# On the graph of the dimension function of the Lagrange and Markov spectra

Carlos Matheus, Carlos Gustavo Moreira and Polina Vytnova  C. Matheus, CNRS &
École Polytechnique, CNRS (UMR 7640), 91128, Palaiseau, France.
<EMAIL_ADDRESS>C. G. Moreira, School of Mathematical Sciences, Nankai
University, Tianjin 300071, P. R.China, and IMPA, Estrada Dona Castorina 110,
CEP 22460-320, Rio de Janeiro, Brazil<EMAIL_ADDRESS>P. Vytnova, Department of
Mathematics, University of Surrey, Guildford, GU2 7XH, UK.
<EMAIL_ADDRESS>
###### Abstract.

We study the graph of the function $d(t)$ encoding the Hausdorff dimensions of
the classical Lagrange and Markov spectra with half-infinite lines of the form
$(-\infty,t)$. For this sake, we use the fact that the Hausdorff dimension of
dynamically Cantor sets drop after erasing an element of its Markov partition
to determine twelve non-trivial plateaux of $d(t)$. Next, we employ rigorous
numerical methods (from our recent joint paper with Pollicott) to produce
approximations of the graph of $d(t)$ between these twelve plateaux. As a
corollary, we prove that the largest ten non-trivial plateaux of $d(t)$ are
exactly those plateaux with lengths $>0.005$.

The second author was partly supported by CNPq, FAPERJ, and INCTMAT project of
J. Palis. The third author was partly supported by EPSRC grant EP/T001674/1.

##  1\. Introduction

The best constants of Diophantine approximations of irrational numbers and
real indefinite binary quadratic forms constitute two subsets of the real line
known as the Lagrange and Markov spectra. More concretely, the _Lagrange
spectrum_ is defined by

$L:=\left\\{\limsup\limits_{\begin{subarray}{c}p,q\to\infty\\\
p\in\mathbb{Z},q\in\mathbb{N}\end{subarray}}\frac{1}{|q^{2}\alpha-
pq|}<\infty:\alpha\in\mathbb{R}\right\\}$

and the _Markov spectrum_ is defined by

$M:=\left\\{\sup\limits_{\begin{subarray}{c}(p,q)\in\mathbb{Z}^{2}\\\
(p,q)\neq(0,0)\end{subarray}}\frac{1}{|ap^{2}+bpq+cq^{2}|}<\infty:ax^{2}+bxy+cy^{2}\textrm{
real indefinite, }b^{2}-4ac=1\right\\}.$

The features of these sets were systematically by several authors since the
foundational works by Markov [8], [9] in 1879–1880, and we refer the reader to
the book of Cusick–Flahive [3] for a nice account on the literature on this
topic before 1989.

The Lagrange and Markov spectra possess fascinating fractal structures between
$3$ and the so-called _Freiman’s constant_ $c_{F}=4.5278\dots$. In particular,
the second author proved in [11] that

$\dim(L\cap(-\infty,t))=\dim(M\cap(-\infty,t))$

for all $t\in\mathbb{R}$ (where $\dim$ stands for the Hausdorff dimension)
and, moreover, the _dimension function_

$d(t):=\dim(L\cap(-\infty,t))=\dim(M\cap(-\infty,t))$

is a _continuous_ function such that $d(3+\varepsilon)>0$ and
$d(\sqrt{12})=1$. Furthermore, it was recently shown in [10] that111Throughout
the manuscript all numbers are truncated and not rounded.

$t_{1}:=\min\\{t\in\mathbb{R}:d(t)=1\\}=3.334384\dots$

In this note, we investigate the graph of the dimension function $d(t)$. More
concretely, it is known that $L\subset M$ are closed subsets of the real line,
so that their complements are countable unions of open intervals consisting of
their gaps. By definition, the dimension function $d(t)$ is constant on the
gaps of $M$, that is, any such gap is included in a _plateau_ of $d(t)$.
Hence, we can _rigorously_ depict a portion of the graph of $d(t)$ by
determining the first few largest plateaux of $d(t)$ between $3$ and $t_{1}$.
The plot of the function $d$ on the complement to the plateaux can be found in
Appendix B.

For the smoothness of exposition, let us denote the two infinite plateaux by
$P_{0}=(-\infty,3)$ and $P_{1}=(t_{1},+\infty)$. Our main result, Theorem 3.1
below, states that the ten largest (non-trivial) plateaux
$P_{2},P_{3},\dots,P_{11}$ of $d(t)$ are precisely the plateaux of length
$0.005$. As a byproduct, we also identify two plateaux of length $0.004185$
and $0.004003$ and this helps to reduce necessary numerical computations.
Interestingly enough, the numbers and sizes of the gaps of $M$ included in
these plateaux are variable: for instance,

  * •

the largest plateau $P_{2}$ is the closure of the union of the largest gap
$G_{1}=(a_{1},b_{1})$ of $M$ before $t_{1}$ and another gap
$J_{1}=(b_{1},c_{1})$ of $M$, where $b_{1}$ is an isolated point of $L$;

  * •

the _second_ largest plateau $P_{3}$ is the closure of the union of the
_third_ largest gap $G_{3}=(a_{3},b_{3})$ of $M$ before $t_{1}$ and another
gap $J_{3}=(b_{3},c_{3})$ of $M$, where $b_{3}$ is an isolated point of $L$;

  * •

the _third_ largest plateau $P_{4}$ is the closure of the union of the
_second_ largest gap $G_{2}=(a_{2},b_{2})$ of $M$ before $t_{1}$ and another
gap $J_{2}=(b_{2},c_{2})$ of $M$, where $b_{2}$ is an isolated point of $L$;

  * •

the seventh largest plateau $P_{8}$ is a _single_ gap of $M$.

######  Remark 1.1.

In general, the structure of gaps of the spectra contained in small plateaux
of $d(t)$ can be quite intricate: for instance, this is the case of any
plateau of $d(t)$ containing a gap of $L$ including a Cantor set of points of
$M\setminus L$.

The main goal of this note is to rigorously justify the drawing of the
dimension function $d(t)$ of the Lagrange and Markov spectra shown in Figure
1. To this end, we first identify twelve plateaux. Then we compute an
approximation to the function on the complement to the plateaux using the same
method as was used in [10] to compute the value of $t_{0}$. It is rooted in
the approach introduced by Bumby in [1] and relies on two Lemmas that connect
the value $d(t)$ to the dimension of a certain Gauss–Cantor sets.

The dimension of the arising Gauss–Cantor sets is computed using the method
for computing the Hausdorff dimension originally developed in [14] and then
improved in [10].
We organise this paper as follows. In §2, we briefly review some classical
facts about the Lagrange and Markov spectra, including their dynamical
characterisations due to Perron. In §3, we state and prove the main result of
this note, namely, Theorem 3.1, about the first few (non-trivial) plateaux of
the dimension function $d(t)$.

Figure 1. A plot of the function $d$. Pink intervals are gaps in the Markov
spectrum identified in [3]

Finally, in §4, we discuss the numerical methods yielding rigorous
approximations to the graph of $d(t)$. The numerical results can be found in
the Appendix. The code is available at the third author’s github
https://github.com/Polevita/.

### Acknowledgments

The authors are very grateful to Prof. David Loeffler for providing access to
one of the computer servers that belong to the Number Theory group at the
University of Warwick. All computations have been carried out on that machine.

##  2\. Preliminaries

###  2.1. Perron’s characterisation of the classical spectra

Let $(\mathbb{N}^{*})^{\mathbb{Z}}$ be the set of bi-infinite sequences of
elements of $\mathbb{N}^{*}=\mathbb{N}\setminus\\{0\\}$. Given
$\underline{\alpha}=(\alpha_{n})_{n\in\mathbb{Z}}\in(\mathbb{N}^{*})^{\mathbb{Z}}$
and $k\in\mathbb{Z}$, we set

$\lambda_{k}(\underline{\alpha}):=[\alpha_{k};\alpha_{k+1},\alpha_{k+2},\ldots]+[0;\alpha_{k-1},\alpha_{k-2},\ldots],$

where $[b_{0};b_{1},b_{2},\ldots]$ stands for the continued fraction expansion

$[b_{0};b_{1},b_{2},\ldots]:=b_{0}+\cfrac{1}{b_{1}+\cfrac{1}{b_{2}+\cfrac{1}{\ddots}}}.$

Let $\sigma:(\mathbb{N}^{*})^{\mathbb{Z}}\to(\mathbb{N}^{*})^{\mathbb{Z}}$,
$\sigma((\alpha_{n})_{n\in\mathbb{Z}})=(\alpha_{n+1})_{n\in\mathbb{Z}}$, be
the Bernoulli shift. The Lagrange value $\underline{\alpha}$ is the limit
superior of values of $\lambda_{0}$ along the $\sigma$-orbit of
$\underline{\alpha}$, i.e.,

$\ell(\underline{\alpha}):=\limsup_{n\to\infty}\lambda_{0}(\sigma^{n}\underline{\alpha})=\limsup_{n\to\infty}\lambda_{n}(\underline{\alpha})$

and the Markov value of $\underline{\alpha}$ is the supremum of values of
$\lambda_{0}$ along the $\sigma$-orbit of $\underline{\alpha}$, i.e.,

$m(\underline{\alpha}):=\sup_{n\in\mathbb{Z}}\lambda_{0}(\sigma^{n}\underline{\alpha})=\sup_{n\in\mathbb{Z}}\lambda_{n}(\underline{\alpha}).$

In 1921, Perron [13] showed that the classical spectra are the collections of
(finite) Lagrange and Markov values:

(2.1)
$L:=\left\\{\ell(\alpha)\in\mathbb{R}\mid\alpha\in(\mathbb{N}^{*})^{\mathbb{Z}}\right\\}\quad\mbox{
and }\quad
M:=\left\\{m(\alpha)\in\mathbb{R}\mid\alpha\in(\mathbb{N}^{*})^{\mathbb{Z}}\right\\}.$

###  2.2. Some known facts about the classical spectra

Perron’s description of $L$ and $M$ allows to check that $L\subset M$ are
closed subsets of $\mathbb{R}$. As it was shown by Markov (circa $1880$),

$L\cap(-\infty,3)=M\cap(-\infty,3)=\\{\sqrt{5}<\sqrt{8}<\sqrt{221}/5<\dots\\}$

is an explicit sequence of quadratic irrationals converging to $3$. Also,
Perron showed in 1921 that $\sqrt{12},\sqrt{13}\in L$,

$L\cap(\sqrt{12},\sqrt{13})=M\cap(\sqrt{12},\sqrt{13})=\varnothing,$

and $m(\underline{\alpha})\leqslant\sqrt{12}$ if and only if
$\underline{\alpha}\in\\{1,2\\}^{\mathbb{Z}}$. Moreover, Hall [6] showed in
1947 that there is a smallest constant $c_{F}\leqslant 6$ such that
$L\cap[c_{F},\infty)=M\cap[c_{F},\infty)=[c_{F},\infty)$, and Freiman [5]
determined the exact value of $c_{F}$ in $1975$.

