ON THE SHORTEST DISTANCE FUNCTION IN CONTINUED FRACTIONS

Abstract

Let $x\in [0,1)$ be an irrational number and let $x=[a_{1}(x),a_{2}(x),\ldots ]$ be its continued fraction expansion with partial quotients $\{a_{n}(x): n\geq 1\}$. Given a natural number m and a vector $(x_{1},\ldots ,x_{m})\in [0,1)^{m},$ we derive the asymptotic behaviour of the shortest distance function

$$ \begin{align*} M_{n,m}(x_{1},\ldots,x_{m})=\max\{k\in \mathbb{N}: a_{i+j}(x_{1})=\cdots= a_{i+j}(x_{m}) \ \text{for}~ j=1,\ldots,k \mbox{ and some } i \mbox{ with } 0\leq i \leq n-k\}, \end{align*} $$

which represents the run-length of the longest block of the same symbol among the first n partial quotients of $(x_{1},\ldots ,x_{m}).$ We also calculate the Hausdorff dimension of the level sets and exceptional sets arising from the shortest distance function.

1 Introduction

Let $T:[0,1)\to [0,1)$ be the Gauss map defined by

$$ \begin{align*} T(0)= 0, \quad T(x)= \frac{1}{x} \pmod1 \quad\text{for } x \in (0,1).\end{align*} $$

Every irrational number $x\in [0,1)$ can be uniquely expanded into an infinite form

(1.1) $$ \begin{align} x:=\frac{1}{a_{1}(x)+\dfrac{1}{a_{2}(x)+\dfrac{1}{\ddots+\dfrac{1}{a_{n}+T^{n}(x)}}}} =\frac{1}{a_{1}(x)+\dfrac{1}{a_{2}(x)+\dfrac{1}{a_{3}(x)+\dfrac{1}{\ddots}}}}, \end{align} $$

where $a_{n}(x)=\lfloor 1/T^{n-1}(x)\rfloor $ are called the partial quotients of $x.$ (Here $\lfloor \cdot \rfloor $ denotes the greatest integer less than or equal to a real number and $T^0$ denotes the identity map.) For simplicity of notation, we write (1.1) as

$$ \begin{align*} x=[a_{1}(x),a_{2}(x),\ldots,a_{n}(x)+T^{n}(x)]=[a_{1}(x),a_{2}(x),a_{3}(x),\ldots]. \end{align*} $$

It is clear that the Gauss transformation T acts as the shift map on the continued fraction system. That is, for each $x=[a_1(x),a_2(x),a_3(x),\ldots ]\in [0,1)\cap \mathbb {Q}^{c},$

$$ \begin{align*}T(x)=T([a_{1}(x),a_{2}(x),a_3(x),\ldots])=[a_{2}(x),a_{3}(x),\ldots].\end{align*} $$

Gauss observed that T is measure-preserving and ergodic with respect to the Gauss measure $\mu $ defined by

$$ \begin{align*}d\mu=\frac{1}{\log 2}\frac{1}{x+1}dx.\end{align*} $$

For more information on the continued fraction expansion, see [3].

The metrical theory of continued fractions, which concerns the properties of the partial quotients for almost all $x\in [0,1),$ is one of the major themes in the study of continued fractions. Wang and Wu [7] considered the metrical properties of the maximal run-length function

$$ \begin{align*}R_{n}(x)=\max\{l\in \mathbb{N}: a_{i+1}(x)=\cdots= a_{i+l}(x) \ \text{for some}\ i \mbox{ with } 0\leq i \leq n-l\},\end{align*} $$

which counts the longest run of the same symbol among the first n partial quotients of x. They proved that, for $\mu $ almost all $x\in [0,1),$

$$ \begin{align*}\lim_{n\rightarrow\infty}\frac{R_{n}(x)}{\log n}=\frac{1}{2\log({(\sqrt{5}+1)}/{2})}.\end{align*} $$

