A246781 Numbers n such that A182134(n) = 3, i.e., there exist only three primes p with prime(n) < p < prime(n)^(1 + 1/n).

12, 13, 16, 18, 20, 21, 27, 31, 34, 39, 44, 53, 59, 60, 65, 96, 97, 98, 99, 136, 154, 202, 214, 215, 220, 221, 280, 324, 325, 326, 365, 366, 736, 780, 2146, 2225, 3792, 5946, 5947, 5948, 6902, 6903, 18524, 22078, 23510, 23511, 23512, 31542, 31544, 33606

Offset: 1

Author

Farideh Firoozbakht, Oct 12 2014

Keywords

• nonn

Comments

Firoozbakht's conjecture states that for every n, there exists at least one prime p with prime(n) < p < prime(n)^(1+1/n).

The only known indices n for which A182134(n) = 1 are {1, 2, 3, 4, 8}.

See A246782 for indices n such that A182134(n) = 2.

This sequence lists numbers n such that A182134(n) = 3.

Examples

			12 is in the sequence since there exists only three primes p where, prime(12) < p < prime(12)^(1 + 1/12). Note that prime(12) = 37, 37^(1 + 1/12) ~ 49.99 and 37 < 41 < 43 < 47 < 49.99.
		

Links

• Robert Price, Table of n, a(n) for n = 1..170
• Carlos Rivera, Conjecture 30. The Firoozbakht Conjecture PrimePuzzles.net.
• A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
• Wikipedia, Firoozbakht's conjecture

Crossrefs

Cf. A000040, A182134, A246782.

Cf. A249566.

Programs

Haskell

a246781 n = a246781_list !! (n-1)
a246781_list = filter ((== 3) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014

Maple

N:= 10^5: # to get all terms where prime(n)^(1+1/n) < N
Primes:= select(isprime,[2,seq(2*i+1,i=1..floor((N+1)/2))]):
filter:= proc(n) local t; t:= Primes[n]^(n+1); Primes[n+3]^n <= t and Primes[n+4]^n > t end proc:
select(filter, [$1..nops(Primes)-4]); # Robert Israel, Mar 23 2015

Mathematica

np[n_] := (a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a + 1, b], PrimeQ]]); Select[Range[10000], np[#] == 3 &]

Extensions

a(43)-a(50) from Robert Price, Oct 24 2014