Measure theoretic properties of large products of consecutive partial quotients

Adam Brown-Sarre , Gerardo González Robert and Mumtaz Hussain Department of Mathematical and Physical Sciences, La Trobe University, Bendigo 3552, Australia. [email protected] [email protected], [email protected] [email protected]

Abstract.

The theory of uniform approximation of real numbers motivates the study of products of consecutive partial quotients in regular continued fractions. For any non-decreasing positive function $\varphi:\mathbb{N}\to\mathbb{R}^{+}$ and $\ell\in\mathbb{N}$ , we determine the Lebesgue measure and Hausdorff dimension of the set $\mathcal{F}_{\ell}(\varphi)$ of irrational numbers $x$ whose regular continued fraction $x\leavevmode\nobreak\ =\leavevmode\nobreak\ [a_{1}(x),a_{2}(x),\ldots]$ is such that for infinitely many $n\in\mathbb{N}$ there are two numbers $1\leq j<k\leq n$ satisfying

a_{k}(x)\cdots a_{k+\ell-1}(x)\geq\varphi(n),\;a_{j}(x)\cdots a_{j+\ell-1}(x)\geq\varphi(n).

One of the consequences of the results is that the strong law of large numbers for products of $\ell$ consecutive partial quotients is impossible even if the block with the largest product is removed.

2020 Mathematics Subject Classification:

Primary 11K55; Secondary 11J83, 28A80

1. Introduction

Regular continued fractions constitute a representation system of real numbers with outstanding approximation properties. Recall that, for each $x\in\mathbb{R}\setminus\mathbb{Q}$ there is a single sequence of integers $(a_{j}(x))_{j\geq 0}$ with $a_{j}(x)\geq 1$ for $j\geq 1$ such that

x=a_{0}(x)+\cfrac{1}{a_{1}(x)+\cfrac{1}{a_{2}(x)+\cfrac{1}{\ddots}}}:=[a_{0}(x);a_{1}(x),a_{2}(x),\ldots].

The numbers $a_{n}(x)$ are known as the partial quotients of $x$ . We have $x\in[0,1)$ if and only if $a_{0}(x)=0$ . In this case, we write the continued fractions of $x$ as $[a_{1}(x),a_{2}(x),\ldots]$ . It is well known that the $n$ -th convergent of $x$ given by

\frac{p_{n}(x)}{q_{n}(x)}=[a_{1}(x),\ldots,a_{n}(x)]\quad\text{ for all }n\in\mathbb{N},

where $q_{n}(x)\geq 1$ and $p_{n}(x)$ are coprime integers, satisfies the relation

\frac{1}{(a_{n+1}(x)+2)q_{n}^{2}(x)}<\left|x-\frac{p_{n}(x)}{q_{n}(x)}\right|<\frac{1}{q_{n}^{2}(x)}\quad\text{ for all }n\in\mathbb{N},

see Equation (34) in [16]. Because $(q_{n}(x))_{n\geq 1}$ is strictly increasing, the rate at which rationals approximate $x$ is closely related to the size of its partial quotients. This leads us to consider, for any function $\varphi:\mathbb{N}\to\mathbb{R}_{>0}$ , the set

\mathcal{E}_{1}(\varphi):=\{x\in[0,1):a_{n}(x)\geq\varphi(n)\text{ for i.m. }n\in\mathbb{N}\}.

Here and in what follows “i.m.” stands for “infinitely many”. The well-known Borel-Bernstein Theorem establishes the Lebesgue measure $\mathfrak{m}$ of $\mathcal{E}_{1}(\varphi)$ :

\mathfrak{m}\left(\mathcal{E}_{1}(\varphi)\right)=\begin{cases}0,\text{ if }\displaystyle\sum_{n\geq 1}\frac{1}{\varphi(n)}<\infty,\\ 1,\text{ if }\displaystyle\sum_{n\geq 1}\frac{1}{\varphi(n)}=\infty.\\ \end{cases}

For proof of this theorem, see for example [2, Theorem 1.11]. Łuczak [21] and Feng et al. [6] calculated the Hausdorff dimension, denoted as $\dim_{\mathrm{H}}$ , of $\mathcal{E}_{1}(\varphi)$ when $\varphi(n)=c^{b^{n}}$ for some fixed $c,b>1$ (see Lemma 2.7). Wang and Wu solved the general case in [27] (see Theorem 1.2 below with $\ell=1$ ).

While studying improvements to the classical Dirichlet theorem, Kleinbock and Wadleigh [18] showed that the set of Dirichlet non-improvable numbers is related to the set

\mathcal{E}_{2}(\varphi):=\{x\in[0,1):a_{n}(x)a_{n+1}(x)\geq\varphi(n)\text{ for i.m. }n\in\mathbb{N}\}.

In fact, they computed the Lebesgue measure of $\mathcal{E}_{2}(\varphi)$ and the Hausdorff measure and dimension were proven in [1, 12]. In [11], Huang, Wu, and Xu determined the Lebesgue measure and the Hausdorff dimension of the general set $\mathcal{E}_{\ell}(\varphi)$ , $\ell\in\mathbb{N}$ , defined as:

\mathcal{E}_{\ell}(\varphi):=\{x\in[0,1):a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)\text{ for i.m. }n\in\mathbb{N}\}.

Theorem 1.1 (Huang, Wu, Xu [11]).

If $\varphi:\mathbb{N}\to\mathbb{R}_{\geq 2}$ , then

\mathfrak{m}\left(\mathcal{E}_{\ell}(\varphi)\right)=\begin{cases}0,&\text{\rm if }\displaystyle\sum_{n=1}^{\infty}\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}<\infty,\\ 1,&\text{\rm if }\displaystyle\sum_{n=1}^{\infty}\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}=\infty.\end{cases}

Define $B$ and $b$ as

(1)

\log B:=\liminf_{n\xrightarrow{}\infty}\frac{\log\varphi(n)}{n},\;\text{ and }\;\log b=\liminf_{n\xrightarrow{}\infty}\frac{\log\log\varphi(n)}{n}.

Theorem 1.2 (Huang, Wu, Xu [11]).

Let $(f_{n})_{n\geq 1}$ be the sequence of real functions on $[0,1]$ given by

f_{1}(s)=s,\;\text{ and }\;f_{n+1}=\frac{sf_{n}(s)}{1-s+f_{n}(s)}\;\text{ for all }n\in\mathbb{N}.

Let $\varphi:\mathbb{N}\to\mathbb{R}_{>0}$ be arbitrary and $B$ , $b$ as in (1). Then, the Hausdorff dimension of $\mathcal{E}_{\ell}(\varphi)$ is

\dim_{\mathrm{H}}\mathcal{E}_{\ell}(\varphi)=\begin{cases}1,&\text{\rm if }B=1,\\ \inf\{s\geq 0:P(T,-f_{\ell}\log B-s\log|T^{\prime}|)\leq 0\},&\text{\rm if }1<B<\infty,\\ \frac{1}{1+b},&\text{\rm if }B=\infty.\end{cases}

The pressure function $P(T,\cdot)$ appearing in the Hausdorff dimension formula above and in all the theorems below will be defined in Section 2.

From a probabilistic perspective, we may interpret partial quotients as random variables defined on $[0,1)$ (their definition of rational numbers is irrelevant for our purposes). These random variables are not independent, hence the Borel-Bernstein Theorem is not a trivial consequence of the Borel-Cantelli lemma. Moreover, the Borel-Bernstein theorem tells us that for almost every number $x$ (with respect to $\mathfrak{m}$ ) the inequality $a_{n}(x)\geq n\log n$ holds infinitely often. This implies that, writing $S_{n}(x):=a_{1}(x)+\cdots+a_{n}(x)$ , we have

\limsup_{n\to\infty}\frac{1}{n}S_{n}(x)=\infty\;\text{ for almost all }x\in[0,1).

Despite not being independent and having infinite expected value, the partial quotients satisfy a weak law of large numbers. Khinchin [17, §. 4] proved that $S_{n}/(n\log n)$ converges in Lebesgue measure to $\frac{1}{\log 2}$ as $n\to\infty$ . Furthermore, Philipp [23, Theorem 1] proved that there is no reasonable function $\sigma$ for which $S_{n}/\sigma(n)$ converges almost everywhere to a finite positive constant. Large partial quotients are to blame for this phenomenon. Certainly, Diamond and Vaaler [4, Corollary 1] showed that convergence almost everywhere does hold for Khinchin’s averaging function when omitting the largest partial quotient: for almost every $x\in[0,1)$ we have

(2)

\lim_{n\to\infty}\frac{1}{n\log n}\left(S_{n}(x)-\max_{1\leq j\leq n}a_{j}(x)\right)=\frac{1}{\log 2}.

This result was extended by Hu, Hussain, and Yu for sums of products of two partial quotients [10, Theorem 1.5]. They showed that almost every $x\in[0,1)$ satisfies

(3)

\lim_{n\to\infty}\frac{1}{n\log^{2}n}\left(\sum_{j=1}^{n}a_{j}(x)a_{j+1}(x)-\max_{1\leq j\leq n}a_{j}(x)a_{j+1}(x)\right)=\frac{1}{2\log 2}.

A key aspect in the proofs by Diamond and Vaaler, and Hu, Hussain, and Yu respectively is proving that, for $c>\frac{1}{2}$ and $d>\frac{3}{2}$ , the sets

\left\{x\in[0,1):\begin{matrix}a_{j}(x)\geq n(\log n)^{c},\\ a_{k}(x)\geq n(\log n)^{c},\end{matrix}\;\text{ for some }1\leq j<k\leq n\text{ for i.m. }n\in\mathbb{N}\right\}

and

\left\{x\in[0,1):\begin{matrix}a_{j}(x)a_{j+1}(x)\geq n(\log n)^{d},\\ a_{k}(x)a_{k+1}(x)\geq n(\log n)^{d},\\ \end{matrix}\;\text{ for some }1\leq j<k\leq n\text{ for i.m. }n\in\mathbb{N}\right\}

are of Lebesgue measure $0$ .

Recently, Tan, Tian, and Wang [25] generalised this result of Diamond and Vaaler to a wider family of functions. Later, Tan and Zhou [26] also extended the result of Hu, Hussain and Yu. More precisely, define

\mathcal{F}_{\ell}(\varphi):=\left\{x\in[0,1):\begin{matrix}a_{j}(x)\cdots a_{j+\ell-1}(x)\geq\varphi(n),\\ a_{k}(x)\cdots a_{k+\ell-1}(x)\geq\varphi(n),\end{matrix}\;\text{ for some }1\leq j<k\leq n\text{ for i.m. }n\in\mathbb{N}\right\}.

In both [25] and [26], $\varphi$ is assumed to be non-decreasing. It is obvious that $\mathcal{F}_{\ell}(\varphi)=[0,1)\setminus\mathbb{Q}$ whenever $\varphi(n)\leq 1$ for all $n\in\mathbb{N}$ . Hence, in what follows we suppose that $\varphi$ is non-decreasing and $\varphi:\mathbb{N}\to[2,\infty)$ .

Theorem 1.3 (Tan, Tian, Wang [25]).

Let $\varphi:\mathbb{N}\to[2,\infty)$ be non-decreasing. Then, the Lebesgue measure of $\mathcal{F}_{1}(\varphi)$ is

\mathfrak{m}(\mathcal{F}_{1}(\varphi))=\begin{cases}0,&\text{\rm if }\displaystyle\sum_{k\geq 1}\frac{n}{\varphi(n)^{2}}<\infty,\\ 1,&\text{\rm if }\displaystyle\sum_{k\geq 1}\frac{n}{\varphi(n)^{2}}=\infty.\end{cases}

Theorem 1.4 (Tan, Zhou, 2024).

Let $\varphi:\mathbb{N}\to[2,\infty)$ be non-decreasing. Then, the Lebesgue measure of $\mathcal{F}_{2}(\varphi)$ is given by

\mathfrak{m}(\mathcal{F}_{2}(\varphi))=\begin{cases}0,&\text{\rm if }\displaystyle\sum_{n\geq 1}\frac{n\log\varphi(n)}{\varphi^{2}(n)}+\frac{1}{\varphi(n)}<\infty,\\ 1,&\text{\rm if }\displaystyle\sum_{n\geq 1}\frac{n\log\varphi(n)}{\varphi^{2}(n)}+\frac{1}{\varphi(n)}=\infty.\end{cases}

Our first result, which includes Theorem 1.4, completes the picture of the Lebesgue measure of $\mathcal{F}_{\ell}(\varphi)$ .

Theorem 1.5.

Let $\varphi:\mathbb{N}\to[2,\infty)$ be non-decreasing. For $\ell\geq 2$ , the Lebesgue measure of $\mathcal{F}_{\ell}(\varphi)$ is given by

\mathfrak{m}(\mathcal{F}_{\ell}(\varphi))=\begin{cases}0&\text{\rm if }\displaystyle\sum_{n\geq 1}\frac{n\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}+\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}<\infty,\\ 1&\text{\rm if }\displaystyle\sum_{n\geq 1}\frac{n\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}+\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}=\infty.\end{cases}

Remark 1.6.

Since all the partial quotients are at least $1$ , for almost every $x\in[0,1)$ we have

\limsup_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}a_{j}(x)\cdots a_{j+\ell}(x)=\infty.

Theorem 1.5 implies that for almost every $x\in[0,1)$ there are infinitely many numbers $n\in\mathbb{N}$ for which there are $1\leq j<k\leq n$ satisfying

a_{j}(x)\cdots a_{j+\ell-1}(x)\geq n\log^{\ell-1}n\;\text{ and }\;a_{k}(x)\cdots a_{k+\ell-1}(x)\geq n\log^{\ell-1}n.

