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\title{\bf New  infinite families of congruences modulo   8
 for partitions with even parts distinct}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address

\author{Ernest X.W. Xia\thanks{Supported by the National Natural Science Foundation of China (No. 11201188).}\\
\small Department of Mathematics\\[-0.8ex]
\small  Jiangsu University\\[-0.8ex]
\small  Zhenjiang, Jiangsu  212013, P. R. China\\
\small\tt ernestxwxia@163.com\\
%\and
%Forgotten Second Author \qquad  Forgotten Third Author\\
%\small School of Hard Knocks\\[-0.8ex]
%\small University of Western Nowhere\\[-0.8ex]
%\small Nowhere, Australasiaopia\\
%\small\tt \{fsa,fta\}@uwn.edu.ao
 }

% \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{Jan 16, 2014}{Sep 24, 2014}\\
\small Mathematics Subject Classifications: 05A17, 11P83}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
% should be informative, clear, and as complete as possible. Phrases
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\begin{abstract}
  Let
 $ped(n)$ denote the number of   partitions of  an integer $n$ wherein
   even  parts are
distinct.  Recently, Andrews, Hirschhorn and Sellers, Chen, and Cui
and Gu have derived a number of interesting congruences modulo 2, 3
and 4 for  $ped(n)$.
 In this paper
  we prove  several new   infinite families
 of
congruences modulo  8 for $ped(n)$. For example, we prove that for
  $ \alpha \geq 0$
  and $n\geq 0$,
\[
 ped\left(3^{4\alpha+4}n+\frac{11\times 3^{4\alpha+3}-1}{8}\right)\equiv 0
 \  ({\rm
mod \ 8}).
\]


  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} partition;
   congruence; regular partition
\end{abstract}

\section{Introduction}

Let $ped(n)$ denote the function
 which
enumerates the number of partitions of $n$ wherein even  parts are
distinct (and odd parts are unrestricted). For a positive integer
$t$ we say that a partition is $t$-regular if no part is divisible
by $t$. Andrews, Hirschhorn and Sellers \cite{Andrews}  found    the
generating function for $ped(n)$:
\begin{align}\label{1-1}
\sum_{n=0}^\infty ped(n)q^n =\prod_{n=1}^\infty
\frac{1+q^{2n}}{1-q^{2n-1}}=\frac{f_4}{f_1},
\end{align}
where  here and throughout this  paper,   and for any positive
 integer $k$,
   $f_k$ is defined  by
    \begin{align}\label{1-2}
f_k:=\prod_{n=1}^{\infty}(1-q^{kn}).
 \end{align}
 From \eqref{1-1}  it is easy to see that
  $ped(n)$ equals
the number of  4-regular partitions of $n$.
 In recent years  many  congruences  for  the number of
 regular partitions   have been
 discovered (see for example, Cui and Gu
 \cite{Cui,Cui-1},
    Dandurand and
Penniston \cite{Dandurand},    Furcy and  Penniston \cite{Furcy},
Gordon and Ono \cite{Gordon}, Keith \cite{Keith}, Lin and Wang
\cite{Lin}, Lovejoy and Penniston \cite{Lovejoy}, Penniston
\cite{Penniston,Penniston-2},
   Webb \cite{Webb}, Xia and Yao \cite{Xia,Xia-1}, and Yao\cite{Yao}).

Numerous congruence properties are known for the function $ped(n)$.
 For example,
Andrews, Hirschhorn and Sellers \cite{Andrews}
  proved that
for $\alpha \geq 1$ and
  $n\geq 0$,
 \begin{align}
ped(3n+2) &\equiv 0 \  ({\rm mod }\ 2), \label{1-3}\\[6pt]
ped(9n+4) &\equiv 0 \  ({\rm mod }\ 4), \label{1-4}\\[6pt]
ped(9n+7) &\equiv 0 \  ({\rm mod }\ 12), \label{1-5}\\[6pt]
ped\left(3^{2\alpha+2}n+\frac{11\times 3^{2\alpha+1}-1}{8}\right) &
\equiv 0 \  ({\rm mod \ 2}),\label{1-6}\\[6pt]
ped\left(3^{2\alpha+1}n+\frac{17\times 3^{2\alpha}-1}{8}\right) &
\equiv 0 \  ({\rm mod \ 6}), \label{1-7}\\[6pt]
ped\left(3^{2\alpha+2}n+\frac{19\times 3^{2\alpha+1}-1}{8}\right) &
\equiv 0 \  ({\rm mod \ 6}). \label{1-8}
 \end{align}
Recently,   Chen  \cite{Chen}
 obtained many interesting congruences   modulo
  2 and  4  for  $ped(n)$  using
the theory of Hecke eigenforms and  Cui and Gu \cite{Cui}
 found   infinite families of wonderful congruences modulo 2 for the function
  $ped(n)$.


