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\def\ga{\gamma}
\def\Ga{\Gamma}
\def\de{\delta}
\def\De{\Delta}
\def\la{\lambda}
\def\te{\theta}
\def\vep{\varepsilon}
\def\vfi{\varphi}
\def\om{\omega}
\def\Om{\Omega}
\def\si{\sigma}

\def\ca{\mathcal{A}}
\def\cb{\mathcal{B}}
\def\cf{\mathcal{F}}
\def\cn{\mathcal{N}}
\def\cg{\mathcal{G}}
\def\cp{\mathcal{P}}
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\begin{document}

\phantom{.}
\vskip 1 true cm
\centerline{\bf EVEN PARTITION FUNCTIONS}
\bigskip
\centerline{\bf by}
\bigskip


\centerline {\bf F. Ben sa\"{\i}d\footnote{Research partially supported by
French-Tunisian exchange program, C.M.C.U. $n^o$ 99/F 1507 and by CNRS,
Institut Girard Desargues, UMR 5028.}}
\centerline {Facult\'e des Sciences de Monastir}
\centerline {Avenue de l'environnement}
\centerline {5000, Monastir, Tunisie}
\centerline {e-mail : {\tt Fethi.BenSaid@fsm.rnu.tn}}

\medskip
\centerline{and}
\medskip

\centerline{\bf J.-L. Nicolas\footnotemark[1]}
\centerline{Institut Girard Desargues, UMR 5028,}
\centerline{B\^at. Doyen Jean Braconnier,}
\centerline{Universit\'e Claude Bernard (Lyon 1),}
\centerline{21 Avenue Claude Bernard,}
\centerline{F-69622 Villeurbanne Cedex, France}
\centerline{e-mail: {\tt jlnicola@in2p3.fr}}

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\font\rms=cmr8 
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\font\bfs=cmbx8
\markright{\its S\accent19eminaire Lotharingien de
Combinatoire \bfs 46 \rms (2002), Article~B46i\hfill}
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\vskip 1 true cm

\ni
{\bf Abstract.} Let ${\cal A}$ be a set of  positive
integers. Let us denote by $p({\cal A}, n)$ the number of
partitions of $n$ with parts in ${\cal A}$.
While the study of the parity of the classical partition function
$p(\N,n)$ (where $\N$ is the set of positive integers) is a deep and
difficult problem, it is easy to construct a set ${\cal A}$ for which 
$p({\cal A},n)$ is even for $n$ large enough : as explained in a paper of
I.Z. Ruzsa, A. \sark and J.-L. Nicolas published in 1998 in the
{\it Journal of Number Theory}, if ${\cal B}$ is a subset of
$\{1,2,\ldots,N\}$, there is a unique set ${\cal A}={\cal A}_0({\cal
B},N)$ such that  
${\cal A}\cap\{1,2,\dots ,N\}={\cal B}$ and $p({\cal A}, n)$ is even for
$n>N$.
 
In this paper we recall some properties of the sets 
${\cal A}_0({\cal B},N)$, 
we describe the factorization into primes
of the elements of the set ${\cal A}_0(\{1,2,3\},3)$, and prove that
the number of elements of this set up to $x$ is asymptotically
equivalent to $\ds c\frac{x}{(\log x)^{3/4}}$, where $c=0.937\ldots$.

\medskip
\ni
{\bf R\'esum\'e.} Si ${\cal A}$ est un ensemble d'entiers positifs,
nous noterons $p({\cal A}, n)$ le nombre de partitions de $n$ dont les
parts sont dans ${\cal A}$. L'\'etude de la parit\'e de la fonction
usuelle de partition $p(\N,n)$ (o\`u $\N$ est l'ensemble des entiers
positifs) est un probl\`eme profond et difficile; mais il est facile
de construire un ensemble ${\cal A}$ tel que le nombre $p({\cal A},
n)$ soit pair pour tout $n$ assez grand: dans un article paru au
{\it Journal of Number Theory} en 1998, I.Z. Ruzsa, A. \sark et
J.-L. Nicolas montrent que si ${\cal B}$ est un sous-ensemble de 
$\{1,2,\ldots,N\}$, il existe un seul ensemble ${\cal A}={\cal A}_0({\cal
B},N)$ tel que ${\cal A}\cap\{1,2,\dots ,N\}={\cal B}$ et $p({\cal A}, n)$
est pair pour $n>N$.

Dans cet article, nous rappelons quelques propri\'et\'es des ensembles 
${\cal A}={\cal A}_0({\cal B},N)$, nous d\'ecrivons la
d\'ecomposition en facteurs premiers des \'el\'ements de ${\cal
A}_0(\{1,2,3\},3)$ et nous montrons que le nombre des \'el\'ements de
cet ensemble inf\'erieurs \`a $x$ est \'equivalent \`a $\ds
c\frac{x}{(\log x)^{3/4}}$, o\`u  $c=0.937\ldots$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ Introduction.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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%\markboth{\SMALL HELMUT PRODINGER}{\SMALL ON A FUNCTIONAL--DIFFERENCE
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$\nb_0$ and $\nb$ denote the set of the non negative integers, resp.
positive integers. ${\cal A}$ will denote a set of positive
integers, and its counting function will be denoted by $A(x)$:
$$\ds A(x)=|\{a:\ a\leq x\ ,\ a\in {\cal A}\}|.$$
%We shall also use the notation:
%\begin{equation}\label{Aodd}
%\ds A^{\textrm{odd}}(x)=|\{a:\ a\leq x\ ,a \textrm{ odd },\ a\in {\cal
%A}\}|. 
%\end{equation}
If ${\cal A}=\{a_1, a_2,\dots \}\subset \N$ (where $a_1<a_2<\cdots $), then
$p({\cal A}, n)$ denotes the number of partitions of $n$ with parts in
${\cal A}$, 
that is the number of solutions of the equation
\begin{equation}\label{part}
 a_1x_1+a_2x_2+\ldots=n
\end{equation}
in non negative integers $x_1,x_2,\dots $. As usual, we shall set
\begin{equation}\label{pA0}
p({\cal A}, 0)=1\quad \textrm{ and } \quad p({\cal A}, n)=0 \textrm{
for } n<0.
\end{equation}
We shall use the generating function: 
\begin{equation}\label{genF}
F(z)=F_\ca (z)=\sum_{n=0}^\infty p(\ca,n)z^n=\prod_{a\in
\ca}\frac{1}{1-z^a}\cdot 
\end{equation}

When $\ca=\N$ it seems highly probable that the number of integers $n\le x$
such that $p(\N,n)$ is even is close to $x/2$ as $x\to\infty$; but the
known results are rather poor (see
\cite{NRS}, \cite{Ono1}, \cite{Ono2}  and the references in them).
That is the reason for which, in \cite{NRS}, it was observed that
there exist sets $\ca$ such that $p(\ca,n)$ is even for $n$ large
enough. In this paper, we want to investigate the properties of such
sets.

For $i=0$ or $1$, if ${\cal A}\subset \N$ and there is a number $N$
such that 
\begin{equation}\label{AiN}
\ds p({\cal A}, n)\equiv i \pmod{2}\quad\hbox{for}\quad n\in
\nb~,~n>N.
\end{equation}
then ${\cal A}$ is said to possess property $\ds P_i(N)$.

If $i=0$ or $1$, ${\cal B}$ is a finite set of positive integers, and
$N\in \N$ satisfying
\begin{equation}\label{BN}
{\cal B}=\{b_1,\dots ,b_k\}\not =\emptyset,\quad 0<b_1<\dots <b_k,
\quad\quad N\geq b_k
\end{equation}  
then there is (cf. \cite{NRS}) a unique set ${\cal
A}\subset \nb$ such that
\begin{equation}\label{AiB}
\ds {\cal A}\cap\{1,2,\dots ,N\}={\cal B}
\end{equation} 
and possessing property $P_i(N)$;
we will denote it by ${\cal A}_i({\cal B}, N)$.

Let us recall the construction of ${\cal A}_i({\cal
B}, N)$ as described in \cite{NRS}, when, for instance, $i=0$. The set
${\cal A}={\cal A}_0({\cal B},N)$ will be defined by recursion.
We write $\ds {\cal A}_n={\cal A}\cap\{1,2,\dots ,n\}$ so that
$$\ds {\cal A}_N={\cal A}\cap\{1,2,\dots ,N\}={\cal B}.$$
Assume that $n\geq N+1$ and ${\cal A}_{n-1}$ has been defined so that $p({\cal A}, m)$
is even for $N+1\leq m\leq n-1$. Then set
$$\ds n\in {\cal A}\quad\hbox{if and only if}\quad p({\cal A}_{n-1},n)\quad\hbox{is odd}.$$
It follows from the construction that for $n\geq N+1$, we have
$$\ds\hbox{if}~n\in {\cal A}~~,~p({\cal A},n)=1+p({\cal A}_{n-1},n)$$
$$\ds\hbox{if}~n\notin {\cal A}~~,~p({\cal A},n)=\quad p({\cal A}_{n-1},n)$$
which shows that $p({\cal A},n)$ is even for $n\geq N+1$.
Note that in the same way, any finite set ${\cal B}=\{b_1,b_2,\dots ,b_k\}$ can
be extended to a set ${\cal A}$ so that ${\cal A}_{b_k}={\cal B}$ and
the parity of $p({\cal A},n)$ is given for $n\geq N+1$ (where $N$ is
any integer such that $N\geq b_k$).

