OFFSET
1,1
COMMENTS
Since a(n) and a(n)+8 are consecutive primes either a(n)+7 or a(n)+1 is a multiple of 100; in addition a(n) must have the form 6k-1. Therefore, every century spanned by a(n) and a(n)+8 is a multiple of 300. It appears that every multiple of 3 occurs as the difference round((a(n+1)-a(n))/100); all multiples of 3 through 432 occur as these differences for a(n) < 1000000000.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
MATHEMATICA
a287049[n_] := Map[Last, Select[Map[{NextPrime[#, 1], NextPrime[#, -1]}&, Range[100, n, 100]], First[#]-Last[#]==8&]]
a287049[60000] (* data *)
Select[Partition[Prime[Range[6100]], 2, 1], #[[2]]-#[[1]]==8&&Floor[#[[1]]/ 100] != Floor[#[[2]]/100]&][[All, 1]] (* Harvey P. Dale, Oct 02 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Hartmut F. W. Hoft, May 18 2017
STATUS
approved