Therefore, we have

\limsup_{n\to\infty}\frac{1}{n}\left(\sum_{j=1}^{n}a_{j}(x)\cdots a_{j+\ell-1}(x)-\max_{1\leq j\leq n}a_{j}(x)\cdots a_{j+\ell-1}(x)\right)=\infty.

Theorem 1.5 is also a crucial step in an extension of the limits (2) and (3). In fact, in a forthcoming paper we rely on Theorem 1.5 to show that for almost every $x\in[0,1)$ we have

\lim_{n\to\infty}\frac{1}{n\log^{\ell}n}\left(\sum_{j=1}^{n}a_{j}(x)\cdots a_{j+\ell}(x)-\max_{1\leq j\leq n}a_{j}(x)\cdots a_{j+\ell-1}(x)\right)=\frac{1}{\ell\log 2}.

Other anticipated results are stated in Section 6.

Remark 1.7.

While proving Theorem 1.5 we must obtain new estimates corresponding to overlapping blocks of consecutive partial quotients with large products. The number of terms overlapping in Theorem 1.5 ranges from $0$ up to $\ell-1$ whereas in Theorem 1.4 the only overlaps are of length $0$ and $1$ and there are no overlaps in Theorem 1.3.

The Hausdorff dimensions of $\mathcal{F}_{1}(\varphi)$ and $\mathcal{F}_{2}(\varphi)$ were computed, respectively, in [25] and [26].

Theorem 1.8 (Tan, Tian, Wang [25]).

Let $\varphi:\mathbb{N}\to[2,\infty)$ be non-decreasing and $B$ and $b$ as in (1). Then, the Hausdorff dimension of $\mathcal{F}_{1}(\varphi)$ is given by

\dim_{\mathrm{H}}\mathcal{F}_{1}(\varphi)=\begin{cases}1,&\text{\rm if }B=1,\\ \inf\{s\geq 0:P(T,-(3s-1)\log B-s\log|T^{\prime}|)\leq 0\},&\text{\rm if }1<B<\infty,\\ \frac{1}{1+b},&\text{\rm if }B=\infty.\end{cases}

Theorem 1.9 (Tan, Zhou [26]).

Let $\varphi:\mathbb{N}\to[2,\infty)$ be non-decreasing and $B$ and $b$ as in (1). Then the Hausdorff dimension of $\mathcal{F}_{2}(\varphi)$ is given by

\dim_{\mathrm{H}}\mathcal{F}_{2}(\varphi)=\begin{cases}1,&\text{\rm if }B=1,\\ \inf\{s\geq 0:P(T,-(3s-1-s^{2})\log B-s\log|T^{\prime}|)\leq 0\},&\text{\rm if }1<B<\infty,\\ \frac{1}{1+b},&\text{\rm if }B=\infty.\end{cases}

The case $1<B<\infty$ is the main substance of theorems 1.8 and 1.9. In both theorems, this case can be obtained as an application of a general criterion established in [14] (see Lemma 2.10). We refer to [13] for the study of the product of partial quotients arising from arithmetic progressions and the metrical study of this set providing information on the size of the set of exceptions to the convergence criteria of multiple ergodic averages for the Gauss dynamical systems.

Our second result provides the Hausdorff dimension of $\mathcal{F}_{3}(\varphi)$ . Let $g_{3}:\mathbb{R}\to\mathbb{R}$ be given by

g_{3}(s):=\frac{3s^{3}-5s^{2}+4s-1}{s^{2}-s+1}.

Theorem 1.10.

Let $\varphi:\mathbb{N}\to[2,\infty)$ be non-decreasing and $B$ , $b$ as in (1). Then, the Hausdorff dimension of $\mathcal{F}_{3}(\varphi)$ is given by

\dim_{\mathrm{H}}\mathcal{F}_{3}(\varphi)=\begin{cases}1,&\text{\rm if }B=1,\\ \inf\{s\geq 0:P(T,-g_{3}(s)\log B-s\log|T^{\prime}|)\leq 0\},&\text{\rm if }1<B<\infty,\\ \frac{1}{1+b},&\text{\rm if }B=\infty.\end{cases}

The paper is organised as follows. Section 2 contains some preliminaries and auxiliary lemmas from measure theory and continued fractions, in particular on the Hausdorff dimension and pressure functions. In Section 3 we estimate certain series that are fundamental for our proofs. The proof of Theorem 1.5 is in Section 4 and the proof of Theorem 1.10 is in Section 5. Lastly, Section 6 contains some final remarks and results developed to further this study.

Notation.

Whenever $(x_{n})_{n\geq 1}$ and $(y_{n})_{n\geq 1}$ are two sequences of positive real numbers, we write $x_{n}\ll y_{n}$ to mean that for some constant $c>0$ and every $n\in\mathbb{N}$ we have $x_{n}\leq cy_{n}$ . When $c$ depends on some parameter $\lambda$ , we write $x_{n}\ll_{\lambda}y_{n}$ . By $x_{n}\asymp y_{n}$ we mean $x_{n}\ll y_{n}$ and $y_{n}\ll x_{n}$ . For the rest of the paper, $\ell$ will denote an arbitrary natural number.

Acknowledgements. GGR and MH were funded by the Australian Research Council discovery project grant number 200100994. We thank Nikita Shulga and Ben Ward for useful discussions.

2. Preliminaries

In this section, we recall some definitions and elementary results on measure theory, Hausdorff dimension, continued fractions, and pressure functions.

2.1. Full and null sets

The Borel-Cantelli Lemma is a fundamental tool for determining the measure of a limsup set. A proof may be found, for example, in [15, Lemma 4.18].

Theorem 2.1 (Borel-Cantelli Lemma).

Let $(E_{k})_{k\geq 1}$ be a sequence of measurable sets in a probability space $(\Omega,\mathscr{A},\mu)$ and $E=\limsup_{n}E_{n}$ .

(i).

If $\sum_{n}\mu(E_{n})<\infty$ , then $\mu(E)=0$ .
(ii).

If the collection $\{E_{n}:n\in\mathbb{N}\}$ is stochastically independent and $\sum_{n}\mu(E_{n})=\infty$ , then $\mu(E)=1$ .

The independence assumption is rather restrictive, it will not hold in our context. A way to circumvent this problem is proving the positive measure of the limsup set and showing by some other means that the measure is either full or null.

We show the full measure statement in Theorem 1.5 using a ubiquity-type property and Lemma 2.2. This lemma is attributed to Knopp [19] by Tan and Zhou [26]. Recall that, given two measurable sets $A,B$ in a probability space $(X,\mathscr{A},\mu)$ , we write $A\equiv B\pmod{\mu}$ if and only if $\mu(A\setminus B)=\mu(B\setminus A)=0$ .

Lemma 2.2.

Let $A\subset[0,1)$ be a Borel set and $\mathcal{C}$ a class of subintervals of $[0,1)$ satisfying both of the following conditions:

(1)

Every open subinterval of $[0,1)$ is an at most countable union $\pmod{\mathfrak{m}}$ of disjoint elements from $\mathcal{C}$ ,
(2)

There is a constant $c>0$ such that for any $B\in\mathcal{C}$ we have $\mathfrak{m}(A\bigcap B)\geq c\mathfrak{m}(B)$ .

Then, $\mathfrak{m}(A)=1$ .

In order to apply Lemma 2.2, we need the following result due to Chung and Erdös [3] (see also [24, Lemma 6] for the formulation stated below).

Lemma 2.3 (Chung, Erdös, [3]).

Let $(E_{k})_{k\geq 1}$ be a sequence of measurable sets in a probability space $(\Omega,\mathscr{A},\mu)$ . If $\sum_{n}\mu(E_{n})=\infty$ , then

\mu\left(\limsup_{n\to\infty}E_{n}\right)\geq\limsup_{N\to\infty}\frac{\left(\displaystyle\sum_{1\leq n\leq N}\mu(E_{n})\right)^{2}}{\displaystyle\sum_{1\leq i\neq j\leq N}\mu(E_{i}\cap E_{j})}.

2.2. Hausdorff dimension

The Hausdorff dimension provides a way to distinguish between the sets of Lebesgue measure $0$ . We recall the definition of the Hausdorff dimension and refer to [5] for its general theory.

Given a non-empty subset $A$ of real numbers, let $|A|$ be its diameter. For $\delta>0$ , by a $\delta$ -cover of $A$ we mean a countable collection $\{B_{i}:i\in\mathbb{N}\}$ such that

\sup_{i\in\mathbb{N}}|B_{i}|\leq\delta\quad\text{ and }\quad A\subseteq\bigcup_{i\in\mathbb{N}}B_{i}.

For $s\geq 0$ and $E\subseteq\mathbb{R}$ we define

\mathcal{H}^{s}_{\delta}(E)=\inf\left\{\sum_{i\in\mathbb{N}}|B_{i}|^{s}:\{B_{i}:i\in\mathbb{N}\}\text{ is a }\delta-\text{cover of }E\right\}.

The $s$ -Hausdorff measure $\mathcal{H}^{s}$ of $E$ is the outer measure given by

\mathcal{H}^{s}(E):=\lim_{\delta\to 0}\mathcal{H}^{s}_{\delta}(E).

It is well known that for each $E\subseteq\mathbb{R}$ there is a number, called the Hausdorff dimension of $E$ and denoted $\dim_{\mathrm{H}}E$ , such that for all $s\geq 0$

\mathcal{H}^{s}(E):=\begin{cases}\infty,&\text{if}\ \ s<\dim_{\mathrm{H}}E,\\ 0,&\text{if}\ \ s>\dim_{\mathrm{H}}E.\end{cases}

Thus, the Hausdorff dimension of $E$ can also be defined as

\dim_{\mathrm{H}}E=\inf\left\{s\geq 0:\mathcal{H}^{s}(E)=0\right\}.

2.3. Continued fractions

We recall some elementary facts of continued fractions. For each real number $t$ , let $[t]$ denote its integer part. Regular continued fractions can be generated with the Gauss map $T:[0,1)\to[0,1)$ given by

T(x)=\begin{cases}x^{-1}-[x^{-1}],&x\neq 0,\\ 0,&x\neq 0.\end{cases}

For any $x\in[0,1)\setminus\mathbb{Q}$ let $a_{1}:[0,1)\setminus\mathbb{Q}\to\mathbb{N}$ be given by $a_{1}(x)=[x^{-1}]$ and for each $n\in\mathbb{N}$ we define $a_{n}(x)=a_{1}(T^{n-1}(x))$ . The numbers $(a_{n}(x))_{n\geq 1}$ are known as the partial quotients of $x$ and they are the single sequence of natural numbers satisfying

x=[a_{1}(x),a_{2}(x),\ldots].

We can extend the definition of partial quotients in an obvious manner for rational numbers. In this case, however, we obtain a finite sequence. For any non-negative integer $n$ , the $n$ -th convergent of $x$ is

\frac{p_{n}(x)}{q_{n}(x)}:=[a_{1}(x),\ldots,a_{n}(x)].

An elementary result (see, for example, [9, Proposition 1.1]) states that the sequences $(p_{n}(x))_{n\geq 0}$ , $(q_{n}(x))_{n\geq 1}$ verify

q_{0}(x)=1,\quad q_{1}(x)=a_{1}(x),\quad p_{0}(x)=0,\quad p_{1}(x)=1

and

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)\;\text{ for all }n\in\mathbb{N}.

The terms of the sequence $(q_{n}(x))_{n\geq 1}$ are called continuants. Observe that the $n$ -th continuant depends only on the first partial $n$ quotients of $x$ . Therefore, any other number $y$ whose first $n$ partial quotients are exactly those of $x$ also has the same continuants as $x$ . When we consider the partial quotients as functions on $[0,1)$ , we can easily determined their inverse images. For each $n\in\mathbb{N}$ and $\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathbb{N}^{n}$ , the fundamental interval $I_{n}(\mathbf{a})$ is

I_{n}(\mathbf{a}):=\left\{x\in[0,1):a_{1}(x)=a_{1},\ldots,a_{n}(x)=a_{n}\right\}.

In the following, we denote by $q_{n}(\mathbf{a})$ the $n$ -th convergent of any number in $I_{n}(\mathbf{a})$ . We omit the argument of $q_{n}$ if there is no risk of ambiguity.

Lemma 2.4 ([16, Section 12]).

There are some absolute constants such that for all $n\in\mathbb{N}$ and all $\mathbf{a}\in\mathbb{N}^{n}$ we have

|I_{n}(\mathbf{a})|\asymp\frac{1}{q_{n}^{2}(\mathbf{a})}.

For any $n,m\in\mathbb{N}$ , $\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathbb{N}^{n}$ , and $\mathbf{b}=(b_{1},\ldots,b_{m})\in\mathbb{N}^{m}$ , we write

(\mathbf{a},\mathbf{b}):=(a_{1},\ldots,a_{n},b_{1},\ldots,b_{m}).

Lemma 2.5 ([9, Proposition 1.1]).

For any $m,n\in\mathbb{N}$ , $\mathbf{a}\in\mathbb{N}^{n}$ , and $\mathbf{b}\in\mathbb{N}^{m}$ ,

q_{n}(\mathbf{a})q_{m}(\mathbf{b})\leq q_{n+m}(\mathbf{a},\mathbf{b})\leq 2q_{n}(\mathbf{a})q_{m}(\mathbf{b}).

Lemma 2.6 ([28, Lemma 2.1]).

For any $n\in\mathbb{N}$ , $\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathbb{N}^{n}$ , and $k\in\{1,\ldots,n\}$ , we have

\frac{a_{k}+1}{2}\leq\frac{q_{n}(a_{1},\ldots,a_{n})}{q_{n-1}(a_{1},\ldots,a_{k-1},a_{k+1},\ldots,a_{n})}\leq a_{k}+1.

