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\title{\bf Arithmetic properties\\ of overcubic partition pairs}

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% the address should include the country, and does not have to include
% the street address

\author{Bernard L.S. Lin\thanks{Supported by the
Training Program Foundation for Distinguished Young Scholars and
Research Talents of Fujian Higher Education, and the Natural Science
Foundation of Fujian Province of China (No. 2013J05011).}\\
\small School of Sciences\\[-0.8ex]
\small Jimei University\\[-0.8ex]
\small Xiamen 361021,
P.R. China\\
\small\tt linlsjmu@163.com\\
}
% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{May 25, 2014}{Aug 27, 2014}{Sep 4, 2014}\\
\small Mathematics Subject Classifications: 05A17, 11P83}


\begin{document}

\maketitle

\begin{abstract} Let $\overline{b}(n)$ denote the number of overcubic partition
pairs of $n$.  In this paper, we establish two Ramanujan type
congruences and several infinite families of congruences modulo $3$
satisfied by $\overline{b}(n)$. For modulus $5$, we obtain one
Ramanujan type congruence and two congruence relations for
$\overline{b}(n)$, from which some strange congruences are derived.

\bigskip\noindent \textbf{Keywords:} overcubic partition pairs; theta
function; congruence
\end{abstract}



\section{Introduction}

In a series of papers \cite{Chan10a, Chan10b, Chan10c}, Chan
investigated congruence properties of cubic partition function
$a(n)$ which is defined by
\begin{equation}
\frac{1}{(q;q)_\infty(q^2;q^2)_\infty}=\sum_{n=0}^\infty a(n)q^n.
\end{equation}
Throughout this paper, we assume $|q|<1$ and adopt the following
customary notation
\[
(a;q)_\infty = \prod_{n=1}^\infty (1-aq^{n-1}).
\]
After Chan's work, many analogous partition functions have been
studied. Kim \cite{Kim10} studied its overpartiton analog in which
the overcubic partition function $\overline{a}(n)$ was given by
\begin{equation}
\frac{(-q;q)_\infty(-q^2;q^2)_\infty}{(q;q)_\infty(q^2;q^2)_\infty}=\sum_{n=0}^\infty
\overline{a}(n)q^n.
\end{equation}
It is worth mentioning that Hirschhorn \cite{Hirschhorn12} has given
an elementary proof of the results satisfied by $\overline{a}(n)$
appeared in \cite{Kim10}. Later, Zhao and Zhong \cite{ZZ11}
established congruences modulo $5$, $7$ and $9$ for the following
partition function
\begin{equation}
\frac{1}{(q;q)_\infty^2(q^2;q^2)_\infty^2}=\sum_{n=0}^\infty
b(n)q^n.
\end{equation}
Kim \cite{Kim11} introduced two partition statistics to explain the
congruences   modulo $5$ and $7$ for $b(n)$. Since $b(n)$ counts a
pair of cubic partitions, Kim \cite{Kim11} christened $b(n)$ the
number of cubic partition pairs. Zhou \cite{Z12} also found
combinatorial interpretations for the congruences modulo $5$ and $7$
satisfied by $b(n)$. Recently, Kim \cite{Kim12} focused on studying
congruence properties of $\overline{b}(n)$ whose generating function
is
\begin{equation}
\frac{(-q;q)_\infty^2(-q^2;q^2)_\infty^2}
{(q;q)_\infty^2(q^2;q^2)_\infty^2}=\sum_{n=0}^\infty
\overline{b}(n)q^n.
\end{equation}
Similarly, Kim named $\overline{b}(n)$ as the number of overcubic
partition pairs of $n$. Using arithmetic properties of quadratic
forms and modular forms, Kim \cite{Kim12} derived the following two
congruences
\begin{eqnarray}
\overline{b}(8n+7) &\equiv&  0\ ({\rm mod\ }64),\\[5pt]
\overline{b}(9n+3) &\equiv&  0\ ({\rm mod\ }3).
\end{eqnarray}

The paper is organized as follows. In Section 2 we introduce
 necessary notation and some preliminary results. In Section
3 we aim to establish two Ramanujan type congruences and several
infinite families of congruences modulo $3$ satisfied by
$\overline{b}(n)$. We obtain some unexpected congruence results for
$\overline{b}(n)$ with modulus $5$ in Section 4.

