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Positions of 1 in A283963; complement of A283964.
3

%I #6 Mar 26 2017 11:51:16

%S 1,3,5,6,8,10,11,13,15,17,19,20,22,24,25,27,29,31,33,34,36,38,39,41,

%T 43,44,46,48,49,51,53,55,57,58,60,62,63,65,67,69,71,72,74,76,77,79,81,

%U 82,84,86,87,89,91,93,95,96,98,100,101,103,105,107,109,110

%N Positions of 1 in A283963; complement of A283964.

%C Conjecture: -1 < n*sqrt(3) - a(n) < 2 for n >= 1.

%H Clark Kimberling, <a href="/A283965/b283965.txt">Table of n, a(n) for n = 1..10000</a>

%e The first 10 letters of the word in A282963 are 1010110101, in which the positions of 1 are 1,3,5,6,8.

%t s = Nest[Flatten[# /. {0 -> {1}, 1 -> {1, 0, 1, 0}}] &, {0}, 10] (* A283963 *)

%t Flatten[Position[s, 0]] (* A283964 *)

%t Flatten[Position[s, 1]] (* A283965 *)

%Y Cf. A283963, A283964.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Mar 25 2017