A078961 Primes p such that the differences between the 5 consecutive primes starting with p are (6,4,2,4).

31, 1291, 1861, 1987, 2677, 4507, 5641, 7867, 13681, 17377, 24097, 35521, 42451, 44257, 55807, 80671, 88651, 88801, 93481, 110557, 113011, 113161, 118891, 134581, 155371, 163981, 198817, 221707, 234181, 266671, 269377, 284731, 290611, 313981, 331537, 332461, 344161

Offset: 1

Author

Labos Elemer, Dec 19 2002

Keywords

• nonn

Comments

Equivalently, primes p such that p, p+6, p+10, p+12 and p+16 are consecutive primes.

Examples

			31 is a term since 31, 37 = 31 + 6, 41 = 31 + 10, 43 = 31 + 12 and 47 = 31 + 16 are consecutive primes.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Crossrefs

Subsequence of A078855. - R. J. Mathar, May 06 2017

Cf. A001223, A078866, A078867, A078946-A078971, A022006, A022007.

Programs

Mathematica

Transpose[Select[Partition[Prime[Range[26000]],5,1],Differences[#]=={6,4,2,4}&]][[1]] (* Harvey P. Dale, Aug 26 2014 *)

PARI

list(lim) = {my(p1 = 2, p2 = 3, p3 = 5, p4 = 7); forprime(p5 = 11, lim, if(p2 - p1 == 6 && p3 - p2 == 4 && p4 - p3 == 2 && p5 - p4 == 4, print1(p1, ", ")); p1 = p2; p2 = p3; p3 = p4; p4 = p5);} \\ Amiram Eldar, Feb 22 2025

Formula

From Amiram Eldar, Feb 22 2025: (Start)

a(n) == 1 (mod 6).

a(n) == 1 or 7 (mod 30). (End)

Extensions

Edited by Dean Hickerson, Dec 20 2002