# Unfocused notes on the Markoff equation and T-Singularities

Markus Perling

###### Abstract.

We consider minimal resolutions of the singularities for weighted projective
planes of type $\mathbb{P}(e^{2},f^{2},g^{2})$, where $e,f,g$ satisfy the
Markoff equation $e^{2}+f^{2}+g^{2}=3efg$. We give a complete classification
of such resolutions in terms of continued fractions similar to classical work
of Frobenius. In particular, we investigate the behaviour of resolutions under
mutations and describe a Cantor set emerging as limits of continued fractions.

###### 2020 Mathematics Subject Classification:

Primary: 14J17, 11J06; Secondary: 28A78, 28A80, 14F08

##  1\. Introduction

The main aim of these notes is to describe the minimal resolutions of
singularities for weighted projective planes $\mathbb{P}(e^{2},f^{2},g^{2})$,
where $e,f,g$ are integral solutions of the Markoff equation:

$e^{2}+f^{2}+g^{2}=3efg.$

It has been observed in [HP10] that these singularities are T-singularities in
the sense of Wahl [Wah81]. Such singularities have been connected to
exceptional vector bundles on rational surfaces by results of Hacking-
Prokhorov [HP10] and Hacking [Hac13]. This aspect has also been investigated
in our earlier work [HP08], [Per18]. We think that the complete understanding
of these resolutions and the fractal structure emerging via the associated
Markoff tree are of independent interest, but our hope is that this also may
lead to further insights into the global structure of exceptional collections.

Consider a T-singularity of type $\frac{1}{n^{2}}(1,nk-1)$, i.e. a cyclic
quotient singularity of order $n^{2}$ with weights $1,nk-1$. The minimal
resolution of such a singularity can be desribed via Hirzebruch-Jung continued
fractions $\llbracket
a_{1},\dots,a_{t}\rrbracket=a_{1}-1/(a_{2}-1/(a_{3}-\cdots(a_{t-1}-1/a_{t})\cdots))$.
These continued fractions have been classified for T-singularities in [KS88]
(see also Section 6). Our first basic tool will be a simplified representation
of these Hirzebruch-Jung continued fractions in terms of integer sequences
$c_{1},\dots,c_{s}$ (see Section 6), which is based on the Kollár–Shepherd-
Barron classification. It turns out that the $c_{i}$ can be recovered from the
regular continued fraction expansion
$n^{2}/(nk-1)=[c_{s}+1,c_{s-1},\dots,c_{2},c_{1}+x,c_{1}+2-x,c_{2},\dots,c_{s-1},c_{s}+1]$,
where $x=2$ if $s$ is even and $x=0$ if $s$ is odd (Theorem 7.8).

Now, for any Markoff triple $(e,f,g)$ with $1<e,f<g$,
$\mathbb{P}(e^{2},f^{2},g^{2})$ has the three T-singularities
$\frac{1}{e^{2}}(1,ew_{e}-1)$, $\frac{1}{f^{2}}(1,fw_{f}-1)$ and
$\frac{1}{g^{2}}(1,gw_{g}-1)$, respectively, where we call $w_{e},w_{f},w_{g}$
the T-weights of this triple. When passing from one Markoff triple to the
next, say $(e,g,E)$ with $E=3fg-e$, we will show in Section 7 how to construct
the continued fraction of $E^{2}/(Ew_{E}-1)$ from those of
$\frac{1}{e^{2}}(1,ew_{e}-1)$, $\frac{1}{f^{2}}(1,fw_{f}-1)$ and
$\frac{1}{g^{2}}(1,gw_{g}-1)$, respectively, and thereby obtain a complete
classification of the minimal resolutions of T-singularities associated to
Markoff triples in terms of their mutations along the Markoff tree.

A strongly related question is the distribution of quotients $g/w_{g}\approx
g^{2}/(gw_{g}-1)$ and whether limits of such quotients exists when mutating
Markoff triples downward paths in the Markoff tree. This is very closely
related to classical work by Frobenius [Fro68]. In loc. cit., Frobenius
derives a complete description of continued fraction expansions of quotients
$g/r_{g}$, where $r_{g}$ is associated to a triple that contains $g$ such that
$r_{g}^{2}\equiv-1\mod g$. Using Frobenius’ description, it is straightforward
to see that limits $\lim_{i\rightarrow\infty}r_{g_{i}}/g_{i}$ exist, for any
sequence $g_{i}$ that follows a path downward the Markoff tree (Proposition
4.3). It turns out that the set of these limit points forms a Cantor set. We
will show that this set has measure and Hausdorff dimension zero (Theorems 4.9
and 4.12). Moreover, we will characterize this Cantor set by its natural
intervals and we will show that the boundary points of these intervals are
quadratic numbers, which are explicitly described in Proposition 4.5.

The quotients $w_{g}/g$ are related to the $r_{g}/g$ simply by the affine
transform $w_{g}/g=3r_{g}/g-1$ for every $g$ (Corollary 5.4). Therefore, their
limit points via the Markoff tree exist as well and the resulting Cantor set
is an affine transform of the one associated to the $r_{g}/g$. Their dimension
and measure therefore coincide (Theorem 7.11). The boundary points are
quadratic as well and their explicit formulas and periodic continued fraction
expansions are computed in Secton 7.

We point out that the quotients $r_{g}/g$ have shown up in earlier work as the
slopes of exceptional vector bundles on $\mathbb{P}^{2}$, see [Rud89], [DL85],
and their fractal nature has already been observed in [DL85].

Overview. In Section 2 we fix some general conventions regarding the Markoff
tree that we will follow for the rest of these notes. In Section 3, we collect
all fundamental facts and computations related to Markoff triples that will be
needed in the subsequent sections. This includes some well-known formulas, but
also some new formulas for which we could not find any reference in the
literature. Also, we review Frobenius’ results on continued fractions related
to Markoff triples. In Section 4 we use the results of Section 3 to analyze
the Cantor set that emerges from the quotients $r_{g}/g$. In Section 5, we
introduce T-weights and perform some calculations similar as in Section 3. In
Section 6, we analyze the continued fraction expansions of T-singularities. We
then apply these results in the final Section 7 in order to classify the
minimal resolutions for T-singularities associated to the Markoff equation and
to describe the Cantor set that emerges from the quotients $w_{g}/g$.

In Appendix A we collect some basic facts about continued fractions and
continuants that we will need throughout the text. In Appendix B we introduce
a new class of polynomials related to T-singularities. These are not needed in
the main body of these notes, but they may be of independent interest. In
Appendix C, we collect some material that we need in order to deal with the
Hausdorff dimension. In Appendix D, we display the first few levels of the
Markoff tree, both only the maximal elements and the triples. We found it
quite useful to be able to occasionally pinpoint a specific example and its
location in the tree in one peek. Appendix E has a computer generated list of
the first 300 Markoff numbers together with their weights and T-weights. If
needed, the table can be extracted from this documents LaTeX source. In
Appendix F we present some simplistic numerical experiments regarding the
growth of Markoff numbers and the uniqueness conjecture.

##  2\. The Markoff Tree

In this section we will state some general facts and fix our general
conventions regarding the Markoff tree (we refer to [Aig13] for an overview).
We are exclusively interested in positive integral solutions $(e,f,g)$ of the
Markoff equation $e^{2}+f^{2}+g^{2}=3efg$, which we will call Markoff triples.
If for any given Markoff triple $(e,f,g)$ we denote $E=3fg-e$, $F=3eg-f$, and
$G=3ef-g$, then we obtain three more Markoff triples $(E,f,g)$, $(e,F,g)$, and
$(e,f,G)$. We call these substitutions mutations of the triple $(e,f,g)$. By
rearranging the formulas $3eg=f+F$, $3fg=e+E$, and $3ef=g+G$, it becomes
obvious that mutations are reversible. All Markoff triples can be obtained by
mutation, starting from the fundamental solution $(1,1,1)$. So we can consider
the set of all Markoff triples as vertices of a trivalent acyclic graph whose
edges are labeled by the mutations, where the fundamental solution is the
natural root. By the symmetry of the Markoff equation, we have a natural
action of the symmetric group $S_{3}$ on the set of Markoff triples. This
action has an almost everywhere trivial stabilizer with exceptions $(1,1,1)$
and the permutations of $(1,2,1)$. These triples, which are the only ones
where $e,f,g$ are not pairwise distinct, are called singular solutions of the
Markoff equation. All other Markoff triples are called regular. The set of
regular Markoff triples up to permutation forms a binary tree, whose root is
represented by the triple $(1,5,2)$. Figure 1 shows the first few levels of
this tree.

$(1,5,2)$$(1,13,5)$$(1,34,13)$$(1,89,34)$$(34,1325,13)$$(13,194,5)$$(13,7561,194)$$(194,2897,5)$$(5,29,2)$$(5,433,29)$$(5,6466,433)$$(433,37666,29)$$(29,169,2)$$(29,14701,169)$$(169,985,2)$
Figure 1. A part of the Markoff tree.

As in Figure 1, we will use from now on the convention that every regular
Markoff triple will be represented by $(e,g,f)$, such that $0<e,f<g$. Because
$g<E,F$, mutation of $e$ or $f$ does mean that we descend downwards the
Markoff tree, where we fix the direction as indicated in Figure 2.

$(e,g,f)$$(e,F,g)$$(g,E,f)$ Figure 2. Left and right mutation.

We will in call the triple $(e,F,g)$ the left mutation and $(g,E,f)$ the right
mutation of $(e,f,g)$, respectively. With this convention, if we go upwards by
mutating $g$, then we will either end up with the triple $(e,f,G)$ (if $e<f$)
or $(G,e,f)$ (if $e>f$). By requiring that the root is represented by
$(1,5,2)$, the ordering of every representative is thus uniquely fixed. In
particular, if we traverse any level below the root from left to right then we
find that alternatingly $e<f$ or $e>f$, with $1=e<f$ for the leftmost vertex
and $e>f=2$ for the rightmost vertex.

It follows that, given a triple $(e,g,f)$, then starting from this triple,
there are precisely four branches where at least one of $e,f,g$ is preserved,
as indicated by Figure 3.

$(e,g,f)$$(e,\cdot,g)$$(e,\cdot,\cdot)$$(e,\cdot,\cdot)$$(\cdot,\cdot,g)$$(\cdot,\cdot,g)$$(g,\cdot,f)$$(g,\cdot,\cdot)$$(g,\cdot,\cdot)$$(\cdot,\cdot,f)$$(\cdot,\cdot,f)$
Figure 3. How the elements in a Markoff triple propagate downwards.

The two outmost branches of the Markoff tree are of the form

$(1,F_{2n+1},F_{2n-1})\text{ for }n\geq
2\quad\text{}\quad(P_{2n-1},P_{2n+1},2)\text{ for }n\geq 1,$

where $F_{i}$ and $P_{i}$ denote the Fibonacci and Pell numbers. Note that
$5=F_{5}=P_{3}$. We call these the Fibonacci and Pell branches, respectively.

##  3\. The Markoff tree and finite continued fractions à la Frobenius

In this section we give an expository overview around some aspects of
Frobenius’ classical account [Fro68] on the Markoff equation which are related
to finite continued fractions. We also refer to [Cas57, Chapter II], [CF89],
[Aig13], [Reu19]. We will paraphrase some of the material to suit our needs
and complement it by a few observations of our own which, though probably
well-known to specialists, we couldn’t locate in the literature.

###  3.1. Weights and coweights

Denote $(e,g,f)$ a regular Markoff triple. Then we set:

$\displaystyle r_{e}$ $\displaystyle:=\begin{cases}0&\text{ if }e=1,\\\
f^{-1}g\bmod e&\text{ else},\end{cases}$ $\displaystyle r_{f}$
$\displaystyle:=g^{-1}e\bmod f,$ $\displaystyle r_{g}$
$\displaystyle:=e^{-1}f\bmod g.$

Note that the choice $r_{e}=0$ for $e=1$ is made for consistency, as we will
see below. Then the Markoff equation implies that

$r_{e}^{2}\equiv-1\bmod e\,\,(\text{for }e>1),\quad r_{f}^{2}\equiv-1\bmod
f,\quad\text{ and }\quad r_{g}^{2}\equiv-1\bmod g.$

We choose to call the triple $(r_{e},r_{g},r_{f})$ the weights associated to
the triple $(e,g,f)$. Moreover, we define the coweights $(s_{e},s_{g},s_{f})$
via the following equations:

$r_{e}^{2}=-1+s_{e}e,\quad r_{f}^{2}=-1+s_{f}f,\quad r_{g}^{2}=-1+s_{g}g.$

Note that in particular $s_{e}=1$ for $e=1$. The weights and coweights satisfy
a number of nice equations, in particular

$\displaystyle gr_{f}-fr_{g}$ $\displaystyle=e,$ (1) $\displaystyle
er_{g}-gr_{e}$ $\displaystyle=f,$ $\displaystyle er_{f}-fr_{e}$
$\displaystyle=G$

(see [Cas57, §II.3] or [Fro68, p. 602]).

######  Remark:

In [Rud89] the weights have been identified with the first Chern classes of
the vector bundles in an exceptional collection on $\mathbb{P}^{2}$ and the
ratios $r_{e}/e,r_{g}/g,r_{f}/f$ with their slopes.

######  Lemma 3.1:

Let $(e,g,f)$ be a regular Markoff triple.

  1. (i)

The weights and coweights transform under mutation as follows:

( |  $e$, |  $g$, | $f$ | )
---|---|---|---|---
( |  $r_{e}$, |  $r_{g}$, | $r_{f}$ | )
( |  $s_{e}$, |  $s_{g}$, | $s_{f}$ | )
( |  $e$, |  $F$, | $g$ | )
---|---|---|---|---
( |  $r_{e}$, |  $r_{F}$, | $r_{g}$ | )
( |  $s_{e}$, |  $s_{F}$, | $s_{g}$ | )
( |  $g$, |  $E$, | $f$ | )
---|---|---|---|---
( |  $r_{g}$, |  $r_{E}$, | $r_{f}$ | )
( |  $s_{g}$, |  $s_{E}$, | $s_{f}$ | )

where

$r_{E}=3fr_{g}-r_{e},\quad s_{E}=3fs_{g}-s_{e},\quad r_{F}=3er_{g}-r_{f},\ \
\text{ and }\ \ s_{F}=3es_{g}-s_{f}.$

  2. (ii)

The slopes of a regular Markoff triple are strictly increasing:

$\frac{r_{e}}{e}<\frac{r_{g}}{g}<\frac{r_{f}}{f}$

(see also [Rud89, Prop. 3.1]).

  3. (iii)

We have $r_{e}<e-r_{e}$, $r_{f}<f-r_{f}$, and $r_{g}<g-r_{g}$ for all $e,g$
and all $f>2$.

###### Proof.

(i) We only consider the left mutation; the right mutation follows
analogously. With $F=3eg-f$ and using the Markoff equation we get
$g^{-1}F\bmod e=-f^{-1}g\bmod e=g^{-1}f\bmod e=r_{e}$. Similarly, we have
$F^{-1}e\bmod g=-f^{-1}e\bmod g=e^{-1}f\bmod g=r_{g}$. Hence, $r_{e}$ and
$r_{g}$ (and thus $s_{e}$ and $s_{g}$) don’t change under left mutation. To
determine $r_{F}$, we write first

$g=3ef-G=3ef+fr_{e}-er_{f}=3e(er_{g}-gr_{e})+fr_{e}-er_{f}=e(3er_{g}-r_{f})-Fr_{e},$

where we use (1) twice. On the other hand, using (1) for the triple $(e,F,g)$,
we have

$g=er_{F}-Fr_{e}.$

Comparing terms then yields $r_{F}=3er_{g}-r_{f}$. Similarly, the equality
$s_{F}=3es_{g}-s_{f}$ follows via evaluation of $r_{F}^{2}=-1+Fs_{F}$ by
direct computation, using (1) and the Markoff equation.

(ii) The statement is trivially true for the triple $(1,5,2)$. Then we use (i)
and induction over the Markoff tree.

(iii) The statement is obviously true for $g=5$. By (ii) we get for left- and
right mutation:

$\frac{r_{e}}{e}<\frac{r_{F}}{F}<\frac{r_{g}}{g}<\frac{1}{2}\quad\text{ and
}\quad\frac{r_{g}}{g}<\frac{r_{E}}{E}<\frac{r_{f}}{f}<\frac{1}{2}$

and by induction over the Markoff tree, the assertion follows for all $g$ and
therefore for all $e>1$ and $f>2$. For $e=1$ the assertion is trivially true.
∎

######  Remark:

Lemma 3.1 (i) can also be inferred from Cohn’s matrices [Coh55] (see [Aig13,
§4.3]).

The slopes indeed contain some partial information about a triple’s position
in the Markoff tree.

######  Corollary 3.2:

  1. (i)

Consider the $n$-th level of the Markoff tree (where the $0$-th level given by
the triple $(1,5,2)$) and denote $m_{1},\dots,m_{2^{n}}$ the maximal elements
of the Markoff triples on this level, enumerated from left to right, with
weights $r_{m_{1}},\dots,r_{m_{2^{n}}}$. Then the slopes $r_{m_{i}}/m_{i}$ are
strictly increasing from left to right:

$0<\frac{r_{m_{1}}}{m_{1}}<\cdots<\frac{r_{m_{2^{n}}}}{m_{2^{n}}}<\frac{1}{2}.$

  2. (ii)

Let $(m_{1},m_{2},m_{3})$ be a regular solution of the Markoff equation with
$2<m_{1}<m_{2}<m_{3}$ and set $r^{\prime}_{i}=m_{i+1}^{-1}m_{i+2}\bmod m_{i}$
for $i=1,2,3$ (where we read the indices modulo $3$). Then this solution shows
up as a left mutation in the Markoff tree (i.e. as regular Markoff triple
$(m_{1},m_{3},m_{2})$) iff one (and therefore all) of the inequalities
$r^{\prime}_{i}<m_{i}-r^{\prime}_{i}$ holds.

Figure 4 shows the Markoff triples, their weights and their coweights of the
first four levels of the Markoff tree.

( | 1, | 5, | 2 | )
---|---|---|---|---
( | 0, | 2, | 1 | )
( | 1, | 1, | 1 | )
( | 1, | 13, | 5 | )
---|---|---|---|---
( | 0, | 5, | 2 | )
( | 1, | 2, | 1 | )
( | 1, | 34, | 13 | )
---|---|---|---|---
( | 0, | 13, | 5 | )
( | 1, | 5, | 2 | )
( | 1, | 89, | 34 | )
---|---|---|---|---
( | 0, | 34, | 13 | )
( | 1, | 13, | 5 | )
( | 34, | 1325, | 13 | )
---|---|---|---|---
( | 13, | 507, | 5 | )
( | 5, | 194, | 2 | )
( | 13, | 194, | 5 | )
---|---|---|---|---
( | 5, | 75, | 2 | )
( | 2, | 29, | 1 | )
( | 13, | 7561, | 194 | )
---|---|---|---|---
( | 5, | 2923, | 75 | )
( | 2, | 1130, | 29 | )
( | 194, | 2897, | 5 | )
---|---|---|---|---
( | 75, | 1120, | 2 | )
( | 29, | 433, | 1 | )
( | 5, | 29, | 2 | )
---|---|---|---|---
( | 2, | 12, | 1 | )
( | 1, | 5, | 1 | )
( | 5, | 433, | 29 | )
---|---|---|---|---
( | 2, | 179, | 12 | )
( | 1, | 74, | 5 | )
( | 5, | 6466, | 433 | )
---|---|---|---|---
( | 2, | 2673, | 179 | )
( | 1, | 1105, | 74 | )
( | 433, | 37666, | 29 | )
---|---|---|---|---
( | 179, | 15571, | 12 | )
( | 74, | 6437, | 5 | )
( | 29, | 169, | 2 | )
---|---|---|---|---
( | 12, | 70, | 1 | )
( | 5, | 29, | 1 | )
( | 29, | 14701, | 169 | )
---|---|---|---|---
( | 12, | 6089, | 70 | )
( | 5, | 2522, | 29 | )
( | 169, | 985, | 2 | )
---|---|---|---|---
( | 70, | 408, | 1 | )
( | 29, | 169, | 1 | )
Figure 4. The first four levels of the Markoff tree with triples $(e,g,f)$,
$(r_{e},r_{g},r_{f})$, $(s_{e},s_{g},s_{f})$.

In the Fibonacci branch, any triple is of the form
$(e,g,f)=(1,F_{2n+1},F_{2n-1})$ for $n\geq 2$ and correspondingly, we have

$(r_{e},r_{g},r_{f})=(0,F_{2n-1},F_{2n-3})\quad\text{ and
}\quad(s_{e},s_{g},s_{f})=(1,F_{2n-3},F_{2n-5}).$

Similarly, in the Pell branch, any triple is of the form
$(e,g,f)=(P_{2n-1},P_{2n+1},2)$ for $n\geq 1$ with

$(r_{e},r_{g},r_{f})=(P_{2n-2},P_{2n},1)\quad\text{ and
}\quad(s_{e},s_{g},s_{f})=(P_{2n-3},P_{2n-1},1).$

######  Definition 3.3:

For any real numbers $x,y\geq 2/3$ we set

  1. (i)

$\Delta_{x}:=9x^{2}-4$

  2. (ii)

$\Delta_{x,y}:=\frac{1}{2}(x\sqrt{\Delta_{y}}+y\sqrt{\Delta_{x}})$.

For a regular Markoff triple $(e,g,f)$, the inequalities $2ef<g<3ef$ hold
trivially. The following lemma sharpens these inequalities.

######  Lemma 3.4:

Let $(e,g,f)$ be a regular Markoff triple.

  1. (i)

The following inequalities hold:

$\Delta_{e,f}-\frac{ef}{g}<g<\Delta_{e,f}.$

In particular,

$g=\lfloor\Delta_{e,f}\rfloor.$

  2. (ii)

The first inequality can be sharpened to:

$\Delta_{e,f}-\frac{2}{3g}<\Delta_{e,f}-\frac{2}{9ef}<g.$

  3. (iii)

The second to:

$g+\frac{2ef}{g\sqrt{\Delta_{e}}\sqrt{\Delta_{f}}}<\Delta_{e,f}.$

###### Proof.

(i) The last assertion follows from the inequalities because $ef/g<1$.

Squaring both sides and rearranging terms, the second inequality is equivalent
to

$2e^{2}+2f^{2}+2g^{2}-9e^{2}f^{2}<ef\sqrt{9e^{2}-4}\sqrt{9f^{2}-4}.$

Using the Markoff equation and dividing by $ef$ this is equivalent to:

$3(2g-3ef)<\sqrt{9e^{2}-4}\sqrt{9f^{2}-4}.$

Again squaring both sides, we obtain equivalently:

$9(2g-3ef)^{2}<(9e^{2}-4)(9f^{2}-4),$

which, using again the Markoff equation, is trivially verified.