Lemma 2.7 ([6, 21]).

For any constants $b,c>1$ , we have

	$\displaystyle\frac{1}{b+1}$	$\displaystyle=\dim_{\mathrm{H}}\left\{x\in[0,1):a_{n}(x)\geq b^{c^{n}}\text{ for all large }n\in\mathbb{N}\right\},$
		$\displaystyle=\dim_{\mathrm{H}}\left\{x\in[0,1):a_{n}(x)\geq b^{c^{n}}\text{ for i.m. }n\in\mathbb{N}\right\}.$

2.4. Pressure functions

For the general theory of pressure functions, we refer the reader to the monograph by Mauldin and Urbański [22]. Pressure function have gained popularity in Diophantine approximation because their zeros are the Hausdorff dimension of certain sets determined by dynamical means. In this paper, we restrict ourselves to the dynamical system $([0,1)\setminus\mathbb{Q},T)$ .

Given a function $\varphi:[0,1]\to\mathbb{R}$ , for $x\in[0,1]$ and $n\in\mathbb{N}$ , we denote by $S_{n}(\varphi;x)$ the ergodic sum

S_{n}(\varphi;x):=\varphi(x)+\varphi(T(x))+\ldots+\varphi\left(T^{n-1}(x)\right).

For each $n\in\mathbb{N}$ the $n$ –th variation of $\varphi$ is given by

\operatorname{Var}_{n}(\varphi):=\sup\left\{|\varphi(x)-\varphi(y)|:I_{n}(x)=I_{n}(y)\right\}.

For any non-empty subset $A\subseteq\mathbb{N}$ we consider

X_{A}:=\left\{x\in[0,1):a_{n}(x)\in A\text{ for all }n\in\mathbb{N}\right\}.

The pressure function $P_{A}(T,\cdot)$ on $\varphi$ is given by

P_{A}(T,\varphi):=\lim_{n\to\infty}\frac{1}{n}\log\left(\sum_{\mathbf{a}\in A^{n}}\sup_{x\in I_{n}(\mathbf{a})\cap X_{A}}\exp\left(S_{n}(\varphi,x)\right)\right).

In this context, we call $\varphi$ a potential. When $A=\mathbb{N}$ we simply write $P(T,\varphi)$ . The existence of $P_{A}(T,\varphi)$ is not immediate, but it does exist under some reasonable regularity conditions. Furthermore, we can approximate $P(T,\varphi)$ by $P_{A}(T,\varphi)$ with $A$ finite.

Lemma 2.8 ([20, Proposition 2.4]).

Let $\varphi:[0,1]\to\mathbb{R}$ be such that $\operatorname{Var}_{1}(\varphi)<\infty$ and $\operatorname{Var}_{n}(\varphi)\to 0$ as $n\to\infty$ . Then, $P_{A}(T,\varphi)$ is well defined and the supremum in the definition of $P_{A}(T,\varphi)$ may be replaced by any point in $I_{n}(\mathbf{a})\cap X_{A}$ .

Lemma 2.9 ([7, Proposition 2]).

Let $\varphi:[0,1]\to\mathbb{R}$ be such that $\operatorname{Var}_{1}(\varphi)<\infty$ and $\operatorname{Var}_{n}(\varphi)\to 0$ as $n\to\infty$ . We have

P(T,\varphi)=\sup\left\{P_{A}(T,\varphi):A\subseteq\mathbb{N}\;\text{\rm is finite}\right\}.

Given a continuous function $f:[0,1]\to\mathbb{R}$ , consider the potential

\varphi_{s}(x):=-f(s)\log B-s\log|T^{\prime}(x)|.

Then, $\varphi$ satisfies the hypotheses of Lemma 2.8 and of Lemma 2.9. Since for every $x\in[0,1)$ with $n$ partial quotients we have

(T^{n})^{\prime}(x)=\frac{(-1)^{n}}{(xq_{n-1}-p_{n-1})^{2}},

the pressure function $P(T,\cdot)$ evaluated on $\varphi_{s}$ looks like this:

P(T,\varphi_{s})=\lim_{n\to\infty}\frac{1}{n}\log\left(\sum_{\mathbf{a}\in A^{n}}\frac{1}{B^{f(s)}q_{n}^{2s}(\mathbf{a})}\right).

In our context, for some $B>1$ , we consider the potentials $\varphi_{s}$ given by

\varphi_{s}(x)=-g_{3}(s)\log B-s\log|T^{\prime}(x)|.

Define

p_{B}:=\inf\{s\geq 0:-g_{3}(s)\log B-s\log|T^{\prime}(x)|\}.

For each $m\in\mathbb{N}$ write

(4)

s_{m}(B):=\inf_{s\geq 0}\sum_{(a_{1},\ldots,a_{m})\in\mathbb{N}^{m}}\left(\frac{1}{q_{m}^{2s}B^{g_{3}(s)}}\right)\leq 1.

The arguments of [27, Section 2] allow us to conclude

\lim_{m\to\infty}s_{m}(B)=s_{B}.

Moreover, relying on Theorem 1.8, [25, Proposition 2.11], and $g_{3}(x)\leq 3x-1$ for $x\geq\frac{2}{5}$ we may show that $p_{B}\geq\frac{1}{2}$ .

The lower estimate for the Hausdorff dimension of sets similar to that in Theorem 1.10 tends to be computationally demanding. A general strategy is to construct a Cantor set $C$ contained in our set of interest, afterwards define a probablity measure on $C$ and apply the Mass Distribution Principle [5, Proposition 4.2]. Recently, Hussain and Shulga [14] computed the Hausdorff dimension of a large family of sets defined by continued fractions restrictions. For $A_{0},\ldots,A_{\ell-1}>1$ , define

S_{\ell}(A_{0},\ldots,A_{\ell-1}):=\left\{x\in[0,1):A_{0}^{n}\leq a_{n}(x)\leq 2A_{0}^{n},\ldots,A_{\ell-1}^{n}\leq a_{n+\ell-1}(x)\leq 2A_{\ell-1}^{n}\text{ for i.m. }n\in\mathbb{N}\right\}.

Put $\beta_{-1}=1$ and $\beta_{i}=A_{i}\beta_{i-1}$ for $i\in\{0,\ldots,\ell-1\}$ and for such $i$ write

d_{i}:=\inf\left\{s\geq 0:P(T,-s\log|T^{\prime}|-s\log\beta_{i}+(1-s)\log\beta_{i-1}\leq 0\right\}.

Lemma 2.10 ([14, Theorem 1.1]).

$\dim_{\mathrm{H}}S_{\ell}(A_{0},\ldots,A_{\ell-1})=\displaystyle\min_{0\leq i\leq\ell-1}d_{i}$ .

3. Series estimates

In this section we estimate certain series appearing in the proofs of theorems 1.5 and 1.10. In all of the sums studied in this section, the variables $a_{1},a_{2},\ldots$ are natural numbers. For readibilty, we omit this restriction. We remind the reader that $\ell$ stands for an arbitrary natural number.

Let us start with the lemmas needed in the proof of Theorem 1.5. The first one can be retrieved from the proof of [11, Theorem 1.5].

Lemma 3.1 ([11]).

For any $M>1$ , we have

(5)

\sum_{a_{1}\cdots a_{\ell}\geq M}\frac{1}{a_{1}^{2}\cdots a_{\ell}^{2}}\asymp_{\ell}\frac{\log^{\ell-1}(M)}{M}.

We recall a result from analytic number theory (see, for example, [8, Section 4]).

Lemma 3.2 (Dirichlet-Piltz formula).

Take $k\in\mathbb{N}$ . For each $x\in\mathbb{N}$ , let $d_{k}(x)$ denote the number of ways we can express $x$ as a product of $k$ natural numbers. For every $M>1$ , define

D_{k}(M):=\sum_{x\leq M}d_{k}(x).

There exist a polynomial $P_{k}$ of degree $k-1$ and a constant $\alpha_{k}<1$ such that for any $\varepsilon>0$ we have

D_{k}(M)=MP_{k}(\log M)+O(M^{\alpha_{k}+\varepsilon})\;\text{ as }\;M\to\infty.

Lemma 3.3.

For every large $M$ (depending on $\ell$ ) and any $r,j\in\mathbb{N}$ such that $r+j=\ell$ we have

(6)

\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}b_{1}\cdots b_{j}\geq M\\ b_{1}\cdots b_{j}c_{1}\cdots c_{r}\geq M\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}b_{1}^{2}\cdots b_{j}^{2}c_{1}^{2}\cdots c_{r}^{2}}\asymp_{\ell}\frac{\log^{j-1}(M)}{M},

(7)

\sum_{a_{1}\cdots a_{\ell}\leq M}\frac{1}{a_{1}\cdots a_{\ell}}\asymp_{\ell}\log^{\ell}(M),

(8)

\sum_{\begin{subarray}{c}a_{1}\cdots a_{\ell}<M\\ a_{2}\cdots a_{\ell+1}\geq M\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{\ell}^{2}a_{\ell+1}^{2}}\asymp_{\ell}\frac{\log^{\ell-1}M}{M}.

Remark 3.4.

The assumption $r+j=\ell$ in (6) is needed for the implied constant to depend on $\ell$ . Otherwise, as the proof makes it clear, this constant depends on $j$ .

Proof.

We start with (6). We split the sum on the left hand side of (6) into two sums according to whether $b_{1}\cdots b_{j}\geq M$ or $b_{1}\cdots b_{j}<M$ . In the first case, we have

	$\displaystyle\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}b_{1}\cdots b_{j}\geq M\\ b_{1}\cdots b_{j}c_{1}\cdots c_{r}\geq M\\ b_{1}\cdots b_{j}\geq M\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}b_{1}^{2}\cdots b_{j}^{2}c_{1}^{2}\cdots c_{r}^{2}}$	$\displaystyle=\sum_{b_{1}\cdots b_{j}\geq M}\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}b_{1}\cdots b_{j}\geq M\\ b_{1}\cdots b_{j}c_{1}\cdots c_{r}\geq M\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}b_{1}^{2}\cdots b_{j}^{2}c_{1}^{2}\cdots c_{r}^{2}}$
		$\displaystyle=\sum_{b_{1}\cdots b_{j}\geq M}\frac{1}{b_{1}^{2}\cdots b_{j}^{2}}\sum_{\begin{subarray}{c}a_{1},\ldots,a_{r}\in\mathbb{N}\\ c_{1},\ldots,c_{r}\in\mathbb{N}\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}c_{1}^{2}\cdots c_{r}^{2}}$
		$\displaystyle\asymp\frac{\pi^{4r}}{6^{2r}}\frac{\log^{j-1}M}{M}.$

For the second case, we split the sum corresponding to $b_{1}\cdots b_{j}<M$ into two further sums:

\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}b_{1}\cdots b_{j}\geq M\\ b_{1}\cdots b_{j}c_{1}\cdots c_{r}\geq M\\ \frac{M}{2}<b_{1}\cdots b_{j}<M\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}b_{1}^{2}\cdots b_{j}^{2}c_{1}^{2}\cdots c_{r}^{2}}+\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}b_{1}\cdots b_{j}\geq M\\ b_{1}\cdots b_{j}c_{1}\cdots c_{r}\geq M\\ b_{1}\cdots b_{j}<\frac{M}{2}\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}b_{1}^{2}\cdots b_{j}^{2}c_{1}^{2}\cdots c_{r}^{2}}

Let us start with the first one. Observe that

\frac{M}{2}<b_{1}\cdots b_{j}<M\quad\text{ if and only if}\quad 1<\frac{M}{b_{1}\cdots b_{j}}<2,

hence

	$\displaystyle\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}b_{1}\cdots b_{j}\geq M\\ b_{1}\cdots b_{j}c_{1}\cdots c_{r}\geq M\\ \frac{M}{2}<b_{1}\cdots b_{j}<M\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}b_{1}^{2}\cdots b_{j}^{2}c_{1}^{2}\cdots c_{r}^{2}}$	$\displaystyle=\sum_{\frac{M}{2}<b_{1}\cdots b_{j}<M}\frac{1}{b_{1}^{2}\cdots b_{j}^{2}}\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}\geq\frac{M}{b_{1}\cdots b_{j}}\\ c_{1}\cdots c_{r}\geq\frac{M}{b_{1}\cdots b_{j}}\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}c_{1}^{2}\cdots c_{r}^{2}}$
		$\displaystyle=\sum_{\frac{M}{2}<b_{1}\cdots b_{j}<M}\frac{1}{b_{1}^{2}\cdots b_{j}^{2}}\left(\left(\sum_{a\geq 1}\frac{1}{a^{2}}\right)^{r}-1\right)^{2}$
		$\displaystyle\asymp_{\ell}\sum_{\frac{M}{2}<b_{1}\cdots b_{j}<M}\frac{1}{b_{1}^{2}\cdots b_{j}^{2}}\ll\frac{\log^{j-1}M}{M},$

Lastly, we have

	$\displaystyle\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}b_{1}\cdots b_{j}\geq M\\ b_{1}\cdots b_{j}c_{1}\cdots c_{r}\geq M\\ b_{1}\cdots b_{j}\leq\frac{M}{2}\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}b_{1}^{2}\cdots b_{j}^{2}c_{1}^{2}\cdots c_{r}^{2}}$	$\displaystyle=\sum_{b_{1}\cdots b_{j}\leq\frac{M}{2}}\frac{1}{b_{1}^{2}\cdots b_{j}^{2}}\sum_{\begin{subarray}{c}a_{1}\cdots a_{r}\geq\frac{M}{b_{1}\cdots b_{j}}\\ c_{1}\cdots c_{r}\geq\frac{M}{b_{1}\cdots b_{j}}\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}c_{1}^{2}\cdots c_{r}^{2}}$
		$\displaystyle=\sum_{b_{1}\cdots b_{j}\leq\frac{M}{2}}\frac{1}{b_{1}^{2}\cdots b_{j}^{2}}\left(\sum_{a_{1}\cdots a_{r}\geq\frac{M}{b_{1}\cdots b_{j}}}\frac{1}{a_{1}^{2}\cdots a_{r}^{2}}\right)^{2}$
		$\displaystyle\asymp\sum_{b_{1}\cdots b_{j}\leq\frac{M}{2}}\frac{1}{b_{1}^{2}\cdots b_{j}^{2}}\left(\frac{1}{\frac{M}{b_{1}\cdots b_{j}}}\right)^{2}$
		$\displaystyle=\frac{1}{M^{2}}\sum_{b_{1}\cdots b_{j}\leq\frac{M}{2}}1\asymp\frac{\log^{j-1}M}{M}.$

The last step follows from Dirichlet-Piltz formula.

