Admissible endpoints of gaps in the Lagrange spectrum

TL;DR

This paper characterizes which numbers in the Lagrange spectrum are admissible, providing a necessary and sufficient condition, and constructs an infinite series of numbers that are not admissible.

Contribution

It establishes a complete criterion for admissibility in the Lagrange spectrum and identifies infinitely many non-admissible numbers.

Findings

Derived a necessary and sufficient condition for admissibility.

Constructed an infinite series of non-admissible numbers.

Enhanced understanding of the structure of the Lagrange spectrum.

Abstract

We call a positive real number $\lambda$ admissible if it belongs to the Lagrange spectrum and there exists an irrational number $\alpha$ such that $\mu(\alpha)=\lambda$. Here $\mu(\alpha)$ denotes the Lagrange constant of $\alpha$ - maximal real number $c$ such that $\forall \varepsilon>0$ the inequality $|\alpha-\frac{p}{q}|\le\frac{1}{(c-\varepsilon)q^2}$ has infinitely many solutions for relatively prime $p$ and $q$. In this paper we establish a necessary and sufficient condition of admissibility of the Lagrange spectrum element and construct an infinite series of not admissible numbers.

Equations

$${\mu }^{-1}(\alpha )=\underset{p\in \mathbb{Z},q\in \mathbb{N}}{\mathrm{liminf}}|q(q\alpha -p)|.$$

$$\biggl{|}\alpha-\frac{p}{q}\biggr{|}\leqslant\frac{1}{\mu(\alpha)q^{2}}$$

$$\mu(\alpha)=\biggl{(}\liminf_{p\in\mathbb{Z},q\in\mathbb{N}}|q(q\alpha-p)|\biggr{)}^{-1}$$

$$\alpha =[a_0;a_1,a_2,\dots ,a_n,\dots ].$$

$${\lambda }_i(\alpha )=[a_i;a_{i+1},a_{i+2},\dots ]+[0;a_{i-1},a_{i-2},\dots ,a_1],$$

$$\mathrm{limsup}{\lambda }_i(\alpha )=\mu (\alpha ).$$

$$\biggl{|}\alpha-\frac{p}{q}\biggr{|}\leqslant\frac{1}{\mu(\alpha)q^{2}}$$

$$B=(\dots ,b_{-n},\dots b_{-1},b_0,b_1,\dots ,b_n,\dots ).$$

$${\lambda }_i(B)=[b_i;b_{i-1},\dots ]+[0;b_{i+1},b_{i+2},\dots ].$$

$$L(B)=\underset{i\to \infty }{\mathrm{limsup}}{\lambda }_i(B),M(B)=\sup {\lambda }_i(B).$$

$${\alpha }_n^{\ast }=2+[0;\underset{2n-2}{\underbrace{1,\dots ,1}},\overline{2,2,1,2}]+[0;\underset{2n-1}{\underbrace{1,\dots ,1}},2,\underset{2n-2}{\underbrace{1,\dots ,1}},\overline{2,2,1,2}]$$

$${\beta }_n=2+2[0;\overline{\underset{2n}{\underbrace{1,\dots ,1}},2}].$$

$${\delta }_n<|\alpha -\beta |<{\epsilon }_n.$$

$$\min ([0;A,B,B,C],[0;A,C])<[0;A,B,C]<\max ([0;A,B,B,C],[0;A,C])$$

$$[0;B,C]>[0;B,B,C].$$

$$\begin{array}{r}{B}_N^1=(b_1,b_2,\dots ,b_{n_1-1},b_{n_2},b_{n_2+1},\dots ,b_N),\\ B_N^2=(b_1,b_2,\dots ,b_{n_1-1},b_{n_1},\dots ,b_{n_2-1},b_{n_1},\dots ,b_{n_2-1},b_{n_2},b_{n_2+1},\dots ,b_N).\end{array}$$

$$\begin{array}{r}{\gamma }^1=[0;b_1,b_2,\dots ,b_{n_1-1},b_{n_2},b_{n_2+1},\dots ,b_N,b_{N+1}\dots ]=[0;B_N^1,b_{N+1},\dots ],\\ {\gamma }^2=[0;b_1,b_2,\dots ,b_{n_1-1},b_{n_1},\dots ,b_{n_2-1},b_{n_1},\dots ,b_{n_2-1},b_{n_2},b_{n_2+1},\dots ,b_N,\dots ]=[0;B_N^2,b_{N+1},\dots ].\end{array}$$

