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\vskip 1cm{\LARGE\bf On the Average Growth of Random\\
\vskip .1in
Fibonacci Sequences
}
\vskip 1cm
\large
Beno\^{\i}t Rittaud\\
Universit\'e Paris-13 \\
Institut Galil\'ee \\
Laboratoire Analyse, G\'eom\'etrie et Applications \\
99, avenue Jean-Baptiste Cl\'ement \\
93~430 Villetaneuse\\
France\\
\href{mailto:rittaud@math.univ-paris13.fr}{\tt rittaud@math.univ-paris13.fr}
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\vskip .2 in
\begin{abstract}
We prove that the average value of the $n$-th term of a sequence
defined by the recurrence relation $g_{n}=|g_{n-1}\pm g_{n-2}|$, where
the $\pm$ sign is randomly chosen, increases exponentially, with a
growth rate given by an explicit algebraic number of degree $3$. The
proof involves a binary tree such that the number of nodes in each row
is a Fibonacci number.
\end{abstract}
%\newtheorem{theorem}{Theorem}[section]
%\newtheorem{proposition}{Proposition}[section]
%\newtheorem{corollary}{Corollary}[section]
%\newtheorem{lemma}{Lemma}[section]
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\def\Z{\mathbb Z}
\def\N{\mathbb N}
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\newtheorem{Def}{Definition}
\newtheorem{Rem}{Remark}
\newtheorem{Prop}{Proposition}
\newtheorem{Cor}{Corollary}
\newtheorem{Notation}{Notation}
\newtheorem{Lemme}{Lemma}
\section{Introduction}
A {\em random Fibonacci sequence} is a sequence $(g_{n})_{n}$ in
which $g_{0}$ and $g_{1}$ are arbitrary nonnegative
real numbers and such that, for any
$n\geq 2$, one has $g_{n}=|g_{n-1}\pm g_{n-2}|$, where the $\pm$ sign
is randomly chosen for each $n$.
In 2000, Divakar Viswanath \cite{Viswanath} proved
that, in the set of random Fibonacci sequences
equipped with the natural probabilistic structure $(1/2,
1/2)^{\otimes\N}$, almost all random Fibonacci sequences are
exponentially growing, with a growth rate equal to $1.13198824\ldots$. Up to
now, no analytic expression for this value was known.
In 2005, Jeffrey McGowan and Eran Makover
\cite{McGowanMakover} used an elementary idea to evaluate the growth
rate of the average
value of the $n$-th term of a Fibonacci sequence, using the formalism of
trees. Thanks to Jensen's
inequality, this second growth rate is necessarily bigger than the
value $1.13198824\ldots$, which appears in Viswanath's study, since
this latter value corresponds to a growth rate in an almost-sure sense.
The first goal of the present article is to give an algebraic
expression for the growth rate of the average of the $n$-th term of a
random Fibonacci sequence. We use the formalism of trees, refining the
McGowan-Makover construction \cite{McGowanMakover}, by considering the full
binary tree of all possible random Fibonacci sequences starting from
two fixed initial values. Our main result is the following
\begin{theorem}\label{Main1} For any fixed $g_{0}$ and $g_{1}$ (not both equal
to zero), let us define $m_{n}$ as the mean value of the $n$-th term of a
random Fibonacci sequence starting from $g_{0}$ and $g_{1}$. Then, the
ratio $m_{n+1}/m_{n}$
tends to $\alpha-1\approx 1.20556943$ as $n$ goes to infinity,
where $\alpha\approx 2.20556943$ is the
only real number for which $\alpha^3=2\alpha^2+1$.
\end{theorem}
Let $a$ and $b$ be nonnegative real numbers.
By the {\em random Fibonacci tree of the pair $(a,b)$},
we mean the binary tree
denoted by ${\mathbf{T}}_{(a,b)}$ and
defined in the following way: $a$ is the root, $b$ its only child; if
$x$ is the parent of $y$, then $y$ has two children, which are $x+y$
and $|x-y|$. In other words, the possible walks in the tree
${\mathbf{T}}_{(a,b)}$
give the full list of random Fibonacci sequences $(g_{n})_{n}$ such
that $g_{0}=a$ and $g_{1}=b$. In this formalism, the sequence $(m_{n})_{n}$
can be characterized by the equality $m_{n}=S_{n}/2^n$, where $S_{n}$
is the sum of all values in the $n$-th row of the tree.
The study of the sequence $(S_{n})_{n}$ is made by considering another
binary tree, which we will call the {\em restricted random
Fibonacci tree}, denoted by
${\mathbf{R}}_{(a,b)}$, which is the subtree of ${\mathbf{T}}_{(a,b)}$
obtained by cutting
all redundant edges (a precise definition and the first few rows of
${\mathbf{R}}_{(1,1)}$ are given in subsection \ref{restricted}).
This subtree, which to our knowledge
has never been studied before, seems to have many
interesting properties -- in fact, it seems that it is even more
interesting than random Fibonacci trees.
The present paper is organized in the following way: in section 1
we introduce basic facts about trees and initiate the study of the
tree ${\mathbf{R}}={\mathbf{R}}_{(1,1)}$ in view of Theorem \ref{Main1}.
Section 2 is devoted to the proof of Theorem \ref{Main1}. In section
3, we investigate more
properties of the tree ${\mathbf{R}}$, which has many arithmetical
aspects that are of interest. In
this section, we
also focus our interest in some other trees derived from ${\mathbf{R}}$.
In section 4, we give some open questions and a heuristic formula which
gives a
link between trees ${\mathbf{T}}$ and ${\mathbf{R}}$. This latter
formula will be proved in another
article \cite{EBT}, where the question of the growth rate of almost
all random Fibonacci sequences is considered.
\section{Definitions and fundamental results about trees}
We start with a few simple relevant facts about trees that are of interest
for us.
\subsection{The tree ${\mathbf{T}}$}
It is easily shown that, for any positive numbers $a$, $b$ and $c$, the
trees ${\mathbf{T}}_{(ca,cb)}$ and ${\mathbf{T}}_{(a,b)}$ have the same
nodes up to the multiplicative constant $c$, so, when $a$ and $b$ are
integers (the only case of interest for us in this section), it is not
restrictive to assume that $a$ and $b$ are relatively prime.
Proposition \ref{Containment} will show that it is, in fact, enough to
focus our attention on ${\mathbf{T}}_{(1,1)}$, which will simply be
denoted by ${\mathbf{T}}$ in the following.
A pair $(a,b)$ of natural numbers is said to {\em appear in
${\mathbf{T}}$} (or, simply, {\em appears}) whenever there exists a node in
${\mathbf{T}}$
of value $a$ with a child of value $b$.
\begin{Lemme}\label{Uns} There exist two walks in ${\mathbf{T}}$, denoted
by $1^-$ and $1^+$, such that $1^-$ is exactly composed of all the
pairs
of the form $(1,2n+1)$ and $(2n,1)$ ($n$ an integer) and $1^+$ composed
of all the pairs of the form $(2n+1,1)$ and $(1,2n)$.
\end{Lemme}
\begin{proof} We start from $g_{0}=g_{1}=1$ and $g_{2}=2$. A
trivial calculation shows that, for all $k\geq 2$, defining $g_{k}$
by:
\[g_{k}=\left\{\begin{array}{ll}g_{k-1}+g_{k-2},&\mbox{if $k\equiv 1$ or
$2$ (mod $3$);}\\
g_{k-1}-g_{k-2},&\mbox{if $k\equiv 0$ (mod $3$);}
\end{array}\right.\]
\noindent gives the walk $1^-$. In the same way, if, for $k\geq 2$,
we define $g_{k}$ as:
\[g_{k}=\left\{\begin{array}{ll}g_{k-1}+g_{k-2},&\mbox{if $k\equiv 0$ or
$2$ (mod $3$);}\\
g_{k-1}-g_{k-2},&\mbox{if $k\equiv 1$ (mod $3$);}
\end{array}\right.\]
\noindent then we get $1^+$.
\end{proof}
\begin{Prop}\label{Containment} A pair of
positive integers $(a,b)$ appears in ${\mathbf{T}}$ if and
only if $a$ and $b$ are relatively prime.
In ${\mathbf{T}}$, the only appearing pair of integers
$(a,b)$ with $ab=0$ are $(0,1)$ and $(1,0)$.
\end{Prop}
\begin{proof} We start by proving the second part. If,
for example,
$a=0$ and $b\neq 0$ are such that $(a,b)$ appears, then, the parent
of $0$ is $b$, the parent of this parent is $b$
again, then either $0$ or $2b$, etc. In any case, we get multiples of
$b$ as successive ancestors; since the beginning of
${\mathbf{T}}$ is $1-1-0$, we must have $b=1$.
Let us prove now the first part. A pair $(a,b)$ appearing in ${\mathbf{T}}$
being given such that
$ab\neq 0$, let $d$ be the greatest common divisor of $a$ and $b$. Let $z$ be the
parent of $a$. Since we have $b=|z-a|$ or $b=z+a$, $d$ is also the
greatest
common divisor of $z$ and $a$. By induction, $d$ is the greatest
common divisor of $1$ (the root
of ${\mathbf{T}}$), and $1$ (the child of the root), so $d=1$.
Conversely, let $a\neq b$ be two relatively prime integers. We input
them to the Euclidean algorithm:
we write $r_{0}$ for $a$, $r_{1}$ for $b$ and, for any $i\geq 0$ such
that $r_{i+1}\neq 0$, we define
$r_{i+2}$ as the only integer in $[0, r_{i+1}[$ for which there exists
an integer $n_{i}$ such that $r_{i}=n_{i}r_{i+1}+r_{i+2}$. Let us
denote by $N$ the index such that $r_{N}=0$. Since $a$ and $b$ are
relatively prime, we have $r_{N-1}=1$ so, thanks to Lemma \ref{Uns}, the
pairs $(r_{N-1},r_{N-2})$ and $(r_{N-2},r_{N-1})$ both appear in
${\mathbf{T}}$.
Let assume now that the pairs $(r_{i+1},r_{i})$ and
$(r_{i},r_{i+1})$ both appear, for an $i\leq N-2$.
