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A022885 Primes p(k) such that p(k) + p(k+3) = p(k+1) + p(k+2). 15
5, 7, 11, 13, 23, 37, 53, 73, 97, 101, 103, 109, 137, 157, 179, 191, 223, 251, 263, 307, 353, 373, 389, 409, 419, 433, 457, 479, 487, 541, 563, 571, 593, 683, 691, 701, 757, 809, 821, 853, 859, 877, 883, 911, 977, 1019, 1039, 1049, 1087, 1103 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

These are primes p for which the subsequent alternate prime gaps are equal, so (p(k+3)-p(k+2))/(p(k+1)-p(k)) = 1. It is conjectured that the most frequent alternate prime gaps ratio is one. - Andres Cicuttin, Nov 07 2016

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

EXAMPLE

Starting from 5, the four consecutive primes are 5, 7, 11, 13; and they satisfy 5 + 13 = 7 + 11. So 5 is in the sequence.

MATHEMATICA

Transpose[Select[Partition[Prime[Range[500]], 4, 1], First[#]+Last[#] == #[[2]]+#[[3]]&]][[1]] (* Harvey P. Dale, May 23 2011 *)

PROG

(PARI) isok(p) = {my(k = primepi(p)); (p == prime(k)) && ((prime(k) + prime(k+3)) == (prime(k+1) + prime(k+2))); } \\ Michel Marcus, Jan 15 2014

(MAGMA) [NthPrime(n): n in [1..200] | (NthPrime(n)+NthPrime(n+3)) eq (NthPrime(n+1)+NthPrime(n+2))]; // Vincenzo Librandi, Nov 08 2016

CROSSREFS

Cf. A022884, A260179.

Sequence in context: A216736 A050541 A098865 * A176831 A263467 A200143

Adjacent sequences:  A022882 A022883 A022884 * A022886 A022887 A022888

KEYWORD

nonn

AUTHOR

Clark Kimberling

EXTENSIONS

Name edited by Michel Marcus, Jan 15 2014

STATUS

approved

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Last modified October 17 22:26 EDT 2019. Contains 328134 sequences. (Running on oeis4.)