 The aim  of this paper
   is to establish several new  infinite families
 of
congruences modulo  8 for $ped(n)$ by
 employing some  results of Andrews, Hirschhorn and
 Sellers \cite{Andrews}, and Cui and Gu \cite{Cui}.
 The main results of this  paper can be stated as the following
theorems.



\begin{theorem} \label{Th-1} For $ \alpha \geq 0$ and    $n\geq 0$,
\begin{align}
 ped\left(3^{2\alpha}n+\frac{3^{2\alpha}-1}{8}\right) &\equiv ped(n)
  \  ({\rm mod
\ 4}), \label{1-9}\\[6pt]
 ped\left(3^{4\alpha}n+\frac{  3^{4\alpha}-1}{8}\right)&\equiv
  5^\alpha ped(n)  \  ({\rm
mod \ 8}), \label{1-12}\\[6pt]
 ped\left(3^{4\alpha+4}n+\frac{11\times
  3^{4\alpha+3}-1}{8}\right)&\equiv 0  \  ({\rm
mod \ 8}),\label{1-13}\\[6pt]
 ped\left(3^{4\alpha+4}n+
 \frac{19\times 3^{4\alpha+3}-1}{8}\right)&\equiv 0  \  ({\rm
mod \ 8}).\label{1-14}
\end{align}

\end{theorem}


In view of  \eqref{1-9} and the facts $ped(1)=1$, $ped(2)=2$,
$ped(3)=3$, $ped(4)=4$,  we obtain the following corollary.

\begin{corollary} For $\alpha\geq 0$ and $i=0,\ 1,\ 2,\ 3$ we have
that
\begin{align}
 ped\left( \frac{t_i\times 3^{2\alpha}-1}{8}\right) &\equiv i\  ({\rm mod
\ 4}), \label{1-15}
\end{align}
where $t_0=33$, $t_1=9$, $t_2=17$ and $t_3=25$.
\end{corollary}

Replacing $\alpha$ by $2\alpha$ in \eqref{1-12}, we  find that for
$\alpha\geq 0$,
\begin{align}
  ped\left(3^{8\alpha}n+\frac{  3^{8\alpha}-1}{8}\right)&\equiv
   ped(n)  \ ({\rm
mod \ 8}). \label{1-16}
\end{align}
Employing  \eqref{1-16} and the facts $ped(1)=1$, $ped(2)=2$,
$ped(3)=3$, $ped(4)=4$, $ped(10)=29$, $ped(5)=6$,
$ped(253)=5178754681431$ and $ped(8)=16$, we obtain the following
congruences modulo 8.

\begin{corollary}For $\alpha\geq 0$ and $0\leq j \leq 7$  we have
that
\begin{align}
 ped\left( \frac{s_j\times 3^{8\alpha}-1}{8}\right) &\equiv j\ ({\rm mod
\ 8}), \label{1-17}
\end{align}
where $s_0=65$, $s_1=9$, $s_2=17$, $s_3=25$, $s_4=33$, $s_5=81$,
 $s_6=41$ and $s_7=2025$.
\end{corollary}


Utilizing the generating functions of $ped(9n+4)$, $ped(9n+7)$
   discovered
     by Andrews,   Hirschhorn and
Sellers \cite{Andrews} and the $p$-dissection identities
 of  two Ramanujan's
theta functions  due to Cui and Gu \cite{Cui}, we will prove the
following theorem.


