100

# Orbits of $N$-expansions with a finite set of digits

Jaap de Jonge  Jaap de Jonge, University of Amsterdam, Korteweg - de Vries
Institute for Mathematics, Science Park 105 - 107, 1098 XG Amsterdam, The
Netherlands<EMAIL_ADDRESS>,  Cor Kraaikamp  Cor Kraaikamp, Delft
University of Technology, department of Electrical Engineering, Mathematics
and Computer Science, Mekelweg 4, 2628 CD Delft, The Netherlands
<EMAIL_ADDRESS>and  Hitoshi Nakada  Hitoshi Nakada, Department of
Mathematics, Keio University, Yokohama, 223-8522 Japan<EMAIL_ADDRESS>
###### Abstract.

For $N\in\mathbb{N}_{\geq 2}$ and $\alpha\in\mathbb{R}$ such that
$0<\alpha\leq\sqrt{N}-1$, we define $I_{\alpha}:=[\alpha,\alpha+1]$ and
$I_{\alpha}^{-}:=[\alpha,\alpha+1)$ and investigate the continued fraction map
$T_{\alpha}:I_{\alpha}\to I_{\alpha}^{-}$, which is defined as
$T_{\alpha}(x):=N/x-d(x),$ where $d:I_{\alpha}\to\mathbb{N}$ is defined by
$d(x):=\left\lfloor N/x-\alpha\right\rfloor$. For $N\in\mathbb{N}_{\geq 7}$,
for certain values of $\alpha$, open intervals $(a,b)\subset I_{\alpha}$ exist
such that for almost every $x\in I_{\alpha}$ there is an $n_{0}\in\mathbb{N}$
for which $T_{\alpha}^{n}(x)\notin(a,b)$ for all $n\geq n_{0}$. These _gaps_
$(a,b)$ are investigated in the square $\Upsilon_{\alpha}:=I_{\alpha}\times
I_{\alpha}^{-}$, where the _orbits_ $T_{\alpha}^{k}(x),k=0,1,2,\ldots$ of
numbers $x\in I_{\alpha}$ are represented as cobwebs. The squares
$\Upsilon_{\alpha}$ are the union of _fundamental regions_ , which are related
to the cylinder sets of the map $T_{\alpha}$, according to the finitely many
values of $d$ in $T_{\alpha}$. In this paper some clear conditions are found
under which $I_{\alpha}$ is gapless. When $I_{\alpha}$ consists of at least
five cylinder sets, it is always gapless. In the case of four cylinder sets
there are usually no gaps, except for the rare cases that there is one, very
wide gap. Gaplessness in the case of two or three cylinder sets depends on the
position of the endpoints of $I_{\alpha}$ with regard to the fixed points of
$I_{\alpha}$ under $T$.

###### Key words and phrases:

Continued fractions, Dynamical systems, Gaps

###### 2010 Mathematics Subject Classification:

Primary 11J70; Secondary 37Mxx

##  1\. Introduction

In 2008, Edward Burger and his co-authors introduced in [2] new continued
fraction expansions, the so-called $N$-expansions, which are nice variations
of the _regular continued fraction_ (RCF) _expansion_. These $N$-expansions
have been studied in various papers since; see [1], [3] and [4]. In [5], a
subclass of these $N$-expansions is introduced, for which the digit set is
always finite. These particular $N$-expanions are defined as follows:

For $N\in\mathbb{N}_{\geq 2}$ and $\alpha\in\mathbb{R}$ such that
$0<\alpha\leq\sqrt{N}-1$, let $I_{\alpha}:=[\alpha,\alpha+1]$ and
$I_{\alpha}^{-}:=[\alpha,\alpha+1)$. Hereafter we denote by $\mathbb{N}_{\geq
k}$ the set of positive integers $n\geq k$. We define the $N$-expansion map
$T_{\alpha}:I_{\alpha}\to I_{\alpha}^{-}$ (or $I_{\alpha}$) as

(1) $T_{\alpha}(x):=\frac{N}{x}-d(x),$

where $d:I_{\alpha}\to\mathbb{N}$ is defined by

$d(x):=\left\lfloor\frac{N}{x}-\alpha\right\rfloor,\quad{\text{if either
$x\in(\alpha,\alpha+1]$ or both $x=\alpha$ and
$N/\alpha-\alpha\not\in\mathbb{Z}$}}$

and

$d(\alpha)=\left\lfloor\frac{N}{\alpha}-\alpha\right\rfloor-1,\quad\text{if
$N/\alpha-\alpha\in\mathbb{Z}$}.$

Note that if $N/\alpha-\alpha\in\mathbb{Z}$, we have that
$T_{\alpha}(\alpha)=\alpha+1$. This is the only case in which the range of
$T_{\alpha}$ is $I_{\alpha}$ and not $I_{\alpha}^{-}$.

For a fixed $\alpha\in(0,\sqrt{N}-1]$ and $x\in I_{\alpha}$ we define for
$n\in\mathbb{N}$

$d_{n}=d_{n}(x):=d(T_{\alpha}^{n-1}(x)).$

Note that for $\alpha\in(0,\sqrt{N}-1]$ fixed, there are only _finitely_ many
possibilities for each $d_{n}$.

Applying (1), we obtain for every $x\in I_{\alpha}$ a continued fraction
expansion of the form

(2)
$x=T_{\alpha}^{0}(x)=\cfrac{N}{d_{1}+T_{\alpha}(x)}=\cfrac{N}{d_{1}+\cfrac{\displaystyle
N}{\displaystyle
d_{2}+T_{\alpha}^{2}(x)}}=\dots=\cfrac{N}{d_{1}+\cfrac{\displaystyle
N}{\displaystyle d_{2}+\cfrac{N}{d_{3}+\ddots}}}\quad,$

which we will throughout this paper write as
$x=[d_{1},d_{2},d_{3},\ldots]_{N,\alpha}$ (note that this expansion is
infinite for every $x\in I_{\alpha}$, since $0\not\in I_{\alpha}$); we will
call the numbers $d_{i},\,i\in\mathbb{N}$, the partial quotients or _digits_
of this $N$-continued fraction expansion of $x$; see [3, 5], where these
continued fractions (also with a finite set of digits) were introduced and
elementary properties were studied (such as the convergence in reference [3]).

In each _cylinder set_ $\Delta_{i}:=\\{x\in I_{\alpha};d(x)=i\\}$ of rank $1$,
with $d_{\min}\leq i\leq d_{\max}$, where $d_{max}:=d(\alpha)$ is the largest
partial quotient, and $d_{min}:=d(\alpha+1)$ the smallest one given $N$ and
$\alpha$, the map $T_{\alpha}$ obviously has one _fixed point_ $f_{i}$ . As of
now we will write simply ‘cylinder set’ for ‘cylinder set of rank $1$’.

It is easy to see that111For reasons of legibility we will usually omit
suffices such as ‘$(N)$’, ‘$(N,\alpha)$’ or ‘$(N,d)$’.

(3)
$f_{i}=f_{i}(N):=\dfrac{\sqrt{4N+i^{2}}-i}{2},\,\,{\text{for}}\,\,d_{\min}\leq
i\leq d_{\max}.$

Note that $N/\alpha-\alpha\in\mathbb{Z}$ if and only if for some
$d\in\mathbb{N}_{\leq 2}$ we have that $d+1=\max d_{i}$ for any
$\alpha_{0}<\alpha$, i.e. $\Delta_{d+1}\neq\emptyset$ and $\alpha=f_{d+1}$.

Given $N\in\mathbb{N}_{\geq 2}$, we let $\alpha_{max}=\sqrt{N}-1$ be the
largest value of $\alpha$ we consider. The reason for this is that for larger
values of $\alpha$ we would have $0$ as a partial quotient as well. Since
$T_{\alpha}^{\prime}(x)=-N/x^{2}$ and because $0<\alpha\leq\sqrt{N}-1$, we
have $|T_{\alpha}^{\prime}(x)|>1$ on $I_{\alpha}^{-}$. From this it follows
that the fixed points act as repellers and that the maps $T_{\alpha}$ are
_expanding_ when $0<\alpha\leq\sqrt{N}-1$. This is equivalent to the
convergence of the $N$-expansion of all $x\in I_{\alpha}$.

Each pair of consecutive cylinders sets $(\Delta_{i},\Delta_{i-1})$ is divided
by a _discontinuity point_ $p_{i}(N,\alpha)$ of $T_{\alpha}$, satisfying
$N/p_{i}-i=\alpha$, so $p_{i}=N/(\alpha+i)$. A cylinder set $\Delta_{i}$ is
called _full_ if $T_{\alpha}(\Delta_{i})=I_{\alpha}^{-}$ (or $I_{\alpha}$).
When a cylinder set is not full, it contains either $\alpha$ (in which case
$T_{\alpha}(\alpha)<\alpha+1$) or $\alpha+1$ (in which case
$T_{\alpha}(\alpha+1)>\alpha$), and is called _incomplete_. On account of our
definition of $T_{\alpha}$, cylinder sets will always be an interval, _never_
consist of one single point.

The main object of this paper is the sequence $T_{\alpha}^{n}(x)$,
$n=0,1,2,\ldots$, for $x\in I_{\alpha}$, which is called the _orbit of $x$
under $T_{\alpha}$_. More specifically, we are interested in subsets of
$I_{\alpha}$ that we will call _gaps_ for such orbits. Before we will give a
proper definition of ‘gap’, we will give an example of orbits of points in
$I_{\alpha}$ for a pair $\\{N,\alpha\\}$. Note that, due to the repellence of
the fixed points, orbits cannot remain in one cylinder set indefinitely when
$x\in I_{\alpha}$ is not a fixed point of $T_{\alpha}$, so any orbit will show
an infinite migration between cylinder sets. A naive approach is to compute
the orbits of many points of $I_{\alpha}$ and obtain a _plot_ of the
asymptotic behaviour of these orbits by omitting the first, say hundred,
iterations. Figure 1 shows such a plot for $N=51$ and $\alpha=6$. It appears
that there are parts of $I_{\alpha}$ (illustrated by dashed line segments)
that are not visited by any orbit after many iterations of $T_{\alpha}$.

$\Delta_{1}$$\Delta_{2}$$\alpha$$\alpha+1$$p_{2}$$f_{1}$$f_{2}$$r_{2}$$r_{1}$$\ell_{1}$$\ell_{2}$
Figure 1. $N=51$, $\alpha=6$

In fact, setting $\ell_{i}=T_{\alpha}^{i}(\alpha)$ and
$r_{i}=T_{\alpha}^{i}(\alpha+1)$, the orbit of any point – apart from the
fixed points $f_{1}$ and $f_{2}$ – after once having left the interval
$(r_{2},r_{1})\subset\Delta_{2}$ or $(\ell_{1},\ell_{2})\subset\Delta_{1}$ of
Figure 1, will never return to it.

In order to get a better understanding of the orbits of $N$-expansions, it is
useful to consider the graphs of $T_{\alpha}$, which are drawn in the square
$\Upsilon_{N,\alpha}:=I_{\alpha}\times I_{\alpha}^{-}$. This square is divided
in rectangular sets of points
$\Box_{i}:=\\{(x,y)\in\Upsilon_{\alpha}:d(x)=i\\}$, which are the two-
dimensional _fundamental regions_ associated with the one-dimensional cylinder
sets we already use. We will call these regions shortly _cylinders_. Now
consider $(x,T_{\alpha}(x))\in\Upsilon_{N,\alpha}$. Then $(x,T_{\alpha}(x))$
goes to $(T_{\alpha}(x),T_{\alpha}^{2}(x))$ under $T_{\alpha}$. Regarding
this, $T_{\alpha}$ has one fixed point $F_{i}:=(f_{i},f_{i})$ in each
$\Box_{i}$. We will denote the dividing line between $\Box_{i}$ and
$\Box_{i-1}$ by $l_{i}$, which is the set
$\\{p_{i}\\}\times[\alpha,\alpha+1)$, with $p_{i}$ the discontinuity point
between $\Delta_{i}$ and $\Delta_{i-1}$. In case
$T_{\alpha}(\Delta_{i})=I_{\alpha}^{-}$, we will call the cylinder $\Box_{i}$
full and the branch of the graph of $T_{\alpha}$ in $\Box_{i}$ complete; if a
cylinder is not full, we will call it and its associated branch of
$T_{\alpha}$ incomplete. We will call the collection of $\Upsilon_{\alpha}$
and its associated branches, fixed points and dividing lines an _arrangement_
of $\Upsilon_{\alpha}$. When $\Upsilon_{\alpha}$ is a union of full cylinders,
we will call the associated arrangement also full.

Figure 2 is an example of such an arrangement, in which a part of the _cobweb_
is drawn associated with the orbit we investigated previously. The
discontinuity point $p_{2}=51/8$ is now visible as a dividing line between
$\Delta_{1}$ and $\Delta_{2}$.

$F_{1}$$F_{2}$$\Box_{2}$$\Box_{1}$$6$$7$$7$$\tfrac{51}{8}$$x$ Figure 2.
$N=51$, $\alpha=6$, $x=6.5$

In [3] and [5] the arrangement for $N=4$ and $\alpha=1$ is studied, consisting
of two full cylinders $\Box_{1}$ and $\Box_{2}$ and not showing any gaps. On
the other hand, the demonstration of the interval $(5/2,13/5)$ being a gap of
the interval $[2,3]$ in the case $(N,\alpha)=(9,2)$ in [5] is done without
referring to such an arrangement. In this paper, and even more so in the next
paper, we will show that arrangements may considerably support the insight in
the occurrence of gaps.

We will now give a formal definition of gaps, which is slightly delicate,
since $T_{\alpha}(I_{\alpha})=I_{\alpha}^{-}$ (or $I_{\alpha}$ when
$N/\alpha-\alpha\in\mathbb{Z}$).

######  Definition 1.

A maximal open interval $(a,b)\subset I_{\alpha}$ is called a _gap_ of
$I_{\alpha}$ if for almost every $x\in I_{\alpha}$ there is an
$n_{0}\in\mathbb{N}$ for which $T_{\alpha}^{n}(x)\notin(a,b)$ for all $n\geq
n_{0}$.

######  Remark 1.

In the example of Figure 1 the intervals $(r_{2},r_{1})$ and
$(\ell_{1},\ell_{2})$ are gaps and for
$x\in(r_{2},r_{1})\cup(\ell_{1},\ell_{2})\setminus\\{f_{1},f_{2}\\}$ there
exists an $n_{0}=n_{0}(x)$ such that
$T_{\alpha}^{n}(x)\not\in(r_{2},r_{1})\cup(\ell_{1},\ell_{2})$ for
$n\in\mathbb{N}_{\geq 2}$. The ‘for almost’222All ‘for all’ statements in this
paper are with respect to Lebesgue measure. formulation in the definition of
‘gap’ is necessary so as to exclude fixed points and pre-images of fixed
points, i.e. points that are mapped under $T_{\alpha}$ to a fixed point, which
may never leave an gap. In Section 5 we even find a class of gaps $(a,b)$ such
that for uncountably many $x\in I_{\alpha}$ and all
$n\in\mathbb{N}\cup\\{0\\}$ we have $T_{\alpha}^{n}(x)\in(a,b)$.

######  Remark 2.

When we use the word ‘gap’ in relation to arrangements, we mean the gap of the
associated interval $I_{\alpha}$.

In [5] computer simulations were used to get a more general impression of
orbits of $N$-expansions. For a lot of values of $\alpha$, plots such as
Figure 1 were stacked, for $0<\alpha\leq\alpha_{\max}$, so as to obtain graphs
such as Figure 3, with the values of $\alpha$ on the vertical axis and at each
height the corresponding interval $I_{\alpha}$ drawn. In the same paper,
similar graphs are given for $N=9,20,36$ and $100$. In all cases it appears
that ‘gaps’ such as in Figure 1 appear for values of $\alpha$ equal to or not
much smaller than $\alpha_{\max}$. Since the plots in [5] are based on
computer simulations, they do not actually show very small gaps (smaller than
pixel size) nor clarify much the connection between the gaps for each $N$.
Still, the suggestion is strong that for $\alpha$ sufficiently small there are
no gaps. It also seems that for $\alpha$ large enough several disjoint gaps
may occur. In Figure 3 we see this for $\alpha$ near
$\alpha_{\max}=\sqrt{50}-1$.

Figure 3. A simulation of intervals $I_{\alpha}$ with gaps if existent, for
$0<\alpha\leq\sqrt{50}-1$ and $N=50$

In this paper we will not only investigate conditions for gaplessness, we will
also show that simulations such as Figure 3 fail to reveal the existence, for
certain $N$ and $\alpha$, of one extremely large gap in plots such as Figure 3
_below_ the last visible gap. In a subsequent paper we will go into another
very interesting property of orbits of $N$-expansions that is hardly revealed
by simulations such as Figure 3: the existence of large numbers of gaps for
large $N$ and $\alpha$ close to $\alpha_{\max}$. But now we will concentrate
on gaplessness.

