A133677 Integers k such that prime(k)*(2*prime(k)-1)/3 is an integer.

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, 49, 51, 52, 54, 55, 56, 57, 60, 62, 64, 66, 69, 71, 72, 76, 77, 79, 81, 83, 86, 87, 89, 91, 92, 94, 96, 97, 98, 102, 103, 104, 107, 108, 109, 113, 116, 118, 119, 120, 123, 124, 126

Offset: 1

Author

Roger L. Bagula, Dec 28 2007

Keywords

• nonn

Comments

Apart from the term "2", the same as A091177. - Stefan Steinerberger, Dec 29 2007

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

The asymptotic density of this sequence is 1/2 (by Dirichlet's theorem). - Amiram Eldar, Feb 28 2021

Examples

			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
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A046530, A091177, A133645.

Programs

Maple

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

Mathematica

Union[Table[If[IntegerQ[Prime[n]*(2*Prime[n] - 1)/3], n, {}], {n, 1, 100}]]
pnQ[n_]:=Module[{pn=Prime[n]},IntegerQ[(pn(2pn-1))/3]]; Select[Range[ 150], pnQ] (* Harvey P. Dale, Oct 02 2011 *)
Sort@ Join[{2}, Select[ Range@ 126, Mod[2*Prime[#], 3] == 1 &]] (* Robert G. Wilson v, Sep 28 2016 *)
Select[Range[126], IntegerQ[Prime[#]*(2 *Prime[#] - 1)/3] &] (* Robert Price, Apr 19 2025 *)

Formula

Integers k such that (prime(k) mod 3) mod 2 = 0. - Gary Detlefs, Dec 06 2011