$$\underset{j\to \infty }{\lim }{\lambda }_{k(j)}(\alpha )={\lambda }_0(B)=\mu (\alpha ).$$

$$a={\lambda }_0(B)<{\lambda }_0(B^{\prime })<a+{\epsilon }_n<b.$$

$$\mu ({\alpha }^{\prime })⩽\mu (\alpha )=a.$$

$$\mu ({\alpha }^{\prime })⩾M(B)⩾{\lambda }_0(B^{\prime })>{\lambda }_0(B)=a.$$

$${\lambda }_{k(j)}(\alpha )⩾\mu (\alpha ).$$

$$\underset{j\to \infty }{\lim }{\lambda }_{k(j)}(\alpha )=\mu (\alpha ).$$

$$B=(\overline{L},b_{-m},\dots ,b_0,\dots ,b_m,\overline{R}).$$

$$\begin{array}{r}[a_{k(j)};a_{k(j+1)},\dots ]⩾[b_0;b_1,\dots ,b_m,\overline{R}],\text{or}\\ [0;a_{k(j-1)},\dots ,a_1]⩾[0;b_{-1},\dots ,b_{-m},\overline{L}]\end{array}$$

$$[a_{k(j)};a_{k(j+1)},\dots ]>[b_0;b_1,\dots ,b_m,\overline{R}]$$

$$a_{k(j)},a_{k(j+1)},\dots ,a_{k(j)+m},\underset{t(j)times}{\underbrace{R,\dots ,R}},\dots ,a_{k(j)+r(j)}$$

$$a_{k(j)},a_{k(j+1)},\dots ,a_{k(j)+m},\underset{ntimes}{\underbrace{R,\dots ,R}},\dots ,a_{k(j)+r(j)}.$$

$$[a_{k(j)};a_{k(j+1)},\dots ,a_{k(j)+m},\underset{t(j)times}{\underbrace{R,\dots ,R}},\dots ,a_{k(j)+r(j)}]>[b_0;b_1,\dots ,b_m,\overline{R}]$$

$$[a_{k(j)};a_{k(j+1)},\dots ,a_{k(j)+m},\underset{ntimes}{\underbrace{R,\dots ,R}},\dots ,a_{k(j)+r(j)}]>[b_0;b_1,\dots ,b_m,\overline{R}]+{\delta }_{m+(n+1)p}.$$

Full text

[TABLE]

The set of all values taken by $\mu(\alpha)$ as $\alpha$ varies is called the Lagrange spectrum $\mathbb{L}$. Irrational $\alpha$ is called attainable if the inequality

[TABLE]

holds for infinitely many integers $p$ and $q$. Throughout this paper we will call a real number $\lambda\in\mathbb{L}$ admissible if there exists an irrational attainable $\alpha$ such that $\mu(\alpha)=\lambda$. In our previous paper [5] we constructed an example of not admissible element in the Lagrange spectrum. In the present paper, we give a necessary and sufficient condition of admissibility of a Lagrange spectrum element. We also prove that all elements of the sequence of $\mathbb{L}$ left endpoints $\alpha_{n}^{*},n\geqslant 2$, first considered by Gbur, are not admissible.

1 Introduction

The Lagrange spectrum $\mathbb{L}$ is usually defined as the set of all values of the Lagrange constants

[TABLE]

as $\alpha$ runs through the set of irrational numbers. Consider the continued fraction expansion of $\alpha$

[TABLE]

For any positive integer $i$ define

[TABLE]

It is well known fact that

[TABLE]

The equation (1) provides an equivalent definition of the Lagrange constant $\mu(\alpha)$.

The following properties of $\mathbb{L}$ are well known. The Lagrange spectrum is a closed set [2] with minimal point $\sqrt{5}$. All the numbers of $\mathbb{L}$ which are less than $3$ form a discrete set. It is well known fact that the Lagrange spectrum contains all elements over $\sqrt{21}$ (see[4],[8]). The complement of $\mathbb{L}$ is a countable union of maximal gaps of the spectrum. The maximal gaps are open intervals $(a,b)$ such that $(a,b)\cap\mathbb{L}=\emptyset$, but $a$ and $b$ both lie in the Lagrange spectrum. There are infinitely many gaps in the non-discrete part of the Lagrange spectrum [6].