######  Remark 3.

When no gaps exist with non-empty intersection with a cylinder set, we call
the cylinder set gapless.

In Section 2 we will consider two classes of arrangements that have no gaps:
full arrangements and specific arrangements with more than three cylinders.
The gaplessness of the latter class, involving the proof of Theorem 4, for
which some preliminary results will be presented shortly, is largely given in
Section 2, but involve some intricacies for small values of $N$ so as to
finish it at the end of this paper. In Section 3 we will consider arrangements
with two cylinders and in Section 4 we will concentrate on arrangements with
three cylinders, but will prove a sufficient condition for gaplessness that is
valid for arrangements with any number of cylinders larger than $2$. Finally,
in Section 5 we will prove a result on gaps in certain arrangements with four
cylinders and we will finish the proof of Theorem 4. After that, it is merely
a matter of checking that for $N\in\\{2,3,4,5,6\\}$ all arrangements are
gapless.

##  2\. Full arrangements and arrangements with more than four cylinders

When $I_{\alpha}$ consists of full cylinder sets only, we obviously have no
gaps. In this situation the mutual relations between $N$, $\alpha$ and
$d(\alpha)$ show a great coherence, as expressed in the following theorem:

######  Theorem 1.

The interval $I_{\alpha}$ consists of $m$ full cylinder sets, with
$m\in\mathbb{N}_{\geq 2}$, if and only if there is a positive integer $k$ such
that

(4) $\begin{cases}\alpha=k,\\\ N=mk(k+1),\\\ d(\alpha)=(m-1)(k+1).\end{cases}$

Proof of Theorem 1: Writing $d:=d(\alpha)$, the interval $I_{\alpha}$ is the
union of $m$ full cylinder sets if and only if

(5) $\begin{cases}\frac{N}{\alpha}-d=\alpha+1,\\\
\frac{N}{\alpha+1}-(d-m+1)=\alpha.\end{cases}$

Note that the first equation in (5) can be written as

$N=\alpha^{2}+(d+1)\alpha,$

while the second equation in (5) equals

$N=\alpha^{2}+(d+2-m)\alpha+d+1-m.$

Subtracting the first of these equations from the last we find

(6) $\alpha=\frac{d+1-m}{m-1}.$

From (5), we have

(7) $\alpha=\frac{-(d+1)+\sqrt{(d+1)^{2}+4N}}{2},$

which yields that $\alpha$ is either a quadratic irrational or a rational
number. Since (6) implies that $\alpha$ is a rational number we find that the
integer $(d+1)^{2}+4N$ must be a square, i.e. there exists a positive integer
$s$ such that $s^{2}=(d+1)^{2}+4N$. Note that $d$ is an even integer if and
only if $s^{2}$ is an odd integer if and only if $s$ is an odd integer.
Consequently we find that the numerator of $\alpha$ in (7) is _always_ even,
and (7) yields that $\alpha$ is a positive integer, say $k$. From the
equations in (5) it follows that not only $\alpha=k$ but also $\alpha+1=k+1$
is a divisor of $N$.

From the definition of $T_{\alpha}$ in (1) (especially the case
$N/\alpha-\alpha\in\mathbb{Z}$) we see that

(8) $d=d(\alpha)=\frac{N}{k}-(k+1).$

On the other hand (6) yields that, since $\alpha=k$,

$d=(m-1)(k+1),$

and from this and (8) we see that

$\frac{N}{k}-(k+1)=(m-1)(k+1),$

i.e. $N=mk(k+1)$.

Conversely, let $k$ be a positive integer such that the relations of (4) hold.
Then both $N/\alpha$ and $N/(\alpha+1)$ are positive integers, implying that
_all_ cylinder sets are full. Moreover, since $d=d(\alpha)=d_{\max}$ is given
by

$d=\frac{N}{\alpha}-\alpha-1=\frac{mk(k+1)}{k}-k-1=(m-1)(k+1),$

and $d_{\min}=d(\alpha+1)$ is given by

$d(\alpha+1)=\left\lfloor\frac{N}{\alpha+1}-\alpha\right\rfloor=mk-k=(m-1)k,$

it follows that there are

$d_{\max}-\left(d_{\min}-1\right)=(m-1)(k+1)-(m-1)k+1=(m-1)+1=m$

full cylinder sets. $\Box$

Theorem 1 serves as a starting point for our investigation of orbits of
$N$-expansions. The first thing we will do is give some preliminary results
(in Subsection 2.1) that we need for proving (in Subsection 2.2) Theorem 3 and
Theorem 4 on gaplessness of arrangements with at least five cylinders.

###  2.1. Preliminary results

The first thing to pay attention to is the way $N$ and $\alpha$ and
$d(\alpha)$, the value of the largest partial quotient, are interdependent,
which is illustrated by the following lemmas:

######  Lemma 1.

Given $N$ and $\alpha$, let $d:=d(\alpha)$ be the largest possible digit. Then

$d\geq N-1\,\,{\text{if and only if}}\,\,\alpha<1.$

The proof of this lemma is left to the reader.

When $\alpha=\alpha_{\max}=\sqrt{N}-1$, we have

(9) $\displaystyle
d(\alpha)=\begin{cases}\left\lfloor\frac{2}{\sqrt{2}-1}-(\sqrt{2}-1)\right\rfloor=4&{\text{for}}\,\,N=2;\vskip
6.0pt plus 2.0pt minus 2.0pt\\\
\left\lfloor\frac{3}{\sqrt{3}-1}-(\sqrt{3}-1)\right\rfloor=3&{\text{for}}\,\,N=3;\vskip
6.0pt plus 2.0pt minus 2.0pt\\\
\left\lfloor\frac{4}{\sqrt{4}-1}-(\sqrt{4}-1)\right\rfloor-1=2&{\text{for}}\,\,N=4;\vskip
6.0pt plus 2.0pt minus 2.0pt\\\
\left\lfloor\frac{N}{\sqrt{N}-1}-(\sqrt{N}-1)\right\rfloor=\left\lfloor
2+\frac{\sqrt{N}+1}{N-1}\right\rfloor=2&{\text{for}}\,\,N\in\mathbb{N}_{\geq
5}.\vskip 6.0pt plus 2.0pt minus 2.0pt\\\ \end{cases}$

On the other hand we have, for $N\in\mathbb{N}$, $N\geq 2$ fixed:

$\lim_{\alpha\downarrow 0}d(\alpha)=\lim_{\alpha\downarrow
0}\left\lfloor\frac{N}{\alpha}-\alpha\right\rfloor=\infty.$

The following lemma provides for a lower bound for the rate of increase of
$d(\alpha)$ compared with the rate of decrease of $\alpha$.

######  Lemma 2.

Let $N\in\mathbb{N}_{\geq 2}$ be fixed and $d:=d(\alpha)$. Then $d$ is
constant for $\alpha\in[f_{d+1},f_{d})$, and $d$ increases overall more than
twice as fast as $\alpha$ decreases.

Proof of Lemma 2: Starting from $\alpha_{\max}$, $d$ increases by $1$ each
time $\alpha$ decreases beyond a fixed point, i.e. when
$N/\alpha-\alpha\in\mathbb{N}$. For the difference between two successive
fixed points $f_{d-1}$ and $f_{d}$ we have

$f_{d-1}-f_{d}=\frac{\sqrt{4N+(d-1)^{2}}-(d-1)}{2}-\frac{\sqrt{4N+d^{2}}-d}{2}=\frac{\sqrt{4N+(d-1)^{2}}-\sqrt{4N+d^{2}}+1}{2}<\frac{1}{2}.$

This finishes the proof. $\Box$

Closely related to the previous lemma is the following one, the proof of which
is left to the reader.

######  Lemma 3.

Let $d\in\mathbb{N}_{\geq 2}$ and $N\in\mathbb{N}_{\geq 2}$ be fixed and let
$f_{d}(N)$ be defined by the equation $N/f_{d}(N)-d=f_{d}(N)$ (so $f_{d}(N)$
is the fixed point of the map $x\mapsto N/x-d$ for $x\in(0,N/d)$). Then

$f_{d-1}(N+1)-f_{d}(N+1)>f_{d-1}(N)-f_{d}(N).$

So, for $d$ fixed, the distance between two consecutive fixed points increases
when $N$ increases. We have, in fact, for $d$ fixed:

$\lim_{N\to\infty}\left(f_{d-1}(N)-f_{d}(N)\right)=\tfrac{1}{2};$

cf. the proof of Lemma 2. For $N$ fixed, on the other hand, we have:

$\lim_{d\to\infty}f_{d}(N)=0.$

While $d(\alpha)$ is a monotonously non-increasing function of $\alpha$, the
number of cylinder sets is not. The reason is obvious: starting from
$\alpha=\alpha_{\max}$, the number of cylinder sets changes every time either
$\alpha$ or $\alpha+1$ decreases beyond the value of a fixed point; in the
first case, the number increases by $1$, and in the second case, it decreases
by $1$. Since $T_{\alpha}^{\prime}(x)=-N/x^{2}<0$ and
$T_{\alpha}^{\prime\prime}(x)=2N/x^{3}>0$ on $I_{\alpha}$, $T_{\alpha}(x)$ is
decreasing and convex on $I_{\alpha}$, implying that a per saldo increase of
the number of cylinder sets. Still, for $N$ and $\alpha$ large enough, it may
take a long time of $\alpha$ decreasing from $\alpha_{\max}$ before the amount
of cylinder sets stops alternating between two successive numbers
$k\in\mathbb{N}_{\geq 2}$ and $k+1$, and starts to alternate between the
numbers $k+1$ and $k+2$. As an example, we take $N=100$. When $\alpha$
decreases from $\alpha_{\max}$, the interval $I_{\alpha}$ consists of two
cylinder sets until $\alpha$ decreases beyond $f_{3}$ and cylinder set
$\Delta_{3}$ emerges; then, when $\alpha+1$ decreases beyond $f_{1}$, cylinder
set $\Delta_{1}$ disappears and so on, until $\alpha$ decreases beyond $f_{8}$
and $\Delta_{9}$ emerges while $\Delta_{6}$ has not yet disappeared.

In order to get a grip on counting the number of cylinder sets, the following
arithmetic will be useful: a full cylinder set counts for 1, an incomplete
left one counts for $N/\alpha-d_{\max}-\alpha$, and an incomplete right one
for $\alpha+1-(N/(\alpha+1)-d_{\min})$, giving rise to the following
definition:

######  Definition 2.

Let $N\in\mathbb{N}_{\geq 2}$ and $\alpha\in\mathbb{R}$ such that
$0<\alpha\leq\sqrt{N}-1$ and $T_{\alpha}$ the $N$-continued fraction map. The
branch number333The word ‘branch’ refers to the part of the graph of
$T_{\alpha}$ on the concerning cylinder set. $b(N,\alpha)$ is defined as

$\displaystyle b(N,\alpha):=$
$\displaystyle\,d_{\max}-d_{\min}-1\,\,{\text{(the number of full cylinder
sets save for the outermost ones)}}$ $\displaystyle+\,\,$
$\displaystyle\frac{N}{\alpha}-d_{\max}-\alpha\,\,{\text{(the length of the
image of the leftmost cylinder set)}}$ $\displaystyle+\,\,$
$\displaystyle\alpha+1-\left(\frac{N}{\alpha+1}-d_{\min}\right)\,\,{\text{(the
length of the image of the rightmost cylinder set),}}$

From this the next lemma follows immediately:

######  Lemma 4.

For $N\in\mathbb{N}_{\geq 2}$ and $0<\alpha\leq\sqrt{N}-1$ we have

$b(N,\alpha)=\frac{N}{\alpha}-\frac{N}{\alpha+1}=\frac{N}{\alpha(\alpha+1)}.$

It follows that for fixed $N$ the branch number $b(N,\alpha)$ is a strictly
decreasing function of $\alpha$.

######  Remark 4.

Applying Lemma 4, we find

(10)
$b(N,\alpha_{\max})=\frac{N}{(\sqrt{N}-1)\sqrt{N}}=1+\frac{1}{\sqrt{N}-1}.$

It follows that $b(N,\alpha)>1$ for all $N\in\mathbb{N}_{\geq 2}$, so the
number of cylinder sets is always at least 2. On the other hand, from Lemma 4
it follows that the number of cylinder sets increases to infinity as $\alpha$
decreases from $\alpha_{max}$ to $0$. Actually, we have infinitely many digits
_only_ when $\alpha=0$. In this case the corresponding $N$-expansion is the
_greedy_ $N$-expansion, studied in [1] and [3].

The relation $N/(\alpha(\alpha+1))=b$ yields

(11) $\alpha=\frac{\sqrt{\frac{4N}{b}+1}-1}{2},$

from which we derive that $d(\alpha)=d$ (or $d(\alpha)=d-1$ in case
$N/\alpha-\alpha\in\mathbb{Z}$), where $d$ is given by

(12) $d=\left\lfloor\frac{(b-1)\sqrt{\frac{4N}{b}+1}+b+1}{2}\right\rfloor.$

###  2.2. Gaplessness when the branch number is large enough

So far, we merely discussed the way $I_{\alpha}$ is divided in cylinder sets,
depending on the values of $N,\alpha,d(\alpha)$ and the branch number $b$. In
order to present some first results on sufficient conditions for gaplessness,
we will zoom in on some ergodic properties of $T_{\alpha}$.

######  Lemma 5.

If $\mu$ is an absolutely continuous invariant probability measure for
$T_{\alpha}$, then there exists a function $h$ of bounded variation such that

$\mu(A)=\int_{A}h\,d\lambda,\,\,\lambda-\mbox{a.e.},\,\,\mbox{with $\lambda$
the Lebesgue measure},$

i.e. any absolutely continuous invariant probability measure has a version of
its density function of bounded variation.

Proof of Lemma 5: Since $\inf|T_{\alpha}^{\prime}|>1$, applying Theorem 1 from
[6] immediately yields the assertion. $\Box$

######  Theorem 2.

Let $N\in\mathbb{N}_{\geq 2}$. Then there is a unique absolutely continuous
invariant probability measure $\mu_{\alpha}$ such that $T_{\alpha}$ is ergodic
with respect to $\mu_{\alpha}$.

Proof444see also page 185, Theorem 1 in [7] of Theorem 2: Let $\mu_{\alpha}$
be a unique absolutely continuous invariant probability measure for
$T_{\alpha}$ and choose its density function $h$ of bounded variation. Then
there exists an open interval $J$ such that $h(x)>0$ for any $x\in J$, since
$h$ has at most countably many discontinuity points. Consider
$\\{T^{n}J\,:\,n\geq 0\\}$. Since $\inf|T^{\prime}|>1$, there exists an
$n_{0}$ such that $T^{n_{0}}(J)$ includes a discontinuity point. (If necessary
we may choose endpoints of $J$ not in the preimages of discontinuity points of
$T_{\alpha}$.) We note that for any measurable subset $A\subset J$ with
$\mu_{\alpha}(A)>0$ equivalently $\lambda(A)>0$, $\mu(T^{n}A)>0$ for any
$n\geq 1$. Now $T^{n_{0}+1}(J)$ includes two intervals $J_{\ell}$ and $J_{r}$
attached to $\alpha$ and $\alpha+1$ respectively. For any measurable subset
$B_{0}\subset J_{\ell}\cup J_{r}$ of positive $\lambda$-measure,
$\mu(B_{0})>0$, since otherwise we have a contradiction; $\mu(B_{0})=0$ and
$\mu(T^{-(n_{0}+1)}(B_{0}))>0$ (since there is a $B_{1}\subset J$ such that
$T^{n_{0}+1}(B_{1})=B_{0},\mu(B_{1})>0$). This shows that any two absolutely
continuous invariant probability measures $\mu_{1}$ and $\mu_{2}$ cannot have
disjoint supports (i.e. they cannot be singular to each other), which is
equivalent to the uniqueness of the absolutely continuous invariant
probability measure and hence its ergodicity. $\Box$

The next result follows directly from Theorem 2:

######  Corollary 1.

If iteration of $T_{\alpha}$ maps all open subintervals of $I_{\alpha}$ to the
interval $I_{\alpha}^{-}$, then $I_{\alpha}$ contains no gaps.

