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\begin{center}
\vskip 1cm{\LARGE\bf 
Sets with Even Partition Functions and \\
\vskip .1in
2-adic Integers, II
}
\vskip 1cm
\large
N. Baccar\footnote{Research supported
DGRST of Tunisia, UR 99/15-18, Facult\'e des Sciences de Tunis.} \\
Universit\'e de Sousse \\
ISITCOM Hammam Sousse \\
D\'ep. de Math Inf. \\
5 Bis Rue 1 Juin 1955 \\
4011 Hammam Sousse \\
Tunisie\\
\href{mailto:naceurbaccar@yahoo.fr}{\tt naceurbaccar@yahoo.fr}\\
\ \\
A.  Zekraoui\\
Universit\'e de Monastir\\
F. S. M. \\
D\'ep. de Math. \\
Avenue de l'environnement \\
5000 Monastir \\
Tunisie \\
\href{mailto:ahlemzekraoui@yahoo.fr}{\tt ahlemzekraoui@yahoo.fr}
\end{center}

\vskip .2 in

\begin{abstract}
For $P\in \mathbb{F}_2[z]$ with $P(0)=1$ and $\deg(P)\geq 1$, let
${\cal A}={\cal A}(P)$ be the unique
subset of $\mathbb{N}$ such that $\sum_{n\geq 0}p({\cal
A},n)z^n\equiv P(z)$ (mod $2$), where $p({\cal A},n)$ is the
number of partitions of $n$ with parts in ${\cal A}$. Let $p$ be
an odd prime number, and let $P$ be irreducible of order $p$ ; i.e.,
$p$ is the smallest positive integer such that $P$ divides $1+z^p$
in $\F_2[z]$.  N. Baccar proved that the elements of
${\cal A}(P)$ of the form $2^km$, where $k\geq 0$ and $m$ is odd,
are given by the $2$-adic expansion of a zero  of some polynomial
$R_m$ with integer coefficients. Let $s_p$ be the order of $2$
modulo $p$, i.e., the smallest positive integer such that
$2^{s_p}\equiv 1$ (mod $p$). Improving on the method  with which $R_m$ was obtained explicitly only when
$s_p=\frac{p-1}{2}$, here we make explicit $R_m$ when
$s_p=\frac{p-1}{3}$. For that, we have used the number of points of the elliptic curve $x^3+ay^3 =1 $ modulo $p$.
\end{abstract}


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\section{Introduction.}

Let $\N$ denote the set of positive integers, and let  ${\cal
A}=\left\{a_1,a_2,\ldots\right\}$ be a non-empty subset of $\N$.
For $n \in \N$, let $p({\cal A},n)$ be the number of partitions of
$n$ with parts in ${\cal A}$, i.e., the number of solutions of the
diophantine equation
\begin{equation}\label{eq1}
a_1x_1+a_2x_2+\cdots=n
\end{equation}
in non-negative integers $x_1,~x_2,\ldots$. By convention, $
p({\cal A},0)=1$ and $p({\cal A},n)=0$ for all $n&lt;0$.  The
generating series of $p({\cal A},n)$ is
\begin{equation}\label{eq3}
F_{\cal A}(z):=\sum_{n=0}^{\infty}p({\cal A},n)z^n=\prod_{a \in
{\cal A}}\frac{1}{1-z^a}.
\end{equation}
Let $\F_2$ be the field with two elements and
$P(z)=1+\epsilon_1z+\cdots +\epsilon_Nz^N \in \F_2\left[z\right]$,
 $N\geq 1$. J.-L. Nicolas, I. Z. Ruzsa and A. S\'ark\"ozy  \cite{NRS}
proved that there exists a unique set ${\cal A}={\cal A}(P)$
satisfying
\begin{equation}\label{eq4}
F_{\cal A}(z)\equiv P(z) \pmod 2,
\end{equation}
which means that
\begin{equation}\label{hhhfffrrr}
   \textrm{$p({\cal A},n) \equiv \epsilon_n \pmod 2$ for
$1\leq n \leq N$ }
\end{equation}
and $p({\cal A},n)$ is even for all $n&gt; N$. Indeed, for $n=1$,
\begin{displaymath}
p({\cal A},1)=\left\{ \begin{array}{ll} 1,      &amp; \textrm{if
$1 \in {\cal A};$}\\ 0,      &amp; \textrm{if  $1\notin {\cal A}$}.
 \end{array} \right.
\end{displaymath}
and so, by (\ref{hhhfffrrr}),
$$1 \in {\cal A}\Leftrightarrow \epsilon_1=1.$$
Further, assume that we Know ${\cal A}_{n-1}={\cal A}\cap \{1,\ldots,n-1\}$; since there exists only one partition of $n$ containing the part $n$, then
$$p({\cal A},n)=p({\cal A}_{n-1},n)+\chi({\cal A},n),$$
where $\chi ({\cal A},.)$ is the characteristic function of the
set ${\cal A}$, i.e.,
\begin{displaymath}
\chi ({\cal A},n)=\left\{ \begin{array}{ll} 1,      &amp; \textrm{if
$n \in {\cal A};$}\\ 0,      &amp; \textrm{if  $n\notin {\cal A}$},
 \end{array} \right.
\end{displaymath}
which with (\ref{eq4}) allow one to decide whether $n$ belongs to ${\cal A}$.\\

Let
$p$ be an odd prime number,  and let $s_p$ be the order of $2$ modulo
$p$, i.e., $s_p$ is the smallest positive integer such that $p$
divides $2^{s_p}-1$. Let $P\in \F_2[z]$ be irreducible of order
$p$ ($\text {ord}(P)=p$); in other words, $p$ is the smallest
positive integer such that $P$ divides $1+z^p$ in $\F_2[z]$.  N. Baccar and F. Ben Sa\"{\i}d
\cite{NF2} determined the sets
${\cal A}(P)$ for all $p$ such that $s_p=\frac{p-1}{2}$. Moreover,
they proved that if $k \geq 0$ and $m$ is an odd positive integer,
then the elements of ${\cal A}(P)$ of the form $2^km$ are given by
the $2$-adic expansion of some zero of a polynomial $R_m$ with
integer coefficients.  N. Baccar  \cite{NF3} extended this last
result to any odd prime number $p$. Unfortunately, the method used
in that paper can make explicit $R_m$ only when
$s_p=\frac{p-1}{2}$. In this paper, we will improve on the method
given by N. Baccar \cite{NF3}, by introducing elliptic curves, to make $R_m$
explicit when $s_p=\frac{p-1}{3}$. In Section 2, some properties
of the polynomial $R_m$ are exposed. In Section 3, we introduce
elliptic curves to compute some cardinalities used in Section 4 to
make $R_1$ explicit, and in Section 5 to get $R_m$ explicitly for
any odd integer $m\geq 3$.\\

Throughout this paper, $p$ is an odd prime number and $P$ is some
irreducible polynomial in $\F_2[z]$ of order $p$. We also denote
by $s_p$ the order of $2$ modulo $p$. For $a\in \Z$ and $b\in \N
$, we should write $a \bmod  b$ for the remainder of the euclidean
division of $a$ by $b$.


\section{Some results on the polynomial $R_m$}
Let $p$ be an odd prime. We denote by $(\mathbb{Z}/{p
\mathbb{Z}})^*$  the group of invertible elements modulo $p$ and
by $&lt;2&gt;$  its subgroup generated by $2$. We consider the action
$\star $ of $&lt;2&gt;$ on the set $\mathbb{Z}/{p \mathbb{Z}}$ given by
$a\star n=an$ for all $a\in &lt;2&gt;$ and all $n\in \mathbb{Z}/{p
\mathbb{Z}}$. The quotient set will be denoted by $(\mathbb{Z}/{p
\mathbb{Z}})/_{&lt;2&gt;}$ and the orbit of some $n\in \mathbb{Z}/{p
\mathbb{Z}}$ by ${\it O}(n)$. So, we can write
\begin{equation*}
\mathbb{Z}/{p \mathbb{Z}}= {\it O}(1)\cup{\it
O}(g)\cup\cdots\cup{\it O}(g^{r-1})\cup{\it O}(p),
\end{equation*}
where $g$ is some generator of $(\mathbb{Z}/{p \mathbb{Z}})^*$,
$r=\frac{p-1}{s_p}$ is the number of invertible orbits of
$\mathbb{Z}/{p \mathbb{Z}}$,
\begin{equation}\label{hdnbcvf}
    {\it
O}(g^i)=\left\{2^jg^i \bmod  p:~0\leq j \leq s_p -1\right\},~~0\leq i \leq
r -1,
\end{equation}
\begin{equation*}
{\it O}(p)=\left\{0\right\}.
\end{equation*}
Note that for any  integer $t$,
\begin{equation}\label{114nnbv1}
{\it O}(g^t)={\it O}(g^{t \bmod r}).
\end{equation}
The orbits ${\it O}(n)$ are defined as parts of $\mathbb{Z}/{p \mathbb{Z}}$; however, by extension, they are also considered as parts of $\N$.\\

