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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\Large\bf 
Congruences Modulo Small Powers of 2 and 3 for \\
\vskip .1in
Partitions into Odd Designated Summands
}
\vskip 1cm
\large
B. Hemanthkumar\\
Department of Mathematics\\
M. S. Ramaiah University of Applied Sciences\\
Bengaluru-560 058\\
India\\
\href{mailto:hemanthkumarb.30@gmail.com}{\tt hemanthkumarb.30@gmail.com}\\
\ \\
H. S. Sumanth Bharadwaj and M. S. Mahadeva Naika\footnote{Corresponding author.}\\
Department of Mathematics\\
Central College Campus\\
Bangalore University\\
Bengaluru-560 001\\
India\\
\href{mailto:sumanthbharadwaj@gmail.com}{\tt sumanthbharadwaj@gmail.com}\\
\href{mailto:msmnaika@rediffmail.com}{\tt msmnaika@rediffmail.com}
\end{center}

\vskip .2 in




\begin{abstract}
Andrews, Lewis and Lovejoy introduced a new class of partitions,
partitions with designated summands. Let $\PD(n)$ denote the number of
partitions of $n$ with designated summands and $\PDO(n)$ denote the
number of partitions of $n$ with designated summands in which all parts
are odd. Andrews et al.\ established many congruences modulo 3 for
$\PDO(n)$ by using the theory of modular forms. Baruah and Ojah
obtained numerous congruences modulo 3, 4, 8 and 16 for $\PDO(n)$ by
using theta function identities. In this paper, we prove several
infinite families of congruences modulo 9, 16 and 32 for $\PDO(n)$.
\end{abstract}

\section{Introduction}\label{S1}
A \textit{partition} of a positive integer $n$ is a non increasing sequence of positive integers $\lambda_1, \lambda_2,\ldots, \lambda_m$ such that $n=\lambda_1+\lambda_2+\cdots+\lambda_m$, where the $\lambda_i$'s $(i=1,2,\ldots,m)$ are called parts of the partition. For example, the partitions of $4$ are
\begin{equation*}
4,\quad 3+1,\quad 2+2,\quad 2+1+1,\quad 1+1+1+1.
\end{equation*}
Let $p(n)$ denote the number of partitions of $n$. Thus $p(4)=5$.

Andrews, Lewis and Lovejoy \cite{A} studied \textit{partitions with designated summands}, which are constructed by taking ordinary partitions and tagging exactly one of each part size. Thus the partitions of $4$ with designated summands are given by
\begin{align*}
&amp;\quad \quad 4',\quad 3'+1',\quad  2'+2,\quad 2+2',\quad 2'+1'+1,\quad 2'+1+1',\\
&amp; 1'+1+1+1,\quad 1+1'+1+1,\quad 1+1+1'+1,\quad 1+1+1+1'.
\end{align*}
Let $\PD(n)$ denote the number of partitions of $n$ with designated summands and $\PDO(n)$ denote the number of partitions of $n$ with designated summands in which all parts are odd. Thus $\PD(4)=10$ and $\PDO(4)=5$.

Recently, Chen et al.\ \cite{CJS} obtained the generating functions for $\PD(3n)$, $\PD(3n+1)$ and $\PD(3n+2)$ and gave a combinatorial interpretation of the congruence $\PD(3n+ 2) \equiv 0$ (mod $3)$. Xia \cite{X} proved infinite families of congruences modulo $9$ and $27$ for $\PD(n)$. For example, for all $n\geq 0$ and $k \geq 1$
\begin{equation*}
\PD(2^{18k-1}(12n + 1)) \equiv 0 \pmod{27}.
\end{equation*}
Throughout this paper, we use the notation
\begin{equation*}
f_k:=(q^k;q^k)_{\infty}\ \ (k=1,2,3,\ldots),\ \text{where}\ \ (a;q)_{\infty} := \prod_{m=0}^{\infty}(1-aq^{m}). \end{equation*}
 The generating function for $\PDO(n)$ satisfies
\begin{equation}
\sum_{n=0}^{\infty} \PDO(n) q^n = \frac{f_4f_6^2}{f_1f_3f_{12}}. \label{E1}
\end{equation}
 Using the theory of $q$-series and modular forms Andrews et al.\ \cite{A} derived
\begin{align}
\sum_{n=0}^{\infty} \PDO(3n) q^n &amp;= \frac{f_2^2f_6^4}{f_1^4f_{12}^2}, \label{E2}\\
\sum_{n=0}^{\infty} \PDO(3n+1) q^n &amp;= \frac{f_2^4f_3^3f_{12}}{f_1^5f_4f_6^2}, \label{E3}
\end{align} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(3n+2) q^n = 2 \frac{f_2^3f_6f_{12}}{f_1^4f_4}. \label{E4}
\end{equation}
They also established, for all $n \geq 0$
\begin{equation*}
\PDO(9n+6) \equiv 0 \pmod{3}
\end{equation*}and
\begin{equation*}
\PDO(12n+6) \equiv 0 \pmod{3}.
\end{equation*}

Baruah and Ojah \cite{BO} proved several congruences modulo 3, 4, 8 and 16 for $\PDO(n)$. For instance,
\begin{equation*}
\PDO(8n+7) \equiv 0 \pmod{8}
\end{equation*}
and
\begin{equation*}
\PDO(12n+9) \equiv 0 \pmod{16}.
\end{equation*}