Let $\alpha$ be an arbitrary irrational number. If the inequality

[TABLE]

has infinitely many solutions for integer $p$ and $q$, we call $\alpha$ attainable. This definition was first given by Malyshev in [7]. One can easily see[5] that $\alpha$ is attainable if and only if $\lambda_{i}(\alpha)\geqslant\mu(\alpha)$ for infinitely many indices $i$. We also call a real number $\lambda\in\mathbb{L}$ admissible if there exists an irrational attainable number $\alpha$ such that $\mu(\alpha)=\lambda$.

Let $B$ denote a doubly infinite sequence of positive integers

[TABLE]

For an arbitrary integer $i$ define

[TABLE]

We will call a doubly infinite sequence $B$ purely periodic if there exists a finite sequence $P$ such that $B=(\overline{P})$. A doubly infinite sequence $B$ is called eventually periodic if there exist $3$ finite sequences $P_{l},R,P_{r}$ such that $B=(\overline{P_{l}},R,\overline{P_{r}})$. One can also consider an equivalent definition of the Lagrange spectrum using the doubly infinite sequences. Denote

[TABLE]

The Lagrange spectrum $\mathbb{L}$ is exactly the set of values taken by $L(B)$ as $B$ runs through the set of doubly infinite sequences of positive integers. The set of values taken by $M(B)$ is called the Markoff spectrum. We will denote this set by $\mathbb{M}$.

We will call a doubly infinite sequence $B$ weakly associated with an irrational number $\alpha=[a_{0};a_{1},\ldots,a_{n},\ldots]$ if the following condition holds:

For any natural $i$ the pattern $(b_{-i},b_{-i+1},\ldots,b_{0},\ldots,b_{i})$ occurs in the sequence $a_{1},a_{2},\ldots,a_{n},\ldots$ infinitely many times.

We will call $B$ strongly associated with $\alpha$ if, additionally, 2. 2.

$\mu(\alpha)=\lambda_{0}(B)=M(B)$.

One can easily see that if $B$ is weakly associated with $\alpha$ then $\mu(\alpha)\geqslant M(B)$. As we will show in Section 4, if $\alpha$ has limited partial quotients, it has at least one strongly associated sequence.

2 Results of paper [5]

**Theorem I

***The quadratic irrationality $\lambda_{0}=[3;3,3,2,1,\overline{1,2}]+[0;2,1,\overline{1,2}]$ belongs to $\mathbb{L}$, but if $\alpha$ is such that $\mu(\alpha)=\lambda_{0}$ then $\alpha$ is not attainable.

***Theorem II

***If $\lambda\in\mathbb{L}$ is not a left endpoint of some maximal gap in the Lagrange spectrum then there exists an attainable $\alpha$ such that $\mu(\alpha)=\lambda$.

*One can easily formulate these theorems using the concept of admissible numbers, introduced behind.

**Theorem I’

***The quadratic irrationality $\lambda_{0}=[3;3,3,2,1,\overline{1,2}]+[0;2,1,\overline{1,2}]$ belongs to $\mathbb{L}$, but is not admissible.

***Theorem II’

***If $\lambda\in\mathbb{L}$ is not a left endpoint of some maximal gap in the Lagrange spectrum then $\lambda$ is an admissible number.

3 Main results

Our first theorem is a small generalization of Theorem 3 in [5]. The proof will be quite similar and use some lemmas from [5].

Theorem 1**.**

Let $a$ be a left endpoint of a gap $(a,b)$ in the Lagrange spectrum and $\alpha$ be an irrational number such that $\mu(\alpha)=a$. Consider a doubly infinite sequence $B$ strongly associated with $\alpha$. Then $B$ is an eventually periodic (i.e. periodic at both sides) sequence.

It follows from theorems I and II that there exist non-admissible elements in the Lagrange spectrum but all such numbers are left endpoints of some maximal gaps in $\mathbb{L}$. The following theorem gives a necessary and sufficient condition of admissibility of a Lagrange spectrum element.

Theorem 2**.**

A Lagrange spectrum left endpoint $a$ is admissible if and only if there exists a quadratic irrationality $\alpha$ such that $\mu(\alpha)=a$.