If $\phi_p$ is the cyclotomic polynomial over $\F_2$ of index $p$,
then
\begin{equation*}
1+z^p =(1+z)\phi_p(z).
\end{equation*}
Moreover, one has
\begin{equation*}
\phi_p(z)=P_0(z)P_1(z)\cdots P_{r-1}(z),
\end{equation*}
where $P_0,P_1,\ldots$ and $P_{r-1}$  are the only  distinct
irreducible polynomials in $\F_2\left[z\right]$ of the same degree
$s_p$ and all of which are of order $p$. For all $l,$ $0\leq l \leq
r-1$, let ${\cal A}_l={\cal A}(P_l)$ be the set obtained from
(\ref{eq4}). If $m$ is an odd positive integer, we define the
$2-$adic integer $y_l(m) $ by
\begin{equation}\label{bndcse}
y_l(m)=\chi({\cal A}_l,m)+2\chi({\cal A}_l,2m)+4\chi({\cal
A}_l,4m)+ \cdots=\sum_{k=0}^{\infty}\chi({\cal A}_l,2^km)2^k.
\end{equation}
By computing $y_l(m) \bmod 2^{k+1}$, one can deduce $\chi({\cal
A}_l,2^jm)$ for all $j,$ $0\leq j\leq k$, and obtain all  the
elements of ${\cal A}_l$ of the form $2^jm$. In \cite{NFA}, some necessary conditions on integers to be in
${\cal A}_l$ were given. For instance:
\begin{equation}\label{bacbszek1}
p^2n\notin {\cal A}_l,~\forall ~n\in \N ,
\end{equation}
\begin{equation}\label{bacbszek2}
\text {if}~ q ~\text {is an odd prime in}~ {\it O}(1), \text
{then}~ qn\notin {\cal A}_l,~\forall n\in \N .
\end{equation}

Let ${\K}$ be some field, and let $u(z)=\sum_{j=0}^{n}u_jz^j$ and
$v(z)=\sum_{j=0}^{t}v_jz^j$ be polynomials in ${\K}[z]$. We denote
the resultant of $u$ and $v$ with respect to $z$ by
$\textrm{res}_z(u(z),v(z))$, and recall the following well known result
\begin{lem}\label{lemee}
(i) The resultant $\textrm{res}_z(u(z),v(z))$ is a homogeneous multivariate
 polynomial with integer coefficients, of degree $n+t$ in the $n+t+2$
 variables  $u_i,$ $v_j$.\\
(ii) If $u(z)$ is written as  $u(z)=
u_n(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n)$ in the splitting
field of $u$ over ${\K}$ then
\begin{equation}\label{hgfd}
\textrm{res}_z(u(z),v(z))=u_n^t\prod_{i=1}^{n}v(\alpha_i).
\end{equation}
\end{lem}
 N. Baccar proved  \cite{NF3} that, for all $l$, $0\leq l\leq
r-1$, the $2$-adic integers $y_l(m)$ defined by (\ref{bndcse}) are
the zeros of some polynomial $R_m$ with integer coefficients and
which can be written as the resultant of two polynomials. We
mention here that the expressions given in that paper to $R_m$,
for $m=1$ and $m\geq 3$, can be encoded in  only one. So that we
have
\begin{thm}\label{sfj14jjjj}(\cite{NF3}) 1) Let $m $ be an odd positive integer such that
$m \notin {\it O}(p)$ (i.e., $\gcd(m,p)=1$), and let $\delta =\delta
(m)$ be the unique integer in $ \left\{0,1,\ldots,r-1\right\}$
such that $m\in {\it O}(g^\delta)$. We define the polynomial $A_m$
by
\begin{equation}\label{deee}
A_m(z)=\sum_{h=0}^{r-1}\alpha_h(m)B_h(z),
\end{equation}
where for all $h,$ $0\leq h \leq r-1$,
\begin{equation}\label{vfrd}
\alpha_h(m)= \sum_{\substack{d\dv \widetilde{m},~d \in {\it
O}(g^h)}} \mu(d),
\end{equation}
$\widetilde{m}=\prod_{q \ \textrm{prime} \ q\mid m}q$ is the radical of $m$ with
$\widetilde{1}=1$, $\mu$ is the M\"obius function and $B_h$ is the
polynomial
\begin{equation}\label{dee}
B_h(z)=B_{h,m}(z)= \sum_{j=0}^{s_p-1}z^{\left(2^jg^{(\delta-h)
\bmod r}\right) \bmod p }.
\end{equation}
Then, the $2$-adic integers $y_0(m),y_1(m),\ldots $ and
$y_{r-1}(m)$ are the zeros of the polynomial $R_m(y)$ of $\Z[y]$
defined by the resultant
\begin{equation}\label{sfre}
R_m(y)=\textrm{res}_z(\phi_p(z),my+A_m(z))
\end{equation}
and we have
\begin{equation}\label{sfre22}
R_m(y)=m^{p-1}\bigl(( y-y_0(m) )(y-y_1(m))\cdots(y-y_{r-1}(m))\bigr)^{s_p}.
\end{equation}
2) The $2$-adic integers $y_0(p),y_1(p),\ldots $ and $y_{r-1}(p)$ are the zeros of the
polynomial $R_1(-py-s_p)$; while if $m=pm'$, $m'\geq 3$ and $\gcd
(m',p)=1$, then $y_0(m),y_1(m),\ldots$ and $y_{r-1}(m)$ are the
zeros of the polynomial $R_{m'}(-py)$ defined by (\ref{sfre}).\\
3) If $ m$ is divisible by $p^2$ or by some prime $q$ belonging to
${\it O}(1)$ then we extend the definition (\ref{sfre}) to
$R_m(y)=m^{p-1}y^{s_p}$; so that $y_0(m),y_1(m),\ldots $ and $
y_{r-1}(m)$ remain zeros of $R_m$ since, from (\ref {bacbszek1})
and (\ref{bacbszek2}), they all vanish.\\
\end{thm}
\begin{remark}\label{hhuuggg} Explicit formulas to the polynomials $R_m$ defined by
(\ref{sfre}), when $s_p=\frac {p-1}{2}$, are given in \cite{NF3}.
Moreover in that paper, it is shown that if   $\theta$ is a certain primitive
$p$-th root of unity over the $2$-adic field $\Q_2$, then for all
$l$, $0\leq l \leq r-1$,
\begin{equation}\label{nbsqz}
y_l(1)=-T_l ,
\end{equation}
where, for all $l \in \Z$,
\begin{equation}\label{hgtrvcdes}
    T_l=T_{l \bmod  3}=\sum_{k=0}^{s_p-1}\theta^{2^kg^l}=\sum_{j\in {\it
O}(g^l)}\theta^j.
\end{equation}
We also mention here that N. Baccar  \cite{NF3}  proved that for all
$m\in \N$,
\begin{equation}\label{Rm(yy)}
R_m(y)=\prod_{l=0}^{r-1}\left( my+A_m(\theta^{g^l})\right)^{s_p}.
\end{equation}
\end{remark}








\section{ Orbits and elliptic curves.}
From now on, we keep the above notation and assume that the prime
number $p$ is such that $s_p=\frac{p-1}{3}$ (the first ones up to $1000$ are:
$p=43,109,157,229,277,283,307,499,643,691,733,739,\\811,997$ ). So the number of invertible
orbits is $r=3$ and
\begin{equation}\label{orbital}
    \mathbb{Z}/{p \mathbb{Z}}= {\it O}(1)\cup{\it
O}(g)\cup{\it O}(g^2)\cup{\it O}(p),
\end{equation}
where $g$ is some generator of the cyclic group $(\mathbb{Z}/{p
\mathbb{Z}})^*$. The order of $2$ is $s_p=\frac{p-1}{3}$; if $2$
were a square modulo $p$, its order should divide $\frac{p-1}{2}$,
which is impossible. Hence $2$ cannot be a square modulo $p$, and
by Euler criterion, $p$ has to satisfy $p\equiv \pm 3 $ (mod
$8$), and, as $p\equiv 1$ (mod $3$), $p\equiv 13,~19 $ (mod
$24$).
\begin{lem}\label{lem orbff}
For all $i,$  $0 \leq i \leq 2$, let ${\it O}(g^i)$ be the orbit
of $g^i$ defined by (\ref{hdnbcvf}). Then
\begin{equation}\label{lknb}
{\it
O}(g^i)=\left\{-g^i,-2g^i,\ldots,-2^{s_p-1}g^i\right\}=\left\{g^i,g^{i+3}.
\ldots,g^{i+3(s_p-1)}\right\}.
\end{equation}
In particular, $2$ is a cube modulo $p$ and the sub-group generated by $2$ is the
sub-group of cubes (generated by $g^3$) and contains $-1$.
\end{lem}
\begin{proof}[Proof] To get the first equality  of (\ref{lknb}),
it suffices to show that $-1 \in {\it O}(1)$. This follows from
$-1=(\frac {2}{p})\equiv 2^{\frac {p-1}{2}}$ (mod $p$).\\ To
prove the second equality of (\ref{lknb}), one just use the fact
that (cf. (\ref{114nnbv1})) $g^3 \in {\it O}(1)$.
\end{proof}

