<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amscd}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics,amsmath,amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{latexsym}
\usepackage{epsf}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}}

\usepackage{tkz-euclide}
\usepackage{tikz-cd}
\usepackage{mathtools}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\begin{center}
\vskip 1cm{\LARGE\bf Divisors on Overlapped Intervals  and\\
\vskip .1in
Multiplicative Functions} \vskip 1cm
\large Jos\'e Manuel Rodr\'iguez Caballero\\
D\'epartement de Math\'ematiques\\
UQAM\\
Case Postale 8888, Succ.\ Centre-ville  \\
Montr\'eal, Qu\'ebec H3C 3P8 \\
Canada\\
\href{mailto:rodriguez\_caballero.jose\_manuel@uqam.ca}{\tt rodriguez\_caballero.jose\_manuel@uqam.ca}\\
\end{center}

\def\DD{\mathcal{D}}
\def\SS{\mathcal{S}}

\vskip .2 in
\begin{abstract}
Reutenauer and Kassel introduced a family $P_n(q)$ of polynomials
defined in terms of divisors of $n$ on overlapped intervals. The
evaluation of $P_n(q)$ at roots of unity of order 2, 3, 4, 6 form
well-known integer sequences related to the number of integer solutions
of the equations $x^2 + y^2 = n$, $x^2 + 2y^2 = n$, and $x^2 + xy + y^2
= n$. Also, $P_n(1)$ is the sum of divisors of $n$. In this paper we
define a new family $L_n(q)$ of polynomials defined in terms of
divisors of $n$ on overlapped intervals, slightly
modifying the
definition of $P_n(q)$. The values of $L_n(q)$ at $q = 1$ and $q = -1$
are related to the sum of divisors of $n$ and to the number of integer
solutions of the equations $x^2 + xy + y^2 = n$ and $x^2 + 3 y^2 = n$.
\end{abstract}



\section{Introduction}
For a given integer $n \geq 1$, consider the two-sided   sequence
$$
p_{n,k} = \ln\left(k+\sqrt{k^2 + 2\,n}\right),
$$
where $k \in \mathbb{Z}$ and define the intervals $$
\mathcal{P}_{n,k} = \left(p_{n,k}-\ln 2, p_{n,k}\right]. $$

Kassel and Reutenauer \cite{kassel2015counting} introduced the  polynomials\footnote{The original definition of $P_n(q)$, which we refer to as \emph{Kassel-Reutenauer polynomials} \cite{kassel2015counting} is rather different, but equivalent, to the one presented here. We preferred to take the logarithm of the divisors in place of the divisors themselves in order to work with intervals $\mathcal{P}_{n,k}$ of constant length.}

$$
 \frac{P_n(q)}{q^{n-1}}  = \sum_{d|n}\sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{P}_{n,k}}\left(\ln d\right) \, q^k,$$
where $\boldsymbol{1}_A(x)$ is the characteristic function of the set $A$, 
i.e., $\boldsymbol{1}_A(x) = 1$ if $x \in A$, otherwise $\boldsymbol{1}_A(x) = 0$. Each polynomial $P_n(q)$ is monic of degree $2n-2$, its coefficients are non-negative integers and it is  self-reciprocal \cite{kassel2016complete}. The evaluations of $P_n(q)$ at some complex roots of $1$ have number-theoretical interpretations \cite{kassel2016complete}, e.g.,

\begin{eqnarray*}
\sigma(n) &amp;=&amp; P_n(1), \\
\frac{r_{1,0,1}(n)}{4}  &amp;=&amp; P_n(-1), \\
\frac{r_{1,0,2}(n)}{2} &amp;=&amp; \left| P_n\left(\sqrt{-1}\right)\right|, \\
\frac{r_{1,1,1}(n)}{6} &amp;=&amp; \textrm{Re } P_n\left( \frac{-1+\sqrt{-3}}{2}\right), 
\end{eqnarray*}
where $\sigma(n)$, $\frac{r_{1,0,1}(n)}{4}$, $\frac{r_{1,0,2}(n)}{2}$ and $\frac{r_{1,1,1}(n)}{6}$ are  multiplicative functions \cite{lorenz1871bidrag} given by 
\begin{eqnarray*}
\sigma(n) &amp;=&amp; \sum_{d|n} d, \\
r_{a,b,c}(n) &amp;=&amp; \# \left\{(x,y)\in \mathbb{Z}^2: \quad a\,x^2 + b\,x\,y + c\,y^2 = n \right\}.
\end{eqnarray*}

