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\begin{center}
\vskip 1cm{\LARGE\bf 
Another Proof of a Conjecture by Hirschhorn \\
\vskip .1in
and Sellers on Overpartitions
}
\vskip 1cm
\large
Liuquan Wang\\
Department of Mathematics\\
National University of Singapore\\
Singapore, 119076\\
Singapore\\
\href{wangliuquan@nus.edu.sg}{\tt wangliuquan@nus.edu.sg}\\
\href{mathlqwang@163.com}{\tt mathlqwang@163.com}
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\begin{abstract}
We provide a new elementary proof for a Ramanujan-type congruence for
the overpartition function modulo 40, which was previously conjectured
by Hirschhorn and Sellers and later proved by Chen and Xia.  We also
find some new congruences for the overpartition function modulo 5 and
9.
\end{abstract}


\section {Introduction and Main Results}
An overpartition of an integer $n$ is a partition in which the first occurrence of a part may be overlined. The number of overpartitions of $n$ is denoted by $\overline{p}(n)$. For example, $\overline{p}(3)=8$ as there are 8 overpartitions of 3: $3$, $\overline{3}$, $2+1$, $\overline{2}+1$, $2+\overline{1}$, $\overline{2}+\overline{1}$, $1+1+1$, $\overline{1}+1+1$.  We define $\overline{p}(0)=1$ for convenience. It is well known (see \cite{Lovejoy}, for example) that the generating function of $\overline{p}(n)$  is
\begin{equation}\label{gen}
\sum\limits_{n\ge 0}{\overline{p}(n){{q}^{n}}}=\frac{(-q;q)_{\infty}}{(q;q)_{\infty}}=\frac{1}{\varphi (-q)},
\end{equation}
where ${{(a;q)}_{\infty }}=(1-a)(1-aq)\cdots (1-a{{q}^{n}})\cdots$ is  
standard $q$-series notation, and $\varphi (q)=\sum\nolimits_{n=-\infty }^{\infty }{{{q}^{{{n}^{2}}}}}$ is one of Ramanujan's theta functions.


Overpartitions were first introduced by MacMahon \cite{MacMahon} and have drawn much attention during the past ten years.  There are numerous results concerning  the arithmetic properties of $\overline{p}(n)$.  For more details, we refer the reader to  \cite{Xia,Lovejoy,Sellers05,Mahlburg}.

In 2005, Hirschhorn and Sellers \cite{Hirschhorn05} proved numerous Ramanujan-type identities for $\overline{p}(n)$. For example,
\begin{equation}\label{p4nid}
\sum\limits_{n\ge 0}{\overline{p}(4n+3)q^{n}}=8\frac{{{({{q}^{2}};{{q}^{2}})}_{\infty }}({{q}^{4}};{{q}^{4}})_{\infty }^{6}}{(q;q)_{\infty }^{8}},
\end{equation}
which clearly implies $\overline{p}(4n+3)\equiv 0 $ (mod $8$). Meanwhile, they proposed a curious conjecture:
\begin{conjecture}\label{conj}
For any integer $n\ge 0$, we have
\[\overline{p}(40n+35)\equiv 0 \pmod{40}.\]
\end{conjecture}


In a recent paper, Chen and Xia \cite{Xia} gave a proof of Conjecture \ref{conj}  using $(p,k)$-parametrization of theta functions. Their proof is relatively long and complicated, and in this paper we give a  shorter proof.


