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\begin{document}
\vspace*{-40pt} 
\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 
(2005), \#A16} 
\vskip 30pt 

\begin{center}
\uppercase{\bf Relations Among Fourier Coefficients of\\ Certain Eta
Products}
 \vskip 20pt
 {\bf Shaun Cooper}\\
 {\smallit Institute of Information and Mathematical Sciences, Massey
University,
Auckland  102904, New Zealand}\\
 {\tt s.cooper@massey.ac.nz}\\
 \vskip 10pt
 {\bf Sanoli Gun}\\
 {\smallit Harish-Chandra Research Institute, Jhusi,
Allahabad 211019, India}\\
 {\tt jhulan@mri.ernet.in}\\ 
 \vskip 10pt
 {\bf Michael Hirschhorn}\\
 {\smallit School of Mathematics, University of New South Wales,
 Sydney 2052,
Australia}\\
 {\tt m.hirschhorn@unsw.edu.au}\\ 
 \vskip 10pt
 {\bf B. Ramakrishnan}\\
 {\smallit Harish-Chandra Research Institute,  Jhusi,
Allahabad 211019, India}\\
 {\tt ramki@mri.ernet.in}\\ 
 \end{center}
 \vskip 20pt
 \centerline{\smallit Received: 3/29/05, Accepted: 7/21/05,
 Published: 7/27/05}
 \vskip 30pt




\centerline{\bf Abstract}
\noindent
Certain arithmetic relations for the coefficients in the expansions of
$(q)_\infty^r$, $(q)_\infty^r(q^t)_\infty^s$, $t=2,3,4$, were studied by
M. Newman, S. Cooper, M. D. Hirschhorn, R. Lewis, S. Ahlgren and R.
Chapman. In this work, we prove similar identities for certain
multi-product expansions using an elementary method.

%\footnotesize
%\noindent
%2000 {\em Mathematics subject classification.} Primary
%11F20, 11P83, 33D67
%\normalsize

\pagestyle{myheadings}
 \markright{\smalltt INTEGERS: 
\smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5
(2005), \#A16\hfill}

 \thispagestyle{empty} 
 \baselineskip=15pt 
 \vskip 20pt





\section*{\normalsize 1. Introduction} \addtocounter{section}{+1}
For an integer $r$, let
\begin{equation}
(q)_\infty^r = \prod_{n=1}^\infty (1- q^n)^r = \sum_{n\ge 0} a_r(n) q^n,
\end{equation}
where $q=e^{2\pi iz}$ and $\Im(z)>0$ and let
\begin{equation}\label{super}
f_j(q) = (q)_\infty^{r_j}(q^2)_\infty^{s_j}(q^4)_\infty^{t_j},
\end{equation}
where $r_j, s_j, t_j$ are certain specific integers (see theorem).
In this article we consider the following products
$$
f_i(q^\ell)f_k(q^m) = \sum_{n=0}^\infty a(n)q^n \quad (l,m>0),
$$
and  prove certain identities involving the Fourier
coefficients $a(n)$ by elementary arguments. Similar identities for eta
powers and products of two eta functions were earlier obtained by several
authors \cite{{scott},{chap},{chl},{newman}}.