\begin{theorem}\label{Th-2}
 Let $p$ be a prime such that
  $p\equiv 5,\ 7 \ ({\rm mod} \ 8)$ and $1\leq i \leq p-1$. Then
   for $n\geq 0$ and  $\alpha \geq 1$,
\begin{align}
ped\left(9p^{2\alpha}n+\frac{(72i+33p)p^{2\alpha-1}-1}{8}\right)
\equiv 0 \ ({\rm mod }\ 8)   \label{1-18}
\end{align}
and
\begin{align}
ped\left(9p^{2\alpha}n+\frac{(72i+57p)p^{2\alpha-1}-1}{8}\right)
\equiv 0 \ ({\rm mod }\ 8)  . \label{1-19}
\end{align}
\end{theorem}














%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof  of Theorem \ref{Th-1}}



Andrews, Hirschhorn and Sellers \cite{Andrews}
 established  the following   results  for $ped(3n+1)$:
\begin{align}
\sum_{n=0}^\infty
ped(9n+1)q^n=\frac{f_2^2f_3^{4}f_{4}}{f_1^{5}f_6^2}
+24q\frac{f_2^3f_3^3f_4f_6^3}{f_1^{10}},\label{2-1}\\[6pt]
\sum_{n=0}^\infty ped(9n+4)q^n= 4\frac{f_2f_3f_4f_6}{f_1^4}
+48q\frac{ f_2^2f_4f_6^6}{f_1^{9}} \label{k-1}
\end{align}
and
\begin{align}
\sum_{n=0}^\infty ped(9n+7)q^n= 12 \frac{f_2^4f_3^6f_4}{f_1^{11}}.
\label{k-2}
\end{align}
 By the binomial theorem  it is easy to see that
  for  all positive integers $m$ and  $k$,
 \begin{align}\label{2-2}
f_k^{2m}\equiv f_{2k}^m  \   ({\rm mod}\ 2).
 \end{align}
By \eqref{2-2}  we see that
\begin{align}
\frac{f_1^2}{f_2}\equiv \frac{f_2}{f_1^2}\equiv 1
 \  ({\rm
mod \ 2}), \label{2-3}
\end{align}
which  yields
\begin{align}
\frac{f_2^2}{f_1^4}\equiv \frac{f_3^4}{f_6^2}\equiv 1 \  ({\rm mod \
4}).\label{2-4}
\end{align}
It follows from \eqref{2-1} and \eqref{2-4} that
\begin{align}
\sum_{n=0}^\infty ped(9n+1)q^n\equiv \frac{f_4}{f_1} \  ({\rm mod \
4}).\label{2-5}
\end{align}
In view of \eqref{1-1} and \eqref{2-5}  we  see that for $n\geq 0$,
\begin{align}
 ped(9n+1)\equiv ped(n)  \  ({\rm mod \ 4}).\label{2-6}
\end{align}
Congruence \eqref{1-9} follows from     \eqref{2-6} and mathematical
induction.






%
%Setting $\alpha =k-1$ in \eqref{1-9}, we have
%\begin{align}
%ped\left(9^{k}n+\frac{57\times 9^{k-1}-1}{8}\right) \equiv 0 \
%({\rm mod \ 6}).\label{2-11}
%\end{align}
%Congruence \eqref{1-5} follows from \eqref{2-9} and \eqref{2-11}.

