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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 should be enough to get the necessity of the conjecture).

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 Hanson & Pledger's theorem.

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 $(\ast)$.

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 $T$ - I would have entirely proved it if I knew that the answer to to this question were yes).

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 Hanson & Pledger's theorem.

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 should be enough to get the necessity of the conjecture).

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 Hanson & Pledger's theorem.

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 ?

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 $T$ - I would have entirely proved it if I knew that the answer to to this question were yes).

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 Hanson & Pledger's theorem.