More recently, as we mentioned in §1, Moreira proved in 2018 that
$d(t)=\dim(L\cap(-\infty,t))=\dim(M\cap(-\infty,t))$ is a continuous function
of $t$ with $d(3+\varepsilon)>0$ for all $\varepsilon>0$. Finally, in [4] it
was shown that near $3$ the dimensions of Markov and Lagrange spectra have
asymptotic

$d\left(3+\varepsilon\right)=2\cdot\frac{W(c|\log\varepsilon|)}{|\log\varepsilon|}+O\left(\frac{\log|\log\varepsilon|}{|\log\varepsilon|^{2}}\right),$

where $W$ is the Lambert function and $c=\frac{1}{\log(3+\sqrt{5})/2}$. Figure
on the right shows the asymptotic estimate (smooth solid black line) in
comparison with the lower and upper bound on the dimension function $d$.
Furthermore, Moreira also proved in [11] the following useful dimension
formula: $d(t)=\min\\{1,2\cdot D(t)\\},$ where
$D(t)=\dim\bigl{(}\bigl{\\{}[0;a_{1},\dots,a_{n},\dots]\mid[a_{k};a_{k+1},\dots]+[0;a_{k-1},\dots,a_{1}]\leqslant
t\,\,\forall\,k\in\mathbb{N}\bigr{\\}}\bigr{)}$.

Figure 2. A plot of the function $d$ and its asymptotic near $3$. Pink
intervals are the plateaux identified in Theorem 3.1.

###  2.3. A useful lemma about continued fractions

In §3 below, the following lemma will be used several times:

######  Lemma 2.1.

Let $\beta$ be a finite word and $\theta$ be a symmetric finite word of even
size on $\\{1,2\\}$. Assume that, under some conditions (e.g., after
forbidding a finite list of finite strings), the Markov value of an infinite
word of the type $\gamma=\omega_{2}^{t}\beta^{t}2\theta 2\beta\omega_{1}$ is
attained at one of the $2$ next to $\theta$, where $\omega_{1},\omega_{2}$ are
infinite words (on $\\{1,2\\}$). Then this Markov value is minimum when
$\omega_{1}=\omega_{2}=:\omega$ and $[0;\beta,\omega]$ is minimum (under these
conditions).

###### Proof.

Let $x=[0;\beta,\omega_{1}]$ and $y=[0;\beta,\omega_{2}]$. If the positions
with the $2$’s after and before $\theta$ are, respectively, $0$ and $2r+1$
(where $2r=|\theta|$), then $\lambda_{0}(\gamma)=2+y+[0;\theta,2+x]=:h(x,y)$
and $\lambda_{2r+1}(\gamma)=2+x+[0;\theta,2+y]=h(y,x)$. Assume without loss of
generality (by symmetry) that the Markov value of $\gamma$ is attained at the
position $0$. Since $h(x,y)$ is increasing in $y$ and decreasing in $x$, then
we should have $y\geq x$ (indeed if $y<x$ then
$\lambda_{0}(\gamma)=h(x,y)<h(x,x)<h(y,x)=\lambda_{2r+1}(\gamma)$, a
contradiction). Consequently, in this case we have $h(x,y)\geq h(y,y)$, so the
minimum value in this case is attained when $x=y$, and is equal to
$h(y,y)=2+y+[0;\theta,2+y]$. Since $f(y)=[0;\theta,2+y]$ is a contraction,
$h(y,y)$ is increasing, so the minimum is attained when
$y=[0;\beta,\omega_{2}]=[0;\beta,\omega]$ is minimum. ∎

###  2.4. A useful lemma about Hausdorff dimension

Let $K$ be a $C^{2}$ dynamical Cantor set, i.e.,

$K=\bigcap\limits_{n\in\mathbb{N}}\psi^{-n}(I_{1}\cup\dots\cup I_{r}),$

where $\psi:I_{1}\cup\dots\cup I_{r}\to\mathbb{R}$ is a topologically mixing,
piecewise expanding $C^{2}$ map whose domain is a disjoint union of compact
intervals forming a Markov partition for $\psi$. The following folklore lemma
(see, e.g., Lemma 2.5 of Lima–Moreira [7]) says that the removal of an element
of the construction of a dynamical Cantor set strictly decreases the Hausdorff
dimension.

######  Lemma 2.2.

Let $K$ be a dynamical Cantor set defined by $\psi:I_{1}\cup\dots\cup
I_{m}\to\mathbb{R}$. Given $m\in\mathbb{N}$, let $\mathcal{P}^{m}$ be the
collection of connected components of $\psi^{-m+1}(I_{1}\cup\dots\cup I_{r})$.
If $J\in\mathcal{P}^{m}$ is an interval of the construction of $K$ such that
$\mathcal{P}^{m}\setminus\\{J\\}$ is a Markov partition with respect to
$\psi$, then the dynamical Cantor set

$K^{\prime}=\bigcap\limits_{n\in\mathbb{N}}\psi^{-nm}\left(\bigcup\limits_{I\in\mathcal{P}^{m}\setminus\\{J\\}}I\right)$

has a Hausdorff dimension $\textrm{dim}(K^{\prime})<\textrm{dim}(K)$.

For our purposes, we shall apply this lemma to the so-called _Gauss–Cantor
sets_ , that is, the dynamical Cantor sets produced by the restrictions of
iterates of the Gauss map
$g([0;a_{1},a_{2},a_{3},\dots]):=[0;a_{2},a_{3},\dots]$ to an appropriate
finite collection of compact subintervals of $(0,1)$. It is convenient to
describe them in terms of the so-called _forbidden subsequences_ of the
sequence $(a_{j})_{j\in\mathbb{N}}$.

##  3\. Plateaux of the function $d(t)$

The largest plateaux of the dimension function of the classical spectra are
described by the following statement.

######  Theorem 3.1.

The ten largest plateaux of the dimension function $d$ are precisely the
plateaux of length bigger than $0.005$.

The proof is constructive and consists of two parts. First, we identify the
ten largest plateaux and two extra auxiliary plateaux. Then, estimating $d$
numerically, we show that there is no other plateaux longer than $0.005$. In
order to specify the plateaux, we use Gauss–Cantor sets that we would like to
introduce now.
Consider the sets $FW_{i}$ of forbidden subsequences given in the table below:

$i$ | $FW_{i}$ | $i$ | $FW_{i}$ | $i$ | $FW_{i}$
---|---|---|---|---|---
$2$ | $121$ | $6$ |  $1212$, $2121$ | $10$ |  $1212$, $2121$, $1112111$
$3$ |  $121$, $212$ | $7$ |  $121$, $212$, $111222$, $222111$ | $11$ |  $121$, $212$, $111222$, $222111$, $21112$
$4$ |  $121$, $21222$, $22212$ | $8$ |  $1212$, $2121$, $211121112$ | $12$ |  $121$, $212$, $12221112$, $21112221$
$5$ |  $1212$, $2121$, $12111$, $11121$ | $9$ |  $21212$, $111212$, $212111$ | $13$ |  $21212$, $111212$, $212111$, $12121122$, $22112121$

Let us define the Gauss–Cantor sets of continued fractions by

$K_{i}=\\{[0;a_{1},a_{2},\dots]:(a_{n})_{n\in\mathbb{N}}\in\\{1,2\\}^{\mathbb{N}}\mbox{
doesn't contain a sequence from }FW_{i}\mbox{ as a subsequence}\\},\
i=2,3,\ldots,13;$

and let $D_{i}=\dim(K_{i})$, $i=2,\dots,13$, be the Hausdorff dimensions of
these Gauss–Cantor sets.

We now can list the twelve (non-trivial) plateaux of $d(t)$ in decreasing
order of lengths, that we were able to identify.

$\displaystyle P_{2}$
$\displaystyle=d^{-1}(2D_{2})=\left[\frac{4\sqrt{30}}{7},\frac{16351+720\sqrt{30}}{6409}\right]=[3.129843\ldots,3.166578\ldots];$
$\displaystyle P_{3}$
$\displaystyle=d^{-1}(2D_{3})=\left[\frac{4\sqrt{210}}{19},\frac{74613+4096\sqrt{210}}{43449}\right]=[3.050816\ldots,3.083377\ldots];$
$\displaystyle P_{4}$
$\displaystyle=d^{-1}(2D_{4})=\left[\frac{58473-745\sqrt{210}}{15422},\frac{1341491634738+14336533\sqrt{210}}{430566005892}\right]=[3.091487\dots,3.116129\dots];$
$\displaystyle P_{5}$
$\displaystyle=d^{-1}(2D_{5})=\left[\frac{15-\sqrt{30}}{3},\frac{332001+3760\sqrt{30}}{110409}\right]=[3.174258\dots,3.193538\dots];$
$\displaystyle P_{6}$
$\displaystyle=d^{-1}(2D_{6})=\left[\frac{4\sqrt{6}}{3},\frac{18879+800\sqrt{30}}{7089}\right]=[3.265986\dots,3.281249\dots];$
$\displaystyle P_{7}$
$\displaystyle=d^{-1}(2D_{7})=\left[\frac{28\sqrt{213378}}{4287},\frac{2497149255+2842763\sqrt{213378}}{1258910718}\right]=[3.017028\dots,3.026666\dots];$
$\displaystyle P_{8}$
$\displaystyle=d^{-1}(2D_{8})=\left[\frac{\sqrt{66045}}{79},\frac{190+4\sqrt{30}}{65}\right]=[3.253066\dots,3.260136\dots];$
$\displaystyle P_{9}$
$\displaystyle=d^{-1}(2D_{9})=\left[\frac{11+35\sqrt{87}}{102},\frac{56508+2716\sqrt{87}}{24687}\right]=[3.308414\dots,3.315152\dots];$
$\displaystyle P_{10}$
$\displaystyle=d^{-1}(2D_{10})=\left[\frac{14775-599\sqrt{30}}{3570},\frac{9537392579+1230099\sqrt{30}}{2959418678}\right]=[3.219647\dots,3.225001\dots];$
$\displaystyle P_{11}$
$\displaystyle=d^{-1}(2D_{11})=\left[\frac{10\sqrt{718341}}{2819},\frac{14264157401+16294182\sqrt{718341}}{9321104530}\right]=[3.006562\dots,3.011906\dots];$
$\displaystyle P_{12}$
$\displaystyle=d^{-1}(2D_{12})=\left[\frac{8\sqrt{1785}}{111},\frac{89042158285+148687392\sqrt{1785}}{31262231449}\right]=[3.044991\dots,3.049177\dots];$
$\displaystyle P_{13}$
$\displaystyle=d^{-1}(2D_{13})=\left[\frac{24\sqrt{35}}{43},\frac{457878845-1713407\sqrt{87}}{133663998}\right]=[3.301998\dots,3.306030\dots].$

######  Remark 3.2.

The numerical values of $D_{i}$ are specified in Appendix A. We would like to
highlight that the plateaux do not occur in the order of increasing length.

The remainder of this section is entirely dedicated to the proof of this
theorem _modulo_ some rigorous Hausdorff dimension estimates explained in §4.
More precisely, each subsection of §3.1 below discusses one of the twelve
plateaux above. Afterwards, in §3.2, we complete the proof of this theorem
(modulo some rigorous numerical computations explained in §4) by showing that
any non-trivial plateaux of $d(t)$ distinct from $P_{i}$, $2\leqslant
i\leqslant 13$, has length $<0.005$.