%--------------------------------------------------------------------------
\section{Preliminaries}

We first recall that Ramanujan's general theta function $f(a,b)$ is
defined by
\begin{equation}\label{eqRamanujan}
f(a,b)=\sum_{n=-\infty}^\infty a^{n(n-1)/2}b^{n(n+1)/2},\quad
|ab|<1.
\end{equation}
In Ramanujan's notation, the Jacobi triple product identity takes
the shape
\begin{equation}\label{eqJacobi}
f(a,b)=(-a;ab)_\infty(-b;ab)_\infty(ab;ab)_\infty.
\end{equation}
Three special cases of $f(a,b)$ are defined by
\begin{eqnarray}
\varphi(q)&:=&f(q,q)=\sum_{n=-\infty}^\infty q^{n^2}=
\frac{(q^2;q^2)_\infty^5}{(q;q)_\infty^2(q^4;q^4)_\infty^2},\label{phiq}\\[5pt]
\psi(q)&:=&f(q,q^3)=\sum_{n=0}^\infty q^{n(n+1)/2}=
\frac{(q^2;q^2)_\infty^2}{(q;q)_\infty},\label{psiq}\\[5pt]
f(-q)&:=&f(-q,-q^2)=\sum_{n=-\infty}^\infty
(-1)^nq^{n(3n-1)/2}=(q;q)_\infty,\label{eqf}
\end{eqnarray}
where the above three product representations follows from
\eqref{eqJacobi}.

We now list the necessary preliminary results in the following
lemmas, which will be used in our later proofs.


\begin{lemma}
\begin{eqnarray}
\varphi(-q)&=&\frac{(q;q)_\infty^2}{(q^2;q^2)_\infty},\label{eqlem0a}\\[5pt]
 \varphi^2(-q^2)&=&\varphi(q)\varphi(-q).\label{eqlem0b}
\end{eqnarray}
\end{lemma}
\begin{proof} Applying \eqref{eqJacobi}, we have
\begin{eqnarray*}
\varphi(-q)&=&(q;q^2)_\infty(q;q^2)_\infty(q^2;q^2)_\infty\\[5pt]
&=&\frac{(q;q)_\infty^2}{(q^2;q^2)_\infty}.
\end{eqnarray*}
Multiplying  \eqref{phiq} by \eqref{eqlem0a}, we immediately get
\eqref{eqlem0b}.

\end{proof}


\begin{lemma}
\begin{eqnarray}
\varphi(q)&=&\varphi(q^9)+2qf(q^3,q^{15}),\label{eqlem1a}\\[5pt]
\varphi(q)&=&\varphi(q^4)+2q\psi(q^8),\label{eqlem1b}\\[5pt]
 \varphi^2(q)&=&\varphi^2(q^2)+4q\psi^2(q^4).\label{eqlem1c}
\end{eqnarray}
\end{lemma}

\begin{proof}
 The detailed proofs can be found in \cite[p. 49]{Berndt91} and
\cite[p. 40]{Berndt91} respectively.
\end{proof}

 In Section 4, we will involve
Ramanujan's congruence modulo $5$ for partition function $p(n)$. It
is well known that the generating function of $p(n)$ satisfies
\[
\sum_{n=0}^\infty p(n)q^n=\frac{1}{(q;q)_\infty}.
\]
The congruence  modulo $5$ for $p(n)$ which we require is stated as
follows.

\begin{lemma} For all $n\geq 0$,
\begin{equation}\label{eqp5n+4}
p(5n+4)\equiv 0\ ({\rm mod\ }5).
\end{equation}
\end{lemma}
\begin{proof} See \cite[p. 31]{Berndt06} for a proof.
\end{proof}



An overpartition of $n$ is a partition of $n$ for which the first
occurrence of a number may be overlined. Let $\overline{p}(n)$
denote the number of overpartitions of $n$. For convenience, define
$\overline{p}(0) = 1$. Since the overlined parts form a partition
into distinct parts and the non-overlined parts form an ordinary
partition, the generating function for overpartitions satisfies
\begin{equation}\label{eqover}
\sum_{n=0}^\infty
\overline{p}(n)q^n=\frac{(-q;q)_\infty}{(q;q)_\infty}=\frac{1}{\varphi(-q)}.
\end{equation}
Recently, Chen and Xia  \cite{CX13} confirmed the following
congruence first conjectured by Hirschhorn and Sellers \cite{HS05}
\begin{equation}\label{eqsellers}
\overline{p}(40n+35)\equiv 0\ ({\rm mod\ }5).
\end{equation}
In \cite{Lin14}, the author presented an alternative proof of
\eqref{eqsellers} by firstly establishing the following congruence
relation.

