

A022885


Primes p=prime(k) such that prime(k) + prime(k+3) = prime(k+1) + prime(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
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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

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 Vincenzo Librandi)


FORMULA

a(n) = A000040(A022884(n)).  Amiram Eldar, May 06 2020


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, A261470.
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



