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\begin{center}
\vskip 1cm{\LARGE\bf Arithmetic Properties of Partition $k$-tuples \\
\vskip .1in
with Odd Parts Distinct
}
\vskip 1cm
\large
M. S. Mahadeva Naika and D. S. Gireesh\footnote{This research was partially supported by DST  Govt.\ of India (SR/S4/MS:739/11).} \\
Department of Mathematics\\
Bangalore University\\
Central College Campus\\ 
Bangalore-560 001 \\
Karnataka\\
India\\
\href{mailto:msmnaika@rediffmail.com}{\tt msmnaika@rediffmail.com} \\
\href{mailto:gireeshdap@gmail.com}{\tt gireeshdap@gmail.com} 
\end{center}

\vskip .2 in

\begin{abstract}
Let $\textnormal{pod}_{-k}(n)$ denote the number of partition $k$-tuples of $n$ wherein odd parts are distinct (and even parts are unrestricted).
We establish some interesting infinite families of congruences and internal congruences modulo 4, 16, and 5 for $\textnormal{pod}_{-2}(n)$, $\textnormal{pod}_{-4}(n)$, and $\textnormal{pod}_{-6}(n)$, respectively. We also find Ramanujan-type congruences modulo 5 for $\textnormal{pod}_{-3}(n)$ and densities of $\textnormal{pod}_{-2}(n)$, $\textnormal{pod}_{-3}(n)$, $\textnormal{pod}_{-4}(n)$, and $\textnormal{pod}_{-6}(n)$ modulo 4, 5, 16, and 5, respectively.
\end{abstract}

\section{Introduction}
For $|q|&lt;1$, Ramanujan's theta functions $\varphi(q)$ and $\psi(q)$ are 
defined by
\begin{equation}\label{phi}
\varphi\left(q\right):=1+2\sum_{n=1}^{\infty}q^{n^2}=\sum\limits_{n=-\infty}^{\infty}q^{n^2}=(-q;q^2)^2_\infty(q^2;q^2)_\infty
\end{equation}
and
\begin{equation}\label{psi}
\psi\left(q\right):=\sum\limits_{n=0}^{\infty}q^{(n^2+n)/2}
=\sum\limits_{n=-\infty}^{\infty}q^{2n^2+n}=\dfrac{(q^2;q^2)_\infty}{(q;q^2)_\infty},
\end{equation}
where $(a;q)_\infty=(1-a)(1-aq)(1-aq^2)\cdots$.

Let $\textnormal{pod}(n)$ denote the number of partitions of $n$ wherein odd parts are distinct (and even parts are unrestricted). The generating function of $\textnormal{pod}(n)$ is
\[\sum\limits_{n=0}^{\infty}\textnormal{pod}(n)q^n=\frac{(-q;q^2)_\infty}{(q^2;q^2)_\infty}=\frac{1}{\psi(-q)}.\]

In 2010, Hirschhorn and Sellers \cite{HSpod} proved that, for all $\alpha\geq 0$ and $n\geq0$,
\[\textnormal{pod}\left(3^{2\alpha+3}n+\frac{23\times 3^{2\alpha+2}+1}{8}\right)\equiv 0\pmod{3}.\]
They also found some internal congruences such as \[\textnormal{pod}(81n + 17)\equiv 5\textnormal{pod}(9n + 2) \pmod{27}.\]

Recently, Wang \cite{WANG1} established new congruences for $\textnormal{pod}(n)$. For example, for each $\alpha \geq 1$ and $n\geq 0$,
\[\textnormal{pod}\left(5^{2\alpha+2}n+\frac{11\times 5^{2\alpha+1}+1}{8}\right)\equiv 0\pmod{5}.\]

Let $\textnormal{pod}_{-k}(n)$ denote the number of partition $k$-tuples of $n$ wherein odd parts are distinct (and even parts are unrestricted). The generating function of $\textnormal{pod}_{-k}(n)$ is
\begin{equation}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-k}(n)q^n=\frac{(-q;q^2)^k_\infty}{(q^2;q^2)^k_\infty}=\frac{1}{\psi(-q)^k}.
\end{equation}

Chen and Lin \cite{CL} established congruences modulo 3 and 5 for $\textnormal{pod}_{-2}(n)$. For example, for $\alpha \geq 1$ and $n\geq 0$,
\[\textnormal{pod}_{-2}\left(5^{\alpha+1}n+\frac{11\times 5^{\alpha}+1}{4}\right)\equiv 0\pmod{5}.\]

Wang \cite{WANG2,WANGbul1}  has established congruences modulo 7, 9, and 11 satisfied by $\textnormal{pod}_{-3}(n)$ and congruences modulo 5, 9, and 81 satisfied by $\textnormal{pod}_{-4}(n)$ by employing theta function identities. For example, for $\alpha \geq 1$ and $n\geq 0$,
\[\textnormal{pod}_{-3}\left(3^{2\alpha+2}n+\frac{23\times 3^{2\alpha+1}+3}{8}\right)\equiv 0\pmod{9}\]
and \[\textnormal{pod}_{-4}\left(3^{\alpha+1}n+\frac{5\times 3^{\alpha}+1}{2}\right)\equiv 0\pmod{9}.\]
He also found some internal congruences such as
\[\textnormal{pod}_{-4}(27n+5)\equiv -\textnormal{pod}_{-4}(9n+2) \pmod{9}.\]