Song and Zhou [6] gave a more subtle characterisation of the function $R_{n}(x).$ In this paper, we continue the study by considering the shortest distance function

$$ \begin{align*} M_{n,m}(x_{1},\ldots,x_{m})=\max\{k\in \mathbb{N}: a_{i+j}(x_{1}) = & \cdots= a_{i+j}(x_{m}) \ \text{for}~ j=1,\ldots,k, \\&\ \text{and some}\ i \mbox{ with } 0\leq i \leq n-k\}. \end{align*} $$

This is motivated by the behaviour of the shortest distance between two orbits,

$$ \begin{align*}S_{n,2}(x,y)=\min_{i=0,\ldots,n-1}(d(T^{i}(x),T^{i}(y)))\end{align*} $$

in the continued fraction system. Shi et al. [5] proved that, for $\mu ^{2}$ almost all $(x,y)\in [0,1)\times [0,1),$

$$ \begin{align*}{H_{2}}\cdot\lim_{n\rightarrow\infty}\frac{M_{n,2}(x,y)}{\log n}=\lim_{n\rightarrow\infty}\frac{-\log S_{n,2}(x,y)}{\log n},\end{align*} $$

where $H_{2}$ is the Rényi entropy defined by (1.2). Investigating the shortest distance between two orbits amounts to estimating the longest common substrings between two sequences of partial quotients. In fact, [5] focused on the asymptotics of the length of the longest common substrings in two sequences of partial quotients.

For $n\geq 1$ and $(a_{1},\ldots ,a_{n})\in \mathbb {N}^{n}$ , we call

$$ \begin{align*}I_{n}(a_{1},\ldots,a_{n})=\{x\in[0,1): a_{1}(x)=a_{1},\ldots, a_{n}(x)=a_{n}\}\end{align*} $$

an nth cylinder. For $m\geq 2$ , we define the generalised Rényi entropy with respect to the Gauss measure $\mu $ by

(1.2) $$ \begin{align} {H}_{m}=\lim_{n\rightarrow\infty}\frac{-\log \sum_{(a_1,\ldots,a_n)\in \mathbb{N}^{n}} \mu(I_{n}(a_1,\ldots,a_n))^{m}}{(m-1)n}. \end{align} $$

The existence of the limit (1.2) for the Gauss measure $\mu $ was established in [2].

Theorem 1.1. For $\mu ^{m}$ -almost all $(x_{1},\ldots ,x_{m})\in [0,1)^{m},$

$$ \begin{align*}\lim_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\frac{1}{(m-1)H_{m}}.\end{align*} $$

Here we use the convention that ${1}/{0}=\infty $ and ${1}/{\infty }=0.$

It is natural to study the exceptional set in this limit theorem. We define the exceptional set as

$$ \begin{align*}\widetilde{E}=\bigg\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: \liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n} <\limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\bigg\}\end{align*} $$

and the level set as

$$ \begin{align*}E(\alpha)=\bigg\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: \lim_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\alpha\bigg\}.\end{align*} $$

Throughout the paper, $\dim _{H}A$ denotes the Hausdorff dimension of the set $A.$

Theorem 1.2. For any $\alpha $ with $0\leq \alpha \leq \infty ,$ $\dim _{H}\widetilde {E}=\dim _{H}E(\alpha )=m.$

In fact, Theorem 1.2 follows immediately from the following more general result. For any $0\leq \alpha \leq \beta \leq \infty ,$ set

$$ \begin{align*} E(\alpha,\beta)=\bigg\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: & \liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\alpha, \\ & \limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\beta\bigg\}. \end{align*} $$

Theorem 1.3. For any $\alpha ,\beta $ with $0\leq \alpha \leq \beta \leq \infty ,$ $\dim _{H}E(\alpha ,\beta )=m.$

2 Preliminaries

In this section, we fix some notation and recall some basic properties of continued fraction expansions. A detailed account of continued fractions can be found in Khintchine’s book [3].