\section*{\normalsize 2. Statement of theorem}\addtocounter{section}{+1}
\label{intro}
\setcounter{equation}{0}
Let $q$ be a complex number satisfying $|q|<1$.
It is readily checked that
$\displaystyle{(-q)_\infty =
\frac{(q^2)_\infty^3}{(q)_\infty(q^4)_\infty}.}$
Let \vskip -30pt
$$ \renewcommand{\arraystretch}{2.5}
\begin{array}{lll}
f_1(q) = (q)_\infty, &
  & \displaystyle{f_7(q) = f_1(-q)
  = \frac{(q^2)_\infty^3}{(q)_\infty(q^4)_\infty}}, \\ 
f_2(q) = (q)_\infty^3, &
  & \displaystyle{f_8(q) = f_2(-q)
  = \frac{(q^2)_\infty^9}{(q)_\infty^3(q^4)_\infty^3}}, \\ 
f_3(q) = \displaystyle{\frac{(q)_\infty^2}{(q^2)_\infty}}, &
  & \displaystyle{f_9(q) = f_3(-q)
  = \frac{(q^2)_\infty^5}{(q)_\infty^2(q^4)_\infty^2}}, \\ 
f_4(q) = \displaystyle{\frac{(q^2)_\infty^2}{(q)_\infty}}, &
  & \displaystyle{f_{10}(q) = f_4(-q)
  = \frac{(q)_\infty(q^4)_\infty}{(q^2)_\infty}}, \\ 
f_5(q) = \displaystyle{\frac{(q)_\infty^5}{(q^2)_\infty^2}}, &
  & \displaystyle{f_{11}(q) = f_5(-q)
  = \frac{(q^2)_\infty^{13}}{(q)_\infty^5(q^4)_\infty^5}}, \\ 
f_6(q) = \displaystyle{\frac{(q^2)_\infty^5}{(q)_\infty^2}}, &
  & \displaystyle{f_{12}(q) = f_6(-q)
  = \frac{(q)_\infty^2(q^4)_\infty^2}{(q^2)_\infty}}. \\ 
\end{array}
$$
Observe that each function $f_i(q)$ has the form
$
f_i(q) = (q)_\infty^{r_i}(q^2)_\infty^{s_i}(q^4)_\infty^{t_i},
$
for certain integers $r_i$, $s_i$, $t_i$.
By the triple product and quintuple product identities,
we have \cite[pp. 64--65 and 306--307]{borwein}, \cite{chl}
\begin{eqnarray}
f_1(q) & = & \sum_{\a \equiv 1 \pmod{6}} (-1)^{(\a-1)/6} q^{(\a^2-1)/24},
\nonumber\\
f_2(q) & = & \sum_{\a \equiv 1 \pmod{4}} \a q^{(\a^2-1)/8}, \nonumber\\
f_3(q) & = & \sum_{\a} (-1)^{\a} q^{\a^2}, \nonumber\\
f_4(q) & = & \sum_{\a \equiv 1 \pmod{4}} q^{(\a^2-1)/8}, \nonumber\\
f_5(q) & = & \sum_{\a \equiv 1 \pmod{6}} \a q^{(\a^2-1)/24}, \nonumber\\
f_6(q) & = & \sum_{\a \equiv 1 \pmod{3}} (-1)^{\a-1}\a q^{(\a^2-1)/3},
\label{series}
\end{eqnarray}
where in each case the sum is over all integers $\a$, positive and negative,
satisfying the
given congruence.

For $1\leq i \leq 12$, let
$d_i =r_i+2s_i+4t_i$ and
$\lambda_{i} =\left\lceil\frac{1}{2}(r_i+s_i+t_i)\right\rceil -1.
$
Let
\begin{eqnarray*}
(e_1,e_2,\cdots,e_{12}) & = & (1,3,24,3,1,8,1,3,24,3,1,8), \\
(n_1,n_2,\cdots,n_{12}) & = & (6,4,1,4,6,3,6,4,1,4,6,3).
\end{eqnarray*}
Observe that $d_i=e_i$ unless $i=3$ or $9$, in which case $d_3=d_9=0$.
For $1\leq i\leq 12$ and $p$ an odd prime, define
$\displaystyle{\epsilon_i(p)=\left(\frac{a_i}{p}\right)}$, where 
$
(a_1,\dots,a_{12})=(3,-1,1,1,-3,-3,6,-2,1,2,-6,-3).
$
The main purpose of this article is to prove the following result.