For the first inequality, squaring both sides and using the Markoff equation
leads to the expression

$9(2g-3ef)^{2}+12(2g-3ef)(2+\frac{ef}{g})+4(2+\frac{ef}{g})^{2}<(9e^{2}-4)(9f^{2}-4),$

which, by using the Markoff equation, reduces to:

$4<3(2g-3ef)(2+\frac{ef}{g})+(2+\frac{ef}{g})^{2}.$

This inequality follows from $\frac{ef}{g}>0$ and $2g-3ef=g-G>1$.

(ii) We need only to show the second inequality. For this, it suffices to show
the squared inequality $\Delta_{e,f}^{2}<(g+2/9ef)^{2}$, for which we get:

$ef\sqrt{\Delta_{e}}\sqrt{\Delta_{f}}<\left(g+\frac{2}{9ef}\right)^{2}-\frac{1}{2}(e^{2}\Delta_{f}+f^{2}\Delta_{e}).$

Squaring again and plugging in the following formula which can easily be
derived from the Markoff equation:

$9e^{2}f^{2}-4(e^{2}+f^{2})=(g-G)^{2},$

we get:

$e^{2}f^{2}(9(g-G)^{2}+16)<\left(2(g+\frac{2}{9ef})^{2}-(g-G)^{2}-2(e^{2}+f^{2})\right)^{2}.$

Now we divide both sides by $e^{2}f^{2}(g-G)$ and first observe for the
l.h.s.:

$9+\frac{16}{e^{2}f^{2}(g-G)}\leq 10$

for every Markoff number $\geq 5$. For the r.h.s., we get by using the Markoff
equation and rearranging terms:

$\displaystyle\frac{1}{e^{2}f^{2}(g-G)}\left(2(g+\frac{2}{9ef})^{2}-(g-G)^{2}-2(e^{2}+f^{2})\right)^{2}$
$\displaystyle=\frac{1}{e^{2}f^{2}(g-G)}\left(3ef(g-G)+R\right)^{2}$
$\displaystyle=9ef(g-G)+\frac{6R}{ef}+\frac{R^{2}}{e^{2}f^{2}(g-G)}$
$\displaystyle>10$

for every regular Markoff triple, where we denote
$R:=8g/9ef+8/81e^{2}f^{2}>0$. So, the second inequality follows.

(iii) We show that the inequality holds for all positive real solutions
$(e,g,f)\in\mathbb{R}^{3}$ of the Markoff equation. By considering the Markoff
equation a quadratic equation in $g$, we get the following expression:

$g=\frac{ef}{2}\left(3\pm\sqrt{9-\frac{4}{e^{2}}-\frac{4}{f^{2}}}\right).$

The boundary of the projection of the positive solutions of the Markoff
equation to the $e$-$f$-plane coincides with the branching points for this
equation given by the curve $9e^{2}f^{2}=4(e^{2}+f^{2})$. With

$\inf_{e\to\infty}\sqrt{\frac{4e^{2}}{9e^{2}-4}}=\inf_{f\to\infty}\sqrt{\frac{4f^{2}}{9f^{2}-4}}=\frac{2}{3}$

we see that both functions

$g+\frac{2ef}{g\sqrt{\Delta_{e}}\sqrt{\Delta_{f}}}=\frac{ef}{2}\left(3+\sqrt{9-\frac{4}{e^{2}}-\frac{4}{f^{2}}}\right)+4/\left((3+\sqrt{9-\frac{4}{e^{2}}-\frac{4}{f^{2}}})\sqrt{\Delta_{e}}\sqrt{\Delta_{f}}\right)$

and $\Delta_{e,f}$ are well-defined functions over the projection of the
positive component of the Markoff variety to the $e$-$f$-plane (note that
without loss of generality we choose the bigger solutions for $g$).

We first consider the case $e=f$. Then the formulas simplify to:

$g+\frac{2e^{2}}{g\Delta_{e}}<e\sqrt{\Delta_{e}}\quad\text{ and }\quad
g=\frac{e}{2}\left(3e+\sqrt{9e^{2}-8}\right),$

where we claim that the inequality holds for $e\geq\frac{2\sqrt{2}}{3}$.
Eliminating $g$, we get:

$\frac{\left(3e+\sqrt{9e^{2}-8}\right)}{2
\sqrt{\Delta_{e}}}+\frac{4}{\Delta_{e}^{3/2}(3e+\sqrt{9e^{2}-8})}<1.$

As $\underset{e\rightarrow\infty}{\lim}(\text{l.h.s.})=1$, it suffices to show
that the l.h.s is strictly monotonously increasing for
$e\geq\frac{2\sqrt{2}}{3}$. This means that for the first derivative of the
l.h.s., after rearranging terms and clearing some denominators, it suffices to
show:

$2(3e-\sqrt{9e^{2}-8})>\frac{1}{3e+\sqrt{9e^{2}-8}}\left(\frac{9e\sqrt{9e^{2}-8}}{\Delta_{e}}+1\right)$

for $e\geq 2\sqrt{2}/3$, which can straightforwardly be verified.

For the general case, we fix $C:=9-\frac{4}{e^{2}}-\frac{4}{f^{2}}$, where we
note that $0\leq C<9$. Then, after eliminating $g$ and
$f^{2}=4(9-C-\frac{4}{e^{2}})$ and some rearrangements, the inequality takes
the following form:

$\sqrt{9-\frac{4}{e^{2}}}\sqrt{C+\frac{4}{e^{2}}}(3+\sqrt{C})\left(\sqrt{9-\frac{4}{e^{2}}}+\sqrt{C+\frac{4}{e^{2}}}-\sqrt{C}-3\right)-\frac{2}{e^{2}}(9-C-\frac{4}{e^{2}})>0$

It is straightforward to verify that the l.h.s. of this inequality has four
local extrema at $e^{2}=\frac{4}{9-C}$ and $e^{2}=\frac{8}{9-C}$ within its
domain of definition, and moreover
$\underset{e\rightarrow\infty}{\lim}(\text{l.h.s.})=0$. As
$e^{2}=\frac{8}{9-C}$ iff $e^{2}=f^{2}$, we have seen before that the
inequality holds in that case and therefore we have maxima at
$e^{2}=\frac{8}{9-C}$. For degree reasons it follows that there are no further
local extrema and hence the inequality holds for all $e^{2}\geq\frac{8}{9-C}$
and we are done. ∎

Consider a subtree of the Markoff tree whose root is a regular Markoff triple
$(e,g,f)$. Then the two outer branches of this tree consist of triples of the
form $(e,F_{i},F_{i-1})$ and $(E_{i},E_{i-1},f)$ with
$E_{i+1}=3fE_{i}-E_{i-1}$ and $F_{i+1}=3eF_{i}-F_{i-1}$ for $i\geq 0$, where
we set $E_{-1}=e$, $F_{-1}=f$, and $F_{0}=E_{0}=g$. Standard methods for
solving recurrences of the form $x_{i+1}=3nx_{i}-x_{i-1}$ for $i\geq 0$ yield:

$x_{i}=\lambda_{+}\left(\frac{3n+\sqrt{\Delta_{n}}}{2}\right)^{i}+\lambda_{-}\left(\frac{3n-\sqrt{\Delta_{n}}}{2}\right)^{i}$

with

$\lambda_{\pm}=\frac{1}{2}\left(x_{0}\pm\frac{x_{1}-x_{-1}}{\sqrt{\Delta_{n}}}\right).$

Using Lemma 3.1 (i), we get the following formulas (compare also [Aig13,
§10.2]) from which we can see that the corresponding weights and coweights
follow the same growth as the Markoff numbers.

######  Lemma 3.5:

Denote $P_{\pm}:=(3f\pm\sqrt{\Delta_{f}})/2$ and
$Q_{\pm}:=(3e\pm\sqrt{\Delta_{e}})/2$. Then for $i\geq 0$ we have:

  1. (i)

$\displaystyle E_{i}$
$\displaystyle=\lambda_{+}P_{+}^{i}+\lambda_{-}P_{-}^{i},\quad\text{where}\quad\lambda_{\pm}=\frac{1}{2}\left(g\pm\frac{E-e}{\sqrt{\Delta_{f}}}\right),$
$\displaystyle F_{i}$
$\displaystyle=\kappa_{+}Q_{+}^{i}+\kappa_{-}Q_{-}^{i},\quad\text{where}\quad\kappa_{\pm}=\frac{1}{2}\left(g\pm\frac{F-f}{\sqrt{\Delta_{e}}}\right).$

  2. (ii)

$\displaystyle r_{E_{i}}$
$\displaystyle=\lambda_{+}P_{+}^{i}+\lambda_{-}P_{-}^{i},\quad\text{where}\quad\lambda_{\pm}=\frac{1}{2}\left(r_{g}\pm\frac{r_{E}-r_{e}}{\sqrt{\Delta_{f}}}\right),$
$\displaystyle r_{F_{i}}$
$\displaystyle=\kappa_{+}Q_{+}^{i}+\kappa_{-}Q_{-}^{i},\quad\text{where}\quad\kappa_{\pm}=\frac{1}{2}\left(r_{g}\pm\frac{r_{F}-r_{f}}{\sqrt{\Delta_{e}}}\right).$

  3. (iii)

$\displaystyle s_{E_{i}}$
$\displaystyle=\lambda_{+}P_{+}^{i}+\lambda_{-}P_{-}^{i},\quad\text{where}\quad\lambda_{\pm}=\frac{1}{2}\left(s_{g}\pm\frac{s_{E}-s_{e}}{\sqrt{\Delta_{f}}}\right),$
$\displaystyle s_{F_{i}}$
$\displaystyle=\kappa_{+}Q_{+}^{i}+\kappa_{-}Q_{-}^{i},\quad\text{where}\quad\kappa_{\pm}=\frac{1}{2}\left(s_{g}\pm\frac{s_{F}-s_{f}}{\sqrt{\Delta_{e}}}\right).$

###  3.2. Continued fractions

Recall the following construction of the Stern-Brocot tree. We start with a
binary tree that contains triples of the form
$(\frac{\lambda}{\lambda^{\prime}},\frac{\nu}{\nu^{\prime}},\frac{\mu}{\mu^{\prime}})$
with $\lambda,\lambda^{\prime},\mu,$
$\mu^{\prime},\nu,\nu^{\prime}\in\mathbb{Z}$, where for the moment we formally
consider the fractions as pairs of integers rather than rational numbers. Its
root is $(\frac{1}{0},\frac{1}{1},\frac{0}{1})$ and descendants are formed by
the rule

$(\frac{\lambda}{\lambda^{\prime}},\frac{\nu}{\nu^{\prime}},\frac{\mu}{\mu^{\prime}})$$(\frac{\lambda}{\lambda^{\prime}},\frac{\lambda+\nu}{\lambda^{\prime}+\nu^{\prime}},\frac{\nu}{\nu^{\prime}})$$(\frac{\nu}{\nu^{\prime}},\frac{\mu+\nu}{\mu^{\prime}+\nu^{\prime}},\frac{\mu}{\mu^{\prime}})$

The middle entry of each triple is a pair of coprime positive integers and in
particular a well-defined rational number. The Stern-Brocot tree now is the
binary tree consisting of these middle entries, containing every positive
rational number precisely once. Note that by abuse of notion, we refer to the
tree formed by above triples as Stern-Brocot tree as well. Now, starting with

$(m_{\frac{1}{0}},m_{\frac{1}{1}},m_{\frac{0}{1}}):=(1,5,2)$

and extending to a bijection of trees, we define a bijection between triples
$(\frac{\lambda}{\lambda^{\prime}},\frac{\nu}{\nu^{\prime}},\frac{\mu}{\mu^{\prime}})$
and regular Markoff triples
$(m_{\frac{\lambda}{\lambda^{\prime}}},m_{\frac{\nu}{\nu^{\prime}}},m_{\frac{\mu}{\mu^{\prime}}})$.
In particular, we get a surjective map $\frac{\mu}{\nu}\mapsto
m_{\frac{\mu}{\nu}}$ from $\mathbb{Q}_{\geq 0}\cup\\{\frac{1}{0}\\}$ to the
set of Markoff numbers (one version of the uniqueness conjecture states that
this map is a bijection). This indexing of Markoff numbers has been introduced
by Frobenius [Fro68].

Now it is a natural question whether a positive rational number $\mu/\nu$
enough to construct the corresponding Markoff triple (and vice versa) without
explicitly referring to the binary trees? To see that this is indeed possible,
we start with the following observation. If we consider any triple of the form
$(\frac{\lambda}{\lambda^{\prime}},\frac{\nu}{\nu^{\prime}},\frac{\mu}{\mu^{\prime}})$
and write the fractions as lattice vectors
$(\lambda,\lambda^{\prime}),(\mu,\mu^{\prime}),(\nu,\nu^{\prime})\in\mathbb{Z}^{2}$,
then by construction, every two of these lattice vectors form a basis of
$\mathbb{Z}^{2}$. Conversely, for any primitive lattice vector
$(\mu,\nu)\in\mathbb{Z}^{2}$ it is easy to see that there exist precisely two
lattice vectors
$(\mu_{1},\nu_{1}),(\mu_{2},\nu_{2})\in[0,\mu]\times[0,\nu]\subset\mathbb{Z}^{2}$
such that $(\mu,\nu),(\mu_{1},\nu_{1})$ and $(\mu,\nu),(\mu_{2},\nu_{2})$ form
bases of $\mathbb{Z}$, respectively. Also it is immediately clear that
$(\mu,\nu)=(\mu_{1},\nu_{1})+(\mu_{2},\nu_{2})$ and hence
$(\mu_{1},\nu_{1}),(\mu_{2},\nu_{2})$ form a basis as well. It follows that
either $(\mu_{1}/\nu_{1},\mu/\nu,\mu_{2}/\nu_{2})$ or
$(\mu_{2}/\nu_{2},\mu/\nu,\mu_{1}/\nu_{1})$ represents a vertex in the Stern-
Brocot tree.

Now, consider any fraction $\mu/\nu$ with corresponding Stern-Brocot triple
$(\mu_{1}/\nu_{1},\mu/\nu,\mu_{2}/\nu_{2})$. We denote
$(m_{\mu_{1}/\nu_{1}},m_{\mu/\nu},m_{\mu_{2}/\nu_{2}})$ the corresponding
Markoff triple and $(r_{\mu_{1}/\nu_{1}},r_{\mu/\nu},r_{\mu_{2}/\nu_{2}})$,
$(s_{\mu_{1}/\nu_{1}},s_{\mu/\nu},s_{\mu_{2}/\nu_{2}})$ the associated weights
and coweights. Frobenius has determined the continued fraction expansion of
$m_{\mu/\nu}/r_{\mu/\nu}$ as follows (regarding the definition of length of a
continued fraction, please check the discussion right after Formula A-3 in
Appendix A).

######  Theorem 3.6 ([Fro68], §’s 10, 11, [CF89], Ch. 2):

With above notation, assume that $m_{\mu/\nu}>2$. Then the continued fraction
expansion of $\frac{m_{\mu/\nu}}{r_{\mu/\nu}}$ has the following properties.

  1. (i)

If we write $\frac{m_{\mu/\nu}}{r_{\mu/\nu}}=[a_{1},\dots,a_{n}]$ with
$a_{1},a_{n}>1$ then $n=2s$ is even.

  2. (ii)

The continued fraction expansion
$\frac{m_{\mu/\nu}}{r_{\mu/\nu}}=[a_{1},\dots,a_{2s}]$ is symmetric, i.e
$a_{i}=a_{2s+1-i}$ for every $1\leq i<2s$.

  3. (iii)

The continued fraction expansion of $\frac{m_{\mu/\nu}}{r_{\mu/\nu}}$ begins
and ends in $2$, i.e.

$\frac{m_{\mu/\nu}}{r_{\mu/\nu}}=[2,S(\mu,\nu),2].$

  4. (iv)

For the case $\nu=1$ with
$\frac{m_{\mu/1}}{r_{\mu/1}}=\frac{F_{2\mu+3}}{F_{2\mu+1}}$ we have:

$S(\mu,1)=1_{2\mu-2},$

where $1_{2\mu-2}$ denotes the $2\mu-2$-fold repetition of $1$.

  5. (v)

For $\nu>1$ and $1\leq i<\nu$ we set

$\kappa(i):=\lfloor
i\frac{\mu}{\nu}\rfloor-\lfloor(i-1)\frac{\mu}{\nu}\rfloor.$

Then:

$S(\mu,\nu)=1_{2\kappa(1)},2,2,1_{2\kappa(2)},2,2,\dots,2,2,1_{2\kappa(\nu-1)},2,2,1_{2\kappa(1)}.$

  6. (vi)

Consider a Markoff triple
$(m_{\mu_{1}/\nu_{1}},m_{\mu/\nu},m_{\mu_{2}/\nu_{2}})$ with $\mu,\nu>1$. Then

$S(\mu,\nu)=S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu_{2},\nu_{2})=S(\mu_{2},\nu_{2}),1,1,2,2,S(\mu_{1},\nu_{1}).$

######  Example 3.7:

For $\mu/\nu=8/5$ we get $\kappa(1)=\kappa(3)=1$ and $\kappa(2)=\kappa(4)=2$,
so

$\frac{m_{8/5}}{r_{8/5}}=[2,1,1,2,2,1,1,1,1,2,2,1,1,2,2,1,1,1,1,2,2,1,1,2]=\frac{4400489}{1701181},$

and thus $s_{8/5}=657658$. The vector $(8,5)$ splits into $(3,2)+(5,3)$ such
that every two of $(8,5)$, $(3,2)$, $(5,3)$ form a basis of $\mathbb{Z}^{2}$.
For $(5,3)$ we get $\kappa(1)=1$, $\kappa(2)=2$, hence

$\frac{m_{5/3}}{r_{5/3}}=[2,1,1,2,2,1,1,1,1,2,2,1,1,2]=\frac{7561}{2923}$

with $s_{3/2}=1130$. For $(3,2)$, we get $\kappa(1)=1$, hence

$\frac{m_{3/2}}{r_{3/2}}=[2,1,1,2,2,1,1,2]=\frac{194}{75}$

with $s_{3/2}=29$. This data corresponds to the Markoff triple

$(7561,4400489,194).$

We observe that we can partition the continued fraction expansion of
$m_{\mu/\nu}/r_{\mu/\nu}$ into segments of the form $2,1_{2\kappa(i)},2$ and

$[2,1_{2\kappa(i)},2]=\frac{F_{2\kappa(i)+5}}{F_{2\kappa(i)+3}},$

so, in a sense, the continued fraction expansions yield a decomposition of a
Markoff number into Fibonacci numbers. For sake of exposition, we choose to
call a sequence of the form $2,1_{k},2$ for $k\geq 0$ a Fibonacci segment. The
following statement then follows by induction and
$\sum_{i=1}^{\nu-1}\kappa(i)=\mu-1$.

######  Corollary 3.8:

  1. (i)

The length of the continued fraction expansion of
$\frac{m_{\mu/\nu}}{r_{\mu/\nu}}$ is $2(\mu+\nu-1)$.

  2. (ii)

The continued fraction expansion of $\frac{m_{\mu/\nu}}{r_{\mu/\nu}}$ consists
of $\nu$ Fibonacci segments. In particular, it contains $2(\mu-1)$ ones and
$2\nu$ twos.

It follows that just from the pair $(m_{\mu/\nu},r_{\mu/\nu})$ (or even just
$\mu/\nu$) we can reconstruct the complete Markoff triple. It has already been
pointed out by Frobenius [Fro68, §11], the composition of the continued
fractions of $\frac{m_{\mu/\nu}}{r_{\mu/\nu}}$ can be interpreted in terms of
Christoffel words and their standard factorizations. We want to add the
observation that this also allows another very nice graphical interpretation.
For this, consider the set $[0,\mu+\nu]\times[0,\nu]\subset\mathbb{Z}^{2}$ as
the set of $(\mu+\nu+1)(\nu+1)$ corner points of a collection of $1\times
1$-boxes in the plane. Then there are precisely $(\mu-1)+2\nu$ boxes that
touch the diagonal from $(0,0)$ to $(\mu+\nu,\nu)$. Now we fill the first and
the last box, as well as every corner box with $2$’s, and the remaining boxes
with pairs $1,1$. Then, if we read off the numbers from left to right, we
obtain the continued fraction expansion of $\frac{m_{\mu/\nu}}{r_{\mu/\nu}}$.

21,1221,11,1221,12 Figure 5. The snake diagram and continued fraction
expansion of $\frac{m_{5/3}}{r_{5/3}}=\frac{7561}{2923}$.

Also, every row in this this diagram contains precisely one Fibonacci segment.
Figure 5 shows the corresponding “snake diagram” for $(\mu,\nu)=(5,3)$. For
more details on Christoffel words and snake diagrams we refer to [Aig13, §7].

######  Remark:

Note that in order to reconstruct a Markoff triple from a pair
$(m_{\mu/\nu},r_{\mu/\nu})$, we need the Euclidean algorithm to compute
$\mu_{1}/\nu_{1}$ and $\mu_{2}/\nu_{2}$. Therefore, while we do not have to
deal explictly with mutations, implicitly we still are required to backtrack
the full path to $(m_{\mu/\nu},r_{\mu/\nu})$ via the Stern-Brocot tree.

######  3.9.

A rational number $\mu/\nu>0$ and its inverse $\nu/\mu$ correspond to opposite
positions in the Stern-Brocot tree with respect to reflection along the middle
axis. So given a pair $(m,r)$ consisting of a Markoff number and a weight $r$,
we have seen above that we can determine $(\mu,\nu)$ such that
$(m,r)=(m_{\mu/\nu},r_{\mu/\nu})$ together with the Markoff triple which gives
rise to $(m,r)$. Moreover, once given $(\mu,\nu)$, we can also compute the
opposite Markoff triple corresponding to $(m_{\nu/\mu},r_{\nu/\mu})$. This is
nicely reflected in the corresponding continued fraction expansions as
follows. Both sequences $S(\mu,\nu)=x_{1},\dots,x_{2(\mu+\nu-2)}$ and
$S(\nu,\mu)=y_{1},\dots,y_{2(\mu+\nu-2)}$ have the same length by Corollary
3.8 and moreover, as has been observed by Frobenius [Fro68, p. 620],
$x_{i}+y_{i}=3$ for every $1\leq i\leq 2(\mu+\nu-2)$. In other words, we
obtain $S(\nu,\mu)$ from $S(\mu,\nu)$ by changing $2$’s to $1$’s and vice
versa.

2221,122221,1222 Figure 6. The snake diagram and continued fraction expansion
of $\frac{m_{3/5}}{r_{3/5}}=\frac{37666}{15571}$.

