A138967 Infinite Fibonacci word on the alphabet {1,2,3,4}.

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

Offset: 1

Author

Clark Kimberling, Apr 04 2008

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 1,4.

(a(n)) is the unique fixed point of the morphism 1->12, 2->3, 3->14, 4->3, obtained by coding the overlapping 3-block morphism of the Fibonacci morphism according to 010<->1, 100<->2, 001<->3, 101<->4. - Michel Dekking, Sep 28 2017

Links

• F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.

Crossrefs

Cf. A000201, A001950, A003849, A101864, A270788, A276757.

Formula

a(n) = 3 for n = 3, 8, 21, 55, ..., F(2*k), where k>1.

a(n) = 4 for n = 5, 13, 34, 89, ..., F(2*k+1), where k>1.

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