The aim of this paper is to prove several new infinite families of congruences modulo $9$, $16$ and $32$ for $\PDO(n)$. In particular, we prove the following
\begin{theorem}\label{T0}
For all nonnegative integers $\alpha, \beta$ and $n$, we have
\begin{equation}
\PDO(2^{\alpha+2}3^{\beta}(72n+66)) \equiv 0 \pmod{144} \label{E543}
\end{equation} and
\begin{equation}
\PDO(2^{\alpha+2}3^{\beta}(144n+138)) \equiv 0 \pmod{288}. \label{E544}
\end{equation}
\end{theorem}

In Section \ref{S2}, we list some preliminary results. We prove several infinite families of congruences modulo $9$ for $\PDO(n)$ in Section \ref{S3}, and Theorem \ref{T0} and many infinite families of congruences modulo $16$ and $32$ for $\PDO(n)$ in Section \ref{S4}.

\section{Definitions and preliminaries}\label{S2}
We will make use of the following definitions, notation and results.

Let $f(a, b)$ be Ramanujan's general theta function \cite[p.\ 34]{B} given by
\begin{equation*}
f(a, b):= \sum_{n=-\infty}^\infty a^{\frac{n(n+1)}{2}} b^{\frac{n(n-1)}{2}}.
\end{equation*}
Jacobi's triple product identity can be stated in Ramanujan's notation as follows:
\begin{equation*}
f(a, b) = (-a; ab)_\infty (-b; ab)_\infty (ab; ab)_\infty.
\end{equation*}
In particular,
\begin{equation} \label{E1021}
\varphi(q):= f(q, q)=\sum_{k=-\infty}^{\infty} q^{k^2} =\frac{f_2^5}{f_1^2f_4^2}, \qquad \varphi(-q):= f(-q, -q)= \frac{f_1^2}{f_2},
\end{equation}
\begin{equation}\label{E1022}
\psi(q):= f(q, q^3) = \sum\limits_{k=0}^{\infty}q^{k(k+1)/2} = \frac{f_2^2}{f_1}
\end{equation} and
\begin{equation}\label{E102}
f(-q):= f(-q, -q^2) = \sum\limits_{k=-\infty}^{\infty}(-1)^k q^{k(3k-1)/2} = f_1.
\end{equation}
For any positive integer $k$, let $k(k+1)/2$ be the $k^{th}$ \textit{triangular number} and $k(3k\pm 1)/2$ be a generalized \textit{pentagonal number}.

\begin{lemma}\label{L1}
The following 2-dissections hold:
\begin{align}
f_1^2 &amp;= \frac{f_2f_8^5}{f_4^2f_{16}^2} - 2q \frac{f_2f_{16}^2}{f_8},\label{L11}\\
\frac{1}{f_1^2} &amp;= \frac{f_8^5}{f_2^5f_{16}^2} + 2q \frac{f_4^2f_{16}^2}{f_2^5f_8},\label{L12}\\
f_1^4 &amp;= \frac{f_4^{10}}{f_2^2f_8^4} - 4q \frac{f_2^2f_8^4}{f_4^2}\label{L13}
\end{align} and
\begin{equation}
\frac{1}{f_1^4} = \frac{f_4^{14}}{f_2^{14}f_8^4} + 4q \frac{f_4^2f_8^4}{f_2^{10}}.\label{L14}
\end{equation}
\end{lemma}
\begin{proof}
Lemma \ref{L1} is an immediate consequence of dissection formulas of Ramanujan, collected in Berndt's book \cite[Entry 25, p.\ 40]{B}.
\end{proof}

\begin{lemma}\label{L2}
The following 2-dissections hold:
\begin{align}
\frac{f_1^3}{f_3} &amp;= \frac{f_4^3}{f_{12}} - 3q \frac{f_2^2f_{12}^3}{f_4f_6^2},\label{L22}\\
\frac{f_3^3}{f_1} &amp;= \frac{f_4^3f_6^2}{f_2^2f_{12}} + q \frac{f_{12}^3}{f_4},\label{L23}\\
\frac{f_3}{f_1^3} &amp;= \frac{f_4^6f_6^3}{f_2^9f_{12}^2} + 3q \frac{f_4^2f_6f_{12}^2}{f_2^7} \label{L21}
\end{align} and
\begin{equation}
\frac{f_1}{f_3^3} = \frac{f_2f_4^2f_{12}^2}{f_6^7} - q \frac{f_2^3f_{12}^6}{f_4^2f_6^9}.\label{L24}
\end{equation}
\end{lemma}
\begin{proof}
Hirschhorn et al.\ \cite{HGB} established \eqref{L22} and \eqref{L23}. Replacing $q$ by $-q$ in \eqref{L22} and \eqref{L23}, and using the relation
\begin{equation*}
(-q;-q)_{\infty} = \frac{f_2^3}{f_1f_4},
\end{equation*}
we obtain \eqref{L21} and \eqref{L24}.
\end{proof}
\begin{lemma}\label{L4}
The following 2-dissections hold:
\begin{align}
f_1f_3 &amp;= \frac{f_2f_8^2f_{12}^4}{f_4^2f_6f_{24}^2} - q \frac{f_4^4f_6f_{24}^2}{f_2f_8^2f_{12}^2},\label{L34}\\
\frac{1}{f_1f_3} &amp;= \frac{f_8^2f_{12}^5}{f_2^2f_4f_6^4f_{24}^2} + q \frac{f_4^5f_{24}^2}{f_2^4f_6^2f_8^2f_{12}}\label{L33}
\end{align} and
\begin{equation}
\frac{f_3^2}{f_1^2} = \frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}} + 2q \frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}.\label{L31}
\end{equation}
\end{lemma}
\begin{proof}
Baruah and Ojah \cite{BO1} derived the above identities.
\end{proof}