The estimate (7) is clear when $\ell=1$ . Assume that it holds for some $\ell\in\mathbb{N}$ . It is easily seen that

(9)

\int\frac{1}{t}\log^{\ell}\left(\frac{M}{t}\right)\mathrm{d}t=\frac{-1}{\ell+1}\log^{\ell+1}\left(\frac{M}{t}\right)+C.

We write the sum for $\ell+1$ as follows:

\sum_{a_{1}\cdots a_{\ell}a_{\ell+1}\leq M}\frac{1}{a_{1}\cdots a_{\ell}}=\sum_{1\leq a_{\ell+1}<\frac{M}{2}}\frac{1}{a_{\ell+1}}\sum_{a_{1}\cdots a_{\ell}\leq\frac{M}{a_{\ell+1}}}\frac{1}{a_{1}\cdots a_{\ell}}+\sum_{\frac{M}{2}\leq a_{\ell+1}\leq M}\frac{1}{a_{\ell+1}}\sum_{a_{1}\cdots a_{\ell}\leq\frac{M}{a_{\ell+1}}}\frac{1}{a_{1}\cdots a_{\ell}}

We use the induction hypothesis and (9) on the first term to get

\sum_{1\leq a_{\ell+1}<\frac{M}{2}}\frac{1}{a_{\ell+1}}\sum_{a_{1}\cdots a_{\ell}\leq\frac{M}{a_{\ell+1}}}\frac{1}{a_{1}\cdots a_{\ell}}\asymp\sum_{1\leq a_{\ell+1}<\frac{M}{2}}\frac{1}{a_{\ell+1}}\log^{\ell}\left(\frac{M}{a_{\ell+1}}\right)\asymp\log^{\ell+1}(M).

The second term is bounded, because of (9):

\sum_{\frac{M}{2}\leq a_{\ell+1}\leq M}\frac{1}{a_{\ell+1}}\sum_{a_{1}\cdots a_{\ell}\leq\frac{M}{a_{\ell+1}}}\frac{1}{a_{1}\cdots a_{\ell}}\asymp\int_{M/2}^{M}\frac{1}{t}\log^{\ell}\frac{M}{t}\mathrm{d}t\asymp_{\ell}1.

Therefore, (7) also holds for $\ell+1$ .

Lastly, we prove (8). First, we rewrite the series in (8) as

\sum_{\begin{subarray}{c}a_{1}\cdots a_{\ell}<M\\ a_{2}\cdots a_{\ell+1}\geq M\end{subarray}}\frac{1}{a_{1}^{2}\cdots a_{\ell+1}^{2}}=\sum_{a_{2}\cdots a_{\ell}\leq M}\sum_{a_{1}\leq\frac{M}{a_{2}\ldots a_{\ell}}}\sum_{a_{\ell+1}\geq\frac{M}{a_{2}\cdots a_{\ell}}}\frac{1}{a_{1}^{2}\cdots a_{\ell+1}^{2}}.

We split the last series into two by conditioning on whether $1\leq a_{2}\ldots a_{\ell}\leq\frac{M}{2}$ or $\frac{M}{2}<a_{2}\cdots a_{\ell}<M$ . In the first case, since $1\leq\sum_{j=1}^{n}j^{-2}<3$ for all $n\in\mathbb{N}$ and because of (7), we have

	$\displaystyle\sum_{a_{2}\cdots a_{\ell}\leq\frac{M}{2}}\sum_{a_{1}\leq\frac{M}{a_{2}\ldots a_{\ell}}}\sum_{a_{\ell+1}\geq\frac{M}{a_{2}\cdots a_{\ell}}}\frac{1}{a_{1}^{2}\cdots a_{\ell+1}^{2}}$	$\displaystyle\asymp\sum_{a_{2}\cdots a_{\ell+1}\leq\frac{M}{2}}\frac{1}{a_{2}^{2}\cdots a_{\ell}^{2}}\sum_{a_{1}\leq\frac{M}{a_{2}\ldots a_{\ell+1}}}\frac{1}{a_{1}^{2}}\sum_{a_{\ell+2}\geq\frac{M}{a_{2}\cdots a_{\ell+1}}}\frac{1}{a_{\ell+1}^{2}}$
		$\displaystyle\asymp\sum_{a_{2}\cdots a_{\ell}\leq\frac{M}{2}}\frac{1}{a_{2}^{2}\cdots a_{\ell}^{2}}\left(\sum_{a_{\ell+1}\geq\frac{M}{a_{2}\cdots a_{\ell}}}\frac{1}{a_{\ell+1}^{2}}\right)$
		$\displaystyle\asymp\frac{1}{M}\sum_{a_{2}\cdots a_{\ell}\leq\frac{M}{2}}\frac{1}{a_{2}\cdots a_{\ell}}$
		$\displaystyle\asymp\frac{\log^{\ell-1}M}{M}.$

When $\frac{M}{2}<a_{2}\cdots a_{\ell}<M$ , the condition $a_{1}\leq\frac{M}{a_{2}\ldots a_{\ell}}$ reduces to $a_{1}=1$ and $a_{\ell+1}\geq\frac{M}{a_{2}\cdots a_{\ell}}$ reduces to $a_{\ell+1}\geq 2$ , then

	$\displaystyle\sum_{\frac{M}{2}\leq a_{2}\cdots a_{\ell}\leq M}\sum_{a_{1}\leq\frac{M}{a_{2}\ldots a_{\ell}}}\sum_{a_{\ell+2}\geq\frac{M}{a_{1}\cdots a_{\ell}}}\frac{1}{a_{1}^{2}\cdots a_{\ell+1}^{2}}$	$\displaystyle\asymp\sum_{a_{2}\cdots a_{\ell}\leq M}\frac{1}{a_{2}^{2}\cdots a_{\ell}^{2}}$
		$\displaystyle\asymp\sum_{a_{2}\cdots a_{\ell}\leq\frac{M}{2}}\frac{1}{a_{2}^{2}\cdots a_{\ell}^{2}}\asymp\frac{\log^{\ell-2}M}{M}.$

This proves (8). ∎

Theorem 1.8 requires similar estimates where the exponents may take non-integer values.

Lemma 3.5 ([11, Lemma 4.2]).

For $\beta>1$ , $0<s<1$ , and $n\in\mathbb{N}$ , we have

(10)

\sum_{a_{1}\cdots a_{\ell}\leq\beta^{n}}\left(\frac{1}{a_{1}\cdots a_{\ell}}\right)^{s}\asymp\frac{(\log\beta^{n})^{{\ell}-1}}{({\ell}-1)!}\beta^{n(1-s)}.

Lemma 3.6.

For any $M>1$ and any $t>1$ we have

(11)

\sum_{ab\geq M}\frac{1}{a^{t}b^{t}}\asymp\frac{M^{1-t}}{t-1}\log M,

(12)

\sum_{abc\geq M}\frac{1}{a^{t}b^{t}c^{t}}\ll\left(\frac{1}{t-1}+\log M\right)M^{1-t}.

Proof.

The estimate (11) is shown as (8) and (12) follows from (11). ∎

4. Proof of Theorem 1.5

For any $n\in\mathbb{N}_{\geq\ell}$ and $k\in\{1,\ldots,n-1\}$ , consider

A_{n,k}(\varphi):=\left\{x\in[0,1):\begin{matrix}a_{k}(x)\cdots a_{k+\ell-1}(x)\geq\varphi(n),\\ a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)\end{matrix}\right\}

and define

A_{n}(\varphi):=\bigcup_{k=1}^{n-1}A_{n,k}(\varphi).

As in [26], we may check that the monotonocity of $\varphi$ implies

\mathcal{F}_{\ell}(\varphi)=\limsup_{n\to\infty}A_{n}(\varphi).

4.1. Convergence case

Take $n\in\mathbb{N}_{\geq\ell}$ . We estimate $\mathfrak{m}(A_{n,k})$ by considering two cases according to whether the restricted blocks overlap or not.

Lemma 4.1.

Let $k\in\{1,\ldots,n-1\}$ .

1.

If $1\leq k\leq n-\ell$ , then

$\mathfrak{m}(A_{n,k})\asymp_{\ell}\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}.$
2.

If $n-\ell+1\leq k\leq n-1$ , then

$\mathfrak{m}(A_{n,k})\asymp_{\ell}\frac{\log^{\ell+k-n-1}\varphi(n)}{\varphi(n)}.$

Proof.

Assume that $1\leq k\leq n-\ell$ , so $k+\ell-1\leq n-1$ . Let $\mathbf{b}=(b_{1},\ldots,b_{n+\ell-1})\in\mathbb{N}^{n+\ell-1}$ such that $A_{n,k}\cap I(\mathbf{b})\neq\varnothing$ . Clearly $I(\mathbf{b})\subseteq A_{n,k}$ . Write

\mathbf{b}^{\prime}=(b_{1},\ldots,b_{k-1},b_{k+\ell},\ldots,b_{n-1}),

that is, $\mathbf{b}^{\prime}$ is $\mathbf{b}$ without the blocks $(b_{k},\ldots,b_{k+\ell-1})$ and $(b_{n},\ldots,b_{n+\ell-1})$ , then

\mathfrak{m}(I_{n+\ell-1}(\mathbf{b}))\asymp_{\ell}\mathfrak{m}(I_{n-\ell-1}(\mathbf{b}^{\prime}))\,\frac{1}{(b_{k}^{2}\cdots b_{k+\ell-1}^{2})(b_{n}^{2}\cdots b_{n+\ell-1}^{2})}.

Therefore, we have

	$\displaystyle\mathfrak{m}(A_{n,k})$	$\displaystyle=\sum_{\mathbf{b}\in\mathbb{N}^{n+\ell-1}}\mathfrak{m}\left(A_{n,k}\cap I_{n+\ell-1}(\mathbf{b})\right)$
		$\displaystyle=\sum_{\begin{subarray}{c}\mathbf{b}\in\mathbb{N}^{n+\ell-1}\\ b_{k}\cdots b_{k+\ell-1}\geq\varphi(n)\\ b_{n}\cdots b_{n+\ell-1}\geq\varphi(n)\end{subarray}}\mathfrak{m}\left(I_{n+\ell-1}(\mathbf{b})\right)$
		$\displaystyle\asymp\sum_{\begin{subarray}{c}\mathbf{b}^{\prime}\in\mathbb{N}^{n-\ell-1}\\ b_{k}\cdots b_{k+\ell-1}\geq\varphi(n)\\ b_{n}\cdots b_{n+\ell-1}\geq\varphi(n)\end{subarray}}\mathfrak{m}(I_{n-\ell-1}(\mathbf{b}^{\prime}))\,\frac{1}{(b_{k}^{2}\cdots b_{k+\ell-1}^{2})(b_{n}^{2}\cdots b_{n+\ell-1}^{2})}.$
		$\displaystyle=\left(\sum_{\mathbf{b}^{\prime}\in\mathbb{N}^{n-\ell-1}}\mathfrak{m}(I_{n-\ell-1}(\mathbf{b}^{\prime}))\right)\left(\sum_{b_{k}\cdots b_{k+\ell-1}\geq\varphi(n)}\frac{1}{b_{k}^{2}\cdots b_{k+\ell-1}^{2}}\right)\left(\sum_{b_{n}\cdots b_{n+\ell-1}\geq\varphi(n)}\frac{1}{b_{n}^{2}\cdots b_{n+\ell-1}^{2}}\right)$
		$\displaystyle=\left(\sum_{b_{1}\cdots b_{\ell}\geq\varphi(n)}\frac{1}{b_{1}^{2}\cdots b_{\ell}^{2}}\right)^{2}\asymp\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}.$

The last estimate follows from (5).

Assume that $n-\ell+1\leq k\leq n-1$ and write $j=\ell+k-n$ . Then, $j$ is the length of the overlap of the blocks appearing in the definition of $A_{n,k}$ . We may argue as in the previous case and apply (6) rather than (5) to conclude

\mathfrak{m}(A_{n,k})\asymp_{\ell}\sum_{\begin{subarray}{c}b_{k}\cdots b_{k+\ell-1}\geq\varphi(n)\\ b_{n}\cdots b_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{b_{k}^{2}\cdots b_{k+\ell-1}^{2}b_{k+\ell}^{2}\cdots b_{n+\ell-1}^{2}}\asymp\frac{\log^{j-1}\varphi(n)}{\varphi(n)}.

∎

Proof of Theorem 1.5. Convergence case..