\noindent {\bf Theorem.}{\it
\label{thm}
Let $\ell$ and $m$ be positive integers, and let $1\leq j,k \leq 12$.
Let $p>3$ be any prime satisfying
$
\left(\frac{-e_je_k\ell m}{p}\right) = -1
$
and put
$\Delta=\frac{p^2-1}{24}.
$
Let
$
f_j(q^\ell)f_k(q^m) = \sum_{n=0}^\infty a(n)q^n.
$
Then the coefficients $a(n)$ satisfy
$$
a(pn+(\ell d_j+md_k)\Delta)
= \epsilon_j(p)\epsilon_k(p) p^{\lambda_j+\lambda_k}
a\left(\frac{n}{p}\right).
$$
}
\vskip -10pt
\noindent
{\bf Example.} ($j=5,\;k=10$)\,\,
We have
$
f_5(q) = (q)_\infty^5 (q^2)_\infty^{-2} \mbox{ and }
f_{10}(q) = (q)_\infty(q^2)_\infty^{-1}(q^4)_\infty,
$
so
$
(r_5,s_5,t_5)=(5,-2,0),\;\;(r_{10},s_{10},t_{10})=(1,-1,1),
d_5 = r_5+2s_5+4t_5 = 1,\;\; d_{10} = r_{10}+2s_{10}+4t_{10} = 3,
\lambda_{5} =
 \left\lceil\frac{1}{2}(r_5+s_5+t_5)\right\rceil -1 = 1,\;\;
\lambda_{10} =
 \left\lceil\frac{1}{2}(r_5+s_5+t_5)\right\rceil -1 = 0.
e_5=1,\;e_{10}=3,
\epsilon_5(p) = \left(\frac{-3}{p}\right),\mbox{ and }
\epsilon_{10}(p) = \left(\frac{2}{p}\right).
$
Let $p$ be any prime satisfying
$\displaystyle{\left(\frac{-e_5e_{10}\ell m}{p}\right)=-1}$,
i.e., $\displaystyle{\left(\frac{-3\ell m}{p}\right)=-1}$.
Let $f_5(q^\ell)f_{10}(q^m) = \sum_{n=0}^\infty a(n)q^n.$
Then the Theorem implies
$$
a(pn+(\ell+3m)\Delta) =
  \left(\frac{-3}{p}\right)\left(\frac{2}{p}\right)p\,
a\left(\frac{n}{p}\right),
$$
i.e.,
$$
a(pn+(\ell+3m)\Delta) =
  \left(\frac{-6}{p}\right)p\, a\left(\frac{n}{p}\right).
$$
\section*{\normalsize 3. Proofs}\addtocounter{section}{+1}
\label{proofs}
\setcounter{equation}{0}
We shall require the following elementary lemma, which we state
without further comment.

\noindent {\bf Lemma.} {\it
Let $\ell$ and $m$ be positive integers and let $p$ be an odd prime
satisfying $\displaystyle{\left(\frac{-\ell m}{p}\right)=-1.}$
Let $\a$ and $\b$ be integers satisfying
$
\ell\a^2+m\b^2\equiv 0 \!\!\!\!\pmod{p}.
$
Then $\a\equiv 0 \!\!\!\!\pmod{p}$ and $\b\equiv 0 \pmod{p}$.}


In order to illustrate the technique, we first prove the example,
before proving the general statement of the theorem.