\begin{lemma}
\begin{equation}\label{eqLin}
\sum_{n=0}^\infty \overline{p}(5n)q^n\equiv
\varphi(-q)\varphi^2(-q^5)+q\frac{(q^{10};q^{10})_\infty^2}{(q^2;q^2)_\infty}
\psi(q^2)\varphi(q)\ ({\rm mod\ }5). \end{equation}
\end{lemma}


At the end of this  section, we introduce the following congruence
relations which will be frequently adopted throughout the paper
without explicitly mentioning it.

\begin{lemma}\label{lem3}
For any prime $p$, we have
\begin{align*}
(q;q)_\infty^p &\equiv (q^p;q^p)_\infty \ ({\rm mod\ }p),\\[5pt]
\varphi(-q)^p &\equiv \varphi(-q^p) \ ({\rm mod\ }p).
\end{align*}
\end{lemma}

\begin{proof} By the binomial theorem, we have
\[
(1-q)^p\equiv 1-q^p\ ({\rm mod\ }p),
\]
which yields the first congruence relation. The second congruence
relation follows from the first congruence relation and the product
representation for $\varphi(-q)$.
\end{proof}

%---------------------------------------------------------------------
\section{Congruences modulo $3$ for $\overline{b}(n)$}

In this section, we aim to establish two Ramanujan type congruences
and several infinite families of congruences modulo $3$ for
$\overline{b}(n)$. We begin by rewriting the generating function for
$\overline{b}(n)$ in the following form
\begin{eqnarray}
\sum_{n=0}^\infty
\overline{b}(n)q^n&=&\frac{(q^4;q^4)_\infty^2}{(q;q)_\infty^4(q^2;q^2)_\infty^2}\nonumber
\\[5pt]
&=&\varphi^2(q)\frac{(q^4;q^4)_\infty^6}{(q^2;q^2)_\infty^{12}}\label{eqbn}.
\end{eqnarray}

First we introduce two Ramanujan type congruences  modulo $3$ for
$\overline{b}(n)$.

\begin{theorem}\label{thm3.1}
For all $n\geq 0$,
\begin{eqnarray}
\overline{b}(12n+10)&\equiv& 0\ ({\rm mod\ }3),\\[5pt]
\overline{b}(24n+16)&\equiv& 0\ ({\rm mod\ }3).
\end{eqnarray}
\end{theorem}

\begin{proof} Invoking \eqref{eqlem1c}, we can reformulate \eqref{eqbn} as
\[
\sum_{n=0}^\infty \overline{b}(n)q^n=
\left(\varphi^2(q^2)+4q\psi^2(q^4)\right)
\frac{(q^{4};q^{4})_\infty^6}{(q^2;q^2)_\infty^{12}}.
\]
Choosing the terms for which the power of $q$ is a multiple of $2$,
replacing $q^2$ by $q$ yields
\begin{equation}\label{eqb2n}
\sum_{n=0}^\infty \overline{b}(2n)q^n =
\varphi^2(q)\frac{(q^2;q^2)_\infty^6}{(q;q)_\infty^{12}},
\end{equation}
and
\[
\sum_{n=0}^\infty \overline{b}(2n)q^n \equiv
\left(\varphi(q^9)+2qf(q^3,q^{15})\right)^2\frac{(q^6;q^6)_\infty^2}{(q^3;q^3)_\infty^4}
\ ({\rm mod\ }3).
\]
If we extract those terms whose power of $q$ is congruent to $2$
modulo $3$, divide by $q^2$, and replace $q^3$ by $q$, we obtain
\begin{equation*}
\sum_{n=0}^\infty \overline{b}(6n+4)q^n\equiv
f^2(q,q^5)\frac{(q^2;q^2)_\infty^2}{(q;q)_\infty^4} \ ({\rm mod\
}3).
\end{equation*}
It is straightforward to check that
\[
f(q,q^5)\equiv
\frac{(q;q)_\infty^2(q^4;q^4)_\infty^2}{(q^2;q^2)_\infty} \ ({\rm
mod\ }3).
\]
Combining the above two identities together, we have
 \begin{equation}\label{eq3.1a}
 \sum_{n=0}^\infty \overline{b}(6n+4)q^n\equiv
(q^4;q^4)_\infty(q^{12};q^{12})_\infty\ ({\rm mod\ }3).
\end{equation}
Equating the coefficients of $q^{2n+1}$ and $q^{4n+2}$ respectively,
we deduce that for all $n\geq 0$,
\[
\overline{b}(12n+10)\equiv \overline{b}(24n+16)\equiv 0\ ({\rm mod\
}3).
\]
 This completes the proof.
\end{proof}