For any $n\in\mathbb{N}_{\geq\ell}$ we have

	$\displaystyle\mathfrak{m}(A_{n})$	$\displaystyle\leq\sum_{k=1}^{n}\mathfrak{m}(A_{n,k})$
		$\displaystyle=\sum_{k=1}^{n-\ell}\mathfrak{m}(A_{n,k})+\sum_{k=n-\ell+1}^{n-1}\mathfrak{m}(A_{n,k})$
		$\displaystyle\asymp(n-\ell)\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}+\frac{1}{\varphi(n)}\sum_{k=n-\ell+1}^{n-1}\log^{k+\ell-n-1}\varphi(n)$
		$\displaystyle\asymp\frac{n\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}+\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}.$

The result follows from the convergence part of the Borel-Cantelli Lemma. ∎

4.2. Divergence case

We may impose two additional assumptions without losing any generality. First, we suppose that

\lim_{n\to\infty}\varphi(n)=\infty.

If this was not the case, then every irrational number with unbounded partial quotients would belong to $\mathcal{F}_{\ell}(\varphi)$ and, thus, $\mathfrak{m}(\mathcal{F}_{\ell}(\varphi))=1$ by the Borel-Bernstein Theorem. Second, we may assume that ¹¹1There seems to be a gap in the proof of this assertion for $\ell=2$ in [26]. Consider $0<\alpha<1$ and $\varphi(n)=n(\log n)^{\alpha}$ . Then, $\sum_{n}\varphi(n)^{-1}=\infty$ and $\varphi(n)<n\log\varphi(n)$ for all large $n$ . According to [26, Section 3.2], we should consider the function $\psi$ given by $\psi(n)=\max\{\varphi(n),n\log\varphi(n)\}$ . However, for all large $n$ we have $n\log\psi(n)=n\log(n\log\varphi(n))>n\log\varphi(n)=\psi(n)$ . This minor gap does not affect validity of that paper at all.

(13)

\varphi(n)\geq n\log^{\ell-1}\varphi(n)\quad\text{ for all }n\in\mathbb{N}.

Certainly, for each $n\in\mathbb{N}$ let $x_{n}$ be the solution $x$ of the equation

\frac{x}{\log^{\ell-1}x}=n.

Then, $(x_{n})_{n\geq 1}$ tends to infinity and it is eventually strictly increasing. Let $\psi:\mathbb{N}\to\mathbb{N}$ be given by

\psi(n)=\max\{\varphi(n),x_{n}\}\quad\text{ for all }n\in\mathbb{N}.

We can trivially modify at most finitely many terms of $\psi$ to ensure that it is non-decreasing. Furthermore, since $\psi\geq\varphi$ , we have $\mathcal{E}_{\ell}(\psi)\subseteq\mathcal{E}_{\ell}(\varphi)$ and $\psi(n)\to\infty$ as $n\to\infty$ . It is also clear that $\psi(n)\geq n\log^{\ell-1}\psi(n)$ for all $n\in\mathbb{N}$ , because $x\mapsto\frac{x}{\log^{\ell-1}x}$ is strictly increasing for large $x$ . Lastly, let us show that

\sum_{n=1}^{\infty}n\left(\frac{\log^{\ell-1}\psi(n)}{\psi(n)}\right)^{2}+\frac{\log^{\ell-2}\psi(n)}{\psi(n)}=\infty.

The divergence is obvious if $\varphi(n)>x_{n}$ for all but finitely many $n\in\mathbb{N}$ . If this is not the case, then there is a strictly increasing sequence of natural numbers $(m_{k})_{k\geq 1}$ such that

m_{k+1}>2m_{k}\text{ and }\varphi(m_{k})<x_{m_{k}}\;\text{ for all }\;k\in\mathbb{N}.

Hence,

	$\displaystyle\sum_{n=1}^{\infty}n\left(\frac{\log^{\ell-1}\psi(n)}{\psi(n)}\right)^{2}+\frac{\log^{\ell-2}\psi(n)}{\psi(n)}$	$\displaystyle\geq\sum_{k=1}^{\infty}\sum_{\left[\frac{m_{k+1}}{2}\right]\leq n\leq m_{k+1}}n\left(\frac{\log^{\ell-1}\psi(n)}{\psi(n)}\right)^{2}$
		$\displaystyle\geq\sum_{k=1}^{\infty}\frac{m_{k+1}}{2}\sum_{\left[\frac{m_{k+1}}{2}\right]\leq n\leq m_{k+1}}\left(\frac{\log^{\ell-1}\psi(n)}{\psi(n)}\right)^{2}$
		$\displaystyle\geq\sum_{k=1}^{\infty}\frac{m_{k+1}}{2}\sum_{\left[\frac{m_{k+1}}{2}\right]\leq n\leq m_{k+1}}\left(\frac{\log^{\ell-1}x_{m_{k+1}}}{x_{m_{k+1}}}\right)^{2}$
		$\displaystyle=\sum_{k=1}^{\infty}\frac{m_{k+1}}{2}\sum_{\left[\frac{m_{k+1}}{2}\right]\leq n\leq m_{k+1}}\left(\frac{1}{m_{k+1}}\right)^{2}$
		$\displaystyle\geq\sum_{k=1}^{\infty}\frac{1}{4}=\infty.$

Take $t\in\mathbb{N}$ , $\mathbf{b}=(b_{1},\ldots,b_{t})\in\mathbb{N}^{t}$ , write $I_{t}=I_{t}(\mathbf{b})$ , and let $M=M(\mathbf{b})\in\mathbb{N}_{\geq t+1}$ be such that $\varphi(M)>\left(\max\{b_{1},\ldots,b_{t}\}\right)^{\ell}$ .

Lemma 4.2.

If $n\in\mathbb{N}_{\geq M}$ , then

\mathfrak{m}\left(A_{n}\cap I_{t}\right)\gg_{\ell}\mathfrak{m}(I_{t})\left(\frac{n\log^{2(\ell-1)}\varphi(n)}{\varphi(n)^{2}}+\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}\right).

Proof.

Take $n\in\mathbb{N}_{\geq M}$ . Define the set

B_{n}:=\left\{x\in I_{t}:\begin{matrix}a_{n-1}(x)\cdots a_{n+\ell-2}(x)\geq\varphi(n),\\ a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)\end{matrix}\right\}.

For each $i\in\mathbb{N}$ such that $M\leq i<n-1$ and $i\equiv n-1\pmod{\ell}$ , let $B_{i}$ be the set of those $x\in I_{t}$ satisfying the next properties:

1.

$a_{i}(x)\cdots a_{i+\ell-1}(x)\geq\varphi(n)$ ,
2.

For each $k\in\mathbb{N}$ such that $0<k<\frac{n-1-i}{\ell}$ we have $a_{i+k\ell}(x)\cdots a_{i+(k+1)\ell-1}(x)<\varphi(n)$ .
3.

$a_{n-1}(x)\cdots a_{n+\ell-2}(x)<\varphi(n)$ and $a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)$ .

Note that we are imposing restrictions on non-overlapping contiguous blocks of length $\ell$ . Clearly, we have

(14)

B_{n}\cup\bigcup_{\begin{subarray}{c}M\leq i<n-1\\ i\equiv n-1(\ell)\end{subarray}}B_{i}\subseteq A_{n}\cap I_{t},

where the union is disjoint. From (6) we obtain

(15)

\mathfrak{m}(B_{n})\asymp_{\ell}\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}\mathfrak{m}(I_{t}).

Take $i$ such that $i\equiv n-1\pmod{\ell}$ and $M\leq i<n-1$ . As in the convergence case, in order to estimate $\mathfrak{m}(B_{i})$ it suffices to estimate the series

\sum_{a_{i}\cdots a_{i+\ell-1}\geq\varphi(n)}\,\sum_{\begin{subarray}{c}a_{i+k\ell}\cdots a_{i+(k+1)\ell-1}<\varphi(n)\\ 0<k<\frac{n-1-i}{\ell}\end{subarray}}\,\sum_{\begin{subarray}{c}a_{n-1}\cdots a_{n+\ell-2}<\varphi(n)\\ a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{i}^{2}\cdots a_{n+\ell-2}^{2}}.

From (5) and (8) we obtain

\sum_{a_{i}\cdots a_{i+\ell-1}\geq\varphi(n)}\frac{1}{a_{i}^{2}\cdots a_{i+\ell-1}^{2}}\;\sum_{\begin{subarray}{c}a_{n-1}\cdots a_{n+\ell-2}<\varphi(n)\\ a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{n-1}^{2}\cdots a_{n+\ell+1}^{2}}\asymp\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}.

Also, for every integer $k$ with $0<k<\frac{n-1-i}{\ell}$ we have

\sum_{a_{i+k\ell}\cdots a_{i+(k+1)\ell-1}<\varphi(n)}\frac{1}{a_{i+k\ell}^{2}\cdots a_{i+(k+1)\ell-1}^{2}}\gg 1-\frac{\log^{\ell-1}\varphi(n)}{2\varphi(n)}\geq 1-\frac{1}{2n}.

Hence, by (14), we arrive at

\mathfrak{m}(B_{i})\gg\mathfrak{m}(I_{t})\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi(n)}\left(1-\frac{1}{2n}\right)^{n-i}

and we conclude

(16)

\mathfrak{m}\left(\bigcup_{\begin{subarray}{c}M\leq i<n-1\\ i\equiv n-1(\ell)\end{subarray}}B_{i}\right)\gg\mathfrak{m}(I_{t})\frac{n\log^{2(\ell-1)}\varphi(n)}{\varphi(n)}.

The lemma follows from (14), (15), and (16). ∎

Lemma 4.3.

For large $N\in\mathbb{N}$ we have

(17)

\sum_{M\leq m<n\leq N}\mathfrak{m}\left(A_{m}\cap A_{n}\cap I_{t}\right)\ll_{\ell}\sum_{n=M}^{N}\mathfrak{m}\left(A_{n}\cap I_{t}\right)+\mathfrak{m}(I_{t})^{-1}\left(\sum_{n=M}^{N}\mathfrak{m}\left(A_{n}\cap I_{t}\right)\right)^{2}.

Proof.

Consider the index set

\mathcal{I}:=\{(m,n)\in\mathbb{N}^{2}:M\leq m<n\leq N\}.

We partition $\mathcal{I}$ into

\mathcal{I}_{1}:=\{(m,n)\in\mathcal{I}:M\leq m\leq n-(2\ell-1)\},\quad\mathcal{I}_{2}:=\{(m,n)\in\mathcal{I}:n-2\ell+2\leq m\leq n-1\}\}

and write

S_{1}:=\sum_{(m,n)\in\mathcal{I}_{1}}\mathfrak{m}\left(A_{m}\cap A_{n}\cap I_{t}\right)\;\text{ and }\;S_{2}:=\sum_{(m,n)\in\mathcal{I}_{2}}\mathfrak{m}\left(A_{m}\cap A_{n}\cap I_{t}\right).

It is clear that

S_{2}\leq 2\ell\sum_{n=1}^{N}\mathfrak{m}\left(A_{n}\cap I_{t}\right).

Now we give an upper bound for $S_{1}$ . For each $(m,n)\in\mathcal{I}_{1}$ define the sets $U_{1}(m,n)$ , $U_{2}(m,n)$ , $U_{3}(m,n)$ as follows:

1.
Let $U_{1}(m,n)$ be the set of irrational numbers $x\in I_{t}$ such that
- •
  
  $a_{m}(x)\cdots a_{m+\ell-1}(x)\geq\varphi(m)$ ,
- •
  
  For some $k<m$ we have $a_{k}(x)\cdots a_{k+\ell-1}(x)\geq\varphi(n)$ ,
- •
  
  $a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)$ .
2.
Let $U_{2}(m,n)$ be the set of irrational numbers $x\in I_{t}$ such that
- •
  
  For some $j<m$ we have $a_{j}(x)\cdots a_{j+\ell-1}(x)\geq\varphi(m)$ ,
- •
  
  $a_{m}(x)\cdots a_{m+\ell-1}(x)\geq\varphi(n)$ ,
- •
  
  $a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)$ .
3.
Let $U_{3}(m,n)$ be the set of irrational numbers $x\in I_{t}$ such that
- •
  
  For some $j<m$ we have $a_{j}(x)\cdots a_{j+\ell-1}\geq\varphi(m)$ ,
- •
  
  $a_{m}(x)\cdots a_{m+\ell-1}(x)\geq\varphi(m)$ ,
- •
  
  For some $m<k<n$ we have $a_{k}(x)\cdots a_{k+\ell-1}\geq\varphi(n)$ ,
- •
  
  $a_{n}(x)\cdots a_{m+\ell-1}(x)\geq\varphi(m)$ .

We claim that

(18)

\mathfrak{m}(U_{1}(m,n))\ll\mathfrak{m}(I_{t})\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)},

(19)

\mathfrak{m}(U_{2}(m,n))\ll\mathfrak{m}(I_{t})\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)},

	$\displaystyle\mathfrak{m}(U_{3}(m,n))$	$\displaystyle\ll\mathfrak{m}(I_{t})\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}+$
(20)			$\displaystyle\quad\;+\mathfrak{m}(I_{t})\left(\frac{n\log\varphi(n)}{\varphi^{2}(n)}+\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}\right)\left(\frac{m\log\varphi(m)}{\varphi^{2}(m)}+\frac{\log^{\ell-2}\varphi(n)}{\varphi(m)}\right).$

Let us show (18). When $M\leq k\leq m-\ell$ , we apply (5) to get

\sum_{\begin{subarray}{c}a_{k}\cdots a_{k+\ell-1}\geq\varphi(n)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(m)\\ a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{k}^{2}\cdots a_{k+\ell-1}^{2}a_{m}^{2}\cdots a_{m+\ell-1}^{2}a_{n}^{2}\cdots a_{n+\ell-1}^{2}}\asymp\frac{\log\varphi(m)}{\varphi(m)}\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}.