 Andrews, Hirschhorn and Sellers  \cite{Andrews}
 also
   established  the  following
    3-dissection formula of the generating function of
   $ped(n)$:
  \begin{align}
\sum_{n=0}^\infty ped(n) q^n=\frac{f_{12}f_{18}^4}{f_3^3f_{36}^2}
+q\frac{f_6^2f_9^3f_{36}}{f_3^4f_{18}^2}
+2q^2\frac{f_6f_{18}f_{36}}{f_3^3}.\label{2-8}
  \end{align}
  Fortin,   Jacob and   Mathieu \cite{Fortin},  and
   Hirschhorn and Sellers \cite{Hirschhorn-1}  independently
   derived
the following  3-dissection formula
 of the generating function of overpartitions:
 \begin{align}\label{2-9}
\frac{f_2}{f_1^2}= \frac{f_6^4f_9^6}{f_3^8f_{18}^3}
+2q\frac{f_6^3f_9^3}{f_3^7}+4q^2\frac{f_6^2f_{18}^3}{f_3^6}.
 \end{align}
Combining \eqref{1-1}, \eqref{2-1},  \eqref{2-8}, \eqref{2-9}  we
deduced
 that
\begin{align}
\sum_{n=0}^\infty &ped (9n+1)q^n \equiv
\frac{f_3^4}{f_6^2}\frac{f_2^2}{f_1^4}\frac{f_4}{f_1} \nonumber\\[6pt]
&\qquad   \   \equiv \frac{f_3^4}{f_6^2}\left(
\frac{f_6^4f_9^6}{f_3^8f_{18}^3}
+2q\frac{f_6^3f_9^3}{f_3^7}+4q^2\frac{f_6^2f_{18}^3}{f_3^6}\right)^2
\left(\frac{f_{12}f_{18}^4}{f_3^3f_{36}^2}
+q\frac{f_6^2f_9^3f_{36}}{f_3^4f_{18}^2}
+2q^2\frac{f_6f_{18}f_{36}}{f_3^3}\right) \nonumber\\[6pt]
&\qquad   \  \equiv
\frac{f_6^6f_9^{12}f_{12}}{f_3^{15}f_{18}^2f_{36}^2}
+q\frac{f_6^8f_9^{15}f_{36}}{f_3^{16}f_{18}^8}
+4q\frac{f_6^5f_9^9f_{12}f_{18}}{f_3^{14}f_{36}^2}
+6q^2\frac{f_6^7f_9^{12}f_{36}}{f_3^{15}f_{18}^5}\nonumber\\[6pt]
&\qquad \qquad   \quad
+4q^2\frac{f_6^4f_9^6f_{12}f_{18}^4}{f_3^{13}f_{36}^2}
+4q^3\frac{f_6^6f_9^9f_{36}}{f_3^{14}f_{18}^2}
 \  ({\rm
mod \ 8}).\label{2-10}
\end{align}
Extracting those  terms  associated with powers $q^{3n+1}$
  on both sides of
\eqref{2-10}, then dividing by $q$ and replacing $q^3$ by $q$,
 we find that
\begin{align}
\sum_{n=0}^\infty ped(27n+10) q^n \equiv
 \frac{f_2^8f_3^{15}f_{12}}{f_1^{16}f_{6}^8}
 +4\frac{f_2^5f_3^9f_{4}f_{6}}{f_1^{14}f_{12}^2}
 \  ({\rm
mod \ 8}). \label{2-11}
\end{align}
By the binomial theorem and \eqref{2-3}  we have
\begin{align}
\frac{f_2^8}{f_1^{16}}\equiv \frac{f_3^{16}}{f_6^8}\equiv 1  \ ({\rm
mod \ 8}), \label{2-12}
\end{align}
which yields
\begin{align}
 \frac{f_2^8f_3^{15}f_{12}}{f_1^{16}f_{6}^8}\equiv
  \frac{f_{12}}{f_3} \  ({\rm mod \ 8}).\label{2-13}
\end{align}
It follows from \eqref{2-2} that
\begin{align}
\frac{f_2^5f_3^9f_{4}f_{6}}{f_1^{14}f_{12}^2}\equiv
\frac{f_{12}}{f_3} \  ({\rm mod \ 2}).\label{2-14}
\end{align}
Substituting  \eqref{2-13} and  \eqref{2-14} into  \eqref{2-11}, we
see that
\begin{align}
\sum_{n=0}^\infty ped(27n+10) q^n \equiv
 5\frac{f_{12}}{f_3}
 \  ({\rm
mod \ 8}), \label{2-15}
\end{align}
which  implies  that
\begin{align}
\sum_{n=0}^\infty ped(81n+10) q^n &\equiv
 5\frac{f_{4}}{f_1}
 \  ({\rm
mod \ 8}) \label{2-16}
\end{align}
and for $n\geq 0$,
\begin{align}
 ped(81n+37)&\equiv  0
 \  ({\rm
mod \ 8}),\label{2-17}\\[6pt]
 ped(81n+64) &\equiv 0
 \  ({\rm
mod \ 8}).\label{2-18}
\end{align}
Thanks to \eqref{1-1}
 and \eqref{2-16}, we see that for $n\geq 0$,
 \begin{align}
ped(81n+10)\equiv 5 ped(n)  \  ({\rm mod \ 8}). \label{2-20}
 \end{align}
 Congruence \eqref{1-12} follows from
\eqref{2-20} and mathematical induction. Replacing $n$ by $81n+37$
 in \eqref{1-12}
and employing \eqref{2-17}, we obtain \eqref{1-13}. Replacing $n$ by
$81n+64$ in \eqref{1-12} and using
 \eqref{2-18}, we deduce \eqref{1-14}. The proof
  is complete.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \section{Proof of Theorem \ref{Th-2}}