Starting from
$(r_{i+1},r_{i})$, we consider the walk obtained by two additions, one
subtraction, again two additions, one subtraction, etc. This gives the
sequence $r_{i+1}$, $r_{i}$, $r_{i}+r_{i+1}$, $2r_{i}+r_{i+1}$,
$r_{i}$, $3r_{i}+r_{i+1}$, etc. so the proof splits in two cases: if
$n_{i-1}$ is odd, then we obtain the pair
$(r_{i}, n_{i-1}r_{i}+r_{i-1})=(r_{i},r_{i-1})$. If $n_{i-1}$ is even,
then we obtain the pair $(r_{i}, (n_{i-1}-1)r_{i}+r_{i+1})$, that
is, $(r_{i}, r_{i-1}-r_{i})$. The next elements of the walk are, then,
$r_{i}+(r_{i-1}-r_{i})=r_{i-1}$, then
$|r_{i-1}-(r_{i-1}-r_{i})|=r_{i}$, so we get the pair
$(r_{i-1},r_{i})$.
Similar reasoning holds when we start from $(r_{i},r_{i+1})$. One
addition, one subtraction, then two additions, one subtraction, two
additions, one subtraction, etc. gives the sequence $r_{i}$,
$r_{i+1}$, $r_{i}+r_{i+1}$, $r_{i}$, $2r_{i}+r_{i+1}$,
$3r_{i}+r_{i+1}$, $r_{i}$, $4r_{i}+r_{i+1}$, etc., so if $n_{i-1}$ is
even we finally get the pair $(r_{i},r_{i-1})$. If $n_{i-1}$ is odd,
we get the pair $(r_{i},r_{i-1}-r_{i})$. An addition followed by a
subtraction then gives the pair $(r_{i-1},r_{i})$.
In conclusion, starting from $(r_{i+1},r_{i})$ and $(r_{i},r_{i+1})$
respectively, we obtain $(r_{i},r_{i-1})$ and $(r_{i-1},r_{i})$
if $n_{i-1}$ is odd, $(r_{i-1},r_{i})$ and $(r_{i},r_{i-1})$
if $n_{i-1}$ is even, and the proposition is thus proved by induction.
\end{proof}
Recall that $\lfloor x\rfloor$ denotes the integer part of $x$.
We can sum up the previous construction in the following way:
\begin{itemize}
\item if $n_{i-1}$ is odd, then
\begin{itemize}
\item starting from
$(r_{i},r_{i+1})$, we attain $(r_{i-1},r_{i})$ in $2+3\cdot
\lfloor n_{i-1}/2\rfloor$ steps;
\item starting from
$(r_{i+1},r_{i})$, we attain $(r_{i},r_{i-1})$ in $1+3\cdot
\lfloor n_{i-1}/2\rfloor$ steps;
\end{itemize}
\item if $n_{i-1}$ is even, then
\begin{itemize}
\item starting from
$(r_{i},r_{i+1})$, we attain $(r_{i},r_{i-1})$ in $3n_{i-1}/2$ steps;
\item starting from
$(r_{i+1},r_{i})$, we attain $(r_{i-1},r_{i})$ in $3n_{i-1}/2$ steps.
\end{itemize}
\end{itemize}
It is easily seen that if the pair $(a,b)$ appears in
${\mathbf{T}}_{(c,d)}$, then the full tree ${\mathbf{T}}_{(a,b)}$ appears in ${\mathbf{T}}_{(c,d)}$,
in an obvious sense. The following shows a ``self-containment'' aspect of
random Fibonacci trees.
\begin{Prop}\label{fractal} For any positive integers $a$ and $b$
relatively prime,
${\mathbf{T}}_{(a,b)}$ appears infinitely many times in ${\mathbf{T}}$.
\end{Prop}
\begin{proof}
Proposition \ref{Containment} shows that any pair of the form $(a,b)$
where $a$ and $b$ are relatively prime appears in ${\mathbf{T}}$. It is
then enough to show that ${\mathbf{T}}$ appears in
${\mathbf{T}}_{(a,b)}$. We consider a random Fibonacci sequence
$(g_{n})_{n}$ such that $g_{0}=g_{1}=1$, $g_{n-1}=a$ and $g_{n}=b$ for
an $n$. For $k>0$, we then define $g_{n+k}$ as
$|g_{n+k-1}-g_{n+k-2}|$.
It is easily seen that, for any integers $u$ and $v$, we
always have $\max(|v-u|,v)\leq \max(v,u)$, with
equality iff these maxima are both equal to $v$. The equality case
cannot, therefore, occur for
two successive pairs of $g_{n+k}$. As a consequence, we
get that there is a $k\leq 2n$ such that $g_{n+k}=g_{n+k+1}=1$, and
we are done.
\end{proof}
As a corollary, we get the following result:
\begin{Cor} Any pair appearing in ${\mathbf{T}}$ appears infinitely many
times in ${\mathbf{T}}$.
\end{Cor}
Therefore, if $a$ and $b$ are relatively prime,
the random Fibonacci tree generated by $a$ with a child $b$
is a subtree of ${\mathbf{T}}$. Recall
that, if this is not the case, and
$d>1$ is the greatest common divisor of $a$ and
$b$, then ${\mathbf{T}}_{(a,b)}$ is homothetic
to ${\mathbf{T}}_{(a/d,a/d)}$ which, by the
previous proposition, is a subtree of ${\mathbf{T}}$.
Here is the beginning of the random Fibonacci tree ${\mathbf{T}}$. For a node
$a$ with child $b$, the right child of $b$ is $b+a$ and the left child
is $|b-a|$.
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\end{graph}
\caption{
The random Fibonacci tree ${\mathbf{T}}={\mathbf{T}}_{(1,1)}$ }
\end{figure}
Let us mention, without further elaboration, that the sequence of
labels in the tree read in breadth-first order
($1$, $1$, $0$, $2$, $1$, $1$, $1$, $3$, $1$, $1$, $1$, $1$,
$1$, $3$, $1$, $5$\ldots), gives an example of a {\em 2-regular sequence} in
the
terminology given by Allouche and Shallit
\cite{AS1,AS2} (see also section \ref{restricted} of the
present article, after Figure 2).
\subsection{Shortest walks in ${\mathbf{T}}$}
By a shortest walk from the root of ${\mathbf{T}}$ to a pair $(a,b)$
appearing in ${\mathbf{T}}$, we mean a random Fibonacci sequence such that there
is an $n$ for which $g_{n}=a$, $g_{n+1}=b$ and $n$ is the smallest
integer for which such a random Fibonacci sequence exists.
\begin{Prop}\label{ShortestWay} The random Fibonacci
sequence constructed
in the proof of Proposition \ref{Containment} gives the only shortest
walk in ${\mathbf{T}}$ from the root to the pair $(a,b)$.
\end{Prop}
\begin{proof} To prove this proposition we need the following
\begin{Lemme}\label{StructAsc} Let the pair $(a,b)$ appear in
${\mathbf{T}}$. Let us consider the set of all pairs $(c,d)$ of
possible
ancestors of $(a+b,a)$ in ${\mathbf{T}}$ such that the
ascending walk from $(a,b)$ to $(c,d)$ does not
show the pair $(a,b)$. For all such
$(c,d)$, there exist four integers, $m$, $m'$, $n$ and $n'$,
such that $c=ma+nb$ and $d=m'a+n'd$,
$|mn'-m'n|= 1$ (in particular,
$m$ and $m'$ are relatively prime, and so are $n$ and $n'$) and such that
$m>m'\Longrightarrow n\geq n'$ and $n>n'\Longrightarrow m\geq m'$.
\end{Lemme}
In this lemma, the ``possible ancestors'' of $(a+b,a)$ are the
pairs of integers appearing in ${\mathbf{T}}$ and having
$(a+b,a)$
as successors. In other words, these are the elements of the set of
all ancestors of all pairs of nodes with value $a+b$ at
parent's position and $a$ for child's position (and, so, $b$
for grandchild).
\begin{proof} The property is routinely
verified for the first nodes of the ``ascending tree'' starting from
$(a+b,a)$. Let us start from the pair
$(ma+nb,m'a+n'b)$. The possible parents of
$ma+nb$ are $(m+m')a+(n+n')b$, for which the
required properties are trivially verified, and
$d:=|(m-m')a+(n-n')b|$. Let us consider this second case.
Since $m>m'$ implies $n\geq n'$ and $n>n'$ implies $m\geq m'$, we have $d=
|m-m'|a+|n-n'|b$. In any case, the property
$|mn'-m'n|= 1$ is verified for $(d,ma+nb)$.
Let us
show now that $|m-m'|>m$ implies $|n-n'|\geq n$ and that $|n-n'|>n$
implies $|m-m'|\geq m$. The symmetry of the problem allows us to prove
only one of these two implications: let us do for example the first one.
We thus assume that $|m-m'|>m$ and wish to prove that $|n-n'|\geq n$.
Since $m$ and $m'$ are integers, the inequality $|m-m'|>m$ implies
that $m'=2m+z$, where $z$ is a positive integer. We thus have $\pm 1=
mn'-m'n=mn'-(2m+z)n=m(n'-n)-(m+z)n$, so $m(n'-n)=\pm 1+(m+z)n$. First,
if $m\neq 0$, then we get $n'-n = n+\frac{\pm 1+zn}{m}$.
Since we can assume $n\neq 0$ (else the relation
$|n-n'|\geq n$ is trivial), we have $\pm 1+zn\geq 0$, so the
equality
$n'-n = n+\frac{\pm 1+zn}{m}$ implies $n'-n\geq n$. Second, if $m=0$,
then the equality $\pm 1=mn'-m'n$ implies $m'=n=1$. The only case for
which the relation $|n-n'|\geq n$ is false is, then, the case $n'=1$,
so we get $(d,ma+nb)=(a,b)$, which contradicts the hypothesis, so
Lemma \ref{StructAsc} is proved.
\end{proof}
Lemma \ref{StructAsc} has this important consequence:
\begin{Cor}\label{PropFond} {\bf (characterization of shortest walks)}
Let $(a,b)$ be a pair appearing in ${\mathbf{T}}$.
The shortest way in ${\mathbf{T}}$ from the root to the
pair $(a,b)$ is characterized by the following property:
for any pair $(c,d)$ appearing in this walk, the parent of
$c$ is $|c-d|$.
\end{Cor}
\begin{proof} Let $(c,d)$ be a pair appearing in ${\mathbf{T}}$ which belongs
to the shortest walk from the root to the pair $(a,b)$. If the parent
of $c$ is $c+d$ then, by the previous Lemma, either all the ancestors
of $c+d$ are of the form $mc+nd$ with $m$ and $n$ positive integers,
or the pair $(c,d)$ appears among the ancestors of $(c+d,d)$. The
first case is impossible since the walk could not thus start from the pair
$(1,1)$ which is at the beginning of the tree; the second case is in
contradiction with the assumption that the walk in consideration is the shortest
from $(1,1)$ to $(a,b)$. So, considering the pair $(c,d)$ appearing in
${\mathbf{T}}$, the parent of $c$ is necessarily $|c-d|$, and the corollary is
proved.
