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%I #23 Sep 25 2015 16:41:14
%S 17,19,24,26,32,33,35,36,37,38,40,42,43,47,50,51,52,58,62,63,64,76,77,
%T 78,79,90,91,93,95,121,123,124,125,126,134,135,137,150,153,185,186,
%U 187,188,189,201,203,213,218,219,238,239,259,263,278,279,289,293
%N Numbers n such that A182134(n) = 4, i.e., there exist exactly four primes p with prime(n) < p < prime(n)^(1+1/n).
%C See A246782 for a more complete description of this sequence.
%C a(1136) > 10^12.
%C 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
%H Robert Price, <a href="/A249566/b249566.txt">Table of n, a(n) for n = 1..1135</a>
%H A. Kourbatov, <a href="http://arxiv.org/abs/1503.01744">Verification of the Firoozbakht conjecture for primes up to four quintillion</a>, arXiv:1503.01744 [math.NT], 2015
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Firoozbakht%E2%80%99s_conjecture">Firoozbakht's conjecture</a>
%t 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}]
%t 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 *)
%o (PARI) for(n=1,9e9,primepi(prime(n)^(1+1/n))-n==4&&print1(n",")) \\ _M. F. Hasler_, Nov 03 2014
%o (Haskell)
%o a249566 n = a249566_list !! (n-1)
%o a249566_list = filter ((== 4) . a182134) [1..]
%o -- _Reinhard Zumkeller_, Nov 17 2014
%Y Cf. A000040, A002386, A005669, A182134, A246781, A246782.
%K nonn
%O 1,1
%A _Robert Price_, Nov 01 2014
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