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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{\bf Arithmetic Properties of a Restricted\\ Bipartition Function}

\author{Jian Liu\\
\small School of Insurance\\[-0.8ex]
\small Central University of Finance and Economics\\[-0.8ex]
\small Beijing 102206, P.R. China\\
\small\tt liujian8210@gmail.com\\
\and
Andrew Y.Z. Wang\thanks{Corresponding author}\\
\small School of Mathematical Sciences\\[-0.8ex]
\small University of Electronic Science and Technology of China\\[-0.8ex]
\small Chengdu 611731, P.R. China\\
\small\tt yzwang@uestc.edu.cn
}

\date{\dateline{Feb 12, 2015}{Jun 15, 2015}\\
\small Mathematics Subject Classifications: 05A17, 11P83}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

\maketitle

\begin{abstract}
A bipartition of $n$ is an ordered pair of partitions $(\lambda,\mu)$ such that the sum of all of the parts
equals $n$. In this article, we concentrate on the function $c_5(n)$, which counts the number of bipartitions $(\lambda,\mu)$ of $n$ subject to the restriction that each part of $\mu$ is divisible by $5$. We explicitly establish four Ramanujan type congruences and several infinite families of congruences for $c_5(n)$ modulo $3$.

\bigskip\noindent \textbf{Keywords:} bipartition, congruence
\end{abstract}

%--------------------------------------------------------------------
\section{Introduction}

In a series of papers \cite{Chan10a, Chan10b, Chan11}, Chan studied
the arithmetic properties of the cubic partition function $a(n)$,
which is defined by
\[
\sum_{n=0}^\infty a(n)q^n=\frac{1}{(q;q)_\infty(q^2;q^2)_\infty}.
\]
Throughout the paper,  we adopt the following standard $q$-series
notation
\[
(a;q)_\infty = \prod_{n=1}^\infty (1-aq^{n-1}).
\]

In \cite{Chan10a}, Chan proved that
\begin{theorem}
For $n\geq 0$,
\begin{align}\label{chan3n2}
a(3n+2)\equiv 0 \pmod 3.
\end{align}
\end{theorem}
Later, Kim \cite{Kim11} gave a combinatorial interpretation of the
congruence \eqref{chan3n2}. Furthermore, Chan \cite{Chan10b} showed
that
\begin{theorem}
For $k\geq 1$ and $n\geq 0$,
\begin{align*}
a(3^{k}n+s_k)\equiv 0 \pmod {3^{k+\delta(k)}},
\end{align*}
where $s_k$ is the reciprocal modulo $3^k$ of $8$ and $\delta(k)=1$
if $k$ is even, and $0$ otherwise.
\end{theorem}

Chan and Toh \cite{HChan10} also established the following nice
congruence, which was also discovered by Xiong \cite{X11}
independently.
\begin{theorem}
If $k\geq 1$ and $n\geq 0$, then
\begin{align*}
a(5^kn+t_k)\equiv 0 \pmod {5^{\lfloor k/2\rfloor}},
\end{align*}
where $t_k$ is the reciprocal modulo $5^k$ of $8$.
\end{theorem}

Inspired by the work of Ramanujan on the standard partition function
$p(n)$, Chan\cite{Chan10b} asked whether there are any other
congruences of the following form
\[
a(\ell n+k)\equiv 0\pmod \ell,
\]
where $\ell$ is prime and $0\leq k<\ell$. Later, Sinick
\cite{Sinick08} answered Chan's question in the negative by
considering the following restricted bipartition function:
\begin{equation}\label{eqdef}
\sum_{n=0}^\infty c_N(n)q^n=\frac{1}{(q;q)_\infty(q^N;q^N)_\infty}.
\end{equation}

A bipartition of $n$ is an ordered pair of partitions
$(\lambda,\mu)$ such that the sum of all of the parts equals $n$.
Then we know that $c_N(n)$ counts the number of bipartitions
$(\lambda,\mu)$ of $n$ subject to the restriction that each part of
$\mu$ is divisible by $N$. Recently, bipartitions with certain
restrictions on each partition have been investigated by many
authors, see \cite{BL08, CM12, Chen14, CL11, CL12, Kim10, Lin14b,
Lin14a, Lin14d, Lin14c, Toh12} for instance.

In this paper, we investigate the bipartition function $c_5(n)$ from
an arithmetic point of view in the spirit of Ramanujan's congruences
for the standard partition function $p(n)$.