In this paper, we establish congruences modulo powers of 2 and modulo 5 for $\textnormal{pod}_{-k}(n)$ for $k\in\{2,3,4,6\}$. In this vein, in Section \ref{2podsec1}, we find infinite family of congruences and internal congruences modulo 4 satisfied by $\textnormal{pod}_{-2}(n)$ and we also find density of $\textnormal{pod}_{-2}(n)$ modulo 4. In Section \ref{3podsec1}, we prove Ramanujan-type congruences modulo 5 for $\textnormal{pod}_{-3}(n)$ and that $\textnormal{pod}_{-3}(n)$ is divisible by 5 at least $1/30$ of the time. In Section \ref{4podsec1}, we establish infinite family of congruences and internal congruences modulo 16 satisfied by $\textnormal{pod}_{-4}(n)$ following density of $\textnormal{pod}_{-4}(n)$ modulo 16. In Section \ref{6podsec1}, we determine infinite family of congruences and internal congruences modulo 5 satisfied by $\textnormal{pod}_{-6}(n)$ and we also determine density of $\textnormal{pod}_{-6}(n)$ modulo 5.

\section{Preliminaries}
The following results are useful in proving our main results.

\begin{lemma}\cite[pp.\ 40--49]{BCB1}
We have
\begin{align}
\varphi(q)&amp;=\varphi(q^4)+2q\psi(q^8),\label{2disphi}\\
\varphi(q)^2&amp;=\varphi(q^2)^2+4q\psi(q^4)^2,\label{2disphi2}\\
\psi(q)&amp;=f(q^3,q^6)+q\psi(q^9)\label{3psi}\\
&amp;=f(q^{10},q^{15})+qf(q^5,q^{20})+q^3\psi(q^{25}),\label{5psi}\\
\psi(q)^2&amp;=\varphi(q)\psi(q^2)\label{2psi}.
\end{align}
\end{lemma}

\begin{lemma}\cite[Eq. 1.6.7, p. 26]{AB}
We have
\begin{equation}\label{f1234}
f(q,q^4)f(q^2,q^3)=\psi(q)^2-q\psi(q^5)^2.
\end{equation}
\end{lemma}

\begin{lemma}\label{mgpsi4l}
Let $\sum\limits_{n=0}^{\infty}h(n)q^n=q\psi(q)^4$. Then
\begin{equation}\label{mgpsi4}
\sum\limits_{n=0}^{\infty}h(5n+3)q^n\equiv \psi(q)^4\pmod{5}.
\end{equation}
\end{lemma}
\begin{proof}
From \eqref{5psi}, it follows that
\begin{align*}
\sum\limits_{n=0}^{\infty}h(n)q^n&amp;=q\psi(q)^4\\&amp;
=q\left(f(q^{10},q^{15})+qf(q^5,q^{20})+q^3\psi(q^{25})\right)^4\\&amp;
=12q^5f(q^{10},q^{15})^2f(q^5,q^{20})\psi(q^{25})+4q^{10}f(q^{10},q^{15})\psi(q^{25})^3\\&amp;
\quad+12q^8f(q^{10},q^{15})f(q^5,q^{20})\psi(q^{25})^2+6q^7f(q^{10},q^{15})^2\psi(q^{25})^2\\&amp;
\quad+4q^2f(q^{10},q^{15})^3f(q^5,q^{20})+4q^4f(q^{10},q^{15})^3\psi(q^{25})\\&amp;
\quad+6q^3f(q^{10},q^{15})^2f(q^5,q^{20})^2+q^5f(q^5,q^{20})^4+q^{13}\psi(q^{25})^4\\&amp;
\quad+4q^4f(q^{10},q^{15})f(q^5,q^{20})^3+6q^9f(q^5,q^{20})^2\psi(q^{25})^2\\&amp;
\quad+4q^7f(q^5,q^{20})^3\psi(q^{25})+12q^6f(q^{10},q^{15})f(q^5,q^{20})^2\psi(q^{25})\\&amp;
\quad+4q^{11}f(q^5,q^{20})\psi(q^{25})^3+qf(q^{10},q^{15})^4,
\end{align*}
which yields
\begin{align*}
\sum\limits_{n=0}^{\infty}h(5n+3)q^n\equiv\,&amp; 2qf(q^{2},q^{3})f(q,q^{4})\psi(q^{5})^2+f(q^{2},q^{3})^2f(q,q^{4})^2\\&amp;+q^{2}\psi(q^{5})^4\pmod{5}.
\end{align*}
%\[\sum\limits_{n=0}^{\infty}h(5n+3)q^n\equiv 2qf(q^{2},q^{3})f(q,q^{4})\psi(q^{5})^2+f(q^{2},q^{3})^2f(q,q^{4})^2+q^{2}\psi(q^{5})^4\pmod{5}.\]
Using \eqref{f1234} in the above equation, we arrive at \eqref{mgpsi4}.
\end{proof}

