A217603 Consider sets of 3 consecutive primes a, b, c such that c - a = 100, then sequence gives the values of b.

58831, 286927, 360653, 404941, 590489, 623107, 651587, 673747, 710119, 740801, 779413, 794831, 795427, 1040311, 1107269, 1185241, 1206869, 1320437, 1392007, 1568771, 1581829, 1599803, 1601953, 1613201, 1721081, 1744927, 1942273, 1951321, 1994299, 2024063

Offset: 1

Author

Zak Seidov, Oct 08 2012

Keywords

• nonn

Examples

			a(1) = 58831 because {58789, 58831, 58889} is the first set of 3 consecutive primes a, b, c with c-a=100.
a(2) = 286927 because {286873, 286927, 286973} is the second set of 3 consecutive primes a, b, c with c-a=100.
a(1000) = 23090087 because {23090059, 23090087, 23090159} is the 1000th set of 3 consecutive primes a, b, c with c-a=100.
		

Links

• Zak Seidov, Table of n, a, b, c for n=1..1000
• Zak Seidov, Table of n, a, b, c for n=1..1000

Crossrefs

Cf. A166251, A217561, A217566, A217577.

Programs

Mathematica

s = {}; a = 2; b = 3; c = 5; Do[If[c - a == 100, AppendTo[s, b]; Print[{a, b, c}]]; a = b; b = c; c = NextPrime[c], {10^5}]; s
Select[Partition[Prime[Range[151000]],3,1],#[[3]]-#[[1]]==100&][[;;,2]] (* Harvey P. Dale, Jul 22 2024 *)

PARI

p=2;q=3;forprime(r=5,1e6,if(r-p==100,print1(q", "));p=q;q=r) \\ Charles R Greathouse IV, Nov 14 2012