Let us define the integers $\ell_{i,j},~0\leq i,~j\leq 2$, by
\begin{equation}\label{vfer}
\ell_{i,j}=\left| \left\{t:~0\leq t\leq s_p-1, ~1+g^{j+3t} \in {\it
O}(g^i) \right\} \right|.
\end{equation}
\begin{remark}\label{remllll} As shown just above, $-1\in {\it O}(1)$, so that there exists one and
only one $t\in \left\{0,1,\ldots,s_p-1\right\}$ such that
$1+g^{3t}\in {\it O}(p)$. Moreover, for all $t\in
\left\{0,1,\ldots,s_p-1\right\}$ and $j \in \{1,2\}$, $1+g^{j+3t}\notin {\it O}(p)$.
Hence the integers $\ell_{i,j}$ defined by (\ref {vfer}) satisfy
\begin{equation}\label{sumlij}
\sum_{i=0}^2\ell_{i,j}=s_p-\delta_{0,j},
\end{equation}
where $\delta_{i,j}$ is the Kronecker symbol given by
\begin{equation*}
\delta_{i,j}=\left\{ \begin{array}{ll} 1,      &amp; \textrm{if
$i=j$};\\ 0,      &amp; \textrm{otherwise}.
 \end{array} \right.
\end{equation*}
\end{remark}
~\\ The integers $\ell_{i,j}$ defined above are  cardinalities of some
curves. Indeed, let us consider the  curve over $\F_p$,
\begin{equation*}
{\cal C}_{i,j}:~~1+g^jX^3=g^iY^3
\end{equation*}
and denote by $c_{i,j}$ its cardinality, $c_{i,j}=\mid {\cal
C}_{i,j}\mid $. Since $-1$ is a cube modulo $p$, it is clear that
\begin{equation*}
c_{j,i}=c_{i,j}.
\end{equation*}
Using  (\ref{vfer}) it follows that
\begin{equation*}
  \ell_{i,j}=|\{(X^3,Y^3):~X\neq0,~Y\neq 0~\textrm{and}~(X,Y) \in {\cal C}_{i,j} \}|.
\end{equation*}
Therefore,
\begin{equation}\label{lij=lji}
\ell_{j,i}=\ell_{i,j}.
\end{equation}
Note that, $(X^3,Y^3)=(X'^3,Y'^3)$ if and only if $X'=Xg^{vs_p}$ and $Y'=Yg^{ws_p}$ for some $v,w\in\{0,1,2\}$. Moreover, if $i\neq 0$  (resp. $j\neq 0$), no point on the curve ${\cal C}_{i,j}$ can be of the form $(0,Y)$ (resp. $(X,0)$). But if $i=0$ (resp. $j=0$), we obtain three points on the curve ${\cal C}_{i,j}$ with $X=0$ (resp. $Y=0$). Consequently, we obtain the
relation
\begin{equation}\label{cijlij}
c_{i,j}=9\ell_{i,j}+3\delta_{i,0}+3\delta_{0,j}.
\end{equation}
Now, let us consider the projective plane cubic  curve
\begin{equation*}
{\cal E}_{i,j}:~~Z^3+g^jX^3=g^iY^3
\end{equation*}
and $e_{i,j}=\mid {\cal E}_{i,j}\mid$ its cardinality. If $i\not=
j$, ${\cal E}_{i,j}$ has no points at infinity; whereas if $i=j$,
it has three points at infinity. Hence
\begin{equation}\label{projectiveeij=cij}
e_{i,j}=c_{i,j}+3\delta_{i,j}.
\end{equation}
By multiplying the equation $Z^3+gX^3=gY^3$ by $g^2$, we get the
curve $g^2Z^3+{X'}^3={Y'}^3$. So, by permuting the variables, we deduce that
$e_{1,1}=e_{2,0}$. Similarly, we obtain  $e_{2,2}=e_{1,0}$. Hence, by (\ref{projectiveeij=cij}) and (\ref{cijlij}) we find that
\begin{equation}\label{ttrfecd}
   \ell_{1,1}=\ell_{2,0},
\end{equation}
\begin{equation}\label{ttrfecd1}
  \ell_{2,2}=\ell_{1,0}.
\end{equation}
Therefore, from (\ref{sumlij}), it follows that
\begin{equation}\label{ttrfeceeed}
 \ell_{2,1}=\ell_{0,0}+1.
\end{equation}
 Furthermore, from  (\ref{projectiveeij=cij}) and (\ref{cijlij}) we have for all $i$, $0\leq i \leq 2$,
 \begin{eqnarray}\label{eijcij}
 % \nonumber to remove numbering (before each equation)
   9\ell_{i,0} &amp;=&amp; c_{i,0}-3\delta_{i,0}-3 \nonumber \\
    &amp;=&amp; e_{i,0}-6\delta_{i,0}-3.
 \end{eqnarray}
Hence, to get all the numbers $\ell_{i,j}$, $0\leq i,j \leq 2$, it suffices to know the values of $e_{i,0}, $   $0  \leq i  \leq 2$.\\
\\
 {\bf Computation of $e_{i,0},~i\in \{0,1,2\}$.}\\
 Here, we are
interested with the curve ${\cal E}_{i,0}:~~Z^3+X^3=g^iY^3$. By
setting $X=9g^iz+2y$, $Y=6x$ and $Z=9g^iz-2y$, we get the
Weierstrass's form
\begin{equation*}
  zy^2=x^3-(27/4)g^{2i}z^3,
\end{equation*}
 which, when divided
by $z^3$, gives the form
\begin{equation*}
y^2=x^3-(27/4)g^{2i}.
\end{equation*}
Let
\begin{equation*}
   y^2=x^3+\alpha x+\beta
\end{equation*}
be the equation of an elliptic curve ${\cal
E}$ defined over $\F_p$. It is well known that
the number of points  of ${\cal
E}$  is equal to
\begin{equation}\label{numberpoints}
\mid {\cal E}\mid =p+1+\sum_{x\in \F_p}\left(\frac {x^3+\alpha
x+\beta}{p}\right),
\end{equation}
where $(\frac {.}{p})$ is the Legendre's symbol.
For $\alpha =0$, the sum $\sum_{x\in \F_p}\left(\frac {x^3+\alpha
x+\beta}{p}\right)$ was investigated by S. A. Katre \cite{Kat85}. He obtained:
\begin{lem}\label{Katre}
Let $p$ be a prime number such that
$p\equiv 1$ (mod $3$). Then there exist a unique  $L$, $L\equiv
1 $ (mod $3$) and a unique $M$ up to a sign such that $4p=L^2+27M^2$. Moreover,
if $\beta $ is an integer $\not=
0$ then
\begin{displaymath}
\sum_{x\in \F_p}\left(\frac {x^3+\beta }{p}\right)=\left\{
\begin{array}{ll} (\frac {\beta }{p})L, &amp; \textrm{if}
 ~4\beta \textrm~{is~ a~ cube~ modulo~ p};\\-\frac {1}{2}(\frac {\beta }{p})(L+9M),&amp; \textrm{otherwise, where $M$  is chosen uniquely  }\\
  &amp;  \textrm{by $\left(4 \beta \right)^{\frac{p-1}{3}}\equiv \frac{L+9M}{L-9M} \pmod p$ }.
\end{array} \right.
\end{displaymath}
\end{lem}
~\\ Thanks to Lemma \ref{Katre}, we can give the values of   $e_{i,0}$ for $0 \leq i \leq 2.$\\
{\bf Computation of $e_{0,0}$.} From (\ref{numberpoints}), since $-27$ is a cube , by using Lemma \ref{Katre} with  $\beta=-27/4$, we obtain
\begin{eqnarray}\label{mmppoo}
% \nonumber to remove numbering (before each equation)
  e_{0,0} &amp;=&amp; p + 1 +\left(\frac{-27/4}{p}\right)L \nonumber \\
   &amp;=&amp; p+1+\left(\frac{-27}{p}\right)L \nonumber\\
   &amp;=&amp; p+1+\left(\frac{-3}{p}\right)^3L\nonumber.
\end{eqnarray}
Since $p\equiv 1$ (mod $3$)  then,  by the quadratic reciprocity law,  $-3$ is a quadratic residue modulo $p$. Hence,
 \begin{equation}\label{mmddppoo}
e_{0,0}=p + 1 +L.
\end{equation}
{\bf Computation of $e_{1,0}$.  } If $\beta=-27g^2/4$ then  $4 \beta=-27g^2$ is not a cube modulo $p$. Hence, by using Lemma \ref{Katre} again, it follows that
\begin{eqnarray}\label{mmlpoiu}
% \nonumber to remove numbering (before each equation)
  e_{1,0} &amp;=&amp; p + 1 -\frac{1}{2}\left(\frac{-27g^2/4 }{p}\right)(L+9M) \nonumber \\
   &amp;=&amp; p + 1 -\frac{1}{2}(L+9M),
\end{eqnarray}
where the sign of $M$ is given by
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
  \left(-27g^{2}\right)^{(p-1)/3} &amp;\equiv &amp;  (g^{2})^{(p-1)/3}  \pmod  p \nonumber \\
 &amp; \equiv &amp; \frac{L+9M}{L-9M}  \pmod  p.
\end{eqnarray*}
{\bf Computation of  $e_{2,0}$.}
Let  $M$ be fixed by last congruence, and let $\beta=-27g^4/4$. Since $\left(g^{4}\right)^{(p-1)/3}\not\equiv \left(g^{2}\right)^{(p-1)/3} $ (mod $p$), Lemma \ref{Katre} implies that
\begin{equation}\label{vkjhgfer}
e_{2,0}=p + 1  -\frac{1}{2}(L-9M).
 \end{equation}










\section{The explicit form of $R_1$ when $s_p=\frac {p-1}{3}$.}
First, we remark that the polynomial $R_1(y)$ given by (\ref{sfre}) can also be defined (cf. (\ref{sfre22}), (\ref{nbsqz})) by
\begin{equation}\label{R1*1/sp(y)}
\left(R_1(y)\right)^{1/s_p}=\prod_{i \in \{0,1,2\}}~\biggl( y+T_i \biggr),
\end{equation}
where $\theta\neq 1 $ is some $p$-th root of unity.
\begin{thm}\label{jhgvcsd} Let $p$ be an odd prime such that $s_p=\frac
{p-1}{3}$. Then the polynomial $R_1$ given by (\ref{sfre}) or  (\ref{R1*1/sp(y)}) is equal
to
\begin{equation*}
R_1(y)=\left(y^3-y^2-s_py+\lambda_p\right)^{s_p},
\end{equation*}
with
\begin{equation*}
\lambda_p=\frac{p(L+3)-1}{27},
\end{equation*}
where $L$ is the unique integer satisfying $4p=L^2+27M^2$ and $L\equiv 1$ (mod $3$).
\end{thm}
\begin{remark} $p(L+3) \equiv 1$ (mod $27$) follows easily from the congruences $L\equiv 1$ (mod $3$) and $p \equiv L^2 $ (mod $27$).
\end{remark}
\begin{proof}[Proof of Theorem \ref{jhgvcsd}]
From (\ref{R1*1/sp(y)}), we have
\begin{equation*}
R_1(y)= \biggl( (y+T_0)(y+T_1)(y+T_2)\biggr)^{s_p}.
\end{equation*}
This can be written as
\begin{equation*}
R_1(y)= \biggl(
y^3+\lambda''_py^2+\lambda'_py+\lambda_p\biggr)^{s_p},
\end{equation*}
with
\begin{equation}\label{lambdap}
\lambda_p=T_0T_1T_2,
\end{equation}
\begin{equation}\label{lambdap'}
\lambda'_p=T_0T_1+T_0T_2+T_1T_2,
\end{equation}
\begin{equation*}
\lambda''_p=T_0+T_1+T_2.
\end{equation*}
We begin by calculating $\lambda''_p$. It follows immediately from (\ref{hgtrvcdes}) and (\ref{orbital}) that
\begin{eqnarray}\label{trefc}
% \nonumber to remove numbering (before each equation)
  \lambda''_p &amp;=&amp; T_0+T_1+T_2 \nonumber\\
   &amp;=&amp; \sum_{i=0}^2\sum_{k=0}^{s_p-1}\theta^{2^kg^i}  \nonumber \\
   &amp;=&amp; \sum_{j=1}^{p-1}\theta^{j}  \nonumber \\
   &amp;=&amp; -1,
\end{eqnarray}
since cf. Remark \ref{hhuuggg}, $\theta $ is a primitive  $p$-th root of unity.\\