Of course, every quadratic irrationality is strongly associated with the unique sequence, which is purely periodic. Therefore Theorem 2 is equivalent to the following statement.

Corollary 3.1**.**

A Lagrange spectrum left endpoint $a$ is not admissible if and only if there does not exist a purely periodic sequence $B$ such that $\lambda_{0}(B)=M(B)=a$.

Theorem 2 provides an instrument to verify not admissible points in $\mathbb{L}$. Define

[TABLE]

and

[TABLE]

The fact that $(\alpha_{n}^{*},\beta_{n})$ is the maximal gap in the Markoff spectrum was proved by Gbur in [6]. It is easy to show that $\alpha_{n}^{*}$ and $\beta_{n}$ belong to $\mathbb{L}$. We will do it in the section 6. Hence, as $\mathbb{L}\subset\mathbb{M}$ [2], the interval $(\alpha_{n}^{*},\beta_{n})$ is the maximal gap in $\mathbb{L}$ too.

Theorem 3**.**

For any integer $n\geqslant 2$ the irrational number $\alpha_{n}^{*}$ is not admissible.

One can easily see that $\alpha_{1}^{*}=2+[0;\overline{2,2,1,2}]+[0;1,2,\overline{2,2,1,2}]=\mu([0;\overline{2,2,1,2}])=M(\overline{2,2,1,2})$. Thus, $\alpha_{1}^{*}$ is an admissible number by Theorem 2.

4 Proof of Theorem 1

The following statement is well known. See proof in ([1], Ch. 1, Lemma 6).

Lemma 4.1**.**

Let $A=\ldots,a_{-1},a_{0},a_{1},\ldots$ be any doubly infinite sequence. If M(A) is finite, then there exists a doubly infinite sequence $B$ such that $M(A)=M(B)=\lambda_{0}(B)$.

Using the same argument for the sequence $A=(a_{1},a_{2},\ldots,a_{n},\ldots)$, one can easily show that

Lemma 4.2**.**

Let $\alpha=[0;a_{1},\ldots,a_{n},\ldots]$ be an arbitrary irrational number and $a_{i}<c\ \forall i\in\mathbb{N}$ for some positive real number $c$. Then there exists a doubly infinite sequence $B$ which is strongly associated with $\alpha$.

As $\alpha\leqslant\sqrt{21}$, all elements of $B$ are bounded by $4$. For any natural $n$ denote $\varepsilon_{n}=2^{-(n-1)},\delta_{n}=5^{-2(n+2)}$. We need the following lemmas from [5].

Lemma 4.3**.**

Suppose $\alpha=[a_{0};a_{1},\ldots,a_{n},b_{1},\ldots]$ and $\beta=[a_{0};a_{1},\ldots,a_{n},c_{1},\ldots]$, where $n\geqslant 0$, $a_{0}$ is an integer, $a_{1},\ldots,a_{n},b_{1},b_{2},\ldots,c_{1},c_{2},\ldots$ are positive integers bounded by $4$ with $b_{1}\neq c_{1}$. Then for $n$ odd, $\alpha>\beta$ if and only if $b_{1}>c_{1}$; for $n$ even, $\alpha>\beta$ if and only if $b_{1}<c_{1}$. Also,

[TABLE]

Lemma 4.4**.**

Let $\gamma=[0;c_{1},c_{2},\ldots,c_{N},\ldots]$ and $\gamma^{\prime}=[0;c^{\prime}_{1},c^{\prime}_{2},\ldots,c^{\prime}_{N},\ldots]$ be two irrational numbers with partial quotients not exceeding $4$. Suppose that every sequence of partial quotients of length $2n+1$ which occurs in the sequence $(c^{\prime}_{1},c^{\prime}_{2},\ldots,c^{\prime}_{N},\ldots)$ infinitely many times also occurs in the sequence $(c_{1},c_{2},\ldots,c_{N},\ldots)$ infinitely many times. Then $\mu(\gamma^{\prime})<\mu(\gamma)+2\varepsilon_{n}$.

The following technical lemma was formulated in [5] for $N=(2n+1)(4^{2n+1}+1)$ and the proof was incorrect. However, this is not crucial for the results of the paper [5] as we just need $N$ to be bounded from above by some growing function of $n$. In this paper, we give a new version of lemma with correct proof.