%-----------------------------------------------------------------
\section{Ramanujan type congruences for $c_5(n)$}

We first introduce a useful lemma which will be used later.

\begin{lemma}\label{lem1} We have
\begin{equation}
(q;q)_\infty^2(q^5;q^5)_\infty^2\equiv
(q^3;q^3)_\infty^{4}+q(q^3;q^3)_\infty^2(q^{15};q^{15})_\infty^2-q^2(q^{15};q^{15})
_\infty^{4}\pmod 3.\label{eqlem1a}
\end{equation}
\end{lemma}
\begin{proof}
From \cite[p.28, Entry 1.6.2]{Berndt00}, we see that
\begin{eqnarray}
(q;q)_\infty^2(q^5;q^5)_\infty^2&=&(\psi^2(q)-q\psi^2(q^5))(\psi^2(q)-5q\psi^2(q^5))
\nonumber\\[5pt]
&\equiv&\psi(q)\psi(q^3)-q^2\psi(q^5)\psi(q^{15})\pmod
3,\label{eqlem11}
\end{eqnarray}
where
\[
\psi(q)=\sum_{n=0}^\infty
q^{n(n+1)/2}=\frac{(q^2;q^2)_\infty^2}{(q;q)_\infty}.
\]
Invoking \cite[p.49, Corollary (ii)]{Berndt91}, we have
\begin{equation}\label{eqpsi}
\psi(q)=A(q^3)+q\psi(q^9),
\end{equation}
where
\[
A(q)=(-q;q^3)_\infty(-q^2;q^3)_\infty(q^3;q^3)_\infty.
\]
Substituting \eqref{eqpsi} into \eqref{eqlem11}, we find that
\begin{eqnarray*}
(q;q)_\infty^2(q^5;q^5)_\infty^2
&\equiv&\psi(q^3)\left(A(q^3)+q\psi(q^9)\right)\\[5pt]
&&{}-q^2\psi(q^{15})\left(A(q^{15})+q^5\psi(q^{45})\right) \pmod 3\\[5pt]
&=&\psi(q^3)A(q^3)-q^2\psi(q^{15})A(q^{15})\\[5pt]
&&{}+q\left(\psi(q^3)\psi(q^9)-q^6\psi(q^{15})\psi(q^{45})\right).
\end{eqnarray*}
On the other hand, applying \eqref{eqlem11} with $q$ replaced by
$q^3$ yields that
\begin{align*}
(q^3;q^3)_\infty^2(q^{15};q^{15})_\infty^2\equiv\psi(q^3)\psi(q^9)
-q^6\psi(q^{15})\psi(q^{45})\pmod 3.
\end{align*}
Therefore, we arrive at
\begin{equation}\label{eqlem12}
(q;q)_\infty^2(q^5;q^5)_\infty^2\equiv
\psi(q^3)A(q^3)-q^2\psi(q^{15})A(q^{15})+q(q^3;q^3)_\infty^2(q^{15};q^{15})_\infty^2
\pmod 3.
\end{equation}
In addition, it is easy to see that
\[
\psi(q)A(q) \equiv (q;q)_\infty^4\pmod 3.
\]
Utilizing the above congruence in \eqref{eqlem12},  we complete the
proof of \eqref{eqlem1a}.
\end{proof}

With Lemma \ref{lem1} in hand, we now move to the dissections of the
generating function for $c_5(n)$  modulo $3$.