\end{proof}
To conclude the proof of Proposition \ref{ShortestWay}, it is then
enough to verify that the walk constructed in the
proof of Proposition \ref{Containment} verifies the previous characterization
property. This fact is routinely verified.
\end{proof}
The definition of the random Fibonacci tree implies that whenever we
see two nodes both equal to $1$, the first being the parent of the
second, the successive children which
appear next show the full tree. To avoid repetition in the tree, we will
now focus our attention to the subtree, denoted by ${\mathbf{R}}$, which
avoid redundances.
\subsection{The tree ${\mathbf{R}}$}\label{restricted}
We consider the tree ${\mathbf{R}}$ defined as the subtree of ${\mathbf{T}}$ made of all
shortest walks. In other words, we start from $1$, with only
child $1$. Then, the $(n-1)$ first rows being
constructed, the $n$-th one is made of the nodes $b$ such that,
denoting by $a$ their parent, the pair $(a,b)$ did not already appear
upper in the subtree (that is no row before the $n$-th one shows the
pair $(a,b)$). The tree ${\mathbf{R}}$ is the
{\em restricted subtree} of ${\mathbf{T}}$. We denote by $r(a,b)$ the value of
the row in which the edge containing $a$ as a parent and $b$ as a child
appears in the tree ${\mathbf{R}}$.
Lemma \ref{NonAmbigu} ensure that the definition of ${\mathbf{R}}$
is non-ambiguous, at a single exception which is treated
in Lemma \ref{Zero}.
\begin{Lemme}\label{NonAmbigu} Any edge $(a,b)$ appearing in ${\mathbf{T}}$
appears only
once in the row $r(a,b)$, apart from
the pair $(0,1)$ which appears twice in the second and third
rows.
\end{Lemme}
\begin{proof} The case of the pair $(0,1)$ is manually treated.
Apart from this case, we know from the characterization of shortest
walks
that the shortest walk in ${\mathbf{T}}$ from the root to
the pair $(a,b)$ is such that, for any pair
$(c,d)$ appearing in this walk, $c$'s parent is
$|d-c|$, so this walk is unique and Lemma \ref{NonAmbigu} is
proved.
\end{proof}
\begin{Lemme}\label{Zero} The value zero appears only once in
${\mathbf{R}}$ and has no grandchild.
\end{Lemme}
\begin{proof} By Proposition \ref{Containment}, the value $0$ appears
only as a child of $1$, so, by definition of ${\mathbf{R}}$, it cannot
appears twice. Its children in ${\mathbf{T}}$ are $1$ and $1$ (this is the
ambiguity case in the definition of ${\mathbf{R}}$), whose children in
${\mathbf{T}}$ are also $1$ and $1$ and have to be exclude from ${\mathbf{R}}$ since
the pair $(1,1)$ already appears in the root of ${\mathbf{R}}$.
\end{proof}
Lemma \ref{Zero} indicates that the node $0$ of ${\mathbf{R}}$ can be
removed. In the following, this node (and its children) are not
considered as elements of ${\mathbf{R}}$. We will call ${\tilde {\mathbf{R}}}$ the tree
obtained by adding to ${\mathbf{R}}$ the node $0$ corresponding to the left child
of the second $1$. The tree ${\tilde {\mathbf{R}}}$ will be useful to investigate
the positions of the $0$ nodes in ${\mathbf{T}}$ (see section \ref{Zeros}).
Here are the first few rows of ${\mathbf{R}}$. As in the case of ${\mathbf{T}}$, a left
child corresponds to a subtraction and a right child to an addition.
When a node has only one child, we use a vertical line; Lemma
\ref{GaucheGauche}
will show that such a vertical edge always corresponds to an
addition.
\begin{figure}[H]
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\edge{L7n10}{L8n15}
\edge{L7n10}{L8n16}
\edge{L7n11}{L8n17}
\edge{L7n11}{L8n18}
\edge{L7n12}{L8n19}
\edge{L7n13}{L8n20}
\edge{L7n13}{L8n21}
\edge{L7n14}{L8n22}
\edge{L7n14}{L8n23}
\edge{L7n15}{L8n24}
\edge{L7n16}{L8n25}
\edge{L7n16}{L8n26}
\edge{L7n17}{L8n27}
\edge{L7n18}{L8n28}
\edge{L7n18}{L8n29}
\edge{L7n19}{L8n30}
\edge{L7n19}{L8n31}
\edge{L7n20}{L8n32}
\edge{L7n21}{L8n33}
\edge{L7n21}{L8n34}
\edge{L8n1}{L9n1}
\edge{L8n2}{L9n2}
\edge{L8n2}{L9n3}
\edge{L8n3}{L9n4}
\edge{L8n3}{L9n5}
\edge{L8n4}{L9n6}
\edge{L8n5}{L9n7}
\edge{L8n5}{L9n8}
\edge{L8n6}{L9n9}
\edge{L8n6}{L9n10}
\edge{L8n7}{L9n11}
\edge{L8n8}{L9n12}
\edge{L8n8}{L9n13}
\edge{L8n9}{L9n14}
\edge{L8n10}{L9n15}
\edge{L8n10}{L9n16}
\edge{L8n11}{L9n17}
\edge{L8n11}{L9n18}
\edge{L8n12}{L9n19}
\edge{L8n13}{L9n20}
\edge{L8n13}{L9n21}
\edge{L8n14}{L9n22}
\edge{L8n14}{L9n23}
\edge{L8n15}{L9n24}
\edge{L8n16}{L9n25}
\edge{L8n16}{L9n26}
\edge{L8n17}{L9n27}
\edge{L8n18}{L9n28}
\edge{L8n18}{L9n29}
\edge{L8n19}{L9n30}
\edge{L8n19}{L9n31}
\edge{L8n20}{L9n32}
\edge{L8n21}{L9n33}
\edge{L8n21}{L9n34}
\edge{L8n22}{L9n35}
\edge{L8n23}{L9n36}
\edge{L8n23}{L9n37}
\edge{L8n24}{L9n38}
\edge{L8n24}{L9n39}
\edge{L8n25}{L9n40}
\edge{L8n26}{L9n41}
\edge{L8n26}{L9n42}
\edge{L8n27}{L9n43}
\edge{L8n27}{L9n44}
\edge{L8n28}{L9n45}
\edge{L8n29}{L9n46}
\edge{L8n29}{L9n47}
\edge{L8n30}{L9n48}
\edge{L8n31}{L9n49}
\edge{L8n31}{L9n50}
\edge{L8n32}{L9n51}
\edge{L8n32}{L9n52}
\edge{L8n33}{L9n53}
\edge{L8n34}{L9n54}
\edge{L8n34}{L9n55}
\end{graph}
\caption{The restricted tree ${\mathbf{R}}={\mathbf{R}}_{(1,1)}$ }
\end{figure}
Let us remark, again without further elaboration, that
the sequence of labels in the tree ${\mathbf{R}}$, read in
breadth-first order ($1$, $1$, $2$, $1$, $3$, $3$, $1$, $5$, $2$,
$4$, $4$, $2$, $8$\ldots) is a {\em $\beta$-regular
sequence}, as defined by Allouche, Scheicher and
Tichy \cite{AST},
where here $\beta$ is the numeration
system defined by the Fibonacci sequence.
\begin{Notation} Let $a$ be a node of ${\mathbf{R}}$. If $a$ has a left
(resp. right) child, it is denoted by $c^-(a)$ (resp. $c^+(a)$).
\end{Notation}
\begin{Lemme}\label{GaucheGauche} In ${\mathbf{R}}$, if $b$ is the left
child of $a$, then $b$ has no left child.
\end{Lemme}
\begin{proof} Let assume that we can find three nodes $x$,
$y$ and $z$ such that
$z=c^-(y)$ and $y=c^-(x)$. By considering
successive parents if necessary, we can suppose that
$x=c^+(w)$.
Let $v$ be the parent of $w$. Then, we have
$x=v+w$, so $y=|x-w|=v$ and
$z=|y-x|=|v-(v+w)|=w$, so
$(y,z)=(v,w)$, which contradicts the
non-redundance in the definition of ${\mathbf{R}}$.
\end{proof}
\begin{Lemme}\label{Ordre} In ${\mathbf{R}}$, if $a$ has $b$ as
a right (resp.
left) child, then $b$ is smaller (resp.
bigger) than $a$. The only case of equality corresponds to the
pair $(1,1)$ in the beginning of the tree.
\end{Lemme}
\begin{proof} Let us denote by $z$ the parent of $a$.
If $b=c^+(a)$, then it is obvious that
$b>a$ since, apart for nodes in the top of the tree, we
have $b=a+z$ and $z> 0$ by Lemma \ref{Zero}
(and the exclusion of the $0$-node we made after this lemma).
If $b=c^-(a)$, then (if $a$ is not the root of ${\mathbf{R}}$),
by Lemma \ref{GaucheGauche}
we have $a=c^+(z)$ so, by the beginning of the proof,
$a>z$, so $b=a-z3$, $G_{n}=2G_{n-1}+G_{n-3}$.
\end{Prop}
\begin{proof} We write $G^-_{n}$ (resp. $G^+_{n}$) for
$S(\rho^-_{n})$ (resp. $S(\rho^+_{n})$). Lemmas \ref{GaucheGauche},
\ref{Ordre} and \ref{DroiteDeux} imply the relation
$G^-_{n}=G^+_{n-1}-G_{n-2}$. We split the edges of $\rho_{n}^+$ in two
subsets: the one, say $\sigma^+_{n}$, is composed by
the edges which are children
of elements of $\rho^+_{n-1}$, the other, $\sigma^-_{n}$, is composed by
the edges which are children
of elements of $\rho^-_{n-1}$. Lemma \ref{Droit} implies that
$S(\sigma^+_n)=G^+_{n-1}+G_{n-2}$, Lemmas \ref{GaucheGauche} and
\ref{DroiteDeux} that $S(\sigma^-_{n})=G^-_{n-1}+G^+_{n-2}$. Thus, we
get
\begin{eqnarray*}
G_{n}=G^-_{n}+G^+_{n}&=&G^-_{n}+S(\sigma^-_{n})+S(\sigma^+_{n})\\
&=&(G^+_{n-1}-G_{n-2})+(G^-_{n-1}+G^+_{n-2})+(G^+_{n-1}+G_{n-2})\\
&=&2G^+_{n-1}+G^-_{n-1}+G^+_{n-2}\\
&=&2G_{n-1}+G^+_{n-2}-G^-_{n-1}\\
&=&2G_{n-1}+G_{n-3},
\end{eqnarray*}
the relation $G^+_{n-2}-G^-_{n-1}=G_{n-3}$ coming from Lemmas
\ref{GaucheGauche}, \ref{Ordre}, \ref{Droit} and \ref{DroiteDeux}.