For any irrational number $x\in [0,1)$ with continued fraction expansion (1.1), we denote by

$$ \begin{align*}\frac{p_{n}(x)}{ q_{n}(x)} :=[a_{1}(x),\ldots,a_{n}(x)]\end{align*} $$

the nth convergent of $x.$ With the conventions

$$ \begin{align*}p_{-1}(x)=1, \quad q_{-1}(x)=0, \quad p_{0}(x)=0, \quad q_{0}(x)=1,\end{align*} $$

we have, for any $n\ge 0,$

$$ \begin{align*}p_{n+1}(x)=a_{n+1}(x) p_{n}(x)+p_{n-1}(x), \quad q_{n+1}(x)=a_{n+1}(x) q_{n}(x)+q_{n-1}(x).\end{align*} $$

Obviously, $q_{n}(x)$ is determined by the first n partial quotients $a_{1}(x),\ldots ,a_{n}(x).$ So we also write $q_{n}(a_{1}(x),\ldots ,a_{n}(x))$ in place of $q_{n}(x).$ If no confusion is likely to arise, we write $a_{n}$ and $q_{n}$ in place of $a_{n}(x)$ and $q_{n}(x),$ respectively.

Proposition 2.1 [3].

For $n\geq 1$ and $(a_{1},\ldots ,a_{n})\in \mathbb {N}^{n}$ :

1. (1) $ q_n\ge 2^{(n-1)/2}$ and

   $$ \begin{align*} \prod^{n}_{k=1}a_{k}\leq q_n \leq \prod^{n}_{k=1}(a_{k}+1)\leq 2^{n}\prod^{n}_{k=1}a_{k};\end{align*} $$

2. (2) the length of $I_{n}(a_{1},\ldots ,a_{n})$ satisfies

   $$ \begin{align*} \frac{1}{2q_n^2} \leq |I_{n}(a_{1},\ldots,a_{n})|=\frac{1}{(q_n+q_{n-1})q_n} \leq \frac{1}{q_n^2}. \end{align*} $$

The following $\psi $ -mixing property is essential in proving Theorem 1.1.

Lemma 2.2 [4].

For any $k\geq 1,$ let $\mathbb {B}^{k}_{1}=\sigma (a_{1},\ldots ,a_{k})$ and let $\mathbb {B}^{\infty }_{k}=\sigma (a_{k},a_{k+1},\ldots )$ denote the $\sigma $ -algebras generated by the random variables $(a_{1},\ldots ,a_{k})$ and $(a_{k},a_{k+1},\ldots )$ respectively. Then, for any $E\in \mathbb {B}^{k}_{1}$ and $F\in \mathbb {B}^{\infty }_{k+n},$

$$ \begin{align*}\mu(E\cap F)=\mu(E)\cdot \mu(F)(1+\theta \rho^n),\end{align*} $$

where $|\theta |\leq K, \rho <1$ and $K, \rho $ are positive constants independent of $E, F, n$ and $k.$

To estimate the measure of a limsup set in a probability space, the following lemma is widely used.

Lemma 2.3 (Borel–Cantelli lemma).

Let $(\Omega , \mathcal {B}, \nu )$ be a finite measure space and let $\{A_n\}_{n\ge 1}$ be a sequence of measurable sets. Define $A=\bigcap _{N=1}^{\infty }\bigcup _{n=N}^{\infty }A_n.$ Then

$$ \begin{align*} \nu(A)=\begin{cases} 0 & \text{if } \displaystyle\sum_{n= 1}^{\infty}\nu(A_n)<\infty, \\[3pt] \nu(\Omega) & \displaystyle \text{if }\sum_{n= 1}^{\infty}\nu(A_n)=\infty \text{ and } \{A_n\}_{n\ge 1}\text{ are pairwise independent}. \end{cases} \end{align*} $$

