A219096 Indices of primes p such that the next two larger primes are p+6 and p+12.

15, 36, 39, 54, 55, 73, 102, 107, 110, 118, 129, 160, 164, 184, 187, 194, 199, 218, 271, 272, 291, 339, 358, 387, 419, 426, 464, 465, 508, 520, 553, 605, 621, 629, 667, 682, 683, 702, 709, 710, 733, 761, 791, 813, 821, 822, 829, 830, 882, 896, 952, 962, 988

Offset: 1

Author

Clark Kimberling, Mar 05 2013

Keywords

• nonn
• easy

Comments

The primes themselves are given by A047948. Conjecture: if k == 0 mod 6 then there exists a prime p such that p-k, p, p+k are consecutive primes. (This would follow from a proof of Dickson's conjecture; see the Comments and References at A186311.)

Examples

			a(1) = 15 since p(15), p(16), p(17) are consecutive primes (47, 53, 59) with common difference 6: 53 - 47 = 6, and 59 - 53 = 6.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000
• Consecutive primes in arithmetic progression

Crossrefs

Cf. A047948, A128940, A000720.

Programs

Mathematica

z = 1000; t = Differences[Prime[Range[z]]];
f[n_] := If[t[[n + 1]] - t[[n]] == 0, t[[n]], 0]
u = Table[f[n], {n, 1, z - 2}]; s = Flatten[Position[u, 6]] (*A219096*)
Prime[s]  (*A047948*)
Flatten[Position[Partition[Prime[Range[1000]],3,1],?(Differences[#] == {6,6}&),{1},Heads->False]] (* _Harvey P. Dale, Jan 01 2015 *)

PARI

n=0;p=2;q=3;forprime(r=5,1e6,n++;if(r-p==12&&q-p==6,print1(n", "));p=q;q=r) \\ Charles R Greathouse IV, Mar 05 2013

Formula

a(n) = A000720(A047948(n)). - M. F. Hasler, Mar 11 2013