Figure 6 shows the case opposite of $(\mu,\nu)=(5,3)$ from Figure 5.

##  4\. Cantor set

Consider the $n$-th level of the Markoff tree, where $n>1$ and denote
$m_{1},\dots,m_{2^{n}}$ the maximal elements of the Markoff triples in this
level, enumerated from left to right. By Corollary 3.2 (i), we have

$\frac{F_{2n+3}}{F_{2n+1}}=\frac{m_{1}}{r_{m_{1}}}>\cdots>\frac{m_{2^{n}}}{r_{m_{2^{n}}}}=\frac{P_{2n+1}}{P_{2n-1}}.$

Moreover, both $F_{2n+3}/F_{2n+1}$ and $P_{2n+1}/P_{2n-1}$ are monotonous,
bounded sequences which converge from below and above, respectively, to their
respective limits. Using

$\displaystyle\lim_{i\to\infty}\frac{F_{2i+1}}{F_{2i-1}}$
$\displaystyle=\lim_{i\to\infty}[2,1_{2(i-1)},2]=\lim_{i\to\infty}[2,1_{2i}]=[2,\bar{1}]=\frac{1}{2}(3+\sqrt{5}),$
$\displaystyle\lim_{i\to\infty}\frac{P_{2i+1}}{P_{2i-1}}$
$\displaystyle=\lim_{i\to\infty}[2_{2i}]=[\bar{2}]=1+\sqrt{2},$

we see that the spectrum of fractions $m/r$ for Markoff numbers $m\geq 5$ is
bounded:

(2) $1+\sqrt{2}<\frac{m}{r}<\frac{1}{2}(3+\sqrt{5}).$

We can consider the Fibonacci- and Pell branch as particular infinite paths in
the Markoff tree and above limits as limits over these paths. More generally,
we can encode every vertex of a binary tree by its path starting from the
root. We may write every such path as a finite word in the alphabet
$\\{L,R\\}$, where the empty word stands for the root and $L$ and $R$
designate left- and right mutation, respectively. Similarly, an infinite word
in this alphabet denotes an infinite path which we can consider as the limit
over its finite starting segments in the obvious sense.

######  Definition 4.1:

For any $X\in\\{L,R\\}$ we denote by $X^{n}$ $n$-fold repetition for any $n>0$
and by $\bar{X}$ we denote infinite repetition.

In particular, words of the form $L^{n}$ and $R^{n}$ for $n\geq 0$ describe
the entries in the Fibonacci- and Pell branches, and
$\bar{L}=\lim_{n\to\infty}L^{n}$, $\bar{R}=\lim_{n\to\infty}R^{n}$,
respectively.

######  Definition 4.2:

We denote by $\mathfrak{P}_{f}$ the set of finite paths, by $\mathfrak{P}$ the
set of infinite paths, and by $\bar{\mathfrak{P}}_{f}\subset\mathfrak{P}$
paths which are of the form $p\bar{X}$ for $p\in\mathfrak{P}_{f}$ and
$X\in\\{L,R\\}$.

Clearly, both $\mathfrak{P}_{f}$, $\bar{\mathfrak{P}}_{f}$ are countable; in
particular, mapping a path $X_{1}X_{2}\cdots X_{n}$ with $n>0$ to
$X_{1}X_{2}\dots X_{n-1}\bar{X}_{n}$ is a surjection from
$\mathfrak{P}_{f}\setminus\\{\emptyset\\}$ to $\bar{\mathfrak{P}}_{f}$. The
set $\mathfrak{P}$ has the same cardinality as the real numbers. Now let
$p=\lim_{n\to\infty}p_{n}\in\mathfrak{P}$, where $p_{n}$ denotes the starting
segment of length $n$ of $p$ which also corresponds to a rational number
$\mu_{n}/\nu_{n}$ in the Stern-Brocot tree.

######  Proposition 4.3:

With above notation, let $p=\lim_{n\to\infty}p_{n}\in\mathfrak{P}$. Then the
limit $\lim_{n\to\infty}m_{\mu_{n}/\nu_{n}}/r_{\mu_{n}/\nu_{n}}$
$=\mathfrak{l}_{p}$ exists. In particular, every
$m_{\mu_{n}/\nu_{n}}/r_{\mu_{n}/\nu_{n}}$ is a rational convergent of the
continued fraction expansion of $\mathfrak{l}_{p}$.

###### Proof.

As we have considered the Fibonacci and Pell branches already above, it
suffices to consider any Stern-Brocot triple
$(\mu_{1}/\nu_{1},\mu/\nu,\mu_{2}/\nu_{2})$ with $\mu,\nu>1$. Then by Theorem
3.6 we have

$\displaystyle[2,S(\mu,\nu),2]=[2,S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu_{2},\nu_{2}),2]=[2,S(\mu_{2},\nu_{2}),1,1,2,2,S(\mu_{1},\nu_{1}),2],$

so we can view the continued fraction expansion of $m_{\mu/\nu}/r_{\mu/\nu}$
as continuation of its predecessors’ (note that
$[2,S(\mu_{2},\nu_{2}),2]=[2,S(\mu_{2},\nu_{2}),1,1]$). In particular, we can
consider mutation as a process of extending continued fractions to the right:

$[2,S(\mu_{1},\nu_{1}),2]$, $[2,S(\mu,\nu),2]$,
$[2,S(\mu_{2},\nu_{2}),2]$$[2,S(\mu_{1},\nu_{1}),2]$,
$[2,S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu,\nu),2]$,
$[2,S(\mu,\nu),2]$$[2,S(\mu,\nu),2]$,
$[2,S(\mu,\nu),2,2,1,1,S(\mu_{2},\nu_{2}),2]$, $[2,S(\mu_{2},\nu_{2}),2]$

where:

$\displaystyle[2,S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu,\nu),2]$
$\displaystyle=[2,S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu_{2},\nu_{2}),2]$
$\displaystyle=[2,S(\mu,\nu),1,1,2,2,S(\mu_{1},\nu_{1}),2].$

It follows that the limit infinite continued fraction exists for any path down
the Markoff tree, hence the assertions are shown. ∎

It is easy to see that the $\mathfrak{l}_{p}$ are contained in the closed
interval $[1+\sqrt{2},\frac{1}{2}(3+\sqrt{5})]$ and we are going to show that
they form a Cantor set. Before we can do this, we consider one more special
case.

######  Proposition 4.4:

  1. (i)

Let $p=L^{n}\bar{R}$, then

$\mathfrak{l}_{p}=[\overline{2,1_{2n},2}]=\frac{1}{2}\frac{3F_{2n+3}-2F_{2n+1}+\sqrt{9F_{2n+3}^{2}-4}}{3F_{2n+1}-F_{2n-1}}.$

  2. (ii)

Let $p=R^{n}\bar{L}$, then

$\mathfrak{l}_{p}=1+[\overline{1,2_{2n},1}]=\frac{1}{2}\frac{3P_{2n+1}+2P_{2n-1}+\sqrt{9P_{2n+1}^{2}-4}}{3P_{2n-1}+P_{2n-3}}.$

###### Proof.

(i) The Markoff triple in position $L^{n}$ is of the form
$(1,F_{2n+5},F_{2n+3})$ and the continued fraction expansions are
$(-,[2,1_{2n},2],[2,1_{2n-2},2])$. Using Theorem 3.6 (vi), a straightforward
induction yields that the continued fractions at position $L^{n}R^{m}$ for
$m>0$ are

$([\langle 2,1_{2n},2\rangle_{m}],[\langle
2,1_{2n},2\rangle_{m+1}],[2,1_{2n-2},2]),$

where $\langle\cdots\rangle_{k}$ denotes $k$-fold repetition. Hence

$\mathfrak{l}_{p}=\lim_{m\to\infty}[\langle
2,1_{2n},2\rangle_{m}]=[\overline{2,1_{2n},2}].$

(ii) The Markoff triple in position $R^{n}$ is of the form
$(P_{2n+1},P_{2n+3},2)$ with continued fraction expansions
$([2_{2n}],[2_{2n+2}],[2])$. Again, using Theorem 3.6 (vi) and induction, we
get for the continued fractions at position $R^{n}L^{m}$ for $m>0$:

$([2_{2n}],[2,\langle 2_{2n},1,1\rangle_{m},2_{2n+1}],[2,\langle
2_{2n},1,1\rangle_{m-1},2_{2n+1}]),$

hence

$\mathfrak{l}_{p}=\lim_{m\to\infty}[2,\langle
2_{2n},1,1\rangle_{m}]=[2,\overline{2_{2n},1,1}]=1+[\overline{1,2_{2n},1}].$

In both cases, using A-5, the translation of the periodic continued fractions
into quadratic irrational numbers follows by straightforward computation (see
also the proof of Proposition 4.5 below). ∎

Given a triple $(m_{\mu_{1}/\nu_{1}},m_{\mu/\nu},m_{\mu_{2}/\nu_{2}})$
corresponding to a finite path $p$, then, as indicated in Figure 3, the set of
triples below this triple wich contain $m_{\mu/\nu}$ are precisely those in
the paths $pL\bar{R}$ and $pR\bar{L}$.

######  Proposition 4.5:

Consider a Markoff triple $(e,g,f)$ with $e>1$, $f>2$ and denote
$(\mu_{1}/\nu_{1},\mu/\nu,\mu_{2}/\nu_{2})$ and $p$ the corresponding Stern-
Brocot triple and finite path, respectively. Moreover, denote
$p_{1}=pL\bar{R}$, $p_{2}=pR\bar{L}$. Then

$\displaystyle\mathfrak{l}_{p_{1}}=[\overline{2,S(\mu,\nu),1,1,2}]$
$\displaystyle=\frac{1}{2}\frac{3g-2r_{g}+\sqrt{9g^{2}-4}}{3r_{g}-s_{g}},$
$\displaystyle\mathfrak{l}_{p_{2}}=1+[\overline{1,S(\mu,\nu),2,2,1}]$
$\displaystyle=\frac{1}{2}\frac{3g+2r_{g}+\sqrt{9g^{2}-4}}{3r_{g}+s_{g}}.$

###### Proof.

We denote

$[2,S(\mu_{1},\nu_{1}),2],[2,S(\mu,\nu),2],[2,S(\mu_{2},\nu_{2}),2]$

the continued fraction expansions of $e/r_{e},g/r_{g},f/r_{f}$. Then it
follows from Theorem 3.6 (vi) that after one left- and one right mutation, we
obtain

$[2,S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu,\nu),2],[2,S(\mu_{1},\nu_{1}),2,2,1,1,S(\mu,\nu),2,2,1,1,S(\mu,\nu),2],[2,S(\mu,\nu),2]$

and by straightforward induction and Theorem 3.6 (vi), we get at the position
$pLR^{n}$ for $n>0$:

$[2,S(\mu_{1},\nu_{1}),2,\langle
2,1,1,S(\mu,\nu),2\rangle_{n}],[2,1,1,S(\mu_{1},\nu_{1}),2,\langle
2,1,1,S(\mu,\nu),2\rangle_{n+1}],[2,S(\mu,\nu),2].$

It follows that the limit of finite continued fractions over $pLR^{n}$ for
$n\to\infty$ exists:

$\lim_{n\to\infty}[2,S(\mu_{1},\nu_{1}),2,\langle
2,1,1,S(\mu,\nu),2\rangle_{n+1}]=[2,S(\mu_{1},\nu_{1}),2,\overline{2,1,1,S(\mu,\nu),2}].$

Using symmetry, we get moreover

$\displaystyle\mathfrak{l}_{p_{1}}=\lim_{n\to\infty}[2,S(\mu_{1},\nu_{1}),2,\langle
2,1,1,S(\mu,\nu),2\rangle_{n+1}]$ $\displaystyle=\lim_{n\to\infty}[\langle
2,S(\mu,\nu),1,1,2\rangle_{n+1},2,S(\mu_{1},\nu_{1}),2]$
$\displaystyle=[\overline{2,S(\mu,\nu),1,1,2}].$

In particular, $\mathfrak{l}_{p_{1}}$ is a quadratic number which we can
determine using (A-5). For this, we first observe

$[2,1,1,S(\mu,\nu),2]=2+\frac{1}{[1,1,S(\mu,\nu),2]}=\frac{K(2,S(\mu,\nu),1,1,2)}{K(2,S(\mu,\nu),1,1)}=\frac{3g-r_{g}}{g},$

where we have used Lemma A.1 and the symmetry of continuants. Moreover, with

$(3r_{g}-s_{g})g=3gr_{g}-r_{g}^{2}-1=(3g-r_{g})r_{g}-1$

and equation (A-4) we get

$K(S(\mu,\nu),1,1,2)=3r_{g}-s_{g}.$

Finally, with $[2,S(\mu,\nu),2]=g/r_{g}$ we get

$K(S(\mu,\nu),1,1)=K(S(\mu,\nu),2)=r_{g}.$

Plugging these four continuants into (A-5), we compute

$\mathfrak{l}_{p_{1}}=\frac{1}{2}\frac{3g-2r_{g}+\sqrt{9g^{2}-4}}{3r_{g}-s_{g}}.$

In a similar fashion, we obtain at the position $pRL^{n}$ for $n>0$:

$[2,S(\mu,\nu),2],1+[\langle
1,S(\mu,\nu),2,2,1\rangle_{n+1},1,S(\mu_{2},\nu_{2}),2],[\langle
1,S(\mu,\nu),2,2,1\rangle_{n},1,S(\mu_{2},\nu_{2}),2].$

Using symmetry again, we get:

$\mathfrak{l}_{p_{2}}=1+[\overline{1,S(\mu,\nu),2,2,1}]=:1+\xi.$

In order to compute the purely periodic continued fraction, we use

$\sigma:=[\overline{2,1,1,S(\mu,\nu),2}]=[2,1,\overline{1,S(\mu,\nu),2,2,1}].$

By symmetry, we can reuse the continuants we have computed above and compute
with A-5:

$\displaystyle\sigma=\frac{1}{2g}(3g-2r_{g}+\sqrt{9g^{2}-4}).$

Then, with $\sigma=2+1/(1+1/\xi)$, we get:

$\xi=[\overline{1,S(\mu,\nu),2,2,1}]=\frac{\sigma-2}{3-\sigma}=\frac{1}{2}\frac{3g-4r_{g}-2s_{g}+\sqrt{9g^{2}-4}}{3r_{g}+s_{g}}$

(see also Formula (A-7)) and the assertion follows. ∎

######  4.6.

$\frac{1}{2}\frac{4+\sqrt{32}}{2}$, $2,2$$\frac{1}{2}(1+\sqrt{5})$,
$1,1$$\frac{1}{2}\frac{11+\sqrt{221}}{5}$$2,1,1,2$$\frac{1}{2}\frac{29+\sqrt{1517}}{13}$$2,1,1,1,1,2$$\frac{1}{2}\frac{68+\sqrt{10400}}{34}$$2,1,1,1,1,1,1,2$$\frac{1}{2}\frac{199+\sqrt{71285}}{89}$$2,1,1,1,1,1,1,1,1,2$$\frac{1}{2}\frac{105+\sqrt{71285}}{115}$$1,1,1,1,1,1,1,2,2,1$$\frac{1}{2}\frac{40+\sqrt{10400}}{44}$$1,1,1,1,1,2,2,1$$\frac{1}{2}\frac{2961+\sqrt{15800621}}{1327}$$2,1,1,1,1,2,2,1,1,1,1,1,1,2$$\frac{1}{2}\frac{1559+\sqrt{15800621}}{1715}$$1,1,1,1,1,2,2,1,1,1,1,2,2,1$$\frac{1}{2}\frac{15+\sqrt{1517}}{17}$$1,1,1,2,2,1$$\frac{1}{2}\frac{432+\sqrt{338720}}{196}$$2,1,1,2,2,1,1,1,1,2$$\frac{1}{2}\frac{16837+\sqrt{514518485}}{15278}$$2,1,1,2,2,1,1,1,1,2,2,1,1,1,1,2$$\frac{1}{2}\frac{8731+\sqrt{514518485}}{19798}$$1,1,1,2,2,1,1,1,1,2,2,1,1,2,2,1$$\frac{1}{2}\frac{224+\sqrt{338720}}{254}$$1,1,1,2,2,1,1,2,2,1$$\frac{1}{2}\frac{6451+\sqrt{75533477}}{2927}$$2,1,1,2,2,1,1,2,2,1,1,1,1,2$$\frac{1}{2}\frac{3345+\sqrt{75533477}}{3793}$$1,1,1,2,2,1,1,2,2,1,1,2,2,1$$\frac{1}{2}\frac{5+\sqrt{221}}{7}$$1,2,2,1$$\frac{1}{2}\frac{63+\sqrt{7565}}{41}$$2,2,2,1,1,2$$\frac{1}{2}\frac{941+\sqrt{1687397}}{463}$$2,2,2,1,1,2,2,1,1,2$$\frac{1}{2}\frac{14052+\sqrt{376282400}}{6914}$$2,2,2,1,1,2,2,1,1,2,2,1,1,2$$\frac{1}{2}\frac{6496+\sqrt{376282400}}{9124}$$1,2,2,1,1,2,2,1,1,2,2,2,2,1$$\frac{1}{2}\frac{435+\sqrt{1687397}}{611}$$1,2,2,1,1,2,2,2,2,1$$\frac{1}{2}\frac{81856+\sqrt{12768548000}}{40276}$$2,2,2,1,1,2,2,2,2,1,1,2,2,1,1,2$$\frac{1}{2}\frac{37840+\sqrt{12768548000}}{53150}$$1,2,2,1,1,2,2,2,2,1,1,2,2,2,2,1$$\frac{1}{2}\frac{29+\sqrt{7565}}{31}$$1,2,2,2,2,1$$\frac{1}{2}\frac{367+\sqrt{257045}}{181}$$2,2,2,2,2,1,1,2$$\frac{1}{2}\frac{31925+\sqrt{1945074605}}{31490}$$2,2,2,2,2,1,1,2,2,2,2,1,1,2$$\frac{1}{2}\frac{14703+\sqrt{1945074605}}{41578}$$1,2,2,2,2,1,1,2,2,2,2,2,2,1$$\frac{1}{2}\frac{169+\sqrt{257045}}{239}$$1,2,2,2,2,2,2,1$$\frac{1}{2}\frac{2139+\sqrt{8732021}}{1055}$$2,2,2,2,2,2,2,1,1,2$$\frac{1}{2}\frac{985+\sqrt{8732021}}{1393}$$1,2,2,2,2,2,2,2,2,1$
Figure 7. The purely periodic spectrum.

Figure 7 shows the purely periodic values $\mathfrak{l}_{p_{1}}$ and
$\mathfrak{l}_{p_{2}}-1=\frac{1}{2}\frac{3g-4r_{g}-2s_{g}+\sqrt{9g^{2}-4}}{3r_{g}+s_{g}}$,
together with their periods, for the first five levels of the Markoff tree.
More precisely, the root shows $\mathfrak{l}_{p_{2}}-1$ for $g=1$ and
$\mathfrak{l}_{p_{1}}-1$ for $g=2$, and any pair of siblings corresponding to
mutations $(e,F,g)$ and $(g,E,f)$, show $\mathfrak{l}_{p_{1}}$ and
$\mathfrak{l}_{p_{2}}-1$ corresponding to $g$, respectively. As remarked in
3.9, for any $\mu/\nu$, the sequences $S(\mu,\nu)$ from $S(\nu,\mu)$ are
symmetric to each other in the sense that they correspond to opposite
locations in the Markoff tree and can be transformed into each other by
exchanging $1$’s and $2$’s. As can easily be seen from Proposition 4.5, in the
representation of Figure 7 this symmetry extends to the purely periodic
spectrum.

In order to arrive at nicer representations, we pass now from the spectrum of
quotients $m/r$ to the slopes $r/m$. By Formula (2), this spectrum is
contained in the interval

$\frac{1}{2}(3-\sqrt{5})<\frac{r}{m}<\sqrt{2}-1.$

######  Definition 4.7:

Let $(e,g,f)$ be a regular Markoff triple located at a path $p$ and denote
$p_{1}=pL\bar{R}$, $p_{2}=pR\bar{L}$. Then we denote

$A_{g}:=1/\mathfrak{l}_{p_{1}},\qquad B_{g}:=1/\mathfrak{l}_{p_{2}}.$

We compute

$\left(\frac{1}{2}\frac{3g\pm 2r_{g}+\sqrt{9g^{2}-4}}{3r_{g}\pm
s_{g}}\right)^{-1}=\frac{r_{g}}{g}\pm\frac{1}{2}\left(3-\sqrt{9-\frac{4}{g^{2}}}\right),$

hence $A_{g}<B_{g}$ such that the closed interval $[A_{g},B_{g}]$ is centered
at $r_{g}/g$ and has length $3-\sqrt{9-4/g^{2}}$. The four branches of Figure
3, from left to right, are given by $p\bar{L}$, $p_{1}$, $p_{2}$, and
$p\bar{R}$, respectively, and it follows that

$1/\mathfrak{l}_{p\bar{L}}=B_{e}<A_{g}<B_{g}<A_{f}=1/\mathfrak{l}_{p\bar{R}}.$

Moreover, for any other infinite path of the form $pq$ it follows that either

$B_{e}<1/\mathfrak{l}_{pq}<A_{g}\quad\text{ or }\quad
B_{g}<1/\mathfrak{l}_{pq}<A_{f}.$

For any regular Markoff triple $(e,g,f)$, we denote the intervals:

$\displaystyle I_{g}$ $\displaystyle=[B_{e},A_{f}],$ $\displaystyle J_{g}$
$\displaystyle=(A_{g},B_{g}),$

and a short computation yields for their lengths:

$\displaystyle|I_{g}|$
$\displaystyle=\frac{1}{ef}\left(\Delta_{e,f}-g\right),$
$\displaystyle|J_{g}|$ $\displaystyle=\frac{1}{g}(3g-\sqrt{\Delta_{g}}).$

With this, we can construct a Cantor set in the interval
$[\frac{1}{2}(3-\sqrt{5}),\sqrt{2}-1]$ as follows. For $n\geq 0$, in the
$n$-th step we denote
$(e^{n}_{1},g^{n}_{1},f^{n}_{1}),\dots,(e^{n}_{2^{n}},g^{n}_{2^{n}},f^{n}_{2^{n}})$
the Markoff triples on the $n$-th level of the Markoff tree, enumerated from
left to right. Then for any $n\geq 0$, the union
$C_{n+1}:=\bigcup_{i=1}^{2^{n+1}}I_{g_{i}^{n+1}}$ coincides with
$\coprod_{i=1}^{2^{n}}I_{g_{i}^{n}}\setminus J_{g^{n}_{i}}$ and, in
particular, is disjoint. It follows that

$\bigcap_{n\geq
0}C_{n}=\bigcap_{g}I_{g}=[\frac{1}{2}(3-\sqrt{5}),\sqrt{2}-1]\setminus\bigcup_{g}J_{g}$

forms a Cantor set in the interval $[\frac{1}{2}(3-\sqrt{5}),\sqrt{2}-1]$,
where $g$ runs over the maximal elements of every regular Markoff triple.