\end{proof}
The same kind of study shows that, more generally,
if we write
$v_{n}=\alpha_{n}a+\beta_{n}b$ for
the sum of the $n$-th row of the general restricted
tree, then
we have
\[\alpha_{0}=1\quad
\alpha_{1}=0\quad\alpha_{2}=1\quad\alpha_{3}=2
\quad\alpha_{n}=2\alpha_{n-1}+\alpha_{n-3}\mbox
{ for $n\geq 4$ (so $\alpha_{n}=G_{n-1}$),}\]
\[\beta_{0}=0\quad\beta_{1}=1\quad\beta_{2}=1\quad\beta_{3}=2\quad
\beta_{n}=2\beta_{n-1}+\beta_{n-3}
\mbox{ for $n\geq 3$}.\]
\begin{Cor}\label{CroissRest} Let us denote by ${\tilde m}_{n}$
the average value of an element of the $n$-th row of ${\mathbf{R}}$. As $n$ goes
to infinity, the ratio
${\tilde m}_{n+1}/{\tilde m}_{n}$ tends to
$\alpha/\varphi\approx 1.363117$, where
$\alpha$ is the only real zero of the equation $x^3=2x^2+1$ and
$\varphi$ the golden ratio.
\end{Cor}
\begin{proof} Classical facts about linear recurring sequences
show that the ratio $G_{n+1}/G_{n}$ tends to $\alpha$ as $n$ tends to
infinity. Since Proposition \ref{Fib} shows that ${\tilde
m}_{n}=G_{n}/F_{n}$, we get ${\tilde m}_{n+1}/{\tilde
m}_{n}=(G_{n+1}/G_{n})\cdot(F_{n+1}/F_{n})$ and the result follows
from the fact that $F_{n+1}/F_{n}$ tends to $\varphi$ as $n$ goes to
infinity.
\end{proof}
\subsection{Projection of walks of ${\mathbf{T}}$ into walks of ${\mathbf{R}}$}\label{TR}
A (finite) walk in ${\mathbf{T}}$ can be represented as a finite sequence
$(w_{i})_{0\leq i\leq N}$ of $+$ and $-$ signs. To such a sequence
is associated the random Fibonacci sequence $(g_{n})_{n}$ defined by
$g_{0}=g_{1}=1$ and, for all $n\geq 2$, $g_{n}=g_{n-1}+g_{n-2}$ iff
$w_{n-2}=+$. A finite walk $(w_{i})_{0\leq i\leq N}$ being
given, we generically denote by
$(g_{i})_{0\leq i\leq N}$ the associated random Fibonacci sequence,
that is the values of the nodes successively
attained in ${\mathbf{T}}$ by this walk.
The following proposition gives a way to
transform a walk in ${\mathbf{T}}$ into a walk on ${\mathbf{R}}$.
\begin{Lemme}\label{Remontee} For $N\geq 2$,
let $(w_{n})_{0\leq n\leq N}$ be a finite
walk in ${\mathbf{T}}$ such that a minus sign is never followed by another minus
sign, apart for $w_{N-1}$ and $w_{N}$ (both minus signs). Denoting by
$(g_{i})_{0\leq i\leq N+2}$ the corresponding node sequence, we have
$(g_{N+2},g_{N+1}) = (g_{N-1},g_{N})$.
\end{Lemme}
\begin{proof} The assumption made on $(w_{i})_{0\leq i 3$, we have
$s(\rho_{n})=\rho_{n+1}+\rho_{n-2}$.
\end{Lemme}
Let us consider now the beginning of the trees ${\mathbf{T}}_{(1,\varphi)}$ and
${\mathbf{T}}_{(1,\varphi^{-1})}$ (where $\rho_{2}=\rho_{2}^{(\varphi^{-1},1)}$
for the left figure and $\rho_{2}=\rho_{2}^{(\varphi^{-2},1)}$ for the
right one, with the previous abuse of notation in this latter case).
\begin{figure}[H]
{\hskip 1.2in}
\begin{graph}(19,7)(5,3)
\newcommand{\node}[4]{%
\textnode #1(#2,#3){#4}[\graphlinecolour{1}]}
\node{Rac}{5}{8}{$1$}
\node{FilsRac}{5}{7}{$\varphi$}
\node{PetitFils1}{3}{6}{$\varphi^{-1}$}
\node{PetitFils2}{7}{6}{$\varphi^{2}$}
\node{L2n1}{2}{4}{$1$}
\node{L2n2}{4}{4}{$\varphi+\varphi^{-1}$}
\node{L2n3}{6}{4}{$1$}
\node{L2n4}{8}{4}{$\varphi^3$}
\edge{Rac}{FilsRac}
\edge{FilsRac}{PetitFils1}
\edge{FilsRac}{PetitFils2}
\edge{PetitFils1}{L2n1}[\graphlinedash{3 3}]
\edge{PetitFils1}{L2n2}
\edge{PetitFils2}{L2n3}
\edge{PetitFils2}{L2n4}
\edgetext{Rac}{FilsRac}{$\rho_{2}$}
\edgetext{PetitFils1}{L2n1}{$\varphi^{-1}\rho_{2}$}
%2
\node{2Rac}{15}{8}{$1$}
\node{2FilsRac}{15}{7}{$\varphi^{-1}$}
\node{2PetitFils1}{13}{6}{$\varphi^{-2}$}
\node{2PetitFils2}{17}{6}{$\varphi$}
\node{2L2n1}{12}{4}{$\varphi^{-3}$}
\node{2L2n2}{14}{4}{$1$}
\node{2L2n3}{16}{4}{$1$}
\node{2L2n4}{18}{4}{$\varphi+\varphi^{-1}$}
\edge{2Rac}{2FilsRac}
\edge{2FilsRac}{2PetitFils1}
\edge{2FilsRac}{2PetitFils2}
\edge{2PetitFils1}{2L2n1}[\graphlinedash{3 3}]
\edge{2PetitFils1}{2L2n2}
\edge{2PetitFils2}{2L2n3}
\edge{2PetitFils2}{2L2n4}
\edgetext{2Rac}{2FilsRac}{$\rho_{2}$}
\edgetext{2PetitFils1}{2L2n1}{$\varphi^{-2}\rho_{2}$}
\end{graph}
\caption{The trees ${\mathbf{T}}_{(1,\varphi)}$ and ${\mathbf{T}}_{(1,\varphi^{-1})}$.}
\end{figure}
We're interested in the sequence $s^n(\rho_{2}^{(|b-a|,a)})$ for
$(a,b)=(1,\varphi)$ and $(1,\varphi^{-1})$. Let us define $z$ as
$z:=|b-a|$: we thus have $z=\varphi^{-1}$ for $b=\varphi$ and
$z=\varphi^{-2}$ for $b=\varphi^{-1}$. Together with
the rules given in Lemma \ref{SuccR}, we have the following
rules for $b=\varphi$ or $\varphi^{-1}$:
\[s(\rho_{2}^{(z,1)})=\rho_{3}^{(z,1)},\]
\[s(\rho_{3}^{(z,1)})=\rho_{4}^{(z,1)}+
z\rho_{2}^{(z,1)}.\]
Let us define, for all integer $m\geq 2$
\[\nu_{m}:=\sum_{i=0}^{\lfloor m/2\rfloor -1}z^{i}\rho_{m-2i}.\]
\begin{Lemme}\label{SuccNu} For $(a,b):=(1,\varphi)$ or
$(1,\varphi^{-1})$, we have $s(\nu_{2})=\nu_{3}$,
$s(\nu_{3})=\nu_{4}$ and, for all $m\geq 4$:
\[s(\nu_{m})=\nu_{m+1}+\nu_{m-2}.\]
\end{Lemme}
\begin{proof} It is a simple application of Lemma \ref{SuccR} and
the previous rules for $s(\rho_{2})$ and $s(\rho_{3})$.
\end{proof}
We then get the desired link between the rows of ${\mathbf{R}}_{(|b-1|,1)}$ and
${\mathbf{T}}_{(1,b)}$ for $b=\varphi$ or $\varphi^{-1}$.
\begin{Prop}\label{LienRT} For all $n\geq 0$ and
$(a,b)=(1,\varphi)$ or $(1,\varphi^{-1})$ we have:
\[\tau_{n+2}=s^n(\rho_{2})=\sum_{m=0}^{\lfloor n/3\rfloor}\left({n\choose m}-
2{n \choose m-1}\right) \nu_{n+2-3m}.\]
\end{Prop}
Before proving it, let us give the following combinatorial interpretation
of this proposition: the
set $\nu_{n+2-3m}$ is the set of nodes attained by walks of length $n$
in the random Fibonacci tree such that the projection of such a walk in the
restricted tree is obtained by the cut of $m$ sequences $+--$. The
value ${n\choose m}-2{n \choose m-1}$ corresponds to the number of
such walks leading to
all nodes whose projection in
the restricted tree (see subsection \ref{TR}) are equal to a fixed
node.
\begin{proof} We write $c_{n,m}$ for
${n\choose m}-2{n \choose m-1}$: a classical relation in Pascal's
triangle shows that
$c_{n,m}+c_{n,m+1}=c_{n+1,m+1}$ for any $n$ and $m$.
The desired property is easily
verified for the first values of $n$.
Let assume for example that the property is true until $n=3k$ where $k$ is a
positive integer. Thus, by Lemma \ref{SuccNu}, and the induction
hypothesis, we get
\begin{eqnarray*}
\tau_{n+3}&=&s(s^{n}(\rho_{2}))\\
&=&s\left(\sum_{m=0}^{k}c_{n,m}
\nu_{n+2-3m}\right)\\
&=&s( c_{n,k} \nu_{2} )+ \sum_{m=0}^{k-1} c_{n,m}
\cdot(\nu_{n-3(m-1)}+\nu_{n-3m})\\
&=&c_{n,k}\nu_{3}+c_{n,0}\nu_{n+3}+c_{n,k-1}\nu_{n-3(k-1)}+
\sum_{m=0}^{k-2}(c_{n,m+1}+c_{n,m})\nu_{n-3m}\\
&=&(c_{n,k}+c_{n,k-1})\nu_{3}+\sum_{m=-1}^{k-2}(c_{n,m+1}+c_{n,m})\nu_{n-3m}\\
&=&c_{n+1,k}\nu_{3}+
\sum_{m=-1}^{k-2}c_{n+1,m+1}\nu_{n-3m}\\
&=&\sum_{m=0}^kc_{n+1,m}\nu_{(n+1)+2-3m}.