Let $\mathbb {K}=\{k_{n}\}_{n\geq 1}$ be a subsequence of $\mathbb {N}$ that is not cofinite. Define a mapping $\phi _{\mathbb {K}}:[0,1)\cap \mathbb {Q}^{c}\rightarrow [0,1)\cap \mathbb {Q}^{c}$ as follows. For each $x=[a_1,a_2,\ldots ]\in [0,1)\cap \mathbb {Q}^{c},$ put $\phi _{\mathbb {K}}(x)=\overline {x}=[c_1,c_2,\ldots ],$ where $[c_1,c_2,\ldots ]$ is obtained by eliminating all the terms $a_{k_{n}}$ from the sequence $a_1, a_2,\ldots .$ Let $\{b_{n}\}_{n\geq 1}$ be a sequence with $b_{n}\in \mathbb {N},n\geq 1.$ Write

$$ \begin{align*}E(\mathbb{K},\{b_{n}\})=\{x\in[0,1)\cap \mathbb{Q}^{c}: a_{k_{n}}(x)= b_{n} \ \text{for all}~ n\geq1\}.\end{align*} $$

Lemma 2.4 [6].

Assume that $\{b_{n}\}_{n\geq 1}$ is bounded. If the sequence $\mathbb {K}$ is of density zero in $\mathbb {N},$ that is,

$$ \begin{align*}\lim_{n\rightarrow\infty}\frac{\sharp\{i\leq n: i\in\mathbb{K}\}}{n}=0,\end{align*} $$

where $\sharp $ denotes the number of elements in a set, then

$$ \begin{align*}\dim_{H}E(\mathbb{K},\{b_{n}\})=\dim_{H}\phi_{\mathbb{K}}(E(\mathbb{K},\{b_{n}\}))=1.\end{align*} $$

We close this section by citing Marstrand’s product theorem.

Lemma 2.5 [1].

If $E,F\subset \mathbb {R}^{d}$ for some $d,$ then $\dim _{H}(E\times F)\geq \dim _{H}E+\dim _{H}F.$

3 Proof of Theorem 1.1

Theorem 1.1 can be proved from the following two propositions.

Proposition 3.1. For $\mu ^{m}$ -almost all $(x_{1},\ldots ,x_{m})\in [0,1)^{m},$

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\leq\frac{1}{(m-1)H_{m}}.\end{align*} $$

Proof. We can assume that $H_{m}>0$ (the case $H_m=0$ is obvious). Fix $s_1< s_2< (m-1)H_{m}.$ By the definition of the $H_{m},$

(3.1) $$ \begin{align} \sum_{(a_{1},\ldots,a_{n})\in\mathbb{N}^{n}}\mu(I_{n}(a_{1},\ldots,a_{n}))^{m}<\exp\bigg\{-\frac{s_1+s_2}{2}n\bigg\} \end{align} $$

for sufficiently large n. Set $u_{n}=\lfloor {\log n}/{s_1}\rfloor .$ Note that, for any $(x_{1},\ldots ,x_{m})\in [0,1)^{m}$ with $M_{n,m}(x_{1},\ldots ,x_{m})=k$ , there exists i with $0\leq i\leq n-k$ such that

$$ \begin{align*} a_{i+j}(x_{1})=\cdots=a_{i+j}(x_{m})\end{align*} $$

for $j=1,\ldots ,k.$ We deduce

$$ \begin{align*} &\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots,x_{m})>u_{n}\})\\ &\quad =\sum_{k=u_{n}+1}^{\infty}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots,x_{m})=k\})\\ &\quad \leq\sum_{k=u_{n}+1}^{\infty}\sum_{i=0}^{n-k}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: a_{i+j}(x_{1})=\cdots=a_{i+j}(x_{m}),\ j=1,\ldots,k\}). \end{align*} $$

By the invariance of $\mu $ under $T,$ it follows that

$$ \begin{align*} &\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots,x_{m})>u_{n}\})\\&\quad\leq n \sum_{k=u_{n}+1}^{\infty}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: a_{j}(x_{1})=\cdots=a_{j}(x_{m}),\ j=1,\ldots,k\})\\&\quad= n\sum_{k=u_{n}+1}^{\infty}\sum_{(a_{1},\ldots,a_{k})\in \mathbb{N}^{k}}\mu(I_{k}(a_{1},\ldots,a_{k}))^{m}\\&\quad\leq Cn\cdot\exp\bigg\{-\frac{s_1+s_2}{2}(u_{n}+1)\bigg\} \qquad (\text{by } (3.1))\\&\quad\leq Cn^{-{(s_2-s_1)}/{2s_1}}, \end{align*} $$