###  3.1. The twelve plateaux

####  3.1.1. The largest plateau: first appearances of $121$

Cusick identified in [2] a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,2,2,2}]+[0;\overline{2,2,1,2}]=\frac{4\sqrt{30}}{7}=3.1298431857...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factor $121$ (indeed $[2;\overline{1,2,2,2}]$ is the largest value
of a continued fraction with coefficients $1$ and $2$ and integer part $2$
without the factor $121$ and $[0;\overline{2,2,1,2}]$ is the largest value of
a continued fraction with coefficients $1$ and $2$ starting with $[0;2]$
without the factor $121$) and
$[2;\overline{1,1,2}]+[0;\overline{1,1,2}]=\sqrt{10}=3.16227766...$, the
smallest element of $M$ corresponding to a sequence with the factor $121$. The
left endpoint of this interval is also the left endpoint of the plateau, since
if the Markov value of an infinite sequence is strictly smaller than
$\frac{4\sqrt{30}}{7}$ then, for $n$ large enough, the sequence $(2212)^{n}$
is forbidden, which decreases the Hausdorff dimension (the regular Cantor set
of continued fractions with coefficients $1$ and $2$ with the factors $121$
and $(2212)^{n}$ forbidden has Hausdorff dimension strictly smaller than the
regular Cantor set where only $121$ is forbidden thanks to Lemma 2.2).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if the word $1212$ (or $2121$) is a factor of a
sequence, its Markov value is at least
$[2;1,2,\overline{2,1}]+[0;1,\overline{1,2}]>3.28$. If $12111$ appears as a
factor of a sequence, its Markov value is at least
$[2;1,1,1,\overline{1,2}]+[0;1,\overline{1,2}]>3.189$. So, if the Markov value
is smaller than $3.189$ and there is a factor $121$ then it must have a
neighbourhood $2112112$. If the factors $21121122$ and $22112112$ are
forbidden then $121$ forces the neighbourhood $121121121$, which begins and
ends by $121$, and repeating the argument we see that the sequence is forced
to be $\overline{121}$, and the Markov value is the right endpoint of the gap.
Let us determine the smallest possible Markov value of a sequence containing
$2112^{*}1122$ as a factor (knowing that $1212$, $2121$, $12111$ and $11121$
are forbidden). It should continue to the left as $22112112^{*}1122$, and so
is $y^{t}22112^{\\#}112^{*}1122x$, where $x$ and $y$ are infinite sequences.
If $a=[0;1,1,2,2,x]$ and $b=[0;1,1,2,2,y]$, by Lemma 2.1 the Markov value is
minimum when $a=b$ are minimum (with the previous constraints), so when
$a=b=[0;1,1,2,2,\overline{2,1,2,2}]$, and the minimum value is
$[2;1,\overline{1,2,2,2}]+[0;1,1,2,1,\overline{1,2,2,2}]=\frac{16351+720\sqrt{30}}{6409}=3.166578626...$.
This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factor $121$ forbidden the
finite word $(2221)^{n}121121(1222)^{n}$, for $n$ large enough, which connects
to any words of the previous regular Cantor set on both sides, so increasing
its Hausdorff dimension (cf. Lemma 2.2).

####  3.1.2. The second plateau: first appearances of $212$

Cusick identified in [2] a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,1,1,2,2,2}]+[0;\overline{2,2,1,1,1,2}]=\frac{4\sqrt{210}}{19}=3.050816157...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $121$ and $212$ (indeed $[2;\overline{1,1,1,2,2,2}]$ is
the largest value of a continued fraction with coefficients $1$ and $2$ and
integer part $2$ without the factors $121$ and $212$ and
$[0;\overline{2,2,1,1,1,2}]$ is the largest value of a continued fraction with
coefficients $1$ and $2$ starting with $[0;2]$ without the factors $121$ and
$212$) and
$[2;\overline{1,2,2}]+[0;\overline{2,1,2}]=\frac{\sqrt{85}}{3}=3.07318148576...$,
the smallest element of $M$ corresponding to a sequence with the factor $212$.
This is the third largest gap before $t_{1}=3.334384...$, but will produce the
second largest plateau, as we will see. The left endpoint of this interval is
also the left endpoint of the plateau, since if the Markov value of an
infinite sequence is strictly smaller than $\frac{4\sqrt{210}}{19}$ then, for
$n$ large enough, the sequence $(221112)^{n}$ is forbidden, which decreases
the Hausdorff dimension (the regular Cantor set of continued fractions with
coefficients $1$ and $2$ with the factors $121$, $212$ and $(221112)^{n}$
forbidden has Hausdorff dimension strictly smaller than the regular Cantor set
where only $121$ and $212$ are forbidden, cf. Lemma 2.2).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if $21222$ appears as a factor of a sequence, its Markov
value is at least $[2;1,2,\overline{2,1}]+[0;2,2,\overline{2,1}]>3.1156$. If
$212211$ appears as a factor of a sequence and $121$ is forbidden then its
Markov value is at least
$[2;1,2,\overline{2,1,2,2}]+[0;2,1,1,\overline{1,2,2,2}]>3.0833$. So, if the
Markov value is smaller than $3.0833$ and there is a factor $212$ then it must
have a neighbourhood $212212212$, which begins and ends by $212$, and
repeating the argument we see that the sequence is forced to be
$\overline{212}$, and its Markov value is the right endpoint of the gap. Let
us determine the smallest possible Markov value of a sequence containing
$12212^{*}211$ as a factor (knowing that $121$, $21222$ and $22212$ are
forbidden). It should continue to the left as $112212212^{*}211$, and so is
$y^{t}1122^{\\#}12212^{*}211x$, where $x$ and $y$ are infinite sequences
(notice that factors $2212^{o}212$ are not responsible for the Markov value
since $[2;1,2,2,\overline{2,1,2,2}]+[0;2,1,2,\overline{2,1,2,2}]<3.078$). If
$a=[0;2,1,1,x]$ and $b=[0;2,1,1,y]$, by Lemma 2.1 the Markov value is minimum
when $a=b$ are minimum (with the previous constraints), so when
$a=b=[0;2,1,1,\overline{1,2,2,2,1,1}]$, and the minimum value is
$[2;2,1,1,\overline{1,2,2,2,1,1}]+[0;1,2,2,1,2,2,1,1,\overline{1,2,2,2,1,1}]=\frac{74613+4096\sqrt{210}}{43449}=3.08337773...$.
This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $121$ and $212$
forbidden the finite word $(222111)^{n}22122122(111222)^{n}$, for $n$ large
enough, which connects to the previous regular Cantor set on both sides, so
increasing its Hausdorff dimension (thanks to Lemma 2.2).

####  3.1.3. The third plateau: first appearances of $21222$

Cusick identified in [2] a gap in Markov (and Lagrange) spectrum between
$[2;1,2,2,\overline{1,1,2,2,2,1}]+[0;2,\overline{1,1,2,2,2,1}]=\frac{58473-745\sqrt{210}}{15422}=3.09148776579...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $121$ and $21222$ (indeed
$[2;1,2,2,\overline{1,1,2,2,2,1}]$ is the largest value of a continued
fraction with coefficients $1$ and $2$ and integer part $2$ without the
factors $121$ and $21222$ and $[0;2,\overline{1,1,2,2,2,1}]$ is the largest
value of a continued fraction with coefficients $1$ and $2$ starting with
$[0;2]$ without the factors $121$ and $21222$) and
$[2;\overline{1,2,2,1,2,2,2,2}]+[0;\overline{2,2,2,1,2,2,1,2}]=\frac{\sqrt{233285}}{155}=3.11610231739...$,
the smallest element of $M$ corresponding to a sequence with the factor
$21222$. This is the second largest gap before $t_{1}=3.334384...$, but
produces the third largest plateau, as we will see. The left endpoint of this
interval is also the left endpoint of the plateau, since if the Markov value
of an infinite sequence is strictly smaller than
$\frac{58473-745\sqrt{210}}{15422}$ then, for $n$ large enough, the sequence
$(111222)^{n}112212211(222111)^{n}$ is forbidden, which decreases the
Hausdorff dimension (the regular Cantor set of continued fractions with
coefficients $1$ and $2$ with the factors $121$, $21222$ and
$(111222)^{n}112212211(222111)^{n}$ forbidden has Hausdorff dimension strictly
smaller than the regular Cantor set where only $121$ and $21222$ are
forbidden).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if $212221$ appears as a factor of a sequence, its
Markov value is at least
$[2;1,2,\overline{2,1}]+[0;2,2,1,\overline{1,2}]>3.1216$. If $2221222$ appears
as a factor of a sequence and $121$ is forbidden then its Markov value is at
least $[2;1,2,2,2,\overline{2,1,2,2}]+[0;2,2,\overline{2,1,2,2}]>3.1198$. If
$2122222$ appears as a factor of a sequence and $121$ is forbidden then its
Markov value is at least
$[2;1,2,\overline{2,1,2,2}]+[0;2,2,2,2,\overline{2,1,2,2}]>3.1173$. If
$11221222$ appears as a factor of a sequence and $121$ is forbidden then its
Markov value is at least
$[2;1,2,2,1,1,\overline{1,2,2,2}]+[0;2,2,\overline{2,1,2,2}]>3.117$. If
$21222211$ appears as a factor of a sequence and $121$ is forbidden then its
Markov value is at least
$[2;1,2,\overline{2,1,2,2}]+[0;2,2,2,1,1,\overline{1,2,2,2}]>3.1164$. If
$1221221222$ appears as a factor of a sequence and $121$ is forbidden then its
Markov value is at least
$[2;1,2,2,1,2,2,1,\overline{1,2,2,2}]+[0;2,2,\overline{2,1,2,2}]>3.11611$. So,
if the Markov value is smaller than $3.11611$ and there is a factor $21222$
then it must have a neighbourhood $2122221221222212$, which begins by $21222$
and ends by its transpose $22212$, and repeating the argument we see that the
sequence is forced to be $\overline{21222212}$, and its Markov value is the
right endpoint of the gap. Let us determine the smallest possible Markov value
of a sequence containing $12212212^{*}22$ as a factor (knowing that $121$,
$212221$, $2221222$, $2122222$, $11221222$, $21222211$ and their transposes
are forbidden). It should continue as $12212212^{*}22212212$, and if it
continue as $12212212^{*}2221221221$ then it is
$x^{t}12212212^{*}222^{\\#}1221221y$, where $x$ and $y$ are infinite
sequences. By Lemma 2.1, the minimum value is attained when
$[0;1,2,2,1,2,2,1,y]=[0;1,2,2,1,2,2,1,x]$ are minimum, so it should correspond
to
$\overline{122211}2212212^{*}222122122\overline{112221}=\overline{112221}12212212^{*}2221221221\overline{122211}$
(since $[0,\overline{1,2,2,2,1,1}]$ is the largest continued fraction in
$[0,1)$ with coefficients $1,2$ with $121$ and $122212$ forbidden), and the
minimum value is

$[2;2,2,2,1,2,2,1,2,2,\overline{1,1,2,2,2,1}]+[0;1,2,2,1,2,2,\overline{1,1,2,2,2,1}]\\\
=\frac{1341491634738+14336533\sqrt{210}}{430566005892}=3.1161294028759882...$

This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $121$, $212221$,
$2221222$, $2122222$, $11221222$, $11221222$, $1221221222$ and their
transposes forbidden the finite word

$(222111)^{n}22211221221222212212211222(111222)^{n},$

for $n$ large enough, which connects to the previous regular Cantor set on
both sides, so increasing its Hausdorff dimension.

