%I #15 Mar 16 2022 05:50:48
%S 2,4,5,7,9,10,12,14,16,17,19,21,22,24,26,28,29,31,33,34,36,38,39,41,
%T 43,45,46,48,50,51,53,55,57,58,60,62,63,65,67,68,70,72,74,75,77,79,80,
%U 82,84,86,87,89,91,92,94,96,98,99,101,103,104,106,108,109
%N Positions of 1 in A286665; complement of A286666.
%C a(n) - a(n-1) is in {2,3} for n>=2. Conjecture: a(n)/n -> 1 + sqrt(1/2).
%C This conjecture follows easily from the fact that (a(n)) is a Beatty sequence, see my comments in A286665. - _Michel Dekking_, Mar 11 2018
%C Numbers with a positive number of trailing 0's in their minimal representation in terms of the positive Pell numbers (A317204). - _Amiram Eldar_, Mar 16 2022
%H Clark Kimberling, <a href="/A286667/b286667.txt">Table of n, a(n) for n = 1..10000</a>
%e As a word, A286665 = 010110101101010110101101..., in which 1 is in positions 2,4,5,7,9,...
%t s = Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, 6] (* A171588 *)
%t w = StringJoin[Map[ToString, s]]
%t w1 = StringReplace[w, {"0" -> "01"}]
%t st = ToCharacterCode[w1] - 48 ; (* A286665 *)
%t p0 = Flatten[Position[st, 0]]; (* A286666 *)
%t p1 = Flatten[Position[st, 1]]; (* A286667 *)
%Y Cf. A171588, A286665, A286666, A317204.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, May 13 2017