where $C=\sum _{k=1}^{\infty }\exp \{-{(s_1+s_2)k}/{2}\}.$ Choose an infinite subsequence of integers $\{n_{k}\}_{k\geq 1},$ where $n_{k}=k^{L}$ and $L\cdot {(s_2-s_1)}/{2s_1}>1.$ Then

$$ \begin{align*} \sum_{k=1}^{\infty}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n_{k},m}(x_{1},\ldots,x_{m})>u_{n_{k}}\})<\infty. \end{align*} $$

From the Borel–Cantelli Lemma 2.3, for almost all $(x_{1},\ldots ,x_{m})\in [0,1)^{m},$

$$ \begin{align*}M_{n_{k},m}(x_{1},\ldots,x_{m})\leq u_{n_{k}} \end{align*} $$

for sufficiently large k. Thus,

$$ \begin{align*} \limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n} \leq&\limsup_{k\rightarrow\infty}\frac{M_{n_{k+1},m}(x_{1},\ldots,x_{m})}{\log n_{k}}\\ \leq&\limsup_{k\rightarrow\infty}\frac{M_{n_{k+1},m}(x_{1},\ldots,x_{m})}{\log n_{k+1}}\cdot\limsup_{k\rightarrow\infty}\frac{n_{k+1}}{n_{k}} \leq\frac{1}{s_{1}}. \end{align*} $$

Therefore, by the arbitrariness of $s_{1},$

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\leq\frac{1}{(m-1)H_{m}}.\end{align*} $$

This completes the proof.

Proposition 3.2. For $\mu ^{m}$ -almost all $(x_{1},\ldots ,x_{m})\in [0,1)^{m},$

$$ \begin{align*}\liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\geq\frac{1}{(m-1)H_{m}}.\end{align*} $$

Proof. We can assume that $H_{m}<\infty $ (the case $H_m=\infty $ is obvious). For $1\le d<n$ , set

$$ \begin{align*} M_{[d,n]}(x_{1},\ldots,x_{m})=\max\{k\in \mathbb{N}: a_{i+j}(x_{1}) & =\cdots= a_{i+j}(x_{m}) \ \text{for}~ j=1,\ldots,k, \\&\ \text{and for some } i \mbox{ with } d-1\leq i \leq n-k\}. \end{align*} $$

We denote $\{(x_{1},\ldots ,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots ,x_{m})<k\}$ by $\{M_{n,m}<k\}$ for brevity.

For any $s>(m-1)H_{m},$ by the definition of the $H_{m},$

(3.2) $$ \begin{align} \sum_{(a_{1},\ldots,a_{n}) \in\mathbb{N}^{n}}\mu(I_{n}(a_{1},\ldots,a_{n}))^{m}>\exp\bigg\{-\frac{s+(m-1)H_{m}}{2}n\bigg\} \end{align} $$

for sufficiently large n. Let $u_{n}=\lfloor {\log n}/{s}\rfloor $ and $l_{n}=\lfloor {n}/{u_{n}^{2}}\rfloor .$ Then

$$ \begin{align*} \begin{aligned} \{M_{n,m}<u_{n}\}&\subset \{M_{[iu_{n}^{2}+1,iu_{n}^{2}+u_{n}]}<u_{n} : 0\leq i <l_{n}\}\\& \subset\{M_{u_{n},m}<u_{n}\}\cap(\underbrace{T\times\cdots \times T}_{m \text{ times }})^{-u_{n}^{2}}\{M_{[iu_{n}^{2}+1,i\mu_{n}^{2}+u_{n}]}<u_{n} : 0\leq i <l_{n}-1\}. \end{aligned} \end{align*} $$

