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%I #12 Oct 14 2019 00:32:16
%S 1,8,12,15,20,23,28,29,36,39,44,45,52,55,57,64,68,71,76,77,84,87,89,
%T 96,100,103,108,109,113,120,124,127,132,135,140,141,148,151,153,160,
%U 164,167,172,173,177,184,188,191,196,199,204,205,212,215,217,224,225
%N Positions of 0 in A101666.
%C 4n - a(n) is in {0,1,3} for n >= 1. The first 20 numbers 4n - a(n) are 3, 0, 0, 1, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 3, 0, 0, 1, 0, 3. In the sequence (4n - a(n)), let r be the sequence of positions occupied by 3; and likewise, s for 0 and t for 1; then r = A287373, s = A287374, and t = A287375.
%C From _Michel Dekking_, Oct 13 2019: (Start)
%C Proof of these conjectures: Let b(n):=4n-a(n). Note first that a simple way to express the r,s,t property above is to say that the sequence b is obtained from the sequence x := A101666 = 0,1,1,2,1,2,1,0,1,2,1,0,1,..., by changing the alphabet from {0,1,2} to {3,0,1}.
%C Let sigma be the defining morphism for x:
%C sigma(0) = 01, sigma(1) = 12, sigma(2) = 10.
%C Then sigma^2 is given by
%C sigma^2(0) = 0112, sigma^2(1) = 1210, sigma^2(2) = 1201.
%C Since sigma^2(x) = x, we immediately see from this that the 0's occur in x at positions 0, 1 and 3 modulo 4.
%C Moreover, we have
%C b(n) = 3 iff 4n - a(n) = 3 iff a(n) = 1 mod 4 iff
%C x(4n+1,...,4n+3) = 0112 iff x(n) = 0.
%C Similarly we obtain b(n) = 0 iff x(n) = 1, and b(n) = 1 iff x(n) = 2. This proves that b is obtained from x by changing the alphabet from {0,1,2} to {3,0,1}.
%C (End)
%H Clark Kimberling, <a href="/A287373/b287373.txt">Table of n, a(n) for n = 1..10000</a>
%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 2}, 2 -> {1,0}}] &, {0}, 12] (* A101666 *)
%t Flatten[Position[s, 0]] (* A287373 *)
%t Flatten[Position[s, 1]] (* A287374 *)
%t Flatten[Position[s, 2]] (* A287375 *)
%Y Cf. A287372, A287374, A287375.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, May 25 2017
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