 Thanks to  \eqref{k-1} and \eqref{2-2}, we have
 \begin{align}
\sum_{n=0}^\infty ped(9n+4) q^n &\equiv 4 f_2\psi(q^3)  \  ({\rm mod
\ 8}),\label{3-1}
 \end{align}
where $\psi(q)$ is defined by
\begin{align}
\psi(q):=\frac{f_2^2}{f_1}. \label{3-2}
\end{align}
In their nice paper \cite{Cui}, Cui and Gu   established
$p$-dissection
 formulas  for     $f_1$ and $\psi(q)$.
 They proved that for  any odd prime $p$,
\begin{align}
\psi(q)=\sum_{k=0}^{\frac{p-3}{2}} q^{\frac{k^2+k}{2}}
f\left(q^{\frac{p^2+(2k+1)p}{2}},q^{\frac{p^2-(2k+1)p}{2}} \right)
+q^{\frac{p^2-1}{8}}\psi(q^{p^2}) \label{3-3}
\end{align}
and for any prime $p\geq 5$,
\begin{align}
f_1=\sum_{k=\frac{1-p}{2},\atop
 k\neq \frac{\pm p-1}{6}}^{\frac{p-1}{2} }
 (-1)^k q^{\frac{3k^2+k}{2}}f\left(-q^{\frac{3p^2+(6k+1)p}{2}},
 -q^{\frac{3p^2-(6k+1)p}{2}}\right)+(-1)^{\frac{\pm p-1}{6}}q^{\frac{p^2-1}{24}}
 f_{p^2}, \label{3-4}
\end{align}
where
 \begin{align} \label{3-5}
\frac{\pm p-1}{6}:& = \left\{ \begin{aligned}
         & \frac{p-1}{6} \  \ \ {
         \rm if }\  p\equiv 1 \ ({\rm mod }\ 6)    , \\
                  &\frac{-p-1}{6} \  {
         \rm if }\  p\equiv -1 \ ({\rm mod }\ 6)
                          \end{aligned} \right.
 \end{align}
 and
 the Ramanujan
  theta function $f(a,b)$ is defined by
    \begin{align}\label{Ramanujan}
f(a,b):=\sum_{n=-\infty}^\infty
 a^{n(n+1)/2}b^{n(n-1)/2},
  \end{align}
where $|ab|<1$.