Furthermore, for $q = \frac{1+\sqrt{-3}}{2}$, the same sequence $n \mapsto P_n(q)$ is related to $r_{1,0,1}(n)$ in three ways \cite{kassel2016fourier}, depending on the congruence class of $n$ in $\mathbb{Z}/3\mathbb{Z}$,

$$
\left|P_n\left(\frac{1+\sqrt{-3}}{2} \right)\right| =  \begin{cases}
r_{1,0,1}(n), &amp; \text{if  $n \equiv 0 \pmod{3}$;} \\
\frac{1}{4}\,r_{1,0,1}(n), &amp; \text{if $n \equiv 1 \pmod{3}$;} \\
\frac{1}{2}\,r_{1,0,1}(n), &amp; \text{if $n \equiv 2 \pmod{3}$.}
\end{cases}
$$


For any integer $n \geq 1$, consider the two-sided sequence

$$
\ell_{n,k} = \ln\left( \tfrac{3}{2}\,k+\sqrt{\left(\tfrac{3}{2}\,k \right)^2 + 3\,n} \right) 
$$
and the intervals 

$$\mathcal{L}_{n,k} = \left(\ell_{n,k}-\ln 3,\ell_{n,k}\right],$$
where $k$ runs over the integers. Define a variation of the polynomials $P_n(q)$ as follows:

$$
\frac{L_n(q)}{q^{n-1}} = \sum_{d|n}\sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) \, q^k.
$$

For example, in order to compute $L_6(q)$ from the definition, we need to consider the intervals $\left(\ell_{6,k}-\ln 3,  \ell_{6,k}\right]$ on the real line and to count the number of values of $\ln d$ inside each interval, where $d$ runs over the divisors of $n$. These data are shown in Figure \ref{Fig:1}, where the numbers $\ell_{6,k}$ are plotted on the line below (the corresponding values of $k$ are labelled) whereas the numbers $\ln d$ are plotted on the line above (the corresponding values of $d$ are labelled). Counting the number of intersections between the horizontal and the vertical lines, we obtain that the coefficients of $\frac{L_{6}(q)}{q^{6-1}}$ are as follows:

$$
\frac{L_{6}(q)}{q^{6-1}} = q^{5} + q^{4} + q^{3} + 2 \, q^{2} + 2 \, q + 2\,q^0 + 2\,q^{-1} + 2 \,q^{-2} + q^{-3} + q^{-4} + q^{-5}.
$$

\begin{figure}
\label{Fig:1}
\begin{center}
\begin{tikzpicture}[scale=2]
\draw[-&gt;, thick] (-1.5,0) -- (3.5,0) node[below] {$\ell_{6,k}$};
\draw[-&gt;, thick] (-1.5,1.2) -- (3.5,1.2) node[below] {$\ln d$};
\foreach \x/\xtext in {-0.051425544115108225/-6,
 0.11050682349650519/-5,
 0.298970044167493/-4,
 0.5215628360911632/-3,
 0.7867069304856741/-2,
 1.0986122886681098/-1,
 1.4451858789480825/0,
 1.791759469228055/1,
 2.1036648274104905/2,
 2.3688089218050012/3,
 2.591401713728671/4,
 2.7798649343996593/5,
 2.941797302011273/6  }
 {
    \draw[dashed] (\x,7/12+\xtext/12) -- (\x,0) node[below] {\xtext};
    \draw (\x-1.09861228867,7/12+\xtext/12) -- (\x,7/12+\xtext/12) ;
    \draw[black, fill = white] (\x-1.09861228867,7/12+\xtext/12) circle (0.03cm) ;
    \draw[black, fill = black] (\x,7/12+\xtext/12) circle (0.03cm) ;
   }
\draw[-&gt;, thick] (-1.5,0) -- (3.5,0);
\foreach \x/\xtext in {0.0/1, 0.6931471805599453/2, 1.0986122886681098/3, 1.791759469228055/6}
 {
%    \draw[thick] (\x,0) -- (\x,\xtext/12);
     \draw[black,fill=gray] (\x,1.2) circle (0.03cm);     
	\draw[gray, thick]  (\x,0) -- (\x, 1.2) node[black, above] {$\xtext$};
     }     
%\draw (0.2,0.5pt) node[above] {$c$};
%\draw[[-, ultra thick, black] (0,0) -- (0.01,0);
%\draw[-), ultra thick, black] (0.19,0) -- (0.2,0);
%\fill[opacity = 0.2, blue,rounded corners=1ex] (0,-.16ex) -- (0.2, -.16ex) -- (0.2, .16ex) -- (0,.16ex) -- cycle;
\end{tikzpicture}
\end{center}
\caption{Representation of $L_6(q)$.}
\end{figure}