\begin{theorem}\label{thm2.1}
We have
\begin{eqnarray}
\sum_{n=0}^\infty c_5(3n)q^n&\equiv&
\frac{(q^3;q^3)_\infty}{(q^5;q^5)_\infty} \pmod 3,\label{eqthm2.1a}\\[5pt]
\sum_{n=0}^\infty c_5(3n+1)q^n&\equiv&
(q;q)_\infty(q^5;q^5)_\infty \pmod 3,\label{eqthm2.1b}\\[5pt]
\sum_{n=0}^\infty c_5(3n+2)q^n&\equiv&-
\frac{(q^{15};q^{15})_\infty}{(q;q)_\infty} \pmod
3.\label{eqthm2.1c}
\end{eqnarray}
\end{theorem}
\begin{proof}
From \eqref{eqdef}, we can easily deduce that
\[
\sum_{n=0}^\infty c_5(n)q^n\equiv
\frac{(q;q)_\infty^2(q^5;q^5)_\infty^2}{(q^3;q^3)_\infty(q^{15};q^{15})_\infty}
\pmod 3.
\]
Applying Lemma \ref{lem1}, we obtain
\begin{eqnarray*}
\sum_{n=0}^\infty c_5(n)q^n&\equiv&
\frac{(q^3;q^3)_\infty^3}{(q^{15};q^{15})_\infty}+q(q^3;q^3)_\infty(q^{15};q^{15})_\infty
-q^2\frac{(q^{15};q^{15})_\infty^3}{(q^3;q^3)_\infty}\\[5pt]
&\equiv&
\frac{(q^9;q^9)_\infty}{(q^{15};q^{15})_\infty}+q(q^3;q^3)_\infty(q^{15};q^{15})_\infty
-q^2\frac{(q^{45};q^{45})_\infty}{(q^3;q^3)_\infty} \pmod 3,
\end{eqnarray*}
frow which we get
\begin{eqnarray*}
\sum_{n=0}^\infty c_5(3n)q^{3n}&\equiv&
\frac{(q^9;q^9)_\infty}{(q^{15};q^{15})_\infty} \pmod 3,\\[5pt]
\sum_{n=0}^\infty c_5(3n+1)q^{3n+1}&\equiv&
q(q^3;q^3)_\infty(q^{15};q^{15})_\infty \pmod 3,\\[5pt]
\sum_{n=0}^\infty c_5(3n+2)q^{3n+2}&\equiv&-
q^2\frac{(q^{45};q^{45})_\infty}{(q^3;q^3)_\infty} \pmod 3,
\end{eqnarray*}
simplification upon which yields the desired results.
\end{proof}

The following is a consequence of Theorem \ref{thm2.1}.
\begin{corollary}\label{thm3.1}
We have
\begin{equation}\label{eq9n+7}
\sum_{n=0}^\infty c_5(9n+7)q^n\equiv
-(q^3;q^3)_\infty(q^{15};q^{15})_\infty \pmod 3.
\end{equation}
\end{corollary}
\begin{proof}
By \eqref{eqthm2.1b}, we find that
\begin{eqnarray*}
\sum_{n=0}^\infty c_5(3n+1)q^n&\equiv&
(q^3;q^3)_\infty(q^{15};q^{15})_\infty \times
\frac{1}{(q;q)_\infty^2(q^5;q^5)_\infty^2} \pmod 3\\[5pt]
&=&(q^3;q^3)_\infty(q^{15};q^{15})_\infty \times
\left(\sum_{n=0}^\infty c_5(3n)q^{3n}\right.\\[5pt]
&&+\left.\sum_{n=0}^\infty c_5(3n+1)q^{3n+1}+\sum_{n=0}^\infty
c_5(3n+2)q^{3n+2}\right)^2.
\end{eqnarray*}
Extracting those terms on each side for which the powers of $q$ are
of the form $3n+2$, dividing by $q^2$, and replacing $q^3$ by $q$,
we obtain
\begin{eqnarray*}
\sum_{n=0}^\infty c_5(9n+7)q^n&\equiv&(q;q)_\infty(q^5;q^5)_\infty
\times \left(\left(\sum_{n=0}^\infty c_5(3n+1)q^n\right)^2\right.\\
&&{}\left.+2\sum_{n=0}^\infty c_5(3n)q^n\sum_{n=0}^\infty
c_5(3n+2)q^n\right) \pmod 3.
\end{eqnarray*}
It follows from Theorem \ref{thm2.1} that
\begin{eqnarray*}
\sum_{n=0}^\infty c_5(9n+7)q^n&\equiv& (q;q)_\infty(q^5;q^5)_\infty
\times
\left((q;q)_\infty^2(q^5;q^5)_\infty^2-2\frac{(q^3;q^3)_\infty(q^{15};q^{15})_\infty}
{(q^5;q^5)_\infty(q;q)_\infty}\right)\\[5pt]
&\equiv& -(q^3;q^3)_\infty(q^{15};q^{15})_\infty \pmod 3.
\end{eqnarray*}
This completes the proof.
\end{proof}

