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\begin{document}
\vspace*{-40pt}
\centerline{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7
(2007), \#A28}
\vskip 40pt


\begin{center}
\textbf{THE NUMBER OF REPRESENTATIONS BY SUMS OF SQUARES AND TRIANGULAR
NUMBERS} \vskip 20pt \textbf{Heung Yeung Lam}\\[0pt]
{\smallit Institute of Information and Mathematical Sciences, Massey
University, Albany Campus, Private Bag 102 904, North Shore Mail Centre,
Auckland, New Zealand}\\[0pt]
\texttt{h.y.lam@massey.ac.nz}\\[0pt]
\end{center}

\vskip30pt \centerline{\smallit Received: 11/27/06,
Revised: 4/12/07, Accepted: 4/22/07, Published: 6/11/07
} % We will fill in the dates
\vskip30pt




\vskip30pt

\centerline{\bf Abstract} \noindent In this paper, we present eighteen
interesting infinite products and their Lambert series expansions. From
these, we deduce formulae for the number of representations of an integer $n$
by eighteen quadratic forms in terms of divisor sums.

\vskip 20pt
\hfill
\textit{--Dedicated to the memory of my grandmother Yuet Kwai Mah.}

\pagestyle{myheadings} 
\markright{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007),
\#A28\hfill} %

\thispagestyle{empty} \baselineskip=15pt \vskip 30pt

\section{Introduction and Statement of Results}

Let $\tau $ be a fixed complex number satisfying Im $(\tau )>0$ and let $%
q=e^{i\pi \tau },$ so that $\left\vert q\right\vert <1.$ \ Let 
\[
\varphi \left( q\right) =\sum_{j=-\infty }^{\infty }q^{j^{2}}, 
\]%
and 
\[
\psi \left( q\right) =\sum_{j=0}^{\infty }q^{j\left( j+1\right) /2}. 
\]%
The purpose of this paper is to study and give proofs of eighteen theorems
in the area of the number of representations by sums of squares and
triangular numbers.\ Most of these results appear to be new.\ The results in
this paper can be found in the author's thesis \cite{Lam03}.\ Section 2
contains preliminary results and will be used as a basis for Section 3.\
In Section 3, we will prove the following results.

\begin{theorem}
\label{tm1} 
\begin{eqnarray}
\varphi \left( q\right) \varphi \left( q^{4}\right) &=&1+2\sum_{j=1}^{\infty
}\frac{\left( -1\right) ^{j}q^{2j}}{1+q^{4j}}-2\sum_{j=1}^{\infty }\frac{%
\left( -1\right) ^{j}q^{2j-1}}{1-q^{2j-1}},  \label{r2t01b} \\
\varphi \left( q\right) \psi \left( q^{8}\right) &=&-\sum_{j=1}^{\infty }%
\frac{\left( -1\right) ^{j}q^{2j-2}}{1-q^{2j-1}}-\sum_{j=1}^{\infty }\frac{%
\left( -1\right) ^{j}q^{2j-1}}{1+q^{4j}},  \label{r2t02b} \\
\varphi \left( q^{4}\right) \psi \left( q^{2}\right) &=&-\frac{1}{2}%
\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}q^{\frac{j-1}{2}}}{1-q^{\frac{%
2j-1}{2}}}-\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}(-q)^{%
\frac{j-1}{2}}}{1-(-q)^{\frac{2j-1}{2}}}.  \label{r2t03b}
\end{eqnarray}
\end{theorem}

\begin{theorem}
\label{tm2} 
\begin{eqnarray}
\varphi ^{3}\left( q\right) \psi \left( q^{8}\right) &=&2\sum_{j=1}^{\infty }%
\frac{jq^{j-1}}{1+(-q)^{j}}-2\sum_{j=1}^{\infty }\frac{\left( -1\right)
^{j}jq^{4j-1}}{1+q^{4j}}  \nonumber \\
&&+\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left( 2j-1\right) q^{2j-2}%
}{1+q^{4j-2}},  \label{r4t01b} \\
\varphi ^{2}\left( q\right) \psi ^{2}\left( q\right) &=&\frac{1}{2}%
\sum_{j=1}^{\infty }\frac{\left( 2j-1\right) q^{\frac{j-1}{2}}}{1-q^{\frac{%
2j-1}{2}}}-\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left(
2j-1\right) q^{\frac{j-1}{2}}}{1+q^{\frac{2j-1}{2}}},  \label{r4t02b} \\
\varphi ^{2}\left( q\right) \psi ^{2}\left( q^{4}\right)
&=&\sum_{j=1}^{\infty }\frac{jq^{j-1}}{1+(-q)^{j}}-\sum_{j=1}^{\infty }\frac{%
\left( -1\right) ^{j}jq^{2j-1}}{1+q^{2j}},  \label{r4t03b} \\
\varphi ^{2}\left( q\right) \psi ^{2}\left( q^{8}\right) &=&\frac{1}{2}%
\sum_{j=1}^{\infty }\frac{jq^{j-2}}{1+(-q)^{j}}-\sum_{j=1}^{\infty }\frac{%
\left( -1\right) ^{j}jq^{4j-2}}{1+q^{4j}}  \nonumber \\
&&+\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left(
2j-1\right) q^{2j-3}}{1+q^{4j-2}}+\frac{1}{2}\sum_{j=1}^{\infty }\frac{%
\left( -1\right) ^{j}jq^{2j-2}}{1+q^{2j}},  \label{r4t04b} \\
\varphi \left( q\right) \psi ^{3}\left( q^{8}\right) &=&\frac{1}{8}%
\sum_{j=1}^{\infty }\frac{jq^{j-3}}{1+\left( -q\right) ^{j}}+\frac{3}{8}%
\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}jq^{2j-3}}{1+q^{2j}} 
\nonumber \\
&&-\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}jq^{4j-3}}{%
1+q^{4j}}+\frac{1}{8}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left(
2j-1\right) q^{2j-4}}{1+q^{4j-2}},  \label{r4t05b} \\
\varphi \left( q^{4}\right) \psi ^{3}\left( q^{2}\right) &=&\frac{1}{16}%
\sum_{j=1}^{\infty }\frac{(2j-1)q^{\frac{j-2}{2}}}{1-q^{\frac{2j-1}{2}}}+%
\frac{1}{16}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}(2j-1)q^{\frac{j-2%
}{2}}}{1-(-q^{\frac{1}{2}})^{2j-1}}  \nonumber \\
&&-\frac{1}{16}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{\frac{j}{2}%
}(2j-1)q^{\frac{j-2}{2}}}{1-(-q)^{\frac{2j-1}{2}}}  \nonumber \\
&&-\frac{1}{16}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{\frac{3j}{2}%
}(2j-1)q^{\frac{j-2}{2}}}{1+(-q)^{\frac{2j-1}{2}}},  \label{r4t06b}
\end{eqnarray}%
\begin{eqnarray}
\varphi \left( q\right) \varphi \left( q^{4}\right) \psi ^{2}\left(
q^{4}\right) &=&\frac{1}{2}\sum_{j=1}^{\infty }\frac{jq^{j-1}}{1+(-q)^{j}}-%
\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}jq^{2j-1}}{1+q^{2j}%
}  \nonumber \\
&&-\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}(2j-1)q^{2j-2}}{%
1+q^{4j-2}},  \label{r4t07b} \\
\varphi \left( q\right) \varphi ^{2}\left( q^{2}\right) \psi \left(
q^{8}\right) &=&\sum_{j=1}^{\infty }\frac{jq^{j-1}}{1+(-q)^{j}}%
+\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}jq^{2j-1}}{1+q^{2j}} 
\nonumber \\
&&-2\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}jq^{4j-1}}{1+q^{4j}},
\label{r4t08b} \\
\varphi \left( q\right) \psi ^{2}\left( q^{4}\right) \psi \left(
q^{8}\right) &=&\frac{1}{4}\sum_{j=1}^{\infty }\frac{jq^{j-2}}{1+(-q)^{j}}-%
\frac{1}{4}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}jq^{2j-2}}{1+q^{2j}%
}  \nonumber \\
&&+\frac{1}{4}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}(2j-1)q^{2j-3}}{%
1+q^{4j-2}}.  \label{r4t09b}
\end{eqnarray}
\end{theorem}