######  Definition 4.8:

We denote by $\mathfrak{M}$ the set of values $\\{1/\mathfrak{l}_{p}\\}$,
where $p$ runs over all infinite paths in the Markoff tree that start at the
root

######  Theorem 4.9:

$\mathfrak{M}$ forms a Cantor set of Lebesgue measure zero in the interval
$[\frac{1}{2}(3-\sqrt{5}),\sqrt{2}-1]$.

###### Proof.

It only remains to show that our Cantor set has measure zero. With above
notation, it equals $\bigcap_{n\geq 0}C_{n}$, where every $C_{n}$ is a finite
disjoint union of intervals in $\mathbb{R}$. By a standard argument, it
suffices to show that $|I_{g}|<3|J_{g}|$ for every regular Markoff triple
$(e,g,f)$. So we want to show that

$\frac{1}{ef}\Delta_{e,f}-\frac{g}{ef}<\frac{3}{g}(3g-\sqrt{\Delta_{g}}).$

With $(3g+\sqrt{\Delta_{g}})/2<3g$ we show the sharper bound:

$\frac{1}{ef}\Delta_{e,f}<\frac{g}{ef}+\frac{1}{2g^{2}}(3g+\sqrt{\Delta_{g}})(3g-\sqrt{\Delta_{g}})=\frac{g}{ef}+\frac{2}{g^{2}}.$

Using $g<3ef$, it suffices to show:

$\frac{1}{ef}\Delta_{e,f}<\frac{g}{ef}+\frac{2}{3efg},$

or, equivalently:

$\Delta_{e,f}<g+\frac{2}{3g}.$

This inequality has been shown in Lemma 3.4 (ii). ∎

With the Cantor set being of measure zero, it follows that the lengths of the
complementary open intervals add up to the length of the interval $I_{5}$:

######  Corollary 4.10 (see also [LT07, §4.5]):

It follows that

$(3-\sqrt{5})+(3-\sqrt{8})+2\sum_{(e,g,f)}\left(3-\sqrt{9-\frac{4}{g^{2}}}\right)=1,$

where the sum runs over the regular Markoff triples.

######  Remark 4.11:

Theorem 4.9 and Corollary 4.10 are well-known to specialists, though [LT07] is
the only prior reference we could find. According to [LT07], Corollary 4.10
can be derived from McShane’s identity (see [Bow96]), whereas our proof is
elementary.

Next we are going to determine the Hausdorff dimension $\dim_{H}\mathfrak{M}$
of $\mathfrak{M}$ (see Appendix C for details and references). Consider any
regular Markoff triple $(e,g,f)$ corresponding to a path
$W\in\mathfrak{P}_{f}$. Then we denote $I_{W}:=I_{g}$ and we set
$d_{WX}=|I_{WX}|/|I_{W}|$ for $X\in\\{L,R\\}$. Now, for any $p\in\mathfrak{P}$
and its corresponding point $x_{p}\in\mathfrak{M}$, we define the local
Hausdorff measure

$h^{s}(x)=\liminf_{k\rightarrow\infty}\prod_{i=0}^{k}(d_{p|_{i}L}^{s}+d_{p|_{i}R}^{s}),$

where $p|_{i}$ denotes the prefix of length $i$ of $p$. We define the local
Hausdorff dimension of $\mathfrak{M}$ at $x$ as:

$h(x)=\inf_{s\geq 0}\\{h^{s}(x)>0\\}.$

######  Theorem 4.12:

$\dim_{H}\mathfrak{M}=0$ and $h(x)=0$ for any $x\in\mathfrak{M}$.

###### Proof.

Using Corollary C.3, $\dim_{H}\mathfrak{M}=0$ is implied by $h(x)=0$ for every
$x\in\mathfrak{M}$. Therefore, it suffices to show the latter. We consider
three cases of points $x_{p}$, depending on the structure of $p$.

The case $p=p|_{k}\bar{L}$ for some $k$.

In order to show that $h^{s}(x_{p})=0$ for any $s>0$, we are going to show
that $d^{s}_{p|_{j}}+d^{s}_{p|_{j}}<1$ for sufficiently large $j\gg 0$ for any
$s>0$. For this, we show that both $d_{p|_{j}L}$ and $d_{p|_{j}}$ are bounded
from above by monotonously decreasing series, where for $d_{p|_{j}L}$ the
bounding series has limit $<1$ and for $d_{p|_{j}R}$ we obtain
$\lim_{j\rightarrow\infty}d_{p|_{j}R}=0$.

Using the same notation as in Lemma 3.5, assume that $p=p|_{k}\bar{L}$ and
denote $(e,g,f)$ the Markov triple corresponding to the path $p|_{k}$. Then we
get:

$d_{p|_{k+i+1}}=\frac{|I_{p|_{k+i+1}}|}{|I_{p|_{k+i}}|}=\frac{F_{i-1}}{F_{i}}\frac{\Delta_{e,F_{i}}-F_{i+1}}{\Delta_{e,F_{i-1}}-F_{i}}\quad\text{and}\quad
d_{p|_{k+i}R}=\frac{|I_{p|_{k+i}R}|}{|I_{p|_{k+i}}|}=\frac{e}{F_{i}}\frac{\Delta_{F_{i},F_{i-1}}-H_{i+1}}{\Delta_{e,F_{i-1}}-F_{i}}$

for $i\geq 0$, where $H_{i}:=3F_{i}F_{i-1}-e$. Using first Lemma 3.4 (iii),
then 3.4 (ii), then $\sqrt{\Delta_{F_{i-1}}}<3F_{i-1}$, we get:

$d_{p|_{k+i+1}}<\frac{\sqrt{\Delta_{e}}\sqrt{\Delta_{F_{i}}}}{2e}(\Delta_{e,F_{i+1}}-F_{i+2})<\frac{\sqrt{\Delta_{e}}}{9e^{2}}\frac{\sqrt{\Delta_{F_{i-1}}}}{F_{i}}<\frac{\sqrt{\Delta_{e}}}{3e^{2}}\frac{F_{i-1}}{F_{i}}.$

From Lemma 3.5 (i), it is straightforward to see that
$F_{i}/F_{i+1}<F_{i-1}/F_{i}$ for every $i$ and moreover,

$\lim_{i\rightarrow\infty}\frac{\sqrt{\Delta_{e}}}{3e^{2}}\frac{F_{i-1}}{F_{i}}=\frac{2}{3e^{2}}\frac{\sqrt{\Delta_{e}}}{3e+\sqrt{\Delta_{e}}}<1.$

Similarly, we get

$d_{p|_{k+i}R}<\frac{\sqrt{\Delta_{e}}}{3H_{i+1}}\xrightarrow[\
i\rightarrow\infty\ ]{}0.$

Note that these inequalities also show that $\mathfrak{M}$ satisfies the
conditions for Corollary C.3.

The case $p=p|_{k}\bar{R}$ for some $k$.

Follows analogously.

The remaining paths.

Let $p$ be any infinite path not in $\mathfrak{P}_{f}$. Then $p$ has
infinitely many continuous segments consisting of repetitions of $L$. Let
denote $p|_{k_{j}},\dots,p|_{k_{j}+t_{j}}$ the $j$-th such segment, i.e. the
last letter of every $p|_{k_{j}+i},0\leq i\leq t_{j}$ is $L$. Denote
$(e_{j},g_{j},f_{j})$ the root of this segment. Then we have seen above that
$d_{p|_{k_{j}+i}}$ is bounded by values (very close to) $1/3e_{j}^{2}$.
Therefore, for $j\rightarrow\infty$, $d_{p|_{n}}\rightarrow 0$ for $k_{j}\leq
n\leq k_{j}+t_{j}$. Similarly, the $d_{p|_{n-1}R}$ are bounded even stronger.
We can argue analogously for the segments consisting of $R$s. Therefore
$\lim_{i\rightarrow\infty}(d^{s}_{p|_{i}L}+d^{s}_{p|_{i}R})=0$ for every
$s>0$, hence $h(x_{p})=0$. ∎

##  5\. Markoff triples and T-singularities

According to [HP10], given a Markoff triple $(e,g,f)$, the singular points of
$\mathbb{P}(e^{2},g^{2},f^{2})$ are cyclic quotient singularities of orders
$e^{2}$, $g^{2}$, and $f^{2}$, respectively. These singularities can locally
be described as the diagonal action of, say, $\mathbb{Z}/g^{2}\mathbb{Z}$ on
$\mathbb{C}^{2}$ with weights $e^{2}$ and $f^{2}$ (shorthand:
$\frac{1}{g^{2}}(e^{2},f^{2})$). By choosing an appropriate $g^{2}$-th root of
unity, this is equivalent to $\frac{1}{g^{2}}(1,e^{-2}f^{2}\mod g^{2})$ and,
using the Markoff equation, to:

$\frac{1}{g^{2}}(1,gw_{g}-1),$

where $w_{g}=3e^{-1}f\mod g$. This is a particular example of a class of
surface singularities called T-singularities which we will review in more
detail in Section 6. Repeating this calculation for the singularities of
orders $e^{2}$ and $f^{2}$ leads to the following definition.

######  Definition 5.1:

Denote $(e,g,f)$ be a regular Markoff triple. Then we set:

$\displaystyle w_{e}$ $\displaystyle:=\begin{cases}-1&\text{ if }e=1,\\\
3f^{-1}g\mod e&\text{ else},\end{cases}$ $\displaystyle w_{f}$
$\displaystyle:=3g^{-1}e\mod f,$ $\displaystyle w_{g}$
$\displaystyle:=3e^{-1}f\mod g.$

We call $(w_{e},w_{g},w_{f})$ the T-weights of the triple $(e,g,f)$.

The Markoff equation implies that $(e^{-1}f)^{2}\equiv-1\mod g$ and hence
$w_{g}^{2}\equiv-9\mod g$. Similarly, $w_{e}^{2}\equiv-9\mod e$ and
$w_{f}^{2}\equiv-9\mod f$.

######  Definition 5.2:

We define the T-coweights $(v_{e},v_{g},v_{f})$ of $(e,g,f)$ to be the unique
integers such that the following equations are satisfied:

$\displaystyle w_{e}^{2}$ $\displaystyle=-9+v_{e}e,$ $\displaystyle w_{f}^{2}$
$\displaystyle=-9+v_{f}f,$ $\displaystyle w_{g}^{2}$
$\displaystyle=-9+v_{g}g.$

As the reader has noticed, the $T$-weights and $T$-coweights are very similar
to the weights and coweights we have considered in Section 3. The following
Lemma is proved in the same way as Lemma 3.1.

######  Lemma 5.3:

Let $(e,g,f)$ be a regular Markoff triple.

  1. (i)

The weights and coweights transform under mutation as follows:

( |  $e$, |  $g$, | $f$ | )
---|---|---|---|---
( |  $w_{e}$, |  $w_{g}$, | $w_{f}$ | )
( |  $v_{e}$, |  $v_{g}$, | $v_{f}$ | )
( |  $e$, |  $F$, | $g$ | )
---|---|---|---|---
( |  $w_{e}$, |  $w_{F}$, | $w_{g}$ | )
( |  $v_{e}$, |  $v_{F}$, | $v_{g}$ | )
( |  $g$, |  $E$, | $f$ | )
---|---|---|---|---
( |  $w_{g}$, |  $w_{E}$, | $w_{f}$ | )
( |  $v_{g}$, |  $v_{E}$, | $v_{f}$ | )

where

$w_{E}=3fw_{g}-w_{e},\quad v_{E}=3fv_{g}-v_{e},\quad w_{F}=3ew_{g}-w_{f},\ \
\text{ and }\ \ v_{F}=3ev_{g}-v_{f}.$

  2. (ii)

The slopes of a regular Markoff triple are strictly increasing:

$\frac{w_{e}}{e}<\frac{w_{g}}{g}<\frac{w_{f}}{f}.$

  3. (iii)

We have $w_{e}<e-w_{e}$, $w_{f}<f-w_{f}$, and $w_{g}<g-w_{g}$ for all $e,g$
and all $f>2$.

Figure 8 shows the Markoff triples, their weights and coweights of the first
four levels of the Markoff tree.

( | 1, | 5, | 2 | )
---|---|---|---|---
( | -1, | 1, | 1 | )
( | 10, | 2, | 5 | )
( | 1, | 13, | 5 | )
---|---|---|---|---
( | -1, | 2, | 1 | )
( | 10, | 1, | 2 | )
( | 1, | 34, | 13 | )
---|---|---|---|---
( | -1, | 5, | 2 | )
( | 10, | 1, | 1 | )
( | 1, | 89, | 34 | )
---|---|---|---|---
( | -1, | 13, | 5 | )
( | 10, | 2, | 1 | )
( | 34, | 1325, | 13 | )
---|---|---|---|---
( | 5, | 196, | 2 | )
( | 1, | 29, | 1 | )
( | 13, | 194, | 5 | )
---|---|---|---|---
( | 2, | 31, | 1 | )
( | 1, | 5, | 2 | )
( | 13, | 7561, | 194 | )
---|---|---|---|---
( | 2, | 1208, | 31 | )
( | 1, | 193, | 5 | )
( | 194, | 2897, | 5 | )
---|---|---|---|---
( | 31, | 463, | 1 | )
( | 5, | 74, | 2 | )
( | 5, | 29, | 2 | )
---|---|---|---|---
( | 1, | 7, | 1 | )
( | 2, | 2, | 5 | )
( | 5, | 433, | 29 | )
---|---|---|---|---
( | 1, | 104, | 7 | )
( | 2, | 25, | 2 | )
( | 5, | 6466, | 433 | )
---|---|---|---|---
( | 1, | 1553, | 104 | )
( | 2, | 373, | 25 | )
( | 433, | 37666, | 29 | )
---|---|---|---|---
( | 104, | 9047, | 7 | )
( | 25, | 2173, | 2 | )
( | 29, | 169, | 2 | )
---|---|---|---|---
( | 7, | 41, | 1 | )
( | 2, | 10, | 5 | )
( | 29, | 14701, | 169 | )
---|---|---|---|---
( | 7, | 3566, | 41 | )
( | 2, | 865, | 10 | )
( | 169, | 985, | 2 | )
---|---|---|---|---
( | 41, | 239, | 1 | )
( | 10, | 58, | 5 | )
Figure 8. The first four levels of the Markoff tree with triples $(e,g,f)$,
$(w_{e},w_{g},w_{f})$, $(v_{e},v_{g},v_{f})$.

With Lemmas 3.1 (iii), 5.5 (i), we get:

######  Corollary 5.4:

For every regular Markoff triple, $w_{e}=3r_{e}-e$, $w_{g}=3r_{g}-g$, and
$w_{f}=3r_{f}-f$.

For later use, we state the following formulas (compare [Cas57, §II.3, Lemma
7]).

######  Lemma 5.5:

Let $(e,g,f)$ be a regular Markoff triple. Then the following equations hold:

  1. (i)

$\displaystyle gw_{f}-fw_{g}$ $\displaystyle=3e,$ $\displaystyle
ew_{g}-gw_{e}$ $\displaystyle=3f,$ $\displaystyle ew_{f}-fw_{e}$
$\displaystyle=3G.$

  2. (ii)

$Gw_{g}=ew_{e}+fw_{f}.$

  3. (iii)

$\displaystyle fv_{e}$ $\displaystyle=w_{e}w_{f}+3(w_{g}-3fw_{e})$
$\displaystyle ev_{f}$ $\displaystyle=w_{e}w_{f}+3(3ew_{f}-w_{g}).$

###### Proof.

(i) We will only prove the the last equation and leave the others for the
reader. First we observe $ew_{f}-fw_{e}\equiv-3g\mod ef$, hence
$ew_{f}-fw_{e}=-3g+xef$ for some integer $x$. From
$ew_{f}-fw_{e}+3g<ew_{f}-fw_{e}+3g+3G=ew_{f}-fw_{e}+9ef<10ef$ we get $x<10$.
Moreover, we have
$ew_{f}-fw_{e}+3g=ew_{f}-fw_{e}+9ef-3G=ew_{f}+f(e-w_{e})+8ef-3G$. The latter
term is greater than $8ef$ (implying x = 9 and therefore the assertion) iff
$ew_{f}+f(e-w_{e})>3G$. Because $G<f$, this inequality will be satisfied if
$3G<f(e-w_{e})$ holds and for this it is enough to consider the cases
$e-w_{e}=1,2$. If $e-w_{e}=1$ then $(e-w_{e})^{2}\equiv-9\equiv 1\mod e$,
hence $e\in\\{1,2,5\\}$, but for $e=1$ we have $e-w_{e}=2$ and for $e=5$ we
have $e-w_{e}=4$, hence $e=2$. In that case, $(e,g,f)=(P_{2n-1},P_{2n+1},2)$.
If $G=P_{2n-3}$ then $3G=3P_{2n-3}<P_{2n+1}$ by the well-known properties of
Pell numbers. Otherwise, if $G<e$ then $G=1$ and $(e,g,f)=(5,29,2)$, thus
$3G<f$.

If $e-w_{e}=2$ then $(e-w_{e})^{2}\equiv-9\equiv 4\mod e$, hence
$e\in\\{1,13\\}$, but for $e=13$ we have $e-w_{e}=11$, hence $e=1$. But then
$(e,g,f)=(1,F_{2n+1},F_{2n-1})$ and we conclude analogously as for the Pell
numbers.

(ii) By (i) we have
$ew_{f}w_{g}+fw_{e}w_{g}-gw_{e}w_{f}=ew_{f}w_{g}-3ew_{e}=fw_{e}w_{g}+3fw_{f}$
which, again by (i), implies the assertion.

(iii) We only show the first equality, the other follows analogously.

$\displaystyle efv_{e}$
$\displaystyle=f(9+w_{e}^{2})=9f+fw_{e}^{2}-ew_{e}w_{f}+ew_{e}w_{f}$
$\displaystyle=9f-3Gw_{e}+ew_{e}w_{f}\quad\text{ by (\ref{cofactorlemmai}) }$
$\displaystyle=ew_{e}w_{f}+3e(w_{g}-3fw_{e})\quad\text{ again by
(\ref{cofactorlemmai}). }$

∎

The following observations regarding the Fibonacci and Pell branches may be of
independent interest:

######  Lemma 5.6:

  1. (i)

Let $(e,g,f)$ be a Markoff triple of the form $(1,F_{2n+1},F_{2n-1})$ for
$n\geq 2$. Then:

$\displaystyle(w_{e},w_{g},w_{f})$ $\displaystyle=(-1,F_{2n-3},F_{2n-5}),$
$\displaystyle(e-w_{e},g-w_{g},f-f_{w})$
$\displaystyle=(2,L_{2n-1},L_{2n-3}),$ $\displaystyle(v_{e},v_{g},v_{f})$
$\displaystyle=(10,F_{2n-7},F_{2n-9}),$

where $L_{i}$ denotes the $i$-th Lucas number.

  2. (ii)

Let $(e,g,f)$ be a Markoff triple of the form $(P_{2n-1},P_{2n+1},2)$ for
$n\geq 2$. Then:

$\displaystyle(w_{e},w_{g},w_{f})$ $\displaystyle=(S_{2n-3},S_{2n-1},1),$
$\displaystyle(e-w_{e},g-w_{g},f-w_{f})$
$\displaystyle=(Q_{2n-3},Q_{2n-1},1),$ $\displaystyle(v_{e},v_{g},v_{f})$
$\displaystyle=(R_{2n-5},R_{2n-3},5),$

where $Q_{i}=2Q_{i-1}+Q_{i-2}$, $R_{i}=2R_{i-1}+R_{i-2}$,
$S_{i}=2S_{i-1}+S_{i-2}$ with initial values $Q_{0}=1$, $Q_{1}=4$, $R_{1}=2$,
$R_{2}=4$, and $S_{0}=1$, $S_{1}=1$, respectively.

######  Remark 5.7:

The sequence $2S_{i}$ are the Pell-Lucas numbers and the sequence $S_{2n+1}$
whose first few terms are $1,7,41,239,1393,8119,47321,\dots$ is known as the
sequence of NSW numbers. Together with some interesting interpretations it can
be found in the Online Encyclopedia of Integer Sequences under reference
A002315 (see [Slo]). The $R_{i}$ can be found under A052542, and the odd
entries of that series under A075870. The $Q_{i}$ are referenced by A048654,
their odd entries by A100525. The sequences $(L_{2n-1}^{2}+9)/F_{2n+1}$ and
$(Q_{2n-1}^{2}+9)/P_{2n+1}$ are referenced by A106729 and A254759,
respectively.

##  6\. T-singularities and continued fractions

In this section we consider T-singularities of type

$\frac{1}{n^{2}}(1,nk-1),$

where $1<n$ is not necessarily a Markoff number and $0<k<n$ is coprime to $n$.
General T-singularities have been introduced by Wahl [Wah81] and play an
important role in the three-dimensional minimal model program. Our main source
is [KS88, § 3].

Given such a singularity, its minimal desingularization can be described with
help of Hirzebruch-Jung continued fractions, i.e. if we write

$\frac{n^{2}}{nk-1}=\llbracket
a_{1},a_{2},\dots,a_{t}\rrbracket:=a_{1}-1/(a_{2}-1/(a_{3}-\cdots(a_{t-1}-1/a_{t})\cdots)),$

then the intersection graph of the minimal resolution is a chain with $t$
vertices which are labelled by $-a_{1},\dots,-a_{t}$, where each $a_{i}\geq
2$. The continued fraction expansions corresponding to the T-singularities of
above type have been described in [KS88, Proposition 3.11] as follows:

  * •

$\llbracket 4\rrbracket$ corresponds to a T-singularity (which obviously is
given by the pair $(n,k)=(2,1)$),

  * •

If $\llbracket a_{1},\dots,a_{t}\rrbracket$ corresponds to a T-singularity,
then so do $\llbracket a_{1}+1,a_{2},\dots,a_{t},2\rrbracket$ and $\llbracket
2,a_{1},\dots,a_{t-1},$ $a_{t}+1\rrbracket$.

This classification suggests a more efficient representation of a
T-singularity as follows.

######  Definition:

The length encoding (LE) of a T-singularity of type $\frac{1}{n^{2}}(1,nk-1)$
is defined as:

  1. (i)

The empty sequence $\emptyset$ for $(n,k)=(2,1)$.

