# Is the density of 1's in the Fibonacci word uniform?

The Fibonacci word is the limit of the sequence of words starting with $$0$$ and satisfying rules $$0 \to 01, 1 \to 0$$. Equivalently, it is obtained from the recursion $$S_n= S_{n-1}S_{n-2}$$ under initial conditions $$S_0 = 0, S_1 = 01$$. Let us denote this Fibonacci word as $$f_k\in\{0,1\}^{\mathbb Z_+}$$.

We define the Fibonacci subshift $$\Omega$$ as the set of all two-sided right limits of the Fibonacci word. More precisely, we extend our Fibonacci word to $$\{0,1\}^{\mathbb Z}$$ by defining $$\{f_j\}_{j=-\infty}^0$$ arbitrarily, and define $$(\mathcal Tf)_k=f_{k+1}$$. The Fibonacci subshift $$\Omega$$ is then the set of limit points of $$\{\mathcal T^n f\}_{n=0}^\infty$$ under the product topology on $$\{0,1\}^{\mathbb{Z}}$$.

It is easy to show that the ratio of zeroes to ones in the Fibonacci word is $$\varphi$$, the golden ratio. Equivalently, the density of ones in the Fibonacci word (that is, the ratio of the number of ones in a subword to the length of the subword) tends to $$\frac{1}{\varphi^2}$$. I would like to know if this is true uniformly. That is,

Is the following true for a.e. $$\omega\in\Omega$$? Given any $$c\in\mathbb Z$$, $$\lim_{m\to\infty} \sum_{j=c-m}^{c+m}\frac{\omega_j}{2m+1}=\frac{1}{\varphi^2}$$ uniformly in $$c$$.

I also suspect this might be true for every $$\omega\in\Omega$$, not just a.e.

• Another answer that the one by David Speyer below is to say that the Fibonacci word is the sequence of 0's and 1's obtained from the orbit of 0 under the dynamical system $R_\phi(t)=t+\phi\bmod 1$, where $\phi$ is the golden mean; $I_0=[0,\phi-1)$ and $I_1=[\phi-1,1)$ (so that the sequence of 0's and 1's records the order in which the order visits the intervals $I_0$ and $I_1$). The uniformity you are asking for follows from unique ergodicity of irrational circle rotations. – Anthony Quas Feb 20 '19 at 6:23

Yes. By Proposition 2.1.10 in Lothaire, Algebraic Combinatorics on Words, if $$u$$ is any substring of the Fibonacci word then $$\left| \frac{\mbox{number of 1's in u}}{\mbox{length of u}} - \frac{1}{\phi^2} \right| \leq \frac{1}{\mbox{length of u}}.$$ Any length $$n$$ substring of any $$\omega \in \Omega$$ is also a length $$n$$ substring of the Fibonacci word, so the same bound applies to it.