Proof of Corollary 1: The assumption implies that the absolutely continuous
invariant probability measure $\mu_{\alpha}$ is equivalent to the Lebesgue
measure, which implies that for any measurable subset $A\subset I_{\alpha}$,
$\mu(A)=0$ if and only if $\lambda(A)=0$. Suppose that there is a gap $J$.
Since $J$ is an open interval, we have $\lambda(J)>0$, thus $\mu(J)>0$. Since
$\mu(I_{\alpha})<\infty$ implies a.e. $x\in J$, there exists infinitely many
positive integers $n$ such that $T^{n}(x)\in J$ (by the Poincaré recurrence
theorem), which contradicts the assumption that there is a gap. $\Box$

Before we present the first of two theorems on gaplessness, we note that in
the case $N=2$, the condition $|T_{\alpha}^{\prime}(x)|>2$ for all $x\in
I_{\alpha}$ is not satisfied for any $\alpha\in(0,\sqrt{2}-1]$.

######  Theorem 3.

Let $N\in\mathbb{N}_{\geq 3}$, and let $0<\alpha\leq\sqrt{N}-1$. Let
$|T_{\alpha}^{\prime}(x)|>2$ for all $x\in I_{\alpha}$. Then $I_{\alpha}$
contains no gaps.

Proof of Theorem 3: The condition implies $N/(\alpha+1)^{2}>2$, yielding
$\alpha<\sqrt{N/2}-1$. From Lemma 4 it follows that

$b(N,\alpha)>\frac{2\sqrt{2N}}{\sqrt{2N}-2},$

which is larger than $2$ for all $N\in\mathbb{N}_{\geq 3}$. So $I_{\alpha}$
consists of at least three cylinder sets. Since $|T_{\alpha}^{\prime}(x)|>2$
for all $x\in I_{\alpha}$, there exists an $\varepsilon>0$ such that for any
open interval $J_{0}$ that is contained in a cylinder set of $T_{\alpha}$ we
have

$\left|T_{\alpha}(J_{0})\right|\geq(2+\varepsilon)|J_{0}|,$

where $|J|$ denotes the length (i.e. Lebesgue measure) of an interval $J$.

If $T_{\alpha}(J_{0})$ contains two consecutive discontinuity points
$p_{i+1},p_{i}$ of $T_{\alpha}$, then

$(p_{i+1},p_{i})\subset T_{\alpha}(J_{0}),$

and we immediately have that

$I^{O}_{\alpha}:=(\alpha,\alpha+1)=T_{\alpha}(p_{i+1},p_{i})\subset
T_{\alpha}^{2}(J_{0}).$

If $T_{\alpha}(J_{0})$ contains only one discontinuity point $p$ of
$T_{\alpha}$, then $T_{\alpha}(J_{0})$ is the disjoint union of two
subintervals located in two adjacent cylinder sets:

$T_{\alpha}(J_{0})=J_{1}^{\prime}\cup J_{2}^{\prime}.$

Obviously,

$|T_{\alpha}(J_{0})|=|J_{1}^{\prime}|+|J_{2}^{\prime}|.$

Now select the larger of these two intervals $J_{1}^{\prime}$,
$J_{2}^{\prime}$, and call this interval $J_{1}$. Then

$|J_{1}|\geq(1+\tfrac{\varepsilon}{2})|J_{0}|.$

In case $T_{\alpha}(J_{0})$ does not contain any discontinuity point of
$T_{\alpha}$, we set $J_{1}=T_{\alpha}(J_{0})$. Induction yields that there
exists an $\ell\in\mathbb{N}$ such that

$|J_{\ell}|\geq\left(1+\frac{\varepsilon}{2}\right)^{\ell}\left|J_{0}\right|,$

whenever $T_{\alpha}(J_{\ell-1})$ includes no more than one discontinuity
point of $T_{\alpha}$. But then there must exist a $k\in\mathbb{N}$ such that
$T_{\alpha}(J_{k})$ contains two (or more) consecutive discontinuity points of
$T_{\alpha}$, and we find that $T_{\alpha}^{2}(J_{k})=I_{\alpha}^{O}$.
Applying Corollary 1, we conclude that there is no gap in $I_{\alpha}$. $\Box$

The next theorem, which is partly a corollary of the previous one, gives an
even more explicit condition for gaplessness.

######  Theorem 4.

Let $I_{\alpha}$ consist of five cylinder sets or more. Then $I_{\alpha}$ has
no gaps.

Proof of Theorem 4, part I: Let $I_{\alpha}$ consist of five cylinder sets or
more. Then $b(N,\alpha)>3$, implying

$\alpha<\frac{1}{2}\sqrt{\tfrac{4N}{3}+1}-\frac{1}{2}\,\,{\text{(cf.~{}(\ref{formula
a in b})), in which
case}}\,\,|T_{\alpha}^{\prime}(\alpha+1)|>3-\frac{3\sqrt{12N+9}-9}{2N}.$

The second inequality yields that for $N\in\mathbb{N}_{\geq 18}$ we have
$|T_{\alpha}^{\prime}(\alpha+1)|>2$ and, applying Theorem 3, $I_{\alpha}$ is
gapless. Now suppose $N\in\\{12,\ldots,17\\}$. Then $b(N\alpha)=3$ involves
arrangements with four cylinders. In each of these cases, the smallest
$\alpha$ such that $I_{\alpha}$ has not yet (i.e. decreasing from
$\alpha_{\max}$) consisted of five cylinder sets is $f_{7}$. In all six cases
(two of which are illustrated in Figure 4) we have
$|T_{\alpha}^{\prime}(f_{7}+1)|>2$, yielding the gaplessness of $I_{\alpha}$
for arrangements with five or more cylinders in case $N\in\\{12,\ldots,17\\}$.
This finishes the proof of Theorem 4 for $N\in\mathbb{N}_{\geq 12}$. For
$N\in\\{2,\ldots,11\\}$ a similar approach does not work. We will use some
ideas that we will introduce and develop in the next sections and will finish
the proof of Theorem 4 at the end of Section 5.

$\Box_{6}$$\Box_{5}$$\Box_{4}$$\Box_{3}$$N=17,\alpha=f_{7};|T^{\prime}(\alpha+1)|=2.0098\cdots$

$\Box_{6}$$\Box_{5}$$\Box_{4}$$\Box_{3}$$N=12,\alpha=f_{7};|T^{\prime}(\alpha+1)|=2.0415\cdots$

Figure 4. Two arrangements illustrating the gaplessness of arrangements with
five cylinders or more on account of Theorem 3

In the following we will go into conditions for gaplessness of arrangements
consisting of less than five cylinders. We will start with two cylinders and
will use the results for arrangements with three and four cylinders.

######  Remark 5.

Since $b(N,\alpha)$ is a strictly decreasing function of $\alpha$ (cf. Lemma
4) and $b(N,\alpha_{\max})=1+1/(\sqrt{N}-1)$, the condition
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$ is never satisfied in case
$N\in\\{2,3\\}$.

##  3\. Gaplessness when $I_{\alpha}$ consists of two cylinder sets

In general, when the branch number is not much larger than $1$ (which is when
$\alpha$ is not much smaller than $\alpha_{\max}$), the overall expanding
power of $T_{\alpha}$, determined by $T_{\alpha}^{\prime}$ (or
$|T_{\alpha}^{\prime}|$, which we will often use), is not enough to exclude
the existence of gaps; we shall elaborate on this in a subsequent article.
However, in the case of two cylinder sets
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$, there is a very clear condition under
which this power suffices:

######  Theorem 5.

Let $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$. If $T_{\alpha}(\alpha)\geq
f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq f_{d}$, then $I_{\alpha}$ is gapless.

Although the statement of 5 is intuitively clear, for the proof of Theorem 5
we need several results and lemmas that we will prove first. Then, immediately
following Remark 10 on page 10, we will prove Theorem 5 itself.

######  Remark 6.

If either $T_{\alpha}(\alpha)<f_{d-1}$ or $T_{\alpha}(\alpha+1)>f_{d}$, it is
easy to see that $(T_{\alpha}(\alpha),T_{\alpha}^{2}(\alpha))$ or
$(T_{\alpha}^{2}(\alpha+1),T_{\alpha}(\alpha+1))$ is a gap, respectively.

Since arrangements under the condition of Theorem 5 play an important role in
this section, we introduce the following notations:

######  Definition 3.

Let $N\in\mathbb{N}_{\geq 4}$ be fixed. For $d\in\mathbb{N}_{\geq 2}$, we
define $\mathcal{F}(d)$ as the family of all arrangements
$\Upsilon_{N,\alpha}$ such that $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$,
$T_{\alpha}(\alpha)\geq f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq f_{d}$. We will
write $\mathcal{F}^{*}(d)$ in case $\alpha$ satisfies the equation
$T_{\alpha}(\alpha)=f_{d-1}$, the root of which we will henceforth denote by
$\alpha(N,d)$.

######  Remark 7.

Note that for each $N\in\mathbb{N}_{\geq 4}$ and $d\in\mathbb{N}_{\geq 2}$ we
have that $\alpha(N,d)$, if it exists, is the only value of $\alpha$ such that
$\mathcal{F}^{*}(d)$ is not void.

If the expanding power of $T_{\alpha}$ is large enough to exclude the
existence of gaps for _the largest_ $\alpha$ for which an arrangement in
$\mathcal{F}(d)$ exists, there will not be gaps in any arrangement in
$\mathcal{F}(d)$. We will now first show how to find these largest $\alpha$,
which takes some effort. When we have finished that, we will go into the
expanding power of $|T_{\alpha}^{\prime}|$ in these arrangements with largest
$\alpha$.

For $4\leq N\leq 8$, with $d=2$, we have
$T_{\alpha_{\max}}(\alpha_{\max})>f_{1}$, while
$T_{\alpha_{\max}}(\alpha_{\max}+1)=\alpha_{\max}<f_{2}$. Hence we see
$\Upsilon_{N,\alpha_{\max}}\in\mathcal{F}(2)$ and
$\mathcal{F}(2)\neq\emptyset$. For $N\in\mathbb{N}_{\geq 9}$ we have
$T_{\alpha_{\max}}(\alpha_{\max})<f_{1}$. When $d=2$ we can find $\alpha$ such
that $\Upsilon_{N,\alpha}$ in $\mathcal{F}^{*}(2)$ for each $9\leq N\leq 17$;
see Figure 5, where ten arrangements in various $\mathcal{F}(d)$ are drawn.
Underneath each arrangement we have mentioned an approximation of
$\sigma(\alpha):=|T_{\alpha}^{\prime}(\alpha+1)|$, which we will later return
to. This $\sigma$ is important, because it is the expanding power on the
rightmost cylinder set that may be too weak to exclude gaps.

When $d=2$ and $N\in\mathbb{N}_{\geq 18}$, the condition
$T_{\alpha}(\alpha)=f_{d-1}$ yields $T_{\alpha}(\alpha+1)>f_{d}$, and $d$ has
to _increase by $1$_ so as to find an arrangement in $\mathcal{F}(3)$. When
$d=3$, for $18\leq N\leq 24$ we find that the largest $\alpha$ is $f_{d-2}-1$,
in which case $T_{\alpha}(\alpha+1)=\alpha<f_{d}$ and
$T_{\alpha}(\alpha)>f_{d-1}$ (so in this case the arrangement with the largest
$\alpha$ is in $\mathcal{F}(3)$ but not in $\mathcal{F}^{*}(3)$); for $25\leq
N\leq 49$, the largest $\alpha$ is such that $T_{\alpha}(\alpha)=f_{2}$. When
$N\in\mathbb{N}_{\geq 50}$, the family $\mathcal{F}(3)$ is empty and $d$ has
to increase further; see Figure 5 once more. In the proof of Lemma 8 this
approach (of exhausting $\mathcal{F}(d)$ for successive values of $N$ and
going to $\mathcal{F}(d+1)$ for larger values of $N$) will be formalised into
a proof by induction. Due to (12) such an increase is always possible, no
matter how large $d$ and $N$ become.

$\Box_{1}$$\Box_{2}$$F_{1}$$F_{2}$$N=9,\sigma\approx 1.01$

$\Box_{1}$$\Box_{2}$$F_{1}$$F_{2}$$N=17,\sigma\approx 1.06$

$\Box_{2}$$\Box_{3}$$F_{2}$$F_{3}$$N=18,\sigma\approx 1.27$

$\Box_{2}$$\Box_{3}$$F_{2}$$F_{3}$$N=24,\sigma\approx 1.23$

$\Box_{2}$$\Box_{3}$$F_{2}$$F_{3}$$N=25,\sigma\approx 1.22$

$\Box_{2}$$\Box_{3}$$F_{2}$$F_{3}$$N=49,\sigma\approx 1.20$

$\Box_{3}$$\Box_{4}$$F_{3}$$F_{4}$$N=50,\sigma\approx 1.33$

$\Box_{3}$$\Box_{4}$$F_{3}$$F_{4}$$N=99,\sigma\approx 1.26$

$\Box_{4}$$\Box_{5}$$F_{4}$$F_{5}$$N=100,\sigma\approx 1.36$

$\Box_{4}$$\Box_{5}$$F_{4}$$F_{5}$$N=165,\sigma\approx 1.29$

Figure 5. Arrangements in $\mathcal{F}(d)$, $d\in\\{2,3,4,5\\}$, where
$\alpha$ is maximal

Note that this inductive approach works since for each $d$ only _finitely_
many $N$ exist such that there are $\alpha$ with
$\Upsilon_{N,\alpha}\in{\mathcal{F}}(d)$. To see why this claim holds, note
that for fixed $N$ and $d$, the smallest $\alpha$ for which
$d=d(\alpha)=d_{\rm max}$ is $\alpha_{d}$, given by

$\alpha_{d}=f_{d+1}=\frac{\sqrt{4N+(d+1)^{2}}-(d+1)}{2};$

cf. (3). For this $\alpha$ it is not necessarily so that
$I_{\alpha_{d}}=\Delta_{d}\cup\Delta_{d-1}$, i.e. that $I_{\alpha_{d}}$
consists of two cylinder sets (e.g. if $N=2$ and $d=5$, there are five
cylinder sets). However, if $b(N,\alpha_{d})\leq 2$ we know that
$I_{\alpha_{d}}$ exists of two cylinder sets, the left one of which is full.
According to Lemma 4, the branch number $b(N,\alpha_{d})$ satisfies

$b(N,\alpha_{d})=\frac{N}{f_{d+1}(f_{d+1}+1)}=\frac{4N}{4N+(d+1)^{2}-2d\sqrt{4N+(d+1)^{2}}+d^{2}-1}.$

Keeping $d$ fixed and letting $N\to\infty$, we find

$\lim_{N\to\infty}b(N,\alpha_{d})=1.$

In view of this and Lemma 3 (and the results mentioned directly thereafter),
we choose $N$ sufficiently large, such that for $\alpha\geq\alpha_{d}$ we have
$b(N,\alpha)<5/4$ and $f_{d-1}-f_{d}>1/4$.

Now suppose that for such a sufficiently large value of $N$ there exists an
$\alpha\geq\alpha_{d}$, such that $\alpha\in{\mathcal{F}}(d)$. Then by
Definition 2 of branch number and the assumption that
$\alpha\in{\mathcal{F}}(d)$, we have that

$b(N,\alpha)\geq 1+f_{d-1}-f_{d}>1\tfrac{1}{4},$

which is _impossible_ since for $N$ sufficiently large, $d$ fixed and
$\alpha\geq\alpha_{d}$ we have

$b(N,\alpha)\leq b(N,\alpha_{d})<1\tfrac{1}{4}.$

It follows that for $d$ fixed and $N$ sufficiently large, ${\mathcal{F}}(d)$
is void.

We will prove (in Lemma 8) that when $N\in\mathbb{N}_{\geq 25}$ there exists a
minimal $d\in\mathbb{N}_{\geq 3}$ such that the arrangement in
$\mathcal{F}(d)$ with $\alpha$ maximal lies in $\mathcal{F}^{*}(d)$. Before we
will prove this, we will explain the relation between $d$ and $N$ for
arrangements in $\mathcal{F}^{*}(d)$.