\begin{lemma}\label{L6}
 The following 3-dissections hold:
\begin{equation}
\varphi(-q) = \varphi(-q^9) - 2q f(-q^3,-q^{15})\label{L42}
\end{equation} and
\begin{equation}
\psi(q) = f(q^3,q^6) + q \psi(q^9). \label{L41}
\end{equation}
\end{lemma}
\begin{proof}
See Berndt's book \cite[p.\ 49]{B} for a proof of \eqref{L42} and \eqref{L41}.
\end{proof}

\begin{lemma}\label{L7}
The following 3-dissection holds:
\begin{align}
f_1f_2 = \frac{f_6f_9^4}{f_3f_{18}^2} - q f_9f_{18} - 2q^2 \frac{f_3f_{18}^4}{f_6f_9^2}. \label{L51}
\end{align}
\end{lemma}
\begin{proof}
Hirschhorn and Sellers \cite{HS} have proved the above identity.
\end{proof}
Let $t$ be a positive integer. A partition of $n$ is called a $t$-\textit{core} partition of $n$ if none of the hook numbers of its associated Ferrers-Young diagram are multiples of $t$. Let $a_t(n)$ denote the number of $t$-\textit{core} partitions of $n$. Then the generating function of $a_t(n)$ satisfies
\begin{equation}\label{E5}
\sum\limits_{n=0}^{\infty} a_t(n) q^n = \frac{f_t^t}{f_1}.
\end{equation}
Many mathematicians have studied arithmetic properties of $a_3(n)$. See for example, Keith \cite{K}, and Lin and Wang \cite{LW}. Hirschhorn and Sellers \cite{HS1} obtained an explicit formula for $a_3(n)$ by using elementary methods and proved
\begin{lemma}\label{L01}
Let $3n+1=\prod\limits_{i=1}^{k} p_i^{\alpha_i}\prod\limits_{j=1}^{m} q_j^{\beta_j}$, where $p_i\equiv 1 \pmod{3}$ and $q_j\equiv 2 \pmod{3}$ with $\alpha_i, \beta_j\geq 0$ be the prime factorization of $3n+1$. Then
\begin{equation*}\label{E6}
a_3(n) = \begin{cases}
\prod\limits_{i=1}^{k}(\alpha_i+1), &amp; \text{if all } \beta_j \text{are even;}\\
0, &amp; \text{otherwise}.
\end{cases}
\end{equation*}
\end{lemma}

\section{Congruences modulo 9}\label{S3}
In this section, we prove the following infinite families of congruences modulo 9 for $\PDO(n)$.
\begin{theorem} \label{T1}
For all nonnegative integers $\alpha, \beta$ and $n$, we have
\begin{align}
\PDO(4^{\alpha}(24n+16)) &amp;\equiv \PDO(24n+16) \pmod{9} \label{E401},\\
\PDO(2^{\alpha}3^{\beta}(24n+24)) &amp; \equiv (-1)^{\alpha}\PDO(24n+24) \pmod{9} \label{E402},\\
\PDO(4^{\alpha}(48n+40)) &amp; \equiv 0 \pmod{9} \label{E428}
\end{align} and
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(144n+120)) \equiv 0 \pmod{9}. \label{E429}
\end{equation}
\end{theorem}
\begin{theorem}\label{T3}
For any nonnegative integer $n$, let $3n+1=\prod\limits_{i=1}^{k} p_i^{\alpha_i}\prod\limits_{j=1}^{m} q_j^{\beta_j}$, where $p_i\equiv 1 \pmod{3}$  and $q_j\equiv 2 \pmod{3}$ are primes with $\alpha_i, \beta_j\geq 0$. Then,
\begin{equation}\label{E301}
\PDO(48n+16) \equiv \begin{cases}
6\prod\limits_{i=1}^{k} (\alpha_i+1) \pmod{9}, &amp; \text{if all } \beta_j \text{ are even;}\\
0 \pmod{9}, &amp; \text{otherwise}.
\end{cases}
\end{equation} and
\begin{equation}\label{E302}
\PDO(72n+24) \equiv \begin{cases}
3\prod\limits_{i=1}^{k} (\alpha_i+1) \pmod{9}, &amp; \text{if all } \beta_j \text{ are even;}\\
0 \pmod{9}, &amp; \text{otherwise}.
\end{cases}
\end{equation}
\end{theorem}
\begin{corollary}\label{C31}
Let $p\equiv 2 \pmod{3}$ be a prime. Then for all nonnegative integers $\alpha$ and $n$ with $p \nmid n$, we have
\begin{equation}\label{E303}
\PDO(48p^{2\alpha+1}n+16p^{2\alpha+2}) \equiv 0 \pmod{9}
\end{equation} and
\begin{equation}\label{E304}
\PDO(72p^{2\alpha+1}n+24p^{2\alpha+2}) \equiv 0 \pmod{9}.
\end{equation}
\end{corollary}
\begin{theorem}\label{T34}
If $n$ cannot be represented as the sum of a triangular number and three times a triangular number, then
\begin{equation*} \PDO(48n+24) \equiv 0 \pmod{9}. \end{equation*}
\end{theorem}
\begin{corollary}\label{C34}
For any positive integer $k$, let $p_j\geq 5,\ 1\leq j\leq k$ be primes. If $(-3/p_j)=-1$ for every $j$, then for all nonnegative integers $n$ with $p_k\nmid n$ we have
\begin{equation}
\PDO(48 p_1^2p_2^2\cdots p_{k-1}^2p_kn+ 24 p_1^2p_2^2\cdots p_k^2) \equiv 0 \pmod{9}. \label{C341}
\end{equation}
\end{corollary}