\begin{lemma}
Let $\sum\limits_{n=0}^{\infty}g(n)q^n=\psi(q)^2$. Then
\begin{equation}\label{mgpsi1}
\sum\limits_{n=0}^{\infty}g(5n+1)q^n=2\psi(q)^2-q\psi(q^5)^2.
\end{equation}
\end{lemma}
\begin{proof}
It follows from \eqref{5psi} that
\begin{align*}
\sum\limits_{n=0}^{\infty}g(n)q^n&amp;=\psi(q)^2\\&amp;
=\left(f(q^{10},q^{15})+qf(q^5,q^{20})+q^3\psi(q^{25})\right)^2\\&amp;
=q^6\psi(q^{25})^2+2q^4\psi(q^{25})f(q^5,q^{20})+2q^3\psi(q^{25})f(q^{10},q^{15})\\&amp;
\quad+q^2f(q^5,q^{20})^2+2qf(q^{10},q^{15})f(q^5,q^{20})+f(q^{10},q^{15})^2,
\end{align*}
which yields
\[\sum\limits_{n=0}^{\infty}g(5n+1)q^n=2f(q^{2},q^{3})f(q,q^{4})+q\psi(q^{5})^2.\]
Using \eqref{f1234} in the above equation, we arrive at \eqref{mgpsi1}.
\end{proof}

\begin{lemma}\cite[Theorem\ 2.1]{CG}\label{ps2}
For any odd prime, $p$,
\begin{equation}\label{ps3}
\psi(q)=\sum\limits_{m=0}^{\frac{p-3}{2}}q^{\frac{m^2+m}{2}}f\left(q^{\frac{p^2+(2m+1)p}{2}},q^{\frac{p^2-(2m+1)p}{2}}\right)+q^{\frac{p^2-1}{8}}\psi(q^{p^2}).
\end{equation}
Furthermore, $\frac{m^2+m}{2}\not\equiv\frac{p^2-1}{8}\pmod{p}$, for $0\leq m\leq\frac{p-3}{2}$.
\end{lemma}

\section{Arithmetic properties of \texorpdfstring{$\textnormal{pod}_{-2}(n)$}{Lg}}\label{2podsec1}
In this section, we prove the infinite family of congruences and internal congruences modulo 4 for $\textnormal{pod}_{-2}(n)$.