\begin{theorem}
\label{tm3} 
\begin{eqnarray}
\varphi ^{4}\left( q\right) \varphi ^{2}\left( q^{2}\right)
&=&1+2\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left( 2j-1\right)
^{2}q^{2j-1}}{1-q^{2j-1}}+8\sum_{j=1}^{\infty }\frac{j^{2}q^{j}}{1+q^{2j}} 
\nonumber \\
&&-2\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left( 2j-1\right)
^{2}q^{2j-1}}{1+q^{2j-1}},  \label{r6t01b} \\
\varphi ^{2}\left( q\right) \varphi ^{4}\left( q^{2}\right)
&=&1+2\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left( 2j-1\right)
^{2}q^{2j-1}}{1-q^{2j-1}}+4\sum_{j=1}^{\infty }\frac{j^{2}q^{j}}{1+q^{2j}} 
\nonumber \\
&&-2\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left( 2j-1\right)
^{2}q^{2j-1}}{1+q^{2j-1}},  \label{r6t02b} \\
\psi ^{4}\left( q\right) \psi ^{2}\left( q^{2}\right) &=&\sum_{j=1}^{\infty }%
\frac{j^{2}q^{j-1}}{1+q^{2j}},  \label{r6t03b} \\
\varphi ^{2}\left( q\right) \psi ^{4}\left( q\right) &=&\sum_{j=1}^{\infty }%
\frac{(2j-1)^{2}q^{j-1}}{1+q^{2j-1}},  \label{r6t04b} \\
\varphi ^{2}\left( q\right) \psi ^{4}\left( q^{4}\right) &=&\frac{1}{4}%
\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left( 2j-1\right)
^{2}q^{2j-3}}{1-q^{4j-2}}+\frac{1}{4}\sum_{j=1}^{\infty }\frac{j^{2}q^{j-2}}{%
1+q^{2j}},  \label{r6t05b} \\
\varphi ^{4}\left( q\right) \psi ^{2}\left( q^{4}\right)
&=&\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j}\left( 2j-1\right)
^{2}q^{2j-2}}{1-q^{4j-2}}+2\sum_{j=1}^{\infty }\frac{j^{2}q^{j-1}}{1+q^{2j}}.
\label{r6t06b}
\end{eqnarray}
\end{theorem}

Formula (\ref{r6t04b}) was given by S. Ramanujan \cite[Chapter 17, Entry 17]%
{Rama} \cite[p. 139]{Berndtp3}.\ Proofs of (\ref{r4t02b}), (\ref{r4t03b}),
and (\ref{r6t04b}) were given by S. H. Chan \cite{SHchan04}.\ In the
author's thesis \cite{Lam03}, a total 51 identities are given; only
identities (\ref{r2t01b})--(\ref{r6t06b})  are stated here because
either they appear to be new or involve both sums of squares and triangular
numbers.

Finally, we will demonstrate an arithmetic interpretation of Theorems \ref%
{tm1}--\ref{tm3} in terms of divisor sums.\ For example, (\ref{r4t01b})
implies that the number of solutions in integers $x_{1},x_{2},x_{3},$ and $y_{1}$
of 
$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\left( 2y_{1}+1\right) ^{2}=m, 
$
is 
\begin{equation}
k\left( m\right) \sum_{{d|m}\atop{d\;\text{odd}}}d,  \label{eqm1}
\end{equation}%
where\textit{\ } 
\[
k\left( m\right) =\left\{ 
\begin{tabular}{l}
$1\qquad :m\equiv 1\pmod4,$ \\ 
$6\qquad :m\equiv 2\pmod4,$ \\ 
$3\qquad :m\equiv 3\pmod4,$ \\ 
$8\qquad :m\equiv 4\pmod8,$ \\ 
$0\qquad :m\equiv 0\pmod8.$%
\end{tabular}%
\ \right. 
\]%
An arithmetic interpretation of identity (\ref{r2t03b}) appeared in M. D.
Hirschhorn \cite{Hirschhorn001}. \ 