\end{eqnarray*}
The same calculation works when $n=3k+1$.
When $n=3k+2$, it also works, with the use of the
complementary fact, routinely proved, that for any integer $N$, the
equality $c_{3N,N}=c_{3N-1,N-1}$ holds (one has to apply then this
equality to $N=k+1$). (To be fully convinced, the reader is
strongly encouraged to
write the first $\tau_{n}$s in terms of $\rho_{i}$s with the help of
Lemma \ref{LienRT} and arrange these terms in a convenient way to
show the $\nu_{i}$s.)
\end{proof}
It is interesting to remark that, as a simple calculation shows,
the set $[1,\lfloor n/3\rfloor]$ in which $i$
varies in the formula given in Proposition \ref{LienRT} is exactly the
set of integers $i$ for which the quantity ${n\choose i}-
2{n \choose i-1}$ is positive. The following proof of
Theorem \ref{Main1} seems to us not as elegant as Proposition
\ref{LienRT}, and we should expect a less technical way to conclude -
a way that, at now, we failed to found.
\subsection{Step 3: an explicit expression of the growth rate of
$S(\tau_{n}^{(1,\varphi)})$ and $S(\tau_{n}^{(1,\varphi^{-1})})$}
We know by elementary theory of linear recurring sequences that any
sequence $(u_{n})_{n}$ such that $u_{n}=2u_{n-1}+u_{n-3}$ for all $n$
verifies that there exists $\mu\in\R$ and $\nu\in\C$ depending only on
the three first terms of $(u_{n})_{n}$ and such that, for any $n$:
\[u_{n}=\mu\alpha^n+\nu\beta^n+\overline{\nu}\overline{\beta}^n,\]
\noindent where $\alpha$ is the real zero of the equation
$x^3=2x^2+1$ and $\beta$ and $\overline{\beta}$ the complex ones. A
calculation shows that $\alpha\approx 2.2055694$ and that
$\beta\approx -0.1027847+0.6654569i$. Thus, the part
$\nu\beta^n+\overline{\nu}\overline{\beta}^n$ in the expression of
$u_{n}$ goes exponentially fast to zero. More precisely we have, up to
a multiplicative constant, the estimation $|\nu\beta^n+
\overline{\nu}\overline{\beta}^n|\leq |\beta|^n\approx 0.673348^n$.
The definition of $\nu_{n}$ given before Lemma \ref{SuccNu} leads,
then, to an expression of $S(\nu_{n})$ composed of
three parts, all of them being obtained by
replacing in the right side of the equality all $\rho_{k}$s by
$\alpha^k$, $\beta^k$ or $\overline{\beta}^k$ and introduce the
multiplicative constant $\mu$, $\nu$ or $\overline{\nu}$. Our first
goal is to prove that the parts with $\beta$ and $\overline{\beta}$
can be removed from the study. For any number $x$, we have, by an
elementary calculation:
\[\sigma_{m}(x):=\sum_{i=0}^{\lfloor m/2\rfloor -1}
z^{i}x^{m-2i}=x^m\cdot\frac{1-(z
x^{-2})^{\lfloor m/2\rfloor}}{1-z x^{-2}}.\]
This sequence is, roughly speaking, exponential with ratio $x$ when
$|zx^{-2}|<1$ and exponential
with ratio $z^{1/2}$ if $|zx^{-2}|>1$
(it is not exactly true in the latter case, where we should
take into account the presence of a complementary factor
$x^{m-2\lfloor m/2\rfloor}$; the
reader can ensure that this abuse is not important, since the cases
for which it occurs will leave to negligible sequences in what
follows).
Let us now define the following sequence:
\[\Sigma_{n}(x):=\sum_{m=0}^{\lfloor n/3\rfloor}c_{n,m}\sigma_{n+2-3m}(x).\]
We can then write, up to a multiplicative constant for each term in
the right side of the equality:
\[S(\tau_{n+2})=\Sigma_{n}(\alpha)+
\Sigma_{n}(\beta)+\Sigma_{n}(\overline{\beta}).\]
For any $x$ we have
\begin{eqnarray*}
\lefteqn{(1-zx^{-2})\Sigma_{n}(x)}\\
&=&\sum_{m=0}^{\lfloor n/3\rfloor}c_{n,m}\cdot x^{n+2-3m}\cdot
(1-(zx^{-2})^{\lfloor(n+2-3m)/2\rfloor})\\
&=& \left(\sum_{m=0}^{\lfloor n/3\rfloor}c_{n,m}x^{n+2-3m}\right)-
\left(\sum_{m=0}^{\lfloor
n/3\rfloor}c_{n,m}x^{n+2-3m-2\lfloor(n+2-3m)/2\rfloor}
z^{\lfloor(n+2-3m)/2\rfloor}\right).
\end{eqnarray*}
We denote by $Z_{n}(x)$ the first sum in the right side, and by
$Z'_{n}(x)$ the second one.
Since $|z|<1$, for any $x\geq 0$ we have
\[Z'_{n}(x)\leq x
\sum_{m=0}^{\lfloor n/3\rfloor} c_{n,m}
\leq x\sum_{m=0}^{\lfloor n/3\rfloor} {n\choose m}
\leq x\sum_{m=0}^{n} {n\choose m}
\leq x\cdot 2^n.\]
The same calculation shows that $|Z_{n}(\beta)|$ and
$|Z_{n}(\overline{\beta})|$ are both upper-bounded by a sequence of
the form $c\cdot 2^n$, where $c$ is a real number independant from $n$.
Writing $S(\tau_{n+2})$ as
$\Sigma_{n}(\alpha)+\Sigma_{n}(\beta)+\Sigma_{n}(\overline{\beta})$,
we get that $S(\tau_{n+2})=Z_{n}(\alpha)+c_{n}\cdot 2^n$, where
$c_{n}$ is uniformly bounded. Let us now study the sequence defined by
$Z_{n}(\alpha)$. We have
\begin{eqnarray*}
\frac{Z_{n}(\alpha)}{\alpha^{n+2}}
&=&\sum_{m=0}^{\lfloor n/3\rfloor}{n\choose
m}(\alpha^{-3})^m-2\alpha^{-3}\sum_{m=0}^{\lfloor n/3\rfloor} {n\choose
m-1}(\alpha^{-3})^{m-1}\\
&=&{n\choose \lfloor n/3\rfloor}(\alpha^{-3})^{\lfloor n/3\rfloor}
+(1-2\alpha^{-3})
\sum_{m=0}^{\lfloor n/3\rfloor-1}{n\choose m}(\alpha^{-3})^m.
\end{eqnarray*}
Up to a multiplicative constant,
the first term of the right side of the latter equality is, by
Stirling's formula, equivalent to
${\displaystyle\frac{1}{\sqrt{n}}\cdot\left(\frac{3}{2^{2/3}\alpha}
\right)^n}$, which tends to zero as $n$ goes to infinity. We can
therefore neglect this term, since the second one is lower-bounded by
a positive constant.
Now, then, we focus our attention to this latter term, denoted by
${\tilde Z}_{n}$. Our aim is to
prove that it is exponentially increasing, with a growth rate equal to
$1+\alpha^{-3}$. Let assume first that $\lfloor(n+1)/3\rfloor=\lfloor
n/3\rfloor =k+1$. Then, we
have
\begin{eqnarray*}
\frac{{\tilde Z}_{n+1}}{{\tilde Z}_{n}} &=&
\frac{\sum_{m=0}^{k}{n+1\choose m}(\alpha^{-3})^m}{\sum_{m=0}^{k}
{n\choose m}(\alpha^{-3})^m}\\
&=&\frac{\sum_{m=0}^{k}({n\choose m}+{n\choose m-1})
(\alpha^{-3})^m}{\sum_{m=0}^{k}
{n\choose m}(\alpha^{-3})^m}\\
&=&1+\frac{\sum_{m=0}^{k}{n\choose m-1}(\alpha^{-3})^m}{\sum_{m=0}^{k}
{n\choose m}(\alpha^{-3})^m}\\
&=&1+\alpha^{-3}\frac{\sum_{m=0}^{k}{n\choose m-1}
(\alpha^{-3})^{m-1}}{\sum_{m=0}^{k}
{n\choose m}(\alpha^{-3})^m}\\
&=&1+\alpha^{-3}\frac{\sum_{m=0}^{k-1}{n\choose m}
(\alpha^{-3})^{m}}{\sum_{m=0}^{k}
{n\choose m}(\alpha^{-3})^m}\\
&=&1+\alpha^{-3}\left(1-\frac{{n\choose k}(\alpha^{-3})^k}{\sum_{m=0}^{k}
{n\choose m}(\alpha^{-3})^m}\right).
\end{eqnarray*}
In the last fraction, the numerator tends to zero (same proof as
before, with Stirling's formula) and the denominator is lower-bounded
by $1$, so the fraction tends to zero as $n$ goes to infinity.
If $\lfloor (n+1)/3\rfloor =\lfloor n/3\rfloor +1=k+2$, then we
simply have to add to the numerator
of the first fraction the expression ${n+1\choose
k+1}(\alpha^{-3})^{k+1}$, which again tends to zero as $n$ goes to
infinity, and can be neglected for that reason.
Thus, we have proved that the sequence $(Z_{n})_{n}$ grows
exponentially to infinity with a growth rate equal to
$\alpha(1+\alpha^{-3})$; so does the sequence $(S(\tau_{n}))_{n}$, since
the term $c_{n}\cdot 2^n$ can be neglected.
Using the relation $\alpha^3=2\alpha^2+1$,
we get that $\alpha(1+\alpha^{-3})=2(\alpha-1)$. It remains then to
divide this value by $2$, which gives $\alpha-1$, to obtain
the growth rate of the average value of the $n$-th term of a
random Fibonacci sequence such that its first term is $1$ and
its second is $\varphi$ or $\varphi^{-1}$.
