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%I #32 Feb 28 2021 03:14:32
%S 1,2,3,5,7,9,10,13,15,16,17,20,23,24,26,28,30,32,33,35,39,40,41,43,45,
%T 49,51,52,54,55,56,57,60,62,64,66,69,71,72,76,77,79,81,83,86,87,89,91,
%U 92,94,96,97,98,102,103,104,107,108,109,113,116,118,119,120,123,124,126
%N Integers k such that prime(k)*(2*prime(k)-1)/3 is an integer.
%C Apart from the term "2", the same as A091177. - _Stefan Steinerberger_, Dec 29 2007
%C Numbers n such that the number of distinct residues r in the congruence x^3 == r (mod p) is equal to p where p = prime(n). See A046530. - _Michel Lagneau_, Sep 28 2016
%C The asymptotic density of this sequence is 1/2 (by Dirichlet's theorem). - _Amiram Eldar_, Feb 28 2021
%H Amiram Eldar, <a href="/A133677/b133677.txt">Table of n, a(n) for n = 1..10000</a>
%F Integers k such that (prime(k) mod 3) mod 2 = 0. - _Gary Detlefs_, Dec 06 2011
%e 4 is not in the sequence since prime(4)*(2*prime(4) - 1)/3 = 7*(2*7 - 1)/3 = 7*13/3 = 91/3 is not an integer, but 5 is in the sequence since prime(5)*(2*prime(5) - 1)/3 = 11*(2*11 - 1)/3 = 11*21/3 = 11*7 = 77 is an integer. - _Michael B. Porter_, Sep 28 2016
%p for n from 1 to 126 do if((ithprime(n) mod 3) mod 2=0) then print(n) fi od; # _Gary Detlefs_, Dec 06 2011
%t Union[Table[If[IntegerQ[Prime[n]*(2*Prime[n] - 1)/3], n, {}], {n, 1, 100}]]
%t pnQ[n_]:=Module[{pn=Prime[n]},IntegerQ[(pn(2pn-1))/3]]; Select[Range[ 150], pnQ] (* _Harvey P. Dale_, Oct 02 2011 *)
%t Sort@ Join[{2}, Select[ Range@ 126, Mod[2*Prime[#], 3] == 1 &]] (* _Robert G. Wilson v_, Sep 28 2016 *)
%Y Cf. A046530, A091177, A133645.
%K nonn
%O 1,2
%A _Roger L. Bagula_, Dec 28 2007
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