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a(n) = A256557(n)/A166133(n+1), n>=3.
2

%I #13 Apr 02 2015 11:36:04

%S 5,1,9,8,7,2,13,12,11,5,17,16,15,8,22,21,20,19,12,25,24,7,170,29,28,

%T 27,16,40,39,38,37,36,35,34,33,32,31,21,39,40,41,42,43,44,45,46,47,48,

%U 49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,23,505,81,80,79,78,77,76,75,74,73

%N a(n) = A256557(n)/A166133(n+1), n>=3.

%C Let A166133 = B; A166133 is defined as: After b(1)=1, b(2)=2, and b(3)=4, b(n+1) is the smallest divisor of b(n)^2-1 that has not yet appeared in the sequence.

%C A256557(n) = A166133(n)^2-1. Therefore, a(n) = (A166133(n)^2-1)/A166133(n+1), n>=3; that is, a(n) is A256557(n) divided by the smallest divisor of A166133(n+1)^2-1 which has not yet appeared in A166133. For example, a(12) = 5 means that 5 is A256557(12) = A166133(12)^2-1 = 80 divided its smallest divisor which has not yet appeared in A166133 (i.e., 16).

%e a(13) = 17 because A256557(13)/A166133(14) = 255/15 = 17.

%t lim = 200; s = {1, 2, 4}; Do[d = Divisors[Last[s]^2 - 1]; i = 1; While[i <= Length[d] && MemberQ[s, d[[i]]], i++]; s = Append[s, d[[i]]], {lim}]; a166133 = Table[s[[k]], {k, 1, lim}]; a256557 = #^2 - 1 & /@ a166133; t = PadLeft[Most@a256557, lim]; Drop[t/a166133, 3] (* _Michael De Vlieger_, Apr 02 2015, after _Hans Havermann_ at A166133 *)

%Y Cf. A166133, A256557.

%K nonn

%O 3,1

%A _Bob Selcoe_, Apr 01 2015