\subsection{Step 4: growth rate of $S(\tau_{n}^{(a,b)})$ for all pairs
$(a,b)$}
For any pair $(a,b)$ such that $a\geq 0$, $b\geq 0$ and $ab\neq 0$,
there exist two numbers $u$ and $v$ such that $uv\neq 0$ and such that
$(a,b)=u(1,\varphi)+v(1,\varphi^{-1})$. Thus, we have
\[S(\tau_{n}^{(a,b)})=uS(\tau_{n}^{(1,\varphi)})+
vS(\tau_{n}^{(1,\varphi^{-1})}).\]
Since the sequences $(S(\tau_{n}^{(1,\varphi)}))_{n}$ and
$(S(\tau_{n}^{(1,\varphi^{-1})}))_{n}$ both increase exponentially with
$2(\alpha-1)$ as growth rate, this is still the case for the sequence
$(S(\tau_{n}^{(a,b)}))_{n}$. Dividing by $2$ the value $2(\alpha-1)$ to
get the average value of the $n$-th term, we obtain the result stated
in Theorem \ref{Main1}.
\section{Interesting properties of ${\mathbf{R}}$ and related trees}
This section is devoted to some properties of ${\mathbf{R}}$ which seem of
interest to us. It concerns the arithmetical properties of the nodes
of the tree, some facts about continued fraction expansion, some
combinatorial properties, and also, for an important part, the link
between the tree and the sets $\SL(2,\N)$ and $\SL(2,\Z)$.
\subsection{Values in the $n$-th row of ${\mathbf{R}}$}
Lemma \ref{Uns} allows to know explicitly the numbers of $1$s in any
row. Our aim is here to extend this result to the other numbers.
Let us recall that the {\em Euler
function} $\phi$ maps each positive integer $n$ to the number,
denoted by $\phi(n)$,
of positive integers $k0$ be an integer. There
exists an integer $N(n)$ such that, for any $N\geq N(n)$, there are
exactly $2\phi(n)$ nodes equal to $n$ among the $N$-th, the $(N+1)$-th and the
$(N+2)$-th row of ${\mathbf{R}}$.
\end{Lemme}
\begin{Lemme}\label{Inclusion} For any positive integer $N$, the
$N$-th row of ${\mathbf{R}}$ is included in the $(N+3)$-th one, that is if the
value $v$ appears $a$ times in the $N$-th row, then it appears at
least $a$ times in the $(N+3)$-th one.
\end{Lemme}
Both of the two previous lemmas are immediate
consequences of the fact that each
walk in Lemma \ref{MarchesEntiers} shows exactly two nodes between each value
$n$.
\begin{Prop}\label{SerieGen} For all integers $k$ and $n$, let us
denote by $a_{k,n}$ the number of nodes of value $n$ in the $k$-th row
of ${\mathbf{R}}$. For all $k\geq 0$,
we define the formal series $S_{k}(X)$ as
$\sum_{n>0}a_{k,n}X^n$. The sequence $(S_{k}(X)+S_{k+1}(X)+S_{k+2}(X))_{k}$
is increasing and admits as a limit the series
\[S(X):=2\sum_{n>0}\phi(n)X^n.\]
\end{Prop}
\begin{proof} The fact that the formal series is increasing is a
consequence of Lemma \ref{Inclusion}. Lemma \ref{PositionsEntiers}
shows that, for any integer $n$, $a_{k,n}=2\phi(n)$ for all $k\geq
N(n)$, so the proposition is proved.
\end{proof}
Simple considerations of parity gives the following easy proposition,
which allows to be a little more precise in the description of the
rows of ${\mathbf{R}}$:
\begin{Prop}\label{PairImpair} The values of the nodes in the
$N$-th row of ${\mathbf{R}}$ are all even (resp. all odd) iff $N$ belongs (resp.
does not belong) to $2+3\N$.
\end{Prop}
\begin{proof} The property is true for the first rows of ${\mathbf{R}}$. If
the nodes of two consecutive rows are all odd, then, since the values
of the nodes of the next row are all of the form $|a\pm b|$
where $a$ and $b$ are odd, all of theses values are even. In
the same way, if all nodes of a row are even and all of the following
(or previous) are odd, then the next row is made of odd values only,
and Proposition \ref{PairImpair} is proved.
\end{proof}
\subsection{Positions of $0$-nodes in ${\mathbf{T}}$}\label{Zeros}
By $0$-nodes of ${\mathbf{T}}$ we mean the nodes with value $0$. The aim of this
section is to determine their position in ${\mathbf{T}}$ in term of finite walks
$(w_{i})_{0\leq i\leq N}$.
\begin{Def} A finite walk $(w_{i})_{0\leq i\leq N}$
such that the corresponding random Fibonacci sequence $(g_{i})_{0\leq
i\leq N+2}$
verifies that $g_{i}= 0$ only for $i=N+2$ will be said
a {\em single-$0$ walk}.
\end{Def}
By Proposition
\ref{PairImpair}, if $(w_{i})_{0\leq i\leq N}$ is a
single-$0$ walk, then $N$ is of the form $3n$, where $n$ is an
integer.
\begin{Lemme}\label{TroisMoins} Let $(w_{i})_{0\leq i\leq N}$ be a
single-$0$ walk such that $N>0$. We have $w_{N}=w_{N-1}=w_{N-2}=-$.
\end{Lemme}
\begin{proof} By Proposition \ref{Containment}, $g_{N}=0$
implies $g_{N-1}=1$, and so $g_{N-2}=1$. Since
$g_{N-3}\neq 0$ by hypothesis, we have $g_{N-3}=2$. Finally, we get
$w_{N}=w_{N-1}=w_{N-2}=-$.
\end{proof}
\begin{Prop}\label{Card} A finite walk $(w_{i})_{0\leq i\leq N}$
is a single-$0$ walk iff the inequality
\[\Card(\{0\leq j*0$, we have $w_{0}=+$.
By Lemma \ref{TroisMoins}, the three last terms of the walk are minus
signs, so there is at least one position in the walk
where we find the succession $+--$, which can be removed thanks to
the considerations made in section \ref{TR}. We then get a single-$0$
walk $({\tilde w}_{i})_{0\leq i\leq N-3}$ which, by induction
hypothesis, verifies the cardinality property stated
in the proposition. The re-introduction of the sequence $+--$ does not
modify the validity of this cardinality property, so the walk
$(w_{i})_{0\leq i\leq N}$ verifies this cardinality property.
Conversely, let assume that a walk $(w_{i})_{0\leq i\leq N}$ (with
$N=3n$) verifies the cardinality property. We thus have $w_{N}=-$, and
there must be exactly $n$ plus signs and $2n$ minus signs among the
$w_{i}$s for $i\in [0,3n-1]$. Consequently, the sequence $+--$ must
appear somewhere. Suppress it: the cardinality property remains true
for the new sequence, which is therefore a single-$0$ walk by the
induction hypothesis.
Re-introducing the $+--$ does not introduce one more $0$ in the random
Fibonacci sequence, so the full
sequence $(w_{i})_{0\leq i\leq N}$ is also a single-$0$ walk, and the
proposition is proved.
\end{proof}
\begin{Cor}\label{FormeZero} A single-$0$ walk $(w_{i})_{0\leq
i\leq 3n}$ contains $n$ plus signs and $2n+1$ minus signs.
\end{Cor}
The property mentioned in the previous proposition has for
consequence the following (see the sequence \seqnum{A001764} in
\cite{EIT} for details and references):
\begin{Prop}\label{NbMarchesZero} Let us denote by $\omega_{n}$ the
number of single-$0$ walks $(w_{i})_{0\leq
i\leq 3n}$. We have
\[\omega_{n}=\frac{1}{2n+1}{3n \choose n}.\]
\end{Prop}
We can deduce from what precedes the following description of walks
$(w_{i})_{0\leq i\leq N}$ such that $w_{N}=0$ (without necessarily
$w_{i}\neq 0$ for $im'$ then $n\geq n'$ and if $n>n'$ then $m\geq m'$. In
particular, if $m>m'$ then $m+n>m'+n'$ and $M$ has no left child (take
$a=b=1$ and use Lemma \ref{Ordre}). For the same reason, if $n>n'$
then $M$ has no left child. So if $M$ has a left child, then
$m\leq m'$ and $n\leq n'$ and, since we cannot have $m=m'$ and $n=n'$
(because of the relation $mn'-m'n=\pm 1$), Lemma \ref{Ordre} shows
that this is a sufficient condition. A
simple calculation shows that, in this case, the
left child of $M$ is given by $GM$, where $G=
\left(\begin{array}{cc} 0&1 \\ -1&
1\end{array}\right)$.
Any matrix $M$ of the $\SL(2,\N)$ tree can
therefore be written as a product of the form $G^{\gamma_{0}}D^{\delta_{0}}
GD^{\delta_{1}}\cdots GD^{\delta_{k-1}}GD^{\delta_{k}}\cdot
\left(\begin{array}{cc} 0&1 \\ 1&
1\end{array}\right)$, where $\gamma_{0}=0$ or $1$, $\delta_{i}>0$
for all $i$ between $0$ and $k-1$ and $\delta_{k}\geq 0$. The matrix
$\left(\begin{array}{cc} 0&1 \\ 1&
1\end{array}\right)$ corresponds to the child of the identity matrix
in the $\SL(2,\N)$ tree and is simply equal to $D$.
We remark that $D^T=D$ and that, as a simple computation shows,
$G^T=D^{-1}GD$, so we get
\begin{eqnarray*}
M^T&=&(G^{\gamma_{0}}D^{\delta_{0}}
GD^{\delta_{1}}\cdots GD^{\delta_{k-1}}GD^{\delta_{k}}\cdot
D)^T\\
&=&D\cdot D^{\delta_{k}}{}^tGD^{\delta_{k-1}}{}^tG\cdots
D^{\delta_{1}}{}^tGD^{\delta_{0}}{}^tG^{\gamma_{0}}\\
&=&D^{\delta_{k}+1}(D^{-1}GD)D^{\delta_{k-1}}(D^{-1}GD)\cdots
D^{\delta_{1}}(D^{-1}GD)D^{\delta_{0}}(D^{-1}GD)^{\gamma_{0}}\\
&=&D^{\delta_{k}}GD^{\delta_{k-1}}G\cdots
D^{\delta_{1}}GD^{1+\delta_{0}-\gamma_{0}}G^{\gamma_{0}}D^{\gamma_{0}}.
\end{eqnarray*}
It is then routinely verified that, in the case $\gamma_{0}=0$ as
well as in the case $\gamma_{0}=1$, this latter expression is equal to
$D^{\delta_{k}}GD^{\delta_{k-1}}G\cdots
D^{\delta_{1}}GD^{\delta_{0}}G^{\gamma_{0}}\cdot D$, and the proposition is
proved.