  2. (ii)

If a sequence $c_{1},\dots,c_{m}$ encodes a T-singularity corresponding to
$\llbracket a_{1},\dots,a_{t}\rrbracket$ with $a_{1}\neq 2$, then for any
$c_{m+1}>0$, $c_{1},\dots,c_{m+1}$ encodes the T-singularity corresponding to
$\llbracket 2,\dots,2,a_{1},\dots,a_{t-1},a_{t}+c_{m+1}\rrbracket$.

  3. (iii)

Similarly, if a sequence $c_{1},\dots,c_{m}$ encodes a T-singularity
corresponding to $\llbracket a_{1},\dots,a_{t}\rrbracket$ with $a_{t}\neq 2$,
then for any $c_{m+1}>0$, $c_{1},\dots,c_{m+1}$ encodes the T-singularity
corresponding to $\llbracket a_{1}+c_{m+1},a_{2},\dots,$
$a_{t},2,\dots,2\rrbracket$.

Note that a continued fraction $\llbracket a_{1},\dots,a_{t}\rrbracket$
corresponds to a T-singularity iff $\llbracket a_{t},\dots,a_{1}\rrbracket$
does. More precisely, if $\llbracket a_{1},\dots,a_{t}\rrbracket$ corresponds
to $\frac{1}{n^{2}}(1,nk-1)$ then $\llbracket a_{t},\dots,a_{1}\rrbracket$
corresponds to $\frac{1}{n^{2}}(1,n(n-k)-1)$. This reflects the fact that
$n(n-k)-1\equiv(nk-1)^{-1}\mod n^{2}$ and that, from a geometric point of
view, there is no natural preference whether to read the terms in our minimal
resolutions from left to right or from right to left. The LE coincides for
both representations.

######  Lemma 6.1:

  1. (i)

A positive integer $c$ is the LE of the pair $(c+2,1)$.

  2. (ii)

For $m>1$ let $c_{1},\dots,c_{m}$ be a sequence of positive integers. If
$c_{1},\dots,c_{m-1}$ is the LE for the pair $(n,k)$, where $k<n-k$, then
$c_{1},\dots,c_{m}$ is the LE of the pair $(n+c_{m}(n-k),n-k)$.

###### Proof.

(i) Consider first the continued fraction $\llbracket 2,\dots,2\rrbracket$
(with $c$ entries). Then it is easy to check that $\llbracket
2,\dots,2\rrbracket=\frac{c+1}{c}$. The LE $c$ represents $\llbracket
c+4,2,\dots,2\rrbracket=c+4-1/\llbracket
2,\dots,2\rrbracket=c+4-c/(c+1)=(c+2)^{2}/(c+1)$.

(ii) We first consider a pair $(n,k)$ with $n^{2}/(nk-1)=\llbracket
a_{1},\dots,a_{t}\rrbracket$, with $a_{1}\neq 2$. Then $\llbracket
2,a_{t},\dots,a_{1}\rrbracket=2-1/\llbracket
a_{t},\dots,a_{1}\rrbracket=2-1/(n^{2}/(n(n-k)-1)))=(n^{2}+nk+1)/n^{2}.$ It is
easy to verify that $k^{2}+nk-1$ is inverse to $n^{2}$ modulo $n^{2}+nk+1$.
Hence, we get $\llbracket
a_{1},\dots,a_{t},2\rrbracket=(n^{2}+nk+1)/(k^{2}+nk-1)=((n+k)n+1)/((n+k)k-1)$.
Then it follows that $\llbracket
a_{1}+1,a_{2},\dots,a_{t},2\rrbracket=1+\llbracket
a_{1},\dots,a_{t},2\rrbracket=(n+k)^{2}/((n+k)k-1)$.

We apply this result to an induction on $c_{m}$. For simplicity, we will allow
the redundant case $c_{m}=0$ for which the statement is obviously true. Now
assume that $c_{m}>0$ and $c_{1},\dots,c_{m-1},c_{m}-1$ encodes the pair
$(n+(c_{m}-1)(n-k),n-k)$. Then the statement follows immediately from above
computation. ∎

Conversely, by the lemma we see that if we are given a pair $(n,k)$ then we
can recover its LE by means of the Euclidean algorithm. Namely, we set
$n_{0}:=n-k$, $n_{1}:=k$ and inductively, for every $i>1$ we define $n_{i}$
via division with remainder $n_{i-2}=p_{i-1}n_{i-1}+n_{i}$ such that $0\leq
n_{i}<n_{i-1}$. Then from the resulting sequence $p_{m},\dots,p_{1}$ we get
the LE $c_{1},\dots,c_{m}$ for $(n,k)$, where $c_{i}=p_{m+1-i}$ for $2\leq
i\leq m$ and $c_{1}=n_{m-1}-1$. Moreover, for each $0\leq i<m$ the partial
sequence $c_{1},\dots,c_{m-i}$ is the LE of $(n_{i}+n_{i+1},n_{i+1})$. As a
consequence, we can recover the length encoding of a T-singularity by means of
an ordinary continued fraction:

######  Corollary 6.2:

Let $m>1$ and $c_{1},\dots,c_{m}$ be the LE of the pair $(n,k)$, where
$k<n-k$. Then $\frac{n}{k}=[c_{m}+1,,c_{m-1},\dots,c_{2},$
$c_{1}+1]=(c_{m}+1)+1/(c_{m-1}+\cdots(c_{2}+1/(c_{1}+1))\cdots))$.

Intuitively, one could consider Corollary 6.2 as a version of

$\frac{n^{2}}{nk-1}\approx\frac{n}{k}.$

A strengthening of this sentiment is the observation below that the continued
fraction expansion of $n^{2}/(nk-1)$ is almost symmetric and we can obtain the
$c_{i}$ above from either half. For the proof of this claim, we start by
reminding the reader that
$[a_{1},\dots,a_{t},1]=[a_{1},\dots,a_{t-1},a_{t}+1]$. So, if we refer to the
length of a continued fraction, we have to make explicit whether the last
element is $>1$.

######  Lemma 6.3:

Consider a pair $(n,k)$ with $k<n-k$. Then:

  1. (i)

$\frac{n^{2}}{nk-1}$ has a continued fraction expansion of even length
$\frac{n^{2}}{nk-1}=[a_{1},\dots,a_{2s}]$ for some $s\geq 1$ such that
$a_{2s}>1$;

  2. (ii)

$\frac{n^{2}}{nk+1}=[a_{2s},\dots,a_{1}]$;

  3. (iii)

$\frac{n^{2}}{n(n-k)+1}=[1,a_{1}-1,a_{2},\dots,a_{2s}]$;

  4. (iv)

$\frac{n^{2}}{n(n-k)-1}=[1,a_{2s}-1,a_{2s-1},\dots,a_{1}]$.

###### Proof.

If we write $\frac{n^{2}}{nk-1}=[a_{1},\dots,a_{t}]$ with $a_{t}>1$ then by
equation (A-4), we have $[a_{t},\dots,a_{1}]=\frac{n^{2}}{X}$ for some
$0<X<g^{2}$ such that $(nk-1)X\equiv(-1)^{t+1}\mod n^{2}$. As remarked
earlier, we have $(nk-1)(n(n-k)-1)\equiv 1\mod n^{2}$, so either $X=n(n-k)-1$
and $t$ is odd, or $X=nk+1$ and $t$ is even. Because $a_{t}>1$, we necessarily
have $\lfloor n^{2}/X\rfloor>1$, hence $X=nk+1$ and we conclude
$(\ref{contfracevenoddi})$ and $(\ref{contfracevenoddii})$.

$(\ref{contfracevenoddiv})$ and $(\ref{contfracevenoddiii})$ then follow from
$(\ref{contfracevenoddi})$, $(\ref{contfracevenoddii})$, and Lemma A.1. ∎

The following statement proves the even-ness of the continued fraction
expansion of $n^{2}/(nk-1)$ and it also shows that these expansions are almost
symmetric.

######  Proposition 6.4:

Consider a pair $(n,k)$ with $k<n-k$ and assume that $n/k=[c_{1},\dots,c_{s}]$
with $c_{1},c_{s}>1$. Then:

$\displaystyle\frac{n^{2}}{nk-1}=\begin{cases}[c_{1},\dots,c_{s-1},c_{s}-1,c_{s}+1,c_{s-1},\dots,c_{1}]&\text{
if $s$ is even},\\\
[c_{1},\dots,c_{s-1},c_{s}+1,c_{s}-1,c_{s-1},\dots,c_{1}]&\text{ if $s$ is
odd}.\end{cases}$

###### Proof.

We have $n/k=[c_{1},\dots,c_{s}]$ and $n/k^{\prime}=[c_{s},\dots,c_{1}]$,
where $kk^{\prime}=(-1)^{1+s}+un$ by equation (A-4). Then
$K(c_{1},\dots,c_{s})=n$, $K(c_{2},\dots,c_{s})=k$,
$K(c_{1},\dots,c_{s-1})=k^{\prime}$, and $K(c_{2},\dots,c_{s-1})=u$ by
Cassini’s identity (A-2). We compute

$\displaystyle[c_{1},\dots,c_{s-1},c_{s}-(-1)^{s}]$
$\displaystyle=\frac{n-(-1)^{s}k^{\prime}}{k-(-1)^{s}u},$
$\displaystyle[c_{s}+(-1)^{s},c_{s-1},\dots,c_{1}]$
$\displaystyle=\frac{n+(-1)^{s}k^{\prime}}{k^{\prime}}.$

Then:

$\displaystyle
K(c_{1},\dots,c_{s-1},c_{s}+(-1)^{s})K(c_{s}-(-1)^{s},c_{s-1},\dots,c_{1})+K(c_{1},\dots,c_{s-1})^{2}$
$\displaystyle=(n-(-1)^{s}k^{\prime})(n+(-1)^{s}k^{\prime})+(k^{\prime})^{2}=n^{2}$

and

$\displaystyle K(c_{2},\dots,c_{s-1},c_{s}+(-1)^{s})$ $\displaystyle
K(c_{s}-(-1)^{s},c_{s-1},\dots,c_{1})+K(c_{2},\dots,c_{s-1})K(c_{1},\dots,c_{s-1})$
$\displaystyle=(k-(-1)^{s}u)(n+(-1)^{s}k^{\prime})+uk^{\prime}$
$\displaystyle=nk+(-1)^{s}(kk^{\prime}-un)$ $\displaystyle=nk-1.$

By the concatenation rule A.2, we conclude:

$\frac{n^{2}}{nk-1}=[c_{1},\dots,c_{s-1},c_{s}-(-1)^{s},c_{s}+(-1)^{s},c_{s-1},\dots,c_{1}],$

hence the assertion. ∎

##  7\. Continued fractions and the Markoff tree

In this section we are going to derive mutation formulas for the LEs
corresponding to T-singularities arising from the Markoff equation.
Unfortunately, it seems not to be possible to do this directly via the
fractions $g/w_{g}$, as would be suggested by Corollory 6.2. Instead, we will
show that the continued fraction expansions of $g^{2}/(gw_{g}-1)$ will exhibit
mutation rules analogous to those described in Section 3. The LEs then can be
derived from these continued fractions with Proposition 6.4 and Corollary 6.2.

Using the mutations for $g^{2}/(gw_{g}-1)$, we will see that the tree of
fractions $w_{g}/g$ has limits in a Cantor set, very similar to the Cantor set
of Section 4, which emerges from the fractions $r_{g}/g$. Indeed, we will see
that these Cantor sets coincide up to an affine transformation.

######  Proposition 7.1:

Consider a Markoff number $g$ with T-weight $w$ and T-coweight $v$
corresponding to some Markoff triple containing $g$. We denote $W:=g-w$,
$V:=v+g-2w$ such that $w<W$ and $g^{2}/(gw-1)=[a_{1},\dots,a_{2s}]$ with
$a_{1},a_{2s}>1$. Then the following equations hold:

  1. (i)

$[a_{1},\dots,a_{2s},8]=\frac{9g^{2}-gW+1}{8gw+gv-17}$,

  2. (ii)

$[a_{1},\dots,a_{2s-1},a_{2s}-1,1,8]=\frac{9g^{2}-gw-1}{9gw-gv}$,

  3. (iii)

$[1,a_{1}-1,a_{2}\dots,a_{2s},8]=\frac{9g^{2}-gW+1}{9gW-gV+18}$,

  4. (iv)

$[1,a_{1}-1,a_{2},\dots,a_{2s-1},a_{2s}-1,1,8]=\frac{9g^{2}-gw-1}{8gW+gV-1}$,

  5. (v)

$[a_{2s},\dots,a_{1},8]=\frac{9g^{2}-gW-1}{8gw+gv-1}$,

  6. (vi)

$[a_{2s},\dots,a_{2},a_{1}-1,1,8]=\frac{9g^{2}-gw+1}{9gw-gv+18}$,

  7. (vii)

$[1,a_{2s}-1,a_{2s-1},\dots,a_{1},8]=\frac{9g^{2}-gW-1}{9gW-gV}$,

  8. (viii)

$[1,a_{2s}-1,a_{2s-1},\dots,a_{2},a_{1}-1,1,8]=\frac{9g^{2}-gw+1}{8gW+gV-17}$.

###### Proof.

We first remark that (iii), (iv), (vii), and (viii) follow from (i), (ii),
(v), and (vi), respectively, by Lemma A.1.

(ii) By Lemma 6.3, we have

$[8,1,a_{2s}-1,a_{2s-1},\dots,a_{1}]=8+\cfrac{1}{\frac{g^{2}}{gW-1}}=\frac{9g^{2}-gw-1}{g^{2}}.$

Moreover, we compute:

$g^{2}(9gw-gv)=-1+(gw-1)(9g^{2}-gw-1),$

hence

$[a_{1},\dots,a_{2s-1},a_{2s}-1,1,8]=\frac{9g^{2}-gw-1}{9gw-gv}$

by equation A-4. Cases (i), (v), (vi) follow from similar computations. ∎

######  Proposition 7.2:

Let $(e,g,f)$ be a regular Markoff triple with $e>1$ and $f>2$. and denote
$e^{2}/(ew_{e}-1)=[a_{1},\dots,a_{2s}]$,
$f^{2}/(fw_{f}-1)=[b_{1},\dots,b_{2t}]$ with $a_{1},a_{2s},b_{1},b_{2t}>1$
then:

$\frac{g^{2}}{gw_{g}-1}=[a_{2s},\dots,a_{1},8,1,b_{2t}-1,b_{2t-1},\dots,b_{1}].$

###### Proof.

By the concatenation rule A.2, it suffices to show that

$\displaystyle g^{2}$
$\displaystyle=K(8,a_{1},\dots,a_{2s})K(b_{1},\dots,b_{2t})+K(a_{1},\dots,a_{2s})K(b_{1},\dots,b_{2t-1},b_{2t}-1),$
$\displaystyle gw_{g}-1$
$\displaystyle=K(8,a_{1},\dots,a_{2s-1})K(b_{1},\dots,b_{2t})+K(a_{1},\dots,a_{2s-1})K(b_{1},\dots,b_{2t-1},b_{2t}-1).$

By Lemma 6.3 and Proposition 7.1 (v) these are equivalent to

$\displaystyle g^{2}$
$\displaystyle=(9e^{2}-e(e-w_{e})-1)f^{2}+e^{2}(f(f-w_{f})-1)$ $\displaystyle
gw_{g}-1$ $\displaystyle=(8ew_{e}+ev_{e}-1)f^{2}+(ew_{e}+1)(f(f-w_{f})-1)$

The following computation proves the first equality:

$\displaystyle(9e^{2}-e(e-w_{e})-1)f^{2}+e^{2}(f(f-w_{f})-1)$
$\displaystyle=9e^{2}f^{2}-e^{2}-f^{2}-ef(ew_{f}-fw_{e})$
$\displaystyle=9e^{2}f^{2}-e^{2}-f^{2}-3efG\quad\text{ by Lemma
\ref{cofactorlemma} (\ref{cofactorlemmai})}$ $\displaystyle=g^{2}\quad\text{
by the Markoff equation}.$

For the second equality:

$\displaystyle(8ew_{e}+ev_{e}-1)f^{2}$
$\displaystyle+(ew_{e}+1)(f^{2}-fw_{f}-1)$
$\displaystyle=9ef^{2}w_{e}+ef^{2}v_{e}-efw_{e}w_{f}-ew_{e}-fw_{f}-1$
$\displaystyle=gw_{g}-1+ef(fv_{e}+3(3fw_{e}-w_{g})-w_{e}w_{f})\quad\text{ by
Lemma \ref{cofactorlemma} (\ref{cofactorlemmaiii})}$
$\displaystyle=gw_{g}-1\quad\text{ by Lemma \ref{cofactorlemma}
(\ref{cofactorlemmav})}.$

∎

By Proposition 7.2, the mutations
$(e,F,g)\longleftarrow(e,g,f)\longrightarrow(g,E,f)$ for $e>1$ and $f>2$
behave as follows:

$[a_{1},\dots,a_{2s}]$, $[c_{1},\dots,c_{2(s+t+1)}]$,
$[b_{1},\dots,b_{2t}]$$[a_{1},\dots,a_{2s}]$$[a_{2s},\dots,a_{1},8,1,c_{2(r+s+1)}-1,c_{2(r+s)+1},\dots,c_{1}]$
$[c_{1},\dots,c_{2(s+t+1)}]$$[c_{1},\dots,c_{2(s+t+1)}]$
$[c_{2(r+s+1)},\dots,c_{1},8,1,b_{2t}-1,b_{2t-1},\dots,b_{1}]$
$[b_{1},\dots,b_{2t}]$

Now it remains to establish the initial data. For this and later use, we
introduce the following notation.

######  Definition 7.3:

Let $a_{0},\dots,a_{m-1}$ be a sequence of integers.

  1. (i)

For $n>0$ we denote $(a_{0},\dots,a_{m-1})_{n}$ the sequence
$b_{0},\dots,b_{n-1}$ such that $b_{k}=a_{k\bmod m}$.

  2. (ii)

For any integer $x$, we denote
$x\overset{k}{\curvearrowright}a_{0},\dots,a_{m-1}=a_{0},\dots,a_{k-2},x,a_{k-1},\dots,a_{m-1}$.

######  Example 7.4:

The expression $[3\overset{5}{\curvearrowright}6,(1,5)_{5},6]$ denotes the
continued fraction $[6,1,5,1,3,5,1,6]=89^{2}/(89\cdot 13-1)$. Similarly, we
write $[6\overset{4}{\curvearrowright}(4,8)_{7}]$ for
$[4,8,4,6,8,4,8,4]=985^{2}/(985\cdot 239-1)$.

######  Proposition 7.5:

  1. (i)

Consider a Markoff triple of the form $(1,F_{2n+1},F_{2n-1})$ for $n\geq 3$
with T-weights $(-1,F_{2n-3},F_{2n-5})$, then

$\frac{F_{2n+1}^{2}}{F_{2n+1}F_{2n-3}-1}=[3\overset{n}{\curvearrowright}6,(1,5)_{2n-5},6]=[3\overset{n}{\curvearrowright}6,(1,5)_{2n-3}].$

  2. (ii)

Consider a Markoff triple of the form $(P_{2n-1},P_{2n+1},2)$ for $n\geq 1$
with T-weights $(S_{2n-3},S_{2n-1},1)$, then

$\frac{P_{2n+1}^{2}}{P_{2n+1}S_{2n-1}-1}=[6\overset{n}{\curvearrowright}(4,8)_{2n-1}].$

###### Proof.

Recall that the T-weights and T-coweights for the Fibonacci and Pell branches
have been described in Lemma 5.6.

(i) For the first triple $(1,13,5)$ with T-weights $(-1,2,1)$ we compute:

$\frac{5^{2}}{5\cdot 1-1}=[6,4],\quad\frac{13^{2}}{13\cdot
2-1}=[6,1,3,6]=[3\overset{3}{\curvearrowright}6,(1,5)_{1},6].$

Now assume for $n>3$ that
$F_{2n-1}^{2}/(F_{2n-1}F_{2n-5}-1)=[a_{1},\dots,a_{2s}]=[3\overset{n-1}{\curvearrowright}6,(1,5)_{2n-7},6]$.
We claim that

$\frac{F_{2n+1}^{2}}{F_{2n+1}F_{2n-3}-1}=[6,1,a_{2s}-1,a_{2s-1},\dots,a_{1}]=[3\overset{n}{\curvearrowright}6,(1,5)_{2n-5},6].$

By the concatenation rule Proposition A.2, it suffices to show

$\displaystyle F_{2n+1}^{2}$
$\displaystyle=K(6)K(1,a_{2s}-1,a_{2s-1},\dots,a_{1})+K(a_{2s}-1,a_{2s-1},\dots,a_{1})$
$\displaystyle F_{2n+1}F_{2n-3}-1$
$\displaystyle=K(1,a_{2s}-1,a_{2s-1},\dots,a_{1})$

for $n>3$. By Lemma 6.3 these are equivalent to

$\displaystyle F_{2n+1}^{2}$
$\displaystyle=6F_{2n-1}^{2}+F_{2n-1}(F_{2n-1}-F_{2n-5})$ $\displaystyle
F_{2n+1}F_{2n-3}-1$ $\displaystyle=F_{2n-1}^{2},$

both of which follow from standard properties of the Fibonacci numbers.