\subsection{Infinite family of congruences modulo 4}

\begin{theorem}\label{2podtm1}
Let  $p$ be any  odd prime such that $\left(\frac{-2}{p}\right)=-1$ and $\alpha\geq 0$. Then
\begin{equation}\label{2podt1}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}\left(2p^{2\alpha}n+\frac{3p^{2\alpha}+1}{4}\right)q^n\equiv 2\psi(q)\psi(q^2)\pmod{4}
\end{equation}
and, for all $n\geq 0$ and $1\leq \xi \leq p-1$,
\begin{align}\label{2podt2}
\textnormal{pod}_{-2}\left(2p^{2\alpha+1}(pn+\xi)+\frac{3p^{2\alpha+2}+1}{4}\right)\equiv 0\pmod{4}.
\end{align}
\end{theorem}
\begin{proof}
We have
\begin{equation}\label{defpod2}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(n)q^n=\frac{1}{\psi(-q)^2}.
\end{equation}
Invoking \eqref{2psi} and \eqref{defpod2},
\begin{align*}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(n)q^n&amp;=\frac{1}{\psi(q^2)\varphi(-q)}\\
&amp;=\frac{\left(1-(1-\varphi(-q))\right)^{-1}}{\psi(q^2)}\\
&amp;=\frac{1+(1-\varphi(-q))+(1-\varphi(-q))^2+\cdots}{\psi(q^2)}\\&amp;\equiv \frac{2-\varphi(-q)}{\psi(q^2)}\pmod{4}\quad\text{from \eqref{phi}}.
\end{align*}
Using \eqref{2disphi} in the above equation, we find that
\begin{align*}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(n)q^n\equiv\frac{2-\varphi(q^4)+2q\psi(q^8)}{\psi(q^2)}\pmod{4},
\end{align*}
which yields
\begin{align}\label{2pod4}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(2n+1)q^n\equiv 2\frac{\psi(q^4)}{\psi(q)}\pmod{4}.
\end{align}
From the binomial theorem, we can see that for any  prime $p$ and for each positive integer $\ell$,
\begin{equation}\label{fp1}
(q;q)^{p^{\ell}}\equiv ({q^{p}};{q^{p}})^{p^{\ell-1}}\pmod{p^{\ell}}.
\end{equation}
In view of \eqref{fp1}, \eqref{2pod4} can be expressed as
\begin{align}\label{2pod5}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(2n+1)q^n\equiv 2\psi(q)\psi(q^2)\pmod{4},
\end{align}
which is the $\alpha=0$ case of \eqref{2podt1}. If we assume that \eqref{2podt1} holds for some $\alpha\geq0$, then, substituting \eqref{ps3} in \eqref{2podt1},
\begin{align}
\nonumber&amp;\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}\left(2p^{2\alpha}n+\frac{3p^{2\alpha}+1}{4}\right)q^n\\&amp;
\nonumber\equiv 2\Biggr(\sum\limits_{m=0}^{\frac{p-3}{2}}q^{\frac{m^2+m}{2}}f\left(q^{\frac{p^2+(2m+1)p}{2}},q^{\frac{p^2-(2m+1)p}{2}}\right)
+q^{\frac{p^2-1}{8}}\psi(q^{p^2})\Biggr)\\&amp;
\quad\times \Biggr(\sum\limits_{m=0}^{\frac{p-3}{2}}q^{m^2+m}f\left(q^{p^2+(2m+1)p},q^{p^2-(2m+1)p}\right)
+q^{\frac{p^2-1}{4}}\psi(q^{2p^2})\Biggr)\pmod{4}.\label{2pod6}
\end{align}
For any odd prime, $p$, and $0\leq m_1,m_2\leq(p-3)/2$, consider the congruence
$$\frac{m_1^2+m_1}{2}+2\times\frac{m_2^2+m_2}{2}\equiv\frac{3p^2-3}{8}\pmod {p},$$
which implies that
\begin{equation}\label{2podJ4}
(2m_1+1)^2+2(2m_2+1)^2\equiv 0\pmod{p}.
\end{equation}
Since $\left(\frac{-2}{p}\right)=-1$, the only solution of the congruence \eqref{2podJ4} is $m_1=m_2=\dfrac{p-1}{2}$.
Therefore, equating the coefficients of $q^{pn+\frac{3p^2-3}{8}}$ from both sides of \eqref{2pod6}, dividing throughout by $q^{\frac{3p^2-3}{8}}$ and then replacing $q^p$ by $q$, we obtain
\begin{equation}\label{2podJ5}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}\left(2p^{2\alpha+1}n+\frac{3p^{2\alpha+2}+1}{4}\right)q^n\equiv 2\psi(q^p)\psi(q^{2p})\pmod{4}.
\end{equation}
Equating the coefficients of $q^{pn}$ on both sides of \eqref{2podJ5} and then replacing $q^p$ by $q$, we obtain
\begin{equation*}\label{2podJ6}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}\left(2p^{2\alpha+2}n+\frac{3p^{2\alpha+2}+1}{4}\right)q^n\equiv 2\psi(q)\psi(q^2)\pmod{4},
\end{equation*}
which is the $\alpha+1$ case of \eqref{2podt1}.

Equating the coefficients of $q^{pn+\xi}$ for $1\leq \xi \leq p-1$ from \eqref{2podJ5}, we arrive at \eqref{2podt2}.
\end{proof}

\begin{corollary}
Let  $p$ be any  odd prime such that $\left(\frac{-2}{p}\right)=-1$. Then $\textnormal{pod}_{-2}(n)$ is divisible by $4$ for at least $\frac{1}{2(p+1)}$ of all nonnegative integers $n$.
\end{corollary}
\begin{proof}
The arithmetic sequences $\left\{2p^{2\alpha+1}(pn+\xi)+\frac{3p^{2\alpha+2}+1}{4}:\alpha\geq 0\right\}$ for $1\leq \xi\leq p-1$, on which $\textnormal{pod}_{-2}(\cdot)$ is $0$ modulo $4$, do not intersect. These sequences account for
$$(p-1)\left(\frac{1}{2p^2}+\frac{1}{2p^4}+\frac{1}{2p^6}+\cdots\right)=\frac{1}{2(p+1)}$$
of all nonnegative integers.
\end{proof}

\subsection{Some internal congruences}

\begin{theorem}
For each $n\geq 0$,
\begin{align}
\textnormal{pod}_{-2}(54n+25)&amp;\equiv \textnormal{pod}_{-2}(6n+3)\pmod{4},\label{2podi1}\\
\textnormal{pod}_{-2}(54n+43)&amp;\equiv \textnormal{pod}_{-2}(6n+5)\pmod{4},\label{2podi2}\\
\textnormal{pod}_{-2}(162n+7)&amp;\equiv 2\textnormal{pod}_{-2}(18n+1)\pmod{4},\label{2podi3}\\
\textnormal{pod}_{-2}(162n+115)&amp;\equiv 2\textnormal{pod}_{-2}(18n+13)\pmod{4}.\label{2podi4}
\end{align}
\end{theorem}
\begin{proof}
If $\sum\limits_{n=0}^{\infty}a(n)q^n=\psi(q)\psi(q^2)$, then the authors \cite{MG} found that
\begin{equation}\label{mgpsi2}
\sum\limits_{n=0}^{\infty}a(3n)q^n=\psi(q)\varphi(q)-q\psi(q^3)\psi(q^6).
\end{equation}
Using \eqref{mgpsi2}, we can express \eqref{2pod5} as
\begin{equation}\label{2pod9}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(6n+1)q^n\equiv 2\psi(q)\varphi(q)-2q\psi(q^3)\psi(q^6)\pmod{4}.
\end{equation}
Invoking \eqref{phi} and  \eqref{2pod9},
\begin{equation}\label{2pod10}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(6n+1)q^n\equiv 2\psi(q)+2q\psi(q^3)\psi(q^6)\pmod{4}.
\end{equation}
Substituting \eqref{3psi} into \eqref{2pod10} and extracting the terms involving $q^{3n+1}$,
\begin{equation*}\label{2pod11}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(18n+7)q^n\equiv 2\psi(q^3)+2\psi(q)\psi(q^2)\pmod{4},
\end{equation*}
which is equivalent to
\begin{equation}\label{2pod12}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(18n+7)q^n\equiv 2\psi(q^3)+\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(2n+1)q^n\pmod{4}.
\end{equation}
Equating the coefficients of $q^{3n+1}$ and $q^{3n+2}$ from \eqref{2pod12}, we arrive at \eqref{2podi1} and \eqref{2podi2}, respectively.