Lemma 4.5**.**

Let $n$ be an arbitrary positive integer. Denote $N=N(n)=(2n+2)(4^{2n+2}+1)$. If $b_{1},b_{2},\ldots,b_{N}$ is an arbitrary integer sequence of length $N$ such that $1\leqslant a_{i}\leqslant 4$ for all $1\leqslant i\leqslant N$, then there exist two integers $n_{1},n_{2}$ such that $b_{n_{1}+i}=b_{n_{2}+i}$ for all $0\leqslant i\leqslant 2n+1$ and $n_{1}\equiv n_{2}\pmod{2}$.

Proof.

There exist only $4^{2n+2}$ distinct sequences of length $2n+2$ with elements $1,2,3,4$. Consider $4^{2n+2}+1$ sequences: $(a_{1},\ldots,a_{2n+2}),(a_{2n+3},\ldots,a_{4n+4}),\ldots,(a_{(2n+2)4^{2n+2}+1},\ldots,a_{(2n+2)4^{2n+2}+2n+2})$. Dirchlet’s principle implies that there exist two coinciding sequences among them. Denote these sequences by $(a_{n_{1}},\ldots,a_{n_{1}+2n+1})$ and $(a_{n_{2}},\ldots,a_{n_{2}+2n+1})$. Note that the index of the first element of each sequence is odd, hence $n_{1}\equiv n_{2}\equiv 1\pmod{2}$, that finishes the proof. ∎

If $n_{1}\equiv n_{2}\pmod{2}$ then the sequence $(a_{n_{1}},a_{n_{1}+1},\ldots,a_{n_{2}-1})$ has even length. This fact will be useful in our argument.

Lemma 4.6**.**

Let $B$ be an arbitrary integer sequence of even length. Let $A$ be an arbitrary finite integer sequence and $C$ — an arbitrary non-periodic infinite sequence. Then

[TABLE]

Proof.

As the sequence $C$ is non-periodic, the continued fractions in (4) are not equal. Without loss of generality, one can say that the sequence $A$ is empty. Suppose that

[TABLE]

As the length of $B$ is even, one can see that $[0;C]>[0;B,C]$, which is exactly the right part of the inequality (4). The case when $[0;B,C]<[0;B,B,C]$ is treated in exactly the same way. ∎

Lemma 4.7**.**

Let $\gamma=[0;b_{1},b_{2},\ldots,b_{N},\ldots]$ be an arbitrary irrational number, not a quadratic irrationality. Consider the sequence $B_{N}=(b_{1},b_{2},\ldots,b_{N})$ and define two numbers $n_{1}$ and $n_{2}$ from Lemma 4.5. Define two new sequences of positive integers

[TABLE]

Let us also define two new irrational numbers:

[TABLE]

Then $\max(\gamma^{1},\gamma^{2})>\gamma$.

Proof.

We apply Lemma 4.6 for $A=(b_{1},b_{2},\ldots,b_{n_{1}-1}),B=(b_{n_{1}},b_{n_{1}+1},\ldots,b_{n_{2}-1}),C=(b_{n_{2}},b_{n_{2}+1},\ldots)$. Here $\gamma=[0;A,B,C],\gamma^{1}=[0;A,C],\gamma^{2}=[0;A,B,B,C]$. Note that as $\gamma$ is not a not a quadratic irrationality, the sequence $C$ is not periodic. ∎

Now we are ready to prove Theorem 1.

Proof.

Suppose that $B$ is not periodic at right side. Consider a growing sequence of indices $k(j)$ such that for any natural $j$ the sequence $(a_{k(j)-j},\ldots,a_{k(j)},\ldots,a_{k(j)+j})$ coincides with the sequence $(b_{-j},\ldots,b_{0},\ldots,b_{j})$. Of course,

[TABLE]