It has been shown in \cite{BenAA} (cf. Proposition 4) that, except the
case $i=1$, $\cb=\{1\}$, the set $\ca_i(\cb,N)$ is always infinite.

If ${\cal A}\subset \nb$, let $\chi ({\cal A}, n)$ denote the
characteristic function of ${\cal A}$, i.e.,
\begin{equation}\label{chi}
\ds\chi ({\cal A}, n)=
\left\{
\begin{array}{lll}
1 &\hbox{if}& n\in {\cal A}\\
0 &\hbox{if}& n\notin {\cal A},
\end{array}
\right.
\end{equation}
and for $n\geq 1$,
\begin{equation}\label{sig}
\ds\sigma ({\cal A}, n)=\sum_{d\dv n}\chi({\cal A},d)d=\sum_{d\dv n,\,d\in
\, {\cal A}}d. 
\end{equation} 
It is relevant to consider $\ds\sigma ({\cal A}, n)$, since, as shown
in \cite{NRS}, taking the logarithmic derivative of $F(z)=F_\ca(z)$
defined by (\ref{genF}) yields
\begin{equation}\label{FF}
z\frac{F'(z)}{F(z)}=\sum_{n=1}^\infty \sigma ({\cal A}, n)z^n.
\end{equation}

It has been proved in \cite{BenAA} that for any positive integer
$k$ and any set $\ca=\ca_i(\cb,N)$, the sequence
\begin{equation}\label{conj}
\left(\sigma ({\cal A}, 2^kn) \bmod 2^{k+1}\right)_{n\ge 1} \textrm{
is periodic.}
\end{equation}
(We denote by $a\bmod b$ the remainder in the Euclidean division of $a$
by $b$.) Note that (\ref{conj}) was already  proved for $k=0$ in \cite{NRS},
and for $k=1$ in \cite{Ben4}. The value of the smallest period in
(\ref{conj}) is recalled in Theorem A below, proved in
\cite{BenAA}. Before stating Theorem A, we need to introduce two
definitions.

\medskip
{\bf Definition 1} {\it Let $\F_2$ be the field with two elements and 
$Q(z)\in \F_2[z]$ be a polynomial satisfying
$Q(0)\ne 0$. The order $\be$ of $Q$ is the least positive
integer such that $Q(z)$ divides $1+z^\be$ in $\F_2[z]$
(cf. \cite{LN}, chap. 3).    }

\medskip

From now on, we shall assume that $i=0$, for simplicity (the case
$i=1$ can be found in \cite{BenAA}).   
For ${\cal B}$ and $N$ as in (\ref{BN}) and $\ca=\ca_0(\cb,N)$,
let us define the polynomial $P$ (already
considered in \cite{NS} and \cite{BenAA}):
\begin{equation}\label{P0}
P(z)=\sum_{0\le n\le J} \vep_nz^n
\end{equation}
where $J$ is the largest
integer such that $p(\ca,J)$ is odd (such a $J$ does exist since, from
(\ref{pA0}), $p(\ca,0)=1$), and  $\vep_n$ is defined by 
\begin{equation}\label{ep0}
p(\ca,n)\equiv \vep_n \pmod{2},\quad \vep_n\in \{0,1\}.
\end{equation}


\medskip
{\bf Theorem A (cf. Theorem 2 of \cite{BenAA}).} {\it Let ${\cal B}$
and $N$ as in (\ref{BN}), 
$\ca=\ca_0(\cb,N)$, and  $P$ be the polynomial defined by (\ref{P0})
and (\ref{ep0}).
Let the factorization of $P$ into irreducible factors over $\F_2[z]$
be
\begin{equation}\label{irr}
P=Q_1^{\al_1}Q_2^{\al_2}\ldots Q_s^{\al_s}.
\end{equation}
We denote by $\be_i$ the order
of $Q_i(z)$ (cf. Definition 1), and for all $k\ge 0$, we set
\begin{equation}\label{Jk}
J_k=\{j; 1\le j\le s, \al_j\equiv 2^k \pmod{2^{k+1}} \},
\end{equation}
\begin{equation}\label{Ik}
I_k=J_0\cup J_1\cup\ldots\cup J_k=
\{j; 1\le j\le s, \al_j\not\equiv 0 \pmod{2^{k+1}} \}
\end{equation}
and
\begin{equation}\label{Tk}
q_k=\lcm_{\ j\in I_k} \;\be_j
\end{equation}
(with $q_k=1$ if $I_k=\emptyset$). Then, for all $k\ge 0$, $q_k$ is
odd and is the smallest period of (\ref{conj}) so that
\begin{equation}\label{congqk}
\sigma ({\cal A}, 2^k(n+q_k))\equiv\sigma ({\cal A}, 2^kn) \pmod{2^{k+1}}. 
\end{equation} 
Note that if $\,2^{k_0}$ is the highest power of $2$ dividing any 
exponent $\,\al_j$ in (\ref{irr}), for $k>k_0$, we have
$J_k=\emptyset$, $I_k=I_{k_0}$,
$$q_k=q\stackrel{def}{=} \text{ lcm } (\be_1,\be_2, \ldots,\be_s)$$
and $q$ is a common period for all the sequences
$\ds \left(\si(\ca,2^kn)\bmod 2^{k+1}\right)_{n\ge 1}$, $\;k\ge 0$.   }

\medskip
{\bf Remark 1.} {\it In \cite{Ben4}, one can find examples of ${\cal
B}$ and $N$ such that $q_0\ne q_1$.    } 

\medskip

Note that (\ref{conj}) was already  proved for $k=0$ in \cite{NRS},
and for $k=1$ in \cite{Ben4}. 

By M\"obius inversion formula, (\ref{sig}) gives
\begin{equation}\label{mob1}
\ds n\chi({\cal A},n)=\sum_{d\dv n}\mu(d)\sigma({\cal A}, n/d)
\end{equation}
where $\mu$ is the M\"obius function.
If $n$ is odd, by (\ref{conj}) with $k=0$, we know the value of $\sigma
({\cal A}, n) \bmod 2$, and this allows us  from
(\ref{mob1}) to determine $\chi({\cal A},n)$ for any set
$\ca=\ca_i(\cb,N)$. This
has been done in \cite{NS} for $\ca=\ca_0(\{1,2,3\},3)$ and in
\cite{Ndeb} for $\ca=\ca_0(\{1,2,3,4,5\},5)$. In \cite{Ben4}, the  
validity of (\ref{conj}) for $k=1$ has  been used to determine the
elements of $\ca=\ca_0(\{1,2,3\},3)$ which are congruent to $2$ modulo
$4$. 

Similarly, it is possible to deduce from (\ref{conj}) the value of $
\chi({\cal A},n)$ where $n$ is any positive integer.
For that, it is convenient for $m$ odd to introduce the sum
\begin{equation}\label{S}
S(m,k)=\chi({\cal A},m)+2\chi({\cal A},2m)+\ldots+2^k\chi({\cal
A},2^km).   
\end{equation}
If $n$ writes $n=2^km$ with $k\ge 0$ and $m$ odd, (\ref{sig}) implies 
\begin{equation}\label{sigS}
\sigma ({\cal A}, n)=\sigma ({\cal A}, 2^km)=\sum_{d\dv m} dS(d,k),
\end{equation}
which, by M\"obius inversion formula, gives
\begin{equation}\label{mob2}
mS(m,k)=\sum_{d\dv m} \mu(d) \sigma(\ca,n/d)=\sum_{d\dv \overline{m}}
\mu(d) \sigma(\ca,n/d), 
\end{equation}
where $\ds \;\overline{m}=\prod_{p\dv m}p\;$ denotes the radical of $m$.
In the above sums, $n/d$ is always a multiple of $2^k$, so that, from
(\ref{conj}) , the value of $\sigma(\ca,n/d)$ and thus the value of
$S(m,k)$ are known modulo $2^{k+1}$. Therefore, from (\ref{S}), we can
deduce the value of $\chi(\ca,2^im)$ for $i\le k$.

In Section 4, by the above described method, the multiplicative
structure of the elements of $\ca=\ca_0(\{1,2,3\},3)$ will be given
(Theorem 2), with the surprising property that, for this set $\ca$, the
$2$-adic expansion:
\begin{equation}\label{2adic}
1+2\chi({\cal A},2)+4\chi({\cal A},4)+\ldots+2^k\chi({\cal A},2^k)+\ldots
\end{equation}
is one of the $2$-adic roots of the equation $x^2-x+2=0$.
From Theorem 2, and from an extension of the so-called Selberg-Delange
formula given in \cite{Ben}, we shall give in Section 5 (Theorem 3) an
asymptotic estimation for $A(x)$:
\begin{equation}\label{Ax}
A(x)\sim c\frac{x}{(\log x)^{3/4}},\quad x\to\infty
\end{equation}
where $c$ is a constant the value of which is approximately $0.937$.

The work done in Sections 3 and 4 for the set
$\ca=\ca_0(\{1,2,3\},3)$ could  be done, in principle, for any set
$\ca=\ca_i(\cb,N)$. But, for technical reasons, the calculation can be
difficult. We hope to return to this subject in an other article.