Equating the coefficients of $q^{3n}$ and then replacing $q^3$ by $q$ from \eqref{2pod12},
\begin{equation}\label{2pod13}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(54n+7)q^n\equiv 2\psi(q)+\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(6n+1)q^n\pmod{4}.
\end{equation}
Invoking \eqref{2pod10} and \eqref{2pod13},
\begin{equation}\label{2pod14}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(54n+7)q^n\equiv 2\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-2}(6n+1)q^n-2q\psi(q^3)\psi(q^6)\pmod{4},
\end{equation}
Equating the coefficients of $q^{3n}$ and $q^{3n+2}$ from \eqref{2pod14}, we arrive at \eqref{2podi3} and \eqref{2podi4}, respectively.
\end{proof}

\section{Ramanujan-type congruences for \texorpdfstring{$\textnormal{pod}_{-3}(n)$}{Lg}}\label{3podsec1}
In this section, we prove the Ramanujan-type congruences modulo 5 for $\textnormal{pod}_{-3}(n)$.

\begin{theorem}
For each $\alpha\geq 1$,
\begin{equation}\label{3podt2}
\textnormal{pod}_{-3}\left(5^{2\alpha+1}n+\frac{\mu\times 5^{2\alpha}+3}{8}\right)\equiv 0\pmod{5},
\end{equation}
where $\mu=13$, $21$, $29$, and $37$.
\end{theorem}
\begin{proof}
We have
\begin{equation*}\label{defpod3}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(n)(-1)^nq^n=\frac{1}{\psi(q)^3}.
\end{equation*}
It follows from \eqref{fp1} that
\begin{align*}%\label{3pod1}
\nonumber\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(n)(-1)^nq^n&amp;\equiv\frac{\psi(q)^2}{\psi(q^5)}\pmod{5}\\&amp;
\equiv\frac{1}{\psi(q^5)}\sum\limits_{n=0}^{\infty}g(n)q^n\pmod{5}.
\end{align*}
Extracting the terms involving $q^{5n+1}$, dividing throughout by $q$ and then replacing $q^5$ by $q$,
\begin{align}\label{3pod2}
\nonumber\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(5n+1)(-1)^{n+1}q^n&amp;
\equiv\frac{1}{\psi(q)}\sum\limits_{n=0}^{\infty}g(5n+1)q^n\pmod{5}\\&amp;
\nonumber\equiv\frac{1}{\psi(q)}\left(2\psi(q)^2-q\psi(q^5)^2\right)\pmod{5}\quad\text{from \eqref{mgpsi1}}\\&amp;
\equiv 2\psi(q)-q\psi(q^5)\psi(q)^4\pmod{5}\quad\text{using \eqref{fp1}}.
\end{align}
Substituting \eqref{5psi} into \eqref{3pod2} and from the Lemma \eqref{mgpsi4l} , we find that
\begin{align*}
\nonumber\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(5n+1)(-1)^{n+1}q^n\equiv\,&amp;
2f(q^{10},q^{15})+2qf(q^5,q^{20})+2q^3\psi(q^{25})\\&amp;
-\psi(q^5)\sum\limits_{n=0}^{\infty}h(n)q^n\pmod{5},
\end{align*}
which implies that
\begin{align}\label{3pod3}
\nonumber\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(25n+16)(-1)^{n}q^n &amp;\equiv
2\psi(q^{5})-\psi(q)\sum\limits_{n=0}^{\infty}h(5n+3)q^n\pmod{5}\\&amp;
\equiv 2\psi(q^{5})-\psi(q)^5\pmod{5}\quad\text{using \eqref{mgpsi4}}.
\end{align}
Using \eqref{fp1}, \eqref{3pod3} can be expressed as
\begin{equation}\label{3pod4}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(25n+16)(-1)^{n}q^n \equiv
\psi(q^{5})\pmod{5}.
\end{equation}
Extracting the terms involving $q^{5n}$ from \eqref{3pod4},
\begin{align*}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(125n+16)(-1)^{n}q^n &amp;\equiv
\psi(q)\pmod{5}\\
&amp;\equiv f(q^{10},q^{15})+qf(q^5,q^{20})+q^3\psi(q^{25})\pmod{5},
\end{align*}
which yields
\begin{equation}\label{3pod5}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}(625n+391)(-1)^{n+1}q^n \equiv
\psi(q^{5})\pmod{5}.
\end{equation}
From \eqref{3pod4}, \eqref{3pod5}, and by induction, we find that for each $\alpha \geq 1$,
\begin{equation}\label{3pod6}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-3}\left(5^{2\alpha}n+\frac{5^{2\alpha+1}+3}{8}\right)(-1)^{n+1}q^n\equiv (-1)^{\alpha}\psi(q^5)\pmod{5}.
\end{equation}
Equating the coefficients of $q^{5n+\xi}$ for $1\leq \xi \leq4$ from \eqref{3pod6}, we arrive at \eqref{3podt2}.
\end{proof}