Like $P_n(q)$, the polynomial $L_n(q)$ is monic of degree $2\,n-2$, self-reciprocal and its coefficients are non-negative integers. The aim of this paper is to express the multiplicative functions \cite[p.\ 421]{lorenz1871bidrag} $\frac{r_{1,1,1}(n)}{6}$ and $\frac{r_{1,0,3}(n)}{2}$ in terms of the evaluations of $L_n(q)$ at roots of the unity. More precisely, we will prove the following result.

\begin{theorem}	\label{Theo349jr9834j9}
For each $n \geq 1$,
\begin{eqnarray}
\seqnum{A002324}(n) &amp;:=&amp; \frac{r_{1,1,1}(n)}{6} = 4\,\sigma(n) - 3\,L_n\left( 1\right), \label{Idenr8934jr9j349}\\
\seqnum{A096936}(n) &amp;:=&amp; \frac{r_{1,0,3}(n)}{2} = L_n\left( -1\right).\label{Idenr438j98j43sdf}
\end{eqnarray}
\label{one}
\end{theorem}

\section{Auxiliary results for the first identity of Theorem~\ref{one}}

For any $n \geq 1$, we will use the notation

$$
 d_{a,m}(n) := \#\left\{d|n: \quad d \equiv a \pmod{m} \right\}.
$$

We will use the following well-known result \cite{hirschhorn2001three}. 
\begin{lemma}\label{Lemur439r9834hr83}
For all integers $n \geq 1$,
$$
\frac{r_{1,1,1}(n)}{6} = d_{1,3}(n)-d_{2,3}(n).
$$
\end{lemma}



\begin{lemma}\label{Lem493j9rj49r3r}
For any integer $n \geq 1$,
\begin{eqnarray*}
3\,\left\lceil 3^{-1}\,n \right\rceil - n &amp;=&amp; \begin{cases}
0, &amp; \text{if $n\equiv 0 \pmod{3}$;} \\
2, &amp; \text{if $n\equiv 1 \pmod{3}$;} \\
1, &amp; \text{if $n\equiv 2 \pmod{3}$.}
\end{cases}  \\
n - 3\,\left\lfloor 3^{-1}\,n \right\rfloor &amp;=&amp;  \begin{cases}
0, &amp; \text{if $n\equiv 0 \pmod{3}$;} \\
1, &amp; \text{if $n\equiv 1 \pmod{3}$;} \\
2, &amp; \text{if $n\equiv 2 \pmod{3}$.}
\end{cases}
\end{eqnarray*}

\end{lemma}

\begin{proof}
It is enough to evaluate $3\,\left\lceil 3^{-1}\,n \right\rceil - n$ and $3\,\left\lceil 3^{-1}\,n \right\rceil - n$ at $n = 3k+r$, for $k \in \mathbb{Z}$ and $r \in \{0, 1, 2\}$. 
\end{proof}


\begin{lemma}\label{Lem98ej398jr934jf9ww}
For any pair of integers $n \geq 1$ and $k$, the inequalities

$$
\ell_{n,k} - \ln 3 &lt; \ln d \leq \ell_{n,k}
$$
hold if and only if the inequalities

$$
3^{-1}\,d - \frac{n}{d}\leq k &lt; d - 3^{-1}\,\frac{n}{d}
$$
hold.
\end{lemma}

\begin{proof}
The inequalities 
$$
\ell_{n,k} - \ln 3 &lt; \ln d \leq \ell_{n,k}
$$
are equivalent to
$$
\ln d \leq \ell_{n,k} &lt; \ln d + \ln 3.
$$