By the binomial theorem, it is easy to see that for any positive integer $m$,
\begin{equation}
f_m^3 \equiv f_{3m} \pmod{3} \label{E403}
\end{equation} and
\begin{equation}
f_m^9 \equiv f_{3m}^3 \pmod{9}.\label{E404}
\end{equation}

\begin{proof} [Proof of Theorem \ref{T1}]
 From \eqref{E404}, it follows that
\begin{equation}
\frac{f_3^3}{f_1^5} \equiv f_1^4 \pmod{9}.\label{E405}
\end{equation}
In view of \eqref{E405}, we rewrite \eqref{E3} as
\begin{equation}
\sum_{n=0}^{\infty} \PDO(3n+1) q^n \equiv  \frac{f_1^4f_2^4f_{12}}{f_4f_6^2} \pmod{9}. \label{E406}
\end{equation}
Substituting \eqref{L13} in \eqref{E406} and extracting the terms containing  odd powers of $q$, we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(6n+4) q^n \equiv -4 \frac{f_1^6f_4^4f_6}{f_2^3f_3^2} \pmod{9}. \label{E408}
\end{equation}
Employing \eqref{L22} in \eqref{E408} and extracting the terms containing even powers of $q$,
we derive
\begin{align}
\sum_{n=0}^{\infty} \PDO(12n+4) q^n \equiv -4 \frac{f_2^{10}f_3}{f_1^3f_6^2} \pmod{9}. \label{E409}
\end{align}
Substituting \eqref{L21} in \eqref{E409} and extracting the terms containing odd powers of $q$, we get
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(24n+16) q^n \equiv -12 \frac{f_1^3f_2^2f_6^2}{f_3} \pmod{9}.
\end{equation*}
 From \eqref{E403},
\begin{equation*}
\frac{f_1^3f_2^2f_6^2}{f_3} \equiv \frac{f_6^3}{f_2} \pmod{3}.
\end{equation*}
In view of the above two identities,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+16) q^n \equiv 6 \frac{f_6^3}{f_2} \pmod{9},\label{E410}
\end{equation}
which implies that
\begin{equation}
\PDO(48n+40) \equiv 0 \pmod{9} \label{E412}
\end{equation}
for all $n \geq 0$ and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+16) q^n \equiv 6 \frac{f_3^3}{f_1} \pmod{9}. \label{E411}
\end{equation}
Invoking \eqref{L23} in \eqref{E411} and extracting the terms containing odd powers of $q$,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(96n+64) q^n \equiv 6 \frac{f_6^3}{f_2} \pmod{9}.\label{E413}
\end{equation}
By \eqref{E410} and \eqref{E413},
\begin{equation}
\PDO(96n+64) \equiv \PDO(24n+16) \pmod{9}. \label{E414}
\end{equation}
Congruence \eqref{E401} follows from \eqref{E414} and mathematical induction.
Congruence \eqref{E428} follows from \eqref{E412} and \eqref{E401}.