As a consequence of \eqref{eq3.1a}, we obtain the following
corollary.

\begin{corollary}\label{cor3.1} We have
\begin{equation}\label{eqcor3.1}
\sum_{n=0}^\infty\overline{b}(24n+4)q^n\equiv
(q;q)_\infty(q^3;q^3)_\infty \ ({\rm mod\ }3).
\end{equation}
\end{corollary}

 With the aid of \eqref{eqcor3.1}, we get the following result.

\begin{theorem}\label{thm3.2}
For any  prime $p\geq 5$, $\left(\frac{-3}{p}\right)=-1$, we have
\begin{equation}
\sum_{n=0}^\infty \overline{b}(24pn+4p^2)q^n\equiv (q^p;q^p)_\infty
(q^{3p};q^{3p})_\infty  \ ({\rm mod\ }3).
\end{equation}

\end{theorem}
\begin{proof} Substituting \eqref{eqf} into \eqref{eqcor3.1}, we obtain
\begin{equation}\label{eq3.2a}
\sum_{n=0}^\infty\overline{b}(24n+4)q^n\equiv
\sum_{m,n=-\infty}^\infty(-1)^{m+n}q^{m(3m+1)/2+3n(3n+1)/2} \ ({\rm
mod\ }3).
\end{equation}
We claim that if
\begin{align}\label{hyp}
\frac{m(3m+1)}{2}+\frac{3n(3n+1)}{2}\equiv \frac{p^2-1}{6}\ ({\rm
mod\ }p),
\end{align}
there exist some $k$ and $l$ such that
\begin{align*}
\frac{m(3m+1)}{2}+\frac{3n(3n+1)}{2}=\frac{p^2-1}{6}+\frac{p^2(3k^2+k)}{2}+
\frac{3p^2(3l^2+l)}{2}
\end{align*}
and $(-1)^{m+n}=(-1)^{k+l}$.

First it follows from \eqref{hyp} that
\[(6m+1)^2+3(6n+1)^2\equiv 0\ ({\rm mod\ }p),\]
which implies that $6m+1\equiv 6n+1 \equiv 0\ ({\rm mod\ }p)$ since
$\left(\frac{-3}{p}\right)=-1$.

{\noindent\bf Case 1.} If $p \equiv 1\ ({\rm mod\ }6)$, then
\[m\equiv n\equiv \frac{p-1}{6}\ ({\rm mod\ }p).\] Let
$m=kp+(p-1)/6$ and $n=lp+(p-1)/6$, we have
\[
\frac{m(3m+1)}{2}+\frac{3n(3n+1)}{2}=\frac{p^2-1}{6}+\frac{p^2(3k^2+k)}{2}+
\frac{3p^2(3l^2+l)}{2}
\]
and
\[(-1)^{m+n}=(-1)^{(k+l)p+(p-1)/3}=(-1)^{(k+l)p}=(-1)^{k+l}.\]

{\noindent\bf Case 2.} If $p \equiv -1\ ({\rm mod\ }6)$, then
\[
m\equiv n\equiv \frac{-p-1}{6}\ ({\rm mod\ }p). \] Let
$m=-kp-\frac{p+1}{6}~\mbox{and}~n=-lp-(p+1)/6,$ we also have
\[
\frac{m(3m+1)}{2}+\frac{3n(3n+1)}{2}=\frac{p^2-1}{6}+\frac{p^2(3k^2+k)}{2}+
\frac{3p^2(3l^2+l)}{2}\]
and
\[(-1)^{m+n}=(-1)^{-(k+l)p-(p+1)/3}=(-1)^{(k+l)p}=(-1)^{k+l}.\]
Hence our claim holds.