We are pleased to thank M. Del\'eglise for computing the values of
$A(x)$ (cf. Section 5), D. Barsky, K. Belabas and A. S\'ark\"ozy for 
several remarks. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ The Graeffe transformation.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let us consider the ring of formal power series $\C[[z]]$. For an
element 
$$f(z)=a_0+a_1z+a_2z^2+\ldots+a_nz^n+\ldots$$
of this ring, the product
$$f(z)f(-z)=b_0+b_1z^2+b_2z^4+\ldots+b_nz^{2n}+\ldots$$
is an even power series with
\begin{equation}\label{Grco}
b_0=a_0^2,\; b_1=2a_0a_2-a_1^2,\ldots,\; 
b_n=2\left(\sum_{i=0}^{n-1}(-1)^ia_ia_{2n-i} \right)+(-1)^na_n^2.
\end{equation}
We shall call $g=\cg(f)$ the series
\begin{equation}\label{G}
g(z)=\cg(f)(z)=b_0+b_1z+b_2z^2+\ldots+b_nz^{n}+\ldots.
\end{equation}
Note that we have
\begin{equation}\label{gG}
g(z^2)=\cg(f)(z^2)=f(z)f(-z).
\end{equation}

{\bf Example.} If $q$ is an odd integer and $f(z)=1-z^q$, we have
$f(z)f(-z)=(1-z^q)(1+z^q)=1-z^{2q}$, and
\begin{equation}\label{exam}
\cg(f)=f.
\end{equation}
If $f$ is  a polynomial of degree $n$ which does not vanish in $0$, and
if $\widetilde f(z)=z^nf(1/z)$ is the reciprocal polynomial of $f$,
then we have 
\begin{equation}\label{reci}
\cg(\widetilde f)=(-1)^n\widetilde{\cg(f)}.
\end{equation}
It is obvious that, for any two series $f$ and $g$, the formulas
\begin{equation}\label{prod}
\cg(fg)=\cg(f)\cg(g)
\end{equation}
and, if $g(0)=1$,
\begin{equation}\label{quo}
\cg(f/g)=\cg(f)/\cg(g)
\end{equation}
hold.
We shall often use the following notation for the iterates of $f$ by
the transformation $\cg$:
\begin{equation}\label{iter}
f_0=f,\;\; f_1=\cg(f),\;\; f_2=\cg(f_1),\;\;\ldots,\;\;
f_k=\cg(f_{k-1})=\cg^{(k)}(f),\ldots
\end{equation}

\medskip
{\bf Proposition 1.} {\it Let $f$ be a polynomial of degree $n$ whose roots
are $z_1$, $z_2,\ldots,z_n$ and leading coefficient is $a_n$. Then the
polynomial $g=\cg(f)$, where $\cg$ is defined by (\ref{G}), has a
leading coefficient equal to $(-1)^na_n^2$ and its roots are  $z_1^2$,
$z_2^2,\ldots,z_n^2$.   } 

\medskip
{\bf Proof.} From the relations
$$f(z)=a_n(z-z_1)(z-z_2)\ldots(z-z_n)$$
and
$$f(-z)=a_n(-z-z_1)(-z-z_2)\ldots(-z-z_n)$$
it follows that
$$f(z)f(-z)=(-1)^na_n^2(z^2-z_1^2)(z^2-z_2^2)\ldots(z^2-z_n^2)$$
and therefore, from (\ref{gG})
\begin{equation}\label{Groot}
g(z)=\cg(f)(z)=(-1)^na_n^2(z-z_1^2)(z-z_2^2)\ldots(z-z_n^2),
\end{equation}
which completes the proof of Proposition 1. $\Box$

In numerical analysis (cf. \cite{Gra}, \cite{Bare} or \cite{Ral}), 
the Graeffe method is used to 
compute an approximate value of the roots of a polynomial
equation $f(x)=0$. The first step of the method is to calculate 
$f_k$ defined by (\ref{iter}) for $k$ large enough. From Proposition
1, the roots of $f_k$ are $z_1^{2^k},\ldots,z_n^{2^k}$, and, if we
assume that $|z_1|>|z_2|>\ldots>|z_n|$, the sum of the roots of $f_k$
is close to $z_1^{2^k}$ which yields an approximate value for $|z_1|$. 
This old method is being revisited in the frame of computer algebra 
(cf. \cite{Gou}).

\medskip
{\bf Proposition 2.} {\it Let $f(z)\in \C[[z]]$, $f(0)\ne 0$,and
\begin{equation}\label{prop21}
z\frac{f'(z)}{f(z)}=\sum_{n=1}^\infty a_nz^n.
\end{equation}
Then, for $k\ge 1$, we have
\begin{equation}\label{fkz}
\sum_{n=1}^\infty a_{2^kn}z^n=z\frac{f_k'(z)}{f_k(z)}=\frac{z}{f_k(z)}\frac{d}{dz}f_k(z),
\end{equation}
where $f_k=\cg^{(k)}(f)$ is defined by (\ref{G}) and (\ref{iter}).   }

\medskip
{\bf Remark 2.} {\it Here and in the sequence, $f_k'$ will denote the
derivative of $f_k$ (and not the $k$-iterate of $f'$).  }

\medskip

{\bf Proof.} We shall prove Proposition 2 by induction on $k$. For
$k=1$ and $z=y^2$, we have from (\ref{prop21}) and (\ref{gG})
\begin{eqnarray}\label{prop2}
\sum_{n=1}^\infty a_{2n}z^{n}&=&
\sum_{n=1}^\infty a_{2n}y^{2n}=\frac 12\sum_{n=1}^\infty
(a_{n}y^n+a_n(-y)^n)\nonumber\\
&=&\frac 12 \left( y\frac{f'(y)}{f(y)}-y\frac{f'(-y)}{f(-y)}\right)\
=\frac y2\frac{f'(y)f(-y)-f(y)f'(-y)}{f(y)f(-y)}\notag\\
&=&\frac{y}{2f_1(y^2)} \frac{d}{dy}f_1(y^2)= z\frac{f_1'(z)}{f_1(z)}.
\end{eqnarray}
Further, the induction on $k$ is easy, by substituting 
$a_{2^kn}$ to $a_{2n}$ and  $f_{k-1}$ to $f$ in (\ref{prop2}). $\Box$

\bigskip
{\bf Definition 2.} {\it We shall say that two power series $f,g$
with integral coefficients are congruent modulo $M$ (where $M$ is any
positive integer) if their
coefficients of same degree are congruent modulo $M$. In other words, if
$$f(z)=a_0+a_1z+a_2z^2+\ldots+a_nz^n+\ldots\quad \in \Z[[z]]$$
and
$$g(z)=b_0+b_1z+b_2z^2+\ldots+b_nz^{n}+\ldots\quad \in \Z[[z]]$$ 
then,
\begin{equation}\label{cong}
f\equiv g\pmod{M} \quad\Longleftrightarrow\quad \forall n\ge 0, \quad
a_n\equiv b_n\pmod{M}.
\end{equation}   }
Congruences of formal power series may be added or multiplied. If
\begin{equation}\label{fg}
f\equiv g\pmod{M}
\end{equation}
and 
$$u\equiv v \pmod{M},\qquad u\in\Z[[z]],\;\; v\in\Z[[z]]$$
then
\begin{equation}\label{mufo}
f+u\equiv g+v \pmod{M} \quad \textrm{ and } \quad fu\equiv gv \pmod{M}.
\end{equation} 
You may derivate (\ref{fg}) and get
\begin{equation}\label{cond}
f'\equiv g'\pmod{M}.
\end{equation}
Moreover, if $f(0)=g(0)=1$, $1/f$ and $1/g$ have integer coefficients
and we have, if (\ref{fg}) holds
\begin{equation}\label{coni}
\frac 1f \equiv \frac 1g \pmod{M}.
\end{equation}
It is also easy to see that, for $f\in\Z[[z]]$ and $\cg$ defined by
(\ref{G}) we have
\begin{equation}\label{Gcon}
\cg(f)\equiv f \pmod{2}.
\end{equation}

\medskip
{\bf Proposition 3.} {\it Let $f$ and $g$ be two formal power series with
integral coefficients  such that $f\equiv g\pmod{2}$. Then, for $k\ge
0$, we have
\begin{equation}\label{con2k}
f_k\equiv g_k \pmod{2^{k+1}},
\end{equation}
where $f_k=\cg^{(k)}(f)$ and $g_k=\cg^{(k)}(g)$ are defined by
(\ref{G}) and (\ref{iter}).   } 

\medskip
{\bf Proof.} Let us start by proving that if $u,v\in\Z[[z]]$ satisfy
\begin{equation}\label{prop31}
u\equiv v \pmod{2M}
\end{equation}
where $M$ is any positive integer, then
$u_1=\cg(u)$ and $v_1=\cg(v)$ satisfy
\begin{equation}\label{prop3}
u_1\equiv v_1\pmod{4M}.
\end{equation} 
It follows from (\ref{prop31}) that there exists $w\in\Z[[z]]$ such that
$$u(z)=v(z)+2Mw(z).$$
Further, from (\ref{gG})
\begin{eqnarray*}
u_1(z^2)&=&u(z)u(-z)=(v(z)+2Mw(z))(v(-z)+2Mw(-z))\\
&=&v_1(z^2)+2M[v(z)w(-z)+w(z)v(-z)]+4M^2w_1(z^2),
\end{eqnarray*}
where $w_1=\cg(w)$. But the above bracket is obviously congruent to
$0$ modulo $2$ so that
$$u_1(z^2)\equiv v_1(z^2)\pmod{4M}$$
which, by substituting $z$ to $z^2$, yields (\ref{prop3}).