Let ${{r}_{k}}(n)$ denote the number of representations of $n$  as sum of $k$ squares. We find the following arithmetic relation.
\begin{theorem}\label{thm1}
For any integer $n\ge 1$, we have
\[\overline{p}(5n)\equiv {{(-1)}^{n}}{{r}_{3}}(n) \pmod{5}.\]
\end{theorem}


We have two remarkable corollaries.
\begin{corollary}\label{thm1cor1}
For any integers $n\ge 0$ and $\alpha \ge 0$, we have
\[\overline{p}({{4}^{\alpha }}(40n+35))\equiv 0 \pmod{5}\]
and
\[{{\overline{p}}}(5\cdot {{4}^{\alpha +1}}n)\equiv {{(-1)}^{n}}{{\overline{p}}}(5n) \pmod{5}.\]
\end{corollary}
By letting $\alpha=0$ in Corollary \ref{thm1cor1}, we get $\overline{p}(40n+35)\equiv 0$ (mod $5$). Together with (\ref{p4nid}), Conjecture~\ref{conj} follows immediately.

\begin{corollary}\label{Tre}
For any prime $p\equiv -1 $ \text{\rm{(mod $5$)}}, we have
\[\overline{p}(5{{p}^{3}}n)\equiv 0  \pmod{5}\]
 for all $n$ coprime to $p$.
\end{corollary}

Corollary \ref{Tre} was first proved by  Treneer (see  \cite[Proposition 1.4]{Treneer}) in 2006 using the theory of modular forms, which is  not elementary.


Furthermore, with Corollary \ref{thm1cor1} in mind, we are ready to generalize Conjecture \ref{conj} to the following result.
\begin{theorem}\label{genthm}
For any integers $n\ge 0$ and $\alpha \ge 0$, we have
\[\overline{p}({{4}^{\alpha }}(40n+35))\equiv 0  \pmod{40}.\]
\end{theorem}

Some miscellaneous congruences can be deduced from Theorem \ref{thm1}, and we list some of them here.
\begin{theorem}\label{5alpha}
For any integers $\alpha \ge 1$ and $n \ge 0$, we have
\[\overline{p}({{5}^{2\alpha +1}}(5n+1))\equiv \overline{p}({{5}^{2\alpha +1}}(5n+4))\equiv 0 \pmod{5}.\]
\end{theorem}

\begin{theorem}\label{2case}
Let $p\ge 3$ be a prime, and $N$ a positive integer which is coprime to $p$. Let $\alpha $ be any nonnegative integer. \\
(1) If $p\equiv 1$ \text{\rm{(mod  $5$)}}, then $\overline{p}(5{{p}^{10\alpha +9}}N)\equiv 0$ \text{\rm{(mod  $5$)}}. \\
(2) If $p\equiv 2,3,4$ \text{\rm{(mod  $5$)}}, then $\overline{p}(5{{p}^{8\alpha +7}}N)\equiv 0$ \text{\rm{(mod   $5$)}}.
\end{theorem}

Using  properties of ${{r}_{3}}(n)$, we can establish some other congruences as corollaries of Theorem \ref{thm1}. Recently, Chen et al. \cite{Chennew} found some Ramanujan-type congruences mainly based  on the theory of modular forms.  Their work also contains a proof of Theorem \ref{genthm}. In addition, they proved some congruences such as  $\overline{p}(5n)\equiv {{(-1)}^{n}}\overline{p}(4 \cdot 5n)$ (mod $5$).

Finally, we mention some congruences for overpartitions modulo 3. In 2011, based on the generating function of $\overline{p}(3n)$ discovered by Hirschhorn and Sellers \cite{Sellers05},  Lovejoy and Osburn \cite{Osburn} proved the following result. For any integer $n\ge 1$, we have
\[\overline{p}(3n)\equiv {{(-1)}^{n}}{{r}_{5}}(n) \pmod{3}.\]
Using the same  method  as in the proof of Theorem \ref{thm1}, we are able to improve this congruence to the following one.
\begin{theorem} \label{Lov}
For any integer $n\ge 1$, we have
\[\overline{p}(3n)\equiv {{(-1)}^{n}}{{r}_{5}}(n) \pmod{9}.\]
\end{theorem}