\begin{corollary}
The function $\textnormal{pod}_{-3}(n)$ is divisible by $5$ for at least $\frac{1}{30}$ of all nonnegative integers $n$.
\end{corollary}
\begin{proof}
The arithmetic sequences $\left\{5^{2\alpha+1}n+\frac{\mu\times 5^{2\alpha}+3}{8}:\alpha\geq 1\right\}$ for $\mu=13$, $21$, $29$, and $37$, on which $\textnormal{pod}_{-3}(\cdot)$ is $0$ modulo $5$, do not intersect. These sequences account for
$$4\left(\frac{1}{5^3}+\frac{1}{5^5}+\frac{1}{5^7}+\cdots\right)=\frac{1}{30}$$
of all nonnegative integers.
\end{proof}

\section{Arithmetic properties of \texorpdfstring{$\textnormal{pod}_{-4}(n)$}{Lg}}\label{4podsec1}
In this section, we prove the infinite family of congruences and internal congruences modulo 16 for $\textnormal{pod}_{-4}(n)$.

\subsection{Infinite family of congruences modulo 16}

\begin{theorem}\label{4podtm1}
Let  $p$ be any prime such that $p\equiv 3\pmod{4}$ and $\alpha\geq 0$. Then
\begin{equation}\label{4podt1}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}\left(2p^{2\alpha}n+\frac{p^{2\alpha}+1}{2}\right)q^n\equiv 4\psi(q)^2\pmod{16}
\end{equation}
and, for all nonnegative integers $n$  and $1\leq \xi \leq p-1$,
\begin{equation}\label{4podt2}
\textnormal{pod}_{-4}\left(2p^{2\alpha+1}(pn+\xi)+\frac{p^{2\alpha+2}+1}{2}\right)\equiv 0\pmod{16}.
\end{equation}
\end{theorem}
\begin{proof}
We have
\begin{equation}\label{defpod4}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(n)q^n=\frac{1}{\psi(-q)^4}.
\end{equation}
Invoking \eqref{2psi} and \eqref{defpod4},
\begin{align*}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(n)q^n&amp;=\frac{1}{\psi(q^2)^2\varphi(-q)^2}\\
&amp;=\frac{\left(1-(1-\varphi(-q)^2)\right)^{-1}}{\psi(q^2)^2}\\
&amp;=\frac{1+(1-\varphi(-q)^2)+(1-\varphi(-q)^2)^2+\cdots}{\psi(q^2)^2}\\
&amp;\equiv \frac{2-\varphi(-q)^2}{\psi(q^2)^2}\pmod{16}\quad\text{using \eqref{phi}}\\
&amp;\equiv \frac{2-\varphi(q^2)^2+4q\psi(q^4)^2}{\psi(q^2)^2}\pmod{16}\quad\text{from \eqref{2disphi2}},
\end{align*}
which implies that
\begin{align}\label{4pod4}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(2n+1)q^n\equiv 4\frac{\psi(q^2)^2}{\psi(q)^2}\pmod{16},
\end{align}
In view of \eqref{fp1}, \eqref{4pod4} can be expressed as
\begin{align}\label{4pod5}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(2n+1)q^n\equiv 4\psi(q)^2\pmod{16},
\end{align}
which is the $\alpha=0$ case of \eqref{4podt1}. If we assume that \eqref{4podt1} holds for some $\alpha\geq0$, then, substituting \eqref{ps3} into \eqref{4podt1},
\begin{align}
\nonumber&amp;\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}\left(2p^{2\alpha}n+\frac{p^{2\alpha}+1}{2}\right)q^n\\&amp;
\equiv4\Biggr(\sum\limits_{m=0}^{\frac{p-3}{2}}q^{\frac{m^2+m}{2}}f\left(q^{\frac{p^2+(2m+1)p}{2}},q^{\frac{p^2-(2m+1)p}{2}}\right)
+q^{\frac{p^2-1}{8}}\psi(q^{p^2})\Biggr)^2.\label{4pod6}
\end{align}
For any odd prime, $p$, and $0\leq m_1,m_2\leq(p-3)/2$, consider the congruence
$$\frac{m_1^2+m_1}{2}+\frac{m_2^2+m_2}{2}\equiv\frac{2p^2-2}{8}\pmod {p},$$
which implies that
\begin{equation}\label{4podJ4}
(2m_1+1)^2+(2m_2+1)^2\equiv 0\pmod{p}.
\end{equation}
Since $\left(\frac{-1}{p}\right)=-1$ for $p\equiv 3\pmod{4}$, the only solution of the congruence \eqref{4podJ4} is $m_1=m_2=\dfrac{p-1}{2}$.
Therefore, equating the coefficients of $q^{pn+\frac{2p^2-2}{8}}$ from both sides of \eqref{4pod6}, dividing throughout by $q^{\frac{2p^2-2}{8}}$ and then replacing $q^p$ by $q$, we obtain
\begin{equation}\label{4podJ5}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}\left(2p^{2\alpha}\left(pn+\frac{2p^2-2}{8}\right)+\frac{p^{2\alpha}+1}{2}\right)q^n\equiv 4\psi(q^p)^2\pmod{16}.
\end{equation}
Equating the coefficients of $q^{pn}$ on both sides of \eqref{4podJ5} and then replacing $q^p$ by $q$, we obtain
\begin{equation*}\label{4podJ6}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}\left(2p^{2\alpha+2}n+\frac{p^{2\alpha+2}+1}{2}\right)q^n\equiv 4\psi(q)^2\pmod{16},
\end{equation*}
which is the $\alpha+1$ case of \eqref{4podt1}.
	