\section{Preliminary Results}

Following \cite[pp. 120--121]{Cooper2}, we define $f_{1}\left(
\theta \right) ,$ $f_{2}\left( \theta \right) ,$ and $f_{3}\left( \theta
\right) $ by 
\begin{eqnarray}
f_{1}\left( \theta \right) &=&f_{1}\left( \theta ;q\right) =\frac{1}{2}\cot 
\frac{\theta }{2}-2\sum_{j=1}^{\infty }\frac{q^{2j}}{1+q^{2j}}\sin j\theta ,
\label{f1f} \\
f_{2}\left( \theta \right) &=&f_{2}\left( \theta ;q\right) =\frac{1}{2}\csc 
\frac{\theta }{2}+2\sum_{j=1}^{\infty }\frac{q^{2j-1}}{1-q^{2j-1}}\sin
\left( j-\frac{1}{2}\right) \theta ,  \label{f2f} \\
f_{3}\left( \theta \right) &=&f_{3}\left( \theta ;q\right) =\frac{1}{2}\csc 
\frac{\theta }{2}-2\sum_{j=1}^{\infty }\frac{q^{2j-1}}{1+q^{2j-1}}\sin
\left( j-\frac{1}{2}\right) \theta .  \label{f3f}
\end{eqnarray}%
It can be shown \cite[p. 121]{Cooper2}\ that the series in (\ref{f1f})--(\ref{f3f}) converge
for $-$Im $(2\pi \tau )<$ Im $\left( \theta
\right) <$ Im $(2\pi \tau )$.
By Ramanujan's $_{1}\psi _{1}$ summation formula \cite[Chapter 16, Entry 17]%
{Rama}, we have \cite[p. 121]{Cooper2}:
\begin{eqnarray}
f_{1}\left( \theta \right) &=&\frac{1}{2i}\prod_{k=1}^{\infty }\frac{\left(
1+q^{2k-2}e^{i\theta }\right) \left( 1+q^{2k}e^{-i\theta }\right) \left(
1-q^{2k}\right) ^{2}}{\left( 1-q^{2k-2}e^{i\theta }\right) \left(
1-q^{2k}e^{-i\theta }\right) \left( 1+q^{2k}\right) ^{2}},  \label{f1pd} \\
f_{2}\left( \theta \right) &=&\frac{e^{i\theta /2}}{i}\prod_{k=1}^{\infty }%
\frac{\left( 1-q^{2k-1}e^{i\theta }\right) \left( 1-q^{2k-1}e^{-i\theta
}\right) \left( 1-q^{2k}\right) ^{2}}{\left( 1-q^{2k-2}e^{i\theta }\right)
\left( 1-q^{2k}e^{-i\theta }\right) \left( 1-q^{2k-1}\right) ^{2}},
\label{f2pd} \\
f_{3}\left( \theta \right) &=&\frac{e^{i\theta /2}}{i}\prod_{k=1}^{\infty }%
\frac{\left( 1+q^{2k-1}e^{i\theta }\right) \left( 1+q^{2k-1}e^{-i\theta
}\right) \left( 1-q^{2k}\right) ^{2}}{\left( 1-q^{2k-2}e^{i\theta }\right)
\left( 1-q^{2k}e^{-i\theta }\right) \left( 1+q^{2k-1}\right) ^{2}}.
\label{f3pd}
\end{eqnarray}%
These are valid for all values of $\theta $ except $\theta =2m\pi +2n\pi
\tau ,$ where there are poles of order 1. \ Equations (\ref{f1pd})--(\ref%
{f3pd}) provide an analytic continuation for the functions $f_{1}$, $f_{2}$, 
$f_{3}.$ \ 
The functions
$f_{1}$, $f_{2}$, $f_{3}$ are the Jacobian elliptic functions cs, ns, and
ds, respectively. \ See \cite[p. 77]{Cooper1}\ for precise identification.

From \cite[p. 124]{Cooper2} we have 
\begin{eqnarray}
f_{1}^{\prime }\left( \theta \right) &=&-f_{2}\left( \theta \right)
f_{3}\left( \theta \right) ,  \label{dff1} \\
f_{2}^{\prime }\left( \theta \right) &=&-f_{1}\left( \theta \right)
f_{3}\left( \theta \right) ,  \label{dff2} \\
f_{3}^{\prime }\left( \theta \right) &=&-f_{1}\left( \theta \right)
f_{2}\left( \theta \right) .  \label{dff3}
\end{eqnarray}%
Letting 
\begin{eqnarray}
z &=&z\left( q\right) =\prod_{k=1}^{\infty }\left( 1+q^{2k-1}\right)
^{4}\left( 1-q^{2k}\right) ^{2},  \label{zprod} \\
x &=&x\left( q\right) =16q\prod_{k=1}^{\infty }\frac{\left( 1+q^{2k}\right)
^{8}}{\left( 1+q^{2k-1}\right) ^{8}},  \label{x} \\
x^{\prime } &=&x^{\prime }\left( q\right) =\prod_{k=1}^{\infty }\frac{\left(
1-q^{2k-1}\right) ^{8}}{\left( 1+q^{2k-1}\right) ^{8}},  \label{xp}
\end{eqnarray}%
we have (\cite[p. 124--134]{Cooper2}):
\begin{eqnarray}
x+x^{\prime } &=&1,  \label{xxp1} \\
\varphi \left( q\right) &=&\sum_{j=-\infty }^{\infty }q^{j^{2}}=\sqrt{z},
\label{sumsq1} \\
\psi \left( q\right) &=&\sum_{j=0}^{\infty }q^{j\left( j+1\right) /2}=\frac{%
\sqrt{z}x^{\frac{1}{8}}}{\sqrt{2}q^{\frac{1}{8}}}.  \label{sumtr1}
\end{eqnarray}%
Using the infinite products for $f_{1},f_{2},f_{3},$ and comparing with (\ref%
{zprod})--(\ref{xp}), we obtain \cite[p. 129]{Cooper2} the values in Table 1.


\begin{center}
{{\bf Table 1.} \ Values of $f_{1}$, $f_{2}$, and $f_{3}$.}
\[
\begin{tabular}{|c|c|c|c|}
\hline
& $\pi $ & $\pi \tau $ & $\pi +\pi \tau $ \\ \hline
&  &  &  \\ 
$f_{1}\left( \theta \right) $ & $0$ & $\dfrac{z}{2i}$ & $\dfrac{z\sqrt{%
x^{\prime }}}{2i}$ \\ 
&  &  &  \\ \hline
&  &  &  \\ 
$f_{2}\left( \theta \right) $ & $\dfrac{z}{2}$ & $0$ & $\dfrac{z\sqrt{x}}{2}$
\\ 
&  &  &  \\ \hline
&  &  &  \\ 
$f_{3}\left( \theta \right) $ & $\dfrac{z\sqrt{x^{\prime }}}{2}$ & $\dfrac{z%
\sqrt{x}}{2i}$ & $0$ \\ 
&  &  &  \\ \hline 
\end{tabular} 
\]
\end{center}
We also summarize 3 transformations of the functions $x^{\prime }$, $x$,
and $z$ from \cite[pp. 125--126]{Berndtp3} in Table 2.

\begin{center}
{\bf Table 2.} \ Three transformations of the functions $x^{\prime }$, $x$, and $z$%
.%
\[
\begin{tabular}{|c|c|c|c|}
\cline{1-3}\cline{2-2}\cline{4-4}
$q$ & $x^{\prime }$ & $x$ & $z$ \\ \hline
&  &  &  \\ 
$q\rightarrow -q$ & $\dfrac{1}{x^{\prime }}$ & $-\dfrac{x}{x^{\prime }}$ & $z%
\sqrt{x^{\prime }}$ \\ 
&  &  &  \\ \hline
&  &  &  \\ 
$q\rightarrow q^{\frac{1}{2}}$ & $\dfrac{\left( 1-\sqrt{x}\right) ^{2}}{%
\left( 1+\sqrt{x}\right) ^{2}}$ & $\dfrac{4\sqrt{x}}{\left( 1+\sqrt{x}%
\right) ^{2}}$ & $z\left( 1+\sqrt{x}\right) $ \\ 
&  &  &  \\ \hline
&  &  &  \\ 
$q\rightarrow q^{2}$ & $\dfrac{4\sqrt{x^{\prime }}}{\left( 1+\sqrt{x^{\prime
}}\right) ^{2}}$ & $\dfrac{\left( 1-\sqrt{x^{\prime }}\right) ^{2}}{\left( 1+%
\sqrt{x^{\prime }}\right) ^{2}}$ & $\dfrac{1}{2}z\left( 1+\sqrt{x^{\prime }}%
\right) $ \\ 
&  &  &  \\ \hline
\end{tabular}%
\]
\end{center}

We shall give some explanation for Table 2. \ For example, if we apply the
transformation $q\rightarrow -q$ to the functions $x^{\prime },$ $x,$ and $z$,
then the second row of Table 2 implies that
$
x^{\prime }\left( -q\right) =\dfrac{1}{x^{\prime }}, 
x\left( -q\right) =-\dfrac{x}{x^{\prime }}, \mbox{ and } 
z\left( -q\right) =z\sqrt{x^{\prime }}.
$
The transformations $q\rightarrow q^{\frac{1}{2}}$ and $q\rightarrow q^{2}$
can be read similarly.

The results in Table 3 can be easily obtained by applying the results of
Table 2.

\begin{center}
{\bf Table 3.} \ Four transformations of the functions $x^{\prime }$, $x$, and $z$.
\footnotesize
\renewcommand{\arraystretch}{.5}
\[
\begin{tabular}{|c|c|c|c|} 
\hline
$q$ & $x^{\prime }$ & $x$ & $z$ \\ \hline
&  &  &  \\
$q\rightarrow -q^{\frac{1}{2}}$ & $\dfrac{\left( 1+\sqrt{x}\right) ^{2}}{%
\left( 1-\sqrt{x}\right) ^{2}}$ & $\dfrac{-4\sqrt{x}}{\left( 1-\sqrt{x}%
\right) ^{2}}$ & $z\left( 1-\sqrt{x}\right) $ \\ 
&  &  &  \\ \hline 
&  &  &  \\
$q\rightarrow -q^{2}$ & $\dfrac{\left( 1+\sqrt{x^{\prime }}\right) ^{2}}{4%
\sqrt{x^{\prime }}}$ & $\dfrac{-\left( 1-\sqrt{x^{\prime }}\right) ^{2}}{4%
\sqrt{x^{\prime }}}$ & $z\sqrt[4]{x^{\prime }}$ \\ 
&  &  &  \\ \hline 
&  &  &  \\
$q\rightarrow iq^{\frac{1}{2}}$ & $\dfrac{\left( \sqrt{x^{\prime }}-i\sqrt{x}%
\right) ^{2}}{\left( \sqrt{x^{\prime }}+i\sqrt{x}\right) ^{2}}$ & $\dfrac{4i%
\sqrt{xx^{\prime }}}{\left( \sqrt{x^{\prime }}+i\sqrt{x}\right) ^{2}}$ & $%
z\left( \sqrt{x^{\prime }}+i\sqrt{x}\right) $ \\ 
&  &  &  \\ \hline 
&  &  &  \\
$q\rightarrow -iq^{\frac{1}{2}}$ & $\dfrac{\left( \sqrt{x^{\prime }}+i\sqrt{x%
}\right) ^{2}}{\left( \sqrt{x^{\prime }}-i\sqrt{x}\right) ^{2}}$ & $\dfrac{%
-4i\sqrt{xx^{\prime }}}{\left( \sqrt{x^{\prime }}-i\sqrt{x}\right) ^{2}}$ & $%
z\left( \sqrt{x^{\prime }}-i\sqrt{x}\right) $ \\ 
&  &  &  \\ \hline
\end{tabular}%
\]
\end{center}
\normalsize
\section{Proofs of Theorems \protect\ref{tm1}--\protect\ref{tm3}}

The following 3 lemmas are required to prove Theorems \ref{tm1}--\ref%
{tm3}.

\begin{lemma}
\label{lem1} 
\begin{eqnarray}
\varphi \left( q\right) \varphi \left( q^{4}\right) &=&f_{2}\left( \pi
\right) +if_{1}\left( \pi +\pi \tau ;q^{2}\right) ,  \label{r2t01a} \\
\varphi \left( q\right) \psi \left( q^{8}\right) &=&\frac{1}{2q}\left[
f_{2}\left( \pi \tau \right) -if_{1}\left( \pi +\pi \tau ;q^{2}\right) %
\right] ,  \label{r2t02a} \\
\varphi \left( q^{4}\right) \psi \left( q^{2}\right) &=&\frac{1}{4q^{\frac{1%
}{4}}}\left[ f_{2}^{\left( 0\right) }\left( \pi +\pi \tau ;q^{\frac{1}{2}%
}\right) +\frac{1}{i^{\frac{1}{2}}}f_{2}^{\left( 0\right) }\left( \pi +\pi
\tau ;iq^{\frac{1}{2}}\right) \right] .  \label{r2t03a}
\end{eqnarray}
\end{lemma}

