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Consider the "Chacon words": $B_0=0$ and $B_{n+1} = B_nB_n1B_n$. The word $B_n$ has $\ell_n := \frac{3^{n+1}-1}{2}$ digits and the number of $1$'s in $B_n$ is $\ell_n - 3^n = \ell_{n-1} \sim \ell_n/3$. There's also a word $B_\infty$ since $B_{n+1}$ is an extension of $B_n$.

While dealing with my own research, I discovered the following conjecture. Let $S \subset \mathbb{N}$ be $S = j\mathbb{Z} + k$ (density $d(S)=1/j$) or $S=\bigl\{n \mid \{n\gamma\} \in [a,b)\bigr\}$ (density $d(S)=b-a$, where $0 \leq a < b \leq 1$ and $\gamma \in (0,1)$ is irrational). Then: $$ \frac{\#\{i \mid 1 \leq i \leq N, i \in S, B_\infty(i)=1 \}}{N} \underset{N \to \infty}{\longrightarrow} \frac{1}{3}d(S). $$ Do you know a proof of this result ?

Update

Let me develop how I got this conjecture. According to Theorem 1 of this paper by Hanson & Pledger, the conjecture is equivalent to $$ (\ast)\colon \quad \frac{1}{\ell_n} \sum_{\substack{k=0 \\ B_n(k+1)=1}}^{\ell_n-1} f(T^k\cdot) \overset{L^2}{\longrightarrow} \frac{1}{3}E(f \mid {\cal I}) $$ for every invertible mpt $T$, every $f \in L^2$, and where ${\cal I}$ is the $T$-invariant $\sigma$-field.

In fact, I have almost proved $(\ast)$ (I have only proved it for ergodic $T$ - but this is pretty enough to get the necessity of the conjecture : the conjecture follows for $(\ast)$ applied to periodic transformations on $\mathbb{Z}/j\mathbb{Z}$ and to ergodic rotations).

Note that $(\ast)$ is very close to an application of the ergodic theorem. By the ergodic theorem applied to $T \times C$ where $C$ is the Chacon map and $T$ is ergodic (hence $T \times C$ is always ergodic), we have $$ (\#)\colon\quad \frac{1}{\ell_n} \sum_{k=0}^{\ell_n-1} f(T^kx){\boldsymbol 1}_{C^k u \in [2/3,1[} \longrightarrow \frac{1}{3}E(f) \quad \text{for almost all $(x,u)$}, $$ and for $u=0$ one has $$ \frac{1}{\ell_n} \sum_{k=0}^{\ell_n-1} f(T^kx){\boldsymbol 1}_{C^k u \in [2/3,1[} = \frac{1}{\ell_n} \sum_{\substack{k=0 \\ B_n(k+1)=1}}^{\ell_n-1} f(T^kx). $$ But I don't know if something allows us to apply $(\#)$ for $u=0$. Anyway I wonder whether there is a proof of the conjecture without using $(\ast)$.

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  • $\begingroup$ Perhaps I haven't understood the question properly: what happens if we take $S$ to be the set of all $i$ such that $B_\infty(i)=0$? $\endgroup$
    – Ian Morris
    Commented Mar 9, 2016 at 18:11
  • $\begingroup$ Do you mean something unusual by "natural density?" Are you only considering sets that are natural in the sense that they are not chosen to make this conjecture fail? $\endgroup$
    – Douglas Zare
    Commented Mar 9, 2016 at 18:27
  • $\begingroup$ @DouglasZare en.wikipedia.org/wiki/Natural_density $\endgroup$
    – Stéphane Laurent
    Commented Mar 9, 2016 at 18:51
  • $\begingroup$ @IanMorris Thanks for this remark. Indeed ! I need to do a correction. $\endgroup$
    – Stéphane Laurent
    Commented Mar 9, 2016 at 18:51
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    $\begingroup$ The Chacon map is uniquely ergodic and therefore by the comment before the product of the Chacon map with an irrational rotation is uniquely ergodic. This implies that for any Riemann integrable function $g$ on $[0,1]\times [0,1]$ the Birkhoff averages (w.r.t. $C\times T$ in your notation) converges at every point. Since the function you are considering is Riemann integrable then (#) holds for all $(x,u)$. $\endgroup$
    – user78465
    Commented Mar 17, 2016 at 17:40

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