A022885 - OEIS

%I #43 May 31 2024 10:35:34

%S 5,7,11,13,23,37,53,73,97,101,103,109,137,157,179,191,223,251,263,307,

%T 353,373,389,409,419,433,457,479,487,541,563,571,593,683,691,701,757,

%U 809,821,853,859,877,883,911,977,1019,1039,1049,1087,1103

%N Primes p=prime(k) such that prime(k) + prime(k+3) = prime(k+1) + prime(k+2).

%C 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

%H Amiram Eldar, <a href="/A022885/b022885.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 Vincenzo Librandi)

%F a(n) = A000040(A022884(n)). - _Amiram Eldar_, May 06 2020

%e 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.

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

%o (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

%o (Magma) [NthPrime(n): n in [1..200] | (NthPrime(n)+NthPrime(n+3)) eq (NthPrime(n+1)+NthPrime(n+2))]; // _Vincenzo Librandi_, Nov 08 2016

%o (Python)

%o from sympy import nextprime

%o from itertools import islice

%o def agen(): # generator of terms

%o     p, q, r, s = [2, 3, 5, 7]

%o     while True:

%o         if p + s == q + r: yield p

%o         p, q, r, s = q, r, s, nextprime(s)

%o print(list(islice(agen(), 50))) # _Michael S. Branicky_, May 31 2024

%Y Cf. A022884, A260179, A261470.

%K nonn

%O 1,1

%A _Clark Kimberling_

%E Name edited by _Michel Marcus_, Jan 15 2014