\noindent
{\it Proof of example.}  We have
$$
f_5(q^\ell)f_{10}(q^m) =
\sum_{\a \equiv 1 \pmod{6}} \a q^{\ell(\a^2-1)/24}\,
\sum_{\b \equiv 1 \pmod{4}} (-q^m)^{(\b^2-1)/8},
$$
so
\begin{eqnarray*}
a(n) & = & \sum_{\a \equiv 1\pmod6,\;\b\equiv 1\pmod4
                  \atop \ell (\a^2-1)/24+m(\b^2-1)/8=n} \a (-1)^{(\b^2-1)/8}
\\
     & = & \sum_{\a \equiv 1\pmod6,\;\b\equiv 1 \pmod4
                  \atop \ell \a^2+3m\b^2=24n+\ell+3m} \a (-1)^{(\b^2-1)/8}.
\end{eqnarray*}
Therefore
\begin{equation}
a(pn+(\ell+3m)\Delta)
= \sum_{\a \equiv 1 \pmod{6},\; \b \equiv 1 \pmod{4}
                  \atop \ell \a^2+3m\b^2=24pn+(\ell+3m)p^2} \a
(-1)^{(\b^2-1)/8}.
\label{back-here}
\end{equation}
Now $\ell\a^2+3m\b^2 \equiv 0 \!\!\!\!\pmod{p}$, and the lemma implies
$p \mid \a$, $p \mid \b$. Let
\begin{equation}
\a = \left(\frac{-3}{p}\right) p \a',\;\;
\b = \left(\frac{-1}{p}\right) p \b'.
\label{cov}
\end{equation}
Then $\a'\equiv 1\pmod6$ and $\b'\equiv 1\pmod4$.
Also, modulo 2,
\begin{eqnarray*}
\frac{\b^2-1}{8}-\frac{\b'^2-1}{8}
& = & \frac{\b^2-{\b'}^2}{8} \\
& = & \frac{(p^2-1){\b'}^2}{8} \\
& \equiv & \frac{p^2-1}{8} \\
& \equiv & \left\{ \begin{array}{cl}
           0 & \mbox{if $p\equiv 1$ or $7\pmod8$} \\
           1 & \mbox{if $p\equiv 3$ or $5\pmod8$} \\
\end{array} \right.
\end{eqnarray*}
Therefore
\begin{equation}
(-1)^{(\b^2-1)/8} = \left(\frac{2}{p}\right)(-1)^{({\b'}^2-1)/8}.
\label{cos}
\end{equation}
Substituting \reff{cov} and \reff{cos} into \reff{back-here} we get
\begin{eqnarray*}
a(pn+(\ell+3m)\Delta)
& = & \sum_{\a' \equiv 1 \pmod{6},\; \b' \equiv 1 \pmod{4}
     \atop \ell {\a'}^2+3m{\b'}^2=24n/p+\ell+3m}
      \left(\frac{-3}{p}\right)p\a'
      \left(\frac{2}{p}\right) (-1)^{({\b'}^2-1)/8} \\
& = & \left(\frac{-6}{p}\right)p\,a\left(\frac{n}{p}\right).
\end{eqnarray*}
This completes the proof of the example.
\hfill
\boxx