Similar to Theorem \ref{2case}, we can deduce the following interesting congruences from Theorem \ref{Lov}.
\begin{theorem}\label{Lovcor}
 Let $p\ge 3$ be a prime and $N$ a positive integer which is coprime to $p$. \\
(1) If $p\equiv 1$ \text{\rm{(mod $3$)}}, then $\overline{p}(3{{p}^{6\alpha +5}}N)\equiv 0$ \text{\rm{(mod  $3$)}}  and $\overline{p}(3{{p}^{18\alpha +17}}N)\equiv 0$  \text{\rm{(mod $9$)}}. \\
(2) If $p\equiv 2$  \text{\rm{(mod  $3$)}}, then $\overline{p}(3{{p}^{4\alpha +3}}N)\equiv 0$  \text{\rm{(mod  $9$)}}.
\end{theorem}


%---------------------------------------------------------------------------------------------------------------------------------------------------------------
%---------------------------------------------------------------------------------------------------------------------------------------------------------------
\section{Preliminaries}
\begin{lemma}\label{palpha}(Cf.\ \cite[Lemma 1.2]{pod}.)
Let $p$ be a prime and $\alpha $ a positive integer. Then
\[(q;q)_{\infty }^{{{p}^{\alpha }}}\equiv ({{q}^{p}};{{q}^{p}})_{\infty }^{{{p}^{\alpha -1}}}  \pmod  {{{p}^{\alpha }}}.\]
\end{lemma}

\begin{lemma}\label{r48}(Cf.\ \cite[Theorem 3.3.1, 3.5.4]{Bruce}.)
For any integer $n \ge 1$, we have
\[{{r}_{4}}(n)=8\sum\limits_{d|n,4 \nmid d}{d}, \, \,{{r}_{8}}(n)=16{{(-1)}^{n}}\sum\limits_{d|n}{{{(-1)}^{d}}{{d}^{3}}}.\]
\end{lemma}

\begin{lemma}\label{r48modp}
For any prime $p\ge 3$, we have
 \[{{r}_{4}}(pn)\equiv {{r}_{4}}(n) \pmod{p}, \, \, {{r}_{8}}(pn)\equiv {{r}_{8}}(n) \pmod{p^3}.\]
\end{lemma}
\begin{proof}
By Lemma \ref{r48}, we have
\begin{displaymath}
 {{r}_{4}}(n)=8\sum\limits_{d|n,4 \nmid d}{d}=8\sum\limits_{\begin{smallmatrix}
 d|n \\
 4 \nmid d,p \nmid d
\end{smallmatrix}}{d}+8\sum\limits_{\begin{smallmatrix}
 d|n \\
 4 \nmid d, p | d
\end{smallmatrix}}{d}\equiv 8\sum\limits_{\begin{smallmatrix}
 d|n \\
 4 \nmid d,p \nmid d
\end{smallmatrix}}{d} \pmod{p},
\end{displaymath}
and
\begin{displaymath}
{{r}_{4}}(pn)=8\sum\limits_{d|pn,4 \nmid d}{d}=8\sum\limits_{\begin{smallmatrix}
 d|pn \\
 4\nmid d,p \nmid d
\end{smallmatrix}}{d}+8\sum\limits_{\begin{smallmatrix}
 d|pn \\
 4 \nmid d,p | d
\end{smallmatrix}}{d}=8\sum\limits_{\begin{smallmatrix}
 d|n \\
 4 \nmid d,p \nmid d
\end{smallmatrix}}{d}+8p\sum\limits_{\begin{smallmatrix}
 d|n \\
 4 \nmid d
\end{smallmatrix}}{d}.
\end{displaymath}
Combining them together,  we deduce that ${{r}_{4}}(pn)\equiv {{r}_{4}}(n)$ \text{\rm{(mod $p$)}}.