\begin{lemma}
\label{lem2} 
\begin{eqnarray}
\varphi ^{3}\left( q\right) \psi \left( q^{8}\right) &=&-\frac{1}{q}%
f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}}\right) +\frac{1}{q}f_{1}^{\prime
}\left( \pi ;q^{2}\right) +\frac{i}{q}f_{3}^{\prime }\left( \pi +\pi \tau
;q^{2}\right) ,  \label{r4t01a} \\
\varphi ^{2}\left( q\right) \psi ^{2}\left( q\right) &=&\frac{1}{2q^{\frac{1%
}{4}}}\left[ f_{2}^{\prime }\left( \pi \tau ;q^{\frac{1}{2}}\right)
-if_{3}^{\prime }\left( \pi +\pi \tau ;q^{\frac{1}{2}}\right) \right] ,
\label{r4t02a} \\
\varphi ^{2}\left( q\right) \psi ^{2}\left( q^{4}\right) &=&\frac{-1}{2q}%
f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}}\right) +\frac{1}{2q}%
f_{1}^{\prime }\left( \pi \right) ,  \label{r4t03a} \\
\varphi ^{2}\left( q\right) \psi ^{2}\left( q^{8}\right) &=&\frac{-1}{4q^{2}}%
f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}}\right) +\frac{1}{2q^{2}}%
f_{1}^{\prime }\left( \pi ;q^{2}\right) +\frac{1}{2q^{2}}if_{3}^{\prime
}\left( \pi +\pi \tau ;q^{2}\right)  \nonumber \\
&&-\frac{1}{4q^{2}}f_{1}^{\prime }\left( \pi \right) ,  \label{r4t04a} \\
\varphi \left( q\right) \psi ^{3}\left( q^{8}\right) &=&-\frac{1}{16q^{3}}%
f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}}\right) -\frac{3}{16q^{3}}%
f_{1}^{\prime }\left( \pi \right) +\frac{1}{4q^{3}}f_{1}^{\prime }\left( \pi
;q^{2}\right)  \nonumber \\
&&+\frac{i}{8q^{3}}f_{3}^{\prime }\left( \pi +\pi \tau ;q^{2}\right) ,
\label{r4t05a} \\
\varphi \left( q^{4}\right) \psi ^{3}\left( q^{2}\right) &=&\frac{1}{16q^{%
\frac{3}{4}}}f_{2}^{\prime }\left( \pi \tau ;q^{\frac{1}{2}}\right) +\frac{i%
}{16q^{\frac{3}{4}}}f_{2}^{\prime }\left( \pi \tau ;-q^{\frac{1}{2}}\right) 
\nonumber \\
&&-\frac{i^{\frac{1}{2}}}{16q^{\frac{3}{4}}}f_{2}^{\prime }\left( \pi \tau
;iq^{\frac{1}{2}}\right) +\frac{1}{16i^{\frac{1}{2}}q^{\frac{3}{4}}}%
f_{3}^{\prime }\left( \pi +\pi \tau ;iq^{\frac{1}{2}}\right) ,  \nonumber \\
&&  \label{r4t06a} \\
\varphi \left( q\right) \psi ^{2}\left( q^{4}\right) \phi \left(
q^{4}\right) &=&-\frac{1}{4q}f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}%
}\right) +\frac{1}{4q}f_{1}^{\prime }\left( \pi \right) -\frac{i}{2q}%
f_{3}^{\prime }\left( \pi +\pi \tau ;q^{2}\right) ,  \nonumber \\
&&  \label{r4t07a} \\
\varphi \left( q\right) \varphi ^{2}\left( q^{2}\right) \psi \left(
q^{8}\right) &=&-\frac{1}{2q}f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}%
}\right) -\frac{1}{2q}f_{1}^{\prime }\left( \pi \right) +\frac{1}{q}%
f_{1}^{\prime }\left( \pi ;q^{2}\right) ,  \nonumber \\
&&  \label{r4t08a} \\
\varphi \left( q\right) \psi ^{2}\left( q^{4}\right) \psi \left(
q^{8}\right) &=&-\frac{1}{8q^{2}}f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}%
}\right) +\frac{1}{8q^{2}}f_{1}^{\prime }\left( \pi \right) +\frac{i}{4q^{2}}%
f_{3}^{\prime }\left( \pi +\pi \tau ;q^{2}\right) .  \nonumber \\
&&  \label{r4t09a}
\end{eqnarray}
\end{lemma}

\begin{lemma}
\label{lem3} 
\begin{eqnarray}
\varphi ^{4}\left( q\right) \varphi ^{2}\left( q^{2}\right)
&=&4f_{2}^{\prime \prime }\left( \pi \right) -4if_{1}^{\prime \prime }\left(
\pi \tau \right) +4f_{3}^{\prime \prime }\left( \pi \right) ,  \label{r6t01a}
\\
\varphi ^{2}\left( q\right) \varphi ^{4}\left( q^{2}\right)
&=&4f_{2}^{\prime \prime }\left( \pi \right) -2if_{1}^{\prime \prime }\left(
\pi \tau \right) +4f_{3}^{\prime \prime }\left( \pi \right) ,  \label{r6t02a}
\\
\psi ^{4}\left( q\right) \psi ^{2}\left( q^{2}\right) &=&-\frac{i}{2q}%
f_{1}^{\prime \prime }\left( \pi \tau \right) ,  \label{r6t03a} \\
\varphi ^{2}\left( q\right) \psi ^{4}\left( q\right) &=&-\frac{2i}{q^{\frac{1%
}{2}}}f_{3}^{\prime \prime }\left( \pi \tau \right) ,  \label{r6t04a} \\
\varphi ^{2}\left( q\right) \psi ^{4}\left( q^{4}\right) &=&\frac{1}{4q^{2}}%
f_{2}^{\prime \prime }\left( \pi \right) -\frac{i}{8q^{2}}f_{1}^{\prime
\prime }\left( \pi \tau \right) -\frac{1}{4q^{2}}f_{3}^{\prime \prime
}\left( \pi \right) ,  \label{r6t05a} \\
\varphi ^{4}\left( q\right) \psi ^{2}\left( q^{4}\right) &=&\frac{1}{q}%
f_{2}^{\prime \prime }\left( \pi \right) -\frac{i}{q}f_{1}^{\prime \prime
}\left( \pi \tau \right) -\frac{1}{q}f_{3}^{\prime \prime }\left( \pi
\right) .  \label{r6t06a}
\end{eqnarray}
\end{lemma}

\noindent {\it Proof.}
The proofs of Lemmas \ref{lem1}--\ref{lem3} follow by using (\ref{dff1})--(%
\ref{dff3}), (\ref{xxp1})--(\ref{sumtr1}), and Tables 1--3. We express both
sides of (\ref{r2t01a})--(\ref{r6t06a}) in terms of $z$ and $x.$ \ We give
complete details for $\varphi \left( q^{4}\right) \psi ^{3}\left(
q^{2}\right) $ only; the other formulae can be proved in a similar way.
  \newline
By employing (\ref{sumsq1}), (\ref{sumtr1}), and Table 2, the left hand side
of (\ref{r4t01a}) can be rewritten as
\begin{equation}
\varphi ^{3}\left( q\right) \psi \left( q^{8}\right) =\frac{1}{4q}%
z^{2}\left( 1-x^{\prime ^{\frac{1}{4}}}\right) .  \label{wl1}
\end{equation}
Next by substituting the values of $\theta =\pi $ and $\pi +\pi \tau $ into (%
\ref{dff1}) and (\ref{dff3}), respectively, and then using Table 1, we find
that
\begin{eqnarray}
f_{1}^{\prime }\left( \pi \right) &=&-\frac{z^{2}\sqrt{x^{\prime }}}{4},
\label{fd1} \\
f_{3}^{\prime }\left( \pi +\pi \tau \right) &=&-\frac{iz^{2}\sqrt{xx^{\prime
}}}{4}.  \label{fd2}
\end{eqnarray}
Now if we employ the results of Table 2, Table 3, (\ref{fd1}), and (\ref{fd2}%
) in the right hand side of (\ref{r4t01a}) we get
\begin{eqnarray}
&&-\frac{1}{q}f_{1}^{\prime }\left( \pi ;iq^{\frac{1}{2}}\right) +\frac{1}{q}%
f_{1}^{\prime }\left( \pi ;q^{2}\right) +\frac{i}{q}f_{3}^{\prime }\left(
\pi +\pi \tau ;q^{2}\right)  \notag \\
&=&\frac{1}{4q}z^{2}-\frac{1}{8q}z^{2}x^{\prime ^{\frac{1}{4}}}\left(
1+x^{\prime ^{\frac{1}{2}}}\right) -\frac{1}{8q}z^{2}x^{\prime ^{\frac{1}{4}%
}}\left( 1-x^{\prime ^{\frac{1}{2}}}\right)  \notag \\
&=&\frac{1}{4q}z^{2}\left( 1-x^{\prime ^{\frac{1}{4}}}\right) .  \label{wl2}
\end{eqnarray}
Combining (\ref{wl1}) and (\ref{wl2}) we obtain (\ref{r4t01a}).
This completes the proof of Lemmas \ref{lem1}--\ref{lem3}. 
\hfill $\Box$

We now prove Theorems \ref{tm1}--\ref{tm3}.