Applying the strictly increasing function $x \mapsto \frac{e^x}{3}
 - n \, e^{-x}$ to the last inequalities we obtain the following equivalent inequalities
$$
3^{-1}\,d - \frac{n}{d}\leq k &lt; d - 3^{-1}\,\frac{n}{d}.
$$

Indeed, $\frac{e^{\ln d}}{3}
 - n \, e^{-\ln d} = 3^{-1}\,d - \frac{n}{d}$, $\frac{e^{\ln d + \ln 3}}{3}
 - n \, e^{-\left(\ln d + \ln 3\right)} = d - 3^{-1}\frac{n}{d}$ and

\begin{eqnarray*}
\frac{e^{\ell_{n,k}}}{3} - n \, e^{-\ell_{n,k}} 
&amp;=&amp; \frac{\tfrac{3}{2}\,k+\sqrt{\left(\tfrac{3}{2}\,k \right)^2 + 3\,n}}{3} - \frac{n}{\tfrac{3}{2}\,k+\sqrt{\left(\tfrac{3}{2}\,k \right)^2 + 3\,n}} \\
&amp;=&amp; \frac{\tfrac{3}{2}\,k+\sqrt{\left(\tfrac{3}{2}\,k \right)^2 + 3\,n}}{3} + \frac{\tfrac{3}{2}\,k-\sqrt{\left(\tfrac{3}{2}\,k \right)^2 + 3\,n}}{3} \\
&amp;=&amp; k.
\end{eqnarray*}

So the lemma is proved. 
\end{proof}

\begin{lemma}\label{lem4u98j39j9}
Let $n \geq 1$ be an integer. For all $d | n$,

$$
\sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) = \left\lceil d - 3^{-1}\,\frac{n}{d} \right\rceil - \left\lceil 3^{-1} \, d - \frac{n}{d}\right\rceil.
$$

\end{lemma}

\begin{proof} For all integers $n \geq 1$ and $k$, we have that
\begin{eqnarray*}
&amp; &amp; \sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) \\
&amp;=&amp; \# \left\{k \in \mathbb{Z}: \quad \ell_{n,k} - \ln 3 &lt; \ln d \leq \ell_{n,k} \right\} \\
&amp;=&amp; \# \left\{k \in \mathbb{Z}: \quad 3^{-1}\,d - \frac{n}{d}\leq k &lt; d - 3^{-1}\,\frac{n}{d}\right\} \quad (\textrm{Lemma } \ref{Lem98ej398jr934jf9ww}) \\
&amp;=&amp; \# \left\{k \in \mathbb{Z}: \quad \left\lceil 3^{-1}\,d - \frac{n}{d} \right\rceil \leq k &lt; \left\lceil d - 3^{-1}\,\frac{n}{d}\right\rceil\right\} \\
&amp;=&amp; \left\lceil d - 3^{-1}\,\frac{n}{d} \right\rceil - \left\lceil 3^{-1} \, d - \frac{n}{d}\right\rceil.
\end{eqnarray*}

So the lemma is proved. 
\end{proof}


\section{Auxiliary results for the second identity of Theorem~\ref{one}}

We will use the following well-known result \cite[p.\ 421]{lorenz1871bidrag}.

\begin{lemma}\label{Lemr89j983jr9}
The function $\frac{r_{1,0,3}(n)}{2}$ is multiplicative.
\end{lemma}


We will use the following well-known result \cite{hirschhorn2001three}.
\begin{lemma}\label{Lemurf93wenuf9h2938}
For all integers $n \geq 1$,
$$
\frac{r_{1,0,3}(n)}{2} = d_{1,3}(n)-d_{2,3}(n) + 2\left(d_{4,12}(n)-d_{8,12}(n) \right).
$$
\end{lemma}



Recall that the nonprincipal Dirichlet character mod $3$ is the $3$-periodic arithmetic function $\chi_3(n)$ given by $\chi_3(0) = 0$, $\chi_3(1) = 1$ and $\chi_3(2) = -1$.



\begin{lemma}\label{Lemrj893j9jewrefef54f}
For all $n \geq 1$,
$$
\frac{\left( -1\right)^{\left\lfloor  3^{-1}\,n \right\rfloor} - \left( -1\right)^{\left\lceil 3^{-1} \, n \right\rceil}}{2} = (-1)^{n-1}\,\chi_3(n).
$$
\end{lemma}

\begin{proof}
It is enough to substitute $n = 3k+r$, with $k\in \mathbb{Z}$ and $r \in \{0, 1, 2\}$, in both sides in order to check that they are equal. 
\end{proof}


\begin{lemma}\label{Lem7g76g78iuho8h9h98}
For all $n \geq 1$,
$$
 \sum_{d | n} \left( -1\right)^{\frac{n}{d}-1}\, \left(-1\right)^{d-1}\,\chi_3(d) = \left( -1\right)^{n-1}\,\frac{r_{1,0,3}(n)}{2}.
$$
\end{lemma}

\begin{proof}
By Lemma \ref{Lemr89j983jr9}, the function $ \frac{r_{1,0,3}(n)}{2}$ is multiplicative. Also, it is easy to check that the functions $\left( -1\right)^{n-1}$ and $\chi_3(n)$ are multiplicative. So the functions $f(n) = \left( -1\right)^{n-1}\,\frac{r_{1,0,3}(n)}{2}$ and $\left( -1\right)^{n-1} \, \chi_3(n)$ are multiplicative, because the multiplicative property is preserved by ordinary product. The function $g(n) =  \sum_{d | n} \left( -1\right)^{\frac{n}{d}-1}\, \left(-1\right)^{d-1}\,\chi_3(d)$ is multiplicative, because Dirichlet convolution preserves the multiplicative property. So it is enough to prove that $f\left(p^k\right) = g\left(p^k\right)$ for each prime power $p^k$.


Considering the case $p = 2$. The following elementary equivalences  hold for any integer $m \geq 0$,

\begin{eqnarray*}
2^{m}  \equiv 1 \pmod{3} &amp;\Longleftrightarrow &amp; m \equiv 0 \pmod{2}, \\
2^{m}  \equiv 2 \pmod{3} &amp;\Longleftrightarrow &amp; m \equiv 1 \pmod{2},  \\
2^{m}  \equiv 4 \pmod{12} &amp;\Longleftrightarrow &amp; m \equiv 0 \pmod{2} \textrm{ and } m \neq 0, \\
2^{m}  \equiv 8 \pmod{12} &amp;\Longleftrightarrow &amp; m \equiv 1 \pmod{2}\textrm{ and } m \neq 1.
\end{eqnarray*}

So, for each integer $k \geq 1$,

\begin{eqnarray*}
d_{1,3}\left( 2^k\right) &amp;=&amp; \# [0,k] \cap 2\mathbb{Z}  =  \left\lfloor\frac{k}{2}\right\rfloor + 1, \\
d_{2,3}\left( 2^k\right) &amp;=&amp; \# [1,k] \cap \left(2\mathbb{Z}+1\right)  = \left\lceil\frac{k}{2}\right\rceil, \\
d_{4,12}\left( 2^k\right) &amp;=&amp;  \# [2,k] \cap 2\mathbb{Z} = \left\lfloor\frac{k}{2}\right\rfloor, \\
d_{8,12}\left( 2^k\right) &amp;=&amp; \# [3,k] \cap \left(2\mathbb{Z}+1\right) =  \left\lceil\frac{k}{2}\right\rceil - 1.
\end{eqnarray*}