(ii) For the first triple $(5,29,2)$ with T-weights $(1,7,1)$ we compute::

$\frac{5^{2}}{5\cdot
1-1}=[6,4]=[6\overset{1}{\curvearrowright}(4,8)_{1}],\quad\frac{29^{2}}{29\cdot
7-1}=[4,6,8,4]=[6\overset{2}{\curvearrowright}(4,8)_{3}].$

We assume for $n>2$ that
$P_{2n-1}^{2}/(P_{2n-1}S_{2n-3}-1)=[a_{1},\dots,a_{2n-2}]=[6\overset{n-1}{\curvearrowright}(4,8)_{2n-3}]$.
We claim that

$\frac{P_{2n+1}^{2}}{P_{2n+1}S_{2n-1}-1}=[a_{2n-2},\dots,a_{1}]=[6\overset{n}{\curvearrowright}(4,8)_{2n-1}].$

Note that $P_{2n+1}Q_{2n-1}-1$ and $P_{2n+1}S_{2n-1}-1$ are inverse to each
other modulo $P_{2n+1}^{2}$. It therefore is sufficient to show that

$\frac{P_{2n+1}^{2}}{P_{2n+1}Q_{2n-1}-1}=[1,3,8,a_{1},\dots,a_{2n-2}]=[6\overset{n+2}{\curvearrowright}1,3,8,(4,8)_{2n-3}].$

Invoking again the concatenation rule, it suffices to show that

$\displaystyle
P_{2n+1}^{2}=K(1,3,8)K(a_{1},\dots,a_{2n-2})+K(1,3)K(a_{1},\dots,a_{2n-3})=33P_{2n-1}^{2}+4(P_{2n-1}S_{2n-3}-1),$
$\displaystyle
P_{2n+1}Q_{2n-1}-1=K(3,8)K(a_{1},\dots,a_{2n-2})+K(3)K(a_{2},\dots,a_{2n-2})=25P_{2n-1}^{2}+3(P_{2n-1}S_{2n-3}-1)$

for $n>2$. These equations follow from successive application of well-known
relations among the Pell and Pell-Lucas numbers, plenty of which can be found,
e.g., in [Kos14], which we leave as an exercise for the reader. ∎

######  Corollary 7.6:

  1. (i)

Consider a Markoff triple of the form $(F_{2n+1},g,F_{2n-1})$ (with $g=$
$3F_{2n-1}F_{2n+1}-1$), then

$\frac{g^{2}}{gw_{g}-1}=[3\overset{n-1}{\curvearrowright}6,(1,5)_{2n-5},6,8,[3\overset{n-1}{\curvearrowright}(1,5)_{2n-5},6]]=[3\overset{3n-2}{\curvearrowright}3\overset{n-1}{\curvearrowright}6,(1,5)_{2n-5},6,8,(1,5)_{2n-5},6].$

  2. (ii)

Consider a Markoff triple of the form $(P_{2n-1},g,P_{2n+1})$ (with
$g=3P_{2n-1}P_{2n+1}-2$), then

$\frac{g^{2}}{gw_{g}-1}=[6\overset{n}{\curvearrowright}(4,8)_{2n-2},[6\overset{n+2}{\curvearrowright}1,3,(8,4)_{2n-2}]]=[6\overset{3n+1}{\curvearrowright}6\overset{n}{\curvearrowright}(4,8)_{2n-2},1,3,(8,4)_{2n-2}].$

###### Proof.

Follows from Proposition 7.5 and direct computation. ∎

Here some small observations on the structure of the continued fractions:

######  Corollary 7.7:

Let $(e,g,f)$ be a Markoff triple such that
$g^{2}/(gw-1)=[a_{1},\dots,a_{2s}]$ with $a_{1},a_{2s}>1$. Then

  1. (i)

$a_{i}\in\\{1,3,4,5,6,8\\}$ for every $i$.

  2. (ii)

$a_{i}\neq a_{i+1}$ for every $i$.

  3. (iii)

A triple $(e,g,f)$ with $g>13$ is a descendant of the Fibonacci branch iff
$a_{i}=5$ for some $i$ iff $a_{i}\in\\{1,3,5,6,8\\}$ for every $i$.

  4. (iv)

A triple $(e,g,f)$ is a descendant of the Pell branch iff $a_{i}=4$ for some
$i$ iff $a_{i}\in\\{1,3,4,6,8\\}$ for every $i$.

Figure 9 shows the continued fraction expansions of $g/(gw_{g}-1)$ for the
first five levels of the Markoff tree.

$[6,4]$$[6,1,3,6]$$[6,1,5,3,1,6]$$[6,1,5,1,3,5,1,6]$$[6,1,5,1,5,3,1,5,1,6]$$[6,1,5,3,1,5,1,6,8,1,5,1,3,5,1,6]$$[6,1,3,5,1,6,8,1,5,3,1,6]$$[6,1,3,5,1,6,8,1,5,1,3,5,1,8,6,1,5,3,1,6]$$[6,1,3,5,1,8,6,1,5,3,1,6,8,1,5,3,1,6]$$[6,3,1,6,8,1,3,6]$$[6,3,1,6,8,1,5,3,1,8,6,1,3,6]$$[6,3,1,6,8,1,5,3,1,6,8,1,3,5,1,8,6,1,3,6]$$[6,3,1,6,8,1,3,5,1,8,6,1,3,6,8,1,5,3,1,8,6,1,3,6]$$[6,3,1,8,6,1,3,6,8,1,3,6]$$[6,3,1,8,6,1,3,6,8,1,5,3,1,8,6,3,1,6,8,1,3,6]$$[6,3,1,8,6,3,1,6,8,1,3,6,8,1,3,6]$$[4,6,8,4]$$[4,6,8,1,3,8,6,4]$$[4,6,8,1,3,6,8,3,1,8,6,4]$$[4,6,8,1,3,6,8,1,3,8,6,3,1,8,6,4]$$[4,6,8,1,3,8,6,3,1,8,6,4,8,1,3,6,8,3,1,8,6,4]$$[4,6,8,3,1,8,6,4,8,1,3,8,6,4]$$[4,6,8,3,1,8,6,4,8,1,3,6,8,3,1,8,4,6,8,1,3,8,6,4]$$[4,6,8,3,1,8,4,6,8,1,3,8,6,4,8,1,3,8,6,4]$$[4,8,6,4,8,4]$$[4,8,6,4,8,1,3,8,4,6,8,4]$$[4,8,6,4,8,1,3,8,6,4,8,3,1,8,4,6,8,4]$$[4,8,6,4,8,3,1,8,4,6,8,4,8,1,3,8,4,6,8,4]$$[4,8,4,6,8,4,8,4]$$[4,8,4,6,8,4,8,1,3,8,4,8,6,4,8,4]$$[4,8,4,8,6,4,8,4,8,4]$
Figure 9. Continued fraction expansions of $g^{2}/(gw_{g}-1)$ for the first
four levels of the Markoff tree.
$[3]$$[1,5]$$[3,1,5]$$[1,5,1,5]$$[3,1,5,1,5]$$[6,1,5,1,3,5,1,5]$$[6,1,5,3,1,5]$$[1,5,1,8,6,1,5,3,1,5]$$[3,1,6,8,1,5,3,1,5]$$[6,1,3,5]$$[3,1,8,6,1,3,5]$$[6,1,3,5,1,8,6,1,3,5]$$[1,6,8,1,5,3,1,8,6,1,3,5]$$[1,6,8,1,3,5]$$[3,1,8,6,3,1,6,8,1,3,5]$$[6,1,3,6,8,1,3,5]$$[6,3]$$[1,8,6,3]$$[6,3,1,8,6,3]$$[1,8,6,3,1,8,6,3]$$[4,8,1,3,6,8,3,1,8,6,3]$$[4,8,1,3,8,6,3]$$[6,3,1,8,4,6,8,1,3,8,6,3]$$[1,8,6,4,8,1,3,8,6,3]$$[4,8,3]$$[1,8,4,6,8,3]$$[4,8,3,1,8,4,6,8,3]$$[6,4,8,1,3,8,4,6,8,3]$$[6,4,8,3]$$[1,8,4,8,6,4,8,3]$$[4,8,4,8,3]$
Figure 10. The LEs for minimal resolutions of $\frac{1}{g^{2}}(1,gw_{g}-1)$
for the first four levels of the Markoff tree.

We can now give a complete description of the LEs correponding to Markoff
triples.

######  Theorem 7.8:

Let $(e,g,f)$ be a regular Markoff triple. Then the LE of the minimal
resolution of the corresponding T-singularity $\frac{1}{g^{2}}(1,gw_{g}-1)$ is
given by

$\displaystyle
a_{s}-2,a_{s-1},\dots,a_{2},a_{1}-1=a_{s+1},\dots,a_{2s-1},a_{2s}-1$
$\displaystyle\text{ if $s$ is odd},$ $\displaystyle
a_{s},\dots,a_{2},a_{1}-1=a_{s+1}+2,a_{s+2},\dots,a_{2s-1},a_{2s}-1$
$\displaystyle\text{ if $s$ is even}.$

where $[a_{1},\dots,a_{2s}]=g^{2}/(gw_{g}-1)$ with $a_{1},a_{2s}>1$

###### Proof.

This is a direct consequence of Corollary 6.2 and Proposition 6.4. ∎

For the Fibonacci and Pell branches, we obtain the following direct
descriptions:

######  Lemma 7.9:

  1. (i)

Consider a Markoff triple of the form $(1,F_{2n+1},F_{2n-1})$ for $n\geq 3$
with weights $(-1,F_{2n-3},F_{2n-5})$. Then the LE of
$F_{2n+1}^{2}/(F_{2n+1}F_{2n-3}-1)$ is given by

$\displaystyle 3,(1,5)_{n-2}$ $\displaystyle\text{ if $n$ is even},$
$\displaystyle(1,5)_{n-1}$ $\displaystyle\text{ if $n$ is odd}.$

  2. (ii)

Consider a Markoff triple of the form $(P_{2n-1},P_{2n+1},2)$ for $n\geq 2$
with weights $(Q_{2n-3},Q_{2n-1},1)$. Then the LE of
$P_{2n+1}^{2}/(P_{2n+1}Q_{2n-1}-1)$ is given by

$\displaystyle 6,(4,8)_{n-2},3$ $\displaystyle\text{ if $n$ is even},$
$\displaystyle(4,8)_{n-1},3$ $\displaystyle\text{ if $n$ is odd}.$

Figure 10 shows the LEs for the first four levels of the Markov trees. The
following Corollary shows that the LEs at least restore some of the symmetry
that we have observed for Frobenius’ description of continued fractions in
Section 3.

######  Corollary 7.10:

Similar as in 3.9, we denote $(g_{\mu/\nu},w_{g_{\mu/\nu}})$ and
$(g_{\nu/\mu},w_{g_{\nu/\mu}})$ two Markoff numbers $>5$ that sit at opposite
positions in the Markoff tree together with their T-weights, and let
$[a_{1},\dots,a_{s}]$ and $[b_{1},\dots,b_{t}]$ be their corresponding LEs.
Then:

$s=t\quad\text{ and }\quad[a_{1}+b_{1},\dots a_{s}+b_{s}]=[7,9,\dots,9,8].$

To conclude, we have now a look at fractions $w_{g}/g$ and their associated
Cantor set. As in Section 4 we denote $p$ a finite path and $(e,g,f)$ the
Markoff triple corresponding to that path. Denote $[a_{1},\dots,a_{2s}]$ the
continued fraction expansion of $g^{2}/(gw_{g}-1)$. Then by the mutation rules
and induction we get for the paths $p_{1}=pL\bar{R}$ and $p_{2}=pR\bar{L}$:

$\displaystyle\mathfrak{l}_{p_{1}}$
$\displaystyle=[\overline{a_{1},\dots,a_{2s-1},a_{2s}-1,1,8}],$
$\displaystyle\mathfrak{l}_{p_{2}}$
$\displaystyle=1+[\overline{a_{2s}-1,a_{2s-1},\dots,a_{1},8,1}],$

where the $\mathfrak{l}_{i}$ denote the limits of the fractions $g/w_{g}$
along the paths $p_{i}$. By direct computation we obtain:

$\displaystyle\mathfrak{l}_{p_{1}}$
$\displaystyle=\frac{1}{2}\frac{9g-2w_{g}+3\sqrt{9g^{2}-4}}{9w_{g}-v_{g}},$
$\displaystyle\mathfrak{l}_{p_{2}}$
$\displaystyle=\frac{1}{2}\frac{9g+2w_{g}+3\sqrt{9g^{2}-4}}{9w_{g}+v_{g}}.$

Figure 11 shows the purely periodic values $\mathfrak{l}_{p_{1}}$ and
$\mathfrak{l}_{p_{2}}-1$ together with their periods for the first four levels
of the Markoff tree. We may compare this spectrum that of Section 4, in
particular Figure 7 and observe the symmetry between opposite positions of the
tree such that the periods pointwise add up to 9.

$\frac{1}{2}\frac{16+3\sqrt{32}}{4}$, $4,8$$\frac{1}{2}(5+3\sqrt{5})$,
$5,1$$\frac{1}{2}\frac{43+3\sqrt{221}}{7}$$6,3,1,8$$\frac{1}{2}\frac{113+3\sqrt{1517}}{17}$$6,1,3,5,1,8$$\frac{1}{2}\frac{296+3\sqrt{10400}}{44}$$6,1,5,3,1,5,1,8$$\frac{1}{2}\frac{775+3\sqrt{71285}}{115}$$6,1,5,1,3,5,1,5,1,8$$\frac{1}{2}\frac{589+3\sqrt{71285}}{119}$$5,1,5,3,1,5,1,6,8,1$$\frac{1}{2}\frac{224+3\sqrt{10400}}{46}$$5,1,3,5,1,6,8,1$$\frac{1}{2}\frac{11533+3\sqrt{15800621}}{1735}$$6,1,3,5,1,6,8,1,5,3,1,5,1,8$$\frac{1}{2}\frac{8731+3\sqrt{15800621}}{1793}$$5,1,3,5,1,8,6,1,5,3,1,6,8,1$$\frac{1}{2}\frac{83+3\sqrt{1517}}{19}$$5,3,1,6,8,1$$\frac{1}{2}\frac{1684+3\sqrt{338720}}{274}$$6,3,1,6,8,1,3,5,1,8$$\frac{1}{2}\frac{65633+3\sqrt{514518485}}{10679}$$6,3,1,6,8,1,5,3,1,8,6,1,3,5,1,8$$\frac{1}{2}\frac{48335+3\sqrt{514518485}}{11065}$$5,3,1,6,8,1,3,5,1,8,6,1,3,6,8,1$$\frac{1}{2}\frac{1240+3\sqrt{338720}}{284}$$5,3,1,8,6,1,3,6,8,1$$\frac{1}{2}\frac{25147+3\sqrt{75533477}}{4093}$$6,3,1,8,6,1,3,6,8,1,3,5,1,8$$\frac{1}{2}\frac{18517+3\sqrt{75533477}}{4241}$$5,3,1,8,6,3,1,6,8,1,3,6,8,1$$\frac{1}{2}\frac{25+3\sqrt{221}}{11}$$3,6,8,1$$\frac{1}{2}\frac{247+3\sqrt{7565}}{61}$$4,6,8,3,1,8$$\frac{1}{2}\frac{3689+3\sqrt{1687397}}{911}$$4,6,8,1,3,8,6,3,1,8$$\frac{1}{2}\frac{55088+3\sqrt{376282400}}{13604}$$4,6,8,1,3,6,8,3,1,8,6,3,1,8$$\frac{1}{2}\frac{32600+3\sqrt{376282400}}{14350}$$3,6,8,1,3,8,6,3,1,8,6,4,8,1$$\frac{1}{2}\frac{2183+3\sqrt{1687397}}{961}$$3,6,8,3,1,8,6,4,8,1$$\frac{1}{2}\frac{320900+3\sqrt{12768548000}}{79250}$$4,6,8,3,1,8,6,4,8,1,3,8,6,3,1,8$$\frac{1}{2}\frac{189896+3\sqrt{12768548000}}{83596}$$3,6,8,3,1,8,4,6,8,1,3,8,6,4,8,1$$\frac{1}{2}\frac{145+3\sqrt{7565}}{65}$$3,8,6,4,8,1$$\frac{1}{2}\frac{1439+3\sqrt{257045}}{359}$$4,8,6,4,8,3,1,8$$\frac{1}{2}\frac{125177+3\sqrt{1945074605}}{31229}$$4,8,6,4,8,1,3,8,4,6,8,3,1,8$$\frac{1}{2}\frac{73523+3\sqrt{1945074605}}{32959}$$3,8,6,4,8,3,1,8,4,6,8,4,8,1$$\frac{1}{2}\frac{845+3\sqrt{257045}}{379}$$3,8,4,6,8,4,8,1$$\frac{1}{2}\frac{8387+3\sqrt{8732021}}{2093}$$4,8,4,6,8,4,8,3,1,8$$\frac{1}{2}\frac{4925+3\sqrt{8732021}}{2209}$$3,8,4,8,6,4,8,4,8,1$
Figure 11. The quadratic T-spectrum

Proceeding as in Section 4, we set

$A_{g}:=1/\mathfrak{l}_{p_{1}},\quad B_{g}:=1/\mathfrak{l}_{p_{2}},\quad
I_{g}=[B_{e},A_{f}],\quad J_{g}:=(A_{g},B_{g}).$

This way, obtain a Cantor set $\mathfrak{T}$ by using the intervals $I_{g}$,
$J_{g}$, where

$\displaystyle|I_{g}|$ $\displaystyle=\frac{3}{ef}(\Delta_{e,f}-g),$
$\displaystyle|J_{g}|$ $\displaystyle=\frac{3}{g}(3g-\sqrt{\Delta_{g}}).$

######  Theorem 7.11:

$\mathfrak{T}$ is a Cantor set which is given by the complement of the open
intervals

$\left(\frac{w_{g}}{g}-\frac{3}{2}\left(3-\sqrt{9-\frac{4}{g^{2}}}\right),\frac{w_{g}}{g}+\frac{3}{2}\left(3-\sqrt{9-\frac{4}{g^{2}}}\right)\right)$

within the closed interval
$[(7-3\sqrt{5})/2,3\sqrt{2}-4]\approx[0.145898,0.242641]$, where $g$ runs over
the maximal elements of all regular Markoff triples. Lebesgue measure and
Hausdorff dimension of $\mathfrak{T}$ both are zero.

###### Proof.

It remains to show the assertions about measure and dimension. For this, we
just observe that by Corollary 5.4, $\mathfrak{T}$ is an affine transformation
of $\mathfrak{M}$, given by $\mathfrak{T}=3\mathfrak{M}-1$. Then we conclude
with Theorems 4.9 and 4.12. ∎

##  Appendix A Continuants

In this appendix we state some general facts about continuants and continued
fractions. For detailed expositions we refer to [GKP94, §6.7] and [Per54, Ch I
— III]. Any sequence of real numbers $a_{1},\dots,a_{n}$ such that the
continued fraction
$[a_{1},\dots,a_{t}]=a_{1}+1/(a_{2}+\cdots(a_{n-1}+1/a_{n})\cdots)$ exists can
be written as a quotient

$\frac{K_{n}(a_{1},\dots,a_{n})}{K_{n-1}(a_{2},\dots,a_{n})}$

where for any $t\geq 0$ and indeterminates $x_{1},\dots,x_{t}$,
$K_{t}(x_{1},\dots,x_{t})$ is the $t$-th continuant polynomial. These can
inductively be defined as follows:

$\displaystyle K_{0}$ $\displaystyle=1,$ $\displaystyle K_{1}(x_{1})$
$\displaystyle=x_{1},$ $\displaystyle K_{i}(x_{1},\dots,x_{i})$
$\displaystyle=x_{i}K_{i-1}(x_{1},\dots,x_{i-1})+K_{i-2}(x_{1},\dots,x_{i-2})\text{
for }i>1.$

For $t\leq 5$, we therefore get:

$\displaystyle K_{2}(x_{1},x_{2})$ $\displaystyle=x_{1}x_{2}+1,$
$\displaystyle K_{3}(x_{1},x_{2},x_{3})$
$\displaystyle=x_{1}x_{2}x_{3}+x_{1}+x_{3},$ $\displaystyle
K_{4}(x_{1},x_{2},x_{3},x_{4})$
$\displaystyle=x_{1}x_{2}x_{3}x_{4}+x_{1}x_{2}+x_{1}x_{4}+x_{3}x_{4}+1,$
$\displaystyle K_{5}(x_{1},x_{2},x_{3},x_{4},x_{5})$
$\displaystyle=x_{1}x_{2}x_{3}x_{4}x_{5}+x_{1}x_{2}x_{3}+x_{1}x_{2}x_{5}+x_{1}x_{4}x_{5}+x_{3}x_{4}x_{5}+x_{1}+x_{3}+x_{5}.$

Euler’s rule states that in order to determine the monomial terms of
$K_{t}(x_{1},\dots,x_{t})$, one starts with $x_{1}\cdots x_{t}$ and adds all
possible monomials obtained from successively dividing $x_{1}\cdots x_{t}$ by
factors of the form $x_{i}x_{i+1}$. This description immediately implies that
continuants are invariant under reversing the order of their arguments:

$K_{t}(x_{1},\dots,x_{t})=K_{t}(x_{t},\dots,x_{1}).$

For $s,t\geq 0$ the following factorization rule holds:

$\displaystyle
K_{s+t}(x_{1},\dots,x_{s},x_{s+1},\dots,x_{s+t})=K_{s}(x_{1},\dots,x_{s})K_{t}(x_{s+1},\dots,x_{s+t})$
(A-1)
$\displaystyle+K_{s-1}(x_{1},\dots,x_{s-1})K_{t-1}(x_{s+2},\dots,x_{s+t}).$

For any $1\leq k\leq t$ we moreover have Cassini’s identity:

$\displaystyle
K_{s+t}(x_{1},\dots,x_{s+t})K_{k}(x_{s+1},\dots,x_{s+k})=K_{s+k}(x_{1},\dots,x_{s+k})K_{t}(x_{s+1},\dots,x_{s+t})$
(A-2)
$\displaystyle+(-1)^{k}K_{s-1}(x_{1},\dots,x_{s-1})K_{t-k-1}(x_{s+k+2},\dots,x_{s+t}).$

For the particular case $s=1$, $k=t-1$, Cassini’s identity yields:

(A-3)
$K_{t+1}(x_{1},\dots,x_{t+1})K_{t-1}(x_{2},\dots,x_{t})=K_{t}(x_{1},\dots,x_{t})K_{t}(x_{2},\dots,x_{t+1})+(-1)^{t-1}.$

Any rational number admits a finite continued fraction expansion
$[a_{1},\dots,a_{n}]$ with integers $a_{1},\dots,a_{n}$, where $a_{i}>0$ for
$i>1$. This representation is almost unique with the exception that we can
absorb the last element if it is equal to $1$, i.e.
$[a_{1},\dots,a_{n},1]=[a_{1},\dots,a_{n-1},a_{n}+1]$. In the case $a_{1}>0$
we can use Cassini’s identity to describe the relation between reverse
continued fractions: let
$[a_{1},\dots,a_{n}]=K_{n}(a_{1},\dots,a_{n})/K_{n-1}(a_{2},\dots,a_{n})=p/q$,
then $p>q$ and:

(A-4)
$[a_{n},\dots,a_{1}]=\frac{K(a_{n},\dots,a_{1})}{K(a_{n-1},\dots,a_{1})}=\frac{K(a_{1},\dots,a_{n})}{K(a_{1},\dots,a_{n-1})}=\frac{p}{q^{\prime}},$

where $0<q^{\prime}<p$ such that $qq^{\prime}\equiv(-1)^{n-1}\mod p$ (see
[Per54, §11]). In particular, if $n$ is odd then $q^{\prime}\equiv q^{-1}\mod
p$. For lighter notation we will drop the indices of the $K$’s from now on.
The following lemma gives us some control over the term $(-1)^{n-1}$.

######  Lemma A.1:

Let $0<q<p$ be coprime integers. If $q>q-p$ and $p/(p-q)=[a_{1},\dots,a_{n}]$
then $p/q=[1,a_{1}-1,a_{2},\dots,a_{n}]$. Conversely, if
$p/q=[1,a_{1},\dots,a_{n}]$ then $q>p-q$ and
$p/(p-q)=[a_{1}+1,a_{2},\dots,a_{n}]$.