By Lemma 2.2, it follows that

$$ \begin{align*} & \mu^{m}(\{M_{n,m}<u_{n}\}) \\& \quad \leq \mu^{m}(\{M_{u_{n},m}<u_{n}\}\cap(\underbrace{T\times\cdots \times T}_{m \text{ times }})^{-u_{n}^{2}}\{M_{[iu_{n}^{2}+1,i\mu_{n}^{2}+u_{n}]}<u_{n}, 0\leq i <l_{n}-1\})\\& \quad \leq\mu^{m}(\{M_{u_{n},m}<u_{n}\})^{l_{n}}(1+\theta \rho^{u_{n}^{2}-u_{n}})^{m\cdot l_{n}}\\& \quad \leq\bigg(1-\sum_{(a_{1},\ldots,a_{u_{n}})\in \mathbb{N}^{u_{n}}} \mu(I_{u_{n}}(a_{1},\ldots,a_{u_{n}}))^{m}\bigg)^{l_{n}}(1+\theta \rho^{u_{n}^{2}-u_{n}})^{m\cdot l_{n}}\\& \quad \leq\exp\bigg\{-\frac{n}{u_{n}^{2}}\cdot n^{-{(s+(m-1)H_{m})}/{2s}}\bigg\}\cdot\exp\bigg\{\theta \rho^{u_{n}^{2}-u_{n}}\frac{m\cdot n}{u_{n}^{2}}\bigg\}\\& \quad \leq M\exp\bigg\{-\frac{n^{{(s-(m-1)H_{m})}/{2s}}}{u_{n}^{2}}\bigg\}, \end{align*} $$

where the penultimate inequality follows from (3.2) and the two facts $ (1-x)<\exp (-x)$ for $0<x<1$ and $\lim _{n\to \infty }(1+1/n)^{n}=e$ , and the last inequality follows because $\theta \rho ^{u_{n}^{2}-u_{n}}{m\cdot n}/{u_{n}^{2}}\rightarrow 0$ as $n\rightarrow \infty .$ Thus,

$$ \begin{align*} \sum_{n=1}^{\infty}\mu^{m}(\{M_{n,m}<u_{n}\})<\infty. \end{align*} $$

From the Borel–Cantelli Lemma, for $\mu ^{m}$ -almost all $(x_{1},\ldots ,x_{m})\in [0,1)^{m},$

$$ \begin{align*}\liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\geq\frac{1}{(m-1)H_{m}}.\end{align*} $$

This completes the proof of Proposition 3.2 and of Theorem 1.1.

4 Proof of Theorem 1.3

This section is devoted to the proof of Theorem 1.3. Our strategy is to construct Cantor-like subsets with full Hausdorff dimension. The proof is divided into several cases according to the values of $\alpha $ and $\beta .$ We give a detailed proof for the case $0<\alpha <\beta <\infty $ and a sketch of the proof for the remaining cases.

Case 1: $0<\alpha <\beta <\infty .$

Choose two positive integer sequences $\{n_{k}\}_{k\geq 1}$ and $\{s_{k}\}_{k\geq 1}$ such that, for each $k\geq 1,$

(4.1) $$ \begin{align} n_{1}=2,\quad n_{k+1}=\lfloor n_{k}^{{\beta}/{\alpha}}\rfloor, \quad s_{k}=\lfloor \,\beta \log n_{k}\rfloor. \end{align} $$

We readily check that

(4.2) $$ \begin{align} \lim_{k\rightarrow\infty}\frac{s_{k}}{n_{k+1}-n_{k}}=0. \end{align} $$

Without loss of generality, we assume that $n_{k+1}-n_{k}>s_{k}$ for all $k\geq 1.$ Otherwise, we consider only sufficiently large k. Put

$$ \begin{align*}n_{k+1}-n_{k}=s_{k}\cdot\iota_{k}+\theta_{k},\end{align*} $$

where

$$ \begin{align*}\iota_{k}=\bigg\lfloor\frac{n_{k+1}-n_{k}}{s_{k}}\bigg\rfloor \quad \text{for } 0\leq\theta_{k}<s_{k}.\end{align*} $$

Define a marked set $\mathbb {K}$ of positive integers by

$$ \begin{align*}\mathbb{K}:=\mathbb{K}(\{n_{k}\}, \{s_{k}\})=\bigcup_{k\ge1}\{n_{k},n_{k}+1,n_{k}+2,\ldots,n_{k}+s_{k}, n_{k}+2s_{k},n_{k}+3s_{k},\ldots, n_{k}+\iota_ks_{k}\}.\end{align*} $$

Now we define m sequences as follows.