For any $k \geq 1$, it follows that


\begin{eqnarray*}
g\left( 2^k\right)
&amp;=&amp;  \sum_{j=0}^k \left( -1\right)^{2^{k-j}-1}\, \left(-1\right)^{2^j-1}\,\chi_3\left(2^j \right) \\
&amp;=&amp; \sum_{j=0}^k \left( -1\right)^{2^{k-j}-1}\, \left(-1\right)^{2^j-1}\,\left(-1 \right)^j \\
&amp;=&amp; -1 - \left(-1 \right)^k + \sum_{j=1}^{k-1} \left(-1 \right)^j \\
&amp;=&amp; -1 - \left(-1 \right)^k + \frac{-1 - \left(-1 \right)^k}{2} \\
&amp;=&amp; -3\,\frac{1 + \left(-1 \right)^k}{2} \\
&amp;=&amp; -3 \,\left( 1+ \left\lfloor\frac{k}{2}\right\rfloor -\left\lceil\frac{k}{2}\right\rceil \right)  \\
&amp;=&amp; -\left(\left( \left\lfloor\frac{k}{2}\right\rfloor + 1\right)-\left\lceil \frac{k}{2}\right\rceil + 2\left(\left\lfloor\frac{k}{2}\right\rfloor-\left( \left\lceil\frac{k}{2}\right\rceil - 1\right) \right) \right) \\
&amp;=&amp; \left(-1 \right)^{2^k - 1} \, \left(d_{1,3}\left( 2^k\right)-d_{2,3}\left( 2^k\right) + 2\left(d_{4,12}\left( 2^k\right)-d_{8,12}\left( 2^k\right) \right) \right) \\
&amp;=&amp; f\left( 2^k\right) \quad (\textrm{Lemma } \ref{Lemurf93wenuf9h2938}).
\end{eqnarray*}

Let $p$ and $k \geq 1$ be an odd prime and an integer respectively. Noticing that $\left(-1\right)^{p^j-1} = 1$ for all $0 \leq j \leq k$. Also, $d_{4,12}\left( p^k\right) = d_{8,12}\left( p^k\right) = 0$, because $p^k$ has not even divisor. So, for any $k \geq 1$,

\begin{eqnarray*}
g\left( p^k\right)
&amp;=&amp;  \sum_{j=0}^k \left( -1\right)^{p^{k-j}-1}\, \left(-1\right)^{p^j-1}\,\chi_3\left(p^j \right) \\
&amp;=&amp;  \sum_{j=0}^k \chi_3\left(p^j \right) \\
&amp;=&amp; d_{1,3}\left( p^k\right)-d_{2,3}\left( p^k\right)  \\
&amp;=&amp; \left( -1\right)^{p^k - 1} \,\left( d_{1,3}\left( p^k\right)-d_{2,3}\left( p^k\right) + 2\left(d_{4,12}\left( p^k\right)-d_{8,12}\left( p^k\right) \right) \right) \\
&amp;=&amp; f\left( p^k\right)  \quad (\textrm{Lemma } \ref{Lemurf93wenuf9h2938}).
\end{eqnarray*}
Therefore, $f(n) = g(n)$ for all $n \geq 1$. 
\end{proof}


\begin{lemma}\label{Lemm9348fj9438jhf9h39w}
For each $d|n$, 

$$
\sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) \, \left( -1\right)^k = \frac{1}{2} \, \left(\left( -1\right)^{\left\lceil 3^{-1} \, d - \frac{n}{d}\right\rceil} -  \left( -1\right)^{\left\lceil d - 3^{-1}\,\frac{n}{d} \right\rceil} \right).
$$
\end{lemma}

\begin{proof}
For any integer $n \geq 1$ and any $d|n$,

\begin{eqnarray*}
\sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) \, \left( -1\right)^k 
&amp;=&amp; \sum_{3^{-1}\,d - \frac{n}{d}\leq k &lt; d - 3^{-1}\,\frac{n}{d}} \left( -1\right)^k \quad (\textrm{Lemma } \ref{Lem98ej398jr934jf9ww}) \\
&amp;=&amp; \sum_{\lceil 3^{-1}\,d - \frac{n}{d} \rceil \leq k &lt; \lceil d - 3^{-1}\,\frac{n}{d}\rceil} \left( -1\right)^k.
\end{eqnarray*}

Substituting $a = \lceil 3^{-1}\,d - \frac{n}{d} \rceil$, $b = \lceil d - 3^{-1}\,\frac{n}{d}\rceil$ and $q = -1$ in the geometric sum

$$
\sum_{a \leq k &lt; b} q^k = \frac{q^a - q^b}{1-q} 
$$
we obtain

\begin{eqnarray*}
\sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) \, \left( -1\right)^k &amp;=&amp; \frac{\left( -1\right)^{\left\lceil 3^{-1} \, d - \frac{n}{d}\right\rceil} -  \left( -1\right)^{\left\lceil d - 3^{-1}\,\frac{n}{d} \right\rceil}}{1 - (-1)} \\
&amp;=&amp; \frac{1}{2} \, \left(\left( -1\right)^{\left\lceil 3^{-1} \, d - \frac{n}{d}\right\rceil} -  \left( -1\right)^{\left\lceil d - 3^{-1}\,\frac{n}{d} \right\rceil} \right).
\end{eqnarray*}