We shall prove Proposition 3 by induction on $k$. For $k=0$, 
from (\ref{iter}), (\ref{con2k}) is just our hypothesis $f\equiv g\pmod{2}$.
Let us assume
that (\ref{con2k}) holds for a non negative value of $k$; then applying
(\ref{prop3}) with $u=f_k$, $v=g_k$ and $M=2^{k}$ will give
$$f_{k+1}\equiv g_{k+1} \pmod{2^{k+2}}$$
and the proof of Proposition 3 is completed. $\Box$



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\section{The set ${\cal A}={\cal A}_0(\{1,2,3\},3)$.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In all this Section, $\ca$ will denote the set ${\cal
A}_0(\{1,2,3\},3)$; we shall write $p(n), \si(n), \chi(n)$ instead of 
$p(\ca,n), \si(\ca,n), \chi(\ca,n)$ respectively. We shall recall in
Lemma 1 some classical results on congruences modulo $2^k$.

\medskip
{\bf Lemma 1.} {\it Let $a,b$ and $c$  be integers.

(i) If $a$ is odd and $k\ge 3$, the congruence
\begin{equation}\label{le11}
x^2\equiv a \pmod{2^k}
\end{equation}
has no solution if $a\not\equiv 1 \pmod{8}$, and has four solutions if 
$a\equiv 1 \pmod{8}$. Two of them satisfy
\begin{equation}\label{le12}
x^2\equiv a \pmod{2^{k+1}}.
\end{equation}
If $\rho_k$ is a solution of (\ref{le11}) which satisfies (\ref{le12})
then the four solutions of (\ref{le11}) are $\pm\rho_k$ and
$\pm\rho_k+2^{k-1}$. By the method of Hensel, $\rho_k$ can be
calculated by induction: if $\rho_k^2\equiv a+\vep 2^{k+1}
\pmod{2^{k+2}}$, $\vep\in\{0,1\}$,  then $\rho_{k+1}=\rho_k+\vep 2^{k+1}$.

(ii) the congruence 
\begin{equation}\label{le13}
x^2-x+a\equiv 0 \pmod{2^k}
\end{equation}
has no solution if $a$ is odd; if $a$ is even, it has two solutions,
one is even and the other one is odd. They can be calculated by
Hensel's method: if $\tau_k$ satisfies (\ref{le13}), then
$\tau_{k+1}=\tau_k+ \vep 2^{k}$ is determined from
\begin{equation}\label{le14}
\vep\equiv \frac{\tau_k^2-\tau_k+a}{2^{k}} \pmod{2}.
\end{equation}

(iii) Any congruence $x^2+bx+c\equiv 0\pmod{2^k}$ can be put on the
form (\ref{le11}) or (\ref{le13}) after a change of variable
$y=x+\lfloor b/2\rfloor$. 

(iv) In the $2$-adic field $\Q_2$, the equation $x^2-a=0$ has two
roots if and only if $a\equiv 1 \pmod{8}$. If $x_1$ and $x_2$ are the
two solutions, we have $x_i\equiv \pm\rho_k \pmod{2^k}$.

(v) In the $2$-adic field $\Q_2$, the equation $x^2-x+a=0$ has two
roots $x_1$ and $x_2$ if and only if $a$ is even. Moreover, $x_i \bmod
2^k$ are solutions of (\ref{le13}).    }

\medskip
{\bf Proof of (i).} It is a classical result that each odd residue
class modulo $2^k$ for $k\ge 3$ can be written $\pm 5^\la$ with $0\le
\la<2^{k-2}$. From this result, it is not difficult to prove (i). We
may observe that, if $\rho$ is a solution of (\ref{le12}), 
$$(\pm\rho+2^{k-1})^2\equiv a+2^{k}\pmod{2^{k+1}}$$
so that $\pm\rho+2^{k-1}$ are solutions of (\ref{le11}) but not of 
(\ref{le12}). 

The Hensel method can be found in \cite{Bar} or
\cite{Mig}. It can be accelerated to calculate
$\rho_{k+1},\ldots,\rho_{2k-1}$ in terms of $\rho_k$.

\medskip
{\bf Proof of (ii).} If $\tau_k$ is a solution of (\ref{le13}),
$\tau_k \bmod 2$ is a solution of $x^2-x+a\equiv 0 \pmod{2}$. But if
$a$ is odd this last congruence has no solution while, if $a$ is even,
it has two solutions $0$ and $1$. Both can be extended to solutions of 
(\ref{le13}), by using (\ref{le14}) with $k=1,2,\ldots$.

An other possibility to solve (\ref{le13}) is to set $y=2x-1$. We have 
$y^2=4x^2-4x+1\equiv 1-4a \pmod{2^{k+2}}$. But the four solutions of
this last congruence give only two solutions to (\ref{le13}).

Since the sum of the two roots is $1$ they must have a different
parity. To show (\ref{le14}), we calculate
$$(\tau_k+\vep2^k)^2-(\tau_k+\vep 2^k)+a\equiv \tau_k^2-\tau_k+a -\vep
2^k\pmod{2^{k+1}}.$$ 

\medskip
{\bf Proof of (iii).} It is obvious.

\medskip
{\bf Proof of (iv).} If
$$x_1=\vep_0+\vep_1 2+\vep_2 2^2+\ldots+\vep_k 2^k+\ldots $$
is a $2$-adic solution of $x^2-a=0$, then, by writing
$x_1=U_k+2^{k}R_k$, we have $U_k=x_1 \bmod 2^{k}$ and $x_1^2=a\equiv
U_k^2 \pmod{2^{k+1}}$, so that $U_k$ satisfies (\ref{le11}) and
(\ref{le12}) and thus, from (i), $U_k=\pm\rho_k$. In the
other way, by the Hensel method, each root of (\ref{le11}) satisfying 
(\ref{le12}) can be extended in a $2$-adic solution.

\medskip
{\bf Proof of (v).} The proof of (v) looks like the one of (iv), but
it is easier since (\ref{le13}) has only two solutions while (\ref{le11})
has four solutions, and the proof of Lemma 1 is completed. $\Box$ 




\medskip
{\bf Theorem 1.} {\it Let ${\cal A}={\cal A}_0(\{1,2,3\},3)$. Then :

(i) For all $k\geq 0$, the sequence $\ds\sigma(
2^kn)~\bmod 2^{k+1}$ is periodic in $n$ with period $q_k=7$.

(ii) If $\ds u_k=\sigma(2^k)\bmod 2^{k+1} $ and $\ds v_k=\sigma(3.2^k)
\bmod 2^{k+1}$ then
\begin{equation}\label{4per}
\ds\sigma( 2^kn)\equiv u_k, v_k,-3~(\bmod~2^{k+1}),~\hbox{respectively
as}~\leg n7 =1,-1,0.
\end{equation}
where $\ds\leg n7$ is the Legendre symbol. 

(iii) $u_k$ and $v_k$ are the two solutions of the congruence 
\begin{equation}\label{congsd}
x^2-x+2\equiv 0 ~(\bmod~2^{k+1})
\end{equation}
and satisfy
\begin{equation}\label{ukvk}
u_k+v_k\equiv 1 \pmod{2^{k+1}}\quad\textrm{  and } \quad
u_k\cdot v_k\equiv 2\pmod{2^{k+1}}. 
\end{equation}
Moreover, if $k\ge r$,
\begin{equation}\label{ukur}
u_k\equiv u_r \pmod{2^{r+1}}.
\end{equation}   

(iv) The equation 
\begin{equation}\label{eqsd}
x^2-x+2=0
\end{equation}
has two solutions in the $2$-adic field $\Q_2$; the odd one is
\begin{equation}\label{x1}
x_1=1+2+2^3+2^4+2^6+2^{13}+2^{14}+2^{18}+2^{19}+
2^{22}+2^{25}+\ldots.
\end{equation}
We have 
\begin{equation}\label{ukx1}
u_k\equiv x_1 \pmod{2^{k+1}}\, .
\end{equation}
Moreover, the elements of $\ca$ which are powers of $2$ are
$1,2,2^3,2^4,2^6,2^{13},\ldots$; they are determined by
\begin{equation}\label{2pk}
x_1=\sum_{k=0}^\infty \chi(2^k)2^k.
\end{equation}     }

\medskip
{\bf Proof of (i).} The polynomial $P$ defined by (\ref{P0}) and
(\ref{ep0}) is  
\begin{equation}\label{P3}
P(z)=1+z+z^3
\end{equation}
since $p(0)=1,p(1)=1,p(2)=2$ and $p(3)=3$. It is irreducible over
$\F_2[z]$ so that, in (\ref{irr}), $s=1$ and $Q_1=P$. So, it follows from
Theorem A  that $\be_1=7$ and  $q_k=7$ for all $k\ge 0$,
which proves (i).

\medskip
{\bf Proof of (ii).}
We shall denote by $H$ the other irreducible  polynomial 
of degree $3$ over $\F_2[z]$, 
\begin{equation}\label{H3}
H(z)=1+z^2+z^3.
\end{equation}
From (\ref{iter}) and Proposition 1, the leading coefficient of
$P_k=\cg^{(k)}(P)$ is 
$-1$, and from (\ref{G}), (\ref{pA0}), (\ref{P0}) and (\ref{ep0})  
\begin{equation}\label{Pk00}
P_k(0)=1
\end{equation}
so that we can write
\begin{equation}\label{Pk4}
P_k(z)=\cg^{(k)}(P)(z)= 1+a_kz+b_kz^2-z^3.
\end{equation} 
Now, we observe that, from (\ref{P3}) and (\ref{H3}) the polynomials
$P$ and $H$ are reciprocal. So, from (\ref{Pk4}) and (\ref{reci}) we
have
\begin{equation}\label{Hk4}
H_k(z)=\cg^{(k)}(H)(z)=1-b_kz-a_kz^2-z^3.
\end{equation} 
But we have
\begin{equation}\label{cyclo}
\frac{1-z^7}{1-z}=1+z+z^2+z^3+z^4+z^5+z^6\equiv P(z)H(z) \pmod{2},
\end{equation}
and this implies, from (\ref{quo}), (\ref{exam}), (\ref{prod}) and
Proposition 3 that 
\begin{equation}\label{cyclok}
\hspace{-3mm}\frac{1-z^7}{1-z}=1+z+z^2+z^3+z^4+z^5+z^6\equiv
P_k(z)H_k(z) \hspace{-3mm}\pmod{2^{k+1}}. 
\end{equation}
By expanding the product $P_kH_k$ from (\ref{Pk4}) and (\ref{Hk4}),
we get from (\ref{cyclok})
\begin{equation}\label{akbk}
a_k-b_k\equiv 1 \pmod{2^{k+1}}\quad\textrm{  and } \quad
a_k\cdot b_k\equiv -2\pmod{2^{k+1}}.
\end{equation}
By applying Proposition 2 to (\ref{FF}) 
(where $F(z)=F_\ca(z)$ is defined by (\ref{genF}), we get:
\begin{equation}\label{FF3k}
\sum_{n=1}^\infty \sigma (2^kn)z^n=z\frac{F_k'(z)}{F_k(z)}
\end{equation}
where $F_k$ is the $k$-iterate of $F$ by the transformation $\cg$
(cf. (\ref{iter})), and $\ds F_k'= \frac{d}{dz}(F_k(z))$.
It follows from (\ref{genF}), (\ref{AiN}), (\ref{P0}) and (\ref{ep0}) that
\begin{equation}\label{FconP}
F\equiv P\pmod{2}
\end{equation}
and  Proposition 3 implies that
\begin{equation}\label{FkPk2k}
F_k\equiv P_k \pmod{2^{k+1}}
\end{equation}
for all $k\ge 0$. We deduce from
(\ref{Pk00}) and (\ref{FkPk2k}) that 
\begin{equation}\label{Pk0}
F_k(0)=P_k(0)=1
\end{equation}
and thus, from (\ref{mufo}), (\ref{cond}) and (\ref{coni}), (\ref{FkPk2k})
implies
\begin{equation}\label{Pk}
z\frac{F_k'(z)}{F_k(z)}\equiv z\frac{P_k'(z)}{P_k(z)} \pmod{2^{k+1}}.
\end{equation} 
Therefore, from (\ref{FF3k}) and (\ref{Pk}), it follows:
\begin{equation}\label{FP3k}
\sum_{n=1}^\infty \sigma (2^kn)z^n\equiv
z\frac{P_k'(z)}{P_k(z)} \pmod{2^{k+1}}.
\end{equation}
From (\ref{FP3k}) and (\ref{cyclok}), we have
\begin{equation}\label{si4}
\sum_{n=1}^\infty \si(2^kn)z^n\equiv z\frac{P_k'(z)}{P_k(z)}
\equiv z\frac{(1-z)P_k'(z)H_k(z)}{1-z^7}\pmod{2^{k+1}}.
\end{equation}
The expansion of the numerator of the right hand side of (\ref{si4}) from
 (\ref{Pk4}), (\ref{Hk4}) and (\ref{akbk}) yields
\begin{eqnarray}\label{expa}
z(1-z)P_k'(z)H_k(z)\equiv\hspace{5cm}& &\\
a_kz+a_kz^2-b_kz^3+a_kz^4-b_kz^5-b_kz^6-3z^7 & &\hspace{-5mm}\pmod{2^{k+1}}.\notag 
\end{eqnarray}
By expanding the denominator of the right hand side of (\ref{si4})
:$\ds \frac{1}{1-z^7}=1+z^7+z^{14}+\ldots$, it follows from
(\ref{si4}) and (\ref{expa}) that 
\begin{equation}\label{cong4}
\left.
\begin{array}{rccccccc}
\si(2^kn)\equiv& a_k&a_k&-b_k&a_k&-b_k&-b_k&-3\pmod{2^{k+1}}\\
\hbox{ resp. as }n\equiv& 1&2&3&4&5&6&0\pmod{7}.
\end{array}\right\}
\end{equation}
The congruences (\ref{cong4}) give for $n=1$ and $n=3$
\begin{equation}\label{con134}
u_k=\si(2^k)\equiv a_k \hspace{-2mm}\pmod{2^{k+1}}\textrm{ and }
v_k\equiv \si(3\cdot 2^k)\equiv -b_k \hspace{-2mm}\pmod{2^{k+1}}. 
\end{equation}
Since the quadratic residue classes modulo $7$ are $1,2$ and $4$,
(\ref{cong4}) and (\ref{con134}) prove (ii).

\medskip
{\bf Proof of (iii).}
Formula (\ref{ukvk}) follows from (\ref{akbk}) and (\ref{con134}), and
yields 
$u_k(1-u_k)\equiv 2\pmod{2^{k+1}}$; so,  $u_k$ is a solution
of the congruence (\ref{congsd}). By the same way $v_k$ can be proved
to be also a solution of (\ref{congsd}).

 Finally, if $k\ge r$, 
$$u_k=\si(2^k)=u_r+\sum_{j=r+1}^k \chi(2^j)2^j \equiv u_r\pmod{2^{r+1}},$$ 
which shows (\ref{ukur}) and completes the proof of (iii).  

\medskip
{\bf Proof of (iv).}
Note that $u_k=\si(2^k)$ is odd (since $1\in\ca$) while $v_k$ is even
(since $1,3 \in\ca$). 
Since its discriminant $-7$ is congruent to $1$ modulo $8$ 
(cf. Lemma 1), the
equation (\ref{eqsd}) has two roots in $\Q_2$, $x_1\equiv 1\pmod{2}$
and $x_2\equiv 0\pmod{2}$. Moreover, $x_1 \bmod 2^{k+1}$ is solution
of (\ref{congsd}), and as it is smaller than $2^{k+1}$ (like $u_k$) it
is equal to $u_k$ (because $u_k\equiv  1\pmod{2}$) and so,
(\ref{ukx1}) is proved. Similarly, $v_k=x_2 \bmod 2^{k+1}$.

The expansion (\ref{x1}) has been calculated with the
function {\it polrootspadic} of PARI. The expansion of $x_2$ is
\begin{equation}\label{x2}
x_2=2+2^2+2^5+2^7+2^8+2^9+2^{10}+2^{11}+2^{12}+2^{15}+2^{16}+\ldots .
\end{equation}
At the exception of $2^1$, the powers of $2$ appear either in
$x_1$ or in $x_2$; the reason is that 
$$x_1+x_2=1=2+\frac{1}{1-2}=2+\sum_{n=0}^\infty 2^n.$$
Since $u_k$ has been defined as equal to $\ds \si(2^k)=\sum_{i=0}^k
\chi(2^i)2^i$, (\ref{2pk}) follows from (\ref{ukx1}) and this completes
the proof of Theorem 1. $\Box$


\medskip
\begin{center} 
{\bf Numerical table} 
\end{center}
$$
\begin{array}{c|cccccccccccccc|}
k=& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9 &10&11&12&13\\
\hline
u_k=&1&3&3&11&27&27&91&91&91&91&91&91&91&8283\\
v_k=&0&2&6&6&6&38&38&166&422&934&1958&4006&8102&8102\\
\hline
\end{array}
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\section{The elements of ${\cal A}={\cal A}_0(\{1,2,3\},3)$.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\medskip
Let us define two classical arithmetic functions. If $n$ is a positive
integer, $\om(n)$ will denote the number of distinct primes dividing
$n$ while $\Om(n)$ will be the number of primes dividing
$n$ according to multiplicity.
We are now ready to state Theorem 2 which gives the multiplicative
structure of the elements of $\ca$. 

\medskip
{\bf Theorem 2.} {\it Let ${\cal A}={\cal A}_0(\{1,2,3\},3)$. Then :

(i) The elements of $\ca$ of the form $2^k$ are determined by
Theorem 1 (iv). Similarly, the elements of $\ca$ of the form $7\cdot 2^k$
are determined by the even solution $x_7$ of the $2$-adic equation
$7x^2+7x+2=0$ 
\begin{eqnarray}\label{x7}
x_7&=&\sum_{k=0}^\infty \chi(7\cdot 2^k)2^k\\
 &=& 2 + 2^2 + 2^3 + 2^6 + 2^7
+ 2^9 + 2^{10} + 2^{12} + 2^{16} 
+ 2^{18} + 2^{20} + 2^{21}+\ldots \notag
%+ 2^22 + 2^23 + 2^24 + 2^26 + 2^27 
%+ 2^28 + 2^29 + 2^31 + 2^32 +\ldots
\end{eqnarray}

(ii) Let $p$ be an odd prime congruent to $1,2$ or
$4$ modulo $7$ (i.e.,$\leg p7 =1$). Then $(p,a)=1$ for all $ a\in \ca$. 

(iii) If $n\in\nb$ and $49$ divides $n$, then $n\notin {\cal A}$.

(iv) Let $n=2^km$, and 
\begin{equation}\label{mh}
m=p_1^{\la_1}p_2^{\la_2}\ldots p_h^{\la_h}
\end{equation}
where $p_i$ are  primes, distinct and congruent to $3,5$ or $6$ modulo
$7$ and $\la_i$ are positive integers. We assume that $h=\om(m)\ge 1$
(i.e. $m\ne 1$) and recall that $\Om(m)=\la_1+\ldots+\la_h$.

$\bullet$ (a) If $k\le h-2$, then $n\notin \ca$ and $7n\notin \ca$.

$\bullet$ (b) If $k= h-1$, then, without any restriction,  $n\in \ca$
and $7n\in \ca$.

$\bullet$ (c) If $k= h-1+r$, with $r\ge 1$, then $n\in \ca$ if and only
   if 
\begin{equation}\label{OM}
m\equiv (-1)^{\Om(m)}(1-2u_r)\ell^{-1} \pmod{2^{r+1}}, \quad
\ell=1,3,5,\ldots, 2^r-1.
\end{equation}
Moreover, the expansion
\begin{equation}\label{xm}
x_m=1+\chi(2^hm)2+\chi(2^{h+1}m)2^2+\ldots +\chi(2^{h+j}m)2^{j+1}+\ldots
\end{equation}
is one of the two $2$-adic solutions of the equation 
\begin{equation}\label{2adeq}
m^2x^2+7=0.
\end{equation}

$\bullet$ (d) If $k= h-1+r$, with $r\ge 1$, then $7n\in \ca$ if and only
   if 
\begin{equation}\label{OM7}
m\equiv (-1)^{\Om(m)}(2u_r-1)7^{-1}\ell^{-1} \hspace{-3mm}\pmod{2^{r+1}},
\; \ell=1,3,5,\ldots, 2^r-1.
\end{equation}
Moreover, the expansion
\begin{equation}\label{xm7}
x_{7,m}=1+\chi(7\cdot 2^hm)2+\ldots +\chi(7\cdot 2^{h+j}m)2^{j+1}+\ldots
\end{equation}
is one of the two $2$-adic solutions of the equation 
\begin{equation}\label{2adeq7}
7m^2x^2+1=0.
\end{equation}
We may observe that for $k=0,1,2$, we have for $m$ odd, $m\ne 1$ 
\begin{equation}\label{iff}
n=2^km\in \ca \Longleftrightarrow 7n=7\cdot 2^km\in \ca.
\end{equation}  }

%\medskip
{\bf Remark 3.} {\it Theorem 2 has been proved for $k=0$ in \cite{NS}
and for $k=1$ in \cite{Ben4}.}

\medskip
{\bf Proof of (i).} From (\ref{S}), we have
\begin{equation}\label{Sk}
S(7,k)=\chi(7)+\chi(7\cdot 2)2+\ldots +\chi(7\cdot2^k)2^k
\end{equation}
and by (\ref{mob2}) and (\ref{4per})
\begin{eqnarray}\label{401}
7S(7,k)&=&\sum_{d\dv 7}\mu(d)\si\left(\frac{7\cdot 2^k}{d}\right)\notag\\
&=&\si(7\cdot 2^k)-\si(2^k)\equiv -3-u_k \pmod{2^{k+1}},
\end{eqnarray}
so that $S(7,k)\equiv (-3-u_k)7^{-1} \pmod{2^{k+1}}$. Since, from Theorem
1 (iii), $u_k$ is solution of (\ref{congsd}), a simple calculation
shows that $S(7,k)$ is a solution of $7x^2+7x+2\equiv 0
\pmod{2^{k+1}}$. From (\ref{4per}), $\si(7)=1+7\chi(7)\equiv
-3\pmod{2}$ so that $\chi(7)=0$ and, from (\ref{Sk}), $S(7,k)$ is
even, which proves (i).

\medskip
{\bf Proof of (ii).}
Let us suppose that $n=2^km\in \ca$, $m$ odd and multiple of a prime
$p\equiv 1,2,4\pmod{7}$. Here we have from (\ref{S})
\begin{equation}\label{Skp}
S(m,k)=\chi(m)+\chi(2m)2+\ldots +\chi(2^km)2^k<2^{k+1}.
\end{equation}
We get from (\ref{mob2}) 
\begin{equation}\label{402}
mS(m,k)=\sum_{d\dv \overline{m}}\mu(d)\si\left(\frac nd\right)=\hspace{-2mm}
\sum_{d\dv (\overline{m}/p)}\mu(d)\left[\si\left(\frac
nd\right)-\si\left(\frac {n}{pd}\right)\right]. 
\end{equation} 
But, from (\ref{4per}), the above bracket is congruent to $0$ modulo
$2^{k+1}$, since $\ds \leg{n/d}{7} =\leg{n/pd}{7}\times\leg
p7=\leg{n/pd}{7}$. This implies that $S(m,k)$ is
congruent to $0$ modulo $2^{k+1}$, and as $0\le S(m,k)<2^{k+1}$, $S(m,k)$
vanishes and, from (\ref{Skp}), $\chi(n)=\chi(2^km)$ also vanishes
which proves (ii). 

\medskip
{\bf Proof of (iii).}
Let us now suppose that $n=2^km\in \ca$, $m$ odd and multiple of
$49$. If we define $S(m,k)$ by (\ref{Skp}), (\ref{402}) still holds if
we replace $p$ by $7$. Now, in the bracket, both $\frac nd$ and
$\frac {n}{7d}$ are multiple of $7$, so that, from (\ref{4per}),
again the bracket is congruent to $0$ modulo $2^{k+1}$. The proof of
(iii) ends in the same terms than the one of (ii) above.

\medskip
{\bf Proof of (iv).} We shall need the following lemma:

\medskip
{\bf Lemma 2.} {\it With $m,n,h,k$ as in Theorem 2 (iv), $u_k$ as defined
in Theorem 1, $S(m,k)$ as defined by (\ref{Skp}) (or \ref{S}), we have
\begin{equation}\label{Smk}
mS(m,k)\equiv 2^{h-1}(2u_k-1)(-1)^{\Om(m)} \pmod{2^{k+1}},
\end{equation}
and
\begin{equation}\label{Smk7}
7mS(7m,k)\equiv -2^{h-1}(2u_k-1)(-1)^{\Om(m)} \pmod{2^{k+1}}.
\end{equation}   }

%\medskip
{\bf Proof of Lemma 2.} With $n=2^km$, we have from (\ref{mob2}),
\begin{equation}\label{le21}
mS(m,k)=\sum_{d\dv \overline{m}}\mu(d)\si\left(\frac nd\right)=
\sum_{\substack{d\dv \overline{m} \\ \mu(d)=1}}\si\left(\frac nd\right)-
\sum_{\substack{d\dv \overline{m} \\ \mu(d)=-1}}\si\left(\frac nd\right)
\end{equation}
with $\overline{m}=p_1\ldots p_h$. Since $\leg 27=1$ and
$\leg{p_i}{7}=-1$, we have
$$\leg n7=(-1)^{\Om(m)}\quad \text{ and } \quad
\leg{n/d}{7}=(-1)^{\Om(m)}\mu(d).$$ 
Let us assume that $\Om(m)$ is even; then we get from (\ref{le21}) and
(\ref{4per})
\begin{equation}\label{le22}
mS(m,k)\equiv u_k\left(\sum_{\substack{d\dv \overline{m} \\ \mu(d)=1}}
1\right)- v_k\left(\sum_{\substack{d\dv \overline{m} \\
\mu(d)=-1}}1\right)\pmod{2^{k+1}}. 
\end{equation}
The first (resp. second) sum is the number of subsets of
$\{1,2,\ldots,h\}$ with an even (resp. odd) cardinal; they are both
equal to $2^{h-1}$ (since $h\ge 1$), and from (\ref{ukvk}),
(\ref{le22}) yields 
\begin{equation}\label{Smmk}
mS(m,k)\equiv 2^{h-1}(2u_k-1) \pmod{2^{k+1}}.
\end{equation}
If $\Om(m)$ is odd, the same calculation leads to a formula looking
like (\ref{Smmk}) where $(2u_k-1)$ is replaced by $(1-2u_k)$ so that,
in both cases, (\ref{Smk}) is proved. 

With $n$ still being equal to $2^km$, we have from (\ref{mob2})
\begin{equation}\label{le23}
7mS(7m,k)=\!\!\!\sum_{d\dv (7\overline{m})}\mu(d)\si\left(\frac{7n}{d}\right)=\hspace{-2mm}
\sum_{d\dv \overline{m}}\mu(d)\si\left(\frac{7n}{d}\right)
\!\!-\hspace{-2mm}\sum_{d\dv \overline{m}}\mu(d) \si\left(\frac
{n}{d}\right).  
\end{equation}
But, $7n/d\equiv 0 \pmod{7}$ so that, by (\ref{4per}), $\ds
\si\left(\frac{7n}{d}\right) \equiv -3 \pmod{2^{k+1}}$ and since
$\overline{m}\ne 1$, $\ds \sum_{d\dv
\overline{m}}\mu(d)=0$. Therefore, it 
follows from (\ref{le23}) and (\ref{le21}) that
\begin{equation}\label{le24}
7mS(7m,k)\equiv
-\sum_{d\dv \overline{m}}\mu(d) \si\left(\frac {n}{d}\right)\equiv
-mS(m,k) \pmod{2^{k+1}}, 
\end{equation}
which, together with (\ref{Smk}) proves (\ref{Smk7}). $\Box$

\medskip
Let us prove now Theorem 2 (iv):

\medskip
{\bf Proof of (iv) (a).} If $k\le h-2$, it follows from Lemma 2
(\ref{Smk}) that 
$$S(m,k)\equiv 0\pmod{2^{k+1}}.$$
From (\ref{Skp}), this implies $S(m,k)=0$ (since $0\le
S(m,k)<2^{k+1}$), and, therefore, 
$\chi(n)=\chi(2^kn)=0$, i.e. $n\notin \ca$.