For $m-\ell+1\leq k\leq m-1$ we have

	$\displaystyle\sum_{\begin{subarray}{c}a_{k}\cdots a_{k+\ell-1}\geq\varphi(n)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(m)\\ a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{k}^{2}\cdots a_{m+\ell-1}^{2}a_{n}^{2}\cdots a_{n+\ell-1}^{2}}$	$\displaystyle\leq\sum_{\begin{subarray}{c}a_{k}\cdots a_{k+\ell-1}\geq\varphi(n)\\ a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{k}^{2}\cdots a_{m+\ell-1}^{2}a_{n}^{2}\cdots a_{n+\ell-1}^{2}}$
		$\displaystyle\ll\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}.$

Then, the above inequalities along with (13) yield

	$\displaystyle\mathfrak{m}(U_{1}(m,n))$	$\displaystyle\leq m\frac{\log^{\ell-1}\varphi(m)}{\varphi(m)}\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}+\ell\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}$
		$\displaystyle\ll\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}.$

This shows (18). Note that we have used that for the values of $k$ considered the blocks $(a_{k},\ldots,a_{k+\ell-1})$ and $(a_{n},\ldots,a_{n+\ell-1})$ do not overlap. Here and in the following, whenever we assert something about blocks of a sequence of natural numbers without quantifiers, it should be understood that the assertion holds for every sequence of natural numbers.

We consider two different series to estimate $\mathfrak{m}(U_{2}(m,n))$ . First, if $M\leq j\leq m-\ell$ then the blocks $(a_{j},\ldots,a_{j+\ell-1})$ and $(a_{m},\ldots,a_{m+\ell-1})$ do not overlap and

	$\displaystyle\sum_{j=M}^{m-\ell}\sum_{\begin{subarray}{c}a_{j}\cdots a_{j+\ell-1}\geq\varphi(m)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{j}^{2}\cdots a_{j+\ell-1}^{2}a_{m}^{2}\cdots a_{m+\ell-1}^{2}}$	$\displaystyle\ll m\frac{\log^{\ell-1}\varphi(m)}{\varphi(m)}\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}$
		$\displaystyle\leq\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}.$

The product $a_{j}^{2}\cdots a_{m+\ell-1}^{2}$ in the series to come should be interpreted as follows: if $j+\ell-1\leq m$ , then $a_{j}^{2}\cdots a_{m+\ell-1}^{2}$ is the product of the products of the blocks $(a_{j},\ldots,a_{j+\ell-1})$ and $(a_{m},\ldots,a_{m+\ell-1})$ ; if $j+\ell-1>m$ , then $a_{j}^{2}\cdots a_{m+\ell-1}^{2}$ is the product of the block $(a_{j},a_{j+1},\ldots,a_{m+\ell-1})$ . We use a similar convention for other combinations of the parameters $j,k,m,n$ .

When $m-\ell+1\leq j\leq m-1$ , the blocks $(a_{j},\ldots,a_{j+\ell-1})$ and $(a_{m},\ldots,a_{m+\ell-1})$ overlap and

	$\displaystyle\sum_{j=m-\ell+1}^{m-1}\sum_{\begin{subarray}{c}a_{j}\cdots a_{j+\ell-1}\geq\varphi(m)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{j}^{2}\cdots a_{m+\ell-1}^{2}}$	$\displaystyle\leq\sum_{j=m-\ell+1}^{m-1}\sum_{a_{m}\cdots a_{m+\ell-1}\geq\varphi(n)}\frac{1}{a_{j}^{2}\cdots a_{m+\ell-1}^{2}}$
		$\displaystyle\ll_{\ell}\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}.$

We may now argue as in the convergence case recalling the restriction $a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)$ to conclude (19).

Next, we show (4.2). First, we consider the values of $k$ for which the block $(a_{k},\ldots,a_{k+\ell-1})$ does not overlap neither $(a_{n},\ldots,a_{n+\ell-1})$ nor $(a_{m},\ldots,a_{m+\ell-1})$ . In other words, we assume $m+\ell\leq k\leq n-\ell$ . This leads us to the series

\sum_{j=M}^{m-1}\sum_{k=m+\ell}^{n-\ell}\sum_{\begin{subarray}{c}a_{j}\cdots a_{j+\ell-1}\geq\varphi(m)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(m)\\ a_{k}\cdots a_{k+\ell-1}\geq\varphi(n)\\ a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{j}^{2}\cdots a_{m+\ell-1}^{2}a_{k}^{2}\cdots a_{k+\ell-1}^{2}a_{n}^{2}\cdots a_{n+\ell-1}^{2}},

which is asymptotically equivalent to

	$\displaystyle\left(\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}\right)^{2}$	$\displaystyle\sum_{j=M}^{m-1}\sum_{\begin{subarray}{c}a_{j}\cdots a_{j+\ell-1}\geq\varphi(m)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(m)\end{subarray}}\frac{1}{a_{j}^{2}\cdots a_{m+\ell-1}^{2}}\ll$
		$\displaystyle\ll\left(\frac{\log^{\ell-1}\varphi(n)}{\varphi(n)}\right)^{2}\left(m\frac{\log^{2(\ell-1)}\varphi(m)}{\varphi(m)}+\frac{\log^{\ell-2}\varphi(m)}{\varphi(m)}\right).$

The last estimate is obtained as in the convergence case. Now suppose that $n-\ell+1\leq k\leq n-1$ . This means that $(a_{k},\ldots,a_{k+\ell-1})$ and $(a_{n},\ldots,a_{n+\ell-1})$ overlap but $(a_{k},\ldots,a_{k+\ell-1})$ and $(a_{m},\ldots,a_{m+\ell-1})$ do not (by the definition of $\mathcal{I}_{1}$ ), hence

	$\displaystyle\sum_{j=M}^{m-1}$	$\displaystyle\sum_{k=n-\ell+1}^{n-1}\sum_{\begin{subarray}{c}a_{j}\cdots a_{j+\ell-1}\geq\varphi(m)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(m)\\ a_{k}\cdots a_{k+\ell-1}\geq\varphi(n)\\ a_{n}\cdots a_{n+\ell-1}\geq\varphi(n)\end{subarray}}\frac{1}{a_{j}^{2}\cdots a_{m+\ell-1}^{2}a_{k}^{2}\cdots a_{n+\ell-1}^{2}}\ll$
		$\displaystyle\ll\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}\sum_{j=M}^{m-1}\sum_{\begin{subarray}{c}a_{j}\cdots a_{j+\ell-1}\geq\varphi(m)\\ a_{m}\cdots a_{m+\ell-1}\geq\varphi(m)\end{subarray}}\frac{1}{a_{j}^{2}\cdots a_{m+\ell-1}^{2}}$
		$\displaystyle\ll\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}\left(m\frac{\log^{2(\ell-1)}\varphi(m)}{\varphi(m)}+\frac{\log^{\ell-2}\varphi(m)}{\varphi(m)}\right).$

Lastly, we deal with the values of $k$ for which the blocks $(a_{m},\ldots,a_{m+\ell-1})$ and $(a_{k},\ldots,a_{k+\ell-1})$ overlap. This happens if and only if $k\leq m+\ell-1$ . In this case, $k+\ell-1\leq m+2(\ell-1)<n$ , hence $(a_{k},\ldots,a_{k+\ell-1})$ and $(a_{n},\ldots,a_{n+\ell-1})$ do not overlap. For any $j\in\{M,\ldots,m-1\}$ , $k\in\{m+1,\ldots,m+\ell-1\}$ define

D_{j,k}:=\left\{x\in I_{t}:\begin{matrix}a_{j}(x)\cdots a_{j+\ell-1}(x)\geq\varphi(m),&a_{m}(x)\cdots a_{m+\ell-1}(x)\geq\varphi(m),\\ a_{k}(x)\cdots a_{k+\ell-1}(x)\geq\varphi(n),&a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)\end{matrix}\right\}.

For such $k$ we have

\bigcup_{j=M}^{m-1}D_{j,k}\subseteq\left\{x\in I_{t}:\begin{matrix}a_{k}(x)\cdots a_{k+\ell-1}(x)\geq\varphi(n),\\ a_{n}(x)\cdots a_{n+\ell-1}(x)\geq\varphi(n)\end{matrix}\right\}=A_{n,k}\cap I_{t}.

We conclude

\mathfrak{m}\left(\bigcup_{k=m+1}^{m+\ell-1}\bigcup_{j=M}^{m-1}D_{j,k}\right)\leq\sum_{k=m+1}^{m+\ell-1}\mathfrak{m}(A_{n,k}\cap I_{t})\ll_{\ell}\frac{\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}\mathfrak{m}(I_{t})

and (4.2) follows.

In view of Lemma 4.2, (18), (19), and (4.2), direct computations give

	$\displaystyle S_{1}$	$\displaystyle\leq\sum_{(m,n)\in\mathcal{I}_{1}}\mathfrak{m}(U_{1}(m,n))+\mathfrak{m}(U_{2}(m,n))+\mathfrak{m}(U_{3}(m,n))$
		$\displaystyle\ll\frac{1}{\mathfrak{m}(I_{t})}\left(\sum_{n=M}^{N}\mathfrak{m}(A_{n}\cap I_{t})\right)^{2}.$

The lemma is now proven. ∎

Proof of Theorem 1.5. Divergence case.

Take any $t\in\mathbb{N}$ and $\mathbf{b}\in\mathbb{N}^{t}$ . Write $I_{t}=I_{t}(\mathbf{b})$ and let $M$ be as above. For each $n\in\mathbb{N}$ put $E_{n}:=A_{M-1+n}\cap I_{t}$ . Lemma 4.2 and the divergence assumption imply $\sum_{n}\mathfrak{m}(E_{n})=\infty$ . Then, Lemma 4.3 and the Chung-Erdös inequality yield

\mathfrak{m}\left(\limsup_{n\to\infty}E_{n}\right)\gg_{\ell}\mathfrak{m}(I_{t}).

Let $\mathcal{C}$ denote the family of open intervals which are the interior some fundamental interval. The result now follows from $\mathcal{F}_{3}(\varphi)\cap I_{t}=\limsup_{n}E_{n}$ and Lemma 2.2. ∎

5. Proof of Theorem 1.10

The cases $B=1$ and $B=\infty$ follow from $\mathcal{F}_{2}(\varphi)\subseteq\mathcal{F}_{3}(\varphi)\subseteq\mathcal{E}_{2}(\varphi)$ and theorems 1.8 and 1.2. Thus, we assume $1<B<\infty$ . Define the functions $X_{1},X_{2},X_{3}:[0,1]\to\mathbb{R}$ by

X_{1}(s)=\frac{(1-s)^{2}}{s^{2}-s+1},\quad X_{2}(s)=\frac{s^{2}}{s^{2}-s+1},\quad X_{3}(s)=1-(X_{1}(s)+X_{2}(s)).

Direct computations tell us that for all $s\in[0,1]$ we have $0\leq X_{1}(s),X_{2}(s),X_{3}(s)\leq 1$ , and

(21)	$\displaystyle g_{3}(s)$	$\displaystyle=(2s-1)+X_{1}(s)s,$
(22)	$\displaystyle g_{3}(s)$	$\displaystyle<(1-s)X_{1}-s,$
(23)	$\displaystyle g_{3}(s)$	$\displaystyle\leq(3s-1)(X_{1}(s)+X_{2}(s))+s.$

Write

x_{1}:=X_{1}(p_{B}),\quad x_{2}:=X_{2}(p_{B}),\quad x_{3}:=X_{3}(p_{B}).

5.1. Upper bound

We cover $\mathcal{F}_{3}(B)$ by three sets determined by the length of the overlap of the restricted blocks of partial quotients. Define

	$\displaystyle\mathcal{F}_{3}^{2}(B)$	$\displaystyle:=\left\{x\in[0,1):\begin{matrix}a_{n-1}(x)a_{n}(x)a_{n+1}(x)\geq B^{n},\\ a_{n}(x)a_{n+1}(x)a_{n+2}(x)\geq B^{n}\end{matrix}\text{ for i.m. }n\in\mathbb{N}\right\},$
	$\displaystyle\mathcal{F}_{3}^{1}(B)$	$\displaystyle:=\left\{x\in[0,1):\begin{matrix}a_{n-2}(x)a_{n-1}(x)a_{n}(x)\geq B^{n},\\ a_{n}(x)a_{n+1}(x)a_{n+2}(x)\geq B^{n}\end{matrix}\text{ for i.m. }n\in\mathbb{N}\right\},$
	$\displaystyle\mathcal{F}_{3}^{0}(B)$	$\displaystyle:=\left\{x\in[0,1):\begin{matrix}a_{k}(x)a_{k+1}(x)a_{k+2}(x)\geq B^{n},\\ a_{n}(x)a_{n+1}(x)a_{n+2}(x)\geq B^{n}\end{matrix}\text{ for some }1\leq k\leq n-3\text{ for i.m. }n\in\mathbb{N}\right\}.$

The family $\{\mathcal{F}_{3}^{2}(B),\mathcal{F}_{3}^{1}(B),\mathcal{F}_{3}^{0}(B)\}$ covers $\mathcal{F}_{3}(B)$ . The result will follow from estimating the Hausdorff dimension of each $\mathcal{F}_{3}^{j}(B)$ , $j=0,1,2$ . The computations in the three cases are alike. For this reason, we only give a fully detailed proof for $\mathcal{F}_{3}^{2}(B)$ and sketch the argument for $\mathcal{F}_{3}^{1}(B)$ and $\mathcal{F}_{3}^{0}(B)$ .