In Figure 5 we see that for $N\in\\{49,99,165\\}$ the arrangements in
$\mathcal{F}^{*}$ are very similar, and that the arrangement for $N=100$ is
more similar to these than the arrangement for $N=50$. Moreover, the last
three arrangements look hardly curved. This is easy to understand, considering
the following equations, where $b(N,\alpha)=b$ is fixed:

$|T_{\alpha}^{\prime}(\alpha)|=\frac{b(\sqrt{4bN+b^{2}}+2N+b)}{2N}\quad{\text{and}}\quad|T_{\alpha}^{\prime}(\alpha)|-|T_{\alpha}^{\prime}(\alpha+1)|=\frac{b\sqrt{4bN+b^{2}}}{N}.$

Since $(b\sqrt{4bN+b^{2}})/N$ is a decreasing function of $N$, approaching $0$
from above as $N\to\infty$, the second equation implies that for a fixed
branch number $b$ the branches become less curved as $N$ increases; i.e., the
curves approach linearity as $N\to\infty$ and $b$ is fixed. Although in
$\mathcal{F}^{*}(d)$ the branch number is not so much fixed as bounded between
$1$ and $2$, we have a similar decrease of curviness as $N$ increases. The
arrangements for $N\in\\{49,99,165\\}$ in Figure 5 suggest that (assuming
$T_{\alpha}(\alpha)=f_{d-1}$, i.e. $\alpha=\alpha(N,d)$)) when $N\to\infty$
(and $d\to\infty$ and $\alpha\to\infty$ accordingly), the difference
$f_{d}-T_{\alpha}(\alpha+1)$ tends to $0$, yielding a ‘limit graph’ of
$T_{\alpha}$ that consists of two parallel line segments (the straightened
branch curves of $T_{\alpha}$); see Figure 6, obtained by translating the
graph over $(-\alpha,-\alpha)$. In this situation we have both
$a:=T_{\alpha}(\alpha)\ (\mathrm{mod}\ \alpha)=f_{d-1}\ (\mathrm{mod}\
\alpha)$ and $T_{\alpha}(\alpha+1)=f_{d}$ (also $\ (\mathrm{mod}\ \alpha)$ in
Figure 6). Because in the limit both parts of the graph are linear with the
same slope, we also have that $(0,a+1)$ lies on the prolonged right line
segment, from which we derive that the line segments have slope $-1/a$. The
line with equation $y=-x/a+a+1$ intersects the line $y=1$ at $(a^{2},1)$ (so
the dividing line is $x=a^{2}$) and intersects the line $x=1$ in
$(1,-1/a+a+1)$, yielding the point $(-1/a+a+1,-1/a+a+1)$ on the line through
$(0,a)$ with equation $y=-x/a+a$ (since $T_{\alpha}(\alpha+1)=f_{d})$. From
this we derive $2a^{2}=1$, so $a=\sqrt{1/2}$.

$(a,a)$$(1,-\tfrac{1}{a}+a+1)$$a$$(a^{2},1)$$0$$1$$1$ Figure 6. The ‘limit
graph’ of $T_{\alpha}$, translated over $(-\alpha,-\alpha)$, under the
conditions $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$ and $N/\alpha-d=f_{d-1}$
for $N\to\infty$ (and $\alpha,d\to\infty$ accordingly)

From Figure 6 we almost immediately find that the branch number for the limit
case is $\sqrt{2}$ and that the dividing line is at $1/2$. We use this
_heuristic_ to find a formula describing the relation between $N$ and $d$ for
arrangements in $\mathcal{F}^{*}(d)$ very precisely. Note that for
arrangements similar to the limit graph we have

$1+f_{d-1}-f_{d}=\frac{\sqrt{4N+(d-1)^{2}}-\sqrt{4N+d^{2}}+1}{2}+1\approx
b(N,\alpha)\approx\sqrt{2},$

from which we derive

(13) $N\approx(4+3\sqrt{2})(d^{2}-d)+2\quad{\text{or}}\quad
d\approx\frac{1}{2}\left(1+\sqrt{(6\sqrt{2}-8)(N-2)+1}\right).$

If, for $d$ fixed, we determine arrangements in $\mathcal{F}^{*}(d)$ such that
the difference $f_{d}-T_{\alpha}(\alpha+1)$ is positive and as small as
possible according to our heuristic, the best function seems to be
$N=(4+3\sqrt{2})(d^{2}-d)$, yielding the right $N$ (after rounding off to the
nearest integer) for
$d\in\\{3,\ldots,500\\}\setminus\\{9,50,52,68,69,80,97,129,$
$167,185,210,231,289,330,416,440,444,479,485\\}$, in all of which cases the
rounding off should have been up instead of down. For $d=2$ we find $N=\lceil
2(4+3\sqrt{2})\rceil=17$, for $d=3$ we find $N=\lfloor
6(4+3\sqrt{2})\rfloor=49$, for $d=4$ we find $N=\lceil
12(4+3\sqrt{2})\rceil=99$ and or $d=5$ we find $N=\lceil
20(4+3\sqrt{2})\rceil=165$; see Figure 5 once more.

Although we do not know generally when rounding off to the nearest integer
yields the right $N$, with (13) we can find a very good overall indication of
the relation between $d$ and $N$ for arrangements in $\mathcal{F}^{*}(d)$ by
looking at the difference between the image of $\alpha(N,d)+1$ and $f_{d}$;
see Definition 3. With some straightforward calculations we find that

(14) $\alpha(N,d)=\frac{N\left(\sqrt{4N+(d-1)^{2}}-(d+1)\right)}{2(N-d)}.$

Applying (14), we write $f_{d}(N)-(N/(\alpha(N,d)+1)-(d-1))$ as

$j_{d}(N):=\frac{(N^{2}+dN+d)\sqrt{4N+d^{2}}-N^{2}\sqrt{4N+(d-1)^{2}}-(N^{2}-d(d-4)N-d(d-2))}{2(N^{2}+dN+d)}$

and, more generally, define

(15)
$j_{d}(x):=\frac{(x^{2}+dx+d)\sqrt{4x+d^{2}}-x^{2}\sqrt{4x+(d-1)^{2}}-(x^{2}-d(d-4)x-d(d-2))}{2(x^{2}+dx+d)}$

for $x\in[25,\infty)$.

We note that $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$ is equivalent to
$N/(\alpha+1)-(d-1)\geq\alpha$. In case $\alpha=\alpha(N,d)$, we have

(16)
$\frac{N}{\alpha(N,d)+1}-(d-1)=\frac{N^{2}\sqrt{4N+(d-1)^{2}}-\left((d-1)N^{2}+2d(d-2)N+2d(d-1)\right)}{2(N^{2}+dN+d)}.$

Applying (16), for the difference
$h_{d}(N):=N/(\alpha(N,d)+1)-(d-1)-\alpha(N,d)$ we write

(17)
$h_{d}(N):=\frac{2N^{3}+4dN^{2}+(2d^{3}-5d^{2}+3d)N+2d^{2}(d-1)-dN(2N+1)\sqrt{4N+(d-1)^{2}}}{2(N-d)(N^{2}+dN+d)},$

and, more generally, define

(18)
$h_{d}(x):=\frac{2x^{3}+4dx^{2}+(2d^{3}-5d^{2}+3d)x+2d^{2}(d-1)-dx(2x+1)\sqrt{4x+(d-1)^{2}}}{2(x-d)(x^{2}+dx+d)},$

for $x\in[25,\infty)$.

Now we can prove the lemma that is illustrated by the arrangements for
$N\in\\{17,49,99,165\\}$ in Figure 5. In order to so, we define for fixed
$d\in\mathbb{N}_{\geq 2}$

$S(d):=\\{N\in\mathbb{N}_{\geq
4}:I_{\alpha}=\Delta_{d}\cup\Delta_{d-1},T_{\alpha}(\alpha)=f_{d-1}(N)\,{\text{and}}\,\,T_{\alpha}(\alpha+1)\leq
f_{d}(N)\\}$

and

$M_{d}:=\max S(d).$

######  Lemma 6.

Let $d\in\mathbb{N}_{\geq 2}$. Then
$M_{d}\in\\{\lfloor(4+3\sqrt{2})(d^{2}-d)\rfloor,\lceil(4+3\sqrt{2})(d^{2}-d)\rceil\\}$.

Proof of Lemma 6: First we note that for $d=2$, we have that
$\lceil(4+3\sqrt{2})(d^{2}-d)\rceil$ equals $17$, which corresponds with what
we had already calculated and drawn in Figure 5. Now let $d\in\mathbb{N}_{\geq
3}$. First we have to show that $h_{d}(M_{d})>0$ for
$M_{d}\in\\{\lfloor(4+3\sqrt{2})(d^{2}-d)\rfloor,\lceil(4+3\sqrt{2})(d^{2}-d)\rceil\\}$,
for this assures us that $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$. We will
leave this to the reader; it is merely very cumbersome to show, while
technically straightforward555We have throughout this paper frequently used
(Wolfram) Mathematica for making intricate calculations, all of which are
nonetheless algebraically basic. In relevant cases we think it will be evident
if we did..

The only thing left to do is showing that

(19) $\begin{cases}j_{d}((4+3\sqrt{2})(d^{2}-d))>0;\\\
j_{d}((4+3\sqrt{2})(d^{2}-d)+1)<0,\end{cases}$

since the first equation implies that
$j_{d}(\lfloor(4+3\sqrt{2})(d^{2}-d)\rfloor)>0$, while the second implies that
$j_{d}(\lceil(4+3\sqrt{2})(d^{2}-d)\rceil+1)<0$. The work to be done is as
cumbersome and straightforward as the previous work to be done for this proof
and is left to the reader as well. $\Box$

Before we will show that for $N\in\mathbb{N}_{\geq 25}$ there are a
$d\in\mathbb{N}_{\geq 3}$ and an $\alpha$ such that
$\Upsilon_{N,\alpha}\in\mathcal{F}^{*}(d)$, we will prove the following lemma:

######  Lemma 7.

Let $d\in\mathbb{N}_{\geq 3}$. Let $N\in\mathbb{N}_{\geq 25}$ be such that
$I_{\alpha(M_{d},d)}=\Delta_{d}\cup\Delta_{d-1}$ and
$T_{\alpha(M_{d},d)}(\alpha(M_{d},d))=f_{d-1}$ for $M_{d}\in\\{N,N+1\\}$. Then

$T_{\alpha(N+1,d)}(\alpha(N+1,d)+1)-\alpha(N+1,d)>T_{\alpha(N,d)}(\alpha(N,d)+1)-\alpha(N,d),$

i.e. $h_{d}(N+1)>h_{d}(N)$.

Proof of Lemma 7: We want to show that $h_{d}(N)$ from (17) is an increasing
sequence, and do so by calculating the derivative of with $x\in[25,\infty)$,
and then showing that $h_{d}^{\prime}(x)>0$ on $[25,\infty)$. Although a
little bit intricate, the work is straightforward and is left to the reader.
$\Box$

Now we can prove the following lemma:

######  Lemma 8.

Let $N\in\\{9,\ldots,17,25,26,\ldots\\}$. Then there are $d\in\mathbb{N}_{\geq
2}$ and $\alpha\in(0,\sqrt{N}-1)$ such that
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$, $T_{\alpha}(\alpha)=f_{d-1}$ and
$T_{\alpha}(\alpha+1)\leq f_{d}$ (i.e. $\alpha=\alpha(N,d)$).

Proof of Lemma 8: We will use induction on $d$. For
$N\in\\{9,\ldots,17,25,26,\ldots,99\\}$ and $d\in\\{2,3,4\\}$ we refer to
Figure 5 and leave the calculations to the reader. Specifically, we have for
$50\leq N\leq 99$ that $\Upsilon_{N,\alpha(N,4)}\in\mathcal{F}^{*}(4)$. It is
easily seen that $\Upsilon_{99,\alpha(99,5)}\in\mathcal{F}^{*}(5)$ as well.
Due to Lemma 6, there is an $N_{5}>99$ such that
$\Upsilon_{N_{5},\alpha(N_{5},5)}\in\mathcal{F}^{*}(5)$. Applying Lemma (7),
we see that for all $N\in\\{99,\ldots,N_{5}\\}$ we have
$\Upsilon_{N,\alpha(N,5)}\in\mathcal{F}^{*}(5)$. For the induction step, let
$d\in\mathbb{N}_{\geq 5}$ be such that there is an $\alpha$ for which
$\Upsilon_{N_{d},\alpha}\in\mathcal{F}^{*}(d)$, where $N_{d}$ is the largest
such $N$ possible, cf. Lemma 6. If we can show that for this $N_{d}$ there is
an $\alpha^{\prime}$ such that
$\Upsilon_{N_{d},\alpha^{\prime}}\in\mathcal{F}^{*}(d+1)$, we are finished.
This can be done by showing that

(20) $h_{d+1}((4+3\sqrt{2})(d^{2}-d)-1)>0,$

for this implies $h_{d+1}(N_{d})>0$, in which case $\alpha^{\prime}$ is such
that $N_{d}/\alpha^{\prime}-(d+1)=f_{d}$, i.e.
$\alpha^{\prime}=\alpha(N,d+1)$. Although intricate, the calculations are
straightforward and are left to the reader. $\Box$

######  Remark 8.

Although Lemma 8 is about $N$ in the first place, our approach is actually
based on increasing $d$ and then determining all $N$ such that arrangements
$\Upsilon_{N,\alpha}\in\mathcal{F}^{*}(d)$ exist. The proof of Lemma 8 yields
the arrangements with the _smallest_ $d$ (and therefore the largest $\alpha$)
for which $\Upsilon_{N,\alpha}\in\mathcal{F}^{*}(d)$, as illustrated by the
last five arrangements of Figure 5.

######  Example 1.

For $d=4$ we have $M_{d}-1=\lfloor(4+3\sqrt{2})(d^{2}-d)\rfloor=98$. Then
$\Upsilon_{M_{d}-1,\alpha(M_{d}-1,4)}\in\mathcal{F}^{*}(4)$ and
$\Upsilon_{M_{d},\alpha(M_{d},4)}\in\mathcal{F}^{*}(4)$, while
$\Upsilon_{M_{d}+1,\alpha(M_{d}+1,5)}\in\mathcal{F}^{*}(5)$; see Figure 5. It
follows immediately from our construction of $\alpha(M_{d}+1,5)$ that this is
the largest $\alpha$ such that $\Upsilon_{M_{d}+1,\alpha}\in\mathcal{F}(5)$.

With manual calculations we can quickly calculate the expanding power of
$T_{\alpha}$ in $\alpha+1$ for arrangements in $\mathcal{F}$ and
$\mathcal{F}^{*}$ and $N$ not too large, say $N\in\mathbb{N}_{\leq 49}$, where
the smallest values are found where $\alpha$ is as large as possible. The next
proposition gives a lower bound for $|T_{\alpha}^{\prime}(\alpha+1)|$ for such
arrangements for most $N$.

######  Proposition 1.

Let $N\in\\{18\\}\cup\\{50,51,\ldots\\}\setminus\\{95,\ldots,99\\}$ and
$\alpha\in(0,\sqrt{N}-1]$ such that $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$
for some $d\in\mathbb{N}$, $d\in\mathbb{N}_{\geq 2}$. Furthermore, suppose
that $T_{\alpha}(\alpha)\geq f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq f_{d}$.
Then $|T_{\alpha}^{\prime}(\alpha+1)|>\sqrt[3]{2}=1.259921\cdots$.

Proof of Proposition 1: Considering Lemma 8, we can confine ourselves to
arrangements in $\mathcal{F}^{*}$ with $\alpha$ as large as possible. For
$\alpha=\alpha(N,d)$ (cf. (14)) we can write
$|T_{\alpha}^{\prime}(\alpha+1)|=N/(\alpha+1)^{2}$ as

(21)
$k_{d}(N)=\frac{2N^{4}+(d-1)^{2}N^{3}+2d(d-1)N^{2}+2d^{2}N+((d-1)N^{3}+2dN^{2})\sqrt{4N+(d-1)^{2}}}{2(N^{4}+2dN^{3}+d(d+2)N^{2}+2d^{2}N+d^{2})}.$

It is not hard to find that, for $d$ _fixed_ , $k_{d}$ is a decreasing
sequence, with $\lim_{N\to\infty}k_{d}(N)=1$. However, from (13) it follows
that if $N\to\infty$ we have that also $d\to\infty$ in a precise manner. Due
to the previous lemmas, for each $d$ we can confine ourselves to considering
only $N/(\alpha+1)^{2}$ for the largest $N$ and $\alpha$ such that
$\Upsilon_{N,\alpha}\in\mathcal{F}^{*}(d)$. Applying Lemma 6, an easy way to
check if indeed $|T_{\alpha(N,d)}^{\prime}(\alpha(N,d)+1)|>\sqrt[3]{2}$ is
considering $k_{d}(x)$, with $x\in[100,\infty)$, and then calculating
$k_{d}((4+3\sqrt{2})(d^{2}-d)+1)$ for $d\in\mathbb{N}_{\geq 5}$, which is
amply larger than $\sqrt[3]{2}=1.2599\cdots$. For the remaining cases $d=3$
and $N=18$ and for $d=4$ and $N\in\\{50,51,\ldots,94\\}$ it is easily checked
manually that indeed $|T_{\alpha(N,d)}^{\prime}(\alpha(N,d)+1)|>\sqrt[3]{2}$.
$\Box$

######  Remark 9.

Considering our previous remarks concerning arrangements in $\mathcal{F}^{*}$,
it may be clear that $\lim_{N\to\infty}N/(\alpha(N,d)+1)^{2}=\sqrt{2}$.

######  Remark 10.

The value $\sqrt[3]{2}$ in the proof of Proposition 1 relates to the proof of
Theorem 5 and also to the proofs of Proposition 9 and Theorem 3, where the
numbers $\sqrt{2}$ and $2$ have a similar importance. Considering the proof of
Proposition 1, we could actually replace $\sqrt[3]{2}$ by the smallest
possible value, given by

$\frac{94}{(\alpha(94)+1)^{2}}=\frac{20480015+320305\sqrt{385}}{21233664}=1.2604\cdots.$

Finally we are ready to prove Theorem 5, stating that
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$, with $d:=d(\alpha)$, is gapless if
$T_{\alpha}(\alpha)\geq f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq f_{d}$.
Considering Remark 10 the value $\sqrt[3]{2}$ in Proposition 1 can be replaced
by $1.26$, the third power of which is $2.000376$. We will use this to stress
that the gaplessness of Theorem 5 is actually relatively ample and does not
require infinitesimal estimations.

Proof of Theorem 5: First we note that the conditions imply
$N\in\mathbb{N}_{\geq 4}$. Now let $\Upsilon_{N,\alpha}\in\mathcal{F}(d)$ and
let $K\subset I_{\alpha}$ be any open interval. Since $K$ expands under
$T_{\alpha}$, there is an $n\in\mathbb{N}\cup\\{0\\}$ such that
$T_{\alpha}^{n}(K)$ contains for the first time a fixed point or the
discontinuity point $p_{d}$, in the former case of which we are finished. So
we assume that $T_{\alpha}^{n}(K)\cap\Delta_{d}=(b,p_{d}]=:L$, with
$f_{d}<b<p_{d}$. Note that
$T_{\alpha}(L)=[\alpha,T_{\alpha}(b))\subset[\alpha,f_{d})$. For
$T_{\alpha}^{2}(L)=(T_{\alpha}^{2}(b),T_{\alpha}(\alpha)]$, we similarly may
assume that $f_{d-1}<T_{\alpha}^{2}(b)<\alpha+1$ (since otherwise $f_{d-1}\in
T_{\alpha}^{2}(L)$, and again we are done).

Now suppose that $T_{\alpha}^{3}(L)$ contains $p_{d}$, excluding $f_{d}\in
T_{\alpha}^{3}(L)$. Then $T_{\alpha}^{3}(L)=L_{1}\cup M_{1}$, with
$L_{1}=[T_{\alpha}^{2}(\alpha),p_{d}]$ and $M_{1}=(p_{d},T_{\alpha}^{3}(b))$.
First we confine ourselves to
$N\in\\{18\\}\cup\\{50,\ldots\\}\setminus\\{95,\ldots,99\\}$. Since then
$|T_{\alpha}^{3}(L)|>2.000376|L|$ (cf. Remark 10), we have certainly
$|L_{1}|>1.001|L|$ or $|M_{1}|>1.001|L|$. If we consider the images of $L_{1}$
and $M_{1}$ under $T_{\alpha}$, $T_{\alpha}^{2}$ and $T_{\alpha}^{3}$
similarly as we did with the images of $L$, we find that due to expansiveness
(see the proof of Theorem 3) there must be an $m$ such that $f_{d}\in
T_{\alpha}^{3m}(L_{1})$ or $f_{d-1}\in T_{\alpha}^{3m}(M_{1})$ and we are
finished. If $T_{\alpha}^{3}(L)$ does _not_ contain $p_{d}$, the expansion of
$L$ will only go on longer, yielding even larger $L_{1}^{\prime}$ and
$M_{1}^{\prime}$ and the reasoning would only be stronger that no gaps can
exist.

For $N\in\\{4,\ldots,17,19,20,\ldots,49,95,96,\ldots,99\\}$ a similar approach
can be taken, but there is no useful general lower bound for
$|T_{\alpha}^{\prime}(x)|$ on $I_{\alpha}$. For these cases, however, the
moderate expanding power in $\Delta_{d-1}$ is easily made up for by a
relatively strong expanding power in $\Delta_{d}$, and the gaplessness is
easily, although tediously, checked by hand (cf. Examples 2 and 3 below). This
finishes the proof of Theorem 5.
$\Box$

######  Example 2.

In case $N=7$, there exist $\alpha\in(0,\sqrt{7}-1]$ for which
$I_{\alpha}=\Delta_{2}\cup\Delta_{1}$. The largest $\alpha$ for which
$\Upsilon_{7,\alpha}\in\mathcal{F}(2)$ is $\alpha_{\max}=\sqrt{7}-1$, in which
case $|T_{\alpha}^{\prime}(\alpha+1)|=1$. However,
$|T_{\alpha}^{\prime}(f_{2})|=2.0938\cdots>2$, and the approach taken above
works if only for the expanding power of $T_{\alpha}$ on $[\alpha,f_{2})$.

######  Example 3.

In case $N=99$, we have $I_{\alpha}=\Delta_{4}\cup\Delta_{3}$, and
$\Upsilon_{99,\alpha}\in\mathcal{F}^{*}$ for
$\alpha=99(\sqrt{405}-5)/190=7.8807\cdots$. Then
$|T_{\alpha}^{\prime}(\alpha+1)|=1.2552\cdots$,
$|T_{\alpha}^{\prime}(f_{3})|=1.3503\cdots$ and
$|T_{\alpha}^{\prime}(f_{4})|=1.4908\cdots$. So for an interval $(p_{4},x)$,
with $x\in(p_{d},f_{3})$, assuming that $f_{4}\not\in
T_{\alpha}^{3}(p_{4},x)$, we have
$|T_{\alpha}^{3}(p_{4},x)|>1.3503\cdots\times 1.2552\cdots\times
1.4908\cdots\times|(p_{4},x)|\gg 2|(p_{4},x)|$, implying enough expanding
power for $T_{\alpha}^{3}$ to exclude the existence of gaps.

######  Remark 11.

We can also prove that $|T_{\alpha}^{\prime}(x)|>\sqrt{2}$ on $\Delta_{d}$ for
all arrangements under the assumptions of Theorem 5, but we cannot do without
knowledge about the slope on $\Delta_{d-1}$.

Next we will make preparations for formulating a sufficient condition for
gaplessness in case $I_{\alpha}$ consists of three cylinder sets. Proving it
involves more subtleties on the one hand, but will have a lot of similarities
with the two-cylinder set case on the other hand. Once we have finished that,
not much work remains to be done for gaplessness in case $I_{\alpha}$ consists
of four or five cylinder sets.

##  4\. A sufficient condition for gaplessness when $I_{\alpha}$ consists of
three or four cylinder sets

When $I_{\alpha}=\Delta_{d}\cup\ldots\cup\Delta_{d-m}$, with $m\in\\{2,3\\}$,
there is a sufficient condition for gaplessness that resembles the condition
for gaplessness in case $I_{\alpha}$ consists of two cylinder sets a lot:

######  Theorem 6.

Let $I_{\alpha}=\Delta_{d}\cup\ldots\cup\Delta_{d-m}$, with $m\in\\{2,3\\}$.
Then $I_{\alpha}$ is gapless if

$T_{\alpha}(\alpha)\geq
f_{d-1}\,\,{\text{\bf{or}}}\,\,T_{\alpha}(\alpha+1)\leq f_{d-m+1}.$

We will prove this theorem in parts. In Subsection 4.1 we will prove Theorem 6
for $m=2$; in Subsection 4.2 we will extend the result of Subsection 4.1 to
$m=3$; considering Theorem 4, extension to larger $m$ is not useful.

######  Remark 12.

The difference between the ‘and’ of Theorem 5 and the ‘or’ of Theorem 6 has to
do with the existence, in the latter case, of at least one full cylinder set.

###  4.1. Gaplessness when $I_{\alpha}$ consists of three cylinder sets

Since we have $m=2$, the condition $T_{\alpha}(\alpha)\geq f_{d-1}$ can be
split in

(22) $\begin{cases}1.\,\,T_{\alpha}(\alpha+1)\leq f_{d-1}\leq
T_{\alpha}(\alpha);\\\ 2.\,\,f_{d-1}\leq T_{\alpha}(\alpha)\leq
T_{\alpha}(\alpha+1);\\\ 3.\,\,f_{d-1}\leq T_{\alpha}(\alpha+1)\leq
T_{\alpha}(\alpha);\end{cases}$

of course the condition $T_{\alpha}(\alpha+1)\leq f_{d-1}$ can be split in a
similar way. We will prove Theorem 6 by proving gaplessness according to this
distinction in three cases, associated with Lemma 9, 10 and 11 respectively.
The first of these is not very hard to prove:

######  Lemma 9.

Let $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}$. If
$T_{\alpha}(\alpha)\geq f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq f_{d-1}$, then
$I_{\alpha}$ is gapless.

Proof of Lemma 9: The assumptions imply that $b(N,\alpha)>2$, yielding
$\sigma(\alpha)=|T_{\alpha}^{\prime}(\alpha+1)|>\sqrt{2}$ for
$N\in\mathbb{N}_{\geq 17}$. If $N\in\mathbb{N}_{\geq 17}$, we let $K\subset
I_{\alpha}$ be any open interval. Since $K$ expands under $T_{\alpha}$, there
is an $n\in\mathbb{N}\cup\\{0\\}$) such that $T_{\alpha}^{n}(K)$ contains a
fixed point or a discontinuity point $p_{d-i}$ (with $i\in\\{0,1\\}$) , in the
former case of which we are finished. So we assume that
$T_{\alpha}^{n}(K)\supset L$, where $L=(b,p_{d-i}]$, with $f_{d-i}<b<p_{d-i}$,
with $i\in\\{0,1\\}$. If $T_{\alpha}(L)$ contains a fixed point, we are
finished. If $T_{\alpha}(L)$ does not contain a fixed point, then it cannot
contain a discontinuity point, and we have that $|T_{\alpha}^{2}(L)|>2|L|$,
implying enough expanding power of $T_{\alpha}$ to ensure gaplessness of at
least one cylinder set (which might be non-full). Since both
$T_{\alpha}(\alpha)\geq f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq f_{d-1}$, it
follows that $I_{\alpha}$ is gapless. For $2\leq N\leq 16$ the slopes on
$I_{\alpha}$ may differ considerably: for some $N$, such as $N=7$ and $N=16$
we also have $\sigma>\sqrt{2}$, but when this is not he case, the steepness
left of $f_{d-2}$ is amply larger then $\sqrt{2}$; see Figure 7 for some
examples where $\alpha$ is as large as possible. This finishes the proof of
Lemma 9 (cf. case 1 in (22)). $\Box$

$\Box_{1}$$\Box_{2}$$\Box_{3}$$N=3,\sigma=1$$|T_{\alpha}^{\prime}(f_{1})|\approx
1.77$$|T_{\alpha}^{\prime}(f_{2})|\approx 3$

$\Box_{1}$$\Box_{2}$$\Box_{3}$$N=4,\sigma\approx
1.06$$|T_{\alpha}^{\prime}(f_{1})|\approx
1.64$$|T_{\alpha}^{\prime}(f_{2})|\approx 2.62$

$\Box_{2}$$\Box_{3}$$\Box_{4}$$N=7,\sigma\approx
1.46$$|T_{\alpha}^{\prime}(f_{2})|\approx
2.09$$|T_{\alpha}^{\prime}(f_{3})|\approx 2.95$

$\Box_{2}$$\Box_{3}$$\Box_{4}$$N=9,\sigma\approx
1.40$$|T_{\alpha}^{\prime}(f_{2})|\approx
1.92$$|T_{\alpha}^{\prime}(f_{3})|\approx 2.62$

$\Box_{5}$$\Box_{4}$$\Box_{3}$$N=16,\sigma\approx
1.64$$|T_{\alpha}^{\prime}(f_{3})|\approx
2.08$$|T_{\alpha}^{\prime}(f_{4})|\approx 2.62$

Figure 7. Arrangements with largest $\alpha$ such that there is a $d$ with
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}$ under the condition
$T_{\alpha}(\alpha)\geq f_{d-1}$ _and_ $T_{\alpha}(\alpha+1)\leq f_{d-1}$

If $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}$ under the condition
$T_{\alpha}(\alpha)\geq f_{d-1}$ and $T_{\alpha}(\alpha+1)>f_{d-1}$ or under
the condition $T_{\alpha}(\alpha)<f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq
f_{d-1}$, $I_{\alpha}$ is gapless as well, but this is much harder to prove.
The following definition will be convenient:

######  Definition 4.

Let $I_{\alpha}=\Delta_{d}\cup\ldots\cup\Delta_{d-m}$, and $1\leq m\leq d-1$.
If $T_{\alpha}(\alpha)\leq f_{d-1}$ or $T_{\alpha}(\alpha+1)\geq f_{d-m+1}$,
the cylinder set $\Delta_{d}$ respectively $\Delta_{d-m}$ is called _small_.

Taking a similar approach as in the proof of Theorem 5, one can show that the
map $T_{\alpha}$ has enough expansive power to ensure that for any open
interval $K\subset I_{\alpha}$ there exists a non-negative integer $n$ such
that $T_{\alpha}^{n}(K)$ contains a fixed point. If this fixed point is in a
non-small or even full cylinder set, we are done (as in the proofs of Theorem
5 and Lemma 9). However, if this fixed point is from the _small_ cylinder set,
then it only follows that every point of the small cylinder set is in the
orbit under $T_{\alpha}$ of some point in $K$. Note this implies that the
_small_ cylinder set is _gapless_. So we may assume that the small cylinder
set is gapless. Let us assume that the left cylinder set is small. We define
$L:=T_{\alpha}(\Delta_{d})\setminus\Delta_{d}$. Since $\Delta_{d}$ is gapless,
we have $L=(p_{d},T(\alpha)]\subset(p_{d},f_{d-1})$, so
$T_{\alpha}(L)=[T_{\alpha}^{2}(\alpha),\alpha+1)$. If
$T_{\alpha}^{2}(\alpha)\leq f_{d-2}$, we are finished, so we assume that
$T_{\alpha}^{2}(\alpha)>f_{d-2}$. We then have
$T_{\alpha}^{2}(L)=(T_{\alpha}(\alpha+1),T_{\alpha}^{3}(\alpha)]$. If
$T_{\alpha}^{3}(\alpha)\geq f_{d-1}$ we are finished, since then $f_{d-1}\in
T_{\alpha}^{2}(L)$.

The question arises whether it is possible to keep avoiding fixed points if we
go on with letting $T_{\alpha}$ work on $L$ and its images (or similarly, when
the right cylinder set is small, some interval
$R:=T_{\alpha}(\Delta_{1})\setminus\Delta_{1}$). We will argue that this is
not possible in the two most plausible cases for gaps to exist, involving the
least expansion.

$\Box_{3}$$\Box_{4}$$\Box_{5}$$N=23,\alpha=2.898$$|T_{\alpha}^{\prime}(\alpha+1)|\approx
1.51$$|T_{\alpha}^{\prime}(\alpha)|\approx 2.74$

$\Box_{1}$$\Box_{2}$$\Box_{3}$$N=11$,
$\alpha=1.873\cdots$$|T_{\alpha}^{\prime}(\alpha+1)|\approx
1.33$$|T_{\alpha}^{\prime}(\alpha)|\approx 3.14$

$\Box_{1}$$\Box_{2}$$\Box_{3}$$N=7,\alpha=1.54$$|T_{\alpha}^{\prime}(\alpha+1)|\approx
1.09$$|T_{\alpha}^{\prime}(\alpha)|\approx 2.95$

Figure 8. Arrangements with one very small cylinder

The first case is illustrated with two arrangements in Figure 8, one where
$N=23$ and one where $N=11$. In both arrangements one outer cylinder is very
small while the other one is full or almost full. In the arrangement where
$N=23$, we see that $L$ is a very narrow strip between $p_{5}$ and
$T_{\alpha}(\alpha)$, $T_{\alpha}^{2}(L)$ is not so narrow anymore, and
$T_{\alpha}^{3}(L)$ is definitely wide enough to make clear that avoiding
fixed points $f_{4}$ and $f_{3}$ is not possible. The middle arrangement,
where $N=11$, is an example of the case where $\Delta_{d-2}$ is small and
$\Delta_{d}$ is actually full. Here we have that
$R:=T_{\alpha}(\Delta_{1})\setminus\Delta_{1}$ is a very narrow strip between
$T_{\alpha}(\alpha+1)$ and $p_{2}$ and that $T_{\alpha}^{2}(R)$ is only
slightly larger than $T_{\alpha}(\Delta_{1})$, whence eventually there will be
an $n\in\mathbb{N}$ such that $f_{3}\in T_{\alpha}^{n}(R)$ or $f_{2}\in
T_{\alpha}^{n}(R)$.