So the lemma is proved. 
\end{proof}

\section{Proof of the main result}

We proceed now with the proof of the main result of this paper.

\begin{proof}[Proof of Theorem \ref{Theo349jr9834j9}]
Identity \eqref{Idenr8934jr9j349} follows from the following transformations,

\begin{eqnarray*}
L_n(1)
&amp;=&amp; \sum_{d | n} \sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) \\
 &amp;=&amp; \sum_{d | n} \left( \left\lceil d - 3^{-1}\,\frac{n}{d} \right\rceil - \left\lceil 3^{-1} \, d - \frac{n}{d}\right\rceil \right) \quad (\textrm{Lemma }\ref{lem4u98j39j9}) \\
&amp;=&amp; \sum_{d | n} \left(  d + \frac{n}{d} \right)  + \sum_{d | n}\left\lceil -3^{-1}\,\frac{n}{d} \right\rceil - \sum_{d | n} \left\lceil 3^{-1}d \right\rceil \\
&amp;=&amp; \sum_{d | n} \left(  d + \frac{n}{d} \right)  - \sum_{d | n} \left\lfloor 3^{-1}\,\frac{n}{d} \right\rfloor - \sum_{d | n} \left\lceil 3^{-1}d \right\rceil 
\end{eqnarray*}
\begin{eqnarray*}
&amp;=&amp; \frac{2}{3}\sum_{d | n} \left(  d + \frac{n}{d} \right)  + \frac{1}{3}\sum_{d | n} \left(\frac{n}{d} - 3\,\left\lfloor 3^{-1}\,\frac{n}{d} \right\rfloor \right) - \frac{1}{3}\sum_{d | n} \left( 3\,\left\lceil 3^{-1}d \right\rceil - d \right) \\
&amp;=&amp; \frac{4\sigma(n)}{3} + \frac{d_{1,3}(n) + 2d_{2,3}(n) }{3} - \frac{2d_{1,3}(n)  + d_{2,3}(n)}{3} \quad (\textrm{Lemma } \ref{Lem493j9rj49r3r}) \\
&amp;=&amp; \frac{4\sigma(n)}{3} - \frac{d_{1,3}(n) - d_{2,3}(n)}{3} \\
&amp;=&amp; \frac{4}{3}\,\sigma(n) - \frac{1}{3} \, \frac{r_{1,1,1}(n)}{6} \quad (\textrm{Lemma } \ref{Lemur439r9834hr83}).
\end{eqnarray*}

Identity \eqref{Idenr438j98j43sdf} follows from the following transformations,

\begin{eqnarray*}
\frac{L_n\left( -1\right)}{\left( -1\right)^{n-1}}  
&amp;=&amp; \sum_{d | n} \sum_{k\in \mathbb{Z}} \boldsymbol{1}_{\mathcal{L}_{n,k}}\left(\ln d\right) \, \left( -1\right)^k \\
&amp;=&amp; \sum_{d | n} \frac{1}{2} \, \left(\left( -1\right)^{\left\lceil 3^{-1} \, d - \frac{n}{d}\right\rceil} -  \left( -1\right)^{\left\lceil \frac{n}{d} - 3^{-1}\,d \right\rceil} \right)\quad (\textrm{Lemma } \ref{Lemm9348fj9438jhf9h39w}) \\
&amp;=&amp; \sum_{d | n} \frac{1}{2} \, \left(\left( -1\right)^{\left\lceil 3^{-1} \, d \right\rceil - \frac{n}{d}} -  \left( -1\right)^{\frac{n}{d} - \left\lfloor  3^{-1}\,d \right\rfloor} \right) \\
&amp;=&amp; \sum_{d | n} \left( -1\right)^{\frac{n}{d}-1}\, \frac{\left( -1\right)^{\left\lfloor  3^{-1}\,d \right\rfloor} - \left( -1\right)^{\left\lceil 3^{-1} \, d \right\rceil}}{2}  \\
&amp;=&amp;  \sum_{d | n} \left( -1\right)^{\frac{n}{d}-1}\, \left(-1\right)^{d-1}\,\chi_3(d) \quad (\textrm{Lemma } \ref{Lemrj893j9jewrefef54f}) \\
&amp;=&amp; \left( -1\right)^{n-1}\,\frac{r_{1,0,3}(n)}{2} \quad (\textrm{Lemma } \ref{Lem7g76g78iuho8h9h98}).
\end{eqnarray*}

