OFFSET
1,1
COMMENTS
Primes p such that p, p+6, p+12, p+18 are consecutive primes.
It is conjectured that there exist arbitrarily long sequences of consecutive primes in arithmetic progression. As of March 2013 the record is 10 primes.
Note that the Green and Tao reference is about arithmetic progressions that are not necessarily consecutive. - Michael B. Porter, Mar 05 2013
Subsequence of A023271. - R. J. Mathar, Nov 04 2006
All terms p == 1 (mod 10) and hence p+24 are always divisible by 5. - Zak Seidov, Jun 20 2015
Subsequence of A054800, with which is coincides up to a(24), but a(25) = A054800(26). - M. F. Hasler, Oct 26 2018
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
Jens Kruse Andersen, The Largest Known CPAP's
Ben Green and Terence Tao, The primes contain arbitrarily long arithmetic progressions, arXiv:math/0404188 [math.NT], 2004-2007.
B. Green and T. Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Math. 167(2008), 481-547.
OEIS wiki, Consecutive primes in arithmetic progression: CPAP with given gap, updated Jan. 2020
FORMULA
EXAMPLE
251, 257, 263, 269 are consecutive primes: 257 = 251 + 6, 263 = 251 + 12, 269 = 251 + 18.
MAPLE
N:=10^5: # to get all terms <= N.
Primes:=select(isprime, [seq(i, i=3..N+18, 2)]):
Primes[select(t->[Primes[t+1]-Primes[t], Primes[t+2]-Primes[t+1],
Primes[t+3]-Primes[t+2]]=[6, 6, 6], [$1..nops(Primes)-3])]; # Muniru A Asiru, Aug 04 2017
MATHEMATICA
A033451 = Reap[ For[p = 2, p < 100000, p = NextPrime[p], p2 = NextPrime[p]; If[p2 - p == 6, p3 = NextPrime[p2]; If[p3 - p2 == 6, p4 = NextPrime[p3]; If[p4 - p3 == 6, Sow[p]]]]]][[2, 1]] (* Jean-François Alcover, Jun 28 2012 *)
Transpose[Select[Partition[Prime[Range[16000]], 4, 1], Union[ Differences[ #]] == {6}&]][[1]] (* Harvey P. Dale, Jun 17 2014 *)
PROG
(PARI) p=2; q=3; r=5; forprime(s=7, 1e4, if(s-p==18 && s-q==12 && s-r==6, print1(p", ")); p=q; q=r; r=s) \\ Charles R Greathouse IV, Feb 14 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved