# Rewriting rules for arithmetics in alternate base systems

Zuzana Masáková, Edita Pelantová and Katarína Studeničová

###### Abstract

For alternate Cantor real base numeration systems we generalize the result of
Frougny and Solomyak on arithmetics on the set of numbers with finite
expansion. We provide a class of alternate bases which satisfy the so-called
finiteness property. The proof uses rewriting rules on the language of
expansions in the corresponding numeration system. The proof is constructive
and provides a method for performing addition of expansions in Cantor real
bases.

##  1 Introduction

Cantor real numeration systems represent a generalization of the – nowadays
well-known – Rényi numeration systems [10]. They have been introduced
independently by two groups of authors, Caalim and Demeglio [3], and Charlier
and Cisternino [4]. In [4] a Cantor real numeration system is given by an
infinite sequence $\boldsymbol{\mathcal{B}}=(\beta_{n})_{n\geq 1}$,
$\beta_{n}>1$, where the $\boldsymbol{\mathcal{B}}$-expansions of numbers in
the interval $[0,1)$ are found by the greedy algorithm. The authors then
provide a description of the language of the
$\boldsymbol{\mathcal{B}}$-expansions using a set of lexicographic
inequalities. They focus on the case of purely periodic Cantor real bases,
which they call alternate bases. In analogy with the result of Bertrand-Mathis
[2] for Rényi numeration systems, which are alternate bases with period 1,
they define a symbolic dynamical system called
$\boldsymbol{\mathcal{B}}$-shift, and show under what conditions the
$\boldsymbol{\mathcal{B}}$-shift is sofic. In [6] it is proven that this
happens precisely for the so-called Parry alternate bases. It is also shown
under what algebraic assumptions on the base $\boldsymbol{\mathcal{B}}$ a
normalization function is computable by a finite Büchi automaton. Other
properties of alternate base systems were studied in [5].

In this paper, we focus on the arithmetical properties of Cantor numeration
systems. We are in particular interested in addition of numbers whose
$\boldsymbol{\mathcal{B}}$-expansions are sequences with finite support. We
define the finiteness and positive finiteness property and provide some
necessary and sufficient conditions on an alternate base
$\boldsymbol{\mathcal{B}}$ for these properties. By that, we generalize the
results for Rényi numeration systems obtained by Frougny and Solomyak [9].
Among other, we prove that the finiteness property of an alternate base
$\boldsymbol{\mathcal{B}}=(\beta_{n})_{n\geq 1}$ with period $p$ implies that
$\delta=\prod_{i=1}^{p}\beta_{i}$ is a Pisot or a Salem number and
$\beta_{i}\in\mathbb{Q}(\delta)$ for $i\in\\{1,2,\dots,p\\}$, see Theorem 6.
On the other hand, we provide a sufficient condition for the finiteness and
positive finiteness property (Theorems 19 and 23). The latter result is proven
using a system of rewriting rules on non-admissible strings. We then give a
family of alternate bases with finiteness property (Theorem 7).

The rewriting system provides a method for performing addition on
$\boldsymbol{\mathcal{B}}$-expansions. It is conceivable that for some classes
of bases this tool will allow us to design an addition automaton, as it was
described by Frougny [8] for the Rényi numeration systems with confluent Parry
bases $\beta>1$. An analogue of confluent bases in the frame of Cantor real
bases is yet to be explored.

##  2 Preliminaries

###  2.1 Finiteness in Cantor real bases

A Cantor real base is a sequence $\boldsymbol{\mathcal{B}}=(\beta_{n})_{n\geq
1}$ of real numbers $\beta_{n}>1$ such that $\prod_{n\geq
1}\beta_{n}=+\infty$. A $\boldsymbol{\mathcal{B}}$-representation of a number
$x\in[0,1]$ is a sequence $\mathbf{x}=x_{1}x_{2}x_{3}\dots$ of non-negative
integers such that

$x=\sum_{n\geq
1}\frac{x_{n}}{\beta_{1}\cdots\beta_{n}}=:\operatorname{val}{\mathbf{x}}.$

Note that $\operatorname{val}\mathbf{x}$ can be defined for any (not only non-
negative) sequence $\mathbf{x}$ of integers. We say that the representation is
finite, if its support
$\operatorname{supp}{\mathbf{x}}=\\{n\in\mathbb{N}:a_{n}\neq 0\\}$ is finite.
Denote

$\mathcal{F}=\\{\mathbf{x}=(x_{n})_{n\geq
1}\,:\,\operatorname{supp}{\mathbf{x}}\text{ is finite and
}x_{n}\in\mathbb{N}\ \text{for each }n\\}$

the set of sequences of non-negative integers with finite support.

A particular $\boldsymbol{\mathcal{B}}$-representation can be obtained by the
greedy algorithm:

For a given $x\in[0,1)$, set $x_{1}=\lfloor\beta_{1}x\rfloor$,
$r_{1}=\beta_{1}x-x_{1}$, and for $n\geq 2$ set
$x_{n}=\lfloor\beta_{n}r_{n-1}\rfloor$, ${r_{n}=\beta_{n}r_{n-1}-x_{n}}$. The
obtained $\boldsymbol{\mathcal{B}}$-representation is denoted by
$d_{\boldsymbol{\mathcal{B}}}(x)=x_{1}x_{2}x_{3}\dots$ and called the
$\boldsymbol{\mathcal{B}}$-expansion of $x$. It follows from the greedy
algorithm that the digits in the $\boldsymbol{\mathcal{B}}$-expansion satisfy
$0\leq x_{i}<\beta_{i}$ and that $d_{\boldsymbol{\mathcal{B}}}(x)$ is
lexicographically greatest among all
$\boldsymbol{\mathcal{B}}$-representations of $x$. Note that the algorithm
works also for $x=1$; the sequence $d_{\boldsymbol{\mathcal{B}}}(1)$ is called
the $\boldsymbol{\mathcal{B}}$-expansion of 1 and it is lexicographically
greatest among all $\boldsymbol{\mathcal{B}}$-representations of 1.

If a $\boldsymbol{\mathcal{B}}$-expansion is finite, i.e. it is of the form
$w_{1}w_{2}\cdots w_{k}0^{\omega}$, where $0^{\omega}$ stands for infinite
repetition of zeros, we sometimes omit the suffix $0^{\omega}$. We define

$\operatorname{Fin}(\boldsymbol{\mathcal{B}})=\left\\{x\in[0,1)\,:\,d_{\boldsymbol{\mathcal{B}}}(x)\in\mathcal{F}\right\\}.$
(1)

######  Definition 1.

We say that a Cantor real base $\boldsymbol{\mathcal{B}}$ satisfies Property
(PF) (the positive finiteness property), if for every
$x,y\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$ we have

$x+y\in[0,1)\implies x+y\in\operatorname{Fin}(\boldsymbol{\mathcal{B}}).$ (2)

We say that a Cantor real base $\boldsymbol{\mathcal{B}}$ satisfies Property
(F) (the finiteness property), if for every
$x,y\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$ we have

$x+y\in[0,1)\implies
x+y\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})\quad\text{and}\quad
x-y\in[0,1)\implies x-y\in\operatorname{Fin}(\boldsymbol{\mathcal{B}}).$ (3)

Property (F) expresses the fact that the set of finite expansions is closed
under addition and subtraction of its elements, provided the sum belong to the
interval $[0,1)$. Property (PF) focuses only on addition.

For a given sequence of real numbers $\beta_{n}>1$, $n\geq 1$, denote by
$\boldsymbol{\mathcal{B}}^{(k)}$ the Cantor real base
${\boldsymbol{\mathcal{B}}^{(k)}=(\beta_{n})_{n\geq k}}$. Realize the
following relation between these bases.

######  Lemma 2.

Let
$\boldsymbol{\mathcal{B}}=\boldsymbol{\mathcal{B}}^{(1)}=(\beta_{n})_{n\geq
1}$ be a Cantor real base. If $\boldsymbol{\mathcal{B}}^{(1)}$ satisfies
Property (PF) or Property (F), then, for every $k\geq 1$, the base
$\boldsymbol{\mathcal{B}}^{(k)}$ satisfies Property (PF) or Property (F),
respectively.

###### Proof.

Consider $z\in[0,1)$. From the greedy algorithm, it is not difficult to show
that

$d_{\boldsymbol{\mathcal{B}}^{(2)}}(z)=z_{2}z_{3}z_{4}\cdots\iff
d_{\boldsymbol{\mathcal{B}}^{(1)}}(z/\beta_{1})=0z_{2}z_{3}z_{4}\cdots.$ (4)

Consequently,

$\frac{1}{\beta_{1}}\operatorname{Fin}\big{(}\boldsymbol{\mathcal{B}}^{(2)}\big{)}\subset\operatorname{Fin}\big{(}\boldsymbol{\mathcal{B}}^{(1)}\big{)}.$

Take $x,y\in\operatorname{Fin}\big{(}\boldsymbol{\mathcal{B}}^{(2)}\big{)}$
such that $x+y<1$. Then $\frac{x}{\beta_{1}}$,
$\frac{y}{\beta_{1}}\in\operatorname{Fin}\big{(}\boldsymbol{\mathcal{B}}^{(1)}\big{)}$,
and, moreover,
${z=\frac{x}{\beta_{1}}+\frac{y}{\beta_{1}}<\frac{1}{\beta}_{1}<1}$. By
Property (PF) of the base $\boldsymbol{\mathcal{B}}^{(1)}$, necessarily
$z\in\operatorname{Fin}(\boldsymbol{\mathcal{B}}^{(1)})$. Moreover, since
$z<\frac{1}{\beta}_{1}$, the greedy algorithm implies that the
$\boldsymbol{\mathcal{B}}^{(1)}$-expansion of $z$ is of the form
$d_{\boldsymbol{\mathcal{B}}^{(1)}}(z/\beta_{1})=0z_{2}z_{3}z_{4}\cdots\in\mathcal{F}$.
Therefore
$d_{\boldsymbol{\mathcal{B}}^{(2)}}(\beta_{1}z)=z_{2}z_{3}z_{4}\cdots\in\mathcal{F}$,
which means that
$x+y\in\operatorname{Fin}\big{(}\boldsymbol{\mathcal{B}}^{(2)}\big{)}$. This
proves that $\boldsymbol{\mathcal{B}}^{(2)}$ satisfies Property (PF). The
proof for subtraction is analogous. For $\boldsymbol{\mathcal{B}}^{(k)}$,
$k\geq 3$, we proceed by induction. ∎

###  2.2 Admissibility

Let $\boldsymbol{\mathcal{B}}=(\beta_{n})_{n\geq 1}$ be a Cantor real base. We
say that a sequence $\mathbf{x}=(x_{n})_{n\geq 1}$ of integers is admissible
in base $\boldsymbol{\mathcal{B}}$, if there exists a real number $x\in[0,1)$
such that $\mathbf{x}=d_{\boldsymbol{\mathcal{B}}}(x)$. The admissible
sequences have been characterized in [4] in terms of the quasi-greedy
expansions of 1. The quasi-greedy expansion of 1 in base
$\boldsymbol{\mathcal{B}}$ is defined as

$d_{\boldsymbol{\mathcal{B}}}^{*}(1)=\lim_{x\to
1-}d_{\boldsymbol{\mathcal{B}}}(x),$

where the limit is taken over the product topology. Note that the quasi-greedy
expansion of 1 is lexicographically greatest among all
${\boldsymbol{\mathcal{B}}}$-representations of 1 with infinitely many non-
zero digits. We obviously have

$d_{\boldsymbol{\mathcal{B}}}^{*}(1)\preceq_{lex}d_{\boldsymbol{\mathcal{B}}}(1),$

with equality precisely if $d_{\boldsymbol{\mathcal{B}}}(1)\notin\mathcal{F}$.

######  Theorem 3 ([4]).

Let ${\boldsymbol{\mathcal{B}}}=(\beta_{n})_{n\geq 1}$ be a Cantor real base.
A sequence of non-negative integers $z_{1}z_{2}z_{3}\cdots$ is admissible in
base ${\boldsymbol{\mathcal{B}}}$ if and only if for each $n\geq 1$ the
following inequality holds true

$z_{n}z_{n+1}z_{n+2}\cdots\prec_{lex}d_{\boldsymbol{\mathcal{B}}^{(n)}}^{*}(1).$

For recognizing admissibility of finite digit strings, we have a condition
formulated using the $\boldsymbol{\mathcal{B}}$-expansions of 1, instead of
the quasi-greedy expansions.

######  Proposition 4.

Let ${\boldsymbol{\mathcal{B}}}=(\beta_{n})_{n\geq 1}$ be a Cantor real base.
A sequence of non-negative integers ${z_{1}z_{2}z_{3}\cdots\in\mathcal{F}}$ is
admissible in base $\boldsymbol{\mathcal{B}}$ if and only if for each $n\geq
1$ the following inequality holds true

$z_{n}z_{n+1}z_{n+2}\cdots\prec_{lex}d_{\boldsymbol{\mathcal{B}}^{(n)}}(1).$

###### Proof.

The implication $\Rightarrow$ follows from Theorem 3, since
$z_{n}z_{n+1}z_{n+2}\cdots\prec_{lex}d_{\boldsymbol{\mathcal{B}}^{*}{(n)}}(1)\preceq_{lex}d_{\boldsymbol{\mathcal{B}}{(n)}}(1)$.
For the opposite implication $\Leftarrow$, we will prove an equivalent
statement:

Let $\mathbf{z}=z_{1}z_{2}z_{3}\cdots$ belong to $\mathcal{F}$. If
$\mathbf{z}$ is not admissible in base $\boldsymbol{\mathcal{B}}$, then there
exists index $\ell\in\mathbb{N},\ell\geq 1$, such that
$d_{\boldsymbol{\mathcal{B}}^{(\ell)}}(1)\preceq
z_{\ell}z_{\ell+1}z_{\ell+2}\dots$.

By Theorem 3, if $\mathbf{z}=z_{1}z_{2}z_{3}\cdots$ is not admissible in
$\boldsymbol{\mathcal{B}}$, then there exists an index $i\geq 1$ such that
$d^{*}_{\boldsymbol{\mathcal{B}}^{(i)}}(1)\preceq_{lex}z_{i}z_{i+1}z_{i+2}\cdots$.
Obviously, $i\leq\max\operatorname{supp}{\mathbf{z}}$. Since
$\mathbf{z}\in\mathcal{F}$ and
$d^{*}_{\boldsymbol{\mathcal{B}}^{(i)}}(1)\notin\mathcal{F}$, we have a strict
inequality

$d^{*}_{\boldsymbol{\mathcal{B}}^{(i)}}(1)\prec_{lex}z_{i}z_{i+1}z_{i+2}\cdots.$
(5)

If $d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)\notin\mathcal{F}$, then
$d^{*}_{\boldsymbol{\mathcal{B}}^{(i)}}(1)=d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)$
and the proof is finished. Consider the other case, namely that
$d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)=w_{1}w_{2}\cdots
w_{n-1}w_{n}0^{\omega}\in\mathcal{F}$, where $w_{n}\geq 1$. It has been shown
in [4] that $d^{*}_{\boldsymbol{\mathcal{B}}^{(i)}}(1)=w_{1}\cdots
w_{n-1}(w_{n}-1)d^{*}_{\boldsymbol{\mathcal{B}}^{(i+n)}}(1)$. Inequality (5)
implies

  1. 1.

either $w_{1}\cdots w_{n-1}(w_{n}-1)\prec_{lex}z_{i}z_{i+1}\cdots z_{i+n-1}$,

  2. 2.

or
$d^{*}_{\boldsymbol{\mathcal{B}}^{(i+n)}}(1)\prec_{lex}z_{i+n}z_{i+n+1}\cdots$.

Case 1. As the lexicographically smallest word of length $n$ strictly greater
than $w_{1}\dots w_{n-1}(w_{n}-1)$ is $w_{1}\dots w_{n-1}w_{n}$, we have
$w_{1}\dots w_{n-1}w_{n}\preceq_{lex}z_{i}z_{i+1}\cdots z_{i+n-1}$. Thus

$d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)\preceq_{lex}z_{i}z_{i+1}\dots
z_{i+n-1}0^{\omega}\preceq_{lex}z_{i}z_{i+1}z_{i+2}\dots,$

as we wanted to show.

Case 2. We have found $i_{new}=i+n>i$ so that (5) is satisfied when
substituting $i_{new}$ in place of $i$. The same idea can be repeated. As
$\operatorname{supp}{\mathbf{z}}$ is bounded, finitely many steps lead to Case
1. ∎

###  2.3 Alternate bases

In case that the Cantor real base $\boldsymbol{\mathcal{B}}=(\beta_{n})_{n\geq
1}$ is a purely periodic sequence with period $p$, we call it an alternate
base and denote

$\boldsymbol{\mathcal{B}}=(\beta_{1},\beta_{2},\dots,\beta_{p}),\qquad\delta=\prod_{i=1}^{p}\beta_{i}.$

For the Cantor real bases
$\boldsymbol{\mathcal{B}}^{(n)}=(\beta_{n},\beta_{n+1},\dots,\beta_{n+p-1})$,
we have $\boldsymbol{\mathcal{B}}^{(n)}=\boldsymbol{\mathcal{B}}^{(n+p)}$ for
every $n\geq 1$. The indices will be therefore considered $\\!\\!\\!\mod p$.
Let us stress that throughout this paper the representatives of the congruence
classes $\\!\\!\\!\mod p$ will be taken in the set
$\mathbb{Z}_{p}=\\{1,2,\dots,p\\}$.

For the Cantor real base $\boldsymbol{\mathcal{B}}^{(n)}$, $n\geq 1$, denote
the $\boldsymbol{\mathcal{B}}^{(n)}$-expansions of 1 by

$\mathbf{t}^{(n)}=t_{1}^{(n)}t_{2}^{(n)}t_{3}^{(n)}\cdots=d_{\boldsymbol{\mathcal{B}}^{(n)}}(1).$
(6)

Note that $\mathbf{t}^{(n)}=\mathbf{t}^{(n+p)}$ for each $n\in\mathbb{N}$. In
fact, we only work with $\mathbf{t}^{(n)}$, $n\in\mathbb{Z}_{p}$.

The specific case when $p=1$ yields the well-known numeration systems in one
real base $\beta>1$ as defined by Rényi [10]. These systems define a sofic
$\beta$-shift precisely when the base $\beta$ is a Parry number. In analogy to
this case we say that the alternate base $\boldsymbol{\mathcal{B}}$ with
period $p\geq 1$ is a Parry alternate base, if all the sequences
$\mathbf{t}^{(\ell)}$, $\ell\in\mathbb{Z}_{p}$, are eventually periodic. If,
moreover, $\mathbf{t}^{(\ell)}\in\mathcal{F}$ for all $\ell\in\mathbb{Z}_{p}$,
we speak about simple Parry alternate base.

Note that (4) for an alternate base $\boldsymbol{\mathcal{B}}$ implies

$d_{\boldsymbol{\mathcal{B}}}(z)=z_{1}z_{2}z_{3}\cdots\iff
d_{\boldsymbol{\mathcal{B}}}(z/\delta)=0^{p}z_{1}z_{2}z_{2}\cdots$ (7)

and

$\frac{1}{\delta}\operatorname{Fin}(\boldsymbol{\mathcal{B}})\subset\operatorname{Fin}(\boldsymbol{\mathcal{B}}).$

We thus have a simple consequence of Lemma 2.

######  Corollary 5.

Let $\boldsymbol{\mathcal{B}}=(\beta_{1},\beta_{2},\dots,\beta_{p})$ be an
alternate base. Then for any $\ell\in\mathbb{Z}_{p}$ the base
$\boldsymbol{\mathcal{B}}^{(\ell)}$ satisfies Property (PF) or Property (F),
if and only if $\boldsymbol{\mathcal{B}}$ satisfies Property (PF) or Property
(F), respectively.

##  3 Main results

We focus on the finiteness property of alternate bases. We first provide
necessary conditions on an alternate base in order to satisfy Property (PF)
and Property (F). For that, we need to recall some number theoretical notions.
An algebraic integer $\delta>1$ is a Pisot number, if all its conjugates other
than $\delta$ belong to the interior of the unit disc; an algebraic integer
$\delta>1$ is a Salem number, if all its conjugates other than $\delta$ belong
to the unit disc and at least one lies on its boundary. If $\delta$ is an
algebraic number of degree $d$, then the minimal subfield of $\mathbb{C}$
containing $\delta$ is of the form

$\mathbb{Q}(\delta)=\left\\{c_{0}+c_{1}\delta+\cdots+c_{d-1}\delta^{d-1}\,:\,c_{i}\in\mathbb{Q}\right\\}.$

The number field $\mathbb{Q}(\delta)$ has precisely $d$ embeddings into
$\mathbb{C}$, namely the field isomorphisms
${\psi:\mathbb{Q}(\delta)\to\mathbb{Q}(\gamma)}$ induced by
$\psi(\delta)=\gamma$, where $\gamma$ is a conjugate of $\delta$.

The necessary condition is formulated as Theorem 6. Its proof is found in
Section 4 separated into several propositions.

######  Theorem 6.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base.

  1. 1.

If $\boldsymbol{\mathcal{B}}$ satisfies Property (PF), then
$\delta=\prod_{i=1}^{p}\beta_{i}$ is a Pisot or a Salem number and
$\beta_{i}\in\mathbb{Q}(\delta)$ for ${i\in\mathbb{Z}_{p}}$.

  2. 2.

If, moreover, $\boldsymbol{\mathcal{B}}$ satisfies Property (F), then
$\boldsymbol{\mathcal{B}}$ is a simple Parry alternate base, and for any non-
identical embedding $\psi$ of $\mathbb{Q}(\delta)$ into $\mathbb{C}$ the
vector $(\psi(\beta_{1}),\dots,\psi(\beta_{p}))$ is not positive.

As the main result of this paper we provide a class of alternate Parry bases
satisfying the finiteness property. It is a generalization of the class given
in [9]. Our proof follows similar ideas as in [1], but requires much deeper
techniques.

######  Theorem 7.

Let $\boldsymbol{\mathcal{B}}$ be an alternate base with period $p$ such that
the corresponding expansions $\mathbf{t}^{(\ell)}$ of 1 satisfy

$t_{1}^{(\ell)}\geq\ t_{2}^{(\ell-1)}\geq\ t_{3}^{(\ell-2)}\geq\
\cdots\quad\text{ for every }\ \ell\in\mathbb{Z}_{p}.$ (8)

Then $\boldsymbol{\mathcal{B}}$ is a Parry alternate base and has property
(PF).

Moreover, if $\boldsymbol{\mathcal{B}}$ is a simple Parry alternate base, then
it satisfies Property (F).

Note that in inequalities (8) we have upper indices in $\mathbb{Z}$, but they
are always counted $\\!\\!\\!\mod p$.

In Section 5 we first show a sufficient condition for Property (PF) in terms
of rewriting rules of non-admissible sequences (Theorem 19). Then, in Section
6, we prove that the class of bases satisfying the inequalities (8) meets the
assumptions of Theorem 19, and thus possesses Property (PF).

Let us illustrate the above results.

######  Example 8.

Consider the favourite example of [4], namely the base
$\boldsymbol{\mathcal{B}}=(\beta_{1},\beta_{2})=\left(\frac{1+\sqrt{13}}{2},\frac{5+\sqrt{13}}{6}\right)$.
Expansions of 1 are of the form

$d_{\boldsymbol{\mathcal{B}}^{(1)}}(1)=t_{1}^{(1)}t_{2}^{(1)}t_{3}^{(1)}0^{\omega}=2010^{\omega}\quad\text{
and }\quad
d_{\boldsymbol{\mathcal{B}}^{(2)}}(1)=t_{1}^{(2)}t_{2}^{(2)}0^{\omega}=110^{\omega}.$

Consequently, $\boldsymbol{\mathcal{B}}$ is a simple Parry alternate base. The
inequalities (8) for $p=2$ are of the form

$\displaystyle t_{1}^{(1)}\geq t_{2}^{(2)}\geq t^{(1)}_{3}\geq
t_{4}^{(2)}\geq\cdots,$ $\displaystyle\text{ in our case }\hskip
30.00005pt2\geq 1\geq 1\geq 0\geq\cdots,$ $\displaystyle t_{1}^{(2)}\geq
t_{2}^{(1)}\geq t^{(2)}_{3}\geq t_{4}^{(1)}\geq\cdots,$ $\displaystyle\text{
in our case }\hskip 30.00005pt1\geq 0\geq 0\geq 0\geq\cdots.$

Therefore, according to Theorem 7, the base $\boldsymbol{\mathcal{B}}$ has
Property (F).

Note that the necessary conditions of Property (F) as presented in Theorem 6
are satisfied. Indeed, we have
$\delta=\frac{1+\sqrt{13}}{2}\cdot\frac{5+\sqrt{13}}{6}=\frac{3+\sqrt{13}}{2}$,
which is the positive root of $x^{2}-3x-1$. The other root of this polynomial
is $\gamma=\frac{3-\sqrt{13}}{2}\approx-0.303$, thus $\delta$ is a Pisot
number.

Obviously,
$\beta_{1},\beta_{2}\in\mathbb{Q}(\delta)=\mathbb{Q}\left(\sqrt{13}\right)$.
The only non-identical embedding $\psi$ of the field
$\mathbb{Q}\left(\sqrt{13}\right)$ into $\mathbb{C}$ is the Galois
automorphism induced by $\sqrt{13}\mapsto-\sqrt{13}$. Thus we have

$\psi(\beta_{1})=\frac{1-\sqrt{13}}{2}\approx-1.303,\hskip
50.00008pt\psi(\beta_{2})=\frac{5-\sqrt{13}}{6}\approx 0.232,$

therefore indeed $\left(\psi(\beta_{1}),\psi(\beta_{2})\right)$ is not a
positive vector.

##  4 Necessary conditions

The proof of Theorem 6 is divided into four propositions. Throughout this
section we have an alternate base
$\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ and
$\delta=\prod_{i=1}^{p}\beta_{i}$. In the demonstration, we will use that
$\prod_{i=1}^{mp}\beta_{i}=\delta^{m}$, which gives a simple expression that
holds for any $n\in\mathbb{N}$ and any sequence of digits
$(c_{k})_{k=1}^{pn}$, namely

$\sum_{k=1}^{pn}\frac{c_{k}}{\prod_{i=1}^{k}\beta_{i}}=\frac{1}{\delta^{n}}\sum_{j=1}^{p}\
\Bigg{(}\prod_{i=j+1}^{p}\beta_{i}\Bigg{)}\sum_{k=0}^{n-1}\delta^{n-1-k}c_{kp+j}.$
(9)

The first step is a rather technical lemma necessary for the proof of
Proposition 10.

######  Lemma 9.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base satisfying Property (PF). Then for any sufficiently large
$n\in\mathbb{N}$ there exist polynomials
$g_{1},g_{2},\ldots,g_{p}\in\mathbb{Z}[X]$ such that

  * •

the degree of $g_{j}$ is at most $n-1$ for every $j\in\\{1,2,\ldots,p-1\\}$;

  * •

$g_{p}$ is monic and its degree is $n$;

  * •

all coefficients of $g_{j}$ are non-negative for every $j\in\\{2,\ldots,p\\}$;

  * •

the product of the row vector
$\bigl{(}g_{1}(\delta),\ldots,g_{p}(\delta)\bigr{)}$ and the column vector
$\vec{v}=\\!\bigg{(}\\!\prod\limits_{i=2}^{p}\beta_{i},\prod\limits_{i=3}^{p}\beta_{i},\ldots,\prod\limits_{i=p}^{p}\beta_{i},1\bigg{)}^{T}$
is zero.

###### Proof.

Choose $m\in\mathbb{N}$ such that
$\delta^{m}\beta_{1}\leq\lceil\delta^{m}\beta_{1}\rceil<\delta^{m}\beta_{p}\beta_{1}$.
Hence
$z:={\lceil\delta^{m}\beta_{1}\rceil}({\delta^{m}\beta_{p}\beta_{1}})^{-1}$
satisfies inequalities $\frac{1}{\beta_{p}}\leq z<1$. It is easy to see that
$({\delta^{m}\beta_{p}\beta_{1}})^{-1}\in\operatorname{Fin}\big{(}\boldsymbol{\mathcal{B}}^{(p)}\big{)}$,
since its $\boldsymbol{\mathcal{B}}^{(p)}$-expansion is of the form
$0^{mp+1}10^{\omega}$. By Corollary 5, Property (PF) of the base
$\boldsymbol{\mathcal{B}}$ implies that also the base
${\boldsymbol{\mathcal{B}}^{(p)}=(\beta_{p},\beta_{1},\ldots,\beta_{p-1})}$
has Property (PF). Thus any integer multiple of
$({\delta^{m}\beta_{p}\beta_{1}})^{-1}$, in particular the number $z$, has a
finite $\boldsymbol{\mathcal{B}}^{(p)}$-expansion. Since
$z={\lceil\delta^{m}\beta_{1}\rceil}({\delta^{m}\beta_{p}\beta_{1}})^{-1}$, we
easily see that ${\frac{1}{\beta_{p}}\leq z<\frac{2}{\beta_{p}}}$, and thus
the greedy algorithm returns the $\boldsymbol{\mathcal{B}}^{(p)}$-expansion of
$z$ of the form ${d_{\boldsymbol{\mathcal{B}}^{(p)}}(z)=1z_{1}z_{2}\cdots
z_{pn}0^{\omega}}$, $z_{i}\in\mathbb{N}$, for sufficiently large
$n\in\mathbb{N},n>m$. Using (9), we derive

$\frac{\lceil\delta^{m}\beta_{1}\rceil}{\delta^{m}\beta_{p}\beta_{1}}=\frac{1}{\beta_{p}}+\frac{1}{\beta_{p}}\sum_{k=1}^{pn}\frac{z_{k}}{\prod_{i=1}^{k}\beta_{i}}=\frac{1}{\beta_{p}}+\frac{1}{\delta^{n}\beta_{p}}\sum_{j=1}^{p}\
\Bigg{(}\prod_{i=j+1}^{p}\beta_{i}\Bigg{)}\sum_{k=0}^{n-1}\delta^{n-1-k}z_{kp+j}.$

In order to simplify notation, denote the $j$-th component of the vector
$\vec{v}$ by $v_{j}=\prod_{i=j+1}^{p}\beta_{i}$. Multiplying the above
equality by $\delta^{n}\beta_{p}$ yields

$\lceil\delta^{m}\beta_{1}\rceil\delta^{n-m-1}v_{1}=\delta^{n}v_{p}+\sum_{j=1}^{p}\
v_{j}\sum_{k=0}^{n-1}\delta^{n-1-k}z_{kp+j}.$ (10)

Now we can define polynomial $g_{j}$, $j\in\mathbb{Z}_{p}$, as follows.

$\displaystyle g_{1}(X)$ $\displaystyle:=-\lceil\delta^{m}\beta_{1}\rceil
X^{n-m-1}+\sum_{k=0}^{n-1}X^{n-k-1}z_{kp+1},$ $\displaystyle g_{j}(X)$
$\displaystyle:=\sum_{k=0}^{n-1}X^{n-k-1}z_{kp+j}\quad\text{ for
}j\in\\{2,3,\ldots,p-1\\},$ $\displaystyle g_{p}(X)$
$\displaystyle:=X^{n}+\sum_{k=0}^{n-1}X^{n-k-1}z_{kp+p}.$

With this notation, equality (10) is of the form
$0=g_{1}(\delta)v_{1}+g_{2}(\delta)v_{2}+\cdots+g_{p}(\delta)v_{p}$. One
easily verifies all required properties of polynomials $g_{j}$,
$j\in\mathbb{Z}_{p}$. ∎

The proof of the following proposition is a modification of the proof of
Theorem 14 in [6].

######  Proposition 10.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base with Property (PF). Then $\delta=\prod_{i=1}^{p}\beta_{i}$ is
an algebraic integer and $\beta_{i}\in\mathbb{Q}(\delta)$ for all
$i\in\mathbb{Z}_{p}$.

###### Proof.

By Corollary 5, $\boldsymbol{\mathcal{B}}^{(\ell)}$ satisfies Property (PF)
for every $\ell\in\mathbb{Z}_{p}$. Choose $n\in\mathbb{N}$ sufficiently large
so that Lemma 9 can be applied for every $\ell\in\mathbb{Z}_{p}$ with the same
value of $n$. By that, we obtain polynomials
$g^{(\ell)}_{1},g^{(\ell)}_{2},\ldots,g^{(\ell)}_{p-1},g^{(\ell)}_{p}\in\mathbb{Z}[X]$
with properties listed in the lemma; in particular, the vectors
$\Big{(}g^{(\ell)}_{1}(\delta),g^{(\ell)}_{2}(\delta),\ldots,g^{(\ell)}_{p}(\delta)\Big{)}$
and
$\vec{v}^{(\ell)}=\bigg{(}\prod\limits_{i=2}^{p}\beta_{i+\ell-1},\,\prod\limits_{i=3}^{p}\beta_{i+\ell-1},\,\ldots,\prod\limits_{i=p}^{p}\beta_{i+\ell-1},\,1\bigg{)}^{T}$
satisfy

$\Big{(}g^{(\ell)}_{1}(\delta),\ldots,g^{(\ell)}_{p}(\delta)\Big{)}\,\vec{v}^{(\ell)}=0\quad\text{for
every }\ell\in\mathbb{Z}_{p}.$ (11)

Now consider $\ell\in\mathbb{Z}_{p}$ fixed. For $k\in\mathbb{Z}_{p}$, the
$k$-th component of the vector $\vec{v}^{(\ell)}$ is equal to

$\vec{v}^{(\ell)}_{k}=\prod_{i=k+1}^{p}\beta_{i+\ell-1}=\prod_{j=k+\ell}^{p+\ell-1}\beta_{j}=\begin{cases}\prod_{j=k+\ell-p}^{\ell-1}\beta_{j}&\text{
for }k+\ell\geq p+2;\\\\[5.69054pt]
\Big{(}\prod_{j=k+\ell}^{p}\beta_{j}\Big{)}\cdot\Big{(}\prod_{j=1}^{\ell-1}\beta_{j}\Big{)}&\text{
for }k+\ell\leq p+1.\end{cases}$

As a result, multiplying the vector $\vec{v}^{(\ell)}$ by the constant
$\prod_{i=\ell}^{p}\beta_{i}$ leads to

$\Bigg{(}\prod_{i=\ell}^{p}\beta_{i}\Bigg{)}\
\vec{v}^{(\ell)}=\Big{(}\delta\vec{v}^{(1)}_{\ell},\dots,\delta\vec{v}^{(1)}_{p},\vec{v}^{(1)}_{1},\dots,\vec{v}^{(1)}_{\ell-1}\Big{)}^{T}=\begin{pmatrix}{\Theta}&\delta
I_{p-\ell+1}\\\ I_{\ell-1}&\Theta\end{pmatrix}\vec{v}^{(1)},$

where $I_{j}$ denotes the identity matrix of order $j$ and $\Theta$ stands for
rectangular zero matrices of suitable sizes. We have thus derived that for
$\ell\in\mathbb{Z}_{p}$

$\Bigg{(}\prod_{i=\ell}^{p}\beta_{i}\Bigg{)}\ \vec{v}^{(\ell)}=R^{p-\ell+1}\
\vec{v}^{(1)}\quad\text{where}\quad R=\begin{pmatrix}{\Theta}&\delta I_{1}\\\
I_{p-1}&\Theta\end{pmatrix}.$ (12)

Combining (11) with (12), we obtain that for $\ell\in\mathbb{Z}_{p}$

$\Big{(}g^{(\ell)}_{1}(\delta),\ldots,g^{(\ell)}_{p}(\delta)\Big{)}\
R^{p-\ell+1}\,\vec{v}^{(1)}=0.$

In other words, the vector $\vec{v}^{(1)}$ satisfies $M\vec{v}^{(1)}=\vec{0}$,
where the matrix $M$ is a square matrix of order $p$ of the form

$M=\begin{pmatrix}\delta g_{1}^{(1)}&\delta g_{2}^{(1)}&\delta
g_{3}^{(1)}&\cdots&\delta g_{p-1}^{(1)}&\delta g_{p}^{(1)}\\\
g_{p}^{(2)}&\delta g_{1}^{(2)}&\delta g_{2}^{(2)}&\cdots&\delta
g_{p-2}^{(2)}&\delta g_{p-1}^{(2)}\\\ g_{p-1}^{(3)}&g_{p}^{(3)}&\delta
g_{1}^{(3)}&\cdots&\delta g_{p-3}^{(3)}&\delta g_{p-2}^{(3)}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
g_{3}^{(p-1)}&g_{4}^{(p-1)}&g_{5}^{(p-1)}&\cdots&\delta g_{1}^{(p-1)}&\delta
g_{2}^{(p-1)}\\\ g_{2}^{(p)}&g_{3}^{(p)}&g_{4}^{(p)}&\cdots&g_{p}^{(p)}&\delta
g_{1}^{(p)}\\\ \end{pmatrix}.$

Note that the components of the matrix $M$ are polynomials in $\delta$ (for
simplicity of notation, we have omitted the dependence on the variable). Their
degree is given by Lemma 9. In particular, the element with strictly highest
degree is $M_{1,p}=\delta g_{p}^{(1)}$ of degree $n+1$.

Equality $M\vec{v}^{(1)}=\vec{0}$ means that $\vec{v}^{(1)}$ is an eigenvector
of the matrix $M$ corresponding to the eigenvalue $0$. Since
$\vec{v}^{(1)}\neq\vec{0}$, necessarily $\det M=0$. We now show that $\det M$
is a monic polynomial with integer coefficients in the variable $\delta$.
Consequently, $\delta$ is an algebraic integer. By Lemma 9, the product

$M_{1,p}M_{p,p-1}M_{p-1,p-2}\cdots
M_{3,2}M_{2,1}=\delta\prod_{\ell=1}^{p}g_{p}^{(\ell)}(\delta)$

is a product of monic polynomials, together of degree $pn+1$. All other
products contributing to the value of $\det M$ are of degree $\leq pn$. Hence,
$\det M$ is a monic polynomial in $\delta$ of degree $pn+1$ with integer
coefficients.

It remains to show that $\beta_{i}\in\mathbb{Q}(\delta)$ for every
$i\in\mathbb{Z}_{p}$. Recall that $\vec{v}^{(1)}$ is an eigenvector of the
matrix $M$ to the eigenvalues $0$. We now show that $0$ is a simple eigenvalue
of $M$. For that, we use Perron-Frobenius theorem. Note that all non-diagonal
components of $M$ are non-negative, since $\delta>1$ and the coefficients of
the polynomials $g^{(\ell)}_{j}$ for $j\geq 2$ are non-negative. Moreover,
$M_{1,p}=\delta g_{p}^{(1)}(\delta)$, $M_{2,1}=g_{p}^{(2)}(\delta)$, …,
$M_{p,p-1}=g_{p}^{(p)}(\delta)\geq\delta^{n}>0$ are positive. Thus for
sufficiently large $c\in\mathbb{N}$, the matrix $M+cI_{p}$ is a non-negative
irreducible matrix. The vector $\vec{v}^{(1)}$ is its positive eigenvector to
the eigenvalue $c$, i.e. $(M+cI_{p})\vec{v}^{(1)}=c\vec{v}^{(1)}$. By Perron-
Frobenius theorem, $c$ is a simple eigenvalue of $M+cI_{p}$ and hence $0$ is a
simple eigenvalue of $M$.

As components of the matrix $M$ belongs to the field $\mathbb{Q}(\delta)$
there exists a vector $\vec{u}$ with components in $\mathbb{Q}(\delta)$ such
that $M\vec{u}=\vec{0}$. We necessarily have $\vec{v}^{(1)}=\alpha\vec{u}$ for
some $\alpha\in\mathbb{R}$. To complete the proof, realize that for every
$i\in\\{2,3,\ldots,p\\}$

$\beta_{i}=\frac{v_{i-1}^{(1)}}{v_{i}^{(1)}}=\frac{u_{i-1}}{u_{i}}\in\leavevmode\nobreak\
\mathbb{Q}(\delta)\text{ and
}\beta_{1}=\frac{\delta}{v_{1}^{(1)}}\in\mathbb{Q}(\delta).$

∎

######  Proposition 11.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base with Property (PF). Then $\delta=\prod_{i=1}^{p}\beta_{i}$ is
either a Pisot or a Salem number.

The proof follows the same ideas as the proof of Theorem 1 in [7].

###### Proof.

According to Proposition 10 the number $\delta$ is an algebraic integer. If
$\delta$ is a rational integer, the proof is finished. Assume that
$\delta\notin\mathbb{N}$. For any sufficiently large $m\in\mathbb{N}$, one has

$\delta^{m}<\lceil\delta^{m}\rceil<\delta^{m}\beta_{p}.$

Since $\frac{1}{\delta^{m+1}}\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$,
Property (PF) implies that
$n\,\frac{1}{\delta^{m+1}}\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$ for
all $n\in\mathbb{N},n<\delta^{m+1}$. In particular,
$x:=\lceil\delta^{m}\rceil\frac{1}{\delta^{m+1}}\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$
as well, i.e. $d_{{\boldsymbol{\mathcal{B}}}}(x)=x_{1}x_{2}x_{3}\cdots
x_{pN}0^{\omega}$ for some $N\in\mathbb{N}$. Realize that $x$ satisfies
$\frac{1}{\delta}<x<\frac{2}{\delta}$, and therefore the greedy algorithm
yields $x_{1}=\dots=x_{p-1}=0$, $x_{p}=1$ and
$r_{p}=\frac{\lceil\delta^{m}\rceil-\delta^{m}}{\delta^{m}}\in\frac{1}{\delta^{m}}(0,1)$.
Hence $x_{p+1}=x_{p+2}=\dots=x_{mp}=0$. When evaluating the string
$d_{{\boldsymbol{\mathcal{B}}}}(x)$, we may group summands in the following
way

$x=\lceil\delta^{m}\rceil\frac{1}{\delta^{m+1}}=\frac{1}{\delta}+\sum_{k=mp+1}^{pN}\frac{x_{k}}{\prod_{i=1}^{k}{\beta_{i}}}=\frac{1}{\delta}+\sum_{j=m+1}^{N}\frac{1}{\delta^{j+1}}\underbrace{\Bigg{(}\sum_{k=1}^{p}x_{jp+k}\prod_{i=k+1}^{p}{\beta_{i}}\Bigg{)}}_{=:d_{j}}.$
(13)

Greedy algorithm for obtaining $d_{\boldsymbol{\mathcal{B}}}(x)$ implies that
the digit $x_{i}$ is bounded from above by $\beta_{i}\leq\delta$, for each
$i\in\mathbb{N},i\geq 1$, and thus every coefficient $d_{j}$ belongs to the
finite alphabet

$\mathcal{D}=\Bigg{\\{}a_{1}\prod_{i=2}^{p}\beta_{i}+a_{2}\prod_{i=3}^{p}\beta_{i}+\cdots+a_{p-1}\prod_{i=p}^{p}\beta_{i}+a_{p}\,:\,a_{j}\in\mathbb{N},a_{j}\leq\lfloor\delta\rfloor\Bigg{\\}}.$

In this notation, equality (13) multiplied by $\delta^{m+1}$ can be rewritten
as

$\lceil\delta^{m}\rceil-\delta^{m}=\sum_{j=m+1}^{N}\frac{d_{j}}{\delta^{j-m}}.$
(14)

In order to show that $\delta$ is a Pisot or a Salem number, denote $\gamma$ a
conjugate of $\delta$, $\gamma\neq\delta$, and $\psi$ the isomorphism between
$\mathbb{Q}(\delta)$ and $\mathbb{Q}(\gamma)$ induced by
$\psi(\delta)=\gamma$. Since $\lceil\delta^{m}\rceil\in\mathbb{N}$, we have
$\psi(\lceil\delta^{m}\rceil)=\lceil\delta^{m}\rceil$. By Proposition 10, all
$\beta_{i}$ belong to the field $\mathbb{Q}(\delta)$, and thus the image of
the isomorphism $\psi$ on the elements of the set $\mathcal{D}$ is well
defined. Application of $\psi$ on (14) yields

$\delta^{m}-\gamma^{m}=\psi\Bigl{(}\lceil\delta^{m}\rceil-\delta^{m}\Bigr{)}-\Bigl{(}\lceil\delta^{m}\rceil-\delta^{m}\Bigr{)}=\sum_{j=m+1}^{N}\frac{\psi(d_{j})}{\gamma^{j-m}}-\sum_{j=m+1}^{N}\frac{d_{j}}{\delta^{j-m}}.$
(15)

Let us show that the assumption $\lvert\gamma\rvert>1$ leads to a
contradiction.

Denote $M=\max\\{|\psi(d)|\,:\,d\in\mathcal{D}\\}$. Obviously

$\Biggl{|}\,\sum_{j=m+1}^{N}\frac{\psi(d_{j})}{\gamma^{j-m}}\,\Biggr{|}\leq
M\frac{1}{|\gamma|-1}\qquad\text{and}\qquad\Biggl{|}\sum_{j=m+1}^{N}\frac{d_{j}}{\delta^{j-m}}\Biggr{|}=\Bigl{|}\lceil\delta^{m}\rceil-\delta^{m}\Bigr{|}<1.$
(16)

Since neither $M$ nor $\gamma$ depends on the choice of $m\in\mathbb{N}$,
relations (15) and (16) imply that the sequence
${\bigl{(}\delta^{m}-\gamma^{m}\bigr{)}_{m\in\mathbb{N}}}$ is bounded. This is
in contradiction with the fact that
$\limsup\limits_{m\to+\infty}|\delta^{m}-\gamma^{m}|=+\infty$. ∎

Propositions 10 and 11 demonstrate Item 1 of Theorem 6. Let us proceed with
the proof of Item 2.

######  Proposition 12.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base. If $\boldsymbol{\mathcal{B}}$ satisfies (F), then
$\boldsymbol{\mathcal{B}}$ is a simple Parry alternate base.

###### Proof.

Denote $\delta=\prod_{i=1}^{p}\beta_{i}$ and
$d_{\boldsymbol{\mathcal{B}}}(1)=t_{1}t_{2}t_{3}\cdots$. It suffices to show
that $d_{\boldsymbol{\mathcal{B}}}(1)\in\mathcal{F}$. The fact that
$d_{\boldsymbol{\mathcal{B}}^{(\ell)}}(1)\in\mathcal{F}$ for every
$\ell\in\mathbb{Z}_{p}$ then follows from Corollary 5.

Clearly,
$\frac{1}{\delta},\frac{1}{\delta\beta_{1}}\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$,
and
$x:=\frac{1}{\delta}-\frac{t_{1}}{\delta\beta_{1}}=\frac{\beta_{1}-\lfloor\beta_{1}\rfloor}{\delta\beta_{1}}\in[0,1)$.
Property (F) implies that $x$ belongs to the set
$\operatorname{Fin}(\boldsymbol{\mathcal{B}})$ as well. Obviously,
$x<\frac{1}{\delta\beta_{1}}$. Therefore, the greedy algorithm for the
$\boldsymbol{\mathcal{B}}$-expansion
$d_{\boldsymbol{\mathcal{B}}}(x)=x_{1}x_{2}x_{3}\cdots$ yields
$x_{1}=x_{2}=\dots=x_{p+1}=0$ and

$r_{p+1}=\delta\beta_{1}x=\beta_{1}-{t_{1}}=\frac{t_{2}}{\beta_{2}}+\frac{t_{3}}{\beta_{2}\beta_{3}}+\cdots\
.$

Consequently, $0^{p+1}t_{2}t_{3}t_{4}\cdots$ is the
$\boldsymbol{\mathcal{B}}$-expansion of $x$. Since
$x\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$, the support of
$d_{\boldsymbol{\mathcal{B}}}(x)=0^{p+1}t_{2}t_{3}t_{4}\cdots$ is finite, and
thus also the support of
$d_{\boldsymbol{\mathcal{B}}}(1)=t_{1}t_{2}t_{3}\cdots$ is finite. ∎

The following proposition completes the proof of Theorem 6.

######  Proposition 13.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base with Property (F) and $\gamma$ be a conjugate of
$\delta=\prod_{i=1}^{p}\beta_{i}$, $\gamma\neq\delta$. Denote $\psi$ the
isomorphism of $\mathbb{Q}(\delta)$ and $\mathbb{Q}(\gamma)$ induced by
$\psi({\delta})=\gamma$. Then the vector
$(\psi(\beta_{1}),\dots,\psi(\beta_{p}))$ is not positive.

###### Proof.

It suffices to consider $\delta\notin\mathbb{Q}$. By Proposition 12, all
$d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)$ have finite support, thus there exist
$n\in\mathbb{N}$ such that we may write

$d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)=t_{1}^{(i)}t_{2}^{(i)}\dots
t_{n}^{(i)}0^{\omega}$

for all $i\in\\{1,\dots,p\\}$. Therefore we have the following relation

$1=\frac{t_{1}^{(i)}}{\beta_{i}}+\frac{t_{2}^{(i)}}{\beta_{i}\beta_{i+1}}+\dots+\frac{t^{(i)}_{n}}{\beta_{i}\cdots\beta_{i+n-1}}.$

for all $i$. We now proceed by contradiction. Assume that $\psi(\beta_{j})>0$
for all $j\in\\{1,\dots,p\\}$. Multiplying the above equation by $\beta_{i}$,
subtracting $t_{1}^{(i)}$ and applying $\psi$ yields

$\psi(\beta_{i})-t_{1}^{(i)}=\frac{t^{(i)}_{2}}{\psi(\beta_{i+1})}+\frac{t_{3}^{(i)}}{\psi(\beta_{i+1})\psi(\beta_{i+2})}+\dots+\frac{t_{n}^{(i)}}{\psi(\beta_{i+1})\cdots\psi(\beta_{i+n-1})}\geq
0.$

Therefore $\psi(\beta_{i})\geq t_{1}^{(i)}$ for all indices
$i\in\\{1,\dots,p\\}$. Consequently,

$\gamma=\psi(\delta)=\psi(\beta_{1})\cdots\psi(\beta_{p})\geq\prod_{i=1}^{p}t_{1}^{(i)}\geq
1,$

and that is a contradiction, because $\gamma=\psi(\delta)\neq 1$ and,
according to Proposition 11, the number $\delta$ is either a Pisot or a Salem
number. ∎

######  Remark 14.

In case that $p=1$, the above statement directly implies that $\delta=\beta$
is not a Salem number, since all Salem numbers have a conjugate in $(0,1)$.
For $p\geq 2$, we were not able to exclude $\delta$ to be a Salem number.

##  5 Sufficient condition

In this section we prove a sufficient condition for an alternate base to
satisfy Property (PF). This property can be reformulated as stated in the
following lemma.

######  Lemma 15.

A Cantor real base $\boldsymbol{\mathcal{B}}$ satisfies Property (PF) if and
only if for any string $\mathbf{z}\in\mathcal{F}$ such that
$\operatorname{val}(\mathbf{z})<1$ we have
$\operatorname{val}(\mathbf{z})\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})$,
i.e.
$d_{\boldsymbol{\mathcal{B}}}(\operatorname{val}\mathbf{z})\in\mathcal{F}$.

In other words, for each string $\mathbf{z}\in\mathcal{F}$ with
$\operatorname{val}(\mathbf{z})<1$, we will find a string
${\mathbf{x}}\in\mathcal{F}$ admissible in base $\boldsymbol{\mathcal{B}}$
such that $\operatorname{val}\mathbf{x}=\operatorname{val}\mathbf{z}$. In the
sequel, we will also repeatedly use the fact that admissible strings are
lexicographically the greatest among all strings representing the same value.

######  Definition 16.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base and denote the corresponding
$\boldsymbol{\mathcal{B}}^{(\ell)}$-expansions of 1 by $\mathbf{t}^{(\ell)}$,
as in (6). Denote the following set of strings with finite support

$\mathcal{S}=\left\\{0^{p+\ell-1}t_{1}^{(\ell)}t_{2}^{(\ell)}\cdots
t_{k-1}^{(\ell)}(t_{k}^{(\ell)}+1)0^{\omega}\ :\ \ell\in\mathbb{Z}_{p},k\geq
1\right\\}\cup\left\\{0^{p+\ell}\mathbf{t}^{(\ell)}\in\mathcal{F}\ :\
\ell\in\mathbb{Z}_{p}\right\\}.$ (17)

We say that the alternate base $\boldsymbol{\mathcal{B}}$ has the rewriting
property, if for any string $\mathbf{a}\in\mathcal{S}$ there exists a string
$T(\mathbf{a})$, such that

$T(\mathbf{a})\in\mathcal{F},\quad\operatorname{val}T(\mathbf{a})=\operatorname{val}\mathbf{a},\quad\text{and}\quad
T(\mathbf{a})\succ_{lex}\mathbf{a}.$ (18)

The following lemma shows that given an alternate base
$\boldsymbol{\mathcal{B}}$ with rewriting property, a non-admissible
$\mathbf{z}\in\mathcal{F}$ can always be replaced by a lexicographically
greater string in $\mathcal{F}$ of the same value. For that, let us introduce
the notation for digit-wise addition of strings. Let
$\mathbf{x}=x_{1}x_{2}x_{3}\cdots$ and $\mathbf{y}=y_{1}y_{2}y_{3}\cdots$ be
two sequences with $x_{n},y_{n}\in\mathbb{Z}$ for every $n\in\mathbb{N},n\geq
1$. Then $\mathbf{x}\oplus\mathbf{y}$ stands for the sequence
$z_{1}z_{2}z_{3}\cdots$, where $z_{n}=x_{n}+y_{n}$ for every $n\in\mathbb{N}$,
$n\geq 1$. Obviously, in this notation, we have

  * •

$\operatorname{val}(\mathbf{x}\oplus\mathbf{y})=\operatorname{val}\mathbf{x}+\operatorname{val}\mathbf{y}$;

  * •

if $\mathbf{x}\prec_{lex}\tilde{\mathbf{x}}$, then for every $\mathbf{y}$ the
inequality
$\mathbf{x}\oplus\mathbf{y}\prec_{lex}\tilde{\mathbf{x}}\oplus\mathbf{y}$
holds true.

######  Lemma 17.

Let $\boldsymbol{\mathcal{B}}$ be an alternate base with rewriting property.
Then for every $\mathbf{z}\in\mathcal{F}$ non-admissible in
$\boldsymbol{\mathcal{B}}$ with value
$\operatorname{val}\mathbf{z}<\frac{1}{\delta}$, there exist
$\mathbf{x},\mathbf{y}\in\mathcal{F}$ and $j\in\mathbb{N}$ such that
$\mathbf{z}=\bigl{(}0^{pj}\mathbf{x}\bigr{)}\oplus\mathbf{y}$ and
$\mathbf{x}\in\mathcal{S}$. Consequently,
$\bigl{(}0^{pj}T(\mathbf{x})\bigr{)}\oplus\mathbf{y}\in\mathcal{F}$ is a
$\boldsymbol{\mathcal{B}}$-representation of $\operatorname{val}\mathbf{z}$
lexicographically strictly greater than $\mathbf{z}$.

###### Proof.

Consider $\mathbf{z}=z_{1}z_{2}\cdots\in\mathcal{F}$ non-admissible in
$\boldsymbol{\mathcal{B}}$. Since
$\operatorname{val}\mathbf{z}<\frac{1}{\delta}$, necessarily
$z_{1}=z_{2}=\dots=z_{p}=0$. By Proposition 4, there exists
$i\in\mathbb{N},i\geq p+1$, such that
$z_{i}z_{i+1}\cdots\succeq_{lex}d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)$.

We distinguish two cases how to determine the string $\mathbf{x}$:

  * a)

Suppose that $z_{i}z_{i+1}\cdots=d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)$.

As $\mathbf{z}\in\mathcal{F}$, we have that
$d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)\in\mathcal{F}$. We can take
$\mathbf{x}\in\mathcal{S}$ and $j\in\mathbb{N}$ such that
$0^{pj}\mathbf{x}=0^{i-1}d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)$.

  * b)

Suppose that
$z_{i}z_{i+1}\cdots\succ_{lex}d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)$.

Choose minimal $n\in\mathbb{N}$ such that $z_{i}z_{i+1}\cdots
z_{i+n}\succ_{lex}d_{\boldsymbol{\mathcal{B}}^{(i)}}(1)$. Then necessarily
$z_{i}z_{i+1}\cdots z_{i+n-1}=t_{1}^{(i)}t_{2}^{(i)}\cdots t_{n}^{(i)}$ and
$z_{i+n}\geq t_{n+1}^{(i)}+1$. In this case we can define
$\mathbf{x}\in\mathcal{S}$ and $j\in\mathbb{N}$ so that
$0^{pj}\mathbf{x}=0^{i-1}t_{1}^{(i)}t_{2}^{(i)}\cdots
t_{n}^{(i)}\left(t_{n+1}^{(i)}+1\right)0^{\omega}$.

It is obvious that the choice of $\mathbf{x}$ allows to find the string
$\mathbf{y}$ with non-negative digits such that
$(0^{jp}\mathbf{x})\oplus\mathbf{y}=\mathbf{z}$. Now replacing
$0^{jp}\mathbf{x}$ by $0^{jp}T(\mathbf{x})$ yields
$\bigl{(}0^{pj}T(\mathbf{x})\bigr{)}\oplus\mathbf{y}\succ_{lex}(0^{jp}\mathbf{x})\oplus\mathbf{y}=\mathbf{z}$.
∎

In rewriting the non-admissible string $\mathbf{z}\in\mathcal{F}$, we need to
ensure that after finitely many steps, the procedure yields the
lexicographically maximal string, i.e. the
$\boldsymbol{\mathcal{B}}$-expansion of $\operatorname{val}\mathbf{z}$.

######  Definition 18.

We say that the base $\boldsymbol{\mathcal{B}}$ has the weight property, if it
has the rewriting property, and, moreover, there exist positive integers
$w_{n}$, $n\geq 1$, such that the weight function
$g:\mathcal{F}\to\mathbb{N}$, $g(\mathbf{a})=\sum_{n\geq 1}w_{n}a_{n}$,
satisfies

  1. 1.

$g(0^{p}\mathbf{a})=g(\mathbf{a})$ for any $\mathbf{a}\in\mathcal{F}$;

  2. 2.

$g(\mathbf{a})\geq g(T(\mathbf{a}))$ for any $\mathbf{a}\in\mathcal{S}$.

The weight property is sufficient to guarantee the positive finiteness
property. We will use the obvious fact that the weight function satisfies
$g(\mathbf{x}\oplus\mathbf{y})=g(\mathbf{x})+g(\mathbf{y})$ for any two digit
strings $\mathbf{x},\mathbf{y}\in\mathcal{F}$.

######  Theorem 19.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be an
alternate base satisfying the weight property. Then $\boldsymbol{\mathcal{B}}$
has Property (PF).

######  Lemma 20.

Let $g:\mathcal{F}\mapsto\mathbb{N}$ be a weight function and let
$\bigl{(}\mathbf{a}^{(k)}\bigr{)}_{k\geq 1}$ be a sequence of infinite strings
satisfying for every $k\geq 1$

  * •

$\mathbf{a}^{(k)}\in\mathcal{F}$;

  * •

$\mathbf{a}^{(k)}\prec_{lex}\mathbf{a}^{(k+1)}$.

Then the integer sequence $\bigl{(}g(\mathbf{a}^{(k)})\bigr{)}_{k\geq 1}$ is
not bounded.

###### Proof.

Let $w_{1},w_{2},\ldots$ be positive integers and $g(\mathbf{a})=\sum_{n\geq
1}w_{n}a_{n}$ for every sequence $\mathbf{a}=(a_{n})_{n\geq 1}\in\mathcal{F}$.

We proceed by contradiction. Assume that there exists $H\in\mathbb{N}$ such
that $g\big{(}\mathbf{a}^{(k)}\big{)}\leq H$ for every $k\geq 1$. Since all
coefficients $w_{n}$ are positive integers, we derive

$a_{n}^{(k)}\in\\{0,1,\ldots,H\\}\qquad\text{and}\qquad\\#\operatorname{supp}\mathbf{a}^{(k)}\leq
H\qquad\text{ for all }n,k\geq 1.$ (19)

The set $\\{0,1,\ldots,H\\}^{\mathbb{N}}$ equipped with the product topology
is a compact space and thus the increasing sequence
$\big{(}\mathbf{a}^{(k)}\big{)}_{k\geq 1}$ has a limit, say
$\mathbf{b}=\lim_{k\to+\infty}\mathbf{a}^{(k)}$. Obviously,
$\mathbf{a}^{(k)}\preceq_{lex}\mathbf{b}$ for each $k\geq 1$.

Let us at first show that $\operatorname{supp}\mathbf{b}$ is infinite. Suppose
the contrary, i.e. that $\mathbf{b}=b_{1}b_{2}\cdots b_{N}0^{\omega}$ for some
$N\geq 1$. Since $\mathbf{b}=\lim_{k\to+\infty}\mathbf{a}^{(k)}$, there exists
$k_{0}\geq 1$ such that $\mathbf{a}^{(k)}$ has a prefix $b_{1}b_{2}\cdots
b_{N}$ for each $k>k_{0}$. The inequality
$\mathbf{a}^{(k)}\preceq_{lex}\mathbf{b}=b_{1}b_{2}\cdots b_{N}0^{\omega}$
implies $\mathbf{a}^{(k)}=\mathbf{b}$ for all $k>k_{0}$, and that is a
contradiction with the fact that $\big{(}\mathbf{a}^{(k)}\big{)}_{k\geq 1}$ is
strictly increasing.

Since $\operatorname{supp}\mathbf{b}$ is infinite, we can choose $M\geq 1$
such that the set

$S_{M}:=\\{n\geq 1:b_{n}\neq 0\text{\ and \ }n\leq
M\\}\subset\operatorname{supp}\mathbf{b}$

has cardinality $\\#S_{M}>H$. Since for all sufficiently large $k$ the string
$\mathbf{a}^{(k)}$ has prefix $b_{1}b_{2}\cdots b_{M}$, it has to be
$S_{M}\subset\operatorname{supp}\mathbf{a}^{(k)}$, and that is a contradiction
with (19). ∎

###### Proof of Theorem 19.

By Lemma 15, it is sufficient to show that if $\mathbf{z}\in\mathcal{F}$ and
$z=\operatorname{val}\mathbf{z}<1$, then the
$\boldsymbol{\mathcal{B}}$-expansion of $z=\operatorname{val}\mathbf{z}$
belongs to $\mathcal{F}$ as well. We distinguish two cases:

  * a)

Suppose that $\operatorname{val}\mathbf{z}<\frac{1}{\delta}$.

If $\mathbf{z}$ is admissible in base $\boldsymbol{\mathcal{B}}$, there is
nothing left to discuss. On the other hand, if $\mathbf{z}$ is not admissible
in base $\boldsymbol{\mathcal{B}}$, then Lemma 17 allows us to find a
lexicographically greater $\boldsymbol{\mathcal{B}}$-representation of $z$,
which also has a finite support. Let us show that after finitely many
applications of Lemma 17 we get the $\boldsymbol{\mathcal{B}}$-expansion of
$z$.

Let us proceed by contradiction. Assume that applying Lemma 17 repeatedly
yields an infinite sequence of strings
$\mathbf{z}=\mathbf{z}^{(0)},\mathbf{z}^{(1)},\mathbf{z}^{(2)},\ldots$ such
that for all $n\in\mathbb{N}$

$\mathbf{z}^{(n)}\ \text{ is a $\boldsymbol{\mathcal{B}}$-representation of
$z$},\quad\mathbf{z}^{(n)}\ \text{ is not admissible in
$\boldsymbol{\mathcal{B}}$,}\quad\text{and}\quad\mathbf{z}^{(n)}\prec_{lex}\mathbf{z}^{(n+1)}.$

Let $g$ be a weight function. To obtain $\mathbf{z}^{(n+1)}$ we use Lemma 17,
i.e., if $\mathbf{z}^{(n)}=\bigl{(}0^{pj}\mathbf{x}\bigr{)}\oplus\mathbf{y}$,
then $\mathbf{z}^{(n+1)}=\bigl{(}0^{pj}T(\mathbf{x})\bigr{)}\oplus\mathbf{y}$.

Properties of $g$ guarantee that $g\big{(}\mathbf{z}^{(n+1)}\big{)}\leq
g\big{(}\mathbf{z}^{(n)}\big{)}$ for all $n\in\mathbb{N}$. In particular, the
values ${g\big{(}\mathbf{z}^{(n)}\big{)}\in\mathbb{N}}$ are bounded by
$g\big{(}\mathbf{z}^{(0)}\big{)}=g(\mathbf{z})$, and that is a contradiction
with Lemma 20.

  * b)

Suppose that $\frac{1}{\delta}\leq\operatorname{val}\mathbf{z}<1$.

Obviously, the string $\mathbf{z}^{\prime}:=0^{p}\mathbf{z}$ belongs to
$\mathcal{F}$ and
$z^{\prime}=\operatorname{val}\mathbf{z}^{\prime}=\frac{1}{\delta}\,z<\frac{1}{\delta}$.
By a), the $\boldsymbol{\mathcal{B}}$-expansion of $z^{\prime}$ has finite
support, and must be of the form $0^{p}z_{p+1}z_{p+2}\cdots\in\mathcal{F}$.
Using (7), we have that
$d_{\boldsymbol{\mathcal{B}}}(z)=z_{p+1}z_{p+2}\cdots\in\mathcal{F}$.

∎

The above theorem gives us a sufficient condition for Property (PF). When
searching for bases with Property (F), by Proposition 12, it suffices to limit
our considerations to simple Parry alternate bases. The following proposition
shows that for such bases Properties (F) and (PF) are equivalent.

######  Proposition 21.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be a
simple Parry alternate base. Then $\boldsymbol{\mathcal{B}}$ has (F) if and
only if $\boldsymbol{\mathcal{B}}$ has (PF).

We first prove an auxiliary statement.

######  Lemma 22.

Let $\boldsymbol{\mathcal{B}}$ be a simple Parry alternate base. For all
$j,k\in\mathbb{N}$, $j\geq 1$, there exists a
$\boldsymbol{\mathcal{B}}$-representation of the number
$\operatorname{val}(0^{j-1}10^{\omega})$ of the form
$0^{j-1}u_{j}u_{j+1}u_{j+2}\cdots\in\mathcal{F}$ such that $u_{j+k}\geq 1$.

###### Proof.

We proceed by induction on $k$. For $k=0$, the statement is trivial. Let
$k\geq 1$. By induction hypothesis, there exists a string
$\mathbf{u}=0^{j-1}u_{j}u_{j+1}u_{j+2}\cdots\in\mathcal{F}$ such that
$\operatorname{val}(0^{j-1}10^{\omega})=\operatorname{val}\mathbf{u}$ and
${u_{j+k}\geq 1}$. From the definition of $\mathbf{t}^{(j+k+1)}$, we derive
that the string $0^{j+k-1}(-1)\mathbf{t}^{(j+k+1)}$ of integer digits has
value 0, i.e.
$\operatorname{val}\big{(}0^{j+k-1}(-1)\mathbf{t}^{(j+k+1)}\big{)}=0$. The
string $\mathbf{u}\oplus 0^{j+k-1}(-1)\mathbf{t}^{(j+k+1)}$ has all digits
non-negative and has finite support. Its digit at position $j+k+1$ is equal to
$u_{j+k+1}+t_{1}^{(j+k+1)}\geq 1$. ∎

###### Proof of Proposition 21.

Clearly, if $\boldsymbol{\mathcal{B}}$ satisfies (F), then it satisfies (PF).
For the opposite implication, assume that a simple Parry alternate base
$\boldsymbol{\mathcal{B}}$ has Property (PF). Therefore, according to Lemma
15, if a number $z\in[0,1)$ has a $\boldsymbol{\mathcal{B}}$-representation in
$\mathcal{F}$, then the $\boldsymbol{\mathcal{B}}$-expansion of $z$ belongs to
$\mathcal{F}$ as well.

In order to show Property (F), it is thus sufficient to verify that for any
$x,y\in\operatorname{Fin}(\boldsymbol{\mathcal{B}})\cap[0,1)$ such that
$z:=x-y\in(0,1)$ we can find a $\boldsymbol{\mathcal{B}}$-representation of
$z$ with finite support. Let $\mathbf{x},\mathbf{y}\in\mathcal{F}$ be the
$\boldsymbol{\mathcal{B}}$-expansions of numbers $x,y\in[0,1)$ such that
$x=\operatorname{val}\mathbf{x}>y=\operatorname{val}\mathbf{y}>0$.

We proceed by induction on the sum of digits in $\mathbf{y}$. If the sum is 0,
then $y=0$ and the statement is trivial. Suppose the sum of digits in
$\mathbf{y}$ is positive. Since
$\operatorname{val}\mathbf{x}>\operatorname{val}\mathbf{y}$, there exist
$j,k\in\mathbb{N}$, $j\geq 1$, such that
$\mathbf{x}=\mathbf{x}^{\prime}\oplus\mathbf{x}^{\prime\prime}$ and
$\mathbf{y}=\mathbf{y}^{\prime}\oplus\mathbf{y}^{\prime\prime}$, where
$\mathbf{x}^{\prime\prime}=0^{j-1}10^{\omega}$ and
$\mathbf{y}^{\prime\prime}=0^{j+k-1}10^{\omega}$. It follows directly from
Lemma 22 that
$\operatorname{val}{\mathbf{x}}^{\prime\prime}-\operatorname{val}{\mathbf{y}}^{\prime\prime}$
has a finite $\boldsymbol{\mathcal{B}}$-representation, say
$\mathbf{z}^{\prime\prime}\in\mathcal{F}$. Hence

$z=\operatorname{val}\mathbf{x}-\operatorname{val}\mathbf{y}=\operatorname{val}\mathbf{x}^{\prime}+\operatorname{val}\mathbf{z}^{\prime\prime}-\operatorname{val}\mathbf{y}^{\prime}.$

Property (PF) guarantees that
$\operatorname{val}\mathbf{x}^{\prime}+\operatorname{val}\mathbf{z}^{\prime\prime}$
has a finite $\boldsymbol{\mathcal{B}}$-expansion, say $\mathbf{x}_{new}$.
Thus

$z=\operatorname{val}\mathbf{x}-\operatorname{val}\mathbf{y}=\operatorname{val}\mathbf{x}_{new}-\operatorname{val}\mathbf{y}^{\prime}.$

The sum of digits in $\mathbf{y}^{\prime}$ is smaller by 1 than the sum of
digits in $\mathbf{y}$. Induction hypothesis implies that $z$ has a finite
$\boldsymbol{\mathcal{B}}$-representation, and, by Property (PF), also a
finite $\boldsymbol{\mathcal{B}}$-expansion. ∎

As a consequence of Proposition 21, we formulate a sufficient condition for
Property (F).

######  Theorem 23.

Let $\boldsymbol{\mathcal{B}}=({\beta_{1},\beta_{2},\dots,\beta_{p}})$ be a
simple Parry alternate base satisfying the weight property. Then
$\boldsymbol{\mathcal{B}}$ has Property (F).

##  6 A class of bases with finiteness property

In this section, we provide the proof of Theorem 7. In particular, we show
that an alternate base $\boldsymbol{\mathcal{B}}$ for which the sequences
$\mathbf{t}^{(\ell)}$ satisfy the set of inequalities (8), has Property (PF).
This will be shown by verifying that a base $\boldsymbol{\mathcal{B}}$
satisfying (8) has the weight property, and thus satisfies the assumptions of
Theorem 19. Firstly, we show the rewritings of the strings in the set
$\mathcal{S}$ required by Definition 16.

######  Proposition 24.

Let $\boldsymbol{\mathcal{B}}$ be an alternate base and
$d_{\boldsymbol{\mathcal{B}}^{(\ell)}}=\mathbf{t}^{(\ell)}$ satisfy (8). Then
${\boldsymbol{\mathcal{B}}}$ is a Parry alternate base and has the rewriting
property.

###### Proof.

Inequalities (8) imply that for every $\ell\in\mathbb{Z}_{p}$, the sequence
$\Big{(}t^{(\ell-n)}_{n+1}\Big{)}_{n\in\mathbb{N}}$ is non-increasing, and
thus eventually constant. Let us denote the constant $s_{\ell}$, i.e. we have
$t^{(\ell-n)}_{n+1}=s_{\ell}$ for all sufficiently large $n$. Therefore, we
can write for some $m\in\mathbb{N}$ that

$\mathbf{t}^{(\ell)}=t^{(\ell)}_{1}\cdots
t^{(\ell)}_{mp}(s_{\ell}s_{\ell+1}\cdots s_{\ell+p-1})^{\omega}\quad\text{for
every }\ell\in\mathbb{Z}_{p},$ (20)

where the index of $s$ is considered $\\!\\!\\!\mod p$. In other words,
inequalities (8) imply, that $\boldsymbol{\mathcal{B}}$ is a Parry alternate
base. Moreover, either $\boldsymbol{\mathcal{B}}$ is a simple Parry base (if
$s_{\ell}=0$ for every $\ell\in\mathbb{Z}_{p}$), or no $\mathbf{t}^{(\ell)}$
has finite support.

In order to show the rewriting property, we have to find, for each string
$\mathbf{a}\in\mathcal{S}$, a string $T(\mathbf{a})$ satisfying the
requirements (18), i.e.

$a)\ \ T(\mathbf{a})\in\mathcal{F},\qquad b)\ \
\operatorname{val}\mathbf{a}=\operatorname{val}T(\mathbf{a})\text{ \ \ \ and \
\ \ }c)\ \ \mathbf{a}\prec_{lex}T(\mathbf{a}).$

The proof will be divided according to the form of the string
$\mathbf{a}\in\mathcal{S}$. The strings are of two types:

Type 1: For $\ell,i\in\mathbb{Z}_{p}$ and $k\in\mathbb{N}$

$\displaystyle\mathbf{x}_{\ell,i,k}$
$\displaystyle=0^{p+\ell-1}t_{1}^{(\ell)}t_{2}^{(\ell)}\cdots
t_{pk+i-1}^{(\ell)}\left(t_{pk+i}^{(\ell)}+1\right)0^{\omega},$ $\displaystyle
T(\mathbf{x}_{\ell,i,k})$
$\displaystyle=0^{p+\ell-2}10^{pk+i}\left(t_{1}^{(\ell+i)}-t_{pk+i+1}^{(\ell)}\right)\left(t_{2}^{(\ell+i)}-t_{pk+i+2}^{(\ell)}\right)\left(t_{3}^{(\ell+i)}-t_{pk+i+3}^{(\ell)}\right)\cdots.$

Type 2: For $\ell\in\mathbb{Z}_{p}$

$\mathbf{x}_{\ell}=0^{p+\ell-1}\mathbf{t}^{(\ell)},\qquad
T(\mathbf{x}_{\ell})=0^{p+\ell-2}10^{\omega}.$

Realize that if the periodic part in (20) is non-vanishing $(s_{1}s_{2}\cdots
s_{p})^{\omega}\neq 0^{\omega}$, i.e. $\boldsymbol{\mathcal{B}}$ is not a
simple Parry base, then $\mathcal{S}$ contains only strings of Type 1.

Type 1: For $\mathbf{x}_{\ell,i,k}$, requirement c) is trivially satisfied. In
order to show a), realize that $t^{(\ell-n)}_{n+1}=s_{\ell}$ for
$\ell\in\mathbb{Z}_{p}$ and sufficiently large $n\in\mathbb{N}$, can be
equivalently rewritten as: $t^{(\ell)}_{j}=s_{\ell+j-1}$ for sufficiently
large $j\in\mathbb{N}$. This implies that for all sufficiently large
$j\in\mathbb{N}$

$t^{(\ell+i)}_{j}-t^{(\ell)}_{pk+i+j}=s_{\ell+i+j-1}-s_{\ell+pk+i+j-1}=0.$

Therefore $T(\mathbf{x}_{\ell,i,k})\in\mathcal{F}$, which proves a). In order
to show b) we need to verify

$w_{p(k+1)+i+\ell-1}+\sum_{n\geq\ell+p}w_{n}t^{(\ell)}_{n-\ell-p+1}-\Biggl{(}w_{p+\ell-1}+\sum_{n\geq
p(k+1)+\ell+i}w_{n}t^{(\ell+i)}_{n-p(k+1)-\ell-i+1}\Biggr{)}=0\,,$

where $w_{n}=\left(\prod_{j=1}^{n}{\beta_{j}}\right)^{-1}$ for all
$n\in\mathbb{N}$, $n\geq 1$. Equivalently,

$\sum_{n\geq 1}w_{n+p+\ell-1}\,t^{(\ell)}_{n}-w_{p+\ell-1}=\sum_{n\geq
1}w_{n+p(k+1)+\ell+i-1}\,t^{(\ell+i)}_{n}-w_{p(k+1)+i+\ell-1}\,.$ (21)

The definition of $\Bigl{(}t_{n}^{(m)}\Bigr{)}_{n\geq 1}$ implies
$1=\sum_{n\geq 1}w^{\prime}_{n}t^{(m)}_{n}$, where
$w^{\prime}_{n}=\Bigl{(}\prod_{j=m}^{n+m-1}{\beta_{j}}\Bigr{)}^{-1}\\!\\!=\frac{w_{n+m-1}}{w_{m-1}}$.
Hence

$w_{m-1}=\sum_{n\geq 1}w_{n+m-1}t^{(m)}_{n}.$ (22)

Using this fact with $m=p+\ell$ on the left-hand side of (21) yields 0.
Similarly, using (22) with $m=p(k+1)+\ell+i$ gives 0 on the right-hand side of
(21).

Type 2: If $\boldsymbol{\mathcal{B}}$ is a simple Parry base, we have to
verify requirements a), b), c) for strings of the form $\mathbf{x}_{\ell}$.
All three requirements are satisfied trivially. ∎

######  Remark 25.

From the proof of the above lemma it follows that finding rewriting rules for
all strings in $\mathcal{S}$ may be unnecessary; sometimes a smaller number of
rules is sufficient. In particular, if
$d_{\boldsymbol{\mathcal{B}}^{(\ell)}}(1)\in\mathcal{F}$, say
$d_{\boldsymbol{\mathcal{B}}^{(\ell)}}(1)=t^{(\ell)}_{1}t^{(\ell)}_{1}\cdots
t^{(\ell)}_{n}0^{\omega}$, then
$\mathbf{x}_{\ell,i,k}=\mathbf{x}_{\ell}\oplus\bigl{(}0^{p(k+1)+\ell+i-2}10^{\omega}\bigr{)}$
for $\ell,i,k$ such that $pk+i\geq n$. As a consequence, in such case we do
not need to use the rewriting rule for the string $\mathbf{x}_{\ell,i,k}$.

On the other hand, note that if
$d_{\boldsymbol{\mathcal{B}}^{(\ell)}}(1)\notin\mathcal{F}$, then
$\mathbf{x}_{\ell}\notin\mathcal{F}$.

######  Example 26.

Let us present rewriting rules for the base as in Example 8. Since
$d_{\boldsymbol{\mathcal{B}}}(1)=2010^{\omega}$ and
$d_{\boldsymbol{\mathcal{B}}^{(2)}}(1)=110^{\omega}$, the necessary rewriting
rules are the following.

$\mathbf{x}_{1,1,0}$ $\mathbf{x}_{1,2,0}$ $\mathbf{x}_{1}$
$\mathbf{x}_{2,1,0}$ $\mathbf{x}_{2}$ $0030^{\omega}$ $00210^{\omega}$
$002010^{\omega}$ $00020^{\omega}$ $000110^{\omega}$ $T(\mathbf{x}_{1,1,0})$
$T(\mathbf{x}_{1,2,0})$ $T(\mathbf{x}_{1})$ $T(\mathbf{x}_{2,1,0})$
$T(\mathbf{x}_{2})$ $01010^{\omega}$ $01001010^{\omega}$ $010^{\omega}$
$00101010^{\omega}$ $0010^{\omega}$

Let us also demonstrate algorithm for addition using rewriting rules. Consider
for example numbers $x,y\in[0,1)$ with
$d_{\boldsymbol{\mathcal{B}}}(x)=000020010^{\omega}$ and
$d_{\boldsymbol{\mathcal{B}}}(y)=000020^{\omega}$. Denote $z=x+y$. The digit-
wise sum of the two $\boldsymbol{\mathcal{B}}$-expansions gives a non-
admissible representation of $z$, namely
$d_{\boldsymbol{\mathcal{B}}}(x)\oplus
d_{\boldsymbol{\mathcal{B}}}(y)={000040010^{\omega}}$. Using rewriting rules
for strings in the set $\mathcal{S}$, as shown in the above table, we can
write

$\displaystyle\underbrace{000040010^{\omega}}_{non-admissible}=$
$\displaystyle\
(000010010^{\omega})\oplus(\underbrace{000030^{\omega}}_{00\mathbf{x}_{1,1,0}})$
$\displaystyle\hskip 85.35826pt\downarrow$ $\displaystyle\
(000010010^{\omega})\oplus(\underbrace{0001010^{\omega}}_{00T(\mathbf{x}_{1,1,0})})=\underbrace{000111010^{\omega}}_{non-
admissible}.$

In this way we have obtained another representation of the same number $z$.
Since the latter is still non-admissible, we continue by rewriting

$\displaystyle\underbrace{000111010^{\omega}}_{non-admissible}=$
$\displaystyle\
(000001010^{\omega})\oplus(\underbrace{000110^{\omega}}_{\mathbf{x}_{2}})$
$\displaystyle\hskip 89.626pt\downarrow$ $\displaystyle\
(000001010^{\omega})\oplus(\underbrace{0010^{\omega}}_{T(\mathbf{x}_{2})})=\underbrace{001001010^{\omega}}_{admissible}.$

The resulting digit string is admissible, and therefore it is the
$\boldsymbol{\mathcal{B}}$-expansion of $z$, i.e.
$d_{\boldsymbol{\mathcal{B}}}(z)=d_{\boldsymbol{\mathcal{B}}}(x+y)=001001010^{\omega}$.

In order to prove Theorem 7, it remains to show that (8) implies the existence
of a weight function for $\boldsymbol{\mathcal{B}}$. The weight function must
satisfy certain inequalities for every string in $\mathcal{S}$. In the
previous proof we have seen that (8) implies that $\boldsymbol{\mathcal{B}}$
is a Parry base. In case that $\boldsymbol{\mathcal{B}}$ is a simple Parry
base, the set $\mathcal{S}$ is richer than in the other situation, and it is
technically more complicated to find the weight function. We will perform the
construction only for a simple Parry alternate base.

For the $p$ sequences ${\textbf{t}}^{(\ell)}\in\mathcal{F}$ for
$\ell\in\mathbb{Z}_{p}$ satisfying (8), denote

$T^{(\ell)}_{j}=\sum_{\mathclap{\begin{subarray}{c}n\geq 1\\\
n=j\\!\\!\\!\\!\mod p\end{subarray}}}t_{n}^{(\ell)}\qquad\text{for all
}j\in\mathbb{Z}_{p}=\\{1,2,\dots,p\\}.$

Note that $T_{j}^{(\ell)}\in\mathbb{N}$ and, moreover, $T_{1}^{(\ell)}\geq
t_{1}^{(\ell)}\geq 1$. Inequalities (8) imply that for all
$\ell\in\mathbb{Z}_{p}$

$T^{(\ell)}_{1}\geq T^{(\ell-1)}_{2}\geq T^{(\ell-2)}_{3}\geq\cdots\geq
T^{(\ell-p+2)}_{p-1}\geq T^{(\ell-p+1)}_{p}.$ (23)

Now consider matrices $I,R,K\in\mathbb{N}^{p\times p}$, where $I$ is the
identity matrix, $R$ is the permutation matrix

$R=\begin{pmatrix}{\Theta}&I_{1}\\\
I_{p-1}&\Theta\end{pmatrix},\qquad\text{and}\qquad
K_{\ell,j}=T^{(\ell+1)}_{j-\ell}\,,\quad\text{for all \ \
}\ell,j\in\\{1,2,\ldots,p\\}.$

The entries of the permutation matrix $R$ can also by written as
$R_{\ell,j}=\delta_{\ell,j+1}$ where $\delta_{i,j}$ is the Kronecker symbol
whose indices are considered $\\!\\!\\!\mod p$, as well as both indices of
$T^{(\ell+1)}_{j-\ell}$ in the above relation.

######  Lemma 27.

There exist positive integers $u_{1},u_{2},\ldots,u_{p}$ such that for the
vector $\vec{u}=(u_{1},u_{2},\ldots,u_{p})^{T}$ it holds that

$(I-R)(K-I)\vec{u}=\vec{0}\quad\text{and}\quad K\vec{u}\geq\vec{u},$

where by the vector inequality we mean inequality in each component.

###### Proof.

First, we show that all non-diagonal elements of the matrix $M=(I-R)(K-I)$ are
non-negative. Indeed, if $\ell\neq j$, then, considering the indices
$\\!\\!\\!\mod p$, we can write

$M_{\ell,j}=K_{\ell,j}-K_{\ell-1,j}+\delta_{\ell,j+1}=\underbrace{T_{j-\ell}^{(\ell+1)}-T_{j-\ell+1}^{(\ell)}}_{\geq
0}+\underbrace{\delta_{\ell,j+1}}_{\geq 0}\geq 0,$

where we have used inequalities (23). Moreover,
$M_{\ell+1,\ell}\geq\delta_{\ell+1,\ell+1}=1$ for all
$\ell\in\\{1,2,\ldots,p\\}$. Therefore, there exists a positive integer $c$
such that the matrix $M+cI$ is non-negative and irreducible. The row vector
$\vec{1}=(1,1,\ldots,1)$ has property $\vec{1}(I-R)=\vec{0}$, and thus
$\vec{{1}}M=\vec{0}$. Therefore, $\vec{{1}}$ is a positive left eigenvector of
$M+cI$ corresponding to the eigenvalue $c$. According to Perron-Frobenius
theorem, $c\in\mathbb{N}$ is equal to the spectral radius of $M+cI$.
Therefore, there exists a positive column vector
$\vec{u}=(u_{1},u_{2},\ldots,u_{p})^{T}$ such that $(M+cI)\vec{u}=c\vec{u}$.
Since $M+cI$ has integer components, the eigenvector $\vec{u}$ corresponding
to $c$ can be chosen to have integer components. Moreover, $M\vec{u}=\vec{0}$.

To prove the second part of the lemma, i.e. $K\vec{u}\geq\vec{u}$, we denote
$\vec{\varepsilon}=(K-I)\vec{u}=(\varepsilon_{1},\dots,\varepsilon_{p})^{T}$.
Then

$\vec{0}=M\vec{u}=(I-R)(K-I)\vec{u}=(I-R)\vec{\varepsilon}=(\varepsilon_{1}-\varepsilon_{p},\varepsilon_{2}-\varepsilon_{1},\varepsilon_{3}-\varepsilon_{2},\ldots,\varepsilon_{p}-\varepsilon_{p-1})^{T}\,.$

Hence all the components of the vector $\vec{\varepsilon}=(K-I)\vec{u}$ have
the same value, or, equivalently,

$\vec{\varepsilon}=\kappa(1,1,\ldots,1)^{T}\quad\text{ for some
}\kappa\in\mathbb{R},$ (24)

and thus

$K\vec{u}=\vec{u}+\kappa(1,1,\ldots,1)^{T}.$ (25)

In order to prove the inequality $K\vec{u}\geq\vec{u}$ it suffices to show
that $\kappa\geq 0$. Recall that all elements of $K$ are non-negative and
$K_{\ell,\ell+1}=T_{1}^{(\ell+1)}\geq t_{1}^{(\ell+1)}\geq 1$ for all
$\ell\in\mathbb{Z}_{p}$. In other words, each column of $K$ contains at least
one element $\geq 1$. Consequently, $\vec{{1}}\leq\vec{{1}}\,K$, and
considering (25) we obtain

$\vec{{1}}\,\vec{u}\leq\vec{{1}}\,K\,\vec{u}=\vec{{1}}\,\bigl{(}\vec{u}+\kappa(1,1,\ldots,1)^{T}\bigr{)}=\vec{{1}}\,\vec{u}+p\kappa.$

Necessarily $\kappa\geq 0$. ∎

######  Proposition 28.

Let $\boldsymbol{\mathcal{B}}$ be an alternate base satisfying (8). Then
$\boldsymbol{\mathcal{B}}$ satisfies the weight property.

###### Proof.

We now find a weight function $g$ satisfying requirements of Definition 18. In
order to define coefficients $w_{1},w_{2},w_{3},\ldots$ for the weight
function $g$, we use positive integers $u_{1},u_{2},\ldots,u_{p}$ found in
Lemma 27 as follows. We set

$w_{n}:=u_{i}\quad\text{if}\quad i=n\\!\\!\mod p.$

Let us verify requirements of Definition 18. The fact that
$g(0^{p}\mathbf{x})=g(\mathbf{x})$ follows directly from the definition of
$(w_{n})_{n\geq 1}$ as a sequence with period $p$.

The second requirement, namely that $g(\mathbf{a})\geq g(T(\mathbf{a}))$ for
each string $\mathbf{a}\in\mathcal{S}$, will be proven separately for the two
types of strings as distinguished in the proof of Proposition 24. In
particular, we show

  1. a)

$g(\mathbf{x}_{\ell,i,k})=g\bigl{(}T(\mathbf{x}_{\ell,i,k}))\bigr{)}$ for all
$\ell,i\in\mathbb{Z}_{p}$ and $k\in\mathbb{N}$;

  2. b)

$g(\mathbf{x}_{\ell})\geq g\bigl{(}T(\mathbf{x}_{\ell})\bigr{)}$ for all
$\ell\in\mathbb{Z}_{p}$.

Let us now focus strings of Type 1. Equality a) is of the form

$\sum_{n\geq 1}w_{n+p+\ell-1}\,t^{(\ell)}_{n}-w_{p+\ell-1}=\sum_{n\geq
1}w_{n+p(k+1)+\ell+i-1}\,t^{(\ell+i)}_{n}-w_{p(k+1)+i+\ell-1}\,.$ (26)

Note that it is of the same form as (21), where now $w_{n}$ are coefficients
in the weight function. We will denote the $j$-th component of some vector
$\vec{z}$ as $(\vec{z})_{j}$. Periodicity of $(w_{n})_{n\geq 1}$ allows us to
rewrite certain summands in (26) as

$\sum_{n\geq
1}w_{n+p+\ell-1}\,t^{(\ell)}_{n}=\sum_{j\in\mathbb{Z}_{p}}w_{j+\ell-1}\sum_{\mathclap{\begin{subarray}{c}n\geq
1\\\ n=j\\!\\!\\!\\!\mod
p\end{subarray}}}t^{(\ell)}_{n}=\sum_{j\in\mathbb{Z}_{p}}w_{j+\ell-1}T^{(\ell)}_{j}=\sum_{j\in\mathbb{Z}_{p}}w_{j}T^{(\ell)}_{j-\ell+1}=\sum_{j\in\mathbb{Z}_{p}}u_{j}K_{\ell-1,j}=(K\vec{u})_{\ell-1},$

and similarly

$\sum_{n\geq 1}w_{n+p(k+1)+\ell+i-1}\,t^{(\ell+i)}_{n}=\sum_{n\geq
1}w_{n+\ell+i-1}\,t^{(\ell+i)}_{n}=\sum_{j\in\mathbb{Z}_{p}}w_{j}T^{(\ell+i)}_{j-\ell-i+1}=\sum_{j\in\mathbb{Z}_{p}}u_{j}K_{\ell+i-1,j}=(K\vec{u})_{\ell+i-1}\,.$

Equality (26) can be rewritten as

$\bigl{(}(K-I)\vec{u}\bigr{)}_{\ell-1}=\bigl{(}(K-I)\vec{u}\bigr{)}_{\ell+i-1}.$
(27)

This follows directly from the proof of Lemma 27, where we show that all
components of the vector $\vec{\varepsilon}=(K-I)\vec{u}$ coincide, cf. (24).
That concludes the proof of a).

For the strings of Type 2, we need to show b), i.e. that $\sum_{n\geq
1}w_{n+\ell-1}t^{(\ell)}_{n}\geq w_{\ell-1}$. As we have shown above, this is
equivalent to $(K\vec{u})_{\ell-1}\geq(\vec{u})_{\ell-1}$. This follows from
$K\vec{u}\geq\vec{u}$, as stated in Lemma 27. ∎

###### Proof of Theorem 7.

Proposition 28 combined with Theorem 19 yields the result. ∎

We have demonstrated the existence of a weight function only in the case that
$\boldsymbol{\mathcal{B}}$ is a simple Parry alternate base. In case that
$\boldsymbol{\mathcal{B}}$ is a Parry base which is not simple, thanks to
Remark 25, the proof is less technical. Let us illustrate the situation on the
following example.

######  Example 29.

Consider $\boldsymbol{\mathcal{B}}=(\beta_{1},\beta_{2})$ with
$\mathbf{t}^{(1)}=34(21)^{\omega}$ and $\mathbf{t}^{(2)}=52(12)^{\omega}$.
Obviously, $\mathbf{t}^{(1)}$ and $\mathbf{t}^{(2)}$ satisfy (8). The
rewriting rules are as follows.

$\ell=1,i,k$ | $1,1,0$ | $1,2,0$ | $1,1,j$ | $1,2,j$
---|---|---|---|---
$\mathbf{x}_{1,i,k}$ | $0040^{\omega}$ | $00350^{\omega}$ | $0034(21)^{j}30^{\omega}$ | $0034(21)^{j}220^{\omega}$
$T(\mathbf{x}_{1,i,k})$ | $01010^{\omega}$ | $0100130^{\omega}$ | $0100(00)^{j}040^{\omega}$ | $0100(00)^{j}00130^{\omega}$

$\ell=2,i,k$ | $2,1,0$ | $2,2,j$ | $2,1,j+1$
---|---|---|---
$\mathbf{x}_{2,i,k}$ | $00060^{\omega}$ | $0005(21)^{j}30^{\omega}$ | $0005(21)^{j}220^{\omega}$
$T(\mathbf{x}_{2,i,k})$ | $0010130^{\omega}$ | $0010(00)^{j}040^{\omega}$ | $0010(00)^{j}00130^{\omega}$

It can be easily seen that for the weight function satisfying
$g(T(\mathbf{x}))<g(\mathbf{x})$, one can take the plain sum-of-digit function
$g(\mathbf{a})=\sum_{n\geq 1}u_{n}a_{n}$ with $u_{n}=1$ for every $n\geq 1$.

## Acknowledgements

This work was supported by CZ.02.1.01/0.0/0.0/16_019/0000778 and
SGS20/183/OHK4/3T/14 of the Czech Technical University in Prague.

## References

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