If we extract those terms whose power of $q$ is congruent to
$(p^2-1)/6$ modulo $p$ from \eqref{eq3.2a}, and employ the above
analysis, we obtain
\begin{align*}
\sum_{n=0}^\infty
\overline{b}(24(pn+(p^2-1)/6)+4)q^{pn+\frac{p^2-1}{6}}\equiv
\sum_{k,l=-\infty}^\infty
(-1)^{k+l}q^{\frac{p^2-1}{6}+\frac{p^2(3k^2+k)}{2}+\frac{3p^2(3l^2+l)}{2}}\
({\rm mod\ }3),
\end{align*}
which can be simplified as
\[
\sum_{n=0}^\infty \overline{b}(24pn+4p^2)q^{n}\equiv
\sum_{k,l=-\infty}^\infty (-1)^{k+l}q^{p(3k^2+k)/2+3p(3l^2+l)/2}\
({\rm mod\ }3).
\]
Applying \eqref{eqf}, we have
\[
\sum_{n=0}^\infty \overline{b}(24pn+4p^2)q^n\equiv (q^p;q^p)_\infty
(q^{3p};q^{3p})_\infty  \ ({\rm mod\ }3),
\]
which finishes the proof.\end{proof}

From Theorem \ref{thm3.2} and by induction,  we obtain the following
theorem.

\begin{theorem}\label{thm3.3}
For any prime $p\geq 5$, $\left(\frac{-3}{p}\right)=-1$, $\alpha\geq
1$, we have
\begin{equation}\label{eq3.3a}
\sum_{n=0}^\infty
\overline{b}(24p^{2\alpha-1}n+4p^{2\alpha})q^n\equiv
(q^p;q^p)_\infty (q^{3p};q^{3p})_\infty  \ ({\rm mod\ }3).
\end{equation}
\end{theorem}

Note that the right-hand side of \eqref{eq3.3a}, when expanded as a
power series, contains only terms of the form $q^{pm}$ for some $m$.
Based on this fact, we  deduce the following corollary.

\begin{corollary}\label{cor3.2}
For any prime $p\geq 5$, $\left(\frac{-3}{p}\right)=-1$, $\alpha\geq
1$, and all $n\geq 0$, we have
\begin{equation}
\overline{b}(24p^{2\alpha}n+24p^{2\alpha-1}i+4p^{2\alpha})\equiv 0\
({\rm mod\ }3),
\end{equation}
where $i=1,2,\ldots,p-1$.
\end{corollary}

To conclude this section,  we present another two infinite families
of congruences modulo $3$ for $\overline{b}(n)$.

\begin{theorem}\label{thm3.4} For   $\alpha\geq 2$ and all $n\geq 0$,
\begin{eqnarray}
\overline{b}(3^{\alpha}(3n+2))&\equiv& 0\ ({\rm mod\ }3),\label{eq3.4a}\\[5pt]
\overline{b}(3^{\alpha}(4n+2))&\equiv& 0\ ({\rm mod\
}3).\label{eq3.4b}
\end{eqnarray}
\end{theorem}

\begin{proof} Putting \eqref{eqlem1a} into \eqref{eqbn}, we find that
\begin{eqnarray*}
\sum_{n=0}^\infty \overline{b}(n)q^n &\equiv&
\left(\varphi(q^9)+2qf(q^3,q^{15})\right)^2
\frac{(q^{12};q^{12})_\infty^2}{(q^6;q^6)_\infty^4}\ ({\rm mod\ }3).
\end{eqnarray*}
Collecting the terms whose power of $q$ is a multiple of $3$,
replacing $q^3$ by $q$, we get
\begin{eqnarray}
\sum_{n=0}^\infty \overline{b}(3n)q^n&\equiv&
\varphi^2(q^3)\frac{(q^4;q^4)_\infty^2}{(q^2;q^2)_\infty^4}\nonumber\\[5pt]
&\equiv&\varphi^2(q^3)\frac{1}{\varphi(-q^2)^2}\nonumber \\[5pt]
&\equiv&\varphi^2(q^3)\frac{\varphi(-q^2)}{\varphi(-q^6)}\nonumber\\[5pt]
&\equiv&
\varphi^2(q^3)\frac{\varphi(-q^{18})-2q^2f(-q^6,-q^{30})}{\varphi(-q^6)}\
({\rm mod\ }3). \label{eqb3n}
\end{eqnarray}
It can be readily seen that no terms on the right-hand of
\eqref{eqb3n} can have its power of $q$ to be congruent to $1$
modulo $3$. Thus, equating the coefficients of $q^{3n+1}$ yields
\[
\overline{b}(9n+3)\equiv 0\ ({\rm mod\ }3),
\]
which is due to Kim \cite{Kim12}.

