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A186311
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Least k such that the interval 100k to 100k+99 has exactly n primes.
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23
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16718, 1559, 3020, 588, 314, 188, 186, 59, 48, 41, 21, 13, 11, 19, 5, 8, 2, 4, 1228537713709, 14688670051164208
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OFFSET
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0,1
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COMMENTS
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It is known that a(21)=1 and a(25)=0. Terms for n = 20, 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
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REFERENCES
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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.
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LINKS
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MATHEMATICA
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t = Differences[PrimePi[100*Range[0, 20000]]]; Flatten[Table[Position[t, n, 1, 1], {n, 0, 17}] - 1]
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PROG
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(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
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CROSSREFS
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Cf. A038822 (number of primes between 100n and 100n+99).
Cf. A181098 (centuries without primes).
Cf. A186509 (centuries having 17 primes).
Cf. A361723 (centuries having 18 primes).
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KEYWORD
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nonn,fini,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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