Employing \eqref{L14} in \eqref{E2},
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(3n) q^n = \frac{f_4^{14}f_6^4}{f_2^{12}f_8^4f_{12}^2} + 4q \frac{f_4^2f_6^4f_8^4}{f_2^8f_{12}^2},
\end{equation*}
which yields
\begin{equation}
\sum_{n=0}^{\infty} \PDO(6n) q^n = \frac{f_2^{14}f_3^4}{f_1^{12}f_4^4f_6^2}\label{E415}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(6n+3) q^n = 4 \frac{f_2^2f_3^4f_4^4}{f_1^8f_6^2}.\label{E416}
\end{equation}
Applying \eqref{L21} and \eqref{E415},
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(6n) q^n = \frac{f_2^{14}}{f_4^4f_6^2}\lb( \frac{f_4^6f_6^3}{f_2^9f_{12}^2} + 3q \frac{f_4^2f_6f_{12}^2}{f_2^7}\rb)^4,
\end{equation*}
which implies that
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(12n) q^n \equiv \frac{f_2^{20}f_3^{10}}{f_1^{22}f_6^8} \pmod{9}.
\end{equation*}
 From \eqref{E404},
\begin{equation*}
\frac{f_2^{20}f_3^{10}}{f_1^{22}f_6^8} \equiv \frac{f_2^2f_3^4}{f_1^4f_6^2} \pmod{9}.
\end{equation*}
In view of the above two identities,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n) q^n \equiv \frac{f_2^2f_3^4}{f_1^4f_6^2} \pmod{9}. \label{E418}
\end{equation}
Substituting \eqref{L23} and \eqref{L21} in \eqref{E418},
and extracting the terms containing even powers of $q$, we have
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n) q^n \equiv \frac{f_2^9f_3^3}{f_1^9f_6^3}+ 3q  \frac{f_2f_6^5}{f_1^5f_3} \pmod{9}. \label{E553}
\end{equation}
Using \eqref{E403} and \eqref{E404} in \eqref{E553}, we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+24) q^n \equiv 3  \frac{f_1f_2f_6^5}{f_3^3} \pmod{9}. \label{E419}
\end{equation}
Substituting \eqref{L24} in \eqref{E419} and using \eqref{E403}, we have
\begin{align*}
\sum\limits_{n=0}^{\infty} \PDO(24n+24) q^n &amp;\equiv 3 \frac{f_2^2f_4^2f_{12}^2}{f_6^2} - 3q \frac{f_2^4f_{12}^6}{f_4^2f_6^4}\\
&amp; \equiv 3 \frac{f_4^2f_{12}^2}{f_2f_6} - 3q \frac{f_2f_4f_{12}^5}{f_6^3} \pmod{9},
\end{align*}
which yields
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+24) q^n \equiv 3 \psi(q) \psi(q^3) \pmod{9} \label{E420}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+48) q^n \equiv -3  \frac{f_1f_2f_6^5}{f_3^3} \pmod{9}. \label{E421}
\end{equation}
By \eqref{E419} and \eqref{E421},
\begin{equation}
\PDO(2(24n+24)) \equiv -\PDO(24n+24) \pmod{9}. \label{E422}
\end{equation}
Employing \eqref{L51} in \eqref{E419} and using \eqref{E403},
\begin{align*}
\sum\limits_{n=0}^{\infty} \PDO(24n+24) q^n &amp; \equiv 3 \frac{f_6^6f_9^4}{f_3^4f_{18}^2} - 3q \frac{f_6^5f_9f_{18}}{f_3^3} - 6q^2 \frac{f_6^4f_{18}^4}{f_3^2f_9^2}\\
&amp; \equiv 3 \frac{f_9^3}{f_3} + 6q \frac{f_{18}^3}{f_6} + 3q^2 \frac{f_3f_6f_{18}^5}{f_9^3} \pmod{9}
\end{align*}
which implies that
\begin{equation}
\sum_{n=0}^{\infty} \PDO(72n+24) q^n \equiv 3 \frac{f_3^3}{f_1} \pmod{9}, \label{E423}
\end{equation}
\begin{equation}
\sum_{n=0}^{\infty} \PDO(72n+48) q^n \equiv 6 \frac{f_6^3}{f_2} \pmod{9} \label{E424}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(72n+72) q^n \equiv 3 \frac{f_1f_2f_6^5}{f_3^3} \pmod{9}. \label{E425}
\end{equation}
 From \eqref{E424},
\begin{equation}
\PDO(144n+120) \equiv 0 \pmod{9}. \label{E426}
\end{equation}
By \eqref{E419} and \eqref{E425},
\begin{equation}
\PDO(3(24n+24)) \equiv \PDO(24n+24) \pmod{9}. \label{E427}
\end{equation}
Congruence \eqref{E402} follows from \eqref{E422}, \eqref{E427},
and mathematical induction.
Congruence \eqref{E429} follows from \eqref{E426} and \eqref{E402}.
\end{proof}
\begin{proof}[Proof of Theorem \ref{T3}]
 From \eqref{E5}, \eqref{E411} and \eqref{E423}, it is clear that for all $n\geq 0$
\begin{equation}\label{E559}
\PDO(48n+16) \equiv 6 a_3(n) \pmod{9},
\end{equation} and
\begin{equation}\label{E560}
\PDO(72n+24) \equiv 3 a_3(n) \pmod{9}.
\end{equation}
Congruence \eqref{E301} follows from Lemma \ref{L01} and \eqref{E559}.  Congruence \eqref{E302} follows from Lemma \ref{L01} and \eqref{E560}.
\end{proof}
For any prime $p$ and any positive integer $N$, let $\upsilon_p(N)$ denote the exponent of the highest power of $p$ dividing $N$.
\begin{proof}[Proof of Corollary \ref{C31}]
 Suppose $\alpha\geq 0,\  p\equiv 2$ (mod $3)$ and $p \nmid n$, then it is clear that
\begin{equation}\label{E573}
\upsilon_p\lb(3\lb(p^{2\alpha+1}n+\frac{p^{2\alpha+2}-1}{3}\rb)+1\rb) = \upsilon_p\lb(3p^{2\alpha+1}n+p^{2\alpha+2}\rb) = 2\alpha+1.
\end{equation}
Congruences \eqref{E303} and \eqref{E304} follow from \eqref{E301}, \eqref{E302} and \eqref{E573}.
\end{proof}
\begin{proof}[Proof of Theorem \ref{T34}]
 From \eqref{E1022} and \eqref{E420}, we have
\begin{equation}\label{E561}
\sum\limits_{n=0}^{\infty} \PDO(48n+24) q^n \equiv 3 \sum\limits_{k=0}^{\infty}\sum\limits_{m=0}^{\infty} q^{k(k+1)/2+3m(m+1)/2} \pmod{9}.
\end{equation}
Theorem \ref{T34} follows from \eqref{E561}.
\end{proof}
\begin{proof}[Proof of Corollary \ref{C34}]
By \eqref{E561},
\begin{equation*}
\sum\limits_{n=0}^{\infty} \PDO(48n+24) q^{48n+24}  \equiv 3 \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} q^{6(2k+1)^2+2(6m+3)^2} \pmod{9}, \label{E}
\end{equation*}
which implies that if $48n+24$ is not of the form $6(2k+1)^2+2(6m+3)^2$, then $\PDO(48n+24)\equiv 0$ (mod $9)$. Let $k\geq 1$ be an integer and let $p_i \geq 5,\ 1\leq i \leq k$ be primes with $(\frac{-3}{p_i})=-1$. If $N$ is of the form $2x^2+6y^2$, then $v_{p_i}(N)$ is even since $(\frac{-3}{p_i})=-1$. Let
\begin{align*}
N &amp; = 48\lb(p_1^2p_2^2\cdots p_{k-1}^2p_kn+\frac{p_1^2p_2^2\cdots p_{k-1}^2p_k^2-1}{2}\rb) + 24\\
&amp; =48p_1^2p_2^2\cdots p_{k-1}^2p_kn+24p_1^2p_2^2\cdots p_{k-1}^2p_k^2.
\end{align*}
If $p_{k}\nmid n$, then $v_{p_k}(N)$ is an odd number and hence $N$ is not of the form $2x^2+6y^2$. Therefore \eqref{C341} holds.
\end{proof}

