A272201 Bisection of primes congruent to 1 modulo 3 (A002476), depending on the corresponding A001479 entry being congruent to 1 modulo 3 or not. Here the second case.

7, 31, 37, 67, 73, 79, 139, 151, 199, 211, 223, 229, 271, 307, 313, 337, 367, 397, 421, 439, 457, 541, 547, 571, 577, 613, 643, 709, 739, 751, 823, 829, 853, 877, 907, 919, 997

Offset: 1

Author

Wolfdieter Lang, Apr 28 2016

Keywords

• nonn
• easy

Comments

The other primes congruent to 1 modulo 3 are given in A272200, where also more details are given.

Each prime == 1 (mod 3) has a unique representation A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) = A001479(m+1) and B(m) = A001480(m+1), m >= 1 (see also A001479). The present sequence gives these primes corresponding to A(m+1) not congruent 1 modulo 3. The ones corresponding to A(m+1) == 1 (mod 3) (the complement) are given in A272200.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000727, A001479, A002476, A001480, A272198, A272200 (complement relative to A002476).

Programs

Maple

filter:= proc(n) local S,x,y;
    if not isprime(n) then return false fi;
    S:= remove(hastype,[isolve(x^2+3*y^2=n)],negative);
    subs(S[1],x) mod 3 <> 1
end proc:
select(filter, [seq(i,i=7..1000,6)]); # Robert Israel, Apr 29 2019

Mathematica

filterQ[n_] := Module[{S, x, y}, If[!PrimeQ[n], Return[False]]; S = Solve[x > 0 && y > 0 && x^2 + 3 y^2 == n, Integers]; Mod[x /. S[[1]], 3] != 1];
Select[Range[7, 1000, 6], filterQ] (* Jean-François Alcover, Apr 21 2020, after Robert Israel *)

Formula

This sequence collects the 1 (mod 3) primes p(m) = A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) not == 1 (mod 3), for m >= 1. A(m) = A001479(m+1).