\noindent {\it Proofs of Theorems 1--3.}
We use the series
expansions of $f_{1}\left( \theta \right) ,f_{2}\left( \theta \right) ,$ and
$f_{3}\left( \theta \right) $ in (\ref{f1f}), (\ref{f2f}), and (\ref{f3f}),
while the right hand sides of the results in Lemmas \ref{lem1}--\ref{lem3}
can be represented explicitly as Lambert series. \ We give complete details
for (\ref{r4t01b}) only; the others can be proved in a similar 
way. 

First, differentiating (\ref{f2f}) and (\ref{f3f}) with respect to $\theta $
we have
\begin{eqnarray}
f_{1}^{\prime }\left( \theta \right) &=&-\frac{1}{4}-\frac{1}{4}\cot ^{2}%
\frac{\theta }{2}-2\sum_{j=1}^{\infty }\frac{jq^{2j}}{1+q^{2j}}\cos j\theta ,
\label{ff22} \\
f_{3}^{\prime }\left( \theta \right) &=&-\frac{1}{4}\csc \frac{\theta }{2}%
\cot \frac{\theta }{2}-\sum_{j=1}^{\infty }\frac{\left( 2j-1\right) q^{2j-1}%
}{1+q^{2j-1}}\cos \left( j-\frac{1}{2}\right) \theta .  \label{ff33}
\end{eqnarray}
Substituting $\theta =\pi $ into (\ref{ff22}) we have
\begin{equation}
f_{1}^{\prime }\left( \pi \right) =-\frac{1}{4}-2\sum_{j=1}^{\infty }\frac{%
\left( -1\right) ^{j}jq^{2j}}{1+q^{2j}}.  \label{pf1}
\end{equation}
Substituting $\theta =\pi +\pi \tau $ into (\ref{ff33}) and recalling that $%
q=e^{i\pi \tau }$, we have
\begin{eqnarray}
f_{3}^{\prime }\left( \pi +\pi \tau \right) &=&\frac{e^{i(\pi +\pi \tau
)/2}+e^{-i(\pi +\pi \tau )/2}}{2\left( e^{i(\pi +\pi \tau )/2}-e^{-i(\pi
+\pi \tau )/2}\right) ^{2}}  \notag \\
&&-\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( 2j-1\right) q^{2j-1}}{%
1+q^{2j-1}}\left( e^{i\left( j-1/2\right) (\pi +\pi \tau )}-e^{-i\left(
j-1/2\right) (\pi +\pi \tau )}\right)  \notag \\
&=&\frac{iq^{1/2}\left( 1-q\right) }{2\left( 1+q\right) ^{2}}-\frac{1}{2}%
\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j-1/2}\left( 2j-1\right)
q^{j-1/2}(1-1+q^{2j-1})}{1+q^{2j-1}}  \notag \\
&&+\frac{1}{2}\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j-1/2}\left(
2j-1\right) q^{j-1/2}}{1+q^{2j-1}}  \notag \\
&=&\frac{iq^{1/2}\left( 1-q\right) }{2\left( 1+q\right) ^{2}}-\frac{1}{2}%
\sum_{j=1}^{\infty }\left( -1\right) ^{j-1/2}\left( 2j-1\right) q^{j-1/2}
\notag \\
&&+\sum_{j=1}^{\infty }\frac{\left( -1\right) ^{j-1/2}\left( 2j-1\right)
q^{j-1/2}}{1+q^{2j-1}}.  \label{fp2}
\end{eqnarray}
We observe that
\begin{eqnarray*}
\sum_{j=1}^{\infty }\left( -1\right) ^{j-1/2}\left( 2j-1\right) q^{j-1/2}
&=&\sum_{j=1}^{\infty }j(-q)^{j/2}-\sum_{j=1}^{\infty }2j(-q)^{j} \\
&=&\frac{(-q)^{1/2}}{\left( 1-(-q)^{1/2}\right) ^{2}}-\frac{2(-q)}{\left(
1-(-q)\right) ^{2}} \\
&=&\frac{iq^{1/2}\left( 1-q\right) }{\left( 1+q\right) ^{2}}.
\end{eqnarray*}
Substituting these into (\ref{fp2}) we obtain
\begin{equation}
f_{3}^{\prime }\left( \pi +\pi \tau \right) =\sum_{j=1}^{\infty }\frac{%
\left( -1\right) ^{j-1/2}\left( 2j-1\right) q^{j-1/2}}{1+q^{2j-1}}.
\label{pf2}
\end{equation}
Using (\ref{pf1}) and (\ref{pf2}) in the right hand side of (\ref{r4t01a}%
) and simplifying the results we obtain (\ref{r4t01b}).\newline
This completes the proofs of Theorems \ref{tm1}--\ref{tm3}. 
\hfill $\Box$

Next we present an arithmetic interpretation of Theorems \ref{tm1}--\ref{tm3}
in terms of divisor sums.

Let $k$ and $m$ be positive integers.  Let $\lambda _{1},\lambda
_{2},\ldots ,\lambda _{k}$ and $\mu _{1},\mu _{2},\ldots ,\mu _{m}$ be
positive integers, where $\lambda _{1}\leq \lambda _{2}\leq \cdots \leq
\lambda _{k}$ and $\mu _{1}\leq \mu _{2}\leq \cdots \leq \mu _{m}$. \ The
function 
\[
r(\lambda _{1}\square +\lambda _{2}\square +\cdots +\lambda _{k}\square +\mu
_{1}\triangle +\mu _{2}\bigtriangleup +\cdots +\mu _{m}\bigtriangleup
)\left( n\right) 
\]%
will denote the number of solutions in integers of 
\begin{eqnarray}
\lambda _{1}x_{1}^{2}+\lambda _{2}x_{2}^{2}+\cdots +\lambda _{k}x_{k}^{2} 
+\mu _{1}\frac{y_{1}\left( y_{1}+1\right) }{2}+\mu _{2}\frac{y_{2}\left(
y_{2}+1\right) }{2}+\cdots +\mu _{m}\frac{y_{m}\left( y_{m}+1\right) }{2} 
=n,  \label{geo1}
\end{eqnarray}%
where $n=0,1,2,3,\ldots .$
We also define $r(\lambda _{1}\square +\cdots +\lambda
_{k}\square +\mu _{1}\triangle  +\cdots +\mu
_{m}\bigtriangleup )\left( 0\right) =1$. \ 