 Now,
let us prove that $\lambda_p'=-s_p$. From (\ref{lambdap'}) and (\ref{hgtrvcdes}), we have
\begin{equation}\label{dscss}
 \lambda'_p=\sum_{0  \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}g^2}+
 \sum_{0  \leq k,k' \leq s_p-1}\theta^{2^kg^2+2^{k'}}+\sum_{0  \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}}.
\end{equation}
To treat the last sum in (\ref{dscss}), let us fix  $k$ and $k'$
in $\left\{0,1,\ldots,s_p-1\right\}$. We have
$\theta^{2^kg+2^{k'}}=\theta^{2^{k'}(1+2^{k-k'}g)}$. Since
$2^{k-k'}g\in {\it O}(g)$ then, from the second equality in
(\ref{lknb}), there exists a unique $t\in \{0,1,\ldots,s_p-1\}$ such
that $2^{k-k'}g=g^{1+3t}$. Hence, the last sum in (\ref{dscss})
becomes
\begin{equation}\label{jhgfv 3}
\sum_{0  \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}}=\sum_{0  \leq
k',t \leq s_p-1}\theta^{2^{k'}(1+g^{1+3t})}.
\end{equation}
For the first and second sums in (\ref{dscss}), arguing as above,
we get
\begin{eqnarray}\label{jhgfv 2}
% \nonumber to remove numbering (before each equation)
  \sum_{0  \leq k,k' \leq s_p-1}\theta^{2^kg^2+2^{k'}} &amp;=&amp; \sum_{0  \leq
k,k' \leq s_p-1}\theta^{2^{k'}(1+2^{k-k'}g^2)} \nonumber \\
   &amp;=&amp; \sum_{0 \leq {k'},t
\leq s_p-1}\theta^{2^{k'}(1+g^{2+3t})}
\end{eqnarray}
and
\begin{eqnarray}\label{jhgfv}
% \nonumber to remove numbering (before each equation)
  \sum_{0  \leq k,k' \leq s_p-1}\theta^{2^kg+2^{k'}g^2} &amp;=&amp; \sum_{0 \leq
k,k' \leq s_p-1}\theta^{2^k(g+2^{k'-k}g^2)} \nonumber \\
   &amp;=&amp; \sum_{0 \leq k,t \leq
s_p-1}\theta^{2^k(g+g^{2+3t})}.
\end{eqnarray}
Now, from (\ref{vfer}), (\ref{ttrfecd}), (\ref{ttrfecd1}), (\ref{hgtrvcdes})   and Remark \ref{remllll}, we obtain
\begin{eqnarray}\label{uytrtg}
% \nonumber to remove numbering (before each equation)
 \sum_{0  \leq k',t \leq
s_p-1}\theta^{2^{k'}(1+g^{1+3t})} &amp;=&amp;\sum_{i=0}^{2}\ell_{i,1}T_i  \nonumber\\
  &amp;=&amp; \ell_{1,0}T_0+ \ell_{2,0}T_1+\ell_{2,1}T_2
\end{eqnarray}
and
\begin{eqnarray}\label{uytrtg 2}
% \nonumber to remove numbering (before each equation)
  \sum_{0  \leq k',t \leq
s_p-1}\theta^{2^{k'}(1+g^{2+3t})} &amp;=&amp; \sum_{i=0}^{2}\ell_{i,2}T_i\nonumber \\
   &amp;=&amp; \ell_{2,0}T_0+\ell_{2,1} T_1+\ell_{1,0}T_2 .
\end{eqnarray}
On the other hand, since for all $v\geq 0$, $0 \leq i,j \leq 2$,
\begin{equation*}
 1+g^{j+3t} \in {\it O}(g^i)\Longleftrightarrow  g^v+g^{v+j+3t} \in {\it O}(g^{v+i}),
\end{equation*}
again by (\ref{vfer}),  (\ref{ttrfecd}), (\ref{hgtrvcdes})  and Remark \ref{remllll}, we get
\begin{eqnarray}\label{uytrtg 3}
% \nonumber to remove numbering (before each equation)
  \sum_{0  \leq k,t \leq s_p-1}\theta^{2^k(g+g^{2+3t})} &amp;=&amp; \sum_{i=0}^{2}\ell_{i,1}T_{1+i}\nonumber \\
   &amp;=&amp; \ell_{2,1}T_0+ \ell_{1,0}T_1+\ell_{2,0}T_2.
\end{eqnarray}
Clearly, from (\ref{sumlij})  and (\ref{ttrfeceeed}), one can deduce that
\begin{equation}\label{l0+l1+l2}
\ell_{1,0}+\ell_{2,0}+\ell_{2,1}=s_p,
\end{equation}
which,  by
(\ref{dscss})-(\ref{uytrtg 3}) and (\ref{trefc}), gives
\begin{eqnarray}\label{lknbv}
% \nonumber to remove numbering (before each equation)
 \lambda'_p &amp;=&amp; s_p\left(T_0+T_1+T_2\right) \nonumber \\
   &amp;=&amp; s_p\sum_{j=1}^{p-1}\theta^j \nonumber \\
  &amp;=&amp; -s_p.
\end{eqnarray}

