A182134 Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n).

1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 2, 2, 3, 4, 3, 4, 3, 3, 2, 2, 4, 5, 4, 3, 2, 2, 2, 3, 4, 4, 3, 4, 4, 4, 4, 3, 4, 5, 4, 4, 3, 2, 2, 4, 5, 5, 4, 4, 4, 3, 5, 6, 5, 5, 4, 3, 3, 2, 4, 4, 4, 3, 2, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 5, 6, 5, 7, 6, 6, 5, 5, 5, 5, 4, 4, 5, 4, 5, 4, 3, 3, 3, 3, 5, 5, 5, 5, 6, 5

Offset: 1

Author

Thomas Ordowski, Apr 20 2012

Keywords

• nonn

Comments

Firoozbakht's conjecture: prime(n+1)^(1/(n+1)) < prime(n)^(1/n), for all n >= 1.

According to Firoozbakht's conjecture, all terms of this sequence are positive. - Jahangeer Kholdi, Jul 30 2014

Conjecture: a(n)=1 only for n = 1, 2, 3, 4, and 8. - Farideh Firoozbakht, Oct 18 2014

See A246782 and A246781 for indices such that a(n)=2 resp. a(n)=3. - M. F. Hasler, Oct 19 2014

Length of n-th row in A244365; a(n) = A001221(A245722(n)). - Reinhard Zumkeller, Nov 18 2014

a(n) = 2 for n = 5, 6, 7, 9, 10, 11, 14, 15, 22, 23, 28, 29, 30, 45, 46, 61, 66, 216, 217, 367, 3793, 1319945, ... = A246782. - Robert G. Wilson v, Feb 20 2015

a(n) = 3 for n = 12, 13, 16, 18, 20, 21, 27, 31, 34, 39, 44, 53, 59, 60, 65, 96, 97, 98, 99, 136, 154, 202, ... = A246781. - Robert G. Wilson v, Feb 20 2015

First occurrence of k: 1, 5, 12, 17, 25, 55, 83, 169, 207, 206, 384, 953, ... = A246810. - Robert G. Wilson v, Feb 20 2015

Conjecture: lim sup n->oo a(n) = oo. - John W. Nicholson, Feb 28 2015

a(n) is unbounded (that is, the above conjecture is true). In particular, there is a constant c > 1 such that a(n) > c log n infinitely often (by Maier's theorem). - Thomas Ordowski and Charles R Greathouse IV, Apr 09 2015

Examples

			a(25) = 5, because p(25) = 97 and there are 5 primes p such that 97 < p < 97^(1 + 1/25) = 121.9299290...: 101, 103, 107, 109, 113.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015.
• Alexei Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht’s Conjecture, arXiv:1506.03042 [math.NT], 2015.
• Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2.
• Wikipedia, Firoozbakht's conjecture

Crossrefs

Cf. A010051, A000040, A249669, A244365, A246810.

Programs

Haskell

a182134 = length . a244365_row  -- Reinhard Zumkeller, Nov 16 2014

Maple

a:= n-> numtheory[pi](ceil(ithprime(n)^(1+1/n))-1)-n:
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 21 2012

Mathematica

Table[i = Prime[n] + 1; j = Floor[Prime[n]^(1 + 1/n)]; Length[Select[Range[i, j], PrimeQ]], {n, 100}] (* T. D. Noe, Apr 21 2012 *)
f[n_] := PrimePi[ Prime[n]^(1 + 1/n)] - n; Array[f, 105] (* Robert G. Wilson v, Feb 20 2015 *)

PARI

A182134(n)=primepi(prime(n)^(1+1/n))-n \\ M. F. Hasler, Nov 03 2014

Python

from sympy import primepi, prime
def a(n): return primepi(prime(n)**(1 + 1/n)) - n # Indranil Ghosh, Apr 23 2017

Formula

a(n) = Sum_{m=A000040(n+1)..A249669(n)} A010051(m). - Reinhard Zumkeller, Nov 16 2014

a(n) = primepi(prime(n)^(1+1/n)) - n (see PARI program). - John W. Nicholson, Feb 11 2015

Extensions

More terms from Alois P. Heinz, Apr 21 2012