5.1.1. Overlaps of length 2

Let us show that

(24)

\dim_{\mathrm{H}}\mathcal{F}_{3}^{2}(B)\leq p_{B}.

Write $\alpha:=B^{x_{1}}$ . Take $n\in\mathbb{N}$ and define

\mathcal{F}_{3,n}^{2}(B):=\left\{x\in[0,1):\begin{matrix}a_{n-1}(x)a_{n}(x)a_{n+1}(x)\geq B^{n},\\ a_{n}(x)a_{n+1}(x)a_{n+2}(x)\geq B^{n}\end{matrix}\right\}.

We cover $\mathcal{F}_{3,n}^{2}(B)$ by the sets $\mathcal{F}_{3,n}^{1,C}(B)$ and $\mathcal{F}_{3,n}^{1,G}(B)$ given by

	$\displaystyle\mathcal{F}_{3,n}^{2,G}(B)$	$\displaystyle:=\left\{x\in[0,1):a_{n-1}(x)\geq\alpha^{n},\,a_{n-1}(x)a_{n}(x)a_{n+1}(x)\geq B^{n},a_{n}(x)a_{n+1}(x)a_{n+2}(x)\geq B^{n}\right\},$
	$\displaystyle\mathcal{F}_{3,n}^{2,C}(B)$	$\displaystyle:=\left\{x\in[0,1):a_{n-1}(x)<\alpha^{n},\,a_{n-1}(x)a_{n}(x)a_{n+1}(x)\geq B^{n}\right\}.$

Let us cover $\mathcal{F}_{3,n}^{2,G}(B)$ with the sets $F^{\prime}$ and $F^{\prime\prime}$ defined as

	$\displaystyle F^{\prime}$	$\displaystyle:=\left\{x\in[0,1):a_{n-1}(x)\geq\alpha^{n},\;\frac{B^{n}}{a_{n-1}(x)}\leq a_{n}(x)a_{n+1}(x)\leq\frac{B^{n}}{\alpha^{n}},a_{n+2}(x)\geq\frac{B^{n}}{a_{n}(x)a_{n+1}(x)}\right\},$
	$\displaystyle F^{\prime\prime}$	$\displaystyle:=\left\{x\in[0,1):a_{n-1}(x)\geq\alpha^{n},\;\frac{B^{n}}{\alpha^{n}}\leq a_{n}(x)a_{n+1}(x)\right\}.$

Put

	$\displaystyle I$	$\displaystyle:=\left\{\mathbf{a}=(a_{1},\ldots,a_{n+1})\in\mathbb{N}^{n+1}:a_{n-1}>\alpha^{n},\;1\leq a_{n}a_{n+1}\leq\frac{B^{n}}{\alpha^{n}}\right\},$
	$\displaystyle I^{\prime}$	$\displaystyle:=\left\{\mathbf{a}=(a_{1},\ldots,a_{n-1})\in\mathbb{N}^{n-1}:a_{n-1}>\alpha^{n}\right\}.$

For any $\mathbf{a}\in I$ the set

J_{n+1}(\mathbf{a}):=\bigcup_{a_{n+2}\geq\frac{B^{n}}{a_{n}a_{n+1}}}I_{n+2}(\mathbf{a},a_{n+2})

satisfies

(25)

|J_{n+1}(\mathbf{a})|\asymp\sum_{a_{n+2}\geq\frac{B^{n}}{a_{n}a_{n+1}}}\frac{1}{q_{n-1}^{2}a_{n}^{2}a_{n+1}^{2}a_{n+2}^{2}}\asymp\frac{1}{q_{n-1}^{2}B^{n}a_{n}a_{n+1}}.

Similarly, for any $\mathbf{a}\in I^{\prime}$ the set

J_{n-1}^{\prime}(\mathbf{a}):=\bigcup_{a_{n}a_{n+1}\geq\frac{B^{n}}{\alpha^{n}}}I_{n+1}(\mathbf{a},a_{n},a_{n+1})

satisfies

(26)

|J_{n-1}^{\prime}(\mathbf{a})|\asymp\frac{1}{q_{n-1}^{2}}\sum_{a_{n}a_{n+1}\geq\frac{B^{n}}{\alpha^{n}}}\frac{1}{a_{n}^{2}a_{n+1}^{2}}\asymp\frac{\alpha^{n}}{q_{n-1}^{2}B^{n}}n\log\frac{\alpha}{B}\asymp\frac{n\alpha^{n}}{q_{n-1}^{2}B^{n}}.

The last estimate follows from the definition of $\alpha$ . Note that

F^{\prime}\subseteq\bigcup_{\mathbf{a}\in I}J_{n+1}(\mathbf{a})\quad\text{ and }\quad F^{\prime\prime}\subseteq\bigcup_{\mathbf{a}\in I^{\prime}}J_{n-1}^{\prime}(\mathbf{a}).

Now define

I^{\prime\prime}:=\left\{(a_{1},\ldots,a_{n-1})\in\mathbb{N}^{n-1}:1\leq a_{n-1}\leq\alpha^{n}\right\}.

For each $\mathbf{a}\in I^{\prime\prime}$ the set

J_{n-1}^{\prime\prime}(\mathbf{a}):=\bigcup_{a_{n}a_{n+1}\geq\frac{B^{n}}{a_{n-1}}}I_{n+1}(\mathbf{a},a_{n},a_{n+1})

satisfies

(27)

|J_{n-1}^{\prime\prime}(\mathbf{a})|\asymp\frac{1}{q_{n-2}^{2}a_{n-1}B^{n}}\log\frac{B^{n}}{a_{n-1}}\asymp\frac{1}{q_{n-2}^{2}a_{n-1}B^{n}}.

The last estimate follows from $a_{n-1}\leq\alpha^{n}$ .

Let $\varepsilon>0$ be arbitrary and write $s=p_{B}+\varepsilon$ . By (21),

(2s-1)+sx_{1}>(2p_{B}-1)+x_{1}p_{B}=g_{3}(p_{B})

Since $g_{3}$ is continuous and strictly increasing, we may pick $\varepsilon_{1},\delta_{1}>0$ satisfying $0<\varepsilon_{1}<\varepsilon$ and

(2s-1)+sx_{1}>g_{3}(p_{B}+\varepsilon_{1})+\delta_{1}.

Let $N_{1}\in\mathbb{N}$ such that $s_{n}<s+\varepsilon_{1}<s+\varepsilon$ whenever $n\in\mathbb{N}_{\geq N_{1}}$ . Then, for such $n$ we have

(2s-1)+sx_{1}>g_{3}(s_{n-2})+\delta_{1}.

As a consequence, taking any $(a_{1},\ldots,a_{n-2})\in\mathbb{N}^{n-2}$ and writing $q_{n-2}=q_{n-2}(a_{1},\ldots,a_{n-2})$ ,

	$\displaystyle\sum_{a_{n-1}\geq\alpha^{n}}\sum_{a_{n}a_{n+1}\leq\frac{B^{n}}{\alpha^{n}}}\left(\frac{1}{q_{n-1}^{2}B^{n}a_{n}a_{n+1}}\right)^{s}$	$\displaystyle\asymp\frac{1}{q_{n-2}^{2s}B^{ns}}\sum_{a_{n-1}\geq\alpha^{n}}\frac{1}{a_{n-1}^{2s}}\sum_{a_{n}a_{n+1}\leq\frac{B^{n}}{\alpha^{n}}}\frac{1}{a_{n}^{s}a_{n+1}^{s}}$
		$\displaystyle\asymp\frac{1}{q_{n-2}^{2s}B^{ns}}\alpha^{n(1-2s)}\left(\frac{B}{\alpha}\right)^{n(1-s)}$
		$\displaystyle\asymp\frac{1}{q_{n-2}^{2s}}B^{n(1-2s)}\alpha^{-ns}$
		$\displaystyle\leq\frac{1}{q_{n-2}^{2s_{n-2}}B^{n(g_{3}(s_{n-2})+\delta_{1})}}.$

This estimate corresponds to (25).

We may use a similar argument relying on (22) rather than on (21) to show the existence of some constants $\varepsilon_{2},\delta_{2}>0$ with $0<\varepsilon_{2}<\varepsilon$ and some $N_{2}\in\mathbb{N}$ such that for any $n\in\mathbb{N}_{\geq N_{2}}$ , any $\mathbf{a}=(a_{1},\ldots,a_{n-2})\in\mathbb{N}^{n-2}$ we have

\sum_{a_{n-1}\geq\alpha^{n}}\left(\frac{n\alpha^{n}}{q_{n-1}^{2}B^{n}}\right)^{s}\asymp\frac{n^{s}}{q_{n-2}^{2s}}\left(\frac{\alpha^{n}}{B^{n}}\right)^{s}\alpha^{n(1-2s)}\leq n^{s}\frac{\alpha^{n(1-s)}}{B^{ns}q_{n-2}^{2s}}\leq\frac{1}{q_{n-2}^{2s_{n-2}}B^{n(g_{3}(s_{n-2})+\delta_{2})}}.

This estimate corresponds to (26).

Likewise, we may show the existence of constants $\varepsilon_{3},\delta_{3}>0$ with $0<\varepsilon_{3}<\varepsilon$ and $N_{3}\in\mathbb{N}$ such that for all $n\in\mathbb{N}_{\geq N_{3}}$ and $(a_{1},\ldots,a_{n-2})\in\mathbb{N}^{n-2}$ we have

\sum_{a_{n-1}\leq\alpha^{n}}\left(\frac{1}{q_{n-2}^{2}a_{n-1}B^{n}}\right)^{s}\ll\frac{1}{q_{n-2}^{2s_{n-2}}B^{n(g_{3}(s_{n-2})+\delta_{3})}}.

This estimate corresponds to (27).

Therefore, if $N=\max\{N_{1},N_{2},N_{3}\}$ and $\delta=\min\{\delta_{1},\delta_{2},\delta_{3}\}$ , for every $n\in\mathbb{N}_{\geq N}$ we have

	$\displaystyle\sum_{a_{1},\ldots,a_{n-2}\in\mathbb{N}}\sum_{a_{n}a_{n+1}\leq\frac{B^{n}}{\alpha^{n}}}\left(\frac{1}{B^{n}a_{n}q_{n-2}^{2}}\right)^{s}$	$\displaystyle+\sum_{a_{n}a_{n+1}\leq\frac{B^{n}}{\alpha^{n}}}\left(\frac{1}{B^{n}a_{n}q_{n-2}^{2}}\right)^{s}+$
		$\displaystyle\quad+\sum_{a_{n-1}\leq\alpha^{n}}\left(\frac{1}{q_{n-2}^{2}a_{n-1}B^{n}}\right)^{s}\ll B^{-n\delta}.$

We conclude

\mathcal{H}^{s}(\mathcal{F}_{3}^{2}(B))\leq\liminf_{N\to\infty}\sum_{n\geq N}B^{-n\delta}=0.

Since $\varepsilon>0$ was arbitrary, $\dim_{\mathrm{H}}\mathcal{F}_{3}^{2}(B)\leq p_{B}$ .

5.1.2. Overlaps of length 1

Write $\beta=B^{x_{1}+x_{2}}$ . Cover $\mathcal{F}_{3}^{1}(B)$ by the sets

	$\displaystyle\mathcal{F}_{3}^{1,C}(B)$	$\displaystyle:=\left\{x\in\mathcal{F}_{3}^{1}(B):a_{n-2}(x)a_{n-1}(x)\leq\beta^{n}\text{ for i.m. }n\in\mathbb{N}\right\}$
	$\displaystyle\mathcal{F}_{3}^{1,G}(B,\beta)$	$\displaystyle:=\left\{x\in\mathcal{F}_{3}^{1}(B):a_{n-2}(x)a_{n-1}(x)>\beta^{n}\text{ for i.m. }n\in\mathbb{N}\right\}.$

We may cover $\mathcal{F}_{3}^{1,G}(B)$ by the sets

	$\displaystyle\mathcal{F}_{3}^{1,G,1}(B)$	$\displaystyle:=\left\{x\in[0,1):a_{n-2}(x)a_{n-1}(x)>\beta^{n},1\leq a_{n}(x)\leq\frac{B^{n}}{\beta^{n}}\text{ for i.m. }n\in\mathbb{N}\right\},$
	$\displaystyle\mathcal{F}_{3}^{1,G,2}(B)$	$\displaystyle:=\left\{x\in[0,1):a_{n-2}(x)a_{n-1}(x)>\beta^{n},a_{n}(x)>\frac{B^{n}}{\beta^{n}}\text{ for i.m. }n\in\mathbb{N}\right\}.$

For each $n\in\mathbb{N}$ and every $\mathbf{a}=(a_{1},\ldots,a_{n-1},a_{n})\in\mathbb{N}^{n}$ such that $a_{n-2}a_{n-1}>\beta^{n}$ and $1\leq a_{n}\leq\frac{B^{n}}{\beta^{n}}$ define

K_{n}^{G,1}(\mathbf{a}):=\bigcup_{a_{n+1}a_{n+2}\geq\frac{B^{n}}{a_{n}}}I_{n+2}(\mathbf{a},a_{n+1},a_{n+2}).

Then, using (11) and $\beta^{n}\leq\frac{B^{n}}{a_{n}}\leq B^{n}$ , we have

|K_{n}^{G,1}(\mathbf{a})|\asymp\frac{n}{q_{n-1}^{2}a_{n}B^{n}}.