Similarly we can prove ${{r}_{8}}(pn)\equiv {{r}_{8}}(n)$ \text{\rm{(mod $p^3$)}}.
\end{proof}


\begin{lemma}\label{r30}(Cf.\ \cite[Theorem 1 in Chapter 4]{Grosswald}.)
For any integers $\alpha  \ge 0$ and $ n\ge 0$, we have ${{r}_{3}}({{4}^{\alpha }}(8n+7))=0$ and ${{r}_{3}}({{4}^{\alpha }}n)={{r}_{3}}(n).$
\end{lemma}

\begin{lemma}\label{r3relation} (Cf.\ \cite{Hsquare}.)
Let $p \ge 3$ be a prime. For any  integers $n \ge 1$  and $\alpha \ge 0$,  we have
\[{{r}_{3}}({{p}^{2 \alpha }}n)=\bigg(\frac{{{p}^{\alpha +1}}-1}{p-1}-\Big(\frac{-n}{p}\Big)\frac{{{p}^{\alpha }}-1}{p-1}\bigg){{r}_{3}}(n)-p\frac{{{p}^{\alpha }}-1}{p-1}{{r}_{3}}(n/{{p}^{2}}).\]
where $(\frac{\cdot }{p})$ denotes the Legendre symbol,  and we take ${{r}_{3}}(n/{{p}^{2}})=0$ unless  ${{p}^{2}} | n$ .
\end{lemma}

\begin{lemma}\label{mod8}(Cf.\ \cite[Theorem 3]{Kim1}.)
Let $n$ be an integer which is neither a square nor twice a square. Then $\overline{p}(n)\equiv 0$  \text{\rm{(mod $8$)}}.
\end{lemma}

\begin{lemma}\label{r5relation}(Cf.\ \cite{Cooper}.)
Let $p\ge 3$ be a prime, and $n$ a positive integer such that ${{p}^{2}} \nmid n$. For any integer $\alpha \ge 0$, we have
\[{{r}_{5}}({{p}^{2\alpha }}n)=\bigg(\frac{{{p}^{3\alpha +3}}-1}{{{p}^{3}}-1}-p\Big(\frac{n}{p}\Big)\frac{{{p}^{3\alpha }}-1}{{{p}^{3}}-1}\bigg){{r}_{5}}(n).\]
\end{lemma}


\section{Proofs of The Theorems}

\begin{proof}[Proof of Theorem \ref{thm1}]
Replacing $q$ by $-q$ in (\ref{gen}), we get
\[\sum\limits_{n\ge 0}{\overline{p}(n){{(-q)}^{n}}}=\frac{1}{\varphi (q)}.\]
Hence we have
\[\varphi {{(q)}^{5}}\sum\limits_{n\ge 0}{\overline{p}(n){{(-q)}^{n}}}=\varphi {{(q)}^{4}}=\sum\limits_{n\ge 0}{{{r}_{4}}(n){{q}^{n}}}.\]
By Lemma \ref{palpha}, we have $\varphi {{(q)}^{5}}\equiv \varphi ({{q}^{5}})$ (mod $5$) and thus
\[\varphi ({{q}^{5}})\sum\limits_{n\ge 0}{\overline{p}(n){{(-q)}^{n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{4}}(n){{q}^{n}}} \pmod{5}.\]
Collecting  all the terms of the form ${{q}^{5n}}$  on both sides, we get
\[\varphi ({{q}^{5}})\sum\limits_{n\ge 0}{\overline{p}(5n){{(-q)}^{5n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{4}}(5n){{q}^{5n}}}  \pmod{5}.\]
Replacing ${{q}^{5}}$ by $q$  and applying Lemma \ref{r48modp} with $p=5$,  we obtain
\[\varphi (q)\sum\limits_{n\ge 0}{\overline{p}(5n){{(-q)}^{n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{4}}(5n){{q}^{n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{4}}(n){{q}^{n}}}=\varphi {{(q)}^{4}} \pmod{5}.\]
Hence we have
\[\sum\limits_{n\ge 0}{\overline{p}(5n){{(-q)}^{n}}}\equiv \varphi {{(q)}^{3}}=\sum\limits_{n\ge 0}{{{r}_{3}}(n){{q}^{n}}} \pmod{5}.\]
Theorem \ref{conj} follows by comparing the coefficients of ${{q}^{n}}$ on both sides.
\end{proof}