We now establish four Ramanujan type congruences for $c_5(n)$.
\begin{theorem}\label{thm1}
For all $n\geq 0$,
\begin{eqnarray}
c_5(15n+6)&\equiv& 0  \pmod 3,\\[5pt]
c_5(15n+10)&\equiv& 0 \pmod 3,\\[5pt]
c_5(15n+12)&\equiv& 0 \pmod 3,\\[5pt]
c_5(15n+13)&\equiv& 0 \pmod 3.
\end{eqnarray}
\end{theorem}
\begin{proof}
Recall that Euler's pentagonal number theorem \cite[p.36, Entry 22]{Berndt91}
\begin{equation}\label{eqEuler}
(q;q)_\infty=\sum_{n=-\infty}^\infty (-1)^nq^{n(3n+1)/2}.
\end{equation}
Substituting \eqref{eqEuler} into \eqref{eqthm2.1a}, we have
\[
\sum_{n=0}^\infty c_5(3n)q^n\equiv
\frac{1}{(q^5;q^5)_\infty}\sum_{n=-\infty}^\infty
(-1)^nq^{3n(3n+1)/2} \pmod 3.
\]
Extracting those terms on each side whose power of $q$ is of the
form $5n+2$ or $5n+4$, and employing the fact that there exist no
integers $n$ such that $3n(3n+1)/2$ is congruent to $2$ or $4$
modulo $5$, we get
\[
\sum_{n=0}^\infty c_5(15n+6)q^{5n+2}\equiv \sum_{n=0}^\infty
c_5(15n+12)q^{5n+4}\equiv 0 \pmod 3,
\]
which means that
\[
c_5(15n+6)\equiv c_5(15n+12)\equiv 0 \pmod 3.
\]
Similarly, from \eqref{eqthm2.1b} and the fact that there are no
integers $n$ with $n(3n+1)/2$ being congruent to $3$ or $4$ modulo
$5$, it is not hard to obtain
\[
c_5(15n+10)\equiv c_5(15n+13)\equiv 0 \pmod 3.
\]
This concludes the proof.
\end{proof}

%--------------------------------------------------------------------
\section{Two infinite families of congruences for $c_5(n)$}

We start with investigating a generalization of the congruences
\eqref{eqthm2.1b} and \eqref{eq9n+7}.
\begin{theorem}\label{thm3.2}
For $\alpha\geq 1$, we have
\begin{eqnarray}
\sum_{n=0}^\infty
c_5\left(3^{2\alpha-1}n+\frac{3^{2\alpha-1}+1}{4}\right)q^n&\equiv&
(-1)^{\alpha+1}(q;q)_\infty(q^5;q^5)_\infty \pmod 3,\\[5pt]
\sum_{n=0}^\infty
c_5\left(3^{2\alpha}n+\frac{3^{2\alpha+1}+1}{4}\right)q^n&\equiv&
(-1)^{\alpha}(q^3;q^3)_\infty(q^{15};q^{15})_\infty \pmod
3.\label{eqthm3.2b}
\end{eqnarray}
\end{theorem}
\begin{proof}
We proceed by induction on $\alpha$. The case $\alpha=1$
corresponds to the congruences \eqref{eqthm2.1b} and \eqref{eq9n+7}.

Assume that
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3^{2\alpha}n+\frac{3^{2\alpha+1}+1}{4}\right)q^n\equiv
(-1)^{\alpha}(q^3;q^3)_\infty(q^{15};q^{15})_\infty \pmod 3
\end{align*}
is true for some fixed integer $\alpha\geq 1$. Since the terms
appearing on the right side of the above congruence are powers of
$q^{3}$, we have
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3^{2\alpha}(3n)+\frac{3^{2\alpha+1}+1}{4}\right)q^{3n}\equiv
(-1)^{\alpha}(q^3;q^3)_\infty(q^{15};q^{15})_\infty \pmod 3,
\end{align*}
which yields that
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3^{2\alpha+1}n+\frac{3^{2\alpha+1}+1}{4}\right)q^{n}\equiv
(-1)^{\alpha+2}(q;q)_\infty(q^{5};q^{5})_\infty \pmod 3.
\end{align*}

Now we suppose that
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3^{2\alpha-1}n+\frac{3^{2\alpha-1}+1}{4}\right)q^n\equiv
(-1)^{\alpha+1}(q;q)_\infty(q^5;q^5)_\infty \pmod 3
\end{align*}
is true for some fixed integer $\alpha\geq 1$, to which applying the
same argument as in the proof of Corollary \ref{thm3.1} yields that
\begin{eqnarray*}
\sum_{n=0}^\infty
c_5\left(3^{2\alpha}n+\frac{3^{2\alpha+1}+1}{4}\right)q^n&\equiv&
(-1)^{\alpha+2}(q^3;q^3)_\infty(q^{15};q^{15})_\infty \pmod 3.
\end{eqnarray*}
The proof is complete.
\end{proof}

