|
%I #36 Jul 03 2018 18:35:57
%S 11,17,23,41,47,53,59,83,89,107,131,167,173,179,227,251,263,269,293,
%T 311,347,353,359,383,389,419,431,443,467,479,503,509,521,557,563,587,
%U 593,599,647,683,719,761,773,797,809,839,863,881,887,929,941,983,1013,1019,1031,1049,1061,1103,1109,1151,1187,1193,1223,1229,1259,1283,1301
%N Primes p >= 7 such that p == 5 (mod 6) and for all r in the range 5 <= r < p - 2 we have p - 2 != r (mod r*(r + 1)/2).
%C The term r*(r + 1)/2 is the triangular number A000217(r).
%C All safe primes q > 7 (A005385) are included. Proof (indirect): Assume a safe prime 7 < q = 2*p + 1 solves 5 <= r < q - 2, q - 2 == r (mod r*(r + 1)/2), q = 6*t - 1. This yields to a Sophie Germain prime (A005384) p = (1/4)*(1 + r)*(2 + r*t). But this is composite in all possible cases for r, t. QED.
%C The set {a(n)} is one of four disjunct classes of primes p >= 7 dependent on all cases of the two conditions p == [1 | 5] (mod 6) and if [at least one | not any] r in the range 5 <= r < p - 2 exists with p - 2 == r (mod r*(r + 1)/2).
%t lst = {}; Do[p = Prime[n]; f = False;
%t If[5 == Mod[p, 6], f = True;
%t Do[If[r == Mod[p - 2, 1/2 r (1 + r)], f = False], {r, 5, p - 3}]];
%t If[f, lst = AppendTo[lst, p]], {n, 4, 500}]; lst
%Y Cf. A007528, A303379, A005385, A000217, A005384.
%K nonn
%O 1,1
%A _Ralf Steiner_, Jun 26 2018
|