By using (\ref{Smk7}) instead of (\ref{Smk}), it can be proved in the same way
that $\chi(7n)=0$, i.e. $7n\notin \ca$. 

\medskip
{\bf Proof of (iv) (b).} If $k= h-1$, (\ref{Smk}) writes
\begin{equation}\label{iva1}
mS(m,k)\equiv 2^{k}(2u_k-1)(-1)^{\Om(m)} \pmod{2^{k+1}}
\end{equation}
so that $S(m,k)$ is a  multiple of $2^k$, and, by dividing
(\ref{iva1}) by $2^k$, $S(m,k)/2^k$ is odd. But then, from
(\ref{Skp}), $S(m,k)=2^k\chi(2^km)=2^k\chi(n)$ and thus $\chi(n)=1$
and $n\in\ca$. A similar proof shows that $7n\in\ca$.


\medskip
{\bf Proof of (iv) (c).} It follows from (\ref{Smk}) that $S(m,k)$ is
a  multiple of $2^{h-1}$ and by dividing (\ref{Smk}) by $2^{h-1}$ and
using (\ref{ukur}), we get
\begin{equation}\label{ivc1}
m\frac{S(m,k)}{2^{h-1}}\equiv (2u_k-1)(-1)^{\Om(m)}\equiv
(2u_r-1)(-1)^{\Om(m)} \pmod{2^{r+1}}. 
\end{equation}
But, from (\ref{Skp}), we have
\begin{equation}\label{ivc2}
\frac{S(m,k)}{2^{h-1}}=1+\chi(2^hm)2+\ldots+\chi(2^km)2^r,
\end{equation}
since, by (b), $\chi(2^{h-1}m)=1$. Let $x$ be the solution of the
congruence 
\begin{equation}\label{ivc3}
mx\equiv -(2u_r-1)(-1)^{\Om(m)} \pmod{2^{r+1}},
\end{equation}
satisfying $0\le x<2^{r+1}$;
then, from (\ref{ivc1}), $2^{r+1}-x$ is equal to $\ds
\frac{S(m,r)}{2^{h-1}}$. So, if $x<2^r$, it follows from (\ref{ivc2}) that
$\chi(n)=\chi(2^km)=1$ while, if $x>2^r$, $\chi(n)=0$.

Since $u_k$ satisfies (\ref{congsd}),
\begin{equation}\label{ivc4}
\left((2u_r-1)(-1)^{\Om(m)}\right)^2=4u_r^2-4u_r+1\equiv -7
\pmod{2^{r+3}}
\end{equation}
and, so, from (\ref{ivc1}), $\ds \frac{S(m,r)}{2^{h-1}}$ is a solution
of $m^2x^2+7\equiv 0 \pmod{2^{r+3}}$, and thus, from (\ref{ivc2}) and
Lemma 1, $x_m$ is a
solution of the $2$-adic equation (\ref{2adeq}). 

By using (\ref{OM}) with $r=1$ we know that $2^hm\in \ca$ if
and only if\\ 
$m(-1)^{\Om(m)}\equiv 3 \pmod{4}$. Thus
\begin{equation}\label{ivc5}
\chi(2^hm)\equiv \frac{(-1)^{\Om(m)}m-1}{2}\pmod{2}
\end{equation}
which allows us to distinguish for $x_m$ between the two roots of
(\ref{2adeq}). 

\medskip
{\bf Proof of (iv) (d).} It is the same proof than the proof of (c).
We just have to show (\ref{iff}), which follows immediately from
(\ref{OM}) and (\ref{OM7}) by noting that  $7^{-1}\equiv
-1\pmod{8}$. In particular, for $r=1$, (\ref{OM}) and (\ref{OM7})
co\"{\i}ncide so that 
$\chi(7\cdot2^hm)=\chi(2^hm)$, and thus, formula (\ref{ivc5}) still
holds when replacing $\chi(2^hm)$ by $\chi(7\cdot2^hm)$. $\Box$ 

\medskip
If $m=p_1^{\la_1}\ldots p_h^{\la_h}$, for $r\le 5$, (\ref{OM}) and
(\ref{OM7}) can be written as

$$\chi(2^hm)=\chi(7\cdot2^hm)=1\Longleftrightarrow\;
(-1)^{\Om(m)}m\equiv 3\pmod{4}$$
$$\chi(2^{h+1}m)=\chi(7\cdot2^{h+1}m)=1\Longleftrightarrow\;
(-1)^{\Om(m)}m\equiv 1,3\pmod{8}$$
$$\chi(2^{h+2}m)=1\Longleftrightarrow\;
(-1)^{\Om(m)}m\equiv 9, 11, 13, 15\pmod{16}$$
$$\chi(7\cdot2^{h+2}m)=1\Longleftrightarrow\;
(-1)^{\Om(m)}m\equiv 1, 3, 5, 7\pmod{16}$$
$$\chi(2^{h+3}m)=1\Longleftrightarrow\;
(-1)^{\Om(m)}m\equiv 1, 5, 11, 15, 19, 23, 25, 29 \pmod{32}$$
$$\chi(7\cdot2^{h+3}m)=1\Longleftrightarrow\;
(-1)^{\Om(m)}m\equiv 1, 3, 5, 7,9,11,13,15\pmod{32}$$
\begin{eqnarray*}
\chi(2^{h+4}m)&=&1\Longleftrightarrow\\
(-1)^{\Om(m)}m&\equiv& 1, 3, 5, 7, 11, 15, 17, 21, 25, 27, 29, 31, 41,
45, 51, 55 \pmod{64}
\end{eqnarray*}
\begin{eqnarray*}
\chi(7\cdot2^{h+4}m)&=&1\Longleftrightarrow\\
(-1)^{\Om(m)}m&\equiv&\hspace{-3mm} 5, 7, 9, 11, 21, 23, 25, 27, 33,
35, 45, 47, 49, 51, 61, 63\hspace{-3mm} \pmod{64}.
\end{eqnarray*}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Estimation of $A(x)$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


In this Section, we want to estimate $A(x)$ where $\ca$ will denote
the 
set ${\cal A}_0(\{1,2,3\},3)$; in view of using the description of the
elements of $\ca$ given in Theorem 2, we need some definitions.

Let us denote by $\cf$ the set of the integers
\begin{equation}\label{6F}
n=2^hp_1^{\la_1}p_2^{\la_2}\ldots p_h^{\la_h},\quad h\ge 1, \quad
p_i\equiv 3,5,6\pmod{7}.
\end{equation}
The elements of $\cf$ are
\begin{equation}\label{6Fs}
\cf=\{6,10,18,26,34,38,50,54,60,62,82,94,\ldots\}.
\end{equation}
In \cite{Ben}, it has been proved that the asymptotic equivalence
\begin{equation}\label{6Fx}
F(x)=|\{n\in \cf; n\le x\}|\sim \ga \frac{x}{(\log x)^{3/4}}
\end{equation}
holds for $x\to\infty$, with 
\begin{equation}\label{6gam}
\ga=\frac{1}{4\Ga(5/4)}\left(\frac{6}{\pi\sqrt7}\right)^{1/4}\ga_1
\end{equation}
and
\begin{equation}\label{6gam1}
\ga_1=
\prod_{p\equiv 3,5,6\hspace{-2mm}\pmod{7}}
\left(\left(1+\frac{1}{2p-2}\right) \left(1-\frac{1}{p}\right)^{1/2}
\left(1-\frac{1}{p^2}\right)^{-1/4}\right).
\end{equation}
Calculating the product (\ref{6gam1}) with the primes up
to $100000$ yields for $\ga_1$ the approximate value $1.07517$.
Note that, with the method described in
\cite{Coh}, it is possible to calculate $\ga_1$ with a very high
precision. From $\Ga(5/4)=0.906402$ and (\ref{6gam}) the approximate
value of $\ga$ is
\begin{equation}\label{6gnum}
\ga=0.2733.
\end{equation}
If $\de(n)$ is the characteristic function of $\cf$, the Dirichlet's
series $\ds \sum_{n=1}^\infty \frac{\de(n)}{n^s}$ has an Eulerian
product for $\Re s>1$ 
\begin{eqnarray}\label{6Dir}
\sum_{n=1}^\infty \frac{\de(n)}{n^s}&=&\prod_{p\equiv
3,5,6\hspace{-2mm} \pmod{7}}
\left(1+\frac{1}{(2p)^s}+\frac{1}{2^sp^{2s}}+\ldots
+\frac{1}{2^sp^{ks}}+\ldots\right)\notag\\
&=&\prod_{p\equiv
3,5,6\hspace{-2mm}\pmod{7}}\left(1+\frac{1}{2^s(p^s-1)}\right).
\end{eqnarray} 
In \cite{Ben}, the proof of (\ref{6Fx}) starts from formula
(\ref{6Dir}). 