The rightmost arrangement in Figure 8 is an is an illustration of the second
plausible case for the existence of gaps: here $\Delta_{3}$ is small, while
$\Delta_{1}$ is incomplete but _not_ small. This arrangement illustrates the
role $p_{d}$ might play in avoiding fixed points: in this case, taking
$L:=T_{\alpha}(\Delta_{3})\setminus\Delta_{3}$, we have
$T_{\alpha}^{3}(L)=M_{1}\cup M_{2}$, with
$M_{1}=[T_{\alpha}^{4}(\alpha),p_{2}]$ and
$M_{2}=(p_{2},T_{\alpha}^{2}(\alpha+1)]$. Since $T_{\alpha}^{3}(L)$ contains a
discontinuity point, the expansion under $T_{\alpha}$ is interrupted. If
$T_{\alpha}(M_{1})$ would be a subset of $T_{\alpha}(\Delta_{3})$ and
$T_{\alpha}(M_{2})$ would be a subset of $T_{\alpha}(L)$, the expansion would
be finished and we would have three gaps:
$(T_{\alpha}(\alpha),T_{\alpha}(\alpha+1))$,
$(T_{\alpha}^{3}(\alpha),T_{\alpha}^{4}(\alpha))$ and
$(T_{\alpha}^{2}(\alpha+1),T_{\alpha}^{2}(\alpha))$ – but this is not the
case, as we will shortly prove.

Of course arrangements exist such that one of the outer cylinders is small,
fixed points are avoided (in the sense we used above) for a long time and it
takes more of $T_{\alpha}$ working on $L$ or $R$ before one of the
discontinuity points is captured. But in these cases the interruption of the
expansion is even weaker than in the cases above. We will first show that
arrangements such as the rightmost one of Figure 8 exclude the existence of
gaps (cf. Lemma 10) and will then consider cases such as the first two
arrangements of Figure 8 (cf. Lemma 11).

######  Lemma 10.

Let $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}$.Then $I_{\alpha}$
is gapless if

$f_{d-1}\leq T_{\alpha}(\alpha)\leq
T_{\alpha}(\alpha+1)\,\,{\text{or}}\,\,T_{\alpha}(\alpha)\leq
T_{\alpha}(\alpha+1)\leq f_{d-1}.$

Proof of Lemma 10: We will confine ourselves to the first case of this lemma,
that is when $f_{d-1}\leq T_{\alpha}(\alpha)\leq T_{\alpha}(\alpha+1)$, since
the second one is proved similarly (in fact, this case is slightly harder due
to the smaller size of the absolute value of the derivatives). Regarding our
observations above, we may assume that the small cylinder set (which in this
case is $\Delta_{d-2}$) is gapless (cf. the remarks after Definition 4). We
will show that this implies the gaplessness of the other cylinder sets as
well. We define $R:=T_{\alpha}(\Delta_{d-2})\setminus\Delta_{d-2}$ and try to
determine $\alpha$ such that $p_{d}\in T_{\alpha}^{3}(R)$ (see the remark
immediately preceding this lemma). Necessary conditions for this are
$T_{\alpha}^{2}(\alpha)<p_{d}<T_{\alpha}^{4}(\alpha+1)$ (assuming that
$f_{d}\not\in T_{\alpha}(R)$ and $f_{d-1}\not\in T_{\alpha}^{2}(R)$, since in
either case we would be done). If these conditions are satisfied, we write
$T_{\alpha}^{3}(R)=V_{1}\cup V_{2}$, with
$V_{1}=[T_{\alpha}^{2}(\alpha),p_{d}]$ and
$V_{2}=(p_{d},T_{\alpha}^{4}(\alpha+1)]$. We will show that we cannot have
both $T_{\alpha}(V_{1})\subset T_{\alpha}(R)$ and $T_{\alpha}(V_{2})\subset
T_{\alpha}(\Delta_{d-2})$, which is necessary for limiting the expansion of
$R$ under $T_{\alpha}$ and so not eventually capturing $f_{d}$ and $f_{d-1}$;
see Figure 9.

$\Box_{2}$$\Box_{3}$$\Box_{4}$$N=30,\alpha=3.6$$|T_{\alpha}^{\prime}(\alpha+1)|\approx
1.42$$V_{2}$$V_{1}$$R$$T_{\alpha}(R)$$T_{\alpha}^{2}(R)$$|T_{\alpha}^{\prime}(\alpha)|\approx
2.31$ Figure 9. Arrangement illustrating Lemma 10

We take an approach that is similar to the proof of Theorem 5, for which
several lemmas and a proposition where used, partially concerning a relation
between $N$ and $d$ in the arrangements involved, partially concerning the
slope in $\alpha+1$. In this proof we will not explicitly formulate similar
statements as lemmas or propositions, nor do we prove them, since they require
similar basic but very intricate calculations that we prefer to omit.

In order to find the relationship between $N$ and $d$ for arrangements with
the conditions $T_{\alpha}^{2}(\alpha)<p_{d}$ and
$T_{\alpha}^{4}(\alpha+1)>p_{d}$ mentioned above, we refer to some more
relevant arrangements, as shown in Figure 10. In both cases in Figure 10,
$\alpha$ is such that $T_{\alpha}^{2}(\alpha)=p_{d}$, which is a value of
$\alpha$ that is only a little larger than the values for which
$T_{\alpha}^{2}(\alpha)<p_{d}$ and $T_{\alpha}^{4}(\alpha+1)>p_{d}$. A ‘limit
arrangement’ (where the third, rightmost cylinder is infinitely small),
similar to the ‘limit arrangement’ used in the proof of Theorem 5, is shown in
Figure 11. The assumptions yield $a^{3}+a^{2}-1=0$, with real root
$a=0.75487\cdots=:\gamma$.

$\Box_{1}$$\Box_{2}$$\Box_{3}$$N=17,\alpha=2.6576$$|T_{\alpha}^{\prime}(\alpha+1)|\approx
1.27$$|T_{\alpha}^{\prime}(\alpha)|\approx 2.41$

$\Box_{2}$$\Box_{3}$$\Box_{4}$$N=35,\alpha=4.00167$$|T_{\alpha}^{\prime}(\alpha+1)|\approx
1.40$$|T_{\alpha}^{\prime}(\alpha)|\approx 2.19$

Figure 10. Two arrangements in which almost
$T_{\alpha}^{2}(\alpha)<p_{d}<T_{\alpha}^{4}(\alpha+1)$
(in fact, in both cases $p_{d}=T_{\alpha}^{2}(\alpha)$).
$(a,a)$$(a,1-a^{2})$$y=-(a+1)x+a+1$$y=-(a+1)x+a$$a$$(\frac{a}{a+1},1)$$0$$1$
Figure 11. The ‘limit graph’ of $T_{\alpha}$, translated over
$(-\alpha,-\alpha)$, under the conditions
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}$ and $N/(N/\alpha-d)-(d-1)=p_{d}$ for
$N\to\infty$ (and $\alpha,d\to\infty$ accordingly). This ‘arrangement’ can be
seen as one with three cylinders, where $\Delta_{d-2}\ (\mathrm{mod}\
\alpha)$, the one on the right, is infinitely small; see also the arrangements
in Figure 10.

Similar to the proof of Theorem 5 we then find that for arrangements as in
Figure 10 we have

$N\approx\frac{(d-1)(d-1+\gamma)(1+\gamma)}{\gamma^{2}}.$

Using this relationship, we can take a similar approach as in the proof of
Proposition 1. We leave out the tedious steps and confine ourselves to
observing that the slope of the line segments in Figure 11 is
$-(\gamma+1)=-1.75487\cdots$ and that in arrangements where
$T_{\alpha}^{2}(\alpha)<p_{d}$ and $T_{\alpha}^{4}(\alpha+1)>p_{d}$, we will
see that the slope $T_{\alpha}^{\prime}(\alpha+1)$ approaches $-(\gamma+1)$ as
$N$ tends to infinity. However, for our proof the inequality
$|T_{\alpha}^{\prime}(\alpha+1)|>1/2(\sqrt{5}+1)=1.61803\cdots=:G$ suffices,
which turns out to hold for $N\in\mathbb{N}_{\geq 273}$. We will use this to
show that for $N\in\mathbb{N}_{\geq 273}$ we have
$|T_{\alpha}^{3}(R)|>|T_{\alpha}(\Delta_{d-2})|+|T_{\alpha}(R)|$. From this it
immediately follows that we cannot have that both $T_{\alpha}(V_{1})\subset
T_{\alpha}(R)$ and $T_{\alpha}(V_{2})\subset T_{\alpha}(\Delta_{d-2})$, and we
are done with the proof of Lemma 11.

Since $|T_{\alpha}^{\prime}(x)|$ is a decreasing function on $I_{\alpha}$, and
writing $\beta:=|\Delta_{d-2}|$, we have

$|T_{\alpha}(\Delta_{d-2})|>|T_{\alpha}^{\prime}(\alpha+1)|\cdot\beta,\,\,{\text{so}}\,\,|R|>(|T_{\alpha}^{\prime}(\alpha+1)|-1)\beta.$

It follows that

$|T_{\alpha}(R)|>(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(p_{d-1})|\beta,$

that

$|T_{\alpha}^{2}(R)|>(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(p_{d-1})|\cdot|T_{\alpha}^{\prime}(f_{d})|\beta,$

and finally that

$|T_{\alpha}^{3}(R)|>(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(p_{d-1})|^{2}\cdot|T_{\alpha}^{\prime}(f_{d})|\beta.$

We also have $|T_{\alpha}(\Delta_{d-2})|<|T_{\alpha}^{\prime}(p_{d-1})|\beta$,
so

$|R|<(|T_{\alpha}^{\prime}(p_{d-1})|-1)\beta\,\,{\text{and}}\,\,|T_{\alpha}(R)|<|T_{\alpha}^{\prime}(f_{d-1})|\cdot(|T_{\alpha}^{\prime}(p_{d-1})|-1)\beta.$

It follows that

$\displaystyle|T_{\alpha}(\Delta_{d-2})|+|T_{\alpha}(R)|$
$\displaystyle<(|T_{\alpha}^{\prime}(p_{d-1})|+|T_{\alpha}^{\prime}(f_{d-1})|\cdot(|T_{\alpha}^{\prime}(p_{d-1})|-1))\beta$
$\displaystyle=(|T_{\alpha}^{\prime}(p_{d-1})|-|T_{\alpha}^{\prime}(f_{d-1})|+|T_{\alpha}^{\prime}(f_{d-1})|\cdot|T_{\alpha}^{\prime}(p_{d-1})|)\beta$
$\displaystyle<|T_{\alpha}^{\prime}(f_{d-1})|\cdot|T_{\alpha}^{\prime}(p_{d-1})|\beta.$

So, although crudely, we certainly have that
$|T_{\alpha}^{3}(R)|>|T_{\alpha}(\Delta_{d-2})|+|T_{\alpha}(R)|$ if

$|T_{\alpha}^{\prime}(f_{d-1})|\cdot|T_{\alpha}^{\prime}(p_{d-1})|<(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(p_{d-1})|^{2}\cdot|T_{\alpha}^{\prime}(f_{d})|,$

that is, if

(23)
$1<(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(p_{d-1})|\cdot\frac{|T_{\alpha}^{\prime}(f_{d})|}{|T_{\alpha}^{\prime}(f_{d-1})|}.$

Since

$(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(p_{d-1})|\cdot\frac{|T_{\alpha}^{\prime}(f_{d})|}{|T_{\alpha}^{\prime}(f_{d-1})|}>(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(p_{d-1})|>(|T_{\alpha}^{\prime}(\alpha+1)|-1)\cdot|T_{\alpha}^{\prime}(\alpha+1)|,$

we know that (23) holds for $|T_{\alpha}^{\prime}(\alpha+1)|>G$, which in turn
holds for all $N\in\mathbb{N}_{\geq 273}$. We remark that this value is quite
a wide upper bound, since we did a rough approximation. Still, checking that
we cannot have both $T_{\alpha}(V_{1})\subset T_{\alpha}(R)$ and
$T_{\alpha}(V_{2})\subset T_{\alpha}(\Delta_{d-2})$ for smaller $N$ is not
that hard and is left to the reader. This finishes the proof of Lemma 10 (cf.
case 2 in (22)). $\Box$

Lemma 10 implies that in case
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}$ and $f_{d-1}\leq
T_{\alpha}(\alpha)\leq
T_{\alpha}(\alpha+1)\,\,{\text{or}}\,\,T_{\alpha}(\alpha)\leq
T_{\alpha}(\alpha+1)\leq f_{d-1}$ the division of an interval containing
$p_{d}$ in two smaller ones cannot prevent an overall expansion that excludes
any gaps. The other plausible case with three cylinder sets in which gaps
might exist is when one outer cylinder set is very small, while the other one
is full or nearly full, such that either $T_{\alpha}^{3}(\alpha+1)\geq
T_{\alpha}(\alpha+1)$ (when $\Delta_{d-2}$ is the small cylinder set) or
$T_{\alpha}^{3}(\alpha)\leq T_{\alpha}(\alpha)$ (when $\Delta_{d}$ is the
small cylinder set). We will show that this is not possible either:

######  Lemma 11.

Let $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}$. Then $I_{\alpha}$
is gapless if

$f_{d-1}\leq T_{\alpha}(\alpha+1)\leq
T_{\alpha}(\alpha)\,\,{\text{or}}\,\,T_{\alpha}(\alpha+1)\leq
T_{\alpha}(\alpha)\leq f_{d-1}.$

Proof of Lemma 11: Taking into account our observations immediately following
Definition 4 and the arrangements of Figure 8 for $N=23$ and $N=11$, we only
have to prove that there are _no_ $\alpha$ such that
$T_{\alpha}^{3}(\alpha)<T_{\alpha}(\alpha)$ is possible when
$T_{\alpha}(\alpha+1)\leq T_{\alpha}(\alpha)\leq f_{d-1}$ (in case
$\Delta_{d}$ is small) or such that
$T_{\alpha}^{3}(\alpha+1)>T_{\alpha}(\alpha+1)$ is possible when $f_{d-1}\leq
T_{\alpha}(\alpha+1)\leq T_{\alpha}(\alpha)$ (in case $\Delta_{d-2}$ is
small). Note that the conditions $T_{\alpha}^{3}(\alpha)<T_{\alpha}(\alpha)$
and $T_{\alpha}^{3}(\alpha+1)>T_{\alpha}(\alpha+1)$ imply that the branch
number is slightly larger than $2$. Now remember that $I_{\alpha}$ consists of
$m$ full cylinder sets if and only if $\alpha=k$, $N=mk(k+1)$ and
$d=(m-1)(k+1)$ for some $k\in\mathbb{N}$, cf. Theorem 1. Figure 12 shows for
increasing values of $N$ a sequence of arrangements where the branch number
$b$ is $2$, from one full arrangement (here for $N=4$) with two cylinders to
the next one (here for $N=12$). Since
$|T_{\alpha}^{\prime}(\alpha)|>|T_{\alpha}^{\prime}(\alpha+1)|$, the
arrangements suggest that in case $b$ is slightly larger than 2, the most
favourable arrangement for $T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$ to
have real roots is when $N=2k^{2}+2k-1$, where $k\in\mathbb{N}_{\geq 2}$,
while for $T_{\alpha}^{3}(\alpha)=T_{\alpha}(\alpha)$ to have real roots is
when $N=2k^{2}+2k+1$, where $k\in\mathbb{N}$. We will confine ourselves to
investigating only the possibility of
$T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$; the calculations for the
other case are similar.

$\Box_{1}$$\Box_{2}$$b(4,\alpha)=2$

$\Box_{1}$$\Box_{2}$$\Box_{3}$$b(5,\alpha)=2$

$\Box_{1}$$\Box_{2}$$\Box_{3}$$b(6,\alpha)=2$

$\Box_{1}$$\Box_{2}$$\Box_{3}$$b(11,\alpha)=2$

$\Box_{2}$$\Box_{3}$$b(12,\alpha)=2$

Figure 12. Arrangements of $\Upsilon_{N,\alpha}$ with $b=2$; in each case
$\alpha=\frac{\sqrt{2N+1}-1}{2}$.

