A246810 a(n) is the smallest number m such that np(m) = n, where np(m) is number of primes p such that prime(m) < p < prime(m)^(1 + 1/m).

1, 5, 12, 17, 25, 55, 83, 169, 207, 206, 384, 953, 1615, 2192, 2197, 3024, 3023, 10709, 10935, 29509, 29508, 62736, 62735, 94333, 94332, 196966, 314940, 608777, 1258688, 1767259, 2448975, 2448973, 7939362, 9373136, 9373134, 16854966, 16854967

Offset: 1

Author

Farideh Firoozbakht and Jahangeer Kholdi, Oct 10 2014

Keywords

• nonn

Comments

Firoozbakht's conjecture says that for every n, there exists at least one prime p where, prime(n) < p < prime(n)^(1 + 1/n). Hence if Firoozbakht's conjecture is true, then there is no m such that np(m) = 0.

Conjecture: For every positive integer n, a(n) exists.

a(65) > 10^12. - Robert Price, Nov 12 2014

Examples

			a(6) = 55 since the number of primes p such that prime(55) < p < prime(55)^(1 + 1/55) is 6 and 55 is the smallest number with this property.
		

Links

• Robert Price, Table of n, a(n) for n = 1..64
• A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
• A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2
• Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399v2 [math.NT], 2010.
• Wikipedia, Firoozbakht's conjecture.

Crossrefs

Cf. A000040, A182134, A246781, A246782, A246783, A246787.

Programs

Mathematica

np[n_]:=(b=Prime[n]; Length[Select[Range[b+1, b^(1 + 1/n)],PrimeQ]]); a[n_]:=(For[m=1, np[m] !=n, m++]; m);
Do[Print[a[n]], {n, 37}]