###### Proof.

The assertions follow from

$\frac{p}{q}=1+\cfrac{1}{\frac{q}{p-q}}\quad\text{ and
}\quad\left\lfloor\frac{q}{p-q}\right\rfloor=\left\lfloor\frac{p}{p-q}\right\rfloor-1$

for $q>p-q$. ∎

Assume now that we have integers $a_{1},\dots,a_{m}$, $b_{1},\dots,b_{n}$
corresponding to continued fractions

$\frac{P}{Q}=[a_{1},\dots,a_{m}],\quad\frac{U}{V}=[b_{1},\dots,b_{n}].$

By the factorization rule (A-1), we have

$\displaystyle K(a_{1},\dots,a_{m},b_{1},\dots,b_{n})$
$\displaystyle=K(a_{1},\dots,a_{m})K(b_{1},\dots,b_{n})+K(a_{1},\dots,a_{m-1})K(b_{2},\dots,b_{n}),$
$\displaystyle K(a_{2},\dots,a_{m},b_{1},\dots,b_{n})$
$\displaystyle=K(a_{2},\dots,a_{m})K(b_{1},\dots,b_{n})+K(a_{2},\dots,a_{m-1})K(b_{2},\dots,b_{n}).$

By the following criterion it is enough to show that these factorizations are
sufficient in order to construct the concatenation of continued fractions.

######  Proposition A.2 (Concatenation Rule):

Let $P,Q,U,V,a_{i},b_{j}$ be as before and denote

$X=PU+Q^{\prime}V,\quad Y=QU+Q^{\prime\prime}V,$

where $Q^{\prime}=K(a_{1},\dots,a_{m-1})$ and
$Q^{\prime\prime}=K(a_{2},\dots,a_{m-1})$. Then:

$\frac{X}{Y}=[a_{1},\dots,a_{m},b_{1},\dots,b_{n}].$

###### Proof.

We write

$[a_{1},\dots,a_{m},b_{1},\dots,b_{n}]=[a_{1},\dots,a_{m},\frac{U}{V}]=\frac{K(a_{1},\dots,a_{m},\frac{U}{V})}{K(a_{2},\dots,a_{m},\frac{U}{V})}=\frac{\frac{U}{V}K(a_{1},\dots,a_{m})+K(a_{1},\dots,a_{m-1})}{\frac{U}{V}K(a_{2},\dots,a_{m})+K(a_{2},\dots,a_{m-1})}.$

Rearranging and comparing terms then yields the assertion. ∎

By a theorem of Lagrange, an irrational number $\xi$ is a quadratic irrational
if and only if its corresponding infinite continued fraction expansion becomes
periodic, i.e. we have $\xi=a+b\sqrt{D}\in\mathbb{Q}[\sqrt{D}]$ for some non-
square integer $D$ if and only if

$\xi=[a_{1},\dots,a_{k},\overline{a_{k+1},\dots,a_{n}}]$

for integers $a_{1},\dots,a_{n}$. A quadratic irrational $\xi$ has a purely
periodic continued fraction expansion $\xi=[\overline{a_{1},\dots,a_{n}}]$ if
and only if it is reduced, i.e. if $\xi>1$ and $-1<\bar{\xi}<0$. Here,
$\bar{\xi}=a-b\sqrt{D}$ is the conjugate of $\xi$. From now we assume that
$\xi=[\overline{a_{1},\dots,a_{n}}]$ is a reduced quadratic irrational number.
Then, following [Per54, §22], we abbreviate for $1\leq i\leq n$:

$\displaystyle A_{i}=K(a_{1},\dots,a_{i}),\quad B_{i}=K(a_{2},\dots,a_{i})$

and get:

$\displaystyle\xi=[a_{1},\dots,a_{n},\xi]=\frac{K(a_{1},\dots,a_{n},\xi)}{K(a_{2},\dots,a_{n},\xi)}=\frac{\xi
A_{n}+A_{n-1}}{\xi B_{n}+B_{n-1}}.$

Solving the resulting quadratic equation $B_{n}\xi^{2}+(B_{n-1}-A_{n})\xi-
A_{n-1}=0$ yields:

(A-5)
$\xi=\frac{1}{2B_{n}}\left(A_{n}-B_{n-1}+\sqrt{(A_{n}-B_{n-1})^{2}+4A_{n-1}B_{n}}\right).$

Consider now a cyclic shift of $\xi$, i.e. let

$\sigma=[\overline{a_{k+1},\dots,a_{n},a_{1},\dots,a_{k}}].$

Then by the relation $\xi=[a_{1},\dots,a_{k},\sigma]$ and the defining
relations for the continuants we get:

$\xi=\frac{K(a_{1},\dots,a_{k},\sigma)}{K(a_{2},\dots,a_{k},\sigma)}=\frac{\sigma
A_{k}+A_{k-1}}{\sigma B_{k}+B_{k-1}}$

and by resolving for $\sigma$ we obtain:

(A-6)
$\sigma=[\overline{a_{k+1},\dots,a_{n},a_{1},\dots,a_{k}}]=\frac{A_{k-1}-\xi
B_{k-1}}{\xi B_{k}-A_{k}}.$

Alternatively, we can write $\sigma=[a_{k+1},\dots,a_{n},\xi]$ and therefore
obtain

(A-7)
$\sigma=[\overline{a_{k+1},\dots,a_{n},a_{1},\dots,a_{k}}]=\frac{K(a_{k+1},\dots,a_{n},\xi)}{K(a_{k+2},\dots,a_{n},\xi)}.$

The reverse period of $\xi$ corresponds to
$\eta:=-1/\bar{\xi}=[\overline{a_{n},\dots,a_{1}}]$ (see [Per54, §23]). Then
for $\tau=[\overline{a_{k},\dots,a_{1},a_{n},\dots,a_{k+1}}]$ we obtain with
(A-6):

(A-8)
$\tau=[\overline{a_{k},\dots,a_{1},a_{n},\dots,a_{k+1}}]=\frac{K(a_{k+2},\dots,a_{n})-\eta
K(a_{k+2},\dots,a_{n-1})}{\eta K(a_{k+1},\dots
a_{n-1})-K(a_{k+1},\dots,a_{n})}=-\frac{K(a_{k+2},\dots,a_{n},\bar{\xi})}{K(a_{k+1},\dots,a_{n},\bar{\xi})}.$

##  Appendix B T-continuants

We introduce a variant of continuants that are related to Hirzebruch-Jung
continued fractions of the type of T-singularities of Section 6. We start by
defining the following sets

$\displaystyle J_{0}$ $\displaystyle=\\{\emptyset\\},$ $\displaystyle J_{1}$
$\displaystyle=\\{\emptyset\\},$ $\displaystyle J_{m}$
$\displaystyle=J_{m-2}\cup\\{X\cup\\{m-1\\}\mid X\in J_{m-1}\\}\text{ for
}m>1,$ $\displaystyle I_{m}$ $\displaystyle=J_{m}\sqcup J_{m+1}\text{ for
}m\geq 0.$

By construction, the cardinality of $J_{m}$ equals the ${m+1}$-st Fibonacci
number $F_{m+1}$ and the cardinality of $I_{m}$ is $F_{m+3}$. Note that
$J_{m}$ and $J_{m+1}$ are are almost disjoint sets, having only the empty set
in common; the disjoint union $I_{m}=J_{m}\sqcup J_{m+1}$ therefore contains
$\emptyset$ twice.

The sets $J_{m}$ and $I_{m}$ can also be nicely described in a non-inductive
way. For this, we consider $1<\cdots<m$ as an ordered set and a subsequence
$i_{1}<i_{2}<\cdots<i_{k}$ of $1<\cdots<m$ is called even-odd if its elements
are alternatingly even or odd (i.e. $0$ or $1\mod$ 2; the smallest element can
be either even or odd — we do not mean to distinguish between even-odd and
odd-even sequences). More precisely, $i_{j}\equiv i_{j+2}\mod 2$ for every
$1\leq j<k-1$ and, if $k>1$, then $i_{1}\not\equiv i_{2}\mod 2$.

######  Lemma B.1:

The non-empty sets in $J_{m+1}$ and $I_{m}$ are even-odd subsequences of
$1,\dots,m$, where:

  1. (i)

$i_{1}<\cdots<i_{k}$ is in $J_{m+1}$ iff $i_{1}\equiv k+m+1\mod 2$.

  2. (ii)

$I_{m}$ contains all even-odd subsequences of $\\{1,\dots,m\\}$.

###### Proof.

(i) We do induction by $m$. The statement is obviously true for $m=0$. For
$m>0$, we have by definition $J_{m+1}=J_{m-1}\cup\\{X\cup\\{m\\}\mid X\in
J_{m})\\}$. Consider any even-odd subsequence $i_{1}<\cdots<i_{k}$ of
$\\{1,\dots,m\\}$. If the sequence is contained in $\\{1,\dots,m-2\\}$ then it
is in $J_{m-1}$ (and therefore in $J_{m+1}$) iff $i_{1}\equiv k+m-1\equiv
k+m+1\mod 2$. If the sequence is contained in $J_{m}$, then it follows
immediately from the induction hypothesis that $i_{1}\equiv k+m\mod 2$ and
$i_{k}\not\equiv m\mod 2$, hence $i_{1}<\cdots<i_{k}<m$ is even-odd and
$i_{1}\equiv(k+1)+m+1\mod 2$. Hence, any subset of $J_{m+1}$ is an even-odd
sequence and satisfies $i_{1}\equiv k+m+1\mod 2$. Now, if $i_{1}<\cdots<i_{k}$
has these properties and is not contained in $J_{m-1}$ then, again, we use
$i_{k}\not\equiv m\mod 2$ and therefore $i_{k}=m$. Invoking again the
induction hypothesis, we get that $i_{1}<\cdots<i_{k-1}$ is contained in
$J_{m}$ and the assertion follows for $m$.

(ii) Follows immediately from (i). ∎

######  Definition B.2:

Let $m\geq 0$ and $x_{1},\dots,x_{m}$ indeterminates. Then we define the
$m$-th T-continuant $T_{m}$ and the $m$-th T-semicontinuant $S_{m}$ as:

$T_{m}=T_{m}(x_{1},\dots,x_{m})=\sum_{I\in I_{m}}x_{I},\quad
S_{m}=S_{m}(x_{1},\dots,x_{m})=\sum_{J\in J_{m}}x_{J},$

where for any subset $I$ of $\\{1,\dots,m\\}$ we write $x_{I}$ for the
monomial $\prod_{i\in I}x_{i}$ and in particular, $x_{\emptyset}=1$.

Here are the first few polynomials $T_{m}$ and $S_{m}$:

$\displaystyle S_{0}$ $\displaystyle=1,$ $\displaystyle T_{0}$
$\displaystyle=2,$ $\displaystyle S_{1}$ $\displaystyle=1,$ $\displaystyle
T_{1}$ $\displaystyle=2+x_{1},$ $\displaystyle S_{2}$ $\displaystyle=1+x_{1},$
$\displaystyle T_{2}$ $\displaystyle=2+x_{1}+x_{2}+x_{1}x_{2},$ $\displaystyle
S_{3}$ $\displaystyle=1+x_{2}+x_{1}x_{2},$ $\displaystyle T_{3}$
$\displaystyle=2+x_{1}+x_{2}+x_{3}+x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{2}x_{3},$
$\displaystyle S_{4}=$ $\displaystyle\
1+x_{1}+x_{3}+x_{2}x_{3}+x_{1}x_{2}x_{3},$ $\displaystyle T_{4}=$
$\displaystyle\
2+x_{1}+x_{2}+x_{3}+x_{4}+x_{1}x_{2}+x_{1}x_{4}+x_{2}x_{3}+x_{3}x_{4}+x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{1}x_{2}x_{3}x_{4},$
$\displaystyle S_{5}=$ $\displaystyle\
1+x_{2}+x_{4}+x_{1}x_{2}+x_{1}x_{4}+x_{3}x_{4}+x_{2}x_{3}x_{4}+x_{1}x_{2}x_{3}x_{4},$
$\displaystyle T_{5}=$ $\displaystyle\
2+x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{1}x_{2}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{5}+x_{3}x_{4}+x_{4}x_{5},$
$\displaystyle+x_{1}x_{2}x_{3}+x_{1}x_{2}x_{5}+x_{1}x_{4}x_{5}+x_{2}x_{3}x_{4}+x_{3}x_{4}x_{5}+x_{1}x_{2}x_{3}x_{4}+x_{2}x_{3}x_{4}x_{5}+x_{1}x_{2}x_{3}x_{4}x_{5}.$

We have seen in Corollary 6.2 that given the LE $c_{1},\dots,c_{m}$ of a pair
$(e,k)$, we have:

$\frac{e}{k}=[c_{m}+1,c_{m-1},\dots,c_{2},c_{1}+1]=\frac{K_{m}(c_{m}+1,c_{m-1},\dots,c_{2},c_{1}+1)}{K_{m-1}(c_{m-1},\dots,c_{2},c_{1}+1)}.$

In the following Lemma we will give a similar representation in terms of
T-(semi-)continuants and explain the relation between continuants and
T-continuants.

######  Lemma B.3:

Let $c_{1},\dots,c_{m}$ be the LE of the pair $(e,k)$ with $k\leq e-k$. Then:

  1. (i)

$k=S_{m}(c_{1},\dots,c_{m-1})$ and
$e=T_{m}(c_{1},\dots,c_{m})=S_{m}(c_{1},\dots,c_{m-1})+S_{m+1}(c_{1},\dots,c_{m})$.

  2. (ii)

For $m>1$ it follows that
$T_{m}(c_{1},\dots,c_{m})=K_{m}(c_{m}+1,c_{m-1},\dots,c_{2},c_{1}+1)$ and
$S_{m}=$ $K_{m-1}(c_{m-1},$ $\dots,c_{2},c_{1}+1)$.

  3. (iii)

$T_{m}(c_{1},\dots,c_{m})=T_{m}(c_{m},\dots,c_{1})$.

  4. (iv)

$c_{m},\dots,c_{1}$ is the LE of $(e,k^{\prime})$, where
$k^{\prime}=S_{m}(c_{m},\dots,c_{2})$ and moreover
$kk^{\prime}\equiv(-1)^{m+1}\mod e$.

###### Proof.

(i) Starting with $c_{1}+2=T_{1}(c_{1})$ and $1=S_{1}$, the assertion follows
from Lemma 6.1 by induction.

(ii) Follows from (i) and the Corollary 6.2.

(iii) Follows from (ii) and the corresponding symmetry of continuants.
Alternatively, one can conclude form Lemma B.1 (ii) that the sets $I_{m}$ are
invariant when reversing the ordering of the set $\\{1,\dots,m\\}$.

(iv) Follows from (ii) and the corresponding property of continuants. ∎

By Lemma B.3, the standard identities of continuants such as Cassini’s carry
over in a natural way to T-continuants. The following Lemma adds to this some
additional relations between T-continuants and T-semicontinuants.

######  Lemma B.4:

  1. (i)

Let $m,n>0$, $c_{1},\dots,c_{m+n}>0$, then

$T_{m+n}(c_{1},\dots,c_{m+n})=S_{m+1}(c_{1},\dots,c_{m})S_{n+1}(c_{m+n},\dots,c_{m+1})+S_{m}(c_{1},\dots,c_{m-1})S_{n}(c_{m+n},\dots,c_{m+2}).$

  2. (ii)

Let $m>0$, $c_{1},\dots,c_{m}>0$, $1\leq i\leq m$, and $c_{i}+d>0$, then:

$T_{m}(c_{1},\dots,c_{i-i},c_{i}+d,c_{i+1},\dots,c_{m})=T_{m}(c_{1},\dots,c_{m})+dS_{i}(c_{1},\dots,c_{i-1})S_{m-i+1}(c_{m},\dots,c_{i+1}),$

in particular:

$\displaystyle T_{m}(c_{1},\dots,c_{m-1},c_{m}+d)$
$\displaystyle=T_{m}(c_{1},\dots,c_{m})+dS_{m}(c_{1},\dots,c_{m-1})$
$\displaystyle T_{m}(c_{1}+d,c_{2},\dots,c_{m})$
$\displaystyle=T_{m}(c_{1},\dots,c_{m})+dS_{m}(c_{m},\dots,c_{2}).$

  3. (iii)

$\displaystyle T_{m}(c_{1},\dots,c_{i-1},c_{i}+2,c_{i+1},\dots,c_{m})=$
$\displaystyle\ S_{m-i+1}(c_{m},\dots,c_{i+1})T_{i}(c_{1},\dots,c_{i})$
$\displaystyle+S_{i}(c_{1},\dots,c_{i-1})T_{m-i}(c_{i+1},\dots,c_{m}).$

  4. (iv)

Let $m\geq i>0$, $c_{1},\dots,c_{m-1}>0$. Then

$S_{m}(c_{1},\dots,c_{m-1})+(-1)^{i}S_{m-i+1}(c_{1},\dots,c_{m-i})=\sum_{j=1}^{i-1}(-1)^{j}T_{m-j}(c_{1},\dots,c_{m-j})$

and in particular, for $i=m$:

$S_{m}(c_{1},\dots,c_{m-1})=\sum_{j=1}^{m-1}(-1)^{j}T_{m-j}(c_{1},\dots,c_{m-j})-(-1)^{m}.$

  5. (v)

$T(c_{1},\dots,c_{m})-T(c_{1},\dots,c_{i})=\sum_{j=i+1}^{m}c_{j}S_{j}(c_{1},\dots,c_{j-1}).$

###### Proof.

(i) Follows by induction with the formulas $S_{i}=S_{i-2}+c_{i-1}S_{i-1}$ and
$T_{i}=S_{i}+S_{i+1}$ for any $i$.

(ii) It follows immediately from the definition that
$S_{m+1}(c_{1},\dots,c_{m-1},c_{m}+d)=S_{m+1}(c_{1},\dots,c_{m})+dS_{m}(c_{1},\dots,c_{m-1})$.
With this the assertion follows by direct computation from (i).

(iii) Direct computation using (ii) and B.3 (iii).

(iv) Induction using $S_{m}=T_{m-1}-S_{m-1}$. ∎

##  Appendix C Local Hausdorff dimension of Cantor sets

In this appendix we collect some material on Hausdorff dimension of Cantor
sets that will be needed in Section 4 (for general reference on Hausdorff
dimension, we refer to [Fal97], but any source on the subject will suffice).
In particular, we will introduce the local Hausdorff dimension, where we
follow some ideas of Baek (see [Bae01], [Bae04]). For our purposes, we will
present a modified version of arguments in [Bae04], in order to allow for more
convenient boundedness conditions.

Recall that for any $s\geq 0$ the Hausdorff measure of a set
$\mathfrak{S}\subseteq\mathbb{R}^{n}$ is defined as

$\mathcal{H}^{s}(\mathfrak{S}):=\underset{\delta\rightarrow
0}{\lim}\left(\inf\\{\sum_{i=1}^{\infty}|B_{i}|^{s}\mid\\{B_{i}\\}_{i=0}^{\infty}\text{
is a $\delta$-cover}\\}\right),$

where a $\delta$-cover is a countable collection of sets $B_{i}$ with
diameters $0<|B_{i}|\leq\delta$ that cover $\mathfrak{S}$. The Hausdorff
dimension of $\mathfrak{S}$ is defined as

$\dim_{H}\mathfrak{S}:=\underset{s\geq
0}{\inf}\\{\mathcal{H}^{s}(\mathfrak{S})=0\\}.$

Now let $\mathfrak{C}\subset\mathbb{R}$ be any Cantor set, where our only
condition is that $\mathfrak{C}$ is totally disconnected, i.e. $\mathfrak{C}$
does not contain a proper interval of $\mathbb{R}$.

As in Section 4, we denote $\mathfrak{P}_{f}$, $\mathfrak{P}$ the sets of
finite and infinite paths, respectively in a binary tree, i.e. finite and
infinite words built from the letters $L$ and $R$. Then $\mathfrak{C}$ is
constructed iteratively by starting with an initial closed interval
$I_{\emptyset}$, where we denote $\emptyset$ the empty word in
$\mathfrak{P}_{f}$. Then, inductively, for any word $W\in\mathfrak{P}_{f}$, a
middle open subinterval of $I_{W}$ is removed, leaving two new intervals
$I_{WL}$ and $I_{WR}$. Then there is a one-to-one correspondence between
elements $p\in\mathfrak{P}$ and points $x_{p}=\bigcap_{i\geq 0}I_{p|_{i}}$,
where $p|_{i}\in\mathfrak{P}_{f}$ denotes the prefix of length $i$ of $p$.

Now, following [Bae01], [Bae04], for any $W\in\mathfrak{P}_{f}$ we set
$d_{WL}=|I_{WL}|/|I_{W}|$ and $d_{WR}=|I_{WR}|/|I_{W}|$. With this, we define
the local Hausdorff measure

$h^{s}(x_{p})=\liminf_{k\rightarrow\infty}\prod_{i=0}^{k}(d_{p|_{i}L}^{s}+d_{p|_{i}R}^{s})$

and the local Hausdorff dimension

$h(x_{p})=\inf_{s\geq 0}\\{h^{s}(x_{p})=0\\}$

for any $p\in\mathfrak{P}$.

######  Lemma C.1:

Let $\mu$ be a finite Borel measure with support on $\mathfrak{C}$ and
$x_{p}=\bigcap_{i\geq 0}I_{p|_{i}}\in\mathfrak{C}$. Assume that there exists a
subsequence $\\{p|_{i_{k}}\mid k\geq 0\\}$ such that
$\lim_{k\rightarrow\infty}\log\mu(I_{p|_{i_{k}}})/\log|I_{p|_{i_{k}}}|=$
$\liminf_{i\rightarrow\infty}$ $\log\mu(I_{p|_{i}})/\log|I_{p|_{i}}|$ and
$2|I_{p|_{i_{k}}}|+|I_{p|_{{i_{k}}-1}X_{i_{k}}}|\leq|I_{p|_{{i_{k}}-1}}|$ for
all $k$ (where $X_{i_{k}}=L$ if the last letter in $p|_{i}$is $R$, and vice
versa). Then

$\liminf_{\epsilon\rightarrow
0}\frac{\log\mu(B_{\epsilon}(x_{p}))}{\log\epsilon}\leq\liminf_{i\rightarrow\infty}\frac{\log\mu(I_{p|_{i}})}{\log|I_{p|_{i}}|},$

where $B_{\epsilon}(x)$ denotes the open $n$-ball of radius $\epsilon$ with
center $x$.