Without loss of generality, on can say that $k(j+1)-k(j)\to\infty$ as $j\to\infty$. Consider an even $n$ such that $\varepsilon_{n}<\frac{b-a}{2}$ and $N=N(n)$ as defined in Lemma 4.5. Define $n_{1}<n_{2}$ from Lemma 4.5 for the sequence $(b_{1},\ldots,b_{N})$. As $B$ is not periodic to the right, define a minimal positive integer $r$ such that $b_{n_{1}+r}\neq b_{n_{2}+r}$. Consider the sequences $B_{N}^{1},B_{N}^{2}$ and the continued fractions $\gamma_{1},\gamma_{2}$ from Lemma 4.7 applied to the continued fraction $[0;b_{1},\ldots,b_{n}\ldots]=\gamma$. If $\gamma_{2}>\gamma$, define $g=2$, otherwise we put $g=1$. Consider the doubly infinite sequence $B^{\prime}=(\ldots,b_{-n},b_{0},B_{N}^{g},b_{N+1},\ldots)$. Note that

[TABLE]

Consider the corresponding continued fraction $\alpha^{\prime}$ which is obtained from the continued fraction $\alpha$ by replacing every segment $(a_{k(j)},\ldots,a_{k(j)+N})=(a_{k(j)},B_{N})$ by the segment $(a_{k(j)},B_{N}^{g})$ for every $j\geqslant n_{2}+r$. One can easily see that $\alpha^{\prime}$ and $\alpha$ satisfy the condition of lemma 4.4 and hence $\mu(\alpha^{\prime})<\mu(\alpha)+2\varepsilon_{n}$. But as $\mu(\alpha)+2\varepsilon_{n}<b$ and $(a,b)$ is the gap in $\mathbb{L}$, we have

[TABLE]

From the other hand, one can easily see that the sequence $B^{\prime}$ is weakly associated with $\alpha^{\prime}$. This means that

[TABLE]

We obtain a contradiction with (5). The case when $B$ is not periodic at left side is considered in exactly the same way. The theorem is proved. ∎

5 Proof of Theorem 2

The following Lemma from [5] immediately implies the $\Leftarrow$ part of the statement of Theorem 2.

Lemma 5.1**.**

Consider an arbitrary point $a$ in the Lagrange spectrum. If there exists a quadratic irrationality $\gamma$ such that $\mu(\gamma)=a$, then $a$ is admissible.

Now it is sufficient to prove that if $a$ is an admissible left endpoint of the Lagrange spectrum, then there exists a quadratic irrationality $\alpha$ such that $\mu(\alpha)=a$.

Proof.

Let $a$ be an admissible left endpoint of the Lagrange spectrum. Let $\alpha=[a_{0};a_{1},\ldots,a_{n},\ldots]$ be an irrational number such that $\mu(\alpha)=a$. Suppose that $\alpha$ is attainable, but not a quadratic irrationality. Let $k(j)$ be a growing sequence of indices such that

[TABLE]

Of course,

[TABLE]

Consider a strongly associated with $\alpha$ sequence $B=(\ldots,b_{-n},\ldots b_{-1},b_{0},b_{1},\ldots,b_{n},\ldots)$ having the following property: the sequence $(b_{-i},\ldots,b_{0},\ldots,b_{i})$ coincides with the sequence $a_{k(j)-i},\ldots,a_{k(j)},\ldots,a_{k(j)+i}$ for infinitely many $j$-s. Theorem 1 implies that $B$ is eventually periodic. That is, there exist a positive integer $m$ and two finite sequences $L$ and $R$ such that

[TABLE]

It follows from (6) that one of the inequalities

[TABLE]

holds for infinitely many $j$-s. Note that $[a_{k(j)};a_{k(j+1)},\ldots]\neq[b_{0};b_{1},\ldots,b_{m},\overline{R}]$, as $\alpha$ is not a quadratic irrationality and, of course, $[0;a_{k(j-1)},\ldots,a_{1}]\neq[0;b_{-1},\ldots,b_{-m},\overline{L}]$. Suppose that

[TABLE]

for infinitely many $j$-s. Denote by $p$ the length of period $R$. Denote by $r(j)$ the minimal positive number such that $a_{k(j)+r(j)}\neq b_{r(j)}$. Without loss of generality, one can say that:

$k(j+1)-k(j)-r(j)\to\infty$ as $j\to\infty$. 2. 2.

$[a_{k(j)};a_{k(j+1)},\ldots]>[b_{0};b_{1},\ldots,b_{m},\overline{R}]$ for every $j\in\mathbb{N}$. 3. 3.