• • For $i=1,$

  $$ \begin{align*}a_{n_{k}}^{(1)}=1,\ a_{n_{k}+1}^{(1)}=\cdots=a_{n_{k}+s_{k}-1}^{(1)}=1,\ a_{n_{k}+s_{k}}^{(1)}=a_{n_{k}+2s_{k}}^{(1)}= \cdots=a_{n_{k}+\iota_ks_{k}}^{(1)}=1.\end{align*} $$

• • For $2\leq i\leq m,$

  $$ \begin{align*}a_{n_{k}}^{(i)}=i,\ a_{n_{k}+1}^{(i)}=\cdots=a_{n_{k}+s_{k}-1}^{(i)}=1,\ a_{n_{k}+s_{k}}^{(i)}=a_{n_{k}+2s_{k}}^{(i)}= \cdots=a_{n_{k}+\iota_ks_{k}}^{(i)}=i.\end{align*} $$

Then, for $i=1,2,\ldots ,m,$ write

$$ \begin{align*}E(\mathbb{K},\{a_{n}^{(i)}\}_{n\geq1})=\{x\in[0,1)\cap \mathbb{Q}^{c}: a_{n}(x)= a_{n}^{(i)} \ \text{for all}~ n\in\mathbb{K}\}.\end{align*} $$

Now we prove $\prod _{i=1}^{m}E(\mathbb {K},\{a_{n}^{(i)}\}_{n\geq 1})\subset E(\alpha ,\beta ).$ Fix $(x_{1},\ldots ,x_{m})\in \prod _{i=1}^{m}E(\mathbb {K},\{a_{n}^{(i)}\}_{n\geq 1})$ for any $n\geq n_{1}$ and let k be the integer such that $n_{k}\leq n< n_{k+1}.$ From the construction of the set $\prod _{i=1}^{m}E(\mathbb {K},\{a_{n}^{(i)}\}_{n\geq 1})$ , we see that

$$ \begin{align*} M_{n,m}(x_{1},\ldots,x_{m})=\begin{cases} s_{k-1}-1 & \text{if}\ n_{k}\leq n<n_{k}+s_{k-1},\\ n-n_{k} & \text{if}\ n_{k}+s_{k-1}\leq n<n_{k}+s_{k},\\ s_{k}-1 & \text{if}\ n_{k}+s_{k}\leq n<n_{k+1}. \end{cases} \end{align*} $$

Further, by (4.1), we deduce that

$$ \begin{align*} \begin{aligned} \liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}&= \liminf_{k\rightarrow\infty} \min\bigg\{\frac{M_{n_{k}+s_{k-1}-1,m}(x_{1},\ldots,x_{m})}{\log(n_{k}+s_{k-1}-1)},\frac{M_{n_{k+1}-1,m}(x_{1},\ldots,x_{m})}{\log(n_{k+1}-1)}\bigg\}\\& =\liminf_{k\rightarrow\infty} \min\bigg\{\frac{s_{k-1}-1}{\log(n_{k}+s_{k-1}-1)},\frac{s_{k}-1}{\log(n_{k+1}-1)}\bigg\}\\& =\alpha \end{aligned} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} \limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}&= \limsup_{k\rightarrow\infty} \max\bigg\{\frac{M_{n_{k},m}(x_{1},\ldots,x_{m})}{\log(n_{k})},\frac{M_{n_{k}+s_{k}-1,m}(x_{1},\ldots,x_{m})}{\log(n_{k}+s_{k}-1)}\bigg\}\\& =\limsup_{k\rightarrow\infty} \max\bigg\{\frac{s_{k-1}-1}{\log(n_{k})},\frac{s_{k}-1}{\log(n_{k}+s_{k}-1)}\bigg\}\\& =\beta. \end{aligned} \end{align*} $$