####  3.1.4. The fourth plateau: first appearances of $12111$

Cusick identified in [2] a gap in Markov (and Lagrange) spectrum between
$[2;1,1,\overline{2,2,1,2}]+[0;1,1,\overline{2,2,1,2}]=\frac{15-\sqrt{30}}{3}=3.174258141649...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $1212$, $12111$ and their transposes (indeed, after a
factor $1,2$, $[0;1,1,\overline{2,2,1,2}]$ is the largest value of a
continuation with coefficients $1$ and $2$ and integer part $0$ without the
factors $1212$, $12111$ and their transposes) and
$[2;\overline{1,1,1,1,2,1,1,2}]+[0;\overline{1,1,2,1,1,1,1,2}]=\frac{\sqrt{9797}}{31}=3.19289664252...$,
the smallest element of $M$ corresponding to a sequence with the factor
$12111$. This is the fourth largest gap before $t_{1}=3.334384...$, and
produces the fourth largest plateau. The left endpoint of this interval is
also the left endpoint of the plateau, since if the Markov value of an
infinite sequence is strictly smaller than $\frac{15-\sqrt{30}}{3}$ then, for
$n$ large enough, the sequence $(2221)^{n}221121122(1222)^{n}$ is forbidden,
which decreases the Hausdorff dimension (the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $1212$, $2121$,
$12111$, $11121$ and $(2221)^{n}221121122(1222)^{n}$ forbidden has Hausdorff
dimension strictly smaller than the regular Cantor set where only $1212$,
$2121$, $12111$ and $11121$ are forbidden).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if $1212$ appears as a factor of a sequence, its Markov
value is at least $[2;1,2,\overline{2,1}]+[0;1,\overline{1,2}]>3.28$. If
$121112$ appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,2,\overline{2,1}]+[0;1,\overline{1,2}]>3.207$. If $1112111$ appears
as a factor of a sequence, then its Markov value is at least
$[2;1,1,1,\overline{1,2}]+[0;1,1,1,\overline{1,2}]>3.22$. If $1211111$ appears
as a factor of a sequence and $1212$ and $2121$ are forbidden then its Markov
value is at least
$[2;1,1,1,1,1,\overline{1,2,1,1}]+[0;1,\overline{1,2,1,1}]>3.197$. If
$22112111$ appears as a factor of a sequence and $1212$ and $2121$ are
forbidden then its Markov value is at least
$[2;1,1,1,\overline{1,2,1,1}]+[0;1,1,2,2,\overline{2,1,2,2}]>3.197$. If
$2112112111$ appears as a factor of a sequence and $1212$ and $2121$ are
forbidden then its Markov value is at least
$[2;1,1,1,\overline{1,2,1,1}]+[0;1,1,2,1,1,2,\overline{2,1,2,2}]>3.1933$. If
$12111122$ appears as a factor of a sequence and $1212$ and $2121$ are
forbidden then its Markov value is at least
$[2;1,1,1,1,2,2,\overline{2,1,2,2}]+[0;1,\overline{1,2,1,1}]>3.19303$. So, if
the Markov value is smaller than $3.19303$ and there is a factor $12111$ then
it must have a neighbourhood $111211211112112111$, which ends by $12111$ and
begins by its transpose $11121$, and repeating the argument we see that the
sequence is forced to be $\overline{12111121}$, and its Markov value is the
right endpoint of the gap. Let us determine the smallest possible Markov value
of a sequence containing $12^{*}111122$ as a factor (knowing that $1212$,
$121112$, $1112111$, $1211111$, $22112111$, $2112112111$ and their transposes
are forbidden). It should continue as $211112112^{*}111122$, and if it
continue as $2211112112^{*}111122$ then it is
$x^{t}2211112^{\\#}112^{*}111122y$, where $x$ and $y$ are infinite sequences.
By Lemma 2.1, the minimum value is attained when
$[0;1,1,1,1,2,2,y]=[0;1,1,1,1,2,2,x]$ are minimum, so it should correspond to
$\overline{2212}2211112112^{*}111122\overline{2122}=\overline{2221}1112112^{*}111\overline{1222}$,
and the minimum value is
$[2;1,1,1,\overline{1,2,2,2}]+[0;1,1,2,1,1,1,\overline{1,2,2,2}]=\frac{332001+3760\sqrt{30}}{110409}=3.1935382818628...$.
This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $1212$, $121112$,
$1112111$, $1211111$, $22112111$, $2112112111$, $12111122$ and their
transposes forbidden the finite word $(2221)^{n}222111121121111222(1222)^{n}$,
for $n$ large enough, which connects to the previous regular Cantor set on
both sides, so increasing its Hausdorff dimension.

####  3.1.5. The fifth plateau: first appearances of $1212$

There is a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,1,1,2}]+[0;\overline{1,1,1,2}]=\frac{4\sqrt{6}}{3}=3.26598632371...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $1212$, $2121$ (indeed, after a factor $1,2$,
$[0;\overline{1,1,1,2}]$ is the largest value of a continuation with
coefficients $1$ and $2$ and integer part $0$ without the factors $1212$ and
$2121$) and
$[2;\overline{1,2,2,1,2,1,1,2}]+[0;\overline{1,1,2,1,2,2,1,2}]=\frac{\sqrt{689}}{8}=3.2811011871...$,
the smallest element of $M$ corresponding to a sequence with the factor
$1212$. This is the fifth largest gap before $t_{1}=3.334384...$, and produces
the fifth largest plateau. The left endpoint of this interval is also the left
endpoint of the plateau, since if the Markov value of an infinite sequence is
strictly smaller than $\frac{4\sqrt{6}}{3}$ then, for $n$ large enough, the
sequence $(1112)^{n}1112111(2111)^{n}$ is forbidden, which decreases the
Hausdorff dimension (the regular Cantor set of continued fractions with
coefficients $1$ and $2$ with the factors $1212$, $2121$ and
$(1112)^{n}1112111(2111)^{n}$ forbidden has Hausdorff dimension strictly
smaller than the regular Cantor set where only $1212$ and $2121$ are
forbidden).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if $21212$ appears as a factor of a sequence, its Markov
value is at least $[2;1,2,\overline{2,1}]+[0;1,2,\overline{1,2}]>3.4$. If
$12121$ appears as a factor of a sequence, then its Markov value is at least
$[2;1,2,1,\overline{1,2}]+[0;1,\overline{1,2}]>3.297$. If $212111$ appears as
a factor of a sequence then its Markov value is at least
$[2;1,2,\overline{2,1}]+[0;1,1,1,\overline{1,2}]>3.314$. If $121222$ appears
as a factor of a sequence then its Markov value is at least
$[2;1,\overline{1,2}]+[0;1,2,2,2,\overline{2,1}]>3.2843$. If $2121122$ appears
as a factor of a sequence then its Markov value is at least
$[2;1,2,\overline{2,1}]+[0;1,1,2,2,\overline{2,1}]>3.288$. If $21211211$
appears as a factor of the sequence then its Markov value is at least
$[2;1,2,\overline{2,1}]+[0;1,1,2,1,1,\overline{1,2}]>3.2826$. If $1212211$
appears as a factor of a sequence then its Markov value is at least
$[2;1,\overline{1,2}]+[0;1,2,2,1,1,\overline{1,2}]>3.2814$. If $12122122$
appears as a factor of a sequence with $12121$, $21212$, $111212$, $212111$,
$222121$ and $121222$ forbidden, then its Markov value is at least
$[2;1,\overline{1,2,1,2,2,1}]+[0;1,2,2,1,2,2,\overline{2,1,2,2}]>3.28121$. So,
if the Markov value is smaller than $3.28121$ and there is a factor $1212$
then it must have a neighbourhood $212112122121$, which begins and ends by
$2121$, and repeating the argument we see that the sequence is forced to be
$\overline{12122121}$, and its Markov value is the right endpoint of the gap.
Let us determine the smallest possible Markov value of a sequence containing
$12^{*}122122$ as a factor (knowing that $21212$, $12121$, $111212$,
$2121122$, $21211211$, $121222$ , $1212211$ and their transposes are
forbidden). It should continue as $212212112^{*}122122$, and if it continues
as $2212212112^{*}122122$ it is $x^{t}2212212^{\\#}112^{*}122122y$, where $x$
and $y$ are infinite sequences. By Lemma 2.1, the minimum value is attained
when $[0;1,2,2,1,2,2,y]=[0;1,2,2,1,2,2,x]$ are minimum, so it should
correspond to
$\overline{2212}2212212112^{*}122122\overline{2122}=\overline{2122}12112^{*}1\overline{2212}$,
and the minimum value is
$[2;1,\overline{2,2,1,2}]+[0;1,1,2,1,\overline{2,2,1,2}]=\frac{18879+800\sqrt{30}}{7089}=3.28124988856557...$.
This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $12121$, $21212$,
$111212$, $2121122$, $21211211$, $121222$, $1212211$ and their transposes
forbidden the finite word $(2221)^{n}222122121121222(1222)^{n}$, for $n$ large
enough, which connects to the previous regular Cantor set on both sides, so
increasing its Hausdorff dimension.

####  3.1.6. The sixth plateau: first appearances of $111222$

We have a gap in Markov (and Lagrange) spectrum between
$[2;2,1,1,\overline{2,2,2,1,1,2,2,1,1,1,2,2,1,1}]+[0;1,1,1,\overline{2,2,1,1,2,2,2,1,1,2,2,1,1,1}]=\frac{28\sqrt{213378}}{4287}=3.017028188796...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $121$, $212$, $111222$ and $222111$ (indeed, if the Markov
value of a sequence with center $2112^{*}22$ is attained at $0$, then it is at
most
$[2;2,2,\overline{1,1,1,2,2,2}]+[0;1,1,2,2,\overline{1,1,1,2,2,2}]<3.007$;
also, $[2;2,1,1,\overline{2,2,2,1,1,2,2,1,1,1,2,2,1,1}]$ is the largest value
of a continued fraction with coefficients $1$ and $2$ starting with $[2;2,1]$
without the factors $121$, $212$, $111222$ and $222111$ and
$[0;1,1,1,\overline{2,2,1,1,2,2,2,1,1,2,2,1,1,1}]$ is the largest value of a
continued fraction with coefficients $1$ and $2$ starting with $[0;1,1]$
without the factors $121$, $212$, $111222$ and $222111$) and
$[2;\overline{2,2,2,1,1,1,1,2}]+[0;\overline{1,1,1,1,2,2,2,2}]=\frac{\sqrt{229}}{5}=3.02654919...$,
the smallest element of $M$ corresponding to a sequence with the factor
$111222$. The left endpoint of this interval is also the left endpoint of the
plateau, since if the Markov value of an infinite sequence is strictly smaller
than $\frac{28\sqrt{213378}}{4287}$ then, for $n$ large enough, the sequence
$(22211221112211)^{n}$ is forbidden, which decreases the Hausdorff dimension
(the regular Cantor set of continued fractions with coefficients $1$ and $2$
with the factors $121$, $212$, $111222$, $222111$ and $(22211221112211)^{n}$
forbidden has Hausdorff dimension strictly smaller than the regular Cantor set
where only $121$, $212$, $111222$ and $222111$ are forbidden).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if the word $1112221$ is a factor of a sequence, its
Markov value is at least
$[2;1,\overline{1,1,1,2,2,2}]+[0;2,2,1,1,\overline{2,2,2,1,1,1}]>3.0322$. If
$2221112$ appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,2,2,\overline{1,1,1,2,2,2}]+[0;\overline{2,2,2,1,1,1}]>3.0433$. If
$22211111$ appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,1,1,\overline{1,2,2,2,1,1}]+[0;\overline{2,2,2,1,1,1}]>3.0305$. If
$22222111$ appears as a factor of a sequence, its Markov value is at least