Extracting those terms with power of $q$ being a multiple of $3$
from \eqref{eqb3n}, then replacing $q^3$ by $q$, we conclude that
\begin{eqnarray}
\sum_{n=0}^\infty \overline{b}(9n)q^n&\equiv&
\frac{\varphi^2(q)\varphi(-q^6)}{\varphi(-q^2)}\nonumber\\[5pt]
&\equiv& \varphi^2(q)\varphi^2(-q^2)\nonumber\\[5pt]
&\equiv&\varphi(q^3)\varphi(-q)\label{eqb9n}
\\[5pt]
&\equiv& \varphi(q^3)\left(\varphi(-q^9)-2qf(-q^3,-q^{15})\right)\
({\rm mod\ }3).\label{eqb9n1}
\end{eqnarray}
Furthermore, if we choose those terms in which the power of $q$ is a
multiple of $3$ from \eqref{eqb9n1}, and replace $q^3$ by $q$, we
obtain
\[
\sum_{n=0}^\infty \overline{b}(27n)q^n\equiv
\varphi(q)\varphi(-q^3)\ ({\rm mod\ }3).
\]
Combing together \eqref{eqb9n} with the above  congruence relation,
we find that
\[
\sum_{n=0}^\infty \overline{b}(9n)q^n\equiv \sum_{n=0}^\infty
\overline{b}(27n)(-q)^n \ ({\rm mod\ }3),
\]
from which we deduce that for $n\geq 0$,
\begin{align}\label{b927}
\overline{b}(9n)\equiv (-1)^n\overline{b}(27n)\ ({\rm mod\ }3).
\end{align}
For each term on the right-hand of \eqref{eqb9n1}, the power of $q$
can not be congruent to $2$ modulo $3$, thus we immediately derive
that for $n\geq 0$,
\begin{align}\label{eqb27n}
\overline{b}(27n+18)\equiv 0\ ({\rm mod\ }3).
\end{align}
On the other hand, substituting \eqref{eqlem1b} into \eqref{eqb9n},
we see that
\[
\sum_{n=0}^\infty \overline{b}(9n)q^n\equiv
\left(\varphi(q^{12})+2q^3\psi(q^{24})\right)\times\left(
\varphi(q^4)-2q\psi(q^8)\right)\ ({\rm mod\ }3).
\]
It follows from the fact there exist no terms of the form $q^{4n+2}$
in the above identity that
\begin{eqnarray}\label{eqb36n}
\overline{b}(36n+18)\equiv 0\ ({\rm mod\ }3).
\end{eqnarray}
Based on \eqref{b927}, \eqref{eqb27n} and \eqref{eqb36n}, and
proceeding by induction on $\alpha$, the desired results
\eqref{eq3.4a} and \eqref{eq3.4b} follows immediately.\end{proof}

%----------------------------------------------------------------
\section{Congruences modulo $5$ for $\overline{b}(n)$}

The goal of this section is devoted to proving the following
unexpected results, from which we obtain some strange congruences
modulo $5$ for $\overline{b}(n)$.

