A249566 Numbers n such that A182134(n) = 4, i.e., there exist exactly four primes p with prime(n) < p < prime(n)^(1+1/n).

17, 19, 24, 26, 32, 33, 35, 36, 37, 38, 40, 42, 43, 47, 50, 51, 52, 58, 62, 63, 64, 76, 77, 78, 79, 90, 91, 93, 95, 121, 123, 124, 125, 126, 134, 135, 137, 150, 153, 185, 186, 187, 188, 189, 201, 203, 213, 218, 219, 238, 239, 259, 263, 278, 279, 289, 293

Offset: 1

Author

Robert Price, Nov 01 2014

Keywords

• nonn

Comments

See A246782 for a more complete description of this sequence.

a(1136) > 10^12.

It is interesting that three consecutive integers n = 20004097201301075, n + 1 and n + 2 are in the sequence. Conjecture: The sequence is infinite. - Farideh Firoozbakht, Nov 01 2014

Links

• Robert Price, Table of n, a(n) for n = 1..1135
• 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, A002386, A005669, A182134, A246781, A246782.

Programs

Haskell

a249566 n = a249566_list !! (n-1)
a249566_list = filter ((== 4) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014

Mathematica

np[n_]:=(a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a+1,b], PrimeQ]]); Do[If[np[n] == 4,Print[n]], {n, 293}]
np[n_]:=(a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a+1,b], PrimeQ]]); Select[Range[293], np[#]==4&] (* Farideh Firoozbakht, Nov 01 2014 *)

PARI

for(n=1,9e9,primepi(prime(n)^(1+1/n))-n==4&&print1(n",")) \\ M. F. Hasler, Nov 03 2014