Equating the coefficients of $q^{pn+\xi}$ for $1\leq \xi \leq p-1$ from \eqref{4podJ5}, we arrive at \eqref{4podt2}.
\end{proof}

\begin{corollary}
Let  $p$ be a prime such that $p\equiv 3\pmod{4}$. Then $\textnormal{pod}_{-4}(n)$ is divisible by 16 for at least $\frac{1}{2(p+1)}$ of all nonnegative integers $n$.
\end{corollary}
\begin{proof}
The arithmetic sequences $\left\{2p^{2\alpha+1}(pn+\xi)+\frac{p^{2\alpha+2}+1}{2}:\alpha\geq 0\right\}$ for $1\leq \xi \leq p-1$, on which $\textnormal{pod}_{-4}(\cdot)$ is $0$ modulo $16$, do not intersect. These sequences account for
$$(p-1)\left(\frac{1}{2p^2}+\frac{1}{2p^4}+\frac{1}{2p^6}+\cdots\right)=\frac{1}{2(p+1)}$$
of all nonnegative integers.
\end{proof}

\subsection{Some internal congruences}

\begin{theorem}\label{4podtm2}
For each $n\geq 0$,
\begin{align*}
\textnormal{pod}_{-4}(50n+3)&amp;\equiv 2\textnormal{pod}_{-4}(10n+1)\pmod{16},\\
\textnormal{pod}_{-4}(50n+23)&amp;\equiv 2\textnormal{pod}_{-4}(10n+5)\pmod{16},\\
\textnormal{pod}_{-4}(50n+33)&amp;\equiv 2\textnormal{pod}_{-4}(10n+7)\pmod{16},\\
\textnormal{pod}_{-4}(50n+43)&amp;\equiv 2\textnormal{pod}_{-4}(10n+9)\pmod{16}.
\end{align*}
\end{theorem}
\begin{proof}
If $\sum\limits_{n=0}^{\infty}g(n)q^n=\psi(q)^2$, then \eqref{4pod5} can be expressed as
\begin{equation*}\label{4pod7}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(2n+1)q^n\equiv 4\sum\limits_{n=0}^{\infty}g(n)q^n\pmod{16},
\end{equation*}
which yields
\begin{equation}\label{4pod8}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(10n+3)q^n\equiv 4\sum\limits_{n=0}^{\infty}g(5n+1)q^n\pmod{16}.
\end{equation}
Invoking \eqref{mgpsi1} and \eqref{4pod8},
\begin{equation}\label{4pod9}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(10n+3)q^n\equiv 8\psi(q)^2-4q\psi(q^5)^2\pmod{16}.
\end{equation}
Substituting \eqref{4pod5} into \eqref{4pod9},
\begin{equation*}\label{4pod10}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(10n+3)q^n\equiv 2\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-4}(2n+1)q^n-4q\psi(q^5)^2\pmod{16},
\end{equation*}
equating the coefficients of $q^{5n+i}$ for $i=0$, $2$, $3$, and $4$ from the above equation, we obtain the desired results.
\end{proof}

\section{Arithmetic properties of \texorpdfstring{$\textnormal{pod}_{-6}(n)$}{Lg}}\label{6podsec1}
In this section, we prove the infinite family of congruences and internal congruences modulo 5 for $\textnormal{pod}_{-6}(n)$.