Then the generating function for $r(\lambda _{1}\square +\lambda _{2}\square
+\cdots +\lambda _{k}\square +\mu _{1}\triangle +\mu _{2}\bigtriangleup
+\cdots +\mu _{m}\bigtriangleup )\left( n\right) $ is 
\begin{eqnarray}
&  \displaystyle\sum_{n=0}^{\infty }r(\lambda _{1}\square +\lambda _{2}\square +\cdots
+\lambda _{k}\square +\mu _{1}\triangle +\mu _{2}\bigtriangleup +\cdots +\mu
_{m}\bigtriangleup )\left( n\right) q^{n}  \nonumber \\
&=\varphi \left( q^{\lambda _{1}}\right) \varphi \left( q^{\lambda
_{2}}\right) \cdots \varphi \left( q^{\lambda _{k}}\right) \psi \left(
q^{\mu _{1}}\right) \psi \left( q^{\mu _{2}}\right) \cdots \psi \left(
q^{\mu _{m}}\right) .  \label{mixst}
\end{eqnarray}%
We remark that since (\ref{geo1}) is equivalent to 
 \footnotesize
\begin{eqnarray}
2\lambda _{1}x_{1}^{2}+2\lambda _{2}x_{2}^{2}+\cdots +2\lambda
_{k}x_{k}^{2} 
+\mu _{1}\left( y_{1}+\frac{1}{2}\right) ^{2}+\mu _{2}\left( y_{2}+\frac{1%
}{2}\right) ^{2}+\cdots +\mu _{m}\left( y_{m}+\frac{1}{2}\right) ^{2} 
=2n+\frac{m}{4},  \label{complete}
\end{eqnarray}%
\normalsize
then geometrically, $2^{m}r(\lambda _{1}\square +\lambda _{2}\square +\cdots
+\lambda _{k}\square +\mu _{1}\triangle +\mu _{2}\bigtriangleup +\cdots +\mu
_{m}\bigtriangleup )\left( n\right) $ counts the number of lattice points on
the $k+m$ dimensional ellipsoid centred at ($0,$ $0,$ $\ldots ,$ $0,$ $-%
\frac{1}{2},$ $-\frac{1}{2},$ $\ldots ,$ $-\frac{1}{2}$), the point whose
first $k$ coordinates are $0$ and remaining $m$ coordinates are $-\frac{1}{2}%
,$ with radius $\sqrt{2n+\frac{m}{4}}.$

Now we give complete details for an arithmetic interpretation of  (%
\ref{r4t01b}) in terms of divisor sums.

\begin{corollary}
For $n\geq 1,$%
\begin{equation}
r\left( \square +\square +\square +8\triangle \right) \left( n\right)
=k\left( n\right) \sum_{{d|n+1} \atop {d\;\text{odd}}}d, 
\label{r4t01c}
\end{equation}
\textit{where } 
\[
k\left( n\right) =\left\{ 
\begin{tabular}{l}
$6\qquad :n\equiv 1\pmod4,$ \\ 
$3\qquad :n\equiv 2\pmod4,$ \\ 
$8\qquad :n\equiv 3\pmod8,$ \\ 
$1\qquad :n\equiv 0\pmod4,$ \\ 
$0\qquad :n\equiv 7\pmod8.$%
\end{tabular}
\right. 
\]
\end{corollary}