So the theorem is proved. 
\end{proof}


\section{Final remarks}
\begin{enumerate}
\item Let $k$ be a field and $\mathcal{R}$ be a $k$-algebra. The \emph{codimension} of an ideal $I$ of $\mathcal{R}$ is the dimension of the quotient $\mathcal{R}/I$ as a vector space over $k$.

We let $\mathbb{Z}\oplus \mathbb{Z}$ denote the free abelian group of
rank $2$.  Let $k = \mathbb{F}_q$ be the finite field with $q$ elements
and $\mathcal{R} = \mathbb{F}_q\left[\mathbb{Z}\oplus
\mathbb{Z}\right]$ be its group algebra. Kassel and Reutenauer
\cite{kassel2015counting} proved that, for any prime power $q$, the
number of ideals of codimension $n \geq 1$ of
$\mathbb{F}_q\left[\mathbb{Z}\oplus \mathbb{Z}\right]$ is $\left(
q-1\right)^2 \, P_n(q)$. So it is natural to look for connections
between the values of $L_n(q)$, when $q$ is a prime power, and the
algebraic structures related to $\mathbb{F}_q$.  

\item The polynomials
$P_n(q)$ are generated by the product \cite{kassel2016complete}
$$
\prod_{m \geq 1}\frac{\left(1-t^m\right)^2}{\left(1-q\, t^m\right)\left(1-q^{-1}\, t^m\right)} = 1 + \left( q + q^{-1} - 2\right) \,\sum_{n=1}^{\infty} \frac{P_n(q)}{q^{n-1}} \, t^n.
$$
It would be interesting to find a similar generating function for $L_n(q)$.
\end{enumerate}


\section{Acknowledgments}
The author deeply thanks Prof. S. Brlek and Prof. C. Reutenauer for reading carefully the paper and
 for providing useful comments. The author gratefully acknowledges the referee.

\begin{thebibliography}{9}

\bibitem{hirschhorn2001three} M. D. Hirschhorn, Three classical results
on representations of a number, in
D. Foata and G.-N. Han, eds., \emph{The Andrews Festschrift},
Springer, 2001, pp.~159--165.

\bibitem{kassel2015counting} C. Kassel and C. Reutenauer, Counting the
ideals of given codimension of the algebra of Laurent polynomials in
two variables, \emph{Michigan Math. J.}, to appear.

\bibitem{kassel2016complete} C. Kassel and C. Reutenauer, Complete
determination of the zeta function of the Hilbert scheme of $ n $
points on a two-dimensional torus, \emph{Ramanujan J.}, to
appear.

\bibitem{kassel2016fourier} C. Kassel and C. Reutenauer, The Fourier
expansion of $\eta(z)$ $\eta(2z)$ $\eta(3z)$ $/$ $\eta(6z)$, {\em
Archiv Math.} {\bf 108} (2017), 453--463.

\bibitem{lorenz1871bidrag} L. V. Lorenz, \emph{\OE{}uvres scientifiques
de L. Lorenz}, Vol.~2, Lehmann \&amp; Stage, 1899.

\bibitem{sloane}
N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences,
available at \url{http://oeis.org}.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11A07, 11A05.

\noindent \emph{Keywords: } 
multiplicative function, representation by quadratic form, polynomial, divisor.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A002324} and \seqnum{A096936}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  September 16 2017;
revised versions received  October 13 2017; October 21 2017.
Published in {\it Journal of Integer Sequences}, October 29 2017.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                


                                                                                
</pre></body></html>