A287049 Prime p1 of consecutive primes p1, p2, where p2 - p1 = 8, and p1, p2 are in different centuries.

1193, 2699, 5399, 5693, 6599, 6899, 7499, 8093, 8699, 12899, 13799, 15299, 17099, 17393, 19793, 20399, 23993, 26099, 26399, 27893, 35099, 35393, 35999, 36299, 36599, 37493, 38699, 39293, 40499, 42299, 43793, 46499, 50093, 50993, 51599, 51899, 53093, 53993, 55799, 56393, 57593, 58199, 59399, 59699

Offset: 1

Author

Hartmut F. W. Hoft, May 18 2017

Keywords

• nonn

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

Crossrefs

Cf. A001359, A158277, A160370, A160440, A160500.

Programs

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 *)