$[a_{k(j)};a_{k(j+1)},\ldots,a_{k(j)+m}]=[b_{0};b_{1},\ldots,b_{m}]$ for every $j\in\mathbb{N}$. 4. 4.

The sequence $(a_{k(j)-j},\ldots,a_{k(j)},\ldots,a_{k(j)+j})$ coincides with the sequence $(b_{-j},\ldots,b_{0},\ldots,b_{j})$

for every $j\in\mathbb{N}$. 5. 5.

Period length $p$ is even.

Denote by $t(j)$ the number of periods $P$ in the sequence $(b_{m+1},\ldots,b_{r(j)}).$ Of course, $t(j)=[\frac{r(j)-m}{p}]$ and $t(j)$ tends to infinity. Denote by $\alpha_{n}$ a continued fraction obtained from the continued fraction $\alpha=[a_{0};a_{1},\ldots,a_{n},\ldots]$ as follows: if $t(j)>n$, then every pattern

[TABLE]

is replaced by the pattern

[TABLE]

Lemma 4.3 implies that since (8)

[TABLE]

and the length of the period $R$ is even, one has

[TABLE]

Since $k(j+1)-k(j)-r(j)\to\infty$ as $j\to\infty$ and the sequence $(a_{k(j)-j},\ldots,a_{k(j)})$ coincides with the sequence $(b_{-j},\ldots,b_{0})\ \forall j\in\mathbb{N}$, one can easily see that $\mu(\alpha_{n})\geqslant\mu(\alpha)+\delta_{m+(n+1)p}$.

Note that

[TABLE]

Indeed, every pattern of length $np$ which occurs in the sequence of partial quotients of $\alpha$ infinitely many times, occurs in the sequence of partial quotients of $\alpha_{n}$ infinitely many times. Similarly, every pattern of length $np$ which occurs in the sequence of partial quotients of $\alpha_{n}$ infinitely many times, occurs in the sequence of partial quotients of $\alpha$ infinitely many times too. Then, by Lemma 4.4,

[TABLE]

as $n\to\infty$. We obtain a contradiction with the fact that $a$ is the left endpoint of the gap $(a,b)$ in the Lagrange spectrum. Indeed, the inequality (9) implies that $\mu(\alpha_{n})>\mu(\alpha)\ \ \forall n\in\mathbb{N}$. In addition, the equality (10) implies that there exists a positive integer $N$ such that for any $n>N$ one has $a=\mu(\alpha)<\mu(\alpha_{n})<b$.

If the inequality (8) does not hold infinitely many times, then the inequality

[TABLE]

holds infinitely many times. This case is treated in exactly the same way. The theorem is proved. ∎

6 Proof of Theorem 3

First of all, let us show that $(\alpha_{n}^{*},\beta_{n})$ is the maximal gap in $\mathbb{L}$. As

[TABLE]

we have $\beta_{n}\in\mathbb{L}$. The proof of the fact that $\alpha_{n}^{*}\in\mathbb{L}$, when $n\geqslant 2$ is little more complicated. Recall that

[TABLE]

Denote by $C_{n}(k)$ the following finite sequence of integers

[TABLE]

Denote by $\zeta_{n}$ the following infinite continued fraction:

[TABLE]

A little calculation shows that $\mu(\zeta_{n})=\alpha_{n}^{*}$ and therefore $\alpha_{n}^{*}$ belongs to the Lagrange spectrum $\mathbb{L}$. By ([6], Lemma 4), $\alpha_{n}^{*}$ is a growing sequence. One can easily see that

[TABLE]

Thus, we have

[TABLE]

The following lemma is compilation of lemmas 3 and 4 from [6]

Lemma 6.1**.**

Consider a doubly infinite sequence $B=(\ldots,b_{-n},\ldots,b_{-1},b_{0},b_{1},\ldots,b_{n},\ldots)$ such that $M(B)<\sqrt{5}+1$. Then all elements of $B$ are bounded by $2$ and $B$ does not contain patterns of the form $(2,1,2,1)$ and $(1,2,1,2)$.

By Lemma 4, without loss of generality one can say that $M(B)=\lambda_{0}(B)$. Denote the continued fractions $[0;b_{1},\ldots,b_{n},\ldots]$ and $[0;b_{-1},\ldots,b_{-n},\ldots]$ by $x$ and $y$ respectively. Then

[TABLE]

Without loss of generality one can say that $x\leqslant y$. Now we need the following lemma from ([6], Theorem 4(i)).

