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A084140
a(n) is the smallest number j such that if x >= j there are at least n primes between x and 2x exclusively.
13
2, 6, 9, 15, 21, 24, 30, 34, 36, 49, 51, 54, 64, 75, 76, 84, 90, 91, 114, 115, 117, 120, 121, 132, 135, 141, 154, 156, 174, 175, 184, 187, 201, 205, 210, 216, 217, 220, 231, 244, 246, 252, 285, 286, 294, 297, 300, 301, 304, 321, 322, 324, 327, 330, 339, 360, 364
OFFSET
1,1
COMMENTS
For all m >= a(n) there are at least n primes between m and 2m exclusively. This calculation relies on the fact that pi(2m) - pi(m) > m/(3*log(m)) for m >= 5. This is one more than the terms of A084139 with offset changed from 0 to 1.
For n > 5889, pi(2n) - pi(n) > f(2, 2n) - f(3, n) where f(k, x) = x/log x * (1 + 1/log x + k/(log x)^2). This may be useful for checking larger terms. The constant 3 can be improved at the cost of an increase in the constant 5889. - Charles R Greathouse IV, May 02 2012
A168421(n) = nextprime(a(n)), where nextprime(x) is the next prime >= x. - John W. Nicholson, Dec 21 2012
a(1) = ceiling((A104272(1)+1)/2) modifies the only even prime, 2; which has been stated, in Formula, as a(1) = A104272(1); for all others, a(n) = (A104272(n)+1)/2 = ceiling ((A104272(n)+1)/2). - John W. Nicholson, Dec 24 2012
Srinivasan's Lemma (2014): previousprime(a(n)) = p_(k-n) < (p_k)/2, where the n-th Ramanujan Prime R_n is the k-th prime p_k, and with n > 1. Proof: By the minimality of R_n, the interval ((p_k)/2,p_k] contains exactly n primes, so p_(k-n) < (p_k)/2. - Copied and adapted from a comment by Jonathan Sondow in A168421 by John W. Nicholson, Feb 17 2015
REFERENCES
Paulo Ribenboim, The Little Book of Big Primes, Springer-Verlag, 1991, p. 140.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag, 2004, p. 181.
LINKS
Vladimir Shevelev, Charles R. Greathouse IV, Peter J. C. Moses, On intervals (kn, (k+1)n) containing a prime for all n>1, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. arXiv version, arXiv:1212.2785 [math.NT], 2012.
Anitha Srinivasan, An upper bound for Ramanujan primes, Integers, 19 (2014), #A19.
Eric Weisstein's World of Mathematics, Bertrand's Postulate.
FORMULA
a(1) = A104272(1); for n >= 2, a(n) = (A104272(n)+1)/2. - Vladimir Shevelev, Dec 07 2012
a(n) = ceiling((A104272(n)+1)/2) for n >= 1. - John W. Nicholson, Dec 24 2012
EXAMPLE
a(11)=51 since there are at least 11 primes between m and 2m for all m >= 51 and this is not true for any m < 51. Although a(100)=720 is not listed, for all m >= 720, there are at least 100 primes between m and 2m.
MATHEMATICA
nn = 60;
R = Table[0, {nn}]; s = 0;
Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--]; If[s < nn, R[[s + 1]] = k], {k, Prime[3 nn]}];
A104272 = R + 1;
Ceiling[(A104272 + 1)/2] (* Jean-François Alcover, Nov 07 2018, after T. D. Noe in A104272 *)
KEYWORD
nonn
AUTHOR
Harry J. Smith, May 15 2003
STATUS
approved