A285977 - OEIS

%I #8 Sep 22 2017 06:07:02

%S 2,3,5,7,8,10,12,13,15,17,19,20,22,23,25,27,28,30,32,33,35,36,38,40,

%T 42,43,45,47,48,50,52,53,55,57,59,60,62,63,65,67,69,70,72,74,75,77,79,

%U 80,82,83,85,87,88,90,92,93,95,97,99,100,102,103,105,107

%N Positions of 1 in A285975; complement of A285976.

%C Conjecture: a(n)/n -> 5/3.

%C This conjecture can be proved in the same way as for A284626, by showing that the frequency of the letter 1 in A285975 is equal to 3/5 - _Michel Dekking_, Sep 22 2017

%H Clark Kimberling, <a href="/A285977/b285977.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A285975 = 011010110101101010110110..., in which 0 is in positions 1,4,6,9,11,...

%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 9] (* Thue-Morse, A010060 *)

%t w = StringJoin[Map[ToString, s]]

%t w1 = StringReplace[w, {"00" -> "0"}]

%t st = ToCharacterCode[w1] - 48 (* A285975  *)

%t Flatten[Position[st, 0]]  (* A285976 *)

%t Flatten[Position[st, 1]]  (* A285977 *)

%Y Cf. A010060, A285975, A285976.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 06 2017