Lemma 6.2**.**

Let $B$ be a doubly infinite sequence such that $M(B)=\lambda_{0}(B)$, then for all $n\geqslant 1$ we have

[TABLE]

It also follows from ([6], Theorem 4(ii)) that

[TABLE]

Denote

[TABLE]

Lemma 6.3**.**

Let $w=[0;a_{1},a_{2},\ldots,a_{n},\ldots]$ be a continued fraction with elements equal to $1$ or $2$. Suppose that the sequence $(a_{1},a_{2},\ldots,a_{n},\ldots)$ does not contain the pattern $(2,1,2,1)$. Then $w\geqslant w_{0}$

Proof.

Denote the elements of the continued fraction $w_{0}=[0;\overline{2,1,2,2}]$ by $[0;a^{\prime}_{1},\ldots,a^{\prime}_{m},\ldots]$. Denote by $r$ the minimal index such that $a_{r}\neq a^{\prime}_{r}$. Suppose that $w<w_{0}$. Then either $r$ is odd, $a_{r}=2,a^{\prime}_{r}=1$ or $r$ is even, $a_{r}=1,a^{\prime}_{r}=2$. However $a^{\prime}_{r}=2$ for any even $r$, thus the first case leads to a contradiction. Consider the second case. Of course, $r\geqslant 4$. Then $a^{\prime}_{r-3}=a_{r-3}=2,a^{\prime}_{r-2}=a_{r-2}=1,a^{\prime}_{r-1}=a_{r-1}=2$. This means that $(a_{r-3},a_{r-2},a_{r-1},a_{r})=(2,1,2,1)$ and we obtain a contradiction. ∎

Lemma 6.4**.**

If $B$ is strongly associated with $\alpha_{n+1}^{*}$, having

[TABLE]

Then $x=x_{0}$ and $y=y_{0}$.

Proof.

By Lemma 6.2 $x=[0;\underbrace{1,\ldots,1}_{2n},2,\ldots]$ and $y=[0;\underbrace{1,\ldots,1}_{2n},\ldots]$. Note that $x\leqslant x_{0}$. Indeed,

[TABLE]

The last equality follows from lemmas 6.3 and 6.1.

Now suppose that

[TABLE]

Denote the elements of the continued fraction $y_{0}$ by $[0;b^{\prime}_{1},\ldots,b^{\prime}_{m},\ldots]$. Denote by $r$ the minimal positive integer such that $b_{-r}\neq b^{\prime}_{r}$. Lemma 6.2 implies that $r>2n$. As $y>y_{0},\ b_{-2n-1}\leqslant b^{\prime}_{2n+1}=1$, therefore $b_{-2n-1}=1$. Similarly, $b_{-2n-2}\geqslant b^{\prime}_{2n+2}=2$, Lemma 6.1 implies that $b_{-2n-2}=2$. Now we have

[TABLE]

Suppose that $r\leqslant 4n+2$. Then $r$ is even, because otherwise $b_{-r}<b^{\prime}_{r}=1$. Hence $2=b_{-r}>b^{\prime}_{r}=1$. Note that $\lambda_{-2n-2}(B)\leqslant\lambda_{0}(B)=M(B)$. As $r-2n-3<2n+1$, the inequality (14) implies that

[TABLE]

Thus, $r>4n+2$. Note that $r\neq 4n+3$, because otherwise by (14) one has

[TABLE]

Hence, we have

[TABLE]

We obtain a contradiction with Lemma 6.3. The lemma is proved. ∎

In other words,

[TABLE]

Now the proof of Theorem 3 is quite simple.

Proof.

Suppose that $\alpha^{*}_{n}$ is admissible for some $n\geqslant 2$. Consider an attainable number $\alpha$ such that $\mu({\alpha})=\alpha^{*}_{n}$. Theorem 2 implies that $\alpha$ is a quadratic irrationality. Then $\alpha$ is associated with the unique purely periodic sequence $(\overline{P})$, where $P$ is the period of continued fraction expansion of $\alpha$. Lemma 6.4 implies that

[TABLE]

which is impossible because the sequence from the right part of (17) is not purely periodic. We obtain a contradiction and the theorem is proved. ∎