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Positions of 0 in A283963; complement of A283965.
3

%I #6 Mar 26 2017 11:51:09

%S 2,4,7,9,12,14,16,18,21,23,26,28,30,32,35,37,40,42,45,47,50,52,54,56,

%T 59,61,64,66,68,70,73,75,78,80,83,85,88,90,92,94,97,99,102,104,106,

%U 108,111,113,116,118,120,122,125,127,130,132,134,136,139,141,144

%N Positions of 0 in A283963; complement of A283965.

%C Conjecture: -1 < n*r - a(n) < 2 for n >= 1, where r = (3+sqrt(3))/2.

%H Clark Kimberling, <a href="/A283964/b283964.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 0 are 2,4,7,9.

%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, A283965.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Mar 25 2017