$[2;1,1,1,\overline{1,2,2,2,1,1}]+[0;2,2,2,2,\overline{2,1,1,1,2,2}]>3.02742.$

So, if the Markov value is smaller than $3.02742$ and there is a factor
$222111$ then it must have a neighbourhood $112222111122$. If $2221111221$
appears as a factor of such a sequence, its Markov value is at least

$[2;1,1,1,1,2,2,1,\overline{1,2,2,2,1,1,2,2,1,1,1,2,2,1}]+[0;2,2,2,1,1,1,1,\overline{2,2,1,1,1,2,2,1,1,2,2,2,1,1}]>3.0266664.$

If $2112222111$ appears as a factor of a sequence, its Markov value is at
least

$[2;1,1,1,1,2,2,2,2,1,1,\overline{2,2,2,1,1,2,2,1,1,1,2,2,1,1}]+[0;2,2,2,1,1,2,\overline{2,1,1,1,2,2,1,1,2,2,2,1,1,2}]>3.0266663.$

If the factors $2221111221$ and $2112222111$ (and their transposes) are
forbidden then $222111$ forces the neighbourhood $1111222211112222$, and
repeating the argument we see that the sequence is forced to be
$\overline{11112222}$, and the Markov value is the right endpoint of the gap.
Let us determine the smallest possible Markov value of a sequence containing
$2112222^{*}111$ as a factor (knowing that $121$, $212$, $1222111$, $2111222$,
$11111222$, $11122222$, $2221111221$ and their transposes are forbidden). It
should continue as $2112222^{*}11112^{\\#}222112$, and so is
$y^{t}2112222^{*}11112^{\\#}222112x$, where $x$ and $y$ are infinite
sequences. If $a=[0;2,2,2,1,1,2,x]$ and $b=[0;2,2,2,1,1,2,y]$, by Lemma 2.1
the Markov value is minimum when $a=b$ are minimum (with the previous
constraints), so when
$a=b=[0;2,2,2,1,1,2,\overline{2,1,1,1,2,2,1,1,2,2,2,1,1,2}]$, and the minimum
value is

$[2;\overline{2,2,2,1,1,2,2,1,1,1,2,2,1,1}]+[0;1,1,1,1,2,\overline{2,2,2,1,1,2,2,1,1,1,2,2,1,1}]\\\
=\frac{2497149255+2842763\sqrt{213378}}{1258910718}=3.026666336477...$

This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $121$, $212$,
$1222111$, $2111222$, $11111222$, $11122222$, $2221111221$, $2112222111$ and
transposes forbidden the finite word
$(22211221112211)^{n}222211112222(11221112211222)^{n}$, for $n$ large enough,
which connects to the previous regular Cantor set on both sides, so increasing
its Hausdorff dimension.

####  3.1.7. The seventh plateau: first appearances of $211121112$

We have a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,1,1,1,1,2,1,1,1,2}]+[0;\overline{1,1,1,2,1,1,1,1,1,2}]=\frac{\sqrt{66045}}{79}=3.25306604786...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $1212$, $2121$ and $211121112$ (indeed,
$[2;\overline{1,1,1,1,1,2,1,1,1,2}]$ is the largest value of a continued
fraction with coefficients $1$ and $2$ starting with $[2;1,1,1,1]$ without the
factors $1212$, $2121$ and $211121112$ and
$[0;\overline{1,1,1,2,1,1,1,1,1,2}]$ is the largest value of a continued
fraction with coefficients $1$ and $2$ preceded by $12$ and starting with $0$
without the factors $1212$, $2121$ and $211121112$) and
$[2;1,1,1,2,\overline{2,1,2,2}]+[0;1,1,1,2,\overline{2,1,2,2}]=\frac{190+4\sqrt{30}}{65}=3.2601369584647...$,
the smallest element of $M$ corresponding to a sequence with the factor
$211121112$. The left endpoint of this interval is also the left endpoint of
the plateau, since if the Markov value of an infinite sequence is strictly
smaller than $\frac{\sqrt{66045}}{79}$ then, for $n$ large enough, the
sequence $(2111211111)^{n}$ is forbidden, which decreases the Hausdorff
dimension (the regular Cantor set of continued fractions with coefficients $1$
and $2$ with the factors $1212$, $2121$, $211121112$ and $(2111211111)^{n}$
forbidden has Hausdorff dimension strictly smaller than the regular Cantor set
where only $1212$, $2121$ and $211121112$ are forbidden, cf. Lemma 2.2).

On the other hand, the right endpoint of this gap is
$[2;1,1,1,2,\overline{2,1,2,2}]+[0;1,1,1,2,\overline{2,1,2,2}]$ since
$[0;1,1,1,2,\overline{2,1,2,2}]$ is the smaller value of a continued fraction
with coefficients $1$ and $2$ starting with $[0;1,1,1,2]$ without the factors
$1212$ and $2121$. This is the right endpoint of the plateau since up to any
value strictly larger than this, we aggregate to the regular Cantor set of
continued fractions with coefficients $1$ and $2$ with the factors $1212$,
$2121$ and $211121112$ forbidden the finite word
$(2212)^{n}211121112(2122)^{n}$, for $n$ large enough, which connects to the
previous regular Cantor set on both sides, so increasing its Hausdorff
dimension (thanks to Lemma 2.2).

####  3.1.8. The eighth plateau: first appearances of $111212$

We have a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,1,2,2,1,2}]+[0;\overline{1,2,1,1,2,2}]=\frac{11+35\sqrt{87}}{102}=3.30841438096...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $21212$, $111212$ and $212111$ (indeed,
$[2;\overline{1,1,2,2,1,2}]$ is the largest value of a continued fraction with
coefficients $1$ and $2$ starting with $[2;1,1,2]$ without the factors
$21212$, $212111$and $111212$ and $[0;\overline{1,2,1,1,2,2}]$ is the largest
value of a continued fraction with coefficients $1$ and $2$ starting with
$[0;1,2,1,1]$ without the factors $21212$, $111212$ and $212111$) and
$[2;\overline{1,2,2,1,2,1,1,1,1,2}]+[0;\overline{1,1,1,1,2,1,2,2,1,2}]=\frac{\sqrt{33245}}{55}=3.31512935564...$,
the smallest element of $M$ corresponding to a sequence with the factor
$111212$. The left endpoint of this interval is also the left endpoint of the
plateau, since if the Markov value of an infinite sequence is strictly smaller
than $\frac{11+35\sqrt{87}}{102}$ then, for $n$ large enough, the sequence
$(112122)^{n}1121211(221211)^{n}$ is forbidden, which decreases the Hausdorff
dimension (the regular Cantor set of continued fractions with coefficients $1$
and $2$ with the factors $21212$, $212111$, $111212$ and
$(112122)^{n}1121211(221211)^{n}$ forbidden has Hausdorff dimension strictly
smaller than the regular Cantor set where only $21212$, $212111$ and $111212$
are forbidden).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if the word $2121112$ is a factor of a sequence, its
Markov value is at least
$[2;1,1,1,2,\overline{2,1}]+[0;1,2,\overline{2,1}]>3.3329$. If $21211111$
appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,1,1,\overline{1,2}]+[0;1,2,\overline{2,1}]>3.32$. If $1212111$
appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,\overline{1,2}]+[0;1,2,1,\overline{1,2}]>3.33$. If $22212111$
appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,\overline{1,2}]+[0;1,2,2,2,\overline{2,1}]>3.3189$. If $212111122$
appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,1,2,2,\overline{2,1,2,1,1,1}]+[0;1,2,\overline{2,1,2,1,1,1}]>3.3162$.
If $112212111$ appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,\overline{1,2,1,2,1,1}]+[0;1,2,2,1,1,\overline{1,2,1,2,1,1}]>3.3161$.
If $2212212111$ appears as a factor of a sequence, its Markov value is at
least
$[2;1,1,1,1,2,1,2,2,1,\overline{1,2,1,2,1,1}]+[0;1,2,2,1,2,2,\overline{2,1,2,1,1,1}]>3.31523$.
If $2121111211$ appears as a factor of a sequence, its Markov value is at
least
$[2;1,1,1,1,2,1,1,\overline{1,2,1,1,1,2}]+[0;1,2,\overline{2,1,2,1,1,1}]>3.31537$.
If $211212212111$ appears as a factor of a sequence, its Markov value is at
least
$[2;1,1,1,1,2,1,2,2,1,\overline{1,2,1,2,1,1,1,1}]+[0;1,2,2,1,2,1,1,2,\overline{2,1,2,1,1,1,1,1}]>3.315145$.
So, if the Markov value is smaller than $3.315145$ and there is a factor
$212111$ then it must have a neighbourhood $1112122121111212$, and repeating
the argument we see that the sequence is forced to be $\overline{2122121111}$,
and the Markov value is the right endpoint of the gap. Let us determine the
smallest possible Markov value of a sequence containing $211212212^{*}111$ as
a factor (knowing that $21212$, $2121112$, $21211111$, $1212111$, $22212111$,
$212111122$, $112212111$, $2212212111$, $2121111211$ and their transposes are
forbidden). It should continue as $211212212^{*}11112^{\\#}122121$, and so is
$y^{t}211212212^{*}11112^{\\#}12212112x$, where $x$ and $y$ are infinite
sequences. If $a=[0;1,2,2,1,2,1,1,2,x]$ and $b=[0;1,2,2,2,1,2,1,1,2,y]$, by
Lemma 2.1 the Markov value is minimum when $a=b$ are minimum (with the
previous constraints), so when
$a=b=[0;1,2,2,1,2,1,1,2,\overline{2,1,2,1,1,2}]$, and the minimum value is
$[2;\overline{1,2,2,1,2,1}]+[0;1,1,1,1,2,\overline{1,2,2,1,2,1}]=\frac{56508+2716\sqrt{87}}{24687}=3.3151521654...$.
This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $21212$, $2121112$,
$21211111$, $1212111$, $22212111$, $212111122$, $112212111$, $2212212111$,
$2121111211$, $211212212111$ and transposes forbidden the finite word
$(221121)^{n}221211112122(121122)^{n}$, for $n$ large enough, which connects
to the previous regular Cantor set on both sides, so increasing its Hausdorff
dimension.