Let $a(n)$ be defined by
\begin{align}
\sum_{n=0}^\infty a(n)q^n:= f_2\psi(q^3). \label{3-6}
\end{align}
It follows from \eqref{3-1} and \eqref{3-6} that for $n\geq 0$,
\begin{align}
ped(9n+4)\equiv 4a(n) \  ({\rm mod \ 8}). \label{3-7}
\end{align}
Substituting  \eqref{3-3} and \eqref{3-4} into \eqref{3-6}, we  see
that for any prime $p\equiv 5,\ 7\ ({\rm mod }\ 8)$,
\begin{align} \label{3-8}
\sum_{n=0}^\infty a(n)q^n=& \left(\sum_{m=\frac{1-p}{2},\atop
 m\neq \frac{\pm p-1}{6}}^{\frac{p-1}{2} }
 (-1)^m q^{ 3m^2+m }f\left(-q^{ 3p^2+(6m+1)p },
 -q^{ 3p^2-(6m+1)p }\right)+(-1)^{\frac{\pm p-1}{6}}q^{\frac{p^2-1}{12}}
 f_{2p^2} \right)
 \nonumber\\[6pt]
 & \  \times
  \left(\sum_{k=0}^{\frac{p-3}{2}} q^{\frac{3(k^2+k)}{2}}
f\left(q^{\frac{3(p^2+(2k+1)p)}{2}},q^{\frac{3(p^2-(2k+1)p)}{2}}
\right) +q^{\frac{3(p^2-1)}{8}}\psi(q^{3p^2})\right).
\end{align}
Now, we consider the congruence
\begin{align}
 3m^2+m +  \frac{3(k^2+k)}{2}\equiv \frac{11(p^2-1)}{24} \ ({\rm
mod  }\ p), \label{3-9}
\end{align}
where  $-\frac{p-1}{2}\leq m \leq \frac{p-1}{2}$ and $0\leq k \leq
\frac{p-1}{2}$. Congruence \eqref{3-9} can be rewritten  as follows
\begin{align}
2(6m+1)^2+(6k+3)^2\equiv 0 \ ({\rm mod  }\ p). \label{3-10}
\end{align}
Since $p\equiv 5,\ 7\ ({\rm mod }\ 8)$, we have  that $-2$ is a
 ratic nonresidue modulo $p$ and hence \eqref{3-10} is equivalent
to
\begin{align}
6m+1\equiv 6k+3\equiv 0 \ ({\rm mod  }\ p). \label{3-11}
\end{align}
Thus, $m=\frac{\pm p-1}{6}$ and $k=\frac{p-1}{2}$. Extracting those
terms  associated with powers $q^{pn+\frac{11(p^2-1)}{24}}$ on
 both sides of \eqref{3-8}
   and
employing
 the fact that Congruence
     \eqref{3-9} holds if and only
  if   $m=\frac{\pm p-1}{6}$ and $k=\frac{p-1}{2}$, we have
  \begin{align}
\sum_{n=0}^\infty
a\left(pn+\frac{11(p^2-1)}{24}\right)q^{pn+\frac{11(p^2-1)}{24}}
=(-1)^{\frac{\pm
p-1}{6}}q^{\frac{11(p^2-1)}{24}}f_{2p^2}\psi(q^{3p^2}). \label{3-12}
  \end{align}
Dividing $q^{\frac{11(p^2-1)}{24}}$ on both sides of \eqref{3-12}
and then replacing $q^p$ by $q$, we get
  \begin{align}
\sum_{n=0}^\infty a\left(pn+\frac{11(p^2-1)}{24}\right)q^n
=(-1)^{\frac{\pm p-1}{6}} f_{2p}\psi(q^{3p}), \label{3-13}
  \end{align}
which  implies that
 \begin{align}
\sum_{n=0}^\infty a\left(p^2n+\frac{11(p^2-1)}{24}\right)q^n
=(-1)^{\frac{\pm p-1}{6}} f_{2}\psi(q^{3}) \label{3-14}
  \end{align}
  and
 \begin{align}
 a\left(p(pn+i)+\frac{11(p^2-1)}{24} \right) =0  \label{3-15}
  \end{align}
   for $n\geq 0$ and $1\leq i \leq p-1$.
Combining  \eqref{3-6} and \eqref{3-14}, we have
\begin{align}
a\left(p^2n+\frac{11(p^2-1)}{24}\right)\equiv a(n)\  ({\rm mod }\
2). \label{3-16}
\end{align}
By \eqref{3-16} and mathematical induction, we find that for $n\geq
0$
 and $\alpha \geq 0$,
 \begin{align}
a\left(p^{2\alpha}n+\frac{11(p^{2\alpha}-1)}{24}\right)\equiv a(n)\
({\rm mod }\ 2). \label{3-17}
 \end{align}
Replacing $n$ by $p(pn+i)+\frac{11(p^2-1)}{24}$ $ (1\leq i \leq
p-1)$ in \eqref{3-17} and using \eqref{3-15}, we deduce that for
$n\geq 0$ and $\alpha\geq 1$,
 \begin{align}
a\left(p^{2\alpha}n+\frac{(24i+11p)p^{2\alpha-1}-11}{24}\right)\equiv
0\  ({\rm mod }\ 2). \label{3-18}
 \end{align}
 Finally, replacing $n$ by $p^{2\alpha}n+\frac{(24i+11p)p^{2\alpha-1}-11}{24}$
  $(1\leq i \leq
p-1)$ in  \eqref{3-7} and using
  \eqref{3-18},
   we get
  \eqref{1-18}.