\end{proof}
The following tree is deduced from the previous one, each matrix
being replaced by the sign of its determinant.
\begin{figure}[H]
{\vskip .3in}{\hskip 0.9in}
\begin{graph}(19,10)(2.5,0)
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\end{graph}
\caption{The determinants tree}
\end{figure}
The structure of the plus and minus signs is given by the following
rules, routinely proved:
\begin{Prop}\label{ReglePlusOuMoins} Let $\varepsilon\in\{-,+\}$
be the label of
an edge. The right edge following it is labelled $-\varepsilon$ and
the left one (if it exists) by $\varepsilon$.
\end{Prop}
As a corollary, we get the number of $+$ and $-$ in each row of
the tree:
\begin{Cor}\label{NbPlusOuMoins} The number of plus signs at the
$n$-th row is $F_{n}/2$ if $n$ is of the form $3k$, $\lfloor
F_{n}/2\rfloor +1$ if
$n$ is of the form $3k+1$, and $\lfloor F_{n}/2\rfloor$ if $n$ is of the form
$3k+2$.
\end{Cor}
\begin{proof} Let us consider the $n$-th row of the tree; we
denote by $d_{n}^+$ (resp. $d_{n}^-$)
its number of right edges labelled $+$ (resp. $-$) and by $g_{n}^+$
(resp. $g_{n}^-$) the number of left edges labelled $+$ (resp. $-$).
Proposition \ref{Fib} easily gives the following relations for all
$n$:
\[d_{n}^++d_{n}^-=F_{n-1}\quad\mbox{and}\quad g_{n}^++g_{n}^-=F_{n-2}.\]
The structure of the tree, together with Proposition
\ref{ReglePlusOuMoins}, also give the following:
\[d_{n+1}^+=g_{n}^-+d_{n}^-\quad d_{n+1}^-=g_{n}^++d_{n}^+,\]
\[g_{n+1}^+=d_{n}^+\quad g_{n+1}^-=d_{n}^-.\]
We deduce from those relations that
$d_{n+1}^++g_{n+1}^+=F_{n-1}+g_{n}^-$ and that
$d_{n+1}^-+g_{n+1}^-=F_{n-1}+g_{n}^+$. If we denote by $e_{n}$ the
difference $g_{n}^+-g_{n}^-$, then we get
\begin{eqnarray*}
e_{n}=g_{n}^+-g_{n}^-&=&d_{n-1}^+-d_{n-1}^-\\
&=&(g_{n-2}^-+d_{n-2}^-)-(g_{n-2}^++d_{n-2}^+)\\
&=&-e_{n-2}-(d_{n-2}^+-d_{n-2}^-)\\
&=&-e_{n-2}-(g_{n-1}^+-g_{n-1}^-)\\
&=&-(e_{n-2}+e_{n-1}).
\end{eqnarray*}
Since an immediate computation shows that $e_{3}=-1$ and $e_{4}=0$, we
easily get that the sequence $(e_{n})_{n\geq 3}$ is simply the sequence
$-1$, $0$, $1$, $-1$, $0$, $1$, etc., and the conclusion follows.
\end{proof}
An immediate consequence of the rules given in Proposition
\ref{ReglePlusOuMoins} is that the infinite
words in $\{-,+\}$ given by walks in the determinant
tree are exactly those which
never show the sequence $- - -$ neither $+++$. In particular:
\begin{Prop}\label{BijMots} In $\{-,+\}^{\N}$ equipped with the
cylinder topology, let us denote by ${\cal D}$ the subset of words who
does not show anywhere the succession $- - -$ or $+++$, and by ${\cal
W}$
the subset of all words which never show the succession $- -$. To any
$w\in {\cal W}$ we associate its corresponding walk in ${\mathbf{R}}$, then make
the codage $d\in {\cal D}$ corresponding to the determinants of the
edges of ${\mathbf{R}}$. Thus, the defined function $\phi$\ :\
${\cal W}\longrightarrow {\cal D}$ is bijective. Moreover, the $n$
first letters of $\phi(w)$ depends only on the $n$ first letters of
$w$, and conversely the $n$ first letters of $d$ depends only on
those of $\phi^{-1}(d)$. In particular, $\phi$
is bicontinuous.
\end{Prop}
Finally, the next tree is obtained by replacing each matrix by its
trace.
\begin{figure}[H]
{\hskip 0.7in}
\begin{graph}(19,10)(2.5,1)
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\edge{L3n1}{L4n1}
\edge{L3n1}{L4n2}
\edge{L3n2}{L4n3}
\edge{L3n3}{L4n4}
\edge{L3n3}{L4n5}
\edgetext{L3n1}{L4n1}{3}
\edgetext{L3n1}{L4n2}{3}
\edgetext{L3n2}{L4n3}{3}
\edgetext{L3n3}{L4n4}{3}
\edgetext{L3n3}{L4n5}{7}
\edge{L4n1}{L5n1}
\edge{L4n2}{L5n2}
\edge{L4n2}{L5n3}
\edge{L4n3}{L5n4}
\edge{L4n3}{L5n5}
\edge{L4n4}{L5n6}
\edge{L4n5}{L5n7}
\edge{L4n5}{L5n8}
\edgetext{L4n1}{L5n1}{3}
\edgetext{L4n2}{L5n2}{3}
\edgetext{L4n2}{L5n3}{5}
\edgetext{L4n3}{L5n4}{3}
\edgetext{L4n3}{L5n5}{5}
\edgetext{L4n4}{L5n6}{5}
\edgetext{L4n5}{L5n7}{5}
\edgetext{L4n5}{L5n8}{11}
\edge{L5n1}{L6n1}
\edge{L5n1}{L6n2}
\edge{L5n2}{L6n3}
\edge{L5n3}{L6n4}
\edge{L5n3}{L6n5}
\edge{L5n4}{L6n6}
\edge{L5n5}{L6n7}
\edge{L5n5}{L6n8}
\edge{L5n6}{L6n9}
\edge{L5n6}{L6n10}
\edge{L5n7}{L6n11}
\edge{L5n8}{L6n12}
\edge{L5n8}{L6n13}
\edgetext{L5n1}{L6n1}{4}
\edgetext{L5n1}{L6n2}{6}
\edgetext{L5n2}{L6n3}{2}
\edgetext{L5n3}{L6n4}{6}
\edgetext{L5n3}{L6n5}{8}
\edgetext{L5n4}{L6n6}{6}
\edgetext{L5n5}{L6n7}{2}
\edgetext{L5n5}{L6n8}{8}
\edgetext{L5n6}{L6n9}{6}
\edgetext{L5n6}{L6n10}{8}
\edgetext{L5n7}{L6n11}{8}
\edgetext{L5n8}{L6n12}{8}
\edgetext{L5n8}{L6n13}{18}
\edge{L6n1}{L7n1}
\edge{L6n2}{L7n2}
\edge{L6n2}{L7n3}
\edge{L6n3}{L7n4}
\edge{L6n3}{L7n5}
\edge{L6n4}{L7n6}
\edge{L6n5}{L7n7}
\edge{L6n5}{L7n8}
\edge{L6n6}{L7n9}
\edge{L6n6}{L7n10}
\edge{L6n7}{L7n11}
\edge{L6n8}{L7n12}
\edge{L6n8}{L7n13}
\edge{L6n9}{L7n14}
\edge{L6n10}{L7n15}
\edge{L6n10}{L7n16}
\edge{L6n11}{L7n17}
\edge{L6n11}{L7n18}
\edge{L6n12}{L7n19}
\edge{L6n13}{L7n20}
\edge{L6n13}{L7n21}
\edgetext{L6n1}{L7n1}{\footnotesize 5}
\edgetext{L6n2}{L7n2}{\footnotesize 5}
\edgetext{L6n2}{L7n3}{\footnotesize 9}
\edgetext{L6n3}{L7n4}{\footnotesize 5}
\edgetext{L6n3}{L7n5}{\footnotesize 5}
\edgetext{L6n4}{L7n6}{\footnotesize 5}
\edgetext{L6n5}{L7n7}{\footnotesize 9}
\edgetext{L6n5}{L7n8}{\footnotesize 13}
\edgetext{L6n6}{L7n9}{\footnotesize 5}
\edgetext{L6n6}{L7n10}{\footnotesize 9}
\edgetext{L6n7}{L7n11}{\footnotesize 5}
\edgetext{L6n8}{L7n12}{\footnotesize 5}
\edgetext{L6n8}{L7n13}{\footnotesize 13}
\edgetext{L6n9}{L7n14}{\footnotesize 9}
\edgetext{L6n10}{L7n15}{\footnotesize 5}
\edgetext{L6n10}{L7n16}{\footnotesize 13}
\edgetext{L6n11}{L7n17}{\footnotesize 9}
\edgetext{L6n11}{L7n18}{\footnotesize 13}
\edgetext{L6n12}{L7n19}{\footnotesize 13}
\edgetext{L6n13}{L7n20}{\footnotesize 13}
\edgetext{L6n13}{L7n21}{\footnotesize 29}
\end{graph}
\caption{The traces tree}
\end{figure}
If we denote by $H_{n}$ the sum of the values of the $n$-th row of this
latter tree, we have the following result:
\begin{Prop}\label{SommeTraces} We have $H_{1}=2$, $H_{2}=1$,
$H_{3}=4$ and, for any $n\geq 3$, $H_{n}=2H_{n-1}+H_{n-3}-(-1)^n\cdot
2$.
\end{Prop}
\begin{proof} We first prove the following statement.
\begin{Prop}\label{SommesGD} With the same notation as in the proof
of Proposition \ref{SumEss}, we have
\[G_{3}^-=G_{4}^-=1\quad G_{n}^-=2G_{n-1}^-+G_{n-3}^-+(-1)^n\cdot
2\mbox{ (for $n\geq 5$)},\]
\[G_{3}^+=3\quad G_{4}^+=8\quad G_{n}^+=2G_{n-1}^++G_{n-3}^+-(-1)^n\cdot
2\mbox{ (for $n\geq 5$)}.\]
\end{Prop}
\begin{proof} The above relations are routinely verified for the
first values of $n$. Let assume that these relations are true for all
indices until
$n-1$. Recalling the relation
$G_{n}^-=G_{n-1}^+-G_{n-2}$ (see the proof of Proposition \ref{SumEss}) we
get
\begin{eqnarray*}
G_{n}^-&=&G_{n-1}^+ -G_{n-2}\\
&=&\bigl(2G_{n-2}^++G_{n-4}^+-(-1)^{n-1}\cdot 2\bigr)-\bigl(
G_{n-2}^++G_{n-2}^-\bigr)\\
&=&G_{n-2}^+-G_{n-2}^-+G_{n-4}^++(-1)^n\cdot 2.