\section{Congruences modulo $2^4$ and $2^5$}\label{S4}
In this section, we establish the following infinite families of congruences modulo 16 and 32 for $\PDO(n)$.
\begin{theorem}\label{T4}
For all nonnegative integers $\alpha$ and $n$, we have
\begin{align}
\PDO(4^{\alpha}(12n+8)) &amp;\equiv \PDO(12n+8) \pmod{2^4}, \label{E434}\\
\PDO(4^{\alpha}(24n+23)) &amp;\equiv 0 \pmod{2^4}, \label{E435}\\
\PDO(4^{\alpha}(48n+14)) &amp;\equiv 0 \pmod{2^4} \label{E436}
\end{align} and
\begin{equation}
\PDO(24n+17) \equiv 0 \pmod{2^4}. \label{E437}
\end{equation}
\end{theorem}

\begin{theorem}\label{T5}
For all nonnegative integers $\alpha$ and $n$, we have
\begin{align}
\PDO(2^{\alpha}(12n)) &amp;\equiv \PDO(12n) \pmod{2^4}, \label{E464}\\
\PDO(2^{\alpha}(72n+42)) &amp;\equiv 0 \pmod{2^4} \label{E467}
\end{align} and
\begin{equation}
\PDO(2^{\alpha}(72n+66)) \equiv 0 \pmod{2^4}. \label{E468}
\end{equation}
\end{theorem}

\begin{theorem}\label{T7}
For all nonnegative integers $\alpha$ and $n$, we have
\begin{align}
\PDO(4^{\alpha}(24n) &amp;\equiv \PDO(24n) \pmod{2^5}, \label{E472}\\
\PDO(9^{\alpha}(6n+3)) &amp;\equiv \PDO(6n+3) \pmod{2^5}, \label{E473}\\
\PDO(24n+9) &amp;\equiv 0 \pmod{2^5}, \label{E477}\\
\PDO(9^{\alpha}(216n+117)) &amp;\equiv 0 \pmod{2^5}, \label{E478}\\
\PDO(2^{\alpha}(72n+69)) &amp;\equiv 0 \pmod{2^5}, \label{E474}\\
\PDO(3^{\alpha}(72n+69)) &amp;\equiv 0 \pmod{2^5} \label{E475}
\end{align} and
\begin{equation}
\PDO(2^{\alpha}(144n+42)) \equiv 0 \pmod{2^5}. \label{E479}
\end{equation}
\end{theorem}

\begin{theorem}\label{T6}
If n cannot be represented as the sum of two triangular numbers, then for all nonnegative integers $\alpha$ and $r\in\{1,6\}$ we have
\begin{equation*}
 \PDO(2^{\alpha}r(12n+3)) \equiv 0 \pmod{2^4}.
\end{equation*}
\end{theorem}

\begin{corollary}\label{C6}
If $p$ is a prime, $p\equiv 3\pmod 4)$, $1\leq j\leq p-1$ and $r\in \{1,6\}$, then for all nonnegative integers $\alpha, \beta$ and $n$, we have
\begin{equation}
\PDO(2^{\alpha}p^{2\beta+1}r(12pn+12j+3p))  \equiv 0 \pmod{2^4}. \label{E471}
\end{equation}
\end{corollary}

For example, taking $p=3$, we deduce that for all $\alpha,\beta,n \geq 0$,
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(216n+126)) \equiv 0 \pmod{2^4} \label{E548}
\end{equation} and
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(216n+198)) \equiv 0 \pmod{2^4}. \label{E549}
\end{equation}
Combining \eqref{E548} and \eqref{E467},
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(72n+42)) \equiv 0 \pmod{2^4}. \label{E550}
\end{equation}
Combining \eqref{E549} and \eqref{E468},
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(72n+66)) \equiv 0 \pmod{2^4}. \label{E551}
\end{equation}

\begin{theorem}\label{T8}
If n cannot be represented as the sum of a triangular number and four times a triangular number, then for all nonnegative integers $\alpha$ and $r\in\{1,3\}$ we have
\begin{equation*} \PDO(2^{\alpha}r(48n+30)) \equiv 0 \pmod{2^5}. \end{equation*}
\end{theorem}

\begin{corollary}\label{C8}
If $p$ is any prime with $p\equiv 3 \pmod 4)$, $1\leq j\leq p-1$ and $r\in \{1,3\}$, then for all nonnegative integers $\alpha, \beta$ and $n$, we have
\begin{equation}
\PDO(2^{\alpha}p^{2\beta+1}r(48pn+48j+30p)) \equiv 0 \pmod{2^5}. \label{E484}
\end{equation}
\end{corollary}