We  conclude the paper by  proving   \eqref{1-19}.  In view of
\eqref{k-2} and \eqref{2-2}, we find that
\begin{align}
\sum_{n=0}^\infty ped(9n+7) q^n \equiv
 4 f_1\psi(q^6) \  ({\rm mod }\
8), \label{3-19}
\end{align}
where $\psi(q)$ is defined by \eqref{3-2}.
  Let $b(n)$ be defined by
\begin{align}
\sum_{n=0}^\infty b(n) q^n := f_1\psi(q^6). \label{3-20}
\end{align}
By  \eqref{3-19} and \eqref{3-20}, we  find that for $n\geq 0$,
\begin{align}
 ped(9n+7)\equiv 4 b(n)  \  ({\rm mod }\
8). \label{3-21}
\end{align}
Substituting \eqref{3-3} and \eqref{3-4} into \eqref{3-20}, we see
that for any prime $p\equiv 5,\ 7\ ({\rm mod } \ 8)$,
\begin{align}
\sum_{n=0}^\infty b(n) q^n=& \left( \sum_{m=\frac{1-p}{2},\atop
 m\neq \frac{\pm p-1}{6}}^{\frac{p-1}{2} }
 (-1)^m q^{\frac{3m^2+m}{2}}f\left(-q^{\frac{3p^2+(6m+1)p}{2}},
 -q^{\frac{3p^2-(6m+1)p}{2}}\right)+(-1)^{\frac{\pm p-1}{6}}q^{\frac{p^2-1}{24}}
 f_{p^2}\right)
 \nonumber\\[6pt]
 &\
  \times
 \left(\sum_{k=0}^{\frac{p-3}{2}} q^{3(k^2+k)}
f\left(q^{ 3(p^2+(2k+1)p) },q^{3(p^2-(2k+1)p) } \right)
+q^{\frac{3(p^2-1)}{4}}\psi(q^{6p^2})\right). \label{3-22}
\end{align}
As above, for any prime $p\equiv 5,\ 7\ ({\rm mod } \ 8)$,
 $-\frac{p-1}{2}\leq m \leq \frac{p-1}{2}$ and $0\leq k \leq
 \frac{p-1}{2}$, the congruence relation
 \begin{align}
\frac{3m^2+m}{2}+3(k^2+k)\equiv \frac{19(p^2-1)}{24} \ ({\rm mod }\
p) \label{3-23}
 \end{align}
holds if and only if $m=\frac{\pm p-1}{6}$ and $k=\frac{p-1}{2}$.
This implies that
\begin{align}
\sum_{n=0}^\infty b\left(pn+\frac{19(p^2-1)}{24}\right)
 q^n =(-1)^{\frac{\pm p-1}{6}}f_p\psi(q^{6p}). \label{3-24}
\end{align}
Thanks to  \eqref{3-24}, we find  that
\begin{align}
\sum_{n=0}^\infty b\left(p^2n+\frac{19(p^2-1)}{24}\right)
 q^n =(-1)^{\frac{\pm p-1}{6}}f_1\psi(q^{6}) \label{3-25}
\end{align}
and
\begin{align}
b\left(p(pn+i)+\frac{19(p^2-1)}{24}\right)=0  \label{3-26}
\end{align}
for $n\geq 0$ and $1\leq i \leq p-1$.  It follows from \eqref{3-20}
and \eqref{3-25}  that for $n\geq 0$,
\begin{align}
b\left(p^2n+\frac{19(p^2-1)}{24}\right)\equiv b(n) \ ({\rm mod } \
2). \label{3-27}
\end{align}
By  \eqref{3-27} and mathematical induction, we deduce that for
$n\geq 0$
 and $\alpha \geq 0$,
\begin{align}
b\left(p^{2\alpha}n+\frac{19(p^{2\alpha}-1)}{24}\right)\equiv b(n) \
({\rm mod } \ 2). \label{3-28}
\end{align}
Replacing $n$ by $p(pn+i)+\frac{19(p^2-1)}{24}$ $(1\leq i \leq p-1)$
in \eqref{3-28} and employing  \eqref{3-26}, we find that
\begin{align}
b\left(p^{2\alpha}n+\frac{(24i+19p)p^{2\alpha-1}-19}{24}\right)\equiv
0 \ ({\rm mod } \ 2)
 \label{3-29}
\end{align}
 for $n\geq
0$,  $\alpha \geq 1$ and $1\leq i \leq p-1$.  Congruence
\eqref{1-19} follows from \eqref{3-21} and \eqref{3-29}.
 This completes the proof of Theorem \ref{Th-2}.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
   We wish to thank    the referee  for    helpful comments.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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