OFFSET

1,1

COMMENTS

From Zak Seidov, Apr 19 2008: (Start)

Proof that all numbers in this sequence are divisible by 3:

if n=(3k+1), then 8n+7=8(3k+1)+7=3(5+8 k) (composite)

if n=(3k+2), then 8n+5=8(3k+2)+5=3(7+8 k) (composite),

so if we require that both 8n+5 and 8n+7 are primes, then n=3k, hence all terms in this sequence are multiples of 3. QED. (End)

LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000

MATHEMATICA

Do[If[PrimeQ[8n + 5] && PrimeQ[8n + 7], Print[n]], {n, 1, 1000}]

Select[Range[3, 6000, 3], AllTrue[8#+{5, 7}, PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Sep 14 2018 *)

CROSSREFS

KEYWORD

nonn

AUTHOR

Artur Jasinski, Dec 10 2006

STATUS

approved