login
Positions of 1 in A285592; complement of A285593.
3

%I #4 Apr 30 2017 22:38:32

%S 1,3,5,7,8,10,12,14,16,18,20,22,24,26,27,29,31,33,35,37,39,41,43,45,

%T 47,49,50,52,54,56,58,60,62,64,66,68,69,71,73,75,76,78,80,82,83,85,87,

%U 89,91,93,95,97,99,101,102,104,106,108,110,112,114,116,118

%N Positions of 1 in A285592; complement of A285593.

%C Conjecture: a(n)/n -> (1+sqrt(4/5).

%H Clark Kimberling, <a href="/A285594/b285594.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A285568 = 1010101101..., in which 1 is in positions 1,3,5,7,8...

%t s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {0, 1, 1, 0}}] &, {0}, 9] (* A285568 *)

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

%t w1 = StringReplace[w, {"11" -> "1", "00" -> "0"}]

%t ss = ToCharacterCode[w1] - 48; (* A285592 *)

%t Flatten[Position[ss, 0]] (* A285593 *)

%t Flatten[Position[ss, 1]] (* A285594 *)

%Y Cf. A285568, A285592, A285593.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Apr 30 2017