\begin{proof}[Proof of Corollary \ref{thm1cor1}]
 This corollary follows immediately from Theorem \ref{thm1} and Lemma \ref{r30}.
\end{proof}
\begin{proof}[Proof of Corollary \ref{Tre}]
Let $\alpha =1$ and we replace $n$ by $np$ in Lemma \ref{r3relation}.  We have
\[{{r}_{3}}({{p}^{3}}n)=(p+1){{r}_{3}}(n)-p{{r}_{3}}(n/p).\]
Since $n$ is coprime to $p$ and $p+1\equiv 0 $ (mod $5$), we get ${{r}_{3}}({{p}^{3}}n)\equiv 0$ (mod $5$). By Theorem \ref{thm1}, we deduce that $\overline{p}(5{{p}^{3}}n)\equiv 0$ (mod $5$).
\end{proof}


 \begin{proof}[Proof of Theorem \ref{genthm}]
Since $40n+35=5(8n+7)$ is an odd number, it cannot be twice a square.  If $5(8n+7)={{x}^{2}}$ is a square where $x$ is an odd number, then we know $5|x$.  Let $x=5y$ where $y$ is an odd number, we get $8n+7=5{{y}^{2}}$. But $5{{y}^{2}}\equiv 5$ (mod $8$), which is a contradiction!  Hence we know ${{4}^{\alpha }}(40n+35)$ is neither a square nor twice a square. By Lemma \ref{mod8}, we have $\overline{p}({{4}^{\alpha }}(40n+35))\equiv 0$ (mod $8$).  Combining this with Corollary \ref{thm1cor1} completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{5alpha}]
Set $p=5$ and $n=5m+r$ where $r\in \{1,4\}$ in Lemma \ref{r3relation}. It is easy to deduce that ${{r}_{3}}({{5}^{2\alpha }}(5m+r))\equiv 0$ (mod $5$) for any integer $\alpha \ge 1$. By applying Theorem \ref{thm1}, we complete the proof.
\end{proof}


\begin{proof}[Proof of Theorem \ref{2case}]
(1)  Let $n=pN$ and then  replace $\alpha $ by $5\alpha +4$ in Lemma \ref{r3relation}. Since
\[\frac{{{p}^{5\alpha +5}}-1}{p-1}=1+p+\cdots +{{p}^{5\alpha +4}}\equiv 0 \pmod{5},\]
we have ${{r}_{3}}({{p}^{10\alpha +9}}N)\equiv 0$  (mod $5$).  By Theorem \ref{thm1}, we deduce that $\overline{p}(5{{p}^{10\alpha +9}}N)\equiv 0$ (mod   $5$). \\
(2)  Let $n=pN$ and then  replace $\alpha $ by $4\alpha +3$ in Lemma \ref{r3relation}. Since  ${{p}^{4\alpha +4}}\equiv 1$ (mod  $5$), we deduce that ${{r}_{3}}({{p}^{8\alpha +7}}N)\equiv 0$ (mod $5$). By Theorem \ref{thm1}, we deduce that $\overline{p}(5{{p}^{8\alpha +7}}N)\equiv 0$ (mod  $5$).
\end{proof}


