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
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
(Python)
from sympy import nextprime
from itertools import islice
def agen(): # generator of terms
p, q, r, s = [2, 3, 5, 7]
while True:
if p + s == q + r: yield p
p, q, r, s = q, r, s, nextprime(s)
print(list(islice(agen(), 50))) # Michael S. Branicky, May 31 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Name edited by Michel Marcus, Jan 15 2014
STATUS
approved