For example, taking $p=3$ we find that for all $\alpha,n \geq 0$ and $\beta \geq 1$,
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(144n+42)) \equiv 0 \pmod{2^5} \label{E554}
\end{equation} and
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(144n+138)) \equiv 0 \pmod{2^5}. \label{E555}
\end{equation}
Combining \eqref{E554} and \eqref{E479}, for all $\alpha, \beta, n \geq 0$,
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(144n+42)) \equiv 0 \pmod{2^5} \label{E556}
\end{equation}
and combining \eqref{E555}, \eqref{E474} and \eqref{E475}, for all $\alpha, \beta, n \geq 0$,
\begin{equation}
\PDO(2^{\alpha}3^{\beta}(72n+69)) \equiv 0 \pmod{2^5}. \label{E557}
\end{equation}

\begin{theorem}\label{T10}
If n cannot be represented as the sum of twice a pentagonal number and three times a triangular number, then for any nonnegative integer $\alpha$ we have
\begin{equation*}
\PDO(4^{\alpha}(24n+11)) \equiv 0 \pmod{2^4}.
\end{equation*}
\end{theorem}

\begin{corollary}\label{C10}
For any positive integer $k$, let $p_j\geq 5,\ 1\leq j\leq k$ be primes. If $(-2/p_j)=-1$ for every $j$, then for all nonnegative integers $\alpha$ and $n$ with $p_k\nmid n$ we have
\begin{equation*}
\PDO(6\cdot 4^{\alpha+1} p_1^2p_2^2\cdots p_{k-1}^2p_kn+ 11\cdot 4^{\alpha} p_1^2p_2^2\cdots p_k^2) \equiv 0 \pmod{2^4}.
\end{equation*}
\end{corollary}

\begin{theorem}\label{T11}
If n cannot be represented as the sum of a pentagonal number and six times a triangular number, then for any nonnegative integer $\alpha$ we have
\begin{equation*}
\PDO(4^{\alpha}(48n+38)) \equiv 0 \pmod{2^4}.
\end{equation*}
\end{theorem}

\begin{corollary}\label{C11}
For any positive integer $k$, let $p_j\geq 5,\ 1\leq j\leq k$ be primes. If $(-2/p_j)=-1$ for every $j$, then for all nonnegative integers $n$ with $p_k\nmid n$ we have
\begin{equation*}
\PDO(3\cdot 4^{\alpha+2} p_1^2p_2^2\cdots p_{k-1}^2p_kn+ 38\cdot 4^{\alpha} p_1^2p_2^2\cdots p_k^2) \equiv 0 \pmod{2^4}.
\end{equation*}
\end{corollary}

\begin{theorem}\label{T12}
If n cannot be represented as the sum of a pentagonal number and four times a pentagonal number, then we have
\begin{equation*}
\PDO(24n+5) \equiv 0 \pmod{2^5}.
\end{equation*}
\end{theorem}

\begin{corollary}\label{C12}
For any positive integer $k$, let $p_j\geq 5,\ 1\leq j\leq k$ be primes. If $(-1/p_j)=-1$ for every $j$, then for all nonnegative integers $n$ with $p_k\nmid n$ we have
\begin{equation*}
\PDO(24 p_1^2p_2^2\cdots p_{k-1}^2p_kn+ 5 p_1^2p_2^2\cdots p_k^2) \equiv 0 \pmod{2^5}.
\end{equation*}
\end{corollary}

\begin{theorem}\label{T13}
If n cannot be represented as the sum of a pentagonal number and sixteen times a pentagonal number, then we have
\begin{equation*}
\PDO(24n+17) \equiv 0 \pmod{2^5}.
\end{equation*}
\end{theorem}

\begin{corollary}\label{C13}
For any positive integer $k$, let $p_j\geq 5,\ 1\leq j\leq k$ be primes. If $(-1/p_j)=-1$ for every $j$, then for all nonnegative integers $n$ with $p_k\nmid n$ we have
\begin{equation*}
\PDO(24 p_1^2p_2^2\cdots p_{k-1}^2p_kn+ 17p_1^2p_2^2\cdots p_k^2) \equiv 0 \pmod{2^5}.
\end{equation*}
\end{corollary}

