A270788 Unique fixed point of the 3-symbol Fibonacci morphism phi-hat_2.

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

Offset: 1

Author

N. J. A. Sloane, Mar 30 2016

Keywords

• nonn
• easy

Comments

Fixed point of the morphism phi-hat_2 given by 1 --> 12, 2 --> 3, 3 --> 12. [Joerg Arndt, Apr 10 2016]

This sequence is the [0->12, 1->3]-transform of the Fibonacci word A003849: if T(0):=12, T(1):=3, then one proves easily with induction that T(phi_1^n(0)) = phi-hat_2^{n+1}(1), and T(phi_1^n(1)) = phi-hat_2^{n+1}(2), where phi_1 denotes the Fibonacci morphism given by 0 --> 01, 1 --> 0. - Michel Dekking, Dec 29 2019

Links

• Joerg Arndt, Table of n, a(n) for n = 1..1000
• F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
• Shuo Li, Zeckendorf expansion, Dirichlet series and infinite series involving the infinite Fibonacci word, arXiv:2106.05672 [math.NT], 2021.

Crossrefs

Cf. A159917 (same sequence if we map 1->2, 2->0, 3->1).

Cf. A000201, A001950, A003849.

Programs

Maple

with(ListTools);
psi:=proc(S)
Flatten(subs( {1=[1,2], 2=[3], 3=[1,2]}, S));
end;
S:=[1];
for n from 1 to 10 do S:=psi(S): od:
S;

Mathematica

m = 121; (* number of terms required *)
S[1] = {1};
S[n_] := S[n] = SubstitutionSystem[{1 -> {1, 2}, 2 -> {3}, 3 -> {1, 2}}, S[n-1]];
For[n = 2, True, n++, If[PadRight[S[n], m] == PadRight[S[n-1], m], Print["n = ", n]; Break[]]];
Take[S[n], m] (* Jean-François Alcover, Feb 15 2023 *)

Python

from math import isqrt
def A270788(n): return (1,3,2)[((m:=(n+2+isqrt(5*(n+2)**2)>>1)-n-2)+isqrt(5*m**2)>>1)-((k:=(n+1+isqrt(5*(n+1)**2)>>1)-n-1)+isqrt(5*k**2)>>1)] # Chai Wah Wu, May 22 2025

Formula

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. Then a(n)=1 if n=A(A(k)) for some k; a(n)=2 if n=B(k) for some k; a(n)=3 if n=A(B(k)) for some k. - Michel Dekking, Dec 27 2016

Extensions

More terms from Joerg Arndt, Apr 10 2016

Offset changed to 1 by Michel Dekking, Dec 27 2016