####  3.1.9. The ninth plateau: first appearances of $1112111$

We have a gap in Markov (and Lagrange) spectrum between
$[2;1,1,1,2,1,1,2,2,\overline{1,2,2,2}]+[0;1,1,2,2,\overline{1,2,2,2}]=\frac{14775-599\sqrt{30}}{3570}=3.21964758558657...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $1212$, $2121$ and $1112111$ (indeed
$[2;1,1,1,2,1,1,2,2,\overline{1,2,2,2}]$ is the largest value of a continued
fraction with coefficients $1$ and $2$ starting with $[2;1,1]$ without the
factors $1212$, $2121$ and $1112111$ and $[0;1,1,2,2,\overline{1,2,2,2}]$ is
the largest value of a continued fraction with coefficients $1$ and $2$
starting with $[0;1,1,2]$ without the factors $1212$, $2121$ and $1112111$)
and
$[2;\overline{1,1,1,1,2}]+[0;\overline{1,1,1,1,2}]=\frac{2\sqrt{65}}{5}=3.2249030993...$,
the smallest element of $M$ corresponding to a sequence with the factor
$1112111$. The left endpoint of this interval is also the left endpoint of the
plateau, since if the Markov value of an infinite sequence is strictly smaller
than $\frac{14775-599\sqrt{30}}{3570}$ then, for $n$ large enough, the
sequence $(2221)^{n}2211211121122(1222)^{n}$ is forbidden, which decreases the
Hausdorff dimension (the regular Cantor set of continued fractions with
coefficients $1$ and $2$ with the factors $1212$, $2121$, $1112111$ and
$(2221)^{n}2211211121122(1222)^{n}$ forbidden has Hausdorff dimension strictly
smaller than the regular Cantor set where only $1212$, $2121$ and $1112111$
are forbidden).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if the word $11121112$ is a factor of a sequence, its
Markov value is at least
$[2;1,1,1,2,\overline{2,1,2,2}]+[0;1,1,1,\overline{1,2,1,1}]>3.24$. If
$111112111$ appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,\overline{1,2,1,1}]+[0;1,1,1,1,1,\overline{1,2,1,1}]>3.229$. If
$2211112111$ appears as a factor of a sequence, its Markov value is at least
$[2;1,1,1,\overline{1,2,1,1}]+[0;1,1,1,1,2,2,\overline{2,1,2,2}]>3.2256$. If
$211211112111$ appears as a factor of a sequence, its Markov value is at least

$[2;1,1,1,\overline{1,2,1,1}]+[0;1,1,1,1,2,1,1,2,\overline{2,1,2,2}]>3.22492.$

So, if the Markov value is smaller than $3.22492$ and there is a factor
$1112111$ then it must have a neighbourhood $11121111211112111$, which begins
and ends by $1112111$, and repeating the argument we see that the sequence is
forced to be $\overline{21111}$, and the Markov value is the right endpoint of
the gap. Let us determine the smallest possible Markov value of a sequence
containing $211211112^{*}111$ as a factor (knowing that $11121112$,
$111112111$, $2211112111$ and their transpose are forbidden). It should
continue as $211211112^{*}1111211112112$, and so is
$y^{t}211211112^{*}11112^{\\#}11112112x$, where $x$ and $y$ are infinite
sequences. If $a=[0;1,1,1,1,2,1,1,2,x]$ and $b=[0;1,1,1,1,2,1,1,2,y]$, by
Lemma 2.1 the Markov value is minimum when $a=b$ are minimum (with the
previous constraints), so when $a=b=[0;1,1,1,1,2,1,1,2,\overline{2,1,2,2}]$,
and the minimum value is

$[2;1,1,1,1,2,1,1,2,\overline{2,1,2,2}]+[0;1,1,1,1,2,1,1,1,1,2,1,1,2,\overline{2,1,2,2}]\\\
=\frac{9537392579+1230099\sqrt{30}}{2959418678}=3.22500164632....$

This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $11121112$,
$111112111$, $2211112111$, $211211112111$ and their transpose forbidden the
finite word $(2221)^{n}221121111211112111121122(1222)^{n}$, for $n$ large
enough, which connects to the previous regular Cantor set on both sides, so
increasing its Hausdorff dimension.

####  3.1.10. The tenth plateau: first appearances of $21112$

We have a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,1,1,1,1,2,2,1,1,2,2,2,1,1,2,2}]+[0;2,\overline{1,1,2,2,2,1,1,2,2,1,1,1,1,1,2,2}]=\frac{10\sqrt{718341}}{2819}=3.006562605623...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $121$, $212$, $111222$, $222111$ and $21112$ (indeed, if
the Markov value of a sequence with center $222^{*}112$ without the factors
$121$, $212$, $111222$, $222111$ is attained at $0$, then it is at most
$[2;2,2,1,1,2,2,\overline{1,1,1,2,2,2}]+[0;1,1,2,2,\overline{1,1,1,2,2,2}]<3.0059$;
also, $[2;\overline{1,1,1,1,1,2,2,1,1,2,2,2,1,1,2,2}]$ is the largest value of
a continued fraction with coefficients $1$ and $2$ with integer part $2$
without the factors $121$, $212$, $111222$, $222111$ and $21112$ and
$[0;2,\overline{1,1,2,2,2,1,1,2,2,1,1,1,1,1,2,2}]$ is the largest value of a
continued fraction with coefficients $1$ and $2$ starting with $[0;2,1]$
without the factors $121$, $212$, $111222$, $222111$ and $21112$) and
$[2;\overline{1,1,1,2,2}]+[0;\overline{2,1,1,1,2}]=\frac{\sqrt{145}}{4}=3.010398644698...$,
the smallest element of $M$ corresponding to a sequence with the factor
$21112$. The left endpoint of this interval is also the left endpoint of the
plateau, since if the Markov value of an infinite sequence is strictly smaller
than $\frac{10\sqrt{718341}}{2819}$ then, for $n$ large enough, the sequence
$(1111122112221122)^{n}$ is forbidden, which decreases the Hausdorff dimension
(the regular Cantor set of continued fractions with coefficients $1$ and $2$
with the factors $121$, $212$, $111222$, $222111$, $21112$ and
$(1111122112221122)^{n}$ forbidden has Hausdorff dimension strictly smaller
than the regular Cantor set where only $121$, $212$, $111222$, $222111$ and
$21112$ are forbidden, cf. Lemma 2.2).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if the word $211221112$ is a factor of a sequence, its
Markov value is at least
$[2;1,1,1,2,\overline{2,1,1,1,2,2}]+[0;2,1,1,2,\overline{2,1,1,1,2,2}]>3.016$.
If $1111221112$ appears as a factor of a sequence, its Markov value is at
least

$[2;1,1,1,2,\overline{2,1,1,1,2,2}]+[0;2,1,1,1,1,1,\overline{2,2,2,1,1,1}]>3.0118.$

So, if the Markov value is smaller than $3.0118$ and there is a factor $21112$
then it must have a neighbourhood $211122111221112$, which begins and ends by
$21112$, and repeating the argument we see that the sequence is forced to be
$\overline{21112}$, and the Markov value is the right endpoint of the gap. Let
us determine the smallest possible Markov value of a sequence containing
$111122^{*}1112$ as a factor (knowing that $121$, $212$, $222111$, $211221112$
and their transpose are forbidden). It should continue as
$111122^{*}11122111221111$, and so is $y^{t}111122^{*}111221112^{\\#}21111x$,
where $x$ and $y$ are infinite sequences. If $a=[0;2,1,1,1,1,x]$ and
$b=[0;2,1,1,1,1,y]$, by Lemma 2.1 the Markov value is minimum when $a=b$ are
minimum (with the previous constraints), so when
$a=b=[0;2,\overline{1,1,1,1,1,2,2,1,1,2,2,2,1,1,2,2}]$, and the minimum value
is
$[2;2,\overline{1,1,1,1,1,2,2,1,1,2,2,2,1,1,2,2}]+[0;1,1,1,2,2,1,1,1,2,2,\overline{1,1,1,1,1,2,2,1,1,2,2,2,1,1,2,2}]=\frac{14264157401+16294182\sqrt{718341}}{9321104530}=3.011906071937...$.
This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $121$, $212$,
$222111$, $211221112$, $1111221112$ and their transpose forbidden the finite
word

$(1111122112221122)^{n}1111122111221112211111(2211222112211111)^{n},$

for $n$ large enough, which connects to the previous regular Cantor set, so
increasing its Hausdorff dimension.

####  3.1.11. The eleventh plateau: first appearances of $21112221$

We have a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,1,1,2,2,2,2,2}]+[0;\overline{2,2,2,2,1,1,1,2}]=\frac{8\sqrt{1785}}{111}=3.0449917368328...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $121$, $212$, $12221112$ and $21112221$ (indeed,
$[2;\overline{1,1,1,2,2,2,2,2}]$ is the largest value of a continued fraction
with coefficients $1$ and $2$ without the factors $121$, $212$, $12221112$ and
$21112221$ and $[0;\overline{2,2,2,2,1,1,1,2}]$ is the largest value of a
continued fraction with coefficients $1$ and $2$ starting with $[0;2,2,2]$
without the factors $121$, $212$, $12221112$ and $21112221$), and

$[2;\overline{2,2,1,1,2,2,2,1,1,1,2,2,1,1,1,2}]+[0;\overline{1,1,1,2,2,1,1,1,2,2,2,1,1,2,2,2}]=\frac{5\sqrt{2059061}}{2353}=3.0491773395772939...,$

the smallest element of $M$ corresponding to a sequence with the factor
$21112221$. The left endpoint of this interval is also the left endpoint of
the plateau, since if the Markov value of an infinite sequence is strictly
smaller than $\frac{8\sqrt{1785}}{111}$ then, for $n$ large enough, the
sequence $(22222111)^{n}22(11122222)^{n}$ is forbidden, which decreases the
Hausdorff dimension (the regular Cantor set of continued fractions with
coefficients $1$ and $2$ with the factors $121$, $212$, $12221112$, $21112221$
and $(22222111)^{n}22(11122222)^{n}$ forbidden has Hausdorff dimension
strictly smaller than the regular Cantor set where only $121$, $212$,
$12221112$ and $21112221$ are forbidden).

On the other hand, the right endpoint of this gap is an isolated point in $M$
(and in $L$). Indeed, if $1222111222$ appears as a factor of a sequence, its
Markov value is at least
$[2;1,1,1,2,2,2,2,\overline{1,1,1,2,2,2}]+[0;2,2,1,1,\overline{2,2,2,1,1,1}]>3.0495$.
If $1112221112$ appears as a factor of a sequence, then its Markov value is at
least
$[2;1,1,1,2,2,\overline{1,1,1,2,2,2}]+[0;2,2,1,1,1,1,\overline{2,2,2,1,1,1}]>3.0498$.
If $122211122112$ appears as a factor of a sequence, its Markov value is at
least
$[2;1,1,1,2,2,1,1,2,2,\overline{1,1,1,2,2,2}]+[0;2,2,1,1,\overline{2,2,2,1,1,1}]>3.0492$.
If $122112221112$ appears as a factor of a sequence, then its Markov value is
at least
$[2;1,1,1,2,2,\overline{1,1,1,2,2,2}]+[0;2,2,1,1,2,2,1,1,\overline{2,2,2,1,1,1}]>3.0492$.
If $1222111221111$ appears as a factor of a sequence, its Markov value is at
least
$[2;1,1,1,2,2,1,1,\overline{1,1,1,2,2,2}]+[0;2,2,1,1,\overline{2,2,2,1,1,1}]>3.04919$.
If $2222112221112$ appears as a factor of a sequence, then its Markov value is
at least
$[2;1,1,1,2,2,\overline{1,1,1,2,2,2}]+[0;2,2,1,1,2,2,\overline{2,2,2,1,1,1}]>3.04918$.
If $122211122111221$ appears as a factor of a sequence, its Markov value is at
least

$[2;1,1,1,2,2,1,1,1,2,2,1,1,2,2,2,\overline{1,1,1,2,2,2,2,2}]+[0;2,2,1,1,2,2,2,\overline{1,1,1,2,2,2,2,2}]>3.0491779.$

If $211222112221112$ appears as a factor of a sequence, then its Markov value
is at least
$[2;1,1,1,2,2,1,1,1,\overline{2,2,2,1,1,1,1,1}]+[0;2,2,1,1,2,2,2,1,1,2,2,\overline{1,1,1,2,2,2}]>3.049178$.
If $1111222112221112$ appears as a factor of a sequence, then its Markov value
is at least
$[2;1,1,1,2,2,1,1,1,\overline{2,2,2,1,1,1,1,1}]+[0;2,2,1,1,2,2,2,1,1,1,1,1,2,2,2,\overline{1,1,1,2,2,2,2,2}]>3.0491775$.
If $1222111221112222$ appears as a factor of a sequence, its Markov value is
at least