Finally, let us calculate $\lambda_p$. From (\ref{lambdap}) and
(\ref{hgtrvcdes}), we have
\begin{eqnarray}\label{lambdap2}
% \nonumber to remove numbering (before each equation)
  \lambda_p &amp;=&amp; \sum_{0\leq k,k',k"\leq
s_p-1}\theta^{2^k+2^{k'}g+2^{k"}g^2} \nonumber\\
   &amp;=&amp; \sum_{0\leq k\leq
s_p-1}\theta^{2^k}\left(\sum_{0\leq k',k"\leq
s_p-1}\theta^{2^{k'}g+2^{k"}g^2}\right).
\end{eqnarray}
Hence, by (\ref{jhgfv}), (\ref{uytrtg 3}) and (\ref{hgtrvcdes}), we get
\begin{eqnarray*}
\lambda_p &amp;=&amp;\sum_{0\leq k\leq s_p-1}\theta^{2^k}\left(\ell_{2,1}\sum_{j
\in {\it O}(1)}\theta^j+\ell_{1,0}\sum_{j \in {\it
O}(g)}\theta^j+\ell_{2,0}\sum_{j \in {\it
O}(g^2)}\theta^j\right)\nonumber\\ &amp;=&amp;\ell_{2,1}\sum_{0\leq k,k'\leq
s_p-1}\theta^{2^k+2^{k'}}+\ell_{1,0}\sum_{0\leq k,k'\leq
s_p-1}\theta^{2^k+2^{k'}g}+\ell_{2,0}\sum_{0\leq k,k'\leq
s_p-1}\theta^{2^k+2^{k'}g^2}.
\end{eqnarray*}
Consequently, from (\ref{jhgfv 3}), (\ref{uytrtg}), (\ref{jhgfv
2}) and (\ref{uytrtg 2}), it happens that
\begin{eqnarray}\label{fgds21}
\lambda_p&amp;=&amp;\ell_{2,1}\sum_{0\leq k,k'\leq
s_p-1}\theta^{2^k+2^{k'}}+(\ell_{1,0}^2+\ell_{2,0}^2)T_0+(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1})T_1\nonumber\\ &amp; &amp; +(\ell_{1,0}\ell_{2,1}+\ell_{1,0}\ell_{2,0})T_2.
\end{eqnarray}
Since $2^{k'-k}\in {\it O}(1)= {\it O}(g^3)$ then, cf.
(\ref{lknb}), there exists a unique  $t \in
\left\{0,1,\ldots,s_p-1\right\}$ such that $2^{k'-k}=g^{3t}$.
Hence
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
  \sum_{0\leq k,k'\leq s_p-1}\theta^{2^k+2^{k'}} &amp;=&amp; \sum_{0\leq
k,k'\leq s_p-1}\theta^{2^k(1+2^{k'-k})} \nonumber \\
   &amp;=&amp;  \sum_{0\leq k,t\leq
s_p-1}\theta^{2^k(1+g^{3t})}.
\end{eqnarray*}
Now, we recall that cf. Remark \ref{remllll}  there exists one and only one  $t\in
\left\{0,1,\ldots,s_p-1\right\}$ satisfying $1+g^{3t} \in {\it
O}(p)$. Consequently, by (\ref{vfer}) and (\ref{hgtrvcdes}), we get
\begin{eqnarray}\label{kjf11}
% \nonumber to remove numbering (before each equation)
  \sum_{0\leq
k,k'\leq s_p-1}\theta^{2^k+2^{k'}} &amp;=&amp; \sum_{0\leq k,t\leq
s_p-1}\theta^{2^k(1+g^{3t})} \nonumber \\
  &amp;=&amp; \ell_{0,0}T_0+ \ell_{1,0}T_1+\ell_{2,0}T_2+s_p.
\end{eqnarray}
Hence, (\ref{fgds21}) gives
\begin{eqnarray}\label{lambdapvaleur1}
\lambda_p&amp;=&amp;(\ell_{1,0}^2+\ell_{2,0}^2+\ell_{0,0}\ell_{2,1})T_0
+(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1}+\ell_{1,0}\ell_{2,1})T_1\nonumber\\
&amp;&amp;+(\ell_{1,0}\ell_{2,0}+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1})T_2+\ell_{2,1}s_p.
\end{eqnarray}
On the other hand, from (\ref{lambdap2}), by changing the order of summation,  $\lambda_p$ can be written   as
\begin{equation*}
\lambda_p=\sum_{0\leq k'\leq
s_p-1}\theta^{2^{k'}g}\left(\sum_{0\leq k,k"\leq
s_p-1}\theta^{2^{k}+2^{k"}g^2}\right)
\end{equation*}
and  we get in the way as above
\begin{eqnarray}\label{lambdapvaleur2}
\lambda_p&amp;=&amp;(\ell_{1,0}\ell_{2,0}+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1})T_0
+(\ell_{1,0}^2+\ell_{2,0}^2+\ell_{0,0}\ell_{2,1})T_1\nonumber\\
    &amp; &amp; +(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1}+\ell_{1,0}\ell_{2,1})T_2+\ell_{2,1}s_p
\end{eqnarray}
Whereas, if we write $\lambda_p$ in the form
\begin{equation*}
\lambda_p=\sum_{0\leq k"\leq
s_p-1}\theta^{2^{k"}g^2}\left(\sum_{0\leq k,k'\leq
s_p-1}\theta^{2^{k}+2^{k'}g}\right),
\end{equation*}
 we get
\begin{eqnarray}\label{lambdapvaleur3}
\lambda_p&amp;=&amp;(\ell_{1,0}\ell_{2,0}+\ell_{2,0}\ell_{2,1}+\ell_{1,0}\ell_{2,1})T_0
+(\ell_{1,0}\ell_{2,0}+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1})T_1\nonumber\\
    &amp; &amp; +(\ell_{1,0}^2+\ell_{2,0}^2+\ell_{0,0}\ell_{2,1})T_2+\ell_{2,1}s_p.
\end{eqnarray}
By summing (\ref{lambdapvaleur1}), (\ref{lambdapvaleur2}) and
(\ref{lambdapvaleur3}), we obtain
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
 3\lambda_p &amp;=&amp; (\ell_{1,0}^2+\ell_{2,0}^2+2\ell_{1,0}\ell_{2,0}+2\ell_{1,0}\ell_{2,1}+2\ell_{2,0}\ell_{2,1}+\ell_{0,0}\ell_{2,1})\times \nonumber \\
&amp; &amp;(T_0+T_1+T_2)+3\ell_{2,1}s_p.
\end{eqnarray*}
But, according to
(\ref{trefc}), $T_0+T_1+T_2=-1$. Hence,
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
 3\lambda_p&amp;=&amp; - (\ell_{1,0}^2+\ell_{2,0}^2+2\ell_{1,0}\ell_{2,0}+2\ell_{1,0}\ell_{2,1}+2\ell_{2,0}\ell_{2,1}+\ell_{0,0}\ell_{2,1})+3\ell_{2,1}s_p \\
 &amp;=&amp; -\left((\ell_{1,0}+\ell_{2,0})^2+\ell_{1,0}\ell_{2,1}+\ell_{2,0}\ell_{2,1}+\ell_{2,1}(\ell_{0,0}+\ell_{1,0}+\ell_{2,0})\right)+3\ell_{2,1}s_p.
\end{eqnarray*}
So that, from (\ref{l0+l1+l2}) and (\ref{sumlij}),  we obtain
$$3\lambda_p=-(s_p-\ell_{2,1})^2-\ell_{2,1}(s_p-\ell_{2,1})-\ell_{2,1}(s_p-1)+3\ell_{2,1}s_p,$$ which
gives $\lambda_p=\frac {(3s_p+1)\ell_{2,1}-s_p^2}{3}=\frac {p\ell_{2,1}-s_p^2}{3}$. Finally, using the value of $\ell_{2,1}$:
 \begin{equation}\label{1kkjjhhggfv}
  \ell_{2,1}=\frac{1}{9}(p+1+L)
\end{equation}
which follows from  (\ref{ttrfeceeed}),   (\ref{eijcij}) and (\ref{mmddppoo}), we complete the
proof of Theorem \ref{jhgvcsd}.
\end{proof}

In the following table we give $L$, $g$, $M$ and  $R_1(y)$ when $p\leq 1000.$
$$\begin{tabular}{|l|l|l|l|l|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
   $p$ &amp; $L$ &amp;$g$ &amp; $M$ &amp;$R_1(y)$ \\ \hline
  $43$ &amp;   $$-8$$ &amp; $$3$$&amp; $$-2$$ &amp;  $(y^3  - y^2  - 14 y - 8)^{14}$ \\ \hline
   $109$ &amp;   $$-2$$ &amp;$$6$$ &amp; $$4$$ &amp;  $(y^3  - y^2  - 36 y +4)^{36}$ \\ \hline
$157$&amp; $$-14$$ &amp; $$5$$&amp; $$4$$ &amp; $ (y^3  - y^2  - 52 y - 64)^{52}$ \\ \hline
 $229$ &amp;  $$22$$ &amp; $$6$$ &amp; $$4$$ &amp;   $(y^3  - y^2  - 76 y +212)^{76}$ \\ \hline
 $ 277$&amp; $$-26$$ &amp; $$5$$&amp;$$-4$$ &amp; $(y^3  - y^2  - 92 y - 236)^{   92}$ \\ \hline
$283$ &amp; $$-32$$ &amp; $$3$$&amp; $$-2$$ &amp; $(y^3  - y^2  - 94 y - 304)^{ 94}$ \\ \hline
$307$ &amp; $$16$$ &amp; $$5$$&amp;$$-6$$ &amp;  $(y^3  - y^2  - 102 y + 216)^{ 102}$ \\ \hline
$499$ &amp;  $$-32$$ &amp; $$7$$ &amp;$$-6$$ &amp; $(y^3  - y^2  - 166 y -536)^{ 166}$ \\ \hline
 $643$ &amp; $$40$$ &amp; $$11$$&amp;$$6$$ &amp; $(y^3  - y^2  - 214 y + 1024)^{ 214}$ \\ \hline
$691$ &amp; $$-8$$ &amp; $$3$$&amp;$$10$$ &amp;  $(y^3  - y^2  - 230 y - 128)^{   230}$ \\ \hline
 $733$ &amp; $$-50$$ &amp; $$6$$&amp;$$4$$ &amp; $(y^3  - y^2  - 244 y - 1276)^{244}$\\ \hline
 $739$ &amp; $$16$$ &amp; $$3$$&amp;$$10$$ &amp; $(y^3  - y^2  - 246 y + 520)^{246}$ \\ \hline
$811$ &amp; $$-56$$ &amp; $$3$$&amp;$$-2$$ &amp;$(y^3 - y^2  - 270 y - 1592)^{270}$ \\ \hline
$997$ &amp;  $$10$$ &amp; $$7$$&amp;$$-12$$ &amp; $(y^3 - y^2  - 332 y + 480)^{  332}$  \\
  \hline
\end{tabular}$$
\begin{center}
xxxx
\end{center}

\section{The explicit form of $R_m$ when $s_p=\frac {p-1}{3}$ and $m\geq 3$.}
We remind that if $m\in {\it O}(p)$ or $m$ is divisible by some
prime $q$ belonging to ${\it O}(1)$, then the polynomial $R_m$ is
given by Theorem \ref{sfj14jjjj}, 2) and 3). Let $m$ be an odd integer $\geq
3$ such that all its prime divisors are in ${\it O}(g)\cup{\it
O}(g^2)$. For $i \in \{1,2\}$, we denote by $\omega_i$ the
arithmetic function which counts the number of distinct prime
divisors belonging to ${\it O}(g^i)$ of an integer, i.e.,
\begin{equation}\label{arith 0}
\omega_i(n)=\sum_{q~\text {prime},\ q\in{\it O}(g^i),~q \dv n}1.
\end{equation}
Let the decomposition of $m$ into irreducible factors be
\begin{equation}\label{jhgvfre}
m=q_{1,1}^{\gamma_{1,1}}q_{1,2}^{\gamma_{1,2}}\cdots
q_{1,\omega_1}^{\gamma_{1,\omega_1}}q_{2,1}^{\gamma_{2,2}}q_{2,2}^{\gamma_{2,2}}\cdots
q_{2,\omega_2}^{\gamma_{2,\omega_2}},
\end{equation}
where $\omega_i=\omega_i(m)$, $\omega =\omega
(m)=\omega_1+\omega_2$ and $q_{i,j}\in {\it O}(g^i)$.\\

 We shall
begin with some result concerning binomial coefficients:
\begin{lem}\label{lemmas d}
For all $n \in \N$ and all $j$ , $0 \leq j \leq 2$,
\begin{equation}\label{cdsew10}
\sum_{k\geq
0}\binom{n}{3k+j}(-1)^{k+j}=2.3^{\frac{n}{2}-1}\cos(\frac{n
\pi}{6}+\frac{2j\pi}{3}).
\end{equation}
\end{lem}
\begin{proof} Let $z_1=e^{(2i\pi)/3}$ and $z_2=e^{(4i\pi)/3}$ be the two cubic primitive roots of
unity, and let $f(z)=\sum_{k\geq 0}a_kz^k$ be some convergent power
series. Since for all $j$, $0 \leq j \leq 2$,
\begin{equation*}
\frac{1+z_1^{n-j}+z_2^{n-j}}{3}=\begin{cases} 1,
~~\text{if}~n\equiv  j \pmod 3;\\ 0,~~\text{otherwise
},\end{cases}
\end{equation*}
it follows that
\begin{equation*}
\frac{f(z) + \frac{1}{z_1^j}f(z_1z) +\frac{1}{z_2^j} f(z_2z)}
{3}=\sum_{k\geq 0}a_{3k+j}z^{3k+j}.
\end{equation*}
Hence, defining  $g_j(z)$, $0 \leq j \leq 2$, by
\begin{equation*}
g_j(z)= \sum_{k\geq 0}\binom{n}{3k+j}z^{3k+j},
\end{equation*}
 and taking $f(z)=(1+z)^n$, we get
\begin{equation*}
g_j(z)=\frac{f(z) + \frac{1}{z_1^j}f(z_1z) +\frac{1}{z_2^j}
f(z_2z)} {3}.
\end{equation*}
By making the substitution $z=-1$, we obtain
\begin{eqnarray*}
g_j(-1)&amp;=&amp;\sum_{k\geq
0}\binom{n}{3k+j}(-1)^{k+j}\\
&amp;=&amp;\frac{\frac{1}{z_1^j}(1-z_1)^n
+\frac{1}{z_2^j} (1-z_2)^n}{3}\nonumber \\
&amp;=&amp;\frac{1}{3}\left\{\frac{1}{z_1^j}\left(\frac{3-i\sqrt{3}}{2}\right)^n+\frac{1}{z_2^j}
\left(\frac{3+i\sqrt{3}}{2}\right)^n\right\}.
\end{eqnarray*}
To get (\ref{cdsew10}), we need only transform the right hand-side
of  the last equality.
\end{proof}
\begin{cor}\label{nbcdse2} Let $m$ be an odd integer $\geq 3$ of
the form (\ref{jhgvfre}), and let  $\alpha_h (m)$ be the quantity
defined by (\ref{vfrd}).  For all $h$, $0\leq h \leq 2$, we have
\begin{equation}\label{sansd 3}
\alpha_h(m)=\eta(m) \cos
\left((\omega_2-\omega_1)\frac{\pi}{6}+4h\frac{\pi}{3}\right)
\end{equation}
where
\begin{equation}\label{sadsqnsd 3}
\eta(m)=2.3^{\frac{\omega }{2}-1}.
\end{equation}
\end{cor}
\begin{proof}
From (\ref{vfrd}), for all $h$, $0\leq h \leq 2$, we have
\begin{equation*}
\alpha_h(m)= \sum_{\substack{d\dv \widetilde{m},~d \in {\it
O}(g^h)}} \mu(d).
\end{equation*}
First, let us  suppose that $\omega_1\neq 0$ and $\omega_2 \neq0$. By (\ref{arith 0}) and (\ref{jhgvfre}), we obtain
that for all  $h$, $0 \leq h \leq 2$,
\begin{equation*}
\alpha_h(m)=\sum_{i_1=
0}^{\omega_1}(-1)^{i_1}\binom{\omega_1}{i_1}\sum_{\substack{i_2=0\\i_2
\equiv i_1+2h \pmod
 3}}^{\omega_2}(-1)^{i_2}\binom{\omega_2}{i_2}.
\end{equation*}
So that, by (\ref{cdsew10}), we get
\begin{eqnarray*}
\alpha_h(m)&amp;=&amp;\sum_{i_1=0}^{\omega_1}(-1)^{i_1}\binom{\omega_1}{i_1}2.3^{\frac{\omega_2}{2}-1}\cos\left(\frac{\omega_2
\pi}{6}+\frac{2(i_1+2h)\pi}{3}\right)\nonumber \\
&amp;=&amp;2.3^{\frac{\omega_2}{2}-1}\sum_{j=0}^{2}\cos\left(\frac{\omega_2
\pi}{6}+\frac{2(j+2h)\pi}{3}\right)\sum_{\substack{i_1=0\\i_1
\equiv j \pmod 3}}^{\omega_1}(-1)^{i_1}\binom{\omega_1}{i_1},
\end{eqnarray*}
which, by (\ref{cdsew10}) again, gives
\begin{eqnarray*}
\alpha_h(m)&amp;=&amp;4.3^{\frac{\omega
}{2}-2}\sum_{j=0}^{2}\cos\left(\frac{\omega_2
\pi}{6}+\frac{2(j+2h)\pi}{3}\right)\cos\left(\frac{\omega_1
\pi}{6}+\frac{2j\pi}{3}\right).
\end{eqnarray*}
Consequently, to get (\ref{sansd 3}), one need only use the
elementary trigonometric formulas
\begin{equation*}
\cos a\cos b=\frac{1}{2}\left( \cos(a+b)+
\cos(a-b)\right)~\textrm{for all}~a ~ \textrm{and }~b ~\textrm{in}
~\R
\end{equation*}
and
\begin{equation*}
\cos
c+\cos(c+\frac{2\pi}{3})+\cos(c+\frac{4\pi}{3})=0,~\textrm{for
all}~c \in \R.
\end{equation*}
In case $\omega_1=$ or $\omega_2=0$, (\ref{sansd 3}) follows immediately from (\ref{cdsew10}).
\end{proof}

\begin{thm}\label{hgfdrce} Let $m$ be an odd integer $\geq 3$ of the form (\ref {jhgvfre}). Let $\eta (m)$ be
as defined in (\ref{sadsqnsd 3}), and let  $R_m$ be the polynomial given
by (\ref{sfre}). Then
\begin{equation}\label{Rm1(y)}
R_m(y)=\left(m^3y^3-\frac{3}{4}pm\eta^2(m)y+\nu_p\right)^{s_p},
\end{equation}
with
\begin{equation}\label{1111444mmmpp}
\nu_p=\left\{%
\begin{array}{lll}
\frac{1}{8}\left(-1\right)^{\frac{\omega_2-\omega_1}{2}}p\eta^3(m)L, &amp;
\hbox{\textrm{if $\omega_2-\omega_1$ is even }}; \\ &amp;   \\
\frac{3\sqrt{3}}{8}\left(-1\right)^{\frac{\omega_2-\omega_1-1}{2}}p\eta^3(m)M,
&amp; \hbox{\textrm{if $\omega_2-\omega_1$ is odd,}} \\
\end{array}%
\right.
\end{equation}
where $L$ and $M$ are the unique integers satisfying $4p=L^2+27M^2$,  $L\equiv 1$ (mod $3$) and $(g^{2})^{(p-1)/3} \equiv \frac{L+9M}{L-9M}$ (mod $p$).
\end{thm}
\begin{proof}
From (\ref{Rm(yy)}), we have
\begin{eqnarray}\label{Rm(y)}
R_m(y)&amp;=&amp;\prod_{l=0}^{2}\left(my+A_m(\theta^{g^l})\right)^{s_p}\nonumber\\
 &amp;=&amp;
\left(m^3y^3+m^2\nu''_py^2+m\nu'_py+\nu_p\right)^{s_p},
\end{eqnarray}
where
\begin{equation}\label{dfcsve, 1}
\nu''_p=A_m(\theta)+A_m(\theta^{g})+A_m(\theta^{g^2}),
\end{equation}
\begin{equation}\label{dfcsve, 2}
\nu'_p=A_m(\theta)A_m(\theta^{g})+A_m(\theta)A_m(\theta^{g^2})+A_m(\theta^{g})A_m(\theta^{g^2})
\end{equation}
and
\begin{equation}\label{dfcsve, 3}
\nu_p=A_m(\theta)A_m(\theta^{g})A_m(\theta^{g^2}).
\end{equation}
Recall that cf. Theorem \ref{sfj14jjjj}, $\delta$ is the unique integer in $
\left\{0,1,2\right\}$ such that $m\in {\it O}(g^\delta)$. So that
from (\ref{deee})-(\ref{dee}) and (\ref{hgtrvcdes}), we get for $i \in\{0,1,2\}$
\begin{equation}\label{hhttrefd}
    A_m(\theta^{g^i})=\sum_{h=0}^{2}\alpha_h(m)T_{\delta -h+i  }.
\end{equation}
{\bf Computation of $\nu''_p$.}\\
From (\ref{hhttrefd}) and  (\ref{dfcsve, 1}) we
deduce that
\begin{equation*}
\nu''_p=\biggl(\alpha_0(m)+\alpha_1(m)+\alpha_2(m)\biggr)\biggl(T_0+T_1+T_2\biggr).
\end{equation*}
Since $\gcd (m,p)=1$ and $m\neq 1$, it follows immediately from
(\ref{vfrd}) and (\ref{orbital}) that
\begin{equation}\label{lkytmlnb}
\alpha_0(m)+\alpha_1(m)+\alpha_2(m)= \sum_{d\dv \widetilde{m}}
\mu(d)=0
\end{equation}
and thus  $\nu''_p=0$.\\
{\bf Computation of $\nu'_p$.}\\
 From (\ref{Rm(y)}), to prove
(\ref{Rm1(y)}) it suffices to show (\ref{1111444mmmpp}) and that
$\nu_p'=\frac {-3}{4}p\eta^2(m)$.
 By (\ref{dfcsve, 2}) and
(\ref{hhttrefd}), we have
\begin{equation*}
  \nu'_p=\sum_{k=0}^{2} \sum_{h=0}^{k}\alpha_h(m)\alpha_k(m) U(h,k)
\end{equation*}
with
\begin{equation*}
  U(h,h)=\sum_{(i,j)\in \{(0,1),(0,2),(1,2)\}}T_{\delta-h+i}T_{\delta-h+j}
\end{equation*}
and, for $h&lt; k$,
\begin{equation*}
  U(h,k)=\sum_{(i,j)\in \{(0,1),(0,2),(1,2)\}}\left( T_{\delta-h+i}T_{\delta-k+j}+ T_{\delta-k+i}T_{\delta-h+j} \right).
\end{equation*}
Observing that, for $0\leq \delta, h, k\leq 2$, $U(h,k)$ does not depend on $\delta$ and is equal to $T_0T_1+T_0T_2+T_1T_2$ when $h=k$ and to $T_0T_1+T_0T_2+T_1T_2+T_0^2+T_1^2+T_2^2$ when $h&lt;k$, we obtain
\begin{equation}\label{bvrtrdydk1}
\nu'_p=\beta(m) \biggl( T_0^2+T_1^2+T_2^2\biggr)+
\beta'(m) \biggl(T_0 T_1+T_0T_2+T_1T_2\biggr),
\end{equation}
where
\begin{equation*}
\beta(m)=
\alpha_0(m)\alpha_1(m)+\alpha_0(m)\alpha_2(m)+\alpha_1(m)\alpha_2(m)
\end{equation*}
and
\begin{equation*}
\beta'(m)=\alpha_0^2(m)+\alpha_1^2(m)+\alpha_2^2(m)+\beta(m).
\end{equation*}
From  (\ref{sansd 3}), it is easy to check that
\begin{equation*}
\beta(m)=-\frac{3}{4}\eta^2(m).
\end{equation*}
By (\ref{lkytmlnb}), we find that
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
  \beta'(m) &amp;=&amp; \left(\sum_{i=0}^2\alpha_i(m)\right)^2-2\beta(m)+\beta(m)\nonumber \\
   &amp;=&amp; -\beta(m)\nonumber \\
   &amp;=&amp; \frac{3}{4}\eta^2(m).
\end{eqnarray*}
On the other hand, using  (\ref{hgtrvcdes}), we get
\begin{equation*}
T_0^2+T_1^2+T_2^2=
\biggl(\sum_{k=0}^{s_p-1}\theta^{2^k}\biggr)^2+\biggl(\sum_{k=0}^{s_p-1}
\theta^{2^kg}\biggr)^2+\biggl(\sum_{k=0}^{s_p-1}\theta^{2^kg^2}\biggr)^2.
\end{equation*}
The first sum in the last equality is, by (\ref{kjf11}), equal to
\begin{eqnarray}\label{hgftrvcds}
% \nonumber to remove numbering (before each equation)
 T_0^2 &amp;=&amp; \sum_{0\leq k,k'\leq
s_p-1}\theta^{2^k+2^{k'}} \nonumber \\
   &amp;=&amp; \ell_{0,0}T_0+\ell_{1,0}T_1+\ell_{2,0}T_2+s_p.
\end{eqnarray}
 Similarly, for the second and third sums, we obtain
\begin{eqnarray}\label{hgftrvcdsppo}
% \nonumber to remove numbering (before each equation)
  T_1^2 &amp;=&amp; \sum_{0\leq k,k'\leq s_p-1}\theta^{2^kg+2^{k'}g} \nonumber \\
   &amp;=&amp; \ell_{0,0}T_1+\ell_{1,0}T_2+\ell_{2,0}T_0+s_p
\end{eqnarray}
and
\begin{eqnarray}\label{hgftrvttrscccds}
% \nonumber to remove numbering (before each equation)
  T_2^2 &amp;=&amp; \sum_{0\leq k,k'\leq s_p-1}\theta^{2^kg^2+2^{k'}g^2} \nonumber \\
   &amp;=&amp; \ell_{0,0}T_2+\ell_{1,0}T_0+\ell_{2,0}T_1+s_p.
\end{eqnarray}
Consequently,
\begin{eqnarray*}
T_0^2+T_1^2+T_2^2=3s_p+(\ell_{0,0}+\ell_{1,0}+\ell_{2,0})(T_0+T_1+T_2).
\end{eqnarray*}
So that, by (\ref{sumlij}) and (\ref{trefc}), we
get
\begin{eqnarray}\label{jqshn111}
T_0^2+T_1^2+T_2^2&amp;=&amp;3s_p-(s_p-1)\nonumber \\
&amp;=&amp; 2s_p+1.
\end{eqnarray}
Therefore, with the use of (\ref{bvrtrdydk1}),
(\ref{lambdap'}) and the fact that $s_p=\frac
{p-1}{3}$, we obtain
\begin{equation*}
\nu'_p=-\frac{3}{4}p\eta^2(m).
\end{equation*}
{\bf Computation of $\nu_p$.}\\
 By (\ref{dfcsve,
3}) and (\ref{hhttrefd}), we obtain
\begin{equation*}
    \nu_p=\sum_{h,k,t \in\{0,1,2\}}\alpha_h(m) \alpha_k(m) \alpha_t(m)T_{\delta-h}T_{\delta-k+1}T_{\delta-t+2}
\end{equation*}
and by observing the $27$ terms of the expansion of the above sum, we find that
\begin{eqnarray}\label{lkjhgf2}
\nu_p&amp;=&amp;\gamma_1(m)\biggl(T_0T_1^2+T_1T_2^2 +T_2 T_0^2\biggr)
+\gamma_2(m)\biggl(T_0T_2^2+T_1T_0^2+T_2T_1^2\biggr)\nonumber\\
&amp;
&amp;+\gamma_3(m)\biggl(T_0^3+T_1^3+T_2^3\biggr)+\gamma_4(m)
T_0T_1T_2,
\end{eqnarray}
where
\begin{equation}\label{nbvcgfd}
\gamma_1(m)=
\alpha_0^2(m)\alpha_1(m)+\alpha_0(m)\alpha_2^2(m)+\alpha_1^2(m)\alpha_2(m),
\end{equation}
\begin{equation}\label{nbvcgsfd}
\gamma_2(m)=\alpha_0^2(m)\alpha_2(m)+\alpha_0(m)\alpha_1^2(m)+\alpha_1(m)\alpha_2^2(m),
\end{equation}
\begin{equation}\label{nbvcgssssfd}
\gamma_3(m)=\alpha_0(m)\alpha_1(m)\alpha_2(m)
\end{equation}
and
\begin{equation}\label{nbvsscgfd}
\gamma_4(m)=\alpha_0^3(m)+\alpha_1^3(m)+\alpha_2^3(m)+3\gamma_3(m).
\end{equation}
Using (\ref{hgftrvcds})-(\ref{hgftrvttrscccds}), we get
\begin{equation*}
T_0^3=\ell_{0,0}T_0^2+\ell_{1,0}T_0T_1+\ell_{2,0}T_0T_2+s_pT_0,
\end{equation*}
\begin{equation*}
T_1^3=\ell_{0,0}T_1^2+\ell_{1,0}T_1T_2+\ell_{2,0}T_0T_1+s_pT_1
\end{equation*}
and
\begin{equation*}
T_2^3=\ell_{0,0}T_2^2+\ell_{1,0}T_0T_2+\ell_{2,0}T_1T_2+s_pT_2.
\end{equation*}
Therefore, $$T_0^3+
T_1^3+T_2^3=s_p\biggl(T_0+T_1+T_2\biggr)+
\ell_{0,0}\biggl(T_0^2+T_1^2+T_2^2\biggr)$$
$$+(\ell_{1,0}+\ell_{2,0})\biggl(T_0T_1+T_0T_2+T_1T_2\biggr).$$
So that, from   (\ref {trefc}), (\ref{jqshn111}) and
(\ref {lambdap'}) , we get $$ T_0^3+
T_1^3+T_2^3=-s_p+\ell_{0,0}(2s_p+1)-(\ell_{1,0}+\ell_{2,0})s_p,$$ which, by
(\ref{sumlij}), gives
\begin{eqnarray}\label{hgbfvrd 1}
T_0^3+ T_1^3+T_2^3&amp;=&amp;\ell_{0,0}(3s_p+1)-s_p^2\nonumber \\
&amp;=&amp;p\ell_{0,0}-s_p^2.
\end{eqnarray}
Similarly, by again using  (\ref {hgftrvcds})-(\ref{hgftrvttrscccds}), we obtain
\begin{equation}\label{mlknvc4}
T_0^2T_1+T_0T_2^2+T_1^2T_2=p\ell_{1,0}-s_p^2
\end{equation}
and
\begin{equation}\label{mlknsdvc4}
T_0^2T_2+T_0T_1^2+T_1T_2^2=p\ell_{2,0}-s_p^2.
\end{equation}
Using (\ref{nbvcgfd})-(\ref{nbvcgssssfd}) and (\ref{sansd 3}),  it is easy to show that
\begin{equation*}
\gamma_1(m)=\frac{3}{4}\eta^3(m)\cos\left((\omega_2-\omega_1)\frac{\pi}{2}+\frac{4\pi}{3}\right),
\end{equation*}
\begin{equation*}
 \gamma_2(m)=\frac{3}{4}\eta^3(m)\cos\left((\omega_2-\omega_1)\frac{\pi}{2}+\frac{2\pi}{3}\right)
\end{equation*}
and
\begin{equation*}
\gamma_3(m)=\frac{1}{4}\eta^3(m)\cos\left((\omega_2-\omega_1)\frac{\pi}{2}\right).
\end{equation*}
Since, cf. (\ref{lkytmlnb}),
$\alpha_0(m)+\alpha_1(m)+\alpha_2(m)=0$, from (\ref {nbvsscgfd})
we find that $$\gamma_4(m)=-
\gamma_1(m)-\gamma_2(m)+3\gamma_3(m).$$
 Therefore,
\begin{equation*}
\gamma_4(m)=\frac{3}{2}\eta^3(m)\cos\left((\omega_2-\omega_1)\frac{\pi}{2}\right).
\end{equation*}
Note that if $w_2-w_1$ is even then
\begin{equation*}
\gamma_1(m)=\gamma_2(m)=-\frac
{3}{2}\gamma_3(m)=-\frac{1}{4}\gamma_4(m)=-\frac
{3}{8}\eta^3(m)\left(-1\right)^{\frac {w_2-w_1}{2}};
\end{equation*}
while if $w_2-w_1$ is odd then
\begin{equation*}
\gamma_1(m)=-\gamma_2(m)=-\frac {3\sqrt{3}}{8}\eta^3(m)(-1)^{\frac
{w_2-w_1+1}{2}},~~\gamma_3(m)=0,~~\gamma_4(m)=0.
\end{equation*}
For $w_2-w_1$ even, from  (\ref{lkjhgf2}), (\ref{hgbfvrd 1})-(\ref{mlknsdvc4})  and (\ref{ttrfeceeed}), we get
\begin{equation*}
  \nu_p=
\frac{1}{8}(-1)^{\frac{\omega_2-\omega_1}{2}}p\eta^3(m)\left(9\ell_{2,1}-p-1\right).
\end{equation*}
For $w_2-w_1$  odd,
from  (\ref{lkjhgf2}) and (\ref{hgbfvrd 1})-(\ref{mlknsdvc4}), we get
\begin{equation*}
  \nu_p=
\frac{3\sqrt{3}}{8}(-1)^{\frac{\omega_2-\omega_1+1}{2}}p\eta^3(m)(\ell_{1,0}-\ell_{2,0}).
\end{equation*}
By (\ref{eijcij}), (\ref{mmlpoiu}) and (\ref{vkjhgfer}), we have
\begin{equation*}
    \ell_{1,0}-\ell_{2,0}=-M.
\end{equation*}
Lastly,  for $w_2-w_1$ even (resp. odd), (\ref{1111444mmmpp})
follows from  (\ref{1kkjjhhggfv}) (resp. the last equality).
\end{proof}
{\bf Example: $p=43$.}\\
As an explicit example,  let us consider the case $p=43$. Then
 $$1+z^{43}= (1+z)P_1(z)P_2(z)P_3(z),$$
where $P_1(z)=z^{14}+z^{12}+z^{10}+z^7+z^4+z^2+1$, $P_2(z)=z^{14}+z^{11}+z^{10}+z^9+z^8+z^7+z^6+z^5+z^4+z^3+1$ and $P_3(z)=z^{14}+z^{13}+z^{11}+z^7+z^3+z+1$ are the only irreducible polynomials over $\F_2[z]$ of order $43$.
For $1\leq l \leq 3$, let ${\cal A}(P_l)$ be the unique set defined by (\ref{eq4}). For $m\geq 1$, let ${\cal A}(P_l)_m$ denote the set of the elements of ${\cal A}(P_l)$ of the form $2^km$. We give bellow the description of the sets ${\cal A}(P_l)_1$ and ${\cal A}(P_l)_3$; $1\leq l \leq 3$.  \\
Since $p=43$ then $g=3$ is a generator of the cyclic group $(\mathbb{Z}/{43
\mathbb{Z}})^*$. Let  $L$ and $M$ be the unique integers satisfying $4p=172=L^2+27M^2$,  $L\equiv 1$ (mod $3$) and $(g^{2})^{(p-1)/3} =(3^2)^{14}\equiv \frac{L+9M}{L-9M}$ (mod $43$). Hence, $L=-8$, $M=-2$, $R_1(y)=(y^3-y^2-14y-8)^{14}$ and $R_3(y)=(27y^3-129y+86)^{14}$.\\
By using the function polrootspadic of PARI, the $2-$adic expansions of the zeros of the polynomial  $R_1(y)$  are\\
$2^2+2^3+2^6+2^{10}+2^{13}+2^{17}+2^{18}+2^{20}+2^{22}+2^{25}+2^{27}+2^{29}+2^{30}+2^{32}+2^{33}+2^{36}+\cdots$\\
$2 + 2^4 + 2^6 + 2^7 + 2^{10} + 2^{15} + 2^{16} +  2^{19} + 2^{20} + 2^{23} + 2^{26} + 2^{27} + 2^{31} + 2^{34} + 2^{35} +\cdots$\\
$1+2+2^5+2^6+2^7+2^9+2^{10}+2^{12}+2^{14}+2^{20}+2^{24}+2^{27}+\cdots$\\
and the $2-$adic expansions of the zeros of the polynomial $R_3(y)$  are\\
$1+2^2+2^3+2^4+2^6+2^7+2^{12}+2^{17}+2^{18}+2^{19}+2^{20}+2^{21} + 2^{25} + 2^{27} + 2^{31} + 2^{32} + 2^{35} + 2^{36} + \cdots$\\
$1 + 2^2 + 2^5 + 2^7 + 2^{10} + 2^{13} + 2^{14} + 2^{19} + 2^{20} + 2^{22} +2^{23}+2^{24}+2^{25}+2^{27}+2^{29}+2^{34}+\cdots$\\
$2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^9 + 2^{1  1} + 2^{15} + 2^{16} + 2^{19} + 2^{21} + 2^{22} + 2^{23} + 2^{24} + 2^{27} + 2^{30} + 2^{33}+\cdots $.\\
After computing some first few elements of the sets ${\cal A}(P_l)$, we deduce that\\
${\cal A}(P_1)_1=\{2, 2^4,  2^6, 2^7,  2^{10}, 2^{15},  2^{16},   2^{19},  2^{20},  2^{23},  2^{26},  2^{27},  2^{31},  2^{34},2^{35},\ldots \} $\\
${\cal A}(P_2)_1=\{2^2,2^3,2^6,2^{10},2^{13},2^{17},2^{18},2^{20},2^{22},2^{25},2^{27},2^{29},2^{30},2^{32},2^{33},2^{36},\ldots\}$ \\
${\cal A}(P_3)_1=\{1,2,2^5,2^6,2^7,2^9,2^{10},2^{12},2^{14},2^{20},2^{24},2^{27},\ldots\}$. \\
${\cal A}(P_1)_3=\{2.3 , 2^2.3,  2^3.3 , 2^4.3 , 2^5.3 , 2^6.3 , 2^9.3 , 2^{1  1}.3,2^{15}.3 , 2^{16}.3 , 2^{19}.3 , 2^{21}.3 , 2^{22}.3 , 2^{23}.3, 2^{24}.3 ,\ldots\} $\\
${\cal A}(P_2)_3=\{3,2^2.3, 2^3.3,2^4.3 ,2^6.3, 2^7.3, 2^{12}.3, 2^{17}.3, 2^{18}.3, 2^{19}.3, 2^{20}.3, 2^{21}.3 , 2^{25}.3 ,  2^{27}.3 , 2^{31}.3,\ldots\}$ \\
${\cal A}(P_3)_3=\{3, 2^2.3 , 2^5.3 , 2^7.3 , 2^{10}.3 , 2^{13}.3 , 2^{14}.3 , 2^{19}.3 ,2^{20}.3 , 2^{22}.3, 2^{23}.3,2^{24}.3,2^{25}.3,2^{27}.3,2^{29}.3,\ldots\}$.
\section{Acknowledgments}
We are pleased to thank professors F. Ben Sa\"{\i}d, F. Morain, C. Delaunay and  J.- L. Nicolas for valuable comments and helpful discussions. We also would like to thank the referees for their valuable remarks that help to improve the initial version of this paper.