Considering $a_{1},\ldots,a_{n-3}$ fixed, we may use again (11) to obtain

n\sum_{a_{n}\leq\frac{B^{n}}{\beta^{n}}}\sum_{a_{n-2}a_{n-1}\leq\beta^{n}}\frac{1}{q_{n-3}^{2s}a_{n-2}^{2s}a_{n-1}^{2s}a_{n}^{2s}B^{ns}}\ll\frac{n}{q_{n-3}^{2s}}B^{n(1-2s)}\beta^{-ns}\leq\frac{n}{q_{n-3}^{2s}}B^{n(1-2s)}\alpha^{-ns}.

As a consequence, arguing as with $\mathcal{F}_{3}^{2}(B)$ , we have $\dim_{\mathrm{H}}\mathcal{F}_{3}^{1,G,1}(B)\leq p_{B}$ . In order to estimate $\dim_{\mathrm{H}}\mathcal{F}_{3}^{1,C,2}(B)$ , for each $\mathbf{a}=(a_{1},\ldots,a_{n-1})\in\mathbb{N}^{n-1}$ such that $a_{n-2}a_{n-1}\geq\beta^{n}$ we consider the set

K_{n-1}^{G,2}(\mathbf{a}):=\bigcup_{a_{n}\geq\frac{B^{n}}{\beta^{n}}}I_{n+2}(\mathbf{a},a_{n}),

which satisfies

|K_{n-1}^{G,2}(\mathbf{a})|\asymp\frac{\beta^{n}}{B^{n}q_{n-1}^{2}}.

Fixing the terms $a_{1},\ldots,a_{n-3}$ , for any $s>0$ we have

\sum_{a_{n-2}a_{n-1}\geq\beta^{n}}\left(\frac{\beta^{n}}{B^{n}q_{n-1}^{2}}\right)^{s}\asymp\frac{\beta^{n(1-3s)}}{q_{n-3}^{2s}B^{ns}}.

We may argue as we did for $F^{\prime\prime}$ using (23) to conclude $\dim_{\mathrm{H}}\mathcal{F}_{3}^{1,G,2}(B)\leq p_{B}$ .

The set $\mathcal{F}_{3}^{1,C}$ is contained in

\mathcal{F}_{3}^{1,C,1}(B):=\left\{x\in[0,1):a_{n-2}(x)a_{n-1}(x)\leq\beta^{n},a_{n}(x)\geq\left(\frac{B}{\beta}\right)^{n}\text{ for i.m. }n\in\mathbb{N}\right\}.

For $n\in\mathbb{N}$ and $(a_{1},\ldots,a_{n-1})\in\mathbb{N}^{n-1}$ such that $a_{n-2}a_{n-1}\leq\beta^{n}$ define

K_{n-1}(a_{1},\ldots,a_{n-1})=\bigcup_{a_{n}>\frac{B^{n}}{\beta^{n}}}I_{n}(a_{1},\ldots,a_{n}),

|K_{n-1}(a_{1},\ldots,a_{n-1})|\asymp\frac{1}{B^{n}q_{n-3}^{s}a_{n-1}a_{n-2}}.

Moreover, fixing $a_{1},\ldots,a_{n-3}$ and assuming that $n$ is large, every $s>0$ verifies

\sum_{a_{n-1}a_{n-2}<\frac{B^{n}}{\beta^{n}}}\left(\frac{1}{B^{n}q_{n-3}^{s}a_{n-1}a_{n-2}}\right)^{s}\asymp\frac{B^{n(1-2s)}\beta^{-ns}}{q_{n-3}^{2s}}<\frac{B^{n(1-2s)}\alpha^{-ns}}{q_{n-3}^{2s}}.

Arguing as above, we conclude that $\dim_{\mathrm{H}}\mathcal{F}_{3}^{1,C,1}(B)\leq p_{B}$ and, thus, $\dim_{\mathrm{H}}\mathcal{F}_{3}^{1,C}(B)\leq p_{B}$ .

5.1.3. No overlaps

We content ourselves with describing the useful covering of $\mathcal{F}_{3}^{0}(B)$ . First, for $n\in\mathbb{N}$ , $k\in\{1,\ldots,n-3\}$ , and $\mathbf{a}\in\mathbb{N}^{n-1}$ such that $a_{k}a_{k+1}a_{k+2}\geq B^{n}$ define

K_{n-1}^{3,G,k}(\mathbf{a}):=\bigcup_{a_{n}\geq\alpha^{n}}I_{n}(\mathbf{a},a_{n}),\quad\text{ so }\quad|K_{n-1}^{3,G,k}(\mathbf{a})|\asymp\frac{1}{\alpha^{n}q_{n-1}^{2}}.

Therefore, letting $a_{k},a_{k+1},a_{k+2}$ run along the pairs satisfying $a_{k}a_{k+1}a_{k+2}>B^{n}$ considering the remaining terms fixed, we have

\sum_{a_{k}a_{k+1}a_{k+2}>B^{n}}\frac{1}{\alpha^{ns}q_{n-1}^{2s}}\ll\frac{B^{n(1-2s)}}{\alpha^{ns}q_{n-4}^{2s}}

and we proceed as in the estimate for $\mathcal{F}_{3}^{2}(B)$ .

Second, for any $n\in\mathbb{N}$ , $k\in\{1,\ldots,n-3\}$ , $\mathbf{a}\in\mathbb{N}^{n}$ such that $a_{k}a_{k+1}a_{k+2}\geq B^{n}$ and $a_{n}\leq\alpha^{n}$ define

K_{n}^{3,C,k}(\mathbf{a}):=\bigcup_{a_{n+1}a_{n+2}\geq\frac{B^{n}}{a_{n}}}I_{n}(\mathbf{a},a_{n+1},a_{n+2}),\quad\text{ so }\quad|K_{n}^{3,C,k}(\mathbf{a})|\asymp\frac{n}{B^{n}a_{n}q_{n-1}^{2}}.

Hence, varying $a_{k},a_{k+1}$ along the pairs satisfying $a_{k}a_{k+1}\geq B^{n}$ and $a_{n}$ along $1\leq a_{n}\leq\alpha^{n}$ , for all $s>0$ we have

\sum_{a_{k}a_{k+1}\geq B^{n}}\sum_{a_{n}\leq\alpha^{n}}\left(\frac{n}{B^{n}a_{n}q_{n-1}^{2}}\right)^{s}\ll\frac{n^{s}}{B^{ns}q_{n-3}^{2s}}B^{n(1-2s)}\alpha^{n(1-s)}\leq\frac{n^{s}}{q_{n-3}^{2s}}B^{n(1-3s)}\alpha^{n(1-s)}.

Direct computations tell us that for $s\geq\frac{1}{2}$ we have

(3s-1)+X_{1}(s)(s-1)>g_{3}(s).

Arguing as we did for $\mathcal{F}_{3}^{2}(B)$ , we may conclude that $\dim_{\mathrm{H}}\mathcal{F}_{3}^{0}(B)\leq p_{B}$ .

5.2. Lower bound

Let $x_{1},x_{2},x_{3}$ be as in the upper bound and define

A_{0}=B^{x_{1}},\quad A_{1}=B^{x_{2}},\quad A_{2}=B^{x_{3}},\quad A_{3}=B^{x_{1}}.

With notation of Lemma 2.10, we have $S_{4}(A_{0},A_{1},A_{2},A_{3})\subseteq\mathcal{F}_{3}(B)$ and

\dim_{\mathrm{H}}S_{4}(A_{0},A_{1},A_{2},A_{3})\leq\dim_{\mathrm{H}}\mathcal{F}_{3}(B).

Also, $A_{0}A_{1}A_{2}=A_{1}A_{2}A_{3}=B$ and

\log\beta_{-1}=0,\quad\log\beta_{0}=x_{1}\log B,\quad\log\beta_{1}=(x_{1}+x_{2})\log B,

\log\beta_{2}=\log B,\quad\log\beta_{3}=(1+x_{1})\log B.

As a consequence, the potentials $\varphi_{i,s}$ defining $d_{i}$ are

	$\displaystyle\varphi_{0,s}(x)$	$\displaystyle=-sx_{1}\log B-s\log\|T^{\prime}(x)\|,$
	$\displaystyle\varphi_{1,s}(x)$	$\displaystyle=-((2x_{1}+x_{2})s-x_{1})\log B-s\log\|T^{\prime}(x)\|,$
	$\displaystyle\varphi_{2,s}(x)$	$\displaystyle=-((1+x_{1}+x_{2})s-x_{1}-x_{2})\log B-s\log\|T^{\prime}(x)\|,$
	$\displaystyle\varphi_{3,s}(x)$	$\displaystyle=-((2+x_{1})s-1)\log B-s\log\|T^{\prime}(x)\|.$

It is not hard to show that $p_{B}=d_{1}=d_{2}=d_{3}\leq d_{0}$ . Certainly, since $sX_{1}(s)\leq(2+X_{1}(s))s-1$ when $\frac{1}{2}\leq s\leq 1$ , we have $d_{3}\leq d_{0}$ . We only explain $p_{B}=d_{1}$ , the other equalities are handled in a similar fashion. First, note that

g_{3}(p_{B})=(2+x_{1})p_{B}-1

and that the functions $g_{3}$ and $s\mapsto(2+x_{1})s-1$ are strictly increasing and continuous on $\left[\frac{1}{2},1\right]$ . If $\varepsilon>0$ and $s=p_{B}+\varepsilon$ , then

(2+x_{1})s-1>g(p_{B})

and we would have $P(T;\varphi_{1,s})\leq 0$ , hence $d_{1}\leq p_{B}+\varepsilon$ . Since $\varepsilon$ was arbitrary, $d_{1}\leq p_{B}$ . We now show that $d_{1}<p_{B}$ leads to a contradiction. If the strict inequality were true, we would be able to pick some number $s$ with $d_{1}<s<p_{B}$ and close enough to $p_{B}$ such that $g(s)=(2+x_{1})s-1<g(p_{B})$ . However, for such $s$ we would have

0>P(T;\varphi_{1,s})=P(T;-g_{3}(s)\log B-s\log|T^{\prime}|)>0.

By virtue of Lemma 2.10, $p_{B}\leq\dim_{\mathrm{H}}\mathcal{F}_{3}(B)$ .

6. Final remarks and further results

We stated in the introduction that Theorem 1.5 is an important piece in the extension of the laws of large numbers due to Diamond and Vaaler (Equation (2)) and Hu, Hussain, and Yu (Equation (3)). In a forthcoming article, we study the asymptotic behaviour of the sums of products of consecutive partial quotients. More precisely, for $n,\ell\in\mathbb{N}$ and any irrational number $x\in[0,1)$ define

S_{n,\ell}(x)=\sum_{j=1}^{n}a_{j}(x)\cdots a_{j+\ell-1}(x).

Theorem 6.1.

For every $\varepsilon>0$ we have

\lim_{n\to\infty}\mathfrak{m}\left(\left\{x\in[0,1):\left|\frac{1}{n\log^{\ell}n}S_{n,\ell}(x)-\frac{1}{\ell\log 2}\right|\geq\varepsilon\right\}\right)=0.

Almost every $x\in[0,1)$ satisfies

\lim_{n\to\infty}\frac{1}{n\log^{\ell}n}\left(S_{n,\ell}(x)-\max_{1\leq j\leq n}a_{j}(x)\cdots a_{j+\ell-1}(x)\right)=\frac{1}{\ell\log 2}.

These results could be extended to products of partial quotients in arithmetic progressions. Given any $d\in\mathbb{N}$ , for $n\in\mathbb{N}$ and any irrational $x\in[0,1)$ define the sum

S_{n,\ell}^{d}(x)=\sum_{j=1}^{n}a_{j}(x)a_{j+d}(x)\cdots a_{j+(\ell-1)d}(x).

Proposition 6.2.

Theorem 6.1 holds if $S_{n,\ell}(x)$ is replaced with $S_{n,\ell}^{d}(x)$ .

With regards to the Hausdorff dimension of certain Lebesgue null sets, for an increasing function $\phi:\mathbb{N}\to\mathbb{R}$ with $\phi(n)\to\infty$ as $n\to\infty$ we consider the set

E_{\ell}(\phi):=\left\{x\in[0,1):\lim_{n\to\infty}\frac{S_{n,\ell}(x)}{\phi(n)}=1\right\}.

Theorem 6.3.

(I)

If $0<\alpha<\frac{1}{2}$ and $\phi(n)=e^{n^{\alpha}}$ , then $\dim_{\mathrm{H}}E_{\ell}(\phi)=1$ .
(II)

If $\frac{1}{2}\leq\alpha$ and $\phi(n)=e^{n^{\alpha}}$ , then $\dim_{\mathrm{H}}E_{\ell}(\phi)=\frac{1}{2}$ .
(III)

If $1<\alpha$ and $\phi(n)=e^{\alpha^{n}}$ , then $\dim_{\mathrm{H}}E_{\ell}(\phi)=\frac{1}{1+\alpha}$ .

Remark 6.4.

Recall that the $n$ -th term of the series in Theorem 1.5 is

\frac{n\log^{2(\ell-1)}\varphi(n)}{\varphi^{2}(n)}+\frac{\log^{\ell-2}\varphi(n)}{\varphi(n)}.

It is clear from the proof that the first term accounts for the large products appearing in non-overlapping blocks and the second term corresponds overlapping blocks. Hence, for $\ell=1$ the second term is non-existent.

Remark 6.5.

We suspect that arguments resembling the proof of Theorem 1.10 can lead to the computation of $\dim_{\mathrm{H}}\mathcal{F}_{\ell}(\varphi)$ for $\ell\geq 4$ . This could provide a recursive definition of the potential giving the Hausdorff dimension when $1<B<\infty$ (see Theorem 1.2). However, we stress that the proofs for $\ell\in\{2,3\}$ do not rely on the result for lower values of $\ell$ .

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