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Positions of 1 in A287801; complement of A287802.
3

%I #4 Jun 04 2017 18:18:54

%S 1,6,7,10,15,16,21,22,25,30,31,34,39,40,45,46,49,54,55,60,61,64,69,70,

%T 73,78,79,84,85,88,93,94,97,102,103,108,109,112,117,118,123,124,127,

%U 132,133,136,141,142,147,148,151,156,157,162,163,166,171,172,175

%N Positions of 1 in A287801; complement of A287802.

%C Then d(n) is in {0, 2} for n >= 1. Indeed,

%C d(n) = 0 if n is in A001950, and d(n) = 1 if n is in A000201.

%H Clark Kimberling, <a href="/A287803/b287803.txt">Table of n, a(n) for n = 1..10000</a>

%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)

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

%t w1 = StringReplace[w, {"0" -> "100", "1" -> "001"}]

%t st = ToCharacterCode[w1] - 48 (* A287801 *)

%t Flatten[Position[st, 0]] (* A287802 *)

%t Flatten[Position[st, 1]] (* A287803 *)

%Y Cf. A000201, A001950, A287801, A287802.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jun 03 2017