 \begin{proof}[Proof of Theorem \ref{Lov}]
We have
\[\varphi {{(q)}^{9}}\sum\limits_{n\ge 0}{\overline{p}(n){{(-q)}^{n}}}=\varphi {{(q)}^{8}}=\sum\limits_{n\ge 0}{{{r}_{8}}(n){{q}^{n}}}.\]
By Lemma \ref{palpha}, we have $\varphi {{(q)}^{9}}\equiv \varphi {{({{q}^{3}})}^{3}}$ (mod  $9$) and thus
\[\varphi {{({{q}^{3}})}^{3}}\sum\limits_{n\ge 0}{\overline{p}(n){{(-q)}^{n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{8}}(n){{q}^{n}}} \pmod{9}.\]
Collecting  all the terms of the form ${{q}^{3n}}$ on both sides, we get
\[\varphi {{({{q}^{3}})}^{3}}\sum\limits_{n\ge 0}{\overline{p}(3n){{(-q)}^{3n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{8}}(3n){{q}^{3n}}} \pmod{9}.\]
Replacing ${{q}^{3}}$ by $q$  and applying Lemma \ref{r48modp} with $p=3$, we obtain
\[\varphi {{(q)}^{3}}\sum\limits_{n\ge 0}{\overline{p}(3n){{(-q)}^{n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{8}}(3n){{q}^{n}}}\equiv \sum\limits_{n\ge 0}{{{r}_{8}}(n){{q}^{n}}}=\varphi {{(q)}^{8}} \pmod{9}.\]
Hence we have
\[\sum\limits_{n\ge 0}{\overline{p}(3n){{(-q)}^{n}}}\equiv \varphi {{(q)}^{5}}=\sum\limits_{n\ge 0}{{{r}_{5}}(n){{q}^{n}}} \pmod{9}.\]
Theorem \ref{Lov} follows by comparing the coefficients of ${{q}^{n}}$ on both sides.
\end{proof}


\begin{proof}[Proof of Theorem \ref{Lovcor}]
(1)  Let $n=pN$ and then replace $\alpha $ by $3\alpha +2$ in Lemma \ref{r5relation}. Since
\[\frac{{{p}^{9\alpha +9}}-1}{{{p}^{3}}-1}=1+{{p}^{3}}+\cdots +{{p}^{3(3\alpha +2)}}\equiv 0 \pmod{3},\]
we have ${{r}_{5}}({{p}^{6\alpha +5}}N)\equiv 0$ (mod $3$). By Theorem \ref{Lov}, we deduce that $\overline{p}(3{{p}^{6\alpha +5}}N)\equiv 0$ (mod $3$).


Similarly, let $n=pN$  and  replace $\alpha $ by $9\alpha +8$ in Lemma \ref{r5relation}.  Since $p\equiv 1$ (mod $3$) implies ${{p}^{3}}\equiv 1$ (mod  $9$), we have
\[\frac{{{p}^{27\alpha +27}}-1}{{{p}^{3}}-1}=1+{{p}^{3}}+\cdots +{{p}}^{3(9\alpha +8)}\equiv 0  \pmod{9}.\]
Hence ${{r}_{5}}({{p}^{18\alpha +17}}N)\equiv 0$ (mod $9$), and we deduce by Theorem \ref{Lov} that
$\overline{p}(3{{p}^{18\alpha +17}}N)\equiv 0$ (mod $9$). \\
(2) Let   $n=pN$ and  replace $\alpha $ by $2\alpha +1$ in Lemma \ref{r5relation}. Note that $p\equiv 2$ (mod $3$) implies  ${{p}^{3}}\equiv -1$ (mod $9$). Since ${{p}^{6\alpha +6}}\equiv 1$ (mod $9$), we have ${{r}_{5}}({{p}^{4\alpha +3}}N)\equiv 0$ (mod $9$).  By Theorem  \ref{Lov}, we deduce that $\overline{p}(3{{p}^{4\alpha +3}}N)\equiv 0$ (mod $9$).
\end{proof}

\section{Acknowledgments}

I would like to express my gratitude to Andrew Hanlon from the University
of California, Berkeley, and Ying Cui from our department for
proofreading my paper. I am also indebted to the editor and anonymous
reviewer for their careful reading and helpful suggestions.


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\end{thebibliography}


\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A17; Secondary 11P83.

\noindent \emph{Keywords: }
overpartition, congruence, sum of squares.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A015128}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 24 2014;
revised version received   August 25 2014; September 6 2014.
Published in {\it Journal of Integer Sequences}, September 7 2014.

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