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

CROSSREFS

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 25 2017

STATUS

approved