

A316188


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).


1



11, 17, 23, 41, 47, 53, 59, 83, 89, 107, 131, 167, 173, 179, 227, 251, 263, 269, 293, 311, 347, 353, 359, 383, 389, 419, 431, 443, 467, 479, 503, 509, 521, 557, 563, 587, 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
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OFFSET

1,1


COMMENTS

The term r*(r + 1)/2 is the triangular number A000217(r).
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.
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).


LINKS

Table of n, a(n) for n=1..67.


MATHEMATICA

lst = {}; Do[p = Prime[n]; f = False;
If[5 == Mod[p, 6], f = True;
Do[If[r == Mod[p  2, 1/2 r (1 + r)], f = False], {r, 5, p  3}]];
If[f, lst = AppendTo[lst, p]], {n, 4, 500}]; lst


CROSSREFS

Cf. A007528, A303379, A005385, A000217, A005384.
Sequence in context: A074207 A063638 A141250 * A096454 A099722 A031505
Adjacent sequences: A316185 A316186 A316187 * A316189 A316190 A316191


KEYWORD

nonn


AUTHOR

Ralf Steiner, Jun 26 2018


STATUS

approved