So we will try and find out if for $N=2k^{2}+2k-1$, $d=k+1$, with
$k\in\mathbb{N}_{\geq 2}$, the positive root of
$T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$ lies in $I_{\alpha}$. To do
this, we solve

$\cfrac{2k^{2}+2k-1}{\cfrac{\displaystyle
2k^{2}+2k-1}{\displaystyle\cfrac{2k^{2}+2k-1}{\alpha+1}-(k-1)}-k}-(k+1)=\frac{2k^{2}+2k-1}{\alpha+1}-(k-1),$

which is reducible to

$(2k^{3}+6k^{2}-k-1)\alpha^{2}+(2k^{4}+5k^{2}+k-2)\alpha-(4k^{5}+6k^{4}+2k^{3}-3k^{2}-k+1)=0,$

yielding

(24)
$\alpha=\frac{\sqrt{36k^{8}+144k^{7}+164k^{6}-12k^{5}-95k^{4}-2k^{3}+21k^{2}-4k}-(2k^{4}+5k^{2}+k-2)}{2(2k^{3}+6k^{2}-k-1)}.$

A straightforward computation shows that this last expression is smaller than
$f_{k+2}$, meaning that the root (24) lies outside $I_{\alpha}$ when
$I_{\alpha}=\Delta_{k+1}\cup\Delta_{k}\cup\Delta_{k-1}$. Since $N=2k^{2}+2k-1$
was the most favourable option for investigation, this finishes our proof (cf.
case 3 in (22)). $\Box$

######  Remark 13.

The arrangement for $N=11$ in Figure 8 illustrates that the difference between
$T_{\alpha}^{3}(\alpha+1)$ and $T_{\alpha}(\alpha+1)$ may be very small.

###  4.2. A sufficient condition for gaplessness in case $I_{\alpha}$ consists
of four cylinder sets

In the previous subsection we proved Theorem 6 for $m=2$, by proving Lemmas 9,
10 and 11. In this subsection we will consider $m=3$ and go into the analogons
of Lemmas 9, 10 and 11.

When $I_{\alpha}$ consists of four cylinder sets, the analogon of Lemma 9 is
that arrangements $I_{\alpha}$ are gapless when
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}\cup\Delta_{d-3}$ while
$T_{\alpha}(\alpha)\geq f_{d-1}$ and $T_{\alpha}(\alpha+1)\leq f_{d-2}$. The
analogon of Lemma 11 is that arrangements $I_{\alpha}$ are gapless when
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}\cup\Delta_{d-3}$ while
$f_{d-2}\leq T_{\alpha}(\alpha+1)\leq
T_{\alpha}(\alpha)\,\,{\text{{\bf{or}}}}\,\,T_{\alpha}(\alpha+1)\leq
T_{\alpha}(\alpha)\leq f_{d-1}$. In both cases branch numbers larger than $3$
are involved, in which case $|T_{\alpha}^{\prime}(\alpha+1)|>2$ when
$N\in\mathbb{N}_{\geq 18}$ (and Theorem 3 yields the desired result). The
cases $2\leq N\leq 17$ can be checked manually and are left to the reader; in
Figure 13 the arrangement for $N=11$, associated with Lemma 11, illustrates
that gaps are out of the question.

$\Box_{2}$$\Box_{3}$$\Box_{4}$$\Box_{5}$$N=11,\alpha=f_{6}$

$\Box_{3}$$\Box_{4}$$\Box_{5}$$N=15,\alpha=f_{6}$and
$T_{\alpha}(\alpha+1)=p_{5}$

$\Box_{4}$$\Box_{5}$$\Box_{6}$$N=24,\alpha=f_{3}-1$and
$T_{\alpha}(\alpha)=p_{5}$

$\Box_{5}$$\Box_{6}$$\Box_{7}$$N=35,\alpha=f_{4}-1$and
$T_{\alpha}(\alpha)=p_{6}$

Figure 13. Four arrangements with two full cylinders

The analogon of Lemma 10 is that arrangements $I_{\alpha}$ are gapless when
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}\cup\Delta_{d-3}$ while
$f_{d-1}\leq T_{\alpha}(\alpha)\leq
T_{\alpha}(\alpha+1)\,\,{\text{{\bf{or}}}}\,\,T_{\alpha}(\alpha)\leq
T_{\alpha}(\alpha+1)\leq f_{d-2}$. The arrangements for $N=15$, $N=24$ and
$N=35$ in Figure 13 are interesting illustrations of the analogon of Lemma 10
in the case of two full cylinder sets instead of one. We will confine
ourselves to the arrangement for $N=15$; the other ones have similar
properties.

$(1,a)$$y=-\frac{1}{a}x+3$$y=-\frac{1}{a}x+2$$y=-\frac{1}{a}x+1$$(a,1)$$0$$1$
Figure 14. The ‘limit graph’ of $T_{\alpha}$, translated over
$(-\alpha,-\alpha)$, under the conditions
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\Delta_{d-2}$ and
$N/(\alpha+1)-(d-2)=p_{d}$ for $N\to\infty$ (and $\alpha,d\to\infty$
accordingly)

The arrangement for $N=15$ is the boundary case for the situation where we
have four cylinders, the left one of which (that would be $\Delta_{6}$ in this
example) is extremely small and the right one is such that almost $p_{5}\in
T_{\alpha}^{2}(T_{\alpha}(\Delta_{6})\setminus\Delta_{6})$. The interesting
thing is that this option would imply a quick interruption of the expansion of
$T_{\alpha}(\Delta_{6})\setminus\Delta_{6}$, involving two large gaps. But it
is not really an option: the arrangement for $N=15$ in Figure 13 is
exceptional among relatively small $N$ (as well as the arrangements for $N=24$
and $N=35$ are), while for $N>36$ we have $|T_{\alpha}^{\prime}(\alpha+1)|>2$
when $I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\Delta_{d-2}$ and
$N/(\alpha+1)-(d-2)=p_{d}$ or $N/\alpha-d=p_{d}$. We derived this in a similar
way as in the proof of Lemma 10 (see Figure 11) or the preparations for
Theorem 5 (see Figure 6). Figure 14 shows the associated ‘limit graph’, from
which it is easily found that $a=1/2(3-\sqrt{5})$, yielding branch number
$2+g$, with $g=1/G$ the small golden section.

With this, we conclude the proof of Theorem 6. $\Box$

In the next section we will prove that if
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}\cup\Delta_{d-3}$ gaps
exist only in very rare cases and if they do, that they are very large. After
that, we will finish the proof of Theorem 4, stating that all arrangements
with five cylinders are gapless.

##  5\. Gaplessness in case $I_{\alpha}$ contains two full cylinder sets

When an arrangement contains three cylinders, and two of them are full, the
arrangement is gapless according to Theorem 6. In this section we will proof
that arrangements of four cylinders generally do not contain a gap either,
save for special values of $N$. The core of this proof rests on values of
$\alpha$ satisfying one of the equations

$T_{\alpha}(\alpha)=T_{\alpha}^{3}(\alpha)\,\,{\text{(with root
$\alpha_{\ell}$)}}\,\,{\text{and}}\,\,T_{\alpha}(\alpha+1)=T_{\alpha}^{3}(\alpha+1)\,\,{\text{(with
root $\alpha_{u}$)}}.$

We will show that for $N$ such that $\alpha_{\ell}<\alpha_{u}$ very large gaps
exist for $\alpha\in[\alpha_{\ell},\alpha_{u}]$.

The central theorem of this section is the following:

######  Theorem 7.

Let $N\in\mathbb{N}_{\geq 2}$ and
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}\cup\Delta_{d-3}$. Then
there is a gap in $I_{\alpha}$ if and only if $N=2k^{2}+2k-i$, with $k>1$ and
$i\in\\{1,2,3\\}$. Moreover, if there is a gap in $I_{\alpha}$, the gap
contains $f_{d-1}$ and $f_{d-2}$, while $\Delta_{d}$ and $\Delta_{d-3}$ are
gapless.

Proof of Theorem 7: Suppose that there is a gap containing $f_{d-1}$ and
$f_{d-2}$ in $I_{\alpha}$ and that $\Delta_{d}$ and $\Delta_{d-3}$ are
gapless. Then, as a sub-interval of a gap, the interval $(f_{d-1},f_{d-2})$ is
a gap. Since $f_{d-1}<f_{d-2}$, $N/(f_{d-1}+d-1)=f_{d-1}$ and
$N/(f_{d-2}+d-2)=f_{d-2}$, we know that

$(f_{d-1},f_{d-2})\subsetneq\left(\frac{N}{f_{d-2}+d-1},\frac{N}{f_{d-1}+d-2}\right),$

where the larger open interval is a gap as well. What is more, the infinite
sequence of intervals

$(f_{d-1},f_{d-2})\subsetneq\left(\frac{N}{f_{d-2}+d-1},\frac{N}{f_{d-1}+d-2}\right)\subsetneq\left(\frac{N}{\frac{N}{f_{d-1}+d-2}+d-1},\frac{N}{\frac{N}{f_{d-2}+d-1}+d-2}\right)\subsetneq\ldots$

consists of the union of $(f_{d-1},f_{d-2})$ with pre-images of
$(f_{d-1},f_{d-2})$ in $\Delta_{d-1}$ and $\Delta_{d-2}$ respectively and
therefore of gaps containing $f_{d-1}$ and $f_{d-2}$. It is contained in the
closed interval $[q,r]$, with

$q=[\overline{d-1,d-2}]_{N,\alpha}\in\Delta_{d-1}\quad{\text{and}}\quad
r=[\overline{d-2,d-1}]_{N,\alpha}\in\Delta_{d-2},$

yielding

(25)
$T_{\alpha}^{2}(q)=q,\,\,T_{\alpha}(q)=r,\,\,T_{\alpha}(r)=q\,\,{\text{and}}\,\,T_{\alpha}^{2}(r)=r.$

Since $\Delta_{d}$ and $\Delta_{d-3}$ are gapless, $T_{\alpha}(\alpha)$ and
$T_{\alpha}(\alpha+1)$ lie outside the interval $(q,r)$, which is to say

$p_{d}<T_{\alpha}(\alpha)\leq q\,\,{\text{and}}\,\,r\leq
T_{\alpha}(\alpha+1)<p_{d-2}.$

For the images of $\alpha$ under $T_{\alpha}$ this means that either
$T_{\alpha}^{2}(\alpha)\in\Delta_{d-3}$ or
$T_{\alpha}^{2}(\alpha)\in\Delta_{d-2}$, in the latter case of which we have,
due to the expansiveness of $T_{\alpha}$ and the equalities of (25),

$|T_{\alpha}(\alpha)-q|\leq|T_{\alpha}^{2}(\alpha)-r|\leq|T_{\alpha}^{3}(\alpha)-q|,$

with equalities only in the case $T_{\alpha}(\alpha)=q$. From this we derive
that

(26)
${\text{either}}\,\,T_{\alpha}^{2}(\alpha)\in\Delta_{d-3}\,\,{\text{or}}\,\,T_{\alpha}^{2}(\alpha)\in\Delta_{d-2}\wedge
T_{\alpha}^{3}(\alpha)\leq T_{\alpha}(\alpha)$

and, similarly, that

(27)
${\text{either}}\,\,T_{\alpha}^{2}(\alpha+1)\in\Delta_{d}\,\,{\text{or}}\,\,T_{\alpha}^{2}(\alpha+1)\in\Delta_{d-1}\wedge
T_{\alpha}^{3}(\alpha+1)\geq T_{\alpha}(\alpha+1).$

In the following we will write $\alpha_{u}(N,m)$ ($u$ for ‘upper’) for the
positive root of the equation $T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$
(so $T_{\alpha}(\alpha+1)=r$) and $\alpha_{\ell}(N,m)$ ($l$ for ‘lower’) for
the positive root of the equation $T_{\alpha}^{3}(\alpha)=T_{\alpha}(\alpha)$
(so $T_{\alpha}(\alpha)=q$), with $m$ the number of full cylinder sets; in the
current case we have $m=2$. Recall that $I_{\alpha}$ consists of $m$ full
cylinder sets if and only if $\alpha=k$, $N=mk(k+1)$ and $d=(m-1)(k+1)$ for
some $k\in\mathbb{N}$, cf. Theorem 1, so when $m=2$, we have arrangements
consisting of two full arrangements only for $\alpha=k$, $N=2k(k+1)$ and
$d=k+1$. If $N$ is $2k(k+1)-n$, with $n\in\mathbb{N}$, and $\alpha=k-x$, with
$x\in\mathbb{R}$, we have

$b(N,\alpha)=\frac{2k^{2}+2k-n}{(k-x)(k+1-x)}=2+\frac{4xk-n+2x-2x^{2}}{(k-1)(k+1-x)}>2+\frac{4xk-n}{k^{2}+x^{2}},$

which is a little bit larger than $2$ provided $x$ and $n$ are relatively
small. For these arrangements we have

$d(\alpha)=\left\lfloor\frac{2k^{2}+2k-n}{k-x}-(k-x)\right\rfloor=\left\lfloor
k+2+3x+\frac{2x^{2}+2x-n}{k-x}\right\rfloor=k+2$

and

$d(\alpha+1)=\left\lfloor\frac{2k^{2}+2k-n}{k+1-x}-(k+1-x)\right\rfloor=\left\lfloor
k-1+3x+\frac{2x^{2}+2x-n}{k+1-x}\right\rfloor=k-1.$

So, for $x$ and $n$ relatively small, the arrangements consist of four
cylinders, while the branch number is only a little bit larger than $2$. We
will now use this to finish the forward implication of Theorem 7.

Since $\Delta_{d-3}$ decreases and $\Delta_{d}$ increases as $\alpha$
decreases, we see that the assumption that there is a gap containing $f_{d-1}$
and $f_{d-2}$ in $I_{\alpha}$ implies $\alpha_{u}(N,2)\geq\alpha_{\ell}(N,2)$.
We will shortly show that the only values of $N$ for which
$\alpha_{u}(N,2)\geq\alpha_{\ell}(N,2)$ are $N=2k^{2}+2k-i$, with $k>1$ and
$i\in\\{1,2,3\\}$; in all cases $d=k+2$. Although we could keep $i$ as a
variable in our calculations, we can limit ourselves to the case $i=3$, since
$i=3$ is the least favourable value of $i$ allowing for a gap, as is
suggested666Note that in Figure 12 we have $d=k+1$, while there is no small
cylinder set $\Delta_{k+2}$ in Figures 12 and 16 through 18. We will show that
for $i=3$ indeed $\alpha_{u}(N,2)\geq\alpha_{\ell}(N,2)$. Subsequently we will
show that for $4\leq i\leq 4k$ no gaps exist; the upper bound is $4k$, since
$2k^{2}+2k-4k=2(k-1)^{2}+2(k-1)$, so as to confine the calculations to the
group of arrangements where $d=k+2$.

$N=11,\alpha=1.8719\cdots$ ($k=2$,
$i=1$)$T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$$F_{4}$$F_{3}$$F_{2}$$F_{1}$$\Box_{1}$$\Box_{2}$$\Box_{3}$$\Box_{4}$
Figure 15.

$F_{4}$$F_{3}$$F_{2}$$F_{1}$$\Box_{1}$$\Box_{2}$$\Box_{3}$$\Box_{4}$$N=11,\alpha=1.8687\cdots\,\,(k=2,i=1)$$T_{\alpha}^{3}(\alpha)=T_{\alpha}(\alpha)$
Figure 16.

So, let $N=2k^{2}+2k-3$ and $d=k+2$. Then
$\alpha_{\ell}(N,2)=[k+2,\overline{k+1,k}]$ and
$\alpha_{u}(N,2)+1=[k-1,\overline{k,k+1}]$777We omit the suffix ‘$N,\alpha$’
behind these expansions not only for eligibility but also because $\alpha$ has
yet to be determined as the root of
$T_{\alpha}^{3}(\alpha)=T_{\alpha}(\alpha)$ or
$T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$.. Omitting straightforward
calculations, we find that

$\alpha_{u}(2k^{2}+2k-3,2)=\frac{(2k^{2}+2k-3)\sqrt{D}-(2k^{4}+3k^{2}+3k-6)}{4k^{3}+12k^{2}-6k-6}$

and

$\alpha_{\ell}(2k^{2}+2k-3,2)=\frac{(2k^{2}+2k-3)(\sqrt{D}-(k^{2}+5k+4))}{4k^{3}-18k-8},$

with

$D=9k^{4}+18k^{3}-3k^{2}-12k=(3k^{2}+3k-2)^{2}-4.$

Since we assume that there is a gap containing $f_{d-1}$ and $f_{d-2}$, we
have $\alpha_{u}\geq\alpha_{\ell}$. Omitting the basic calculations, we find
that this inequality holds for $k\in\mathbb{N}_{\geq 2}$. In the case $k=3$
(and so $N=21$), we have indeed

$\alpha_{u}(2k^{2}+2k-3,2)=\frac{\sqrt{508032}-192}{192}=2.7123\cdots>2.7122\cdots=\frac{\sqrt{508032}-588}{46}=\alpha_{\ell}(2k^{2}+2k-3,2);$

see Figure 20.