As a consequence of Theorem \ref{thm3.2}, we have the following
result.
\begin{corollary}
If $\alpha\geq 1$ and $n\geq 0$,
\begin{eqnarray}
c_5\left(3^{2\alpha+1}n+\frac{7\times
3^{2\alpha}+1}{4}\right)&\equiv& 0\pmod 3,
\label{eqthm2a}\\[5pt]
c_5\left(3^{2\alpha+1}n+\frac{11\times
3^{2\alpha}+1}{4}\right)&\equiv& 0\pmod 3.\label{eqthm2b}
\end{eqnarray}
\end{corollary}
\begin{proof}
Note that all the terms on the right hand side of
\eqref{eqthm3.2b} are of the form $q^{3n}$. We can immediately
obtain \eqref{eqthm2a} and \eqref{eqthm2b} by equating the
coefficients of $q^{3n+1}$ and $q^{3n+2}$ on both sides of
\eqref{eqthm3.2b}.
\end{proof}

%----------------------------------------------------------------------------
\section{More infinite families of congruences for $c_5(n)$}

To establish new congruences for $c_5(n)$, we need the following
lemma.
\begin{lemma}\label{lem2}
Let
\begin{equation}\label{eqdefb(n)}
\sum_{n=0}^\infty b(n)q^n=(q;q)_\infty(q^5;q^5)_\infty.
\end{equation}
Then, for a given prime $p\geq 5$ with
$\left(\frac{-5}{p}\right)=-1$, we have
\begin{equation}
\sum_{n=0}^\infty b\left(pn+\frac{p^2-1}{4}\right)q^n=
(q^p;q^p)_\infty (q^{5p};q^{5p})_\infty.
\end{equation}
\end{lemma}
\begin{proof}
Applying Euler's pentagonal number theorem, we have
\begin{equation}\label{eqlem2temp1}
\sum_{n=0}^\infty
b(n)q^n=\sum_{m,n=-\infty}^\infty(-1)^{m+n}q^{m(3m+1)/2+5n(3n+1)/2}.
\end{equation}
We now consider
\[
\frac{m(3m+1)}{2}+\frac{5n(3n+1)}{2}\equiv \frac{p^2-1}{4}\pmod p,
\]
namely,
\[
(6m+1)^2+5(6n+1)^2\equiv 0\pmod p.
\]
Since $\left(\frac{-5}{p}\right)=-1$, we deduce that
\[
6m+1\equiv 6n+1 \equiv 0\pmod p.
\]
If $p \equiv 1\pmod 6$, then $m\equiv n\equiv \frac{p-1}{6}\pmod p$.
Let
\[m=kp+\frac{p-1}{6}~\mbox{and}~n=lp+\frac{p-1}{6},\] we have
\[
m(3m+1)/2+5n(3n+1)/2=(p^2-1)/4+p^2(3k^2+k)/2+5p^2(3l^2+l)/2.
\]
If $p \equiv -1\pmod 6$, then $m\equiv n\equiv \frac{-p-1}{6}\pmod
p$. Let
\[m=-kp-\frac{p+1}{6}~\mbox{and}~n=-lp-(p+1)/6,\] we also have
\[
m(3m+1)/2+5n(3n+1)/2=(p^2-1)/4+p^2(3k^2+k)/2+5p^2(3l^2+l)/2.
\]
Extracting the terms whose power of $q$ is congruent to
$\frac{p^2-1}{4}$ modulo $p$ from \eqref{eqlem2temp1}, and employing
the above analysis, we obtain
\begin{eqnarray*}
\sum_{n=0}^\infty
b\left(pn+\frac{p^2-1}{4}\right)q^{pn+\frac{p^2-1}{4}}=
\sum_{k,l=-\infty}^\infty
(-1)^{k+l}q^{(p^2-1)/4+p^2(3k^2+k)/2+5p^2(3l^2+l)/2},
\end{eqnarray*}
which can be simplified to
\begin{align*}
\sum_{n=0}^\infty b\left(pn+\frac{p^2-1}{4}\right)q^{n}=
\sum_{k,l=-\infty}^\infty (-1)^{k+l}q^{p(3k^2+k)/2+5p(3l^2+l)/2}.
\end{align*}
Applying Euler's pentagonal number theorem again, we derive that
\[
\sum_{n=0}^\infty b\left(pn+\frac{p^2-1}{4}\right)q^n=
(q^p;q^p)_\infty (q^{5p};q^{5p})_\infty,
\]
which completes the proof.
\end{proof}

