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Numbers k such that (3*k^2 + 1)/4 is prime.
7

%I #50 Dec 09 2023 15:28:56

%S 3,5,7,9,13,19,21,23,27,29,35,47,49,51,55,57,61,65,69,75,77,83,85,91,

%T 97,99,105,111,117,125,127,133,135,149,161,163,173,177,181,183,187,

%U 191,211,217,239,247,251,257,259,273,281,285,295,307,313,315,317,329,331,341

%N Numbers k such that (3*k^2 + 1)/4 is prime.

%H Michel Marcus, <a href="/A121259/b121259.txt">Table of n, a(n) for n = 1..8144</a>

%F a(n) = sqrt((4*A002407(n) - 1)/3). [corrected by _Rémi Guillaume_, Dec 07 2023]

%F a(n) = 2*A002504(n) - 1. - _Hugo Pfoertner_, Oct 07 2023

%F a(n) = 2*A111251(n) + 1. - _Rémi Guillaume_, Dec 06 2023

%e (3*5^2 + 1)/4 = 19 is the 2nd prime of this form, so a(2) = 5.

%e (3*13^2 + 1)/4 = 127 is the 5th prime of this form, so a(5) = 13.

%e (3*19^2 + 1)/4 = 271 is the 6th prime of this form, so a(6) = 19.

%t Select[Range[400],PrimeQ[(3#^2+1)/4]&] (* _Harvey P. Dale_, Mar 24 2011 *)

%o (PARI) is(n) = my(p=(3*n^2+1)/4); (denominator(p)==1) && isprime(p); \\ _Charles R Greathouse IV_, Jun 13 2017; edited by _Michel Marcus_, Oct 12 2023

%Y Cf. comment by Michael Somos in A002407.

%Y Cf. A002504, A111251, A111051 (simpler variant).

%K nonn,easy

%O 1,1

%A _Zak Seidov_, Aug 23 2006