###### Proof.

The conditions imply that for every $k\geq 0$, we can find a ball
$B_{\epsilon_{k}}(x_{p})$ such that
$B_{\epsilon_{k}}(x_{p})\cap\mathfrak{C}=I_{p_{k}}\cap\mathfrak{C}$. Hence,
for any measure $\mu$ that is supported on $\mathfrak{C}$ we have
$\mu(B_{\epsilon_{k}})=\mu(I_{p|_{i_{k}}})$. Moreover,
$\epsilon_{k}\geq|I_{p|_{i_{k}}}|$ for all $k$, hence
$\log\mu(B_{\epsilon_{k}}(x_{p}))/\log\epsilon_{k}\leq\log\mu(I_{p|_{i_{k}}})/\log|I_{p|_{i_{k}}}|$
for $k\gg 0$, and the statement follows. ∎

######  Theorem C.2:

Let $\mathfrak{C}$ be a Cantor set that satisfies the conditions of Lemma C.1
for every $x\in\mathfrak{C}$. Then
$\dim_{H}\mathfrak{C}\leq\sup_{x\in\mathfrak{C}}h(x)$.

###### Proof.

For any $s\geq 0$ and any $W\in\mathfrak{P}_{f}$ of length $n$, we define

$\mu_{s}(I_{W}\cap\mathfrak{C})=\frac{|I_{W}|^{s}}{\prod_{k=0}^{n}(d^{s}_{W|_{k}L}+d^{s}_{W|_{k}R})},$

which in a natural way extends to a finite Borel measure on $\mathfrak{C}$
(see [Bae01], [Bae04]). If $h^{s}(x_{p})=0$ for some $x_{p}\in\mathfrak{C}$,
we get by Lemma C.1:

$\liminf_{\epsilon\rightarrow
0}\frac{\log\mu(B_{\epsilon}(x_{p}))}{\log\epsilon}\leq
s-\limsup_{i\rightarrow\infty}\frac{\log\prod_{k=0}^{i-1}(d^{s}_{p|_{k}L}+d^{s}_{p|_{k}R})}{\log|I_{p|_{i}}|}.$

Now, $\log|I_{p|_{i}}|<0$ for $i\gg 0$, and from $h^{s}(x_{p})=0$ it follows
that $\log\prod_{k=0}^{i-1}(d^{s}_{p|_{k}L}+d^{s}_{p|_{k}R})<0$ for infinitely
many $i$. Hence $\liminf_{\epsilon\rightarrow
0}\log\mu(B_{\epsilon}(x_{p}))/\log\epsilon\leq s$.

Moreover, it follows that $\liminf_{\epsilon\rightarrow
0}\log\mu(B_{\epsilon}(x_{p}))/\log\epsilon\leq h(x)$. Using [Fal97,
Proposition 10.1], we conclude the proof. ∎

The following statement is needed in Section 4:

######  Corollary C.3:

Let $\mathfrak{C}$ be a Cantor set that satisfies the conditions of Lemma C.1
for every $x\in\mathfrak{C}$. If $h(x)=0$ for every $x\in\mathfrak{C}$, then
$\dim_{H}\mathfrak{C}=0$.

In [JS95], the local Hausdorff dimension at a point $x\in\mathfrak{C}$ was
introduced as

$l^{\dim_{H}}_{\mathfrak{C}}(x)=\underset{\varepsilon\rightarrow
0}{\lim}\dim_{H}(B_{\varepsilon}(x)\cap\mathfrak{C}),$

and it is a natural question how $h(x)$ relates to
$l^{\dim_{H}}_{\mathfrak{C}}$. The following corollary provides a partial
answer.

######  Corollary C.4:

Let $x\in\mathfrak{C}$. If $h(x)=\lim_{\epsilon\rightarrow 0}\sup_{y\in
B_{\epsilon}(x)\cap\mathfrak{C}}h(y)$, then

$l^{\dim_{H}}_{\mathfrak{C}}(x)\leq h(x).$

##  Appendix D The first few levels of the Markoff Tree

$1,5,2$$1,13,5$$1,34,13$$1,89,34$$1,233,89$$1,610,233$$233,62210,89$$89,9077,34$$89,2423525,9077$$9077,925765,34$$34,1325,13$$34,135137,1325$$34,13782649,135137$$135137,537169541,1325$$1325,51641,13$$1325,205272962,51641$$51641,2012674,13$$13,194,5$$13,7561,194$$13,294685,7561$$13,11485154,294685$$294685,6684339842,7561$$7561,4400489,194$$7561,99816291793,4400489$$4400489,2561077037,194$$194,2897,5$$194,1686049,2897$$194,981277621,1686049$$1686049,14653451665,2897$$2897,43261,5$$2897,375981346,43261$$43261,646018,5$$5,29,2$$5,433,29$$5,6466,433$$5,96557,6466$$5,1441889,96557$$96557,1873012681,6466$$6466,8399329,433$$6466,162930183509,8399329$$8399329,10910721905,433$$433,37666,29$$433,48928105,37666$$433,63557570729,48928105$$48928105,5528778008357,37666$$37666,3276509,29$$37666,370238963953,3276509$$3276509,285018617,29$$29,169,2$$29,14701,169$$29,1278818,14701$$29,111242465,1278818$$1278818,56399710225,14701$$14701,7453378,169$$14701,328716329765,7453378$$7453378,3778847945,169$$169,985,2$$169,499393,985$$169,253191266,499393$$499393,1475706146,985$$985,5741,2$$985,16964653,5741$$5741,33461,2$
$5$$13$$34$$89$$233$$610$$1597$$426389$$62210$$43484701$$16609837$$9077$$2423525$$647072098$$65995009186$$925765$$25209506681$$94418953$$1325$$135137$$13782649$$1405695061$$5587637513705$$537169541$$217774440785026$$2135248790338$$51641$$205272962$$815959972309$$31801503090601$$2012674$$311809494089$$78442645$$194$$7561$$294685$$11485154$$447626321$$10153507819457$$6684339842$$5909324059011749$$151620880341401$$4400489$$99816291793$$2264132942340130$$1317721482167652770$$2561077037$$33809973988413085$$1490542435045$$2897$$1686049$$981277621$$571101889373$$4963446454828093$$14653451665$$74119312578961858$$127353146734466$$43261$$375981346$$3267653834825$$48795987025021$$646018$$83842154089$$9647009$$29$$433$$6466$$96557$$1441889$$21531778$$417673428514$$1873012681$$542557456311485$$36332699889481$$8399329$$162930183509$$3160519691308253$$4105512645967389917$$10910721905$$274928228722804802$$14173019355266$$37666$$48928105$$63557570729$$82561235448866$$9329254482520315202$$5528778008357$$811537892743746482789$$624740857339396181$$3276509$$370238963953$$41836262445484585$$3639273892628002565$$285018617$$2801598191304130$$24793343170$$169$$14701$$1278818$$111242465$$9676815637$$426776599819081$$56399710225$$216374893891527449$$2487396418774357$$7453378$$328716329765$$14497376284172417$$7350141181533573809$$3778847945$$84495546415824461$$1915868454737$$985$$499393$$253191266$$128367472469$$379325837704445$$1475706146$$2210871958107149$$4360711162037$$5741$$16964653$$50130543874$$292182217634$$33461$$576298801$$195025$

##  Appendix E The 300 Smallest Markoff Numbers

We present a computer generated list of the smallest 300 Markoff numbers
together with their weights and coweights.

Table 1. The 300 smallest Markoff numbers and their weights $n$ |  $m_{n}$ |  $r_{m_{n}}$ |  $s_{m_{n}}$ |  $w_{m_{n}}$ |  $v_{m_{n}}$
---|---|---|---|---|---
1 |  1 |  0 |  1 |  -1 |  10
|  2 |  1 |  1 |  1 |  5
|  5 |  2 |  1 |  1 |  2
|  13 |  5 |  2 |  2 |  1
|  29 |  12 |  5 |  7 |  2
|  34 |  13 |  5 |  5 |  1
|  89 |  34 |  13 |  13 |  2
|  169 |  70 |  29 |  41 |  10
|  194 |  75 |  29 |  31 |  5
10 |  233 |  89 |  34 |  34 |  5
|  433 |  179 |  74 |  104 |  25
|  610 |  233 |  89 |  89 |  13
|  985 |  408 |  169 |  239 |  58
|  1325 |  507 |  194 |  196 |  29
|  1597 |  610 |  233 |  233 |  34
|  2897 |  1120 |  433 |  463 |  74
|  4181 |  1597 |  610 |  610 |  89
|  5741 |  2378 |  985 |  1393 |  338
|  6466 |  2673 |  1105 |  1553 |  373
20 |  7561 |  2923 |  1130 |  1208 |  193
|  9077 |  3468 |  1325 |  1327 |  194
|  10946 |  4181 |  1597 |  1597 |  233
|  14701 |  6089 |  2522 |  3566 |  865
|  28657 |  10946 |  4181 |  4181 |  610
|  33461 |  13860 |  5741 |  8119 |  1970
|  37666 |  15571 |  6437 |  9047 |  2173
|  43261 |  16725 |  6466 |  6914 |  1105
|  51641 |  19760 |  7561 |  7639 |  1130
|  62210 |  23763 |  9077 |  9079 |  1325
30 |  75025 |  28657 |  10946 |  10946 |  1597
|  96557 |  39916 |  16501 |  23191 |  5570
|  135137 |  51709 |  19786 |  19990 |  2957
|  195025 |  80782 |  33461 |  47321 |  11482
|  196418 |  75025 |  28657 |  28657 |  4181
|  294685 |  113922 |  44041 |  47081 |  7522
|  426389 |  162867 |  62210 |  62212 |  9077
|  499393 |  206855 |  85682 |  121172 |  29401
|  514229 |  196418 |  75025 |  75025 |  10946
|  646018 |  249755 |  96557 |  103247 |  16501
40 |  925765 |  353702 |  135137 |  135341 |  19786
|  1136689 |  470832 |  195025 |  275807 |  66922
|  1278818 |  529673 |  219385 |  310201 |  75245
|  1346269 |  514229 |  196418 |  196418 |  28657
|  1441889 |  596067 |  246410 |  346312 |  83177
|  1686049 |  651838 |  252005 |  269465 |  43066
|  2012674 |  770133 |  294685 |  297725 |  44041
|  2423525 |  925943 |  353770 |  354304 |  51797
|  2922509 |  1116300 |  426389 |  426391 |  62210
|  3276509 |  1354498 |  559945 |  786985 |  189026
50 |  3524578 |  1346269 |  514229 |  514229 |  75025
|  4400489 |  1701181 |  657658 |  703054 |  112325
|  6625109 |  2744210 |  1136689 |  1607521 |  390050
|  7453378 |  3087111 |  1278649 |  1807955 |  438553
|  8399329 |  3472225 |  1435394 |  2017346 |  484525
|  9227465 |  3524578 |  1346269 |  1346269 |  196418
|  9647009 |  3729600 |  1441889 |  1541791 |  246410
|  11485154 |  4440035 |  1716469 |  1834951 |  293165
|  13782649 |  5273811 |  2017978 |  2038784 |  301585
|  16609837 |  6344632 |  2423525 |  2424059 |  353770
60 |  16964653 |  7026989 |  2910674 |  4116314 |  998785
|  20031170 |  7651227 |  2922509 |  2922511 |  426389
|  21531778 |  8901089 |  3679649 |  5171489 |  1242085
|  24157817 |  9227465 |  3524578 |  3524578 |  514229
|  38613965 |  15994428 |  6625109 |  9369319 |  2273378
|  43484701 |  16610303 |  6344810 |  6346208 |  926173
|  48928105 |  20226717 |  8361658 |  11752046 |  2822725
|  63245986 |  24157817 |  9227465 |  9227465 |  1346269
|  78442645 |  30015427 |  11485154 |  11603636 |  1716469
|  94418953 |  36074136 |  13782649 |  13803455 |  2017978
70 |  111242465 |  46075462 |  19083973 |  26983921 |  6545450
|  137295677 |  52442283 |  20031170 |  20031172 |  2922509
|  144059117 |  55694245 |  21531778 |  23023618 |  3679649
|  165580141 |  63245986 |  24157817 |  24157817 |  3524578
|  205272962 |  78545995 |  30054973 |  30365023 |  4491749
|  225058681 |  93222358 |  38613965 |  54608393 |  13250218
|  253191266 |  104875077 |  43440605 |  61433965 |  14906249
|  285018617 |  117825755 |  48708778 |  68458648 |  16443089
|  298045301 |  113843800 |  43484701 |  43486099 |  6344810
|  321534781 |  132920268 |  54948325 |  77226023 |  18548098
80 |  375981346 |  145356973 |  56196005 |  60089573 |  9603553
|  433494437 |  165580141 |  63245986 |  63245986 |  9227465
|  447626321 |  173047443 |  66898250 |  71516008 |  11425913
|  537169541 |  205543262 |  78649345 |  79460245 |  11754074
|  576298801 |  238710779 |  98877242 |  139833536 |  33929305
|  647072098 |  247223313 |  94455265 |  94597841 |  13829605
|  780291637 |  298046521 |  113844266 |  113847926 |  16610905
|  941038565 |  359444748 |  137295677 |  137295679 |  20031170
|  981277621 |  379368596 |  146666477 |  156828167 |  25064338
|  1134903170 |  433494437 |  165580141 |  165580141 |  24157817
90 |  1311738121 |  543339720 |  225058681 |  318281039 |  77227930
|  1405695061 |  537877013 |  205813970 |  207935978 |  30758713
|  1475706146 |  611256455 |  253190281 |  358063219 |  86879945
|  1873012681 |  774290566 |  320086397 |  449859017 |  108046858
|  2151239746 |  831684075 |  321534781 |  343812479 |  54948325
|  2561077037 |  990084419 |  382755826 |  409176220 |  65372957
|  2971215073 |  1134903170 |  433494437 |  433494437 |  63245986
|  3057250481 |  1169831520 |  447626321 |  452244079 |  66898250
|  3778847945 |  1565159188 |  648272521 |  916629619 |  222345506
|  4434764269 |  1693992981 |  647072098 |  647214674 |  94455265
100 |  4801489937 |  1984902931 |  820545226 |  1153218856 |  276979385
|  5348189873 |  2042828390 |  780291637 |  780295297 |  113844266
|  6449974274 |  2463670947 |  941038565 |  941038567 |  137295677
|  6684339842 |  2584092721 |  998982001 |  1067938321 |  170621525
|  7645370045 |  3166815962 |  1311738121 |  1855077841 |  450117362
|  7778742049 |  2971215073 |  1134903170 |  1134903170 |  165580141
|  9629807441 |  3679208170 |  1405695061 |  1407817069 |  205813970
|  9676815637 |  4008035521 |  1660086266 |  2347290926 |  569378905
|  10910721905 |  4510417602 |  1864575701 |  2620530901 |  629397602
|  14001740009 |  5348193067 |  2042829610 |  2042839192 |  298048097
110 |  14653451665 |  5665123983 |  2190175426 |  2341920284 |  374286601
|  17445941365 |  6744410242 |  2607315281 |  2787289361 |  445317442
|  19577194573 |  8109139505 |  3358915562 |  4750223942 |  1152597601
|  20365011074 |  7778742049 |  2971215073 |  2971215073 |  433494437
|  24793343170 |  10249486187 |  4237103741 |  5955115391 |  1430359717
|  25209506681 |  9631659149 |  3679915642 |  3685470766 |  538792565
|  30395743789 |  11610578034 |  4435013113 |  4435990313 |  647393602
|  32124537073 |  12419566880 |  4801489937 |  5134163567 |  820545226
|  44208781349 |  16886251875 |  6449974274 |  6449974276 |  941038565
|  44560482149 |  18457556052 |  7645370045 |  10812186007 |  2623476242
120 |  50130543874 |  20764750117 |  8601040685 |  12163706477 |  2951409337
|  53316291173 |  20365011074 |  7778742049 |  7778742049 |  1134903170
|  56399710225 |  23360168307 |  9675536650 |  13680794696 |  3318530233
|  63557570729 |  26274489812 |  10861787305 |  15265898707 |  3666717602
|  65995009186 |  25214353799 |  9633510857 |  9648052211 |  1410484105
|  71700814274 |  29640623697 |  12253230065 |  17221056817 |  4136142677
|  83842154089 |  32413953163 |  12531457130 |  13399705400 |  2141549281
|  95969331289 |  36657026960 |  14001740009 |  14001749591 |  2042829610
|  99816291793 |  38587888548 |  14917656385 |  15947373851 |  2547867970
|  119154326114 |  45593413853 |  17445941365 |  17625915445 |  2607315281
130 |  128367472469 |  53171457184 |  22024301053 |  31146899083 |  7557438842
|  139583862445 |  53316291173 |  20365011074 |  20365011074 |  2971215073
|  143367113573 |  54858181515 |  20991006962 |  21207430972 |  3137087141
|  162930183509 |  67354220371 |  27843772738 |  39132477604 |  9398815925
|  172765826641 |  66007698628 |  25219201985 |  25257269243 |  3692452738
|  208333239010 |  79576653333 |  30395743789 |  30396720989 |  4435013113
|  251250963713 |  95969339651 |  36657030154 |  36657055240 |  5348197193
|  259717522849 |  107578520350 |  44560482149 |  63018038201 |  15290740090
|  292182217634 |  121025831139 |  50130538133 |  70895275783 |  17202073997
|  303011495165 |  115740092172 |  44208781349 |  44208781351 |  6449974274
140 |  311809494089 |  119311314754 |  45653484253 |  46124450173 |  6822963842
|  328716329765 |  136150856363 |  56392256818 |  79736239324 |  19341502949
|  365435296162 |  139583862445 |  53316291173 |  53316291173 |  7778742049
|  370238963953 |  153055564992 |  63272665105 |  88927731023 |  21359559946
|  417673428514 |  172663323955 |  71377831109 |  100316543351 |  24093964765
|  479716816349 |  185461819125 |  71700814274 |  76668641026 |  12253230065
|  571101889373 |  220791871034 |  85359637609 |  91273723729 |  14587401650
|  665048316673 |  275472032399 |  114104251874 |  161367780524 |  39154389145
|  679944086914 |  262858951995 |  101618397709 |  108632769071 |  17355954325
|  815959972309 |  312220310365 |  119468510114 |  120700958786 |  17854701145
150 |  841771717954 |  348653014865 |  144408421169 |  204187326641 |  49529419285
|  956722026041 |  365435296162 |  139583862445 |  139583862445 |  20365011074
|  982145940029 |  375243159204 |  143367113573 |  143583537583 |  20991006962
|  1070710724173 |  442624452524 |  182977905749 |  257162633399 |  61765160770
|  1184065449986 |  452289781295 |  172765826641 |  172803893899 |  25219201985
|  1427933269321 |  545424970563 |  208334944570 |  208341642368 |  30397947073
|  1490542435045 |  576227430677 |  222763233074 |  238139856986 |  38046948649
|  1513744654945 |  627013566048 |  259717522849 |  367296043199 |  89120964298
|  1722099665665 |  657783551488 |  251250963713 |  251250988799 |  36657030154
|  1915868454737 |  793532621205 |  328672889498 |  464729408878 |  112728732989
160 |  2076871684802 |  793294393323 |  303011495165 |  303011495167 |  44208781349
|  2135248790338 |  817034414741 |  312631126589 |  315854453885 |  46722441193
|  2156735837173 |  891587472514 |  368579316689 |  518026580369 |  124424852290
|  2504730781961 |  956722026041 |  365435296162 |  365435296162 |  53316291173
|  3099893879221 |  1184098670126 |  452302470737 |  452402131157 |  66024095098
|  3267653834825 |  1263297435618 |  488399472989 |  522238472029 |  83464478018
|  4360711162037 |  1806262617670 |  748177194673 |  1058076690973 |  256730208074
|  4508515437145 |  1722099687557 |  657783559850 |  657783625526 |  95969350453
|  4643961467965 |  1776973308747 |  679944086914 |  686958458276 |  101618397709
|  5528778008357 |  2285578567387 |  944850630610 |  1327957693804 |  318962279525
170 |  5587637513705 |  2138060991308 |  818110478953 |  826545460219 |  122265876434
|  6557470319842 |  2504730781961 |  956722026041 |  956722026041 |  139583862445
|  7163627708162 |  2769507719995 |  1070710724173 |  1144895451823 |  182977905749
|  8115549747397 |  3099980848801 |  1184131890266 |  1184392799006 |  172851666985
|  8822750406821 |  3654502875938 |  1513744654945 |  2140758220993 |  519435045698
|  9787184545081 |  3738374837400 |  1427933269321 |  1427939967119 |  208334944570
|  9925594216162 |  4111315732857 |  1702962733225 |  2408352982409 |  584364418045
|  10153507819457 |  3925235141920 |  1517453001793 |  1622197606303 |  259173984074
|  14173019355266 |  5859028992773 |  2422082400205 |  3404067623053 |  817587000473
|  14235090298445 |  5437320661083 |  2076871684802 |  2076871684804 |  303011495165
180 |  14622039889385 |  5594996637517 |  2140876896154 |  2162950023166 |  319952129669
|  15988960048321 |  6609726164163 |  2732415356170 |  3840218444168 |  922341268873
|  17167680177565 |  6557470319842 |  2504730781961 |  2504730781961 |  365435296162
|  18696424380481 |  7228166198398 |  2794458744005 |  2988074214713 |  477555886138
|  21246581423810 |  8115777435463 |  3100067821177 |  3100750882579 |  452527201625
|  22592065572301 |  9357939962069 |  3876185648162 |  5481754313906 |  1330096633345
|  26500373448281 |  10244754717563 |  3960510195370 |  4233890704408 |  676436901233
|  30901824368813 |  11803446623878 |  4508515437145 |  4508515502821 |  657783559850
|  31801503090601 |  12168581182878 |  4656206581885 |  4704240458033 |  695875230298
|  36332699889481 |  15019688359352 |  6209035912505 |  8726365188575 |  2095892945914
190 |  44945570212853 |  17167680177565 |  6557470319842 |  6557470319842 |  956722026041
|  46127828641049 |  17623808310363 |  6733432474730 |  6743596290040 |  985871051441
|  48795987025021 |  18864864025739 |  7293286116482 |  7798605052196 |  1246377918925
|  51422757785981 |  21300003689580 |  8822750406821 |  12477253282759 |  3027489309890
|  57850602535042 |  23962504125979 |  9925594182701 |  14036909842895 |  3405925423477
|  65082055350517 |  26957823917211 |  11166277193266 |  15791416401116 |  3831606586645
|  67082333460610 |  25623191867697 |  9787196235121 |  9787242142481 |  1427948370517
|  73224462646361 |  30328804257734 |  12561872555437 |  17761950126841 |  4308490098890
|  80902026460669 |  30901824426127 |  11803446645770 |  11803446817712 |  1722099715837
|  82561235448866 |  34130542039071 |  14109453347537 |  19830390668347 |  4763063342273