Based on Lemma \ref{lem2}, we can easily obtain the following
congruence.
\begin{theorem}\label{thm4.1}
If $p\geq 5$ is a prime with $\left(\frac{-5}{p}\right)=-1$, we have
\begin{equation}
\sum_{n=0}^\infty c_5\left(3pn+\frac{3p^2+1}{4}\right)q^n\equiv
(q^p;q^p)_\infty(q^{5p};q^{5p})_\infty \pmod 3.
\end{equation}
\end{theorem}
\begin{proof}
It follows from \eqref{eqthm2.1b} and \eqref{eqdefb(n)} that
\[c_5(3n+1)\equiv b(n) \pmod 3.\]
Applying Lemma \ref{lem2}, we deduce that
\[
\sum_{n=0}^\infty
c_5\left(3\left(pn+\frac{p^2-1}{4}\right)+1\right)q^n\equiv
(q^p;q^p)_\infty (q^{5p};q^{5p})_\infty \pmod 3,
\]
which finishes the proof.
\end{proof}

One can generalize the above congruence to the form as we show
below.
\begin{theorem}\label{thm4.2}
Given a prime $p\geq 5$ with $\left(\frac{-5}{p}\right)=-1$, then
for all $\alpha\geq 1$, we have
\begin{equation}\label{eqthm4.2a}
\sum_{n=0}^\infty
c_5\left(3p^{2\alpha-1}n+\frac{3p^{2\alpha}+1}{4}\right)q^n\equiv
(q^p;q^p)_\infty(q^{5p};q^{5p})_\infty \pmod 3.
\end{equation}
\end{theorem}
\begin{proof} The proof follows by induction on $\alpha$. The case $\alpha=1$
is given in Theorem \ref{thm4.1}. Assuming the result holds for a
positive integer $\alpha=t$, namely,
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3p^{2t-1}n+\frac{3p^{2t}+1}{4}\right)q^n\equiv
(q^p;q^p)_\infty(q^{5p};q^{5p})_\infty \pmod 3.
\end{align*}
Choosing  those terms on each side whose power of $q$ is of the form
$pn$, and replacing $q^p$ by $q$, we obtain
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3p^{2t}n+\frac{3p^{2t}+1}{4}\right)q^n\equiv
(q;q)_\infty(q^{5};q^{5})_\infty \pmod 3,
\end{align*}
which implies that
\begin{align*}
c_5\left(3p^{2t}n+\frac{3p^{2t}+1}{4}\right)\equiv b(n)\pmod 3.
\end{align*}
Furthermore, from Lemma \ref{lem2} we see that
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3p^{2t}\left(pn+\frac{p^2-1}{4}\right)+\frac{3p^{2t}+1}{4}\right)q^n\equiv
(q^p;q^p)_\infty(q^{5p};q^{5p})_\infty \pmod 3,
\end{align*}
which upon simplification completes the induction on $\alpha$.
\end{proof}

As an immediate consequence of Theorem \ref{thm4.2}, we obtain the
following infinite families of congruences for $c_5(n)$.
\begin{corollary}\label{thm3}
Given a prime $p\geq 5$ with $\left(\frac{-5}{p}\right)=-1$, if
$\alpha\geq 1$ and $n\geq 0$, we have
\begin{equation}
c_5\left(3p^{2\alpha}n+3p^{2\alpha-1}i+\frac{3p^{2\alpha}+1}{4}\right)\equiv
0 \pmod 3,
\end{equation}
where $i=1,2,\ldots,p-1$.
\end{corollary}
\begin{proof}
Collecting those terms on each side of \eqref{eqthm4.2a} for which
the powers of $q$ are of the form $pn+i$, dividing by $q^i$, and
replacing $q^p$ by $q$, we obtain that for $i=1,2,\ldots,p-1$,
\begin{align*}
\sum_{n=0}^\infty
c_5\left(3p^{2\alpha-1}(pn+i)+\frac{3p^{2\alpha}+1}{4}\right)q^n\equiv
0\pmod 3,
\end{align*}
which proves the claim in the corollary.
\end{proof}

%--------------------------------------------------------------------
\subsection*{Acknowledgements}
The authors would like to thank an anonymous referee for a very careful review and many valuable comments on the manuscript. The first author was supported by the Young Doctor Development Foundation of 121 Talent Project of Central University of Finance and Economics (No. QBJ1402) and the second author was supported by the National Natural Science Foundation of China (No. 11401080).

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