\noindent
{\it Proof of Theorem.}
Writing
$$
f_j(q^\ell)f_k(q^m) = \sum_{n=0}^\infty a(n)q^n,
$$
we have, using \reff{series},
$$
a(n) = \sum_{\a \equiv 1\pmod{n_j},\;\b\equiv 1 \pmod{n_k}
                  \atop e_j\ell \a^2+e_km\b^2=24n+d_j\ell+d_km}
                  \phi_j(\a)\phi_k(\b),
$$
where
$$
\begin{array}{ll}
\phi_1(\a) = (-1)^{(\a-1)/6},  &
\phi_7(\a) = (-1)^{(\a-1)/6+(\a^2-1)/24}, \\
\phi_2(\a) = \a,  &
\phi_8(\a) = \a (-1)^{(\a^2-1)/8}, \\
\phi_3(\a) = (-1)^{\a}, &
\phi_9(\a) = 1, \\
\phi_4(\a) = 1, &
\phi_{10}(\a) = (-1)^{(\a^2-1)/8}, \\
\phi_5(\a) = \a,  &
\phi_{11}(\a) = \a (-1)^{(\a^2-1)/24}, \\
\phi_6(\a) = (-1)^{\a-1}\a,  &
\phi_{12}(\a) = (-1)^{\a-1+(\a^2-1)/3}\a.
\end{array}
$$
Therefore
$$
a(pn+(\ell d_j+md_k)\Delta) = \sum_{\a \equiv 1\pmod{n_j},\;\b\equiv 1
\pmod{n_k}
                  \atop e_j\ell \a^2+e_km\b^2=24pn+p^2(d_j\ell+d_km)}
                  \phi_j(\a)\phi_k(\b).
$$
Observe that $e_j\ell \a^2+e_km\b^2 \equiv 0 \pmod p$. The Lemma implies
$p \mid \a$, $p \mid \b$. Let
\begin{eqnarray*}
\a & = & \left\{ \begin{array}{ll}
   \displaystyle{\left(\frac{-3}{p}\right) p \a'}
        & \mbox{if $j=1,\;5,\;6,\;7,\;11$ or $12$} \\ \\
   \displaystyle{\left(\frac{-1}{p}\right) p \a'} & \mbox{if $j=2,\;4,\;8$
or $10$} \\ \\
   p\a' & \mbox{if $j=3$ or $9$}, \\
   \end{array}
\right. \\ \\
\b & = & \left\{ \begin{array}{ll}
   \displaystyle{\left(\frac{-3}{p}\right) p \b'} & \mbox{if
$k=1,\;5,\;6,\;7,\;11$ or $12$} \\ \\
   \displaystyle{\left(\frac{-1}{p}\right) p \b'} & \mbox{if $k=2,\;4,\;8$
or $10$} \\ \\
   p\b' & \mbox{if $k=3$ or $9$}. \\
   \end{array}
\right.
\label{covg}
\end{eqnarray*}
Then it is easily verified that $\a'\equiv 1 \!\!\!\!\pmod{n_j}$ and
$\b'\equiv 1\!\!\!\!\pmod{n_k}$,
and that
$
\phi_j(\a)= \epsilon_j(p)\,p^{\lambda_j}\,\phi_j(\a')$ and
$\phi_k(\b) = \epsilon_k(p)\,p^{\lambda_k}\,\phi_k(\b')$.
Consequently,

$$\begin{array}{lll} \displaystyle
{a(pn+(\ell d_j+m d_k)\Delta)} 
& = & \displaystyle \sum_{\a' \equiv 1\pmod{n_j},\;\b' \equiv 1 \pmod{n_k}
                  \atop e_j\ell {\a'}^2+e_km{\b'}^2=24n/p+d_j\ell+d_km}
                  \epsilon_j(p) \epsilon_k(p) p^{\lambda_j+\lambda_k}
                  \phi_j(\a')\phi_k(\b') \\ \\
& = & \epsilon_j(p)
\epsilon_k(p)p^{\lambda_j+\lambda_k}\,a\left(\frac{n}{p}\right).
\end{array}
$$
This completes the proof of the theorem.
\hfill
\boxx


\noindent {\bf Remark.} {
Though our theorem can be proved using the theory of lacunary 
modular forms, we prefer to present an elementary proof for 
its simplicity.}




\noindent {\bf Acknowledgement}
The first author thanks the staff and students at Harish-Chandra Research
Institute
for warm hospitality during his visit.

\begin{thebibliography}{99} \footnotesize

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S. Ahlgren, {\em Multiplicative relations in powers of Euler's product},
J. Number Theory {\bf 89} (2001), 222-233.

\bibitem{borwein}
J. M. Borwein and P. B. Borwein,
{\em Pi and the AGM},
Wiley, New York, 1987.
 
\bibitem{chap}
  R. Chapman,
  {\em Coefficients of Products of Powers of Eta Functions},
  Ramanujan J. {\bf 5} (2001), 271--279.

\bibitem{chl}
  S. Cooper, M. Hirschhorn and R. Lewis,
  {\em Powers of Euler's product and related identities,}
  Ramanujan J. {\bf 4}, (2000), 137--155.

\bibitem{newman}
M. Newman, {\em An identity for the coefficients of certain modular
forms}, J. London Math. Soc. {\bf 30} (1955), 488--493.

\end{thebibliography}




\end{document}