A287373 Positions of 0 in A101666.

1, 8, 12, 15, 20, 23, 28, 29, 36, 39, 44, 45, 52, 55, 57, 64, 68, 71, 76, 77, 84, 87, 89, 96, 100, 103, 108, 109, 113, 120, 124, 127, 132, 135, 140, 141, 148, 151, 153, 160, 164, 167, 172, 173, 177, 184, 188, 191, 196, 199, 204, 205, 212, 215, 217, 224, 225

Offset

1,2

Comments

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.

From Michel Dekking, Oct 13 2019: (Start)

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}.

Let sigma be the defining morphism for x:

sigma(0) = 01, sigma(1) = 12, sigma(2) = 10.

Then sigma^2 is given by

sigma^2(0) = 0112, sigma^2(1) = 1210, sigma^2(2) = 1201.

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.

Moreover, we have

b(n) = 3 iff 4n - a(n) = 3 iff a(n) = 1 mod 4 iff

x(4n+1,...,4n+3) = 0112 iff x(n) = 0.

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}.

(End)

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 2}, 2 -> {1, 0}}] &, {0}, 12] (* A101666 *)
Flatten[Position[s, 0]]  (* A287373 *)
Flatten[Position[s, 1]]  (* A287374 *)
Flatten[Position[s, 2]]  (* A287375 *)

Crossrefs

Cf. A287372, A287374, A287375.

Sequence in context: A180690 A194592 A175132 * A031034 A152758 A160258

Adjacent sequences: A287370 A287371 A287372 * A287374 A287375 A287376

Keyword

nonn,easy

Author

Clark Kimberling, May 25 2017

Status

approved