\def\refname{References}
\begin{thebibliography}{99}

\bibitem{NF3} N. Baccar, Sets with even partition function and 2-adic
integers, {\it Periodica Math. Hungar} {\bf 55}  (2007), 177--193.

\bibitem{NF2} N. Baccar and F. Ben Sa\"{\i}d, On sets such that the
partition function is even from a certain point on,  {\it Internat.
J. Number Theory} {\bf  5} (2009), 1--22.

\bibitem{NFA} N. Baccar, F. Ben Sa\"{\i}d and A. Zekraoui, On the
divisor function of  sets with even partition functions, {\it Acta
Math. Hungar} {\bf 112} (2006), 25--37.

\bibitem{Ben2} F. Ben Sa\"{\i}d, On some sets with even valued partition
function, {\it Ramanujan J. } {\bf 9} (2005), 63--75.

\bibitem{Ben4} F. Ben Sa\"{\i}d and J.-L. Nicolas, Even partition
functions, {\it S\'eminaire Lotharingien de Combinatoire}
(http//www.mat.univie.ac.at/~slc/), \bf 46 \rm(2002), B 46i.

\bibitem{FJZE}F. Ben Sa\"{\i}d, J.-L. Nicolas and A. Zekraoui, On
the parity of generalised partition function III, to appear in {\it J. 
Th\'eorie Nombres Bordeaux}.

\bibitem{F lalouar}F. Ben Sa\"{\i}d, H. Lahouar and J.-L. Nicolas, On the counting function of the sets of
parts such that the partition function takes even values for $n$ large enough,
{\it Discrete Math.} {\bf 306} (2006), 1115--1125.

\bibitem{Kat85}  S. A. Katre,
Jacobsthal sums in terms of quadratic partitions of a
prime. In K. Alladi, editor, {\it Number Theory}, volume 1122 of {\it Lecture
Notes in Math.}, Springer-Verlag, 
1985, pp.\ 153--162. 

\bibitem{min} M. Mignotte, {\it Math\'ematiques pour le Calcul Formel,}
Presses Universitaires de France, 1989.

\bibitem{NRS} J.-L. Nicolas, I.Z. Ruzsa and A. S\'ark\"ozy, On the
parity of additive representation functions, {\it J. Number Theory } {\bf
73} (1998), 292--317.

\bibitem{GENE}H. S. Wilf, {\it Generatingfunctionology}, Academic Press,
Second Edition,   1994.

\end{thebibliography}


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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11P83; Secondary 11B50, 11D88, 11G20.

\noindent \emph{Keywords: } Partitions, periodic sequences, order of a
polynomial, cyclotomic polynomials, resultant, 2-adic integers, elliptic curves.

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\vspace*{+.1in}
\noindent
Received July 16 2009;
revised version received  December 23 2009.
Published in {\it Journal of Integer Sequences}, December 31 2009.

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