Some more basic calculations show that the cases $N=2k^{2}+2k-1$ and
$N=2k^{2}+2k-2$ allow for larger intervals $[\alpha_{\ell},\alpha_{u}]$ where
large gaps exist; see the next examples.

$\displaystyle\alpha_{u}(11,2)=\frac{\sqrt{9075}-26}{37}=1.8719\cdots\quad{\text{and}}\quad\alpha_{\ell}(11,2)=\frac{99-\sqrt{9075}}{2}=1.8686\cdots$
$\displaystyle\alpha_{u}(10,2)=\frac{\sqrt{1725}-12}{17}=1.7372\cdots\quad{\text{and}}\quad\alpha_{\ell}(10,2)=\frac{45-\sqrt{1725}}{2}=1.7334\cdots$
$\displaystyle\alpha_{u}(9,2)=\frac{\sqrt{5103}-22}{31}=1.5946\cdots\quad{\text{and}}\quad\alpha_{\ell}(9,2)=\frac{27-\sqrt{567}}{2}=1.5941\cdots$
$\displaystyle\alpha_{u}(8,2)=\frac{\sqrt{228}-5}{7}=1.4428\cdots\quad{\text{and}}\quad\alpha_{\ell}(8,2)=9-\sqrt{57}=1.4501\cdots$

We see that the intervals $\alpha_{u}-\alpha_{\ell}$ decrease as $N$
decreases, until (for $N=8$) the ‘interval’ would have negative length, hence
does not exist.

$N=10,\alpha=1.7372\cdots$ ($k=2$,
$i=2$)$T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$$F_{4}$$F_{3}$$F_{2}$$F_{1}$$\Box_{1}$$\Box_{2}$$\Box_{3}$$\Box_{4}$
Figure 17.

$F_{4}$$F_{3}$$F_{2}$$F_{1}$$\Box_{1}$$\Box_{2}$$\Box_{3}$$\Box_{4}$$N=9,\alpha=1.5946\cdots$
($k=2$, $i=3$)$T_{\alpha}^{3}(\alpha+1)=T_{\alpha}(\alpha+1)$ Figure 18.

Now suppose $N=2k^{2}+2k-4$ (note that voor $k=2$ we have $N=8$). Then

$\alpha_{u}(2k^{2}+2k-4,2)=\frac{(k+2)\sqrt{D}-(k^{3}+k^{2}+2k+4)}{2(k^{2}+4k+2)}$

and

$\alpha_{\ell}(2k^{2}+2k-4,2)=\frac{(k-1)\sqrt{D}-(k^{3}+4k^{2}-k-4)}{2(k^{2}-2k-1)},$

with

$D=9k^{4}+18k^{3}-7k^{2}-16k.$

There are no gaps provided
$\alpha_{\ell}(2k^{2}+2k-4,2)-\alpha_{u}(2k^{2}+2k-4,2)>0$. Once more we omit
the calculations, finding that this inequality holds for $k\in\mathbb{N}_{\geq
2}$, so we conclude that there are no gaps in case $N=2k^{2}+2k-4$. When we
replace the number $4$ in $N=2k^{2}+2k-4$ by larger integers (if possible),
there will not be any gaps either: the length of the ‘interval’
$[\alpha_{\ell},\alpha_{u}]$ would only become more negative. This concludes
the proof that if
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}\cup\Delta_{d-3}$ and
there is a gap containing $f_{d-1}$ and $f_{d-2}$ in $I_{\alpha}$, then
$N=2k^{2}+2k-i$, with $k>1$ and $i\in\\{1,2,3\\}$.

For the _converse statement_ , we assume that $N=2k^{2}+2k-i$, with
$k\in\mathbb{N}$ and $i\in\\{1,2,3\\}$. If also $d=k+2$, then earlier in this
proof we showed that only then $\alpha_{\ell}(N,2)\leq\alpha_{u}(N,2)$. We
will show that for $\alpha$ such that
$\alpha_{\ell}(N,2)\leq\alpha\leq\alpha_{u}(N,2)$ there is a gap in
$I_{\alpha}=\Delta_{d}\cup\Delta_{d-1}\cup\Delta_{d-2}\cup\Delta_{d-3}$
containing both $f_{d-1}$ and $f_{d-2}$.

As earlier in this proof, we set

$q=[\,\overline{d-1,\,d-2}\,]_{N,\alpha}\quad\text{and}\quad
r=[\,\overline{d-2,\,d-1}\,]_{N,\alpha}.$

Set $G=(q,r)$, then clearly both $f_{d-1}\in G$ and $f_{d-2}\in G$.
Furthermore, by definition of $\alpha_{\ell}(N,2)$ and $\alpha_{u}(N,2)$ we
have that for every $\alpha\in[\alpha_{\ell}(N,2),\,\alpha_{u}(N,2)]$ that

$T_{\alpha}(\alpha)\leq q\quad(\text{and therefore $T_{\alpha}^{2}(\alpha)\geq
r$})$

and that

$T_{\alpha}(\alpha+1)\geq r\quad(\text{and therefore
$T_{\alpha}^{2}(\alpha+1)\leq q$}).$

Note that $T_{\alpha}((p_{d},q)=(r,\alpha+1)$ and that
$T_{\alpha}((r,p_{d-2})=(\alpha,q)$. Then we have that
$T_{\alpha}(G^{c})=G^{c}$, where $G^{c}$ is the complement of $G$ in
$I_{\alpha}$. We are left to show that $G=(q,r)$ is a gap; i.e. that for
almost all $x\in G$ there exists an $n=n(x)$ such that $T_{\alpha}^{n}(x)\in
G^{c}$.

To show this, consider the map $T:I_{\alpha}\to I_{\alpha}$, defined by

(28)
$T(x)=\begin{cases}\frac{-x}{p_{d}-\alpha}+\frac{(\alpha+1)p_{d}-\alpha^{2}}{p_{d}-\alpha},&\text{if
$x\in\Delta_{d}$};\\\ \frac{N}{x}-(d-1),&\text{if $x\in\Delta_{d-1}$};\\\
\frac{N}{x}-(d-2),&\text{if $x\in\Delta_{d-2}$};\\\
\frac{-x}{\alpha+1-p_{d-2}}+\frac{(\alpha+1)^{2}-\alpha
p_{d-2}}{\alpha+1-p_{d-2}},&\text{if $x\in\Delta_{d-3}$}.\end{cases}$

So on $\Delta_{d}$ and on $\Delta_{d-3}$ we have that $T$ is a straight line
segment with negative slope, through $(\alpha,\alpha+1)$ and $(p_{d},\alpha)$
on $\Delta_{d}$, resp. through $(p_{d-2},\alpha+1)$ and $(\alpha+1,\alpha)$ on
$\Delta_{d-3}$. For $x\in\Delta_{d-1}\cup\Delta_{d-2}$ we have that
$T(x)=T_{\alpha}(x)$. To show that $G$ is a gap, it is enough to show the
ergodicity of $T$. Then the maximality of $G$ follows from the fact that the
support of the absolutely continuous invariant measure is $G^{c}$, since
$T_{\alpha}(G^{c})=G^{c}$. The proof of the existence of the absolutely
continuous invariant measure for $T$ and its ergodicity is similar to the
proof of Theorem 2. Here all branches are complete and the proof is rather
simpler. Once we have the ergodicity of $T$, it is obvious that for a.e. $x\in
G$ there exists $n_{0}=n_{0}(x)$ such that $z=T^{n_{0}}(x)\in G^{c}$. Then $z$
never returns in $G$ under iterations of $T_{\alpha}$. This finishes the proof
of Theorem 7. $\Box$

We stress that the in case of $N=2k^{2}+2k-3$ the intervals
$[\alpha_{\ell},\alpha_{u}]$ on which gaps exist may be very small; see Figure
20. On the other hand, in case $N=2k^{2}+2k-4$, the gaplessness may be a very
close call; see Figure 20. Table 1 illustrates how fast these differences
between $\alpha_{\ell}$ and $\alpha_{u}$ decrease as $N$ increases:

|  $\alpha_{\ell}(N,2)$xxx |  $\alpha_{u}(N,2)$xxx
---|---|---
$N=9$ | $1.594119\cdots$ | $1.594686\cdots$
$N=21$ | $2.712252\cdots$ | $2.712310\cdots$
$N=37$ | $3.776839\cdots$ | $3.776851\cdots$
$N=57$ | $4.817672\cdots$ | $4.817675\cdots$
$N=8$ | $1.450165\cdots$ | $1.442809\cdots$
$N=20$ | $2.613247\cdots$ | $2.611575\cdots$
$N=36$ | $3.700989\cdots$ | $3.700407\cdots$
$N=56$ | $4.756087\cdots$ | $4.755832\cdots$
Table 1. The thin thread between having a gap or not

######  Remark 14.

While a fixed point $f_{i}$ is repellent for points within $\Delta_{i}$, the
fixed points in two adjacent cylinder sets behave mutually contracting for all
other points in these cylinder sets. As a consequence, it may take quite some
time before the orbit of points in the full cylinders of gap arrangements with
four cylinders leave these full cylinders for the first time. As an example we
take the gap arrangement for $k=50$ (according to the notations used above).
Then $N=2\cdot 50^{2}+2\cdot 50-3=5097$, $d=52$ and
$\alpha\approx\alpha_{u}\approx\alpha_{\ell}\approx 49.98019737$. Table 2
shows for ten values of $x$ between $\alpha$ and $\alpha+1$ the smallest $n$
such that $T_{\alpha}^{n}(x)\not\in\Delta_{51}\cup\Delta_{50}$. What is more,
there are _uncountably many_ $x$ in the gap $(a,b)$ that contains $f_{d-1}$
and $f_{d-2}$ such that $T_{\alpha}^{n}(x)\in(a,b)$ for all
$n\in\mathbb{N}\cup\\{0\\}$. Indeed, for any sequence
$(d_{1},d_{2},\ldots,d_{n},\ldots)$ such that $d_{n}\in\\{d-1,d-2\\}$, with
$n\in\mathbb{N}$, we have that
$x=[d_{1},d_{2},d_{3},\ldots]_{N,\alpha}\in(a,b)$.

$x$ | $50$ | $50.1$ | $50.2$ | $50.3$ | $50.4$ | $50.5$ | $50.6$ | $50.7$ | $50.8$ | $50.9$
---|---|---|---|---|---|---|---|---|---|---
$n$ | $5417$ | $2090$ | $3568$ | $1123$ | $4776$ | $185$ | $5816$ | $16231$ | $5646$ | $7604$
Table 2. The difficulty of leaving the gap: with $N=5097$,
$\alpha=49.98019737$, for each of ten values of $x\in[\alpha,\alpha+1]$ the
smallest $n$ is given such that
$T_{\alpha}^{n}(x)\not\in\Delta_{51}\cup\Delta_{50}$.

$F_{5}$$F_{4}$$F_{3}$$F_{2}$$\Box_{2}$$\Box_{3}$$\Box_{4}$$\Box_{5}$$N=20,\alpha=2.6124$$\alpha_{\ell}=2.6132\cdots$
and $\alpha_{u}=2.6115\cdots$ Figure 19.

$F_{5}$$F_{4}$$F_{3}$$F_{2}$$\Box_{2}$$\Box_{3}$$\Box_{4}$$\Box_{5}$$N=21,\alpha=2.7123$$\alpha_{\ell}=2.7122\cdots$
and $\alpha_{u}=2.7123\cdots$ Figure 20.

For the final, second part of the proof of Theorem 4, we will consider one by
one all cases left, that is $N\in\\{2,\ldots,11\\}$. When $N=11$ and
$\alpha\geq f_{7}$, $I_{\alpha}$ consists of five cylinder sets if and only if
$\alpha\in(f_{2}-1,f_{6})$; see the left arrangement of Figure 21, which we
already saw in Figure 13. Since $\alpha_{\ell}(11,3)>\alpha_{u}(11,3)$ (cf.
page 5), we conclude on similar grounds as in the proof of Theorem 7, that the
arrangement is gapless. When $\alpha\in[f_{7},f_{2}-1]$, the interval
$I_{\alpha}$ consists of four cylinder sets, implying gaplessness because of
Theorem 7. Since $|T_{\alpha}^{\prime}(f_{7}+1)|=2.04\cdots$, gaps are also
excluded for all $\alpha\leq f_{7}$. A similar approach works for $N=10$ (with
$|T_{\alpha}^{\prime}(f_{7}+1)|=2.03\cdots$), $N=9$ (with
$|T_{\alpha}^{\prime}(f_{7}+1)|=2.02\cdots$) and even $N=8$, in which case
$f_{7}=1$, $|T_{\alpha}^{\prime}(f_{7}+1)|=2$, and the arrangement with four
cylinders is full.

For $N\in\\{3,\ldots,7\\}$ we take a different approach, confining ourselves
to the case $N=7$; the cases $N\in\\{3,\ldots,6\\}$ are done similarly. We
will omit most calculations, which are generally quite tedious and do hardly
elucidate anything. So let $N=7$. Then $I_{\alpha}$ consists of at least five
cylinder sets if and only if $\alpha<f_{6}$; see the second arrangement of
Figure 21. We have $|T_{\alpha}^{\prime}(\alpha+1)|=2$ for
$\alpha=\frac{1}{2}\sqrt{14}-1$, in which case $d_{\min}=2$; see the third
arrangement of Figure 21. Now suppose
$\frac{1}{2}\sqrt{14}-1\leq\alpha<f_{6}=1$. We have
$|T_{\alpha}^{\prime}(f_{6}+1)|=\frac{7}{4}$ and
$|T_{\alpha}^{\prime}(\alpha)|>7$. Regarding these relatively large values, it
is not hard to understand that $\Delta_{2}$ is gapless. The part of the orbit
of $\alpha+1$ under $T_{\alpha}$ in the third arrangement of Figure 21
illustrates that even in the case of $\alpha=\frac{1}{2}\sqrt{14}$, the
expansion of $[T_{\alpha}(\alpha+1),p_{3}]$ under $T_{\alpha}$ clearly
excludes the existence of gaps.

Finally, let $N=2$. We have $|T_{\alpha}^{\prime}(f_{1})|=2$, indicating the
rapid increase of $|T_{\alpha}^{\prime}|$ on $I_{\alpha}$ when $\alpha$
decreases. The large expansiveness of $T_{\alpha}$ left of $f_{1}$ assures the
gaplessness of $\Delta_{1}$. We will show that for any
$\alpha\in(0,\sqrt{2}-1]$ the image of $[T_{\alpha}(\alpha+1),p_{2}])$
contains most of the fixed points, implying the gaplessness of $I_{\alpha}$;
see the last arrangement of Figure 21 for an illustration of this. When
$T_{\alpha}(\alpha+1)\leq f_{2}$ this is quite obvious, so we assume
$T_{\alpha}(\alpha+1)>f_{2}$. Suppose that $T_{\alpha}^{2}(\alpha+1)=f_{s}$,
for some $s\in\mathbb{N}_{\geq 2}$. Then, omitting some basic calculations, we
have $\alpha=(s+1-\sqrt{s^{2}+8})/(2s-7)$, whence

$d=d(\alpha)=\left\lfloor\frac{4s^{2}-11s+(4s-13)\sqrt{s^{2}+8}-15}{2s-7}\right\rfloor\geq\frac{4s^{2}-13s+(4s-13)\sqrt{s^{2}+8}-8}{2s-7},$

from which we derive that $d\geq 4s$.

This finishes the proof of Theorem 4. $\Box$

$\Box_{2}$$\Box_{3}$$\Box_{4}$$\Box_{5}$$N=11,\alpha=f_{6}$

$\Box_{2}$$\Box_{3}$$\Box_{4}$$\Box_{5}$$N=7,\alpha=f_{6}=1$

$\Box_{2}$$\Box_{3}$$\Box_{4}$$\Box_{5}$$\Box_{6}$$\Box_{7}$$N=7,\alpha=\tfrac{1}{2}\sqrt{14}-1$

$\Box_{13}$$\cdots$$\Box_{2}$$\Box_{1}$$N=2,\alpha=0.15$

Figure 21. Borderline cases for part II of the proof of Theorem 4

## Acknowledgements

The first author of this paper is very thankful to the Dutch organisation for
scientific research Nederlandse Organisatie voor Wetenschappelijk Onderzoek,
which funded his research for this paper with a so-called Promotiebeurs voor
Leraren, with grant number 023.003.036. We also want to thank Julian Lyczak,
then working at Leiden University, for his help with some programming in the
preparatory phase of Section 5. Finally, we thank Niels Langeveld for
providing us with Figure 3. The third author’s research was partially
supported by the Japan Society for the Promotion of Science with Grant-in-Aid
for Scientific Research (C) 20K03661.

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