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Primes of the form (1 + k + k^3)/3.
2

%I #10 Feb 03 2018 09:53:24

%S 23,337,2293,3557,9941,21347,39233,87403,251221,333367,895253,1080647,

%T 1217473,1696207,2076563,2626933,2999707,4493837,6203297,6857033,

%U 8045953,8299127,8821297,11442817,13807361,14538187,17298497,21333467

%N Primes of the form (1 + k + k^3)/3.

%C k must be congruent to 1 (mod 3) for (k^3 + k + 1)/3 to be an integer. - _Michael B. Porter_, Apr 07 2010

%H J. Mulder, <a href="/A163512/b163512.txt">Table of n, a(n) for n = 1..40000</a>

%e (1 + 4 + 4^3)/3 = 23.

%t f[n_]:=(1+n+n^3)/3; lst={};Do[If[PrimeQ[f[n]],AppendTo[lst,f[n]]],{n,7!}];lst

%t Select[Table[(n^3+n+1)/3,{n,400}],PrimeQ] (* _Harvey P. Dale_, Aug 28 2012 *)

%o (PARI) for(n=0,200,m=3*n+1;if(isprime((m^3+m+1)/3),print((m^3+m+1)/3))) \\ _Michael B. Porter_, Apr 07 2010

%K nonn

%O 1,1

%A _Vladimir Joseph Stephan Orlovsky_, Jul 29 2009