A276757 Infinite Fibonacci word on the alphabet {1,2,3,4,5}.

1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2

Offset: 1

Author

Michel Dekking, Sep 17 2016

Keywords

• nonn

Comments

Start with the infinite Fibonacci word A003849, which is 0100101001001010010... and replace each 0 by 1,2,3 and each 1 by 4,5.

The unique fixed point of the 4-block Fibonacci substitution 1 -> 12, 2 -> 3, 3 -> 45, 4 -> 12, 5 -> 3. Here the 4-blocks are coded as 0100 <-> 1, 1001 <-> 2, 0010 <-> 3, 0101 <-> 4, 1010 <-> 5.

Links

• F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
• Michel Dekking and Michael Keane, On the conjugacy class of the Fibonacci dynamical system, arXiv preprint arXiv:1608.04487 [math.DS], 2016.

Crossrefs

Cf. A000201, A001950, A003849, A134859, A035336, A003623, A134860, A101864, A270788, A138967.

Formula

Let A(n) = floor(n*phi), B(n) = n + floor(n*phi), i.e., A and B are the lower and upper Wythoff sequences, A = A000201, B = A001950. Then a(n) = 1 if n = A(A(A(k))) for some k; a(n) = 2 if n = B(A(k)) for some k; a(n) = 3 if n = A(B(k)) for some k; a(n) = 4 if n = A(A(B(k))) for some k; a(n) = 5 if n = B(B(k)) for some k.