\noindent {\it Proof.}
First use (\ref{mixst}) and expand the right hand side using the geometric
series in (\ref{r4t01b}) to get
\begin{eqnarray*}
&&\sum_{n=0}^{\infty }r\left( \square +\square +\square +8\triangle \right)
\left( n\right) q^{n} \\
&=&2\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }\left( -1\right)
^{mj}(j+1)q^{\left( j+1\right) \left( m+1\right) -1} \\
&&+2\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }\left( -1\right)
^{m+j}(j+1)q^{4\left( j+1\right) \left( m+1\right) -1} \\
&&-\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }\left( -1\right)
^{m+j}(2j+1)q^{\left( 2j+1\right) \left( 2m+1\right) -1} \\
&=&2\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }(2j+1)q^{\left( 2j+1\right)
\left( m+1\right) -1}+2\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }\left(
-1\right) ^{m}(2j+2)q^{\left( 2j+2\right) \left( m+1\right) -1} \\
&&-2\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }\left( -1\right)
^{(m+1)}(2j+1)q^{4\left( 2j+1\right) \left( m+1\right) -1} \\
&&+2\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }\left( -1\right)
^{m+1}(2j+2)q^{4\left( 2j+2\right) \left( m+1\right) -1} \\
&&-\sum_{j=0}^{\infty }\sum_{m=0}^{\infty }\left( -1\right)
^{m+j}(2j+1)q^{\left( 2j+1\right) \left( 2m+1\right) -1}
\end{eqnarray*}
\begin{eqnarray*}
&=&\sum_{n=0}^{\infty }\left[ 2\sum_{\left( 2j+1\right) \left( m+1\right)
=n+1}(2j+1)+2\sum_{\left( 2j+2\right) \left( m+1\right) =n+1}\left(
-1\right) ^{m}(2j+1)\right. \\
&&-2\sum_{4\left( 2j+1\right) \left( m+1\right) =n+1}\left( -1\right)
^{(m+1)}(2j+1) \\
&&+2\sum_{4\left( 2j+2\right) \left( m+1\right) =n+1}\left( -1\right)
^{(m+1)}(2j+2) \\
&&\left. -\sum_{\left( 2j+1\right) \left( 2m+1\right) =n+1}\left( -1\right)
^{(m+j)}(2j+1)\right] q^{n} \\
&=&\sum_{n=0}^{\infty }\left[ 2\sum_{{d|n+1} \atop {d\;\text{odd}}}d+4\sum_{
{d|n+1} \atop {n,d\;\text{odd}}}d\,-\,2\!\!\!\!\!\!\!\!\!\!\sum_{{d|n+1} \atop {n\equiv
3\!\!\!\!\!\pmod4,d\;
\text{odd}}}\!\!\!\!\!\!\!\!\!\!\left( -1\right) ^{\frac{n+1}{4d}}d\,-\,4
\!\!\!\!\!\!\!\!\!\!\sum_{{d|n+1} \atop
{n\equiv 7\!\!\!\!\!\pmod8,d\;\text{odd}}}\!\!\!\!\!\!\!\!\!\!d\right. \\
&&\left. -\!\!\!\!\!\sum_{{d|n+1} \atop {n\text{ even, }d\;\text{odd}}}
\!\!\!\!\!\left( -1\right)
^{\frac{n}{2}}d\right] q^{n}.
\end{eqnarray*}
By comparing coefficients of $q^{n}$ on both sides we obtain
\begin{eqnarray*}
r\left( \square +\square +\square +8\triangle \right) \left( n\right)
&=&2\sum_{{d|n+1} \atop {d\;\text{odd}}}d\,+\,4\!\!\sum_{{d|n+1} \atop {n,d\;\text{odd}%
}}d\,-\,2\!\!\!\!\!\!\!\!\!\!\sum_{{d|n+1} \atop {n\equiv 3\!\!\!\!\!\pmod4,d\;\text{odd}}}
\!\!\!\!\!\!\!\!\!\!\left(
-1\right) ^{%
\frac{n+1}{4d}}d \\
&&-\,4\!\!\!\!\!\!\!\!\!\!\sum_{{d|n+1} \atop {n\equiv 7
\!\!\!\!\!\pmod8,d\;\text{odd}}}\!\!\!\!\!\!\!\!\!\!d\,\,\,-\sum_{{d|n+1}
\atop {n\text{ even, }d\;\text{odd}}}\!\!\!\!\!\left( -1\right) ^{\frac{n}{2}}d.
\end{eqnarray*}
If $n\equiv 1\pmod4$, then $n+1=4k+2$ and so
\begin{eqnarray}
r\left( \square +\square +\square +8\triangle \right) \left( n\right)
&=&2\sum_{{d|n+1} \atop {d\;\text{odd}}}d+4\sum_{{d|4k+2}
\atop {d\;\text{odd}}}d  \notag \\
&=&2\sum_{{d|n+1} \atop {d\;\text{odd}}}d+4\sum_{{d|n+1} \atop {d\;\text{odd}}}d  \notag \\
&=&6\sum_{{d|n+1} \atop {d\;\text{odd}}}d.  \label{pt1}
\end{eqnarray}
Similarly, if $n\equiv 2\pmod4$, then $n+1=4k+3$ and so
\begin{eqnarray}
r\left( \square +\square +\square +8\triangle \right) \left( n\right)
&=&2\sum_{{d|n+1} \atop
{d\;\text{odd}}}d-\sum_{{d|4k+3}\atop{d\;\text{odd}}}\left( -1\right)^{\frac{4k+2}{2}}d  \notag
\\ &=&2\sum_{{d|n+1} \atop {d\;\text{odd}}}d+\sum_{{d|n+1} \atop {d\;\text{odd}}}d
\notag \\
&=&3\sum_{{d|n+1} \atop {d\;\text{odd}}}d.  \label{pt2}
\end{eqnarray}
If $n\equiv 3\pmod8$, then $n+1=8k+4$ and so
\begin{eqnarray}
r\left( \square +\square +\square +8\triangle \right) \left( n\right)
&=&2\sum_{{d|n+1} \atop
{d\;\text{odd}}}d+4\sum_{{d|8k+4}\atop{d\;\text{odd}}%
}d-2\sum_{{d|8k+4}\atop{d\;\text{odd}}}\left( -1\right)
^{\frac{8k+4}{4d}}d
\notag \\
&=&2\sum_{{d|n+1} \atop {d\;\text{odd}}}d+4\sum_{{d|n+1} \atop {d\;\text{odd}}%
}d+2\sum_{{d|n+1} \atop {d\;\text{odd}}}d  \notag \\
&=&8\sum_{{d|n+1} \atop {d\;\text{odd}}}d.  \label{pt3}
\end{eqnarray}
If $n\equiv 0\pmod4$, then $n+1=4k+1$ and so
\begin{eqnarray}
r\left( \square +\square +\square +8\triangle \right) \left( n\right)
&=&2\sum_{{d|n+1} \atop
{d\;\text{odd}}}d-\sum_{{d|4k+1}\atop{d\;\text{odd}}}\left( -1\right)^{\frac{4k}{2}}d  \notag
\\ &=&2\sum_{{d|n+1} \atop {d\;\text{odd}}}d-\sum_{{d|n+1} \atop {d\;\text{odd}}}d
\notag \\
&=&\sum_{{d|n+1} \atop {d\;\text{odd}}}d.  \label{pt4}
\end{eqnarray}
If $n\equiv 7\pmod8$, then $n+1=8k+8$ and so
\begin{eqnarray}
r\left( \square +\square +\square +8\triangle \right) \left( n\right)
&=&2\sum_{{d|n+1} \atop
{d\;\text{odd}}}d+4\sum_{{d|8k+8}\atop{d\;\text{odd}}%
}d-2\sum_{{d|8k+8}\atop{d\;\text{odd}}}\left( -1\right)
^{\frac{8k+8}{4d}}d-4\sum_{{d|8k+8}\atop{d\;\text{odd}}}d  \notag \\
&=&0.  \label{pt5}
\end{eqnarray}
Combining (\ref{pt1})--(\ref{pt5}), we obtain (\ref{r4t01c}). 
\hfill $\Box$

By (\ref{complete}), formula (\ref{r4t01c}) is equivalent to (\ref{eqm1}). \ 

\section{Remarks}

The results in this paper can also be proved using the theory of modular
forms. \ I thank the referee for his/her permission to reproduce the
following remark.

Consider identity (\ref{r6t03b}). \ We have that $q\psi ^{4}\left( q\right)
\psi ^{2}\left( q^{2}\right) $ is a modular form of weight 3 on $\Gamma
_{0}\left( 4\right) $. \ On the other hand, it is easy to check that%
\[
F\left( q\right) =\sum_{j=-\infty }^{\infty }\frac{j^{2}q^{j}}{1+q^{2j}} 
\]%
is an Eisenstein series in that same space. \ To see this, let $\chi $ be
the non-trivial Dirichlet character mod 4, and let $\sigma \left( n\right)
:=\sum_{d|n}\chi \left( n/d\right) d^{2}.$ \ Define%
\[
E\left( q\right) =\sum_{n=1}^{\infty }\sigma \left( n\right) q^{n}. 
\]%
It is known that $E\left( q\right) $ is in the space (see \cite[chapter 4]{Diamond1}
for a complete discussion). \ It is not hard to check that $E\left( q\right) =F\left( q\right) $
directly. \ The identity $q\psi ^{4}\left( q\right) \psi ^{2}\left( q^{2}\right) =F\left(
q\right) $ then follows by checking that the first few terms agree.

An arithmetic interpretation of other identities can be found in the
author's thesis \cite{Lam03}. \ We shall remark that Hirschhorn \cite%
{Hirschhorn001} also presented many others which give the number of
representations of an integer $n$ by various quadratic forms in terms of
divisor sums. \ 

\section*{Acknowledgements}

The author would like to thank Professors Shaun Cooper and Mike Hirschhorn
for their encouragement and for uncovering several misprints in an earlier
version of this article. \ The author also would like to thank the anonymous
referee for his/her valuable suggestions.

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