (R). It has been given as rhobar in the Schinzel - Sierpiński link, Théorème 1, p. 201.
Note that there are also admissible k-tuples from members of [0, 1, ..., n-1] which do not start with 0. Such tuples are translations of the ones starting with 0. E.g., [1, 3] is an admissible 2-tuple for any [0, 1,..., n-1] interval with n >= 3, but it is a translation of the considered [0, 2] tuple.
For the multiplicities of k see A047947(k), for k >= 1.
For the smallest k such that a(k) = n see A020497(n), for k >= 1.
For the number of all admissible k-tuples from the interval I_n starting with 0 see the array A292224(n, k), with k = 1..a(n), which has been given in the Engelsma link, Table 2, p. 27.
One of the Hardy-Littlewood conjectures (the prime tuple conjecture, see also conjecture (B) given by [H-R] and [R], and Ribenboim, hypothesis (D_1), p. 373, from the Dickson conjecture) is that there are infinitely many primes with gaps defined by any admissible B_k tuple, that is, all p, p + b_2, ..., p + b_k are prime for infinitely many primes p, for k >= 2. For k = 1 this is well known.
(End)

References

  • Douglas Hensley and Ian Richards, "On the incompatibility of two conjectures concerning primes". Analytic number theory (Proc. Sympos. Pure Math., Vol. XXIV, St. Louis Univ., St. Louis, Mo., 1972), pp. 123-127.
  • P. Ribenboim, The New Book of Prime Number Records, Springer-Verlag, NY, 1996, ch. 6, I, pp. 372-386.

Crossrefs

Cf. A008407 (minimal difference of first and last prime in a prime k-tuplet), A047947 (multiplicities), A066081 (weaker binary conjectures), A062571.
Least inverse is A020497.

Formula

Conjecturally, a(n) = lim sup pi(x+n)-pi(x), where pi = A000720. This would follow from the k-tuple conjecture. - David W. Wilson, May 23 2005
a(n) = minimum m such that A008407(m) >= n. - Max Alekseyev, Nov 03 2008
Richards shows that a(n) > n/log n + kn/log^2 n + o(n/log^2 n), where k = 1 + log 2 = 1.69... . In particular, a(n) > pi(n) for large enough n. Hensley & Richards 1974 cite a result of Montgomery & Vaughan "to appear" that a(n) <= 2*pi(n) for n >= 2. - Charles R Greathouse IV, Apr 16 2013

Extensions

Name corrected by Wolfdieter Lang, Oct 10 2017

A261571 Number of possible prime patterns for centuries having exactly n primes.

Original entry on oeis.org

1, 40, 780, 7528, 47878, 225044, 830270, 2459376, 5900602, 11555200, 18634704, 24942742, 27836859, 25913910, 20053913, 12815608, 6699888, 2829786, 948729, 245756, 47150, 6276, 518, 20
Offset: 0

Views

Author

Tim Johannes Ohrtmann, Aug 27 2015

Keywords

Comments

The index of the final term is A364678(100) = 23. - Peter Munn, Sep 04 2023

Crossrefs

Cf. A010956.
Cf. A038822 (number of primes between 100n and 100n+99).
Cf. A181098 (centuries without primes).
Cf. A186393-A186408 (centuries having 1 to 16 primes).
Cf. A186509 (centuries having 17 primes).
Cf. A186311 (first occurrences).
Cf. A364678.
Showing 1-4 of 4 results.