\end{eqnarray*}
We thus have to verify the relation
$G_{n-2}^+-G_{n-2}^-+G_{n-4}^+=2G_{n-1}^-+G_{n-3}^-$. We write, using again
the relation $G_{n}^-=G_{n-1}^+-G_{n-2}$:
\begin{eqnarray*}
2G_{n-1}^-+G_{n-3}^-&=&2(G_{n-2}^+-G_{n-3})+(G_{n-4}^+-G_{n-5})\\
&=&(G_{n-2}^+-G_{n-2}^-+G_{n-4}^+)+(G_{n-2}^++G_{n-2}^--2G_{n-3}-G_{n-5})\\
&=&(G_{n-2}^+-G_{n-2}^-+G_{n-4}^+)+(G_{n-2}-G_{n-2}),
\end{eqnarray*}
\noindent so the desired relation for $G_{n}^-$ is verified.
We can then write
\begin{eqnarray*}
G_{n}^+&=& G_{n}-G_{n}^-\\
&=&G_{n}-\bigl(2G_{n-1}^-+G_{n-3}^-+(-1)^n\cdot 2)\\
&=&G_{n}-2G_{n-1}^--G_{n-3}^--(-1)^n\cdot 2\\
&=&G_{n}-2G_{n-1}+2G_{n-1}^+-G_{n-3}^- -(-1)^n\cdot 2\\
&=&G_{n-3}+2G_{n-1}^+-G_{n-3}^--(-1)^n\cdot 2\\
&=&2G_{n-1}^++G_{n-3}^+-(-1)^n\cdot 2,
\end{eqnarray*}
\noindent so Proposition \ref{SommesGD} is proved.
\end{proof}
We now come back to the proof of Proposition \ref{SommeTraces}. Recall
(see comments after Proposition \ref{SumEss}) that if we write
$v_{n}=\alpha_{n}a+\beta_{n}b$ for
the sum of the $n$-th row of the general restricted
tree, then
we have
\[\alpha_{0}=1\quad
\alpha_{1}=0\quad\alpha_{2}=1\quad\alpha_{3}=2
\quad\alpha_{n}=2\alpha_{n-1}+\alpha_{n-3}\mbox
{ for $n\geq 4$,}\]
\[\beta_{0}=0\quad\beta_{1}=1\quad\beta_{2}=1\quad\beta_{3}=2\quad
\beta_{n}=2\beta_{n-1}+\beta_{n-3}
\mbox{ for $n\geq 3$}.\]
It is clear that $\beta_{n}$ corresponds to the sum of
all fourth coefficients of the matrices of the $n$-th
row. The sum $\alpha_{n-1}$ is {\em not} exactly the sum of the first
coefficients of these matrices, since some $a$-coefficients in the
$(n-1)$-th row have to be counted twice. We can consider
$\alpha_{n-1}$ as the sum of the first coefficients of all matrices
which are right children. It remains, then, to evaluate the sum of
$a$-coefficients of elements of the $(n-1)$-th row which are left
children. For this, let us consider the particular case $a=1$ and $b=0$
in the general restricted tree. Let us denote by $\alpha_{n}^-$ the
sum of left children in the $n$-th row of this tree: we then have
$H_{n}=\alpha_{n-1}+\alpha_{n-1}^-+\beta_{n}$. We know by Proposition
\ref{SommesGD} and by the fact mentioned in the begining of the
present section that, for all $n\geq 3$, we have
$\alpha_{n-1}^{-}=2\alpha_{n-2}^-+\alpha_{n-4}^-+(-1)^{n-1}\cdot 2$. We thus
obtain the right recurrence relation for the sequence $(H_{n})_{n}$, and a
simple calculation gives the first values of this sequence.
\end{proof}
The following shows that the structure of the trace tree is more
difficult to apprehend than these of the other trees considered here.
\begin{Lemme}\label{Ilot} Let us denote by $t_{n}$ the label of the $n$-th edge
given by a walk $(w_{n})_{n}$ in the trace tree.
\begin{itemize}
\item If $w_{n}=w_{n+1}$
(thus a $+$ sign, by Lemma \ref{GaucheGauche}), then $t_{n+2}=t_{n+1}+t_{n}$.
\item If $w_{n}=w_{n+2}$ and $w_{n+1}=w_{n+3}$ (then $w_{n}=-w_{n+1}$,
again by Lemma \ref{GaucheGauche}), then $t_{n+4}=t_{n+2}+t_{n}$.
\end{itemize}
\end{Lemme}
\begin{proof} Immediate.
\end{proof}
\begin{Lemme}\label{TraceAie} The walks
\[+-\overbrace{+\cdots +}^{k}-\quad\mbox{and}\quad
-+\overbrace{+\cdots +}^{k}-\]
\noindent show the same sequence of labels in the trace tree, except
for the first and the last ones.
\end{Lemme}
\begin{proof} A simple calculation together with Lemma \ref{Ilot}
show that the walk
$+-\overbrace{+\cdots +}^{k}$ shows the successive traces $3$,
$2$, $3$, $5$,\ldots, $F_{k+3}$ and that the walk $-+\overbrace{+\cdots +}^{k}$
shows the traces $1$, $2$, $3$, $5$,\ldots, $F_{k}$, so the
sequence of traces are the same for the two walks, except for the
first one.
To prove that the last traces shown by the walks considered in the statement of
the Lemma are different, we need to be a little more precise. A simple
calculation shows that the last element of $\SL(2,\N)$ associated with
$+-\overbrace{+\cdots +}^{k}$ is $\left(\begin{array}{cc}
F_{k-1}&L_{k}\\ F_{k}&L_{k+1}\end{array}\right)$, where $(L_{n})_{n}$
is the {\em Lucas sequence}, defined as $L_{1}=1$, $L_{2}=3$ and
$L_{n}=L_{n-1}+L_{n-2}$ for $n\geq 2$. In the same way, the last element of
$\SL(2,\N)$ associated with the walk $-+\overbrace{+\cdots +}^{k}$ is
$\left(\begin{array}{cc}
F_{k+2}&F_{k}\\ F_{k+3}&F_{k+1}\end{array}\right)$ (an easy recurrence
shows that, for all $k$, $F_{k-1}+L_{k+1}=F_{k+2}+F_{k+1}$, which
confirms what precedes about equality of traces). Adding the minus
sign to each of the two walks leaves to the matrix
$\left(\begin{array}{cc}
F_{k}&L_{k+1}\\ F_{k-2}&L_{k-1}\end{array}\right)$ for the first walk
and $\left(\begin{array}{cc}
F_{k+3}&F_{k+1}\\ F_{k+1}&F_{k-1}\end{array}\right)$ for the second.
The traces are $F_{k}+L_{k-1}$ and $F_{k-1}+F_{k+3}$, which are
different numbers for any $k$.
\end{proof}
\begin{Cor}\label{TraceTypeInfini} The previous function
$(w_{n})_{n}\longrightarrow (t_{n})_{n}$ is not of finite type.
\end{Cor}
\subsection{Size of walks in ${\mathbf{R}}$}
\begin{Notation} For $a$ and $b$ relatively prime, we denote by
$\rho(a, b)$ the smallest
integer $n$ such that the $n$-th row of edges in ${\mathbf{T}}$ contains an edge
with $a$ as
parent and $b$ as child. As established in
Proposition \ref{ShortestWay} and Proposition \ref{Containment},
this shortest walk shows the pairs $(r_{i},r_{i+1})$ or
$(r_{i+1},r_{i})$ for all $i$ (same notation as in the proof of
Proposition \ref{Containment}). The relatively prime integers $a$
and $b$ being given, any pair of the form $(r_{i},r_{i+1})$
(resp. $(r_{i+1},r_{i})$)
is said in {\em ascending} (resp. {\em descending}) {\em order}. When
after $(r_{i},r_{i+1})$ comes $(r_{i},r_{i-1})$, or after $(r_{i+1},r_{i})$
comes $(r_{i-1},r_{i})$, we say that there is an {\em inversion}.
If $[n_{0},n_{1},\ldots, n_{N-1}]$ is the continued fraction expansion of
$a/b$, then let define $\gamma(a,b)$ as the total number
of even and positive
partial quotients $n_{i}$ with $ib$:
\[\omega^\pm(a,b)=\left\{\begin{array}{ll} \omega^-,&\mbox{if $\gamma$ is
odd;}\\
\omega^+,&\mbox{if $\gamma$ is even.}\end{array}\right.
\quad\mbox{and}\quad
\varepsilon=\left\{\begin{array}{ll} 0,&\mbox{if $n_{N-1}$ is
odd;}\\
2,&\mbox{if $n_{N-1}$ is even and $\gamma$ odd;}\\
1,&\mbox{if $n_{N-1}$ is even and $\gamma$ even.}
\end{array}\right.\]
\end{itemize}
\end{Prop}
\noindent Note that we do not know if there exists
a significantly more elegant way of writing the value of
$\rho(a,b)$.
\begin{proof} We use the construction made in the proof of Lemma
\ref{Uns} and Proposition \ref{Containment}.
By a simple application of Lemma \ref{Uns},
the number of steps to go from the edge $(1,1)$ to the edge
$(1,n_{N-1})$ is $3\cdot
\lfloor n_{N-1}/2\rfloor$ if $n_{N-1}$ is odd and $3\cdot \lfloor
n_{N-1}/2\rfloor -2$ if
$n_{N-1}$ is even. In the same way, it takes $3\cdot \lfloor
n_{N-1}/2\rfloor$
steps to go from $(1,1)$ to $(n_{N-1},1)$ if $n_{N-1}$ is odd and
$3\cdot \lfloor n_{N-1}/2\rfloor -1$ if $n_{N-1}$ is even.
From $(1,n_{N-1})$ (or $(n_{N-1},1)$) to $(a,b)$, the number of
inversions is equal to $\gamma(a,b)$. For clarity, let assume $a***b$ works in the same way.
\end{proof}
\begin{Cor}\label{DiffRho} For any $a$ and $b$ relatively prime, we have
\[|\rho(a,b)-\rho(b,a)|=
|\omega^+(a,b)-\omega^-(a,b)|\pm{\tilde \varepsilon}\]
\noindent where $|{\tilde\varepsilon}|\leq 1$.
\end{Cor}
\begin{proof} Immediate.
\end{proof}
\begin{Cor}\label{SumRho} With the previous notation,
we have the equality
\[\rho(a,b)+\rho(b,a)=3\left(\sum_{i**