\subsection{Infinite family of congruences modulo 5}

\begin{theorem}
Let  $p$ be any prime such that $p\equiv 3\pmod{4}$ and $\alpha\geq 0$. Then
\begin{equation*}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-6}\left(5p^{2\alpha}n+\frac{5p^{2\alpha}+3}{4}\right)q^n\equiv\psi(q)^2\pmod{5}
\end{equation*}
and, for each $n\geq 0$  and $1\leq \xi \leq p-1$,
\begin{equation*}
\textnormal{pod}_{-6}\left(5p^{2\alpha+1}(pn+\xi)+\frac{5p^{2\alpha+2}+3}{4}\right)\equiv 0\pmod{5}.
\end{equation*}
\end{theorem}
\begin{proof}
We have
\begin{equation}\label{defpod6}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-6}(n)q^n=\frac{1}{\psi(-q)^6}.
\end{equation}
In view of \eqref{fp1}, \eqref{defpod6} can be expressed as
\begin{equation}\label{6pod1}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-6}(n)(-1)^nq^n\equiv\frac{\psi(q)^4}{\psi(q^5)^2}\pmod{5}.
\end{equation}
Substituting \eqref{5psi} into \eqref{6pod1},
\begin{equation*}\label{6pod2}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-6}(n)(-1)^nq^n\equiv\frac{(f(q^{10},q^{15})+qf(q^5,q^{20})+q^3\psi(q^{25}))^4}{\psi(q^5)^2}\pmod{5},
\end{equation*}
which yields
\begin{align}
\nonumber\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-6}(5n+2)(-1)^nq^n\equiv\,&amp;\frac{f(q^2,q^3)^2f(q,q^4)^2}{\psi(q)^2}+\frac{2qf(q^2,q^3)f(q,q^4)\psi(q^5)^2}{\psi(q)^2}\\
&amp;+\frac{q^2\psi(q^5)^4}{\psi(q)^2}\pmod{5}.\label{6pod3}
\end{align}
Invoking \eqref{f1234} and \eqref{6pod3},
\begin{align}
\nonumber\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-6}(5n+2)(-1)^nq^n\equiv\,&amp;\psi(q)^2+q^2\frac{\psi(q^5)^4}{\psi(q)^2}-2q\psi(q^5)^2+2q\psi(q^5)^2\\&amp;
-2q^2\frac{\psi(q^5)^4}{\psi(q)^2}+q^2\frac{\psi(q^5)^4}{\psi(q)^2}\pmod{5},\label{6pod4}
\end{align}
which implies that
\begin{align}\label{6pod5}
\sum\limits_{n=0}^{\infty}\textnormal{pod}_{-6}(5n+2)(-1)^nq^n\equiv\psi(q)^2\pmod{5}.
\end{align}
The remainder of the proof is similar to that of Theorem \ref{4podtm1}, but rather than \eqref{4pod5}, we use \eqref{6pod5}.
\end{proof}

\begin{corollary}
Let  $p$ be a prime such that $p\equiv 3\pmod{4}$. Then $\textnormal{pod}_{-6}(n)$ is divisible by $5$ for at least $\frac{1}{5(p+1)}$ of all nonnegative integers $n$.
\end{corollary}
\begin{proof}
The arithmetic sequences $\left\{5p^{2\alpha+1}(pn+\xi)+\frac{5p^{2\alpha+2}+3}{4}:\alpha\geq 0\right\}$ for $1\leq \xi \leq p-1$, on which $\textnormal{pod}_{-6}(\cdot)$ is $0$ modulo $5$, do not intersect. These sequences account for
$$(p-1)\left(\frac{1}{5p^2}+\frac{1}{5p^4}+\frac{1}{5p^6}+\cdots\right)=\frac{1}{5(p+1)}$$
of all nonnegative integers.
\end{proof}

\subsection{Some internal congruences}

\begin{theorem}
For each $n\geq 0$,
\begin{align*}
\textnormal{pod}_{-6}(125n+7)&amp;\equiv 3\textnormal{pod}_{-6}(25n+2)\pmod{5},\\
\textnormal{pod}_{-6}(125n+57)&amp;\equiv 3\textnormal{pod}_{-6}(25n+12)\pmod{5},\\
\textnormal{pod}_{-6}(125n+82)&amp;\equiv 3\textnormal{pod}_{-6}(25n+17)\pmod{5},\\
\textnormal{pod}_{-6}(125n+107)&amp;\equiv 3\textnormal{pod}_{-6}(25n+22)\pmod{5}.
\end{align*}
\end{theorem}
\begin{proof}
The proof is similar to that of Theorem \ref{4podtm2}, but rather than \eqref{4pod5}, we use \eqref{6pod5}.
\end{proof}

\section{Acknowledgments}
The authors would like to thank anonymous referee of his/her helpful suggestions and comments.

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\end{thebibliography}



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

\noindent \emph{Keywords: } 
congruence, partition $k$-tuple, partition with odd parts distinct.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A006950},
\seqnum{A273225},
\seqnum{A273226},
\seqnum{A273228}, and
\seqnum{A273236}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  November 4 2015;
revised version received  April 18 2016; May 28 2016.
Published in {\it Journal of Integer Sequences}, June 4 2016.

\bigskip
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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