Hence, $(x_{1},\ldots ,x_{m})\in E(\alpha ,\beta ).$

It remains to prove that the density of $\mathbb {K}\subset \mathbb {N}$ is zero. For $n_{k}\leq n<n_{k+1}$ with some $k\geq 1{:}$

• • if $n_{k}\leq n< n_{k}+s_{k},$ then $\sharp \{i\leq n : i\in \mathbb {K}\}=\sum _{j=1}^{k-1}(m_{j}+\iota _{j})+n-n_{k}+1;$

• • if $n_{k}+ls_{k}\leq n<n_{k}+(l+1)s_{k}$ for some $l$ with $0< l<\iota _{k},$ then we see that ${\sharp \{i\leq n : i\in \mathbb {K}\}=\sum _{j=1}^{k-1} (s_{j}+\iota _{j})+s_{k}+l};$

• • if $n_{k}+\iota _{k}s_{k}\leq n<n_{k+1},$ then $\sharp \{i\leq n : i\in \mathbb {K}\}=\sum _{j=1}^{k}(s_{j}+\iota _{j}).$

Consequently,

$$ \begin{align*} \begin{aligned} \limsup_{n\rightarrow\infty}\frac{\sharp\{i\leq n : i\in\mathbb{K}\}}{n}&\leq \limsup_{k\rightarrow\infty} \max_{0\leq l<\iota_{k}}\bigg\{\frac{\sum_{j=1}^{k-1}(s_{j}+\iota_{j})+s_{k}+l}{n_{k}+ls_{k}}\bigg\}\\& \leq\limsup_{k\rightarrow\infty}\bigg\{\frac{\sum_{j=1}^{k-1}(s_{j}+\iota_{j})+s_{k}+\iota_{k})}{n_{k}}\bigg\}\\& =0, \end{aligned} \end{align*} $$

where the last equality follows by the Stolz–Cesàro theorem and (4.2). By Lemmas 2.4 and 2.5,

$$ \begin{align*}\dim_{H}E_{\alpha,\beta}\geq\dim_{H}\bigg(\prod\limits_{i=1}^{m}E(\mathbb{K},\{a_{n}^{(i)}\}_{n\geq1})\bigg)\geq\sum_{i=1}^{m}\dim_{H}E(\mathbb{K},\{a_{n}^{(i)}\}_{n\geq1})=m.\end{align*} $$

Similar arguments apply to the remaining cases. We only give the constructions for the proper sequences $\{n_k\}_{k\geq 1}$ and $\{s_k\}_{k\geq 1}.$

Case 2: $0<\alpha =\beta <\infty .$ Take $n_{k}=2^{k}$ and $s_{k}=\lfloor \alpha \log n_{k}\rfloor \text { for } k\geq 1.$

Case 3: $\alpha =0<\beta <\infty .$ Take $n_{k}=2^{2^{2^{k}}}$ and $s_{k}=\lfloor \,\beta \log n_{k}\rfloor \text { for } k\geq 1.$

Case 4: $\alpha =0, \beta =\infty .$ Take $n_{k}=2^{2^{2^{k}}}$ and $s_{k}=\lfloor k\log n_{k}\rfloor ~\text { for } k\geq 1.$

Case 5: $0<\alpha <\beta =\infty .$ Take $n_{k}=2^{k!}$ and $s_{k}=\lfloor \alpha k\log n_{k}\rfloor \text { for } k\geq 1.$

Case 6: $\alpha =\beta =0.$ Take $n_{k}=2^{k}$ and $s_{k}=\lfloor \log \log n_{k}\rfloor \text { for } k\geq 1.$

Case 7: $\alpha =\beta =\infty .$ Take $n_{k}=2^{k}$ and $s_{k}=\lfloor k\log n_{k}\rfloor \text { for } k\geq 1.$