\begin{theorem}\label{thm4.1}
For all  $n\geq 0$, we have $\overline{b}(20n+10)\equiv 0\ ({\rm
mod\ }5)$ and
\begin{eqnarray}
\overline{b}(40n)&\equiv& \overline{b}(80n)\ ({\rm mod\ }5),\label{eqb40n}\\[5pt]
\overline{b}(20n)&\equiv& \overline{b}(100n)\ ({\rm mod\
}5).\label{eqb20n}
\end{eqnarray}
\end{theorem}
\begin{proof} From \eqref{eqb2n}, we have
\begin{eqnarray*}
\sum_{n=0}^\infty
\overline{b}(2n)q^n&=&\frac{\varphi^2(q)}{\varphi^6(-q)}\\[5pt]
&\equiv&\frac{\varphi^2(q)\varphi^4(-q)}{\varphi^2(-q^5)}\ ({\rm
mod\ }5).
\end{eqnarray*}
Thus,
\begin{equation}\label{eq4.1}
\sum_{n=0}^\infty \overline{b}(2n)(-q)^n\equiv
\frac{\varphi^2(-q)\varphi^4(q)}{\varphi^2(q^5)}\ ({\rm mod\ }5).
\end{equation}
Recall Ramanujan's identity \cite[p. 28, Entry 1.6.2]{AB00}
\[
16q(q^2;q^2)_\infty^2(q^{10};q^{10})_\infty^2=(\varphi^2(q)-\varphi^2(q^5))(5\varphi^2(q^5)
-\varphi^2(q)).
\]
It follows that
\begin{equation}\label{eq4.2}
-q(q^2;q^2)_\infty^2(q^{10};q^{10})_\infty^2\equiv
\varphi^4(q)-\varphi^2(q)\varphi^2(q^5)\ ({\rm mod\ }5),
\end{equation}
and
\begin{eqnarray*}
\varphi^2(-q)\varphi^4(q)&\equiv&
\varphi^2(-q)\left(\varphi^2(q)\varphi^2(q^5)
-q(q^2;q^2)_\infty^2(q^{10};q^{10})_\infty^2\right)\\[5pt]
&=&
\varphi^4(-q^2)\varphi^2(q^5)-q(q;q)_\infty^4(q^{10};q^{10})_\infty^2\\[5pt]
&\equiv&
\frac{\varphi^2(q^5)\varphi(-q^{10})}{\varphi(-q^2)}-q\frac{(q^5;q^5)_\infty
(q^{10};q^{10})_\infty^2}{(q;q)_\infty}\ ({\rm mod\ }5).
\end{eqnarray*}
Invoking the above identity, we can rewrite  \eqref{eq4.1} in the
following form
\[
\sum_{n=0}^\infty \overline{b}(2n)(-q)^n\equiv
\frac{\varphi(-q^{10})}{\varphi(-q^2)}-\frac{(q^5;q^5)_\infty
(q^{10};q^{10})_\infty^2}{\varphi^2(q^5)}\sum_{n=0}^\infty
p(n)q^{n+1} \ ({\rm mod\ }5).
\]
If we extract the terms whose power of $q$ is divisible by $5$,
replace $q^5$ by $q$, we find that
\[
\sum_{n=0}^\infty \overline{b}(10n)(-q)^n\equiv
\varphi(-q^2)\sum_{n=0}^\infty
\overline{p}(5n)q^{2n}-\frac{(q;q)_\infty(q^2;q^2)_\infty^2}{\varphi^2(q)}\sum_{n=0}^\infty
p(5n+4)q^n.
\]
 Applying \eqref{eqp5n+4}, we have
\begin{equation}\label{eq4.3}
\sum_{n=0}^\infty \overline{b}(10n)(-q)^n\equiv
\varphi(-q^2)\sum_{n=0}^\infty \overline{p}(5n)q^{2n}\ ({\rm mod\
}5).
\end{equation}
Note that the right-hand side of \eqref{eq4.3}, when expanded as a
power series, only contains terms of the form $q^{2m}$ for some $m$.
Thus, we get
\[
\overline{b}(20n+10)\equiv 0\ ({\rm mod\ }5).
\]
Selecting the terms for which the power of $q$ is even from
\eqref{eq4.3}, replacing $q^2$ by $q$, we obtain
\begin{equation}\label{eq4.4}
\sum_{n=0}^\infty \overline{b}(20n)q^n\equiv
\varphi(-q)\sum_{n=0}^\infty \overline{p}(5n)q^n\ ({\rm mod\ }5).
\end{equation}
Employing \eqref{eqLin} and \eqref{eq4.2}, we deduce that
\begin{eqnarray}
\sum_{n=0}^\infty \overline{b}(20n)q^n &\equiv&
\varphi^2(-q)\varphi^2(-q^5)+q\frac{(q^{10};q^{10})_\infty^2}{(q^2;q^2)_\infty}\psi(q^2)
\varphi^2(-q^2)\label{eq4.5}\\[5pt]
&\equiv&\varphi^4(-q)-q(q^2;q^2)_\infty^2(q^{10};q^{10})_\infty^2+
q\frac{(q^{10};q^{10})_\infty^2}{(q^2;q^2)_\infty}\cdot
\frac{(q^4;q^4)_\infty^2}{(q^2;q^2)_\infty}\cdot
\frac{(q^2;q^2)_\infty^4}{(q^4;q^4)_\infty^2}\nonumber\\[5pt]
&\equiv& \varphi(-q^5)\sum_{n=0}^\infty \overline{p}(n)q^n\ ({\rm
mod\ }5).\label{eq4.6}
\end{eqnarray}
Applying \eqref{eqlem1c} to \eqref{eq4.5}, we have
\begin{eqnarray*}
\sum_{n=0}^\infty \overline{b}(20n)q^n &\equiv&
\left(\varphi^2(q^2)-4q\psi^2(q^4)\right)\times
\left(\varphi^2(q^{10})-4q^5\psi^2(q^{20})\right)\\[5pt]
&+&q\frac{(q^{10};q^{10})_\infty^2}{(q^2;q^2)_\infty}\psi(q^2)
\varphi^2(-q^2)\ ({\rm mod\ }5).
\end{eqnarray*}
If we extract the terms with even power of $q$ from the above
identity and then replace $q^2$ by $q$, we find that
\begin{equation}\label{gm4.9}
\sum_{n=0}^\infty \overline{b}(40n)q^n\equiv
\varphi^2(q)\varphi^2(q^5)+q^3\psi^2(q^2)\psi^2(q^{10})\ ({\rm mod\
}5).
\end{equation}
Using the same argument to \eqref{gm4.9} yields
\begin{equation}\label{gm4.10}
\sum_{n=0}^\infty \overline{b}(80n)q^n\equiv
\varphi^2(q)\varphi^2(q^5)+q^3\psi^2(q^2)\psi^2(q^{10})\ ({\rm mod\
}5).
\end{equation}
Combining \eqref{gm4.9} with \eqref{gm4.10} and equating the
coefficients of $q^n$, we obtain
\[
\overline{b}(40n)\equiv \overline{b}(80n)\ ({\rm mod\ }5).
\]
It follows from \eqref{eq4.6} that
\begin{equation}
\sum_{n=0}^\infty \overline{b}(100n)q^n\equiv
\varphi(-q)\sum_{n=0}^\infty \overline{p}(5n)q^n\ ({\rm mod\ }5).
\end{equation}
Combining the above identity and \eqref{eq4.4} together, we conclude
that
\[
\overline{b}(20n)\equiv \overline{b}(100n)\ ({\rm mod\ }5).
\]
This completes the proof.\end{proof}