$[2;1,1,1,2,2,1,1,1,2,2,2,2,2,\overline{1,1,1,2,2,2,2,2}]+[0;2,2,1,1,2,2,2,\overline{1,1,1,2,2,2,2,2}]>3.0491774.$

So, if the Markov value is smaller than $3.0491774$ and there is a factor
$21112221$ then it must have a neighbourhood $211122211222111221112221$, which
begins and ends by $21112221$, and repeating the argument we see that the
sequence is forced to be $\overline{2111222112221112}$, and its Markov value
is the right endpoint of the gap. Let us determine the smallest possible
Markov value of a sequence containing $1222^{*}111221112222$ as a factor
(knowing that $212$, $121$, $1222111222$, $1112221112$, $122211122112$,
$122112221112$, $1222111221111$, $2222112221112$, $122211122111221$,
$211222112221112$, $1111222112221112$ and their transposes are forbidden). It
should continue as $2221112211122211222^{*}111221112222$, and if it continues
as $22221112211122211222^{*}111221112222$ it is
$x^{t}2222111221112^{\\#}2211222^{*}111221112222y$, where $x$ and $y$ are
infinite sequences. By Lemma 2.1, the minimum value is attained when
$[0;1,1,1,2,2,1,1,1,2,2,2,2,y]=[0;1,1,1,2,2,1,1,1,2,2,2,2,x]$ are minimum, so
it should correspond to

$\overline{22222111}222221112211122211222^{*}1112211122222\overline{11122222},$

and the minimum value is

$\displaystyle[2;1,1,1,2,2,\overline{1,1,1,2,2,2,2,2}]+[0;2,2,1,1,2,2,2,1,1,1,2,2,\overline{1,1,1,2,2,2,2,2}]$
$\displaystyle=$
$\displaystyle\frac{89042158285+148687392\sqrt{1785}}{31262231449}$
$\displaystyle=$ $\displaystyle 3.049177432380786...$

This is the right endpoint of the plateau since up to any value strictly
larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $212$, $121$,
$1222111222$, $1112221112$, $122211122112$, $122112221112$, $1222111221111$,
$2222112221112$, $122211122111221$, $211222112221112$, $1111222112221112$,
$1222111221112222$ and their transposes forbidden the finite word
$(22222111)^{n}221112221122211122(11122222)^{n}$, for $n$ large enough, which
connects to the previous regular Cantor set on both sides, so increasing its
Hausdorff dimension.

####  3.1.12. The twelfth plateau: first appearances of $12121122$

We have a gap in Markov (and Lagrange) spectrum between
$[2;\overline{1,1,2,1,1,2,1,2}]+[0;\overline{1,2,1,1,2,1,1,2}]=\frac{24\sqrt{35}}{43}=3.3019980184742...$,
the largest element of $M$ corresponding to a sequence of $1$’s and $2$’s
without the factors $21212$, $111212$, $212111$, $12121122$ and $22112121$
(indeed, $[2;\overline{1,1,2,1,1,2,1,2}]$ is the largest value of a continued
fraction with coefficients $1$ and $2$ starting with $[2;1,1,2,1]$ without the
factors $21212$, $111212$, $212111$, $12121122$ and $22112121$ and
$[0;\overline{1,2,1,1,2,1,1,2}]$ is the largest value of a continued fraction
with coefficients $1$ and $2$ starting with $[0;1,2,1,1,2,1]$ without the
factors $21212$, $111212$, $212111$, $12121122$ and $22112121$) and
$[2;1,2,1,1,2,1,2,1,1,2,2,\overline{2,1,2,1,1,2}]+[0;1,1,2,2,\overline{2,1,2,1,1,2}]=\frac{457878845-1713407\sqrt{87}}{133663998}=3.306030457347...$,
the smallest element of $M$ corresponding to a sequence with the factor
$12121122$. The left endpoint of this interval is also the left endpoint of
the plateau, since if the Markov value of an infinite sequence is strictly
smaller than $\frac{24\sqrt{35}}{43}$ then, for $n$ large enough, the sequence
$(11211212)^{n}$ is forbidden, which decreases the Hausdorff dimension (the
regular Cantor set of continued fractions with coefficients $1$ and $2$ with
the factors $21212$, $111212$, $212111$, $12121122$, $22112121$ and
$(11211212)^{n}$ forbidden has Hausdorff dimension strictly smaller than the
regular Cantor set where only $21212$, $111212$, $212111$, $12121122$ and
$22112121$ are forbidden).

On the other hand, the right endpoint of this gap is
$[2;1,2,1,1,2,1,2,1,1,2,2,\overline{2,1,2,1,1,2}]+[0;1,1,2,2,\overline{2,1,2,1,1,2}]$
since the minimum Markov value of a sequence with center $1212^{*}1122$ is
attained at a sequence of the type $x^{t}22112^{\\#}1211212^{*}1122y$, where
$x$ and $y$ are infinite sequences. By Lemma 2.1, the minimum value is
attained when $[0;1,1,2,2,y]=[0;1,1,2,2,x]$ are minimum, so it should
correspond to $\overline{211212}221121211212^{*}1122\overline{212112}$, and
the minimum value is
$[2;1,1,2,2,\overline{2,1,2,1,1,2}]+[0;1,2,1,1,2,1,2,1,1,2,2,\overline{2,1,2,1,1,2}]=\frac{457878845-1713407\sqrt{87}}{133663998}=3.3060304573471179...$.
It coincides with the right endpoint of the plateau since up to any value
strictly larger than this, we aggregate to the regular Cantor set of continued
fractions with coefficients $1$ and $2$ with the factors $21212$, $111212$,
$212111$, $12121122$ and $22112121$ forbidden the finite word
$(221121)^{n}222112121121211222(121122)^{n}$, for $n$ large enough, which
connects to the previous regular Cantor set on both sides, so increasing its
Hausdorff dimension.

###  3.2. Between the largest plateaux: end of proof of Theorem 3.1 modulo
dimension estimates at checkpoints

For two sets $A,B\subset\mathbb{R}$ we say that $A\prec B$ if $x<y$ for all
$x\in A$ and $y\in B$. We would like now to list the twelve plateaux of $d(t)$
in order of their _occurrence_ on the real line (from left to right)

$P_{0}\prec P_{11}\prec P_{7}\prec P_{12}\prec P_{3}\prec P_{4}\prec
P_{2}\prec P_{5}\prec P_{10}\prec P_{8}\prec P_{6}\prec P_{13}\prec P_{9}\prec
P_{1}.$

In order to show that $P_{i}$, $2\leqslant i\leqslant 11$, are indeed the $10$
largest plateaux we provide estimates on the values of the dimension function
$d$ on the intervals in between. To setup the computation, we choose the
intervals between the two plateaux $P_{i}=(a_{i},b_{i})\prec
P_{j}=(a_{j},b_{j})$ given by

$S_{(i;j)}=[10^{-4}([10^{4}b_{i}]-1),10^{-4}([10^{4}a_{j}]+1)],$

where $[x]$ stands for the integer part of the number $x$. We obtain the
intervals $S_{(i;j)}$ by straightforward computation: it gives, for example,
$S_{(0;11)}=[2.9999,3.0066]$, $S_{(11;7)}=[3.0117,3.0171]$, and so on. In
Appendix A we list lower and upper bounds on the dimension function at certain
“checkpoints”, chosen at the distance $0.0024$ apart (with a few exceptions),
to show that there is no other plateaux larger than $0.005$. This completes
the proof of Theorem 3.1.

In the last §4 we give an algorithm for computing the dimension function $d$.

##  4\. Numerical method

Computation of the dimension function $d$ is done by obtaining lower and upper
bounds on the values of $d$ at a large number of points $s_{k}\in(3,t_{1})$
and is based on the following two observations [10, §2.1].
First, we fix the maximal length of forbidden strings $n$, say, $n=16$ (larger
value of $n$ will result in a more accurate approximation of $d$). Then for a
point $s\in(a,b)$ we build recursively a finite set of finite “forbidden”
sequences
$\beta_{-n}\beta_{-n+1}\ldots\beta_{-1}\beta_{0}\beta_{1}\ldots\beta_{n}$ so
that any infinite extension $\alpha\in\\{1,2\\}^{\mathbb{Z}}$ with the
property that $\alpha_{j}=\beta_{j}$ for all $-n\leq j\leq n$ we have that
$\lambda_{0}(\alpha)=[0;\alpha_{-1},\alpha_{-2},\ldots]+[\alpha_{0};\alpha_{1},\alpha_{2},\ldots]>s.$
Clearly, after excluding from $E_{2}$ all irrational numbers whose continued
fraction expansion contains a “forbidden” string, we obtain a Cantor set
$K\subsetneq E_{2}$ such that $M\cap(-\infty,s)\subset 2+K+K$. Recall that
$\dim(K+K)\leq\dim K+\dim_{B}K$, where $\dim_{B}K$ denotes the upper box
dimension. It is known that $\dim_{B}K=\dim K$ for these types of sets (cf.
Chapter 4 of Palis–Takens book [12]) and hence an upper bound on $\dim K$
gives us an upper bound

$d(s)=\dim((-\infty,s)\cap M)\leq 2\dim K.$

Once the set of forbidden sequences is constructed, we may compute the maximal
Markov value of strings which do not contain a substring from $K$ and denote
it by $s^{\prime}$. It was shown in [11, proof of Lemma 3] that the dimension
of $K$ gives us a lower bound on $d(s^{\prime})$ via the inequality

$\min\\{2\dim K,1\\}\leq\dim((-\infty,s^{\prime})\cap M)=d(s^{\prime}).$

Repeating the procedure described above for $n$ sufficiently large and for a
sequence of values $\\{s_{k}\\}\in(a,b)$, we obtain an upper bound for
$d(s_{k})$ and a lower bound for $d(s_{k}^{\prime})$. Monotonicity of $d$
implies that $\max_{s_{k}^{\prime}<s}d(s_{k}^{\prime})$ is a lower bound for
$d(s)$. This allows us to obtain an accurate approximation to the dimension
function. The pseudocode for the recursive function that constructs the set of
forbidden words is given in Algorithm 1 below.

Algorithm 1 Recursive computation of the set of forbidden strings.

1function ForbiddenWords($\beta$, $lmax$, $mmax$, $s$, $n$ )

2 $s,n$: threshold and the maximal length of the string; $\beta$: the current
string; $mmax$: the current largest Markov number of the strings that do not
contain forbidden substrings; $lmax$: the current maximal length of the
forbidden strings.

3 if $|\beta|>0$ then

4 $(a,b)$ $\leftarrow$ convex hull of $\lambda_{0}$ obtained from
continuations of $\beta$;

5 if $a>s$ then

6 Save $\beta$ and its reverse; $\triangleright$ Markov values are larger than
$s$

7 if $lmax<|\beta|$ then

8 $lmax$ $\leftarrow$ $|\beta|$

9 end if

10 else if $b<s$ then $\triangleright$ Markov values are smaller than $s$

11 if $b>mmax$ then

12 $mmax$ $\leftarrow$ $b$

13 end if

14 else if $|\beta|<2n+1$ then

15 if $|\beta|$ is even then $\triangleright$ add one symbol on the right

16 ForbiddenWords($\beta 1$, $lmax+1$, $mmax$, $s$, $n$ )

17 ForbiddenWords($\beta 2$, $lmax+1$, $mmax$, $s$, $n$ )

18 else$\triangleright$ add one symbol on the left

19 ForbiddenWords($1\beta$, $lmax+1$, $mmax$, $s$, $n$ )

20 ForbiddenWords($2\beta$, $lmax+1$, $mmax$, $s$, $n$ )

21 end if

22 else if $b>mmax$ then $\triangleright$ $|\beta|=2n+1$ and $s$ is in the
convex hull of $\lambda_{0}$.

