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

1,2

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

Table of n, a(n) for n=1..112.

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.

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.

Crossrefs

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

Sequence in context: A190595 A338863 A280502 * A290397 A322425 A053841

Adjacent sequences: A276754 A276755 A276756 * A276758 A276759 A276760

Keyword

nonn

Author

Michel Dekking, Sep 17 2016

Status

approved