%I
%S 3,6,7,9,12,15,16,18,21,24,27,28,30,33,36,37,39,42,45,48,49,51,54,57,
%T 58,60,63,66,67,69,72,75,78,79,81,84,87,88,90,93,96,99,100,102,105,
%U 108,109,111,114,117,118,120,123,126,129,130,132,135,138,139,141
%N Positions of 0 in A286801; complement of A286803.
%C a(n)  a(n1) is in {1,2,3} for n>=2, and a(n)/n > (12 + 3*sqrt(2))/7.
%H Clark Kimberling, <a href="/A286802/b286802.txt">Table of n, a(n) for n = 1..10000</a>
%e As a word, A286801 = 11011001011011001011011..., in which 0 is in positions 3,6,7,9,12,...
%t s = Nest[Flatten[# /. {0 > {0, 0, 1}, 1 > {0}}] &, {0}, 16]; (* Pell word, A171588 *)
%t w = StringJoin[Map[ToString, s]];
%t w1 = StringReplace[w, {"0" > "110", "1" > "010"}];
%t st = ToCharacterCode[w1]  48 ; (* A286801 *)
%t Flatten[Position[st, 0]]; (* A286802 *)
%t Flatten[Position[st, 1]]; (* A286803 *)
%Y Cf. A171588, A286801, A286803.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, May 16 2017