23 $mmax$ $\leftarrow$ $b$

24 end if

25 else$\triangleright$ $\beta$ is the empty word

26 ForbiddenWords($1$, $1$, $0$, $s$, $n$ )

27 ForbiddenWords($2$, $1$, $0$, $s$, $n$ )

28 end if

29end function

In order to confirm that all continuations of a certain string result in a
Markov value greater than the chosen threshold, the following fact [10, Lemma
2.4] can be used.

######  Lemma 4.1.

For any sequence $\alpha\in\\{1,2\\}^{\mathbb{Z}}$ we have an upper bound

$\lambda_{0}(\alpha)\leq\begin{cases}[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{12}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{12}],&\mbox{
if }k\mbox{ and }j\mbox{ are even,}\\\
[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{21}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{12}],&\mbox{
if }k\mbox{ is even and }j\mbox{ is odd,}\\\
[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{12}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{21}],&\mbox{
if }k\mbox{ is odd and }j\mbox{ is even,}\\\
[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{21}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{21}],&\mbox{
if }k\mbox{ and }j\mbox{ are odd;}\end{cases}$

and a lower bound

$\lambda_{0}(\alpha)\geq\begin{cases}[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{21}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{21}],&\mbox{
if }k\mbox{ and }j\mbox{ are even,}\\\
[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{12}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{21}],&\mbox{
if }k\mbox{ is even and }j\mbox{ is odd,}\\\
[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{21}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{12}],&\mbox{
if }k\mbox{ is odd and }j\mbox{ is even,}\\\
[\alpha_{0};\alpha_{1}\dots\alpha_{j}\overline{12}]+[0;\alpha_{-1}\dots\alpha_{-k}\overline{12}],&\mbox{
if }k\mbox{ and }j\mbox{ are odd.}\end{cases}$

We now want to illustrate the recursive procedure for constructing the set of
forbidden strings using.

######  Example 4.2.

Let us fix $n=6$ and consider a point $s=3.2658$. Then we take $\beta_{0}=1$
and compute lower and upper bounds on $\lambda_{0}$ using Lemma 4.1, which
gives us $1.73<\lambda_{0}(\beta)<2.47$. Since both values less than $s$, we
make a record of the upper bound, and consider $\beta_{0}=2$. This choice
gives us $2.73<\lambda_{0}(\beta)<3.47$, so we consider continuation
$\beta_{-1}\beta_{0}=12$ that gives $2.94<\lambda_{0}(\beta)<3.47$ and then
add $\beta_{1}=1$ that gives $3.15<\lambda_{0}(\beta)<3.47$. We proceed,
adding at odd steps a digit on the left and at even steps a digit on the
right, until either we get an interval for $\lambda_{0}$ that lies to the
right of $s$ or to the left of $s$, or we obtain a string of length $2n+1$
continuations of which contain $s$.
$|\beta|$ $\beta$ $13$ $1121112^{*}$ $\kern-11.0pt111212$ $13$ $2121112^{*}$
$\kern-11.0pt111211$ $12$ $2121112^{*}$ $\kern-11.0pt11121$ $12$ $121112^{*}$
$\kern-11.0pt111212$ $5$ $112^{*}$ $\kern-11.0pt12$ $5$ $212^{*}$
$\kern-11.0pt11$ $4$ $212^{*}$ $\kern-11.0pt1$ $4$ $12^{*}$ $\kern-11.0pt12$
For $s=3.2658$ and $n=6$ the function in Algorithm 1 gives $8$ forbidden
words, that are displayed on the right in the order they have been
constructed. It also gives a maximum Markov value for non-excluded strings
$s^{\prime}=3.2660$. We see that the first $6$ strings are actually redundant,
since they contain the strings of length $4$ as substrings. After further
reduction we are left with only $2$ forbidden strings of length $4$, $2121$
and $1212$. The algorithm developed in [14] gives an estimate $\dim K\leq
0.46933$. We therefore conclude that $d(3.2658)\leq 0.93866\leq d(3.2660)$.

##  Appendix A Estimates on the dimension function at checkpoints

Plateau or an | Length | Checkpoints | Hausdorff dimension
---|---|---|---
intermediate interval | of the plateau | in the interval | lower bound | upper bound
$S_{(0;11)}=[2.9999,3.0066]$ |  | $3.0015$ | $0.432861$ | $0.432861$
| $3.0031$ | $0.455261$ | $0.455261$
| $3.0055$ | $0.510071$ | $0.510071$
$P_{11}$ | $0.00534347$ |  | 0.536334
$S_{(11;7)}=[3.0117,3.0171]$ |  | $3.0126$ | $0.545441$ | $0.545441$
| $3.0150$ | $0.551178$ | $0.551178$
$P_{7}$ | $0.00963815$ |  | 0.569770
$S_{(7;12)}=[3.0265,3.0451]$ |  | $3.0273$ | $0.577393$ | $0.577393$
| $3.0297$ | $0.589935$ | $0.589935$
| $3.0321$ | $0.617218$ | $0.620728$
| $3.0345$ | $0.659302$ | $0.659302$
| $3.0369$ | $0.662750$ | $0.665924$
| $3.0393$ | $0.686493$ | $0.686493$
| $3.0417$ | $0.692932$ | $0.692932$
| $3.0441$ | $0.700073$ | $0.700073$
$P_{12}$ | $0.0041857$ |  | 0.709914
$S_{(12;3)}=[3.0490,3.0509]$ | No estimate needed: $|S_{(12;3)}|=0.0019<0.005$. $R_{1}^{2}$
$P_{3}$ | $0.03256158$ |  | 0.728108
$S_{(3;4)}=[3.0832,3.0915]$ |  | $3.0856$ | $0.730499$ | $0.730499$
| $3.0880$ | $0.739929$ | $0.739929$
| $3.0904$ | $0.742004$ | $0.742004$
$P_{4}$ | $0.02464164$ |  | 0.750628
$S_{(4;2)}=[3.1160,3.1299]$ |  | $3.1184$ | $0.765778$ | $0.766235$
| $3.1208$ | $0.775299$ | $0.775299$
| $3.1232$ | $0.783356$ | $0.784485$
| $3.1256$ | $0.796478$ | $0.796478$
| $3.1280$ | $0.807770$ | $0.807770$
$P_{2}$ | $0.03673544$ |  | 0.812150
$S_{(2;5)}=[3.1664,3.1743]$ |  | $3.1672$ | $0.812927$ | $0.813019$
| $3.1696$ | $0.815735$ | $0.815735$
| $3.1720$ | $0.818542$ | $0.818542$
$P_{5}$ | $0.01928014$ |  | 0.827194
$S_{(5;10)}=[3.1934,3.2170]$ |  | $3.1950$ | $0.830841$ | $0.830841$
| $3.1974$ | $0.830841$ | $0.831421$
| $3.1998$ | $0.843140$ | $0.846527$
| $3.2022$ | $0.851990$ | $0.851990$
| $3.2046$ | $0.859192$ | $0.861786$
| $3.2070$ | $0.868286$ | $0.869415$
| $3.2094$ | $0.869629$ | $0.869629$
| $3.2118$ | $0.876587$ | $0.877350$
| $3.2142$ | $0.878082$ | $0.878082$
| $3.2166$ | $0.882294$ | $0.883118$
| $3.2190$ | $0.888153$ | $0.888794$
$P_{10}$ | $0.00535406$ |  | 0.889660
$S_{(10;8)}=[3.2248,3.2531]$ |  | $3.2265$ | $0.890961$ | $0.891022$
| $3.2289$ | $0.892151$ | $0.892151$
| $3.2313$ | $0.895325$ | $0.897522$
| $3.2337$ | $0.901459$ | $0.901825$
| $3.2361$ | $0.903961$ | $0.905518$
| $3.2385$ | $0.907532$ | $0.909607$
| $3.2409$ | $0.910797$ | $0.910797$
| $3.2433$ | $0.912079$ | $0.913177$
| $3.2457$ | $0.917023$ | $0.917480$
| $3.2481$ | $0.919891$ | $0.921661$
| $3.2505$ | $0.926117$ | $0.928192$
| $3.2529$ | $0.931702$ | $0.932281$
$P_{8}$ | $0.00707091$ |  | 0.932262
$S_{(8;6)}=[3.2600,3.2660]$ |  | $3.2624$ | $0.934113$ | $0.934113$
| $3.2648$ | $0.937103$ | $0.937103$
$P_{6}$ | $0.01526356$ |  | 0.938646
$S_{(6;13)}=[3.2811,3.3021]$ |  | $3.2827$ | $0.938995$ | $0.939240$
| $3.2851$ | $0.944000$ | $0.945496$
| $3.2875$ | $0.949036$ | $0.950256$
| $3.2899$ | $0.954529$ | $0.956268$
| $3.2923$ | $0.960846$ | $0.960846$
| $3.2947$ | $0.965393$ | $0.965973$
| $3.2971$ | $0.966095$ | $0.966095$
| $3.2995$ | $0.966187$ | $0.966187$
| $3.3019$ | $0.967072$ | $0.967834$
$P_{13}$ | $0.00403244$ |  | 0.967812
$S_{(13;9)}=[3.3059,3.3085]$ | No estimate needed: $|S_{(13;9)}|=0.0036<0.005$. $R_{1}^{2}$
$P_{9}$ | $0.00673778$ |  | 0.971588
$S_{(9;1)}=[3.3150,3.3344]$ |  | $3.3158$ | $0.971619$ | $0.972382$
| $3.3182$ | $0.976440$ | $0.976929$
| $3.3206$ | $0.979279$ | $0.980682$
| $3.3230$ | $0.985535$ | $0.986542$
| $3.3254$ | $0.990051$ | $0.991913$
| $3.3278$ | $0.994781$ | $0.995270$
| $3.3302$ | $0.995544$ | $0.995575$
| $3.3326$ | $0.995605$ | $0.995605$

##  Appendix B Plot of the dimension function

Figure 3. A sketch of the graph of the dimension function $d$ on the intervals
between the plateaux. The blue curve corresponds to the upper bound, the red
curve corresponds to the lower bound. Estimates are obtained using a mesh of
size $0.00005$. We would like to point out that all plots are not to scale
and, moreover, scale on axis is different on every plot.

(a) $t\in S_{(0;11)}$

(b) $t\in S_{(11;7)}$

(c) $t\in S_{(7;12)}$

(d) $t\in S_{(12;3)}$

(e) $t\in S_{(3;4)}$

(f) $t\in S_{(4;2)}$

(g) $t\in S_{(2;5)}$

(a) $t\in S_{(5;10)}$

(b) $t\in S_{(10;8)}$

(c) $t\in S_{(8;6)}$

(a) $t\in S_{(6;13)}$

(b) $t\in S_{(13;9)}$

(c) $t\in S_{(9;1)}$

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