Employing \eqref{eqb40n} and \eqref{eqb20n}, and by induction, we
have the following corollary.

\begin{corollary}
For all $k\geq 0$ and $n\geq 0$,
\begin{eqnarray}
\overline{b}(40\times 2^kn)&\equiv& \overline{b}(40n) \ ({\rm mod\
}5),\\[5pt]
\overline{b}(20\times 5^kn)&\equiv& \overline{b}(20n) \ ({\rm mod\
}5).
\end{eqnarray}
\end{corollary}
Utilizing the above result and the known fact that
\[\overline{b}(20)\equiv 2 \ ({\rm mod\ }5),~ \overline{b}(40)\equiv 4 \ ({\rm mod\
}5),~ \overline{b}(380)\equiv 0 \ ({\rm mod\ }5),\] we obtain the
following strange congruences modulo $5$ for $\overline{b}(n)$.

\begin{corollary}
For all $k\geq 0$,
\begin{eqnarray}
\overline{b}(20\cdot 5^k)&\equiv& 2\ ({\rm mod\ }5),\\[5pt]
\overline{b}(40\cdot 2^k)&\equiv& 4\ ({\rm mod\ }5),\\[5pt]
\overline{b}(380\cdot 5^k)&\equiv& 0\ ({\rm mod\ }5).
\end{eqnarray}
\end{corollary}

%--------------------------------------------------------------------
\subsection*{Acknowledgements} The author would like
to thank the referee for helpful suggestions.

%--------------------------------------------------------------------
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