For $i=0$ or $1$, $\ell$ odd, and $r\ge 0$ we define
\begin{equation}\label{6Filr}
\cf_{i;\ r,\ \ell}= \{2^hm\in \cf,\quad \Om(m)\equiv i\pmod{2},\quad
m\equiv \ell \pmod{2^{r+1}}\}. 
\end{equation}
It has been proved in \cite{Ben} that for $r$ fixed, and $x$ going to
infinity, we have for any $i$ and $\ell$
\begin{equation}\label{6Fix}
F_{i;\ r,\ \ell}(x)=|\{n\in \cf_{i;\ r,\ \ell}\:\:  , \;\; n\le x\}|\sim
\frac{\ga}{2^{r+1}}\frac{x}{(\log x)^{3/4}}\cdot 
\end{equation}

From (\ref{6Fx}), (\ref{6Fix}) and Theorem 2, we shall prove

\medskip
{\bf Theorem 3.} {\it Let $\ca={\cal A}_0(\{1,2,3\},3)$. The number
$A(x)$ of elements of $\ca$ up to $x$ satisfies, when $x$ tends to
infinity: 
\begin{equation}\label{6t}
A(x)\sim \frac{24\ga}{7} \frac{x}{(\log x)^{3/4}}
\end{equation}
where $\ga$ is defined by (\ref{6gam}); an approximate value of
$24\ga/7$ is $0.937$.     }

\medskip
{\bf Proof. }
Let $\ca'$ be the subset of $\ca$ whose elements are of the form $2^k$ or
$7\cdot 2^k$. The number of such elements up to $x$ is clearly 
\begin{equation}\label{6A'}
A'(x)=O(\log (x)),\quad x\to\infty.
\end{equation}
Now, it follows from Theorem 2 (i) and (iv) that 
\begin{equation}\label{6Ar}
\ca=\ca'\cup\;\;\bigcup_{r=0}^\infty \left(\ca^{(r)}\cup\ca^{(r,7)}\right)
\end{equation}
where the elements of $\ca^{(r)}$ are the $n$'s, $n\in \ca$,
$n=2^{h-1+r}p_1^{\la_1}\ldots p_h^{\la_h}$ and similarly, the elements
of $\ca^{(r,7)}$ are the $n$'s, $n\in \ca$,
$n=7\cdot 2^{h-1+r}p_1^{\la_1}\ldots p_h^{\la_h}$.

From Theorem 2 (iv) (b) it follows that for $r=0$,
\begin{equation}\label{6rz}
\ca^{(0)}=\frac 12 \cf \quad \textrm{ and } \quad A^{(0)}(x)=F(2x),
\end{equation} 
and
\begin{equation}\label{6rz7}
\ca^{(0,7)}=\frac 72 \cf \quad \textrm{ and } \quad
A^{(0,7)}(x)=F\left(\frac{2x}{7}\right).
\end{equation}
In the same way, if  $a^{-1}$ denotes an inverse of $a$ modulo
$2^{r+1}$,   Theorem 2 (iv) (c) and (d) implies for $r\ge 1$
\begin{equation}\label{6r}
\hspace{-2mm}A^{(r)}(x)=\hspace{-2mm}\sum_{\ell\in \{1,3,\ldots, 2^r-1\}}
\hspace{-2mm}\left(
F_{0;\ r,\ (1-2u_r)\ell^{-1} }\left(\frac{x}{2^{r-1}}\right)
\!\!+\!\!F_{1;\ r,\ (2u_r-1)\ell^{-1} }\left(\frac{x}{2^{r-1}}\right)
\!\right)
\end{equation}
and
\begin{eqnarray}\label{6r7}
A^{(r,7)}(x)&=&\sum_{\ell\in \{1,3,\ldots, 2^r-1\}}
F_{0;\ r,\ (2u_r-1)(7\ell)^{-1} }\left(\frac{x}{7\cdot
2^{r-1}}\right)\notag\\ 
&+&\sum_{\ell\in \{1,3,\ldots, 2^r-1\}}
F_{1;\ r,\ (1-2u_r)(7\ell)^{-1} }\left(\frac{x}{7\cdot
2^{r-1}}\right).
\end{eqnarray}

It also follows from Theorem 2 that, for any $r\ge 0$, $\ca^{(r)}\subset
2^{r-1}\cf$ and $\ca^{(r,7)}\subset 7\cdot 2^{r-1}\cf$ so that 
\begin{equation}\label{6rM}
A^{(r)}(x)\le F\left(\frac{x}{2^{r-1}}\right)\quad \textrm{ and }
\quad A^{(r,7)}(x)\le F\left(\frac{x}{7\cdot
2^{r-1}}\right)\le F\left(\frac{x}{2^{r-1}}\right).
\end{equation}
Moreover, from (\ref{6Fx}) and (\ref{6Fs}), there exists an absolute
constant $K$ such that
\begin{equation}\label{6FxM}
F(x)\le K \frac{x}{(\log x)^{3/4}}\quad \text{ for } x>1 \textrm{
and } F(x)=0 \quad \text{ for } x< 6. 
\end{equation}
It follows from (\ref{6rM}) and (\ref{6FxM}) that $A^{(r)}(x)=
A^{(r,7)}(x)=0$ for $\ds r>\frac{\log(x/3)}{\log 2}$ so that  
(\ref{6Ar}) implies
\begin{equation}\label{6Aso}
A(x)=A'(x)+\sum_{r=0}^{R''} \left( A^{(r)}(x)+A^{(r,7)}(x)\right), \quad
R''=\left\lfloor\frac{\log(x/3)}{\log 2}\right\rfloor. 
\end{equation}
Now, we cut the sum (\ref{6Aso}) in three parts:
\begin{equation}\label{6par}
A(x)=A'(x)+\sum_{r=0}^R+\sum_{r=R+1}^{R'}+\sum_{r=R'+1}^{R''}
\stackrel{def}{=} A'(x)+S+S'+S'', 
\end{equation}
where $R$ is a large but fixed  integer, and $R'$ is defined by
$2^{R'-1}\le \sqrt x<2^{R'}$. We have from (\ref{6rM})
$$S''=\sum_{r=R'+1}^{R''} \left( A^{(r)}(x)+A^{(r,7)}(x)\right)\le
2\sum_{r=R'+1}^{R''} F\left(\frac{x}{2^{r-1}}\right)\; ;$$
and, by observing that, for $r\le R''$, $\ds \frac{x}{2^{r-1}}\ge 6$,
we get from (\ref{6FxM}) 
\begin{equation}\label{6S''}
S''\le 2K\sum_{r=R'+1}^{R''}\frac{x}{(2^{r-1})(\log 6)^{3/4}}
\le \frac{2Kx}{2^{R'-1}}\le 4K\sqrt x.
\end{equation}
Similarly, we get
\begin{eqnarray}\label{6S'}
S'&=&\sum_{r=R+1}^{R'} \left( A^{(r)}(x)+A^{(r,7)}(x)\right)\le
2\sum_{r=R+1}^{R'} F\left(\frac{x}{2^{r-1}}\right)\notag\\
&\le& 2K\sum_{r=R+1}^{R'}\frac{x}{(2^{r-1})(\log(x/2^{R'-1}))^{3/4}}
\le 2K\sum_{r=R+1}^{R'}\frac{x}{2^{r-1}(\log\sqrt x)^{3/4}}  \notag\\
&\le& 2^{1+3/4}K\frac{x}{(\log
x)^{3/4}}\sum_{r=R+1}^{\infty}\frac{1}{2^{r-1}}
\le \frac{8K}{2^R}\frac{x}{(\log x)^{3/4}}\cdot
\end{eqnarray}
We deduce from (\ref{6r}), (\ref{6r7}) and (\ref{6Fix}) that, for $r\ge
1$ and  $r$ fixed,
\begin{equation}\label{6req}
A^{(r)}(x)\sim\frac{\ga}{2^{r}}\frac{x}{(\log x)^{3/4}}
\;\; \textrm{ and } \;\; 
A^{(r,7)}(x)\sim\frac{\ga}{7\cdot2^{r}}\frac{x}{(\log x)^{3/4}}
\;\; (x\to\infty).
\end{equation}
Since $R$ is fixed, we get from (\ref{6rz}), (\ref{6rz7}),
(\ref{6Fx}) and (\ref{6req}) for $x$ tending to infinity  
\begin{eqnarray}\label{6S}
S&=&A^{(0)}(x)+A^{(0,7)}(x)+\sum_{r=1}^{R} \left(
A^{(r)}(x)+A^{(r,7)}(x)\right)\notag\\
&\sim& \frac {8\ga }{7} \frac{x}{(\log x)^{3/4}}
\left(2+\sum_{r=1}^{R}\frac{1}{2^{r}}\right)=
\frac {8\ga }{7}\left(3-\frac{1}{2^{R}}\right)
\frac{x}{(\log x)^{3/4}}.
\end{eqnarray}
By making $R$ going to infinity, (\ref{6t}) follows from 
(\ref{6par}), (\ref{6A'}),
(\ref{6S''}), (\ref{6S'}) and (\ref{6S}) and the proof of Theorem 3 is
completed. $\Box$

The following table has been calculated by M. Del\'eglise by using 
the multiplicative structure of the elements of $\ca=\ca_0(\{1,2,3\}),3)$
given by Theorem 2. The last line is the asymptotic result of Theorem 3.
\medskip
\begin{center} 
{\bf Numerical table} 
\end{center}
$$
\begin{array}{|ccc|}
\hline
x&A(x)&A(x)\left(\log x\right)^{3/4}\left/x\right.\\
\hline
10^2&47&1.48\\
10^3&293&1.25\\
10^4&2\ 204&1.16\\
10^5&17\ 604&1.100\\
10^6&148\ 834&1.066\\
10^7&1\ 297\ 167&1.043\\
10^8&11\ 562\ 386&1.028\\
.\;.\;.\;.&.\;.\;.\;.\;.\;.&.\;.\;.\\
\infty&  &0.937\\
\hline
\end{array}
$$











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\end{document}




 



