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

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

Offset: 0

Author

T. D. Noe, Feb 22 2011

Keywords

• nonn
• fini
• more
• hard

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.

Links

• Chris Caldwell, Prime Glossary: Dickson's Conjecture
• L. E. Dickson, A new extension of Dirichlet's theorem on prime numbers, Messenger of Math., 33 (1904), 155-161.

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).

Cf. A020497, A261571, A364678.

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