By the binomial theorem, it is easy to see that for all positive integers $k$ and $m$,
\begin{equation}
f_m^{2^k} \equiv f_{2m}^{2^{k-1}} \pmod{2^{k}}. \label{E438}
\end{equation}

\begin{proof} [Proof of Theorem \ref{T4}]
Using \eqref{L14}, we can rewrite \eqref{E4} as
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(3n+2) q^n = 2 \frac{f_4^{13}f_6f_{12}}{f_2^{11}f_8^4} + 8q \frac{f_4f_6f_8^4f_{12}}{f_2^7},
\end{equation*}
which yields
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(6n+2) q^n = 2 \frac{f_2^{13}f_3f_6}{f_1^{11}f_4^4} \label{E439}
\end{equation*} and
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(6n+5) q^n = 8  \frac{f_2f_3f_4^4f_6}{f_1^7}. \label{E440}
\end{equation*}
 From \eqref{E438} with $k=3$ and $k=2$, we have
\begin{equation*}
\frac{f_2^{13}f_3f_6}{f_1^{11}f_4^4} \equiv \frac{f_2f_3f_6}{f_1^3} \pmod{2^3} \label{E441}
\end{equation*} and
\begin{equation*}
\frac{f_2f_3f_4^4f_6}{f_1^7} \equiv \frac{f_3f_4^4f_6}{f_1^3f_2} \pmod{2^2}. \label{E442}
\end{equation*}
Thus,
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(6n+2) q^n \equiv 2  \frac{f_2f_3f_6}{f_1^3} \pmod{2^4}
\end{equation*} and
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(6n+5) q^n \equiv 8 \frac{f_3f_4^4f_6}{f_1^3f_2} \pmod{2^5}.
\end{equation*}
Substituting \eqref{L21} in the above two congruences and extracting the terms containing even and odd powers of $q$, we get
\begin{align}
\sum_{n=0}^{\infty} \PDO(12n+2) q^n &amp; \equiv 2 \frac{f_2^6f_3^4}{f_1^8f_6^2} \equiv 2 \frac{f_2^2f_3^4}{f_6^2} \pmod{2^4}, \label{E445}\\
\sum_{n=0}^{\infty} \PDO(12n+8) q^n &amp; \equiv 6 \frac{f_2^2f_3^2f_6^2}{f_1^6} \pmod{2^4}, \label{E446}\\
\sum_{n=0}^{\infty} \PDO(12n+5) q^n &amp; \equiv 8 \frac{f_2^{10}f_3^4}{f_1^{10}f_6^2} \equiv 8 \frac{f_2^6}{f_1^2} \pmod{2^5} \label{E447}
\end{align} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n+11) q^n  \equiv 24 \frac{f_2^6f_3^2f_6^2}{f_1^8} \equiv 24 f_4f_6^3 \pmod{2^4}. \label{E448}
\end{equation}
It follows from \eqref{E448} that
\begin{equation}
\sum\limits_{n=0}^{\infty} \PDO(24n+11) q^n  \equiv 8 f_2f_3^3 \equiv 8 f_2\frac{f_6^2}{f_3} \pmod{2^4} \label{E562}
\end{equation} and
\begin{equation}
\PDO(24n+23) \equiv 0 \pmod{2^4}. \label{E449}
\end{equation}
Employing \eqref{L13} in \eqref{E445} and extracting the terms containing odd powers of $q$,
\begin{equation*}
\sum_{n=0}^{\infty} \PDO(24n+14) q^n  \equiv -8q \frac{f_1^2f_{12}^4}{f_6^2} \equiv 8q f_2f_{12}^3 \pmod{2^4},
\end{equation*}
which implies that
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+38) q^n  \equiv 8 f_1f_{6}^3 \equiv 8 f_1 \frac{f_{12}^2}{f_6} \pmod{2^4} \label{E565}
\end{equation} and
\begin{equation}
\PDO(48n+14) \equiv 0 \pmod{2^4}. \label{E450}
\end{equation}
Substituting \eqref{L14} and \eqref{L31} in \eqref{E446},
and extracting the terms containing even and odd powers of $q$, we have
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+8) q^n  \equiv 6 \frac{f_2^{18}f_3^3f_6^2}{f_1^{17}f_4^5f_{12}} \equiv 6  \frac{f_2^2f_3^3f_6^2}{f_1f_4f_{12}} \pmod{2^4} \label{E451}
\end{equation} and
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(24n+20) q^n  &amp; \equiv 12 \frac{f_2^{15}f_3^4f_{12}}{f_1^{16}f_4^3f_6} + 24 \frac{f_2^6f_3^3f_4^3f_6^2}{f_1^{13}f_{12}}\\ &amp;\equiv 12 \frac{f_2^7f_6f_{12}}{f_4^3} + 24 \frac{f_3^3f_4^3}{f_1} \pmod{2^4}. \label{E452}
\end{align}
Substituting \eqref{L23} in \eqref{E451} and \eqref{E452}, and extracting even and odd powers of $q$, we have
\begin{align}
\sum_{n=0}^{\infty} \PDO(48n+8) q^n &amp; \equiv 6 \frac{f_2^2f_3^4}{f_6^2} \pmod{2^4}, \label{E453}\\
\sum_{n=0}^{\infty} \PDO(48n+32) q^n &amp; \equiv 6  \frac{f_1^2f_3^2f_6^2}{f_2^2} \pmod{2^4} \label{E454}
\end{align} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+44) q^n  \equiv 8 f_2^2 f_6^3 \equiv 8 \frac{f_4 f_{12}^2}{f_6} \pmod{2^4}. \label{E455}
\end{equation}
Again employing \eqref{L13} in \eqref{E453} and extracting the terms containing odd powers of $q$,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(96n+56) q^n \equiv 8q \frac{f_1^2f_{12}^4}{f_6^2} \equiv 8q  f_2f_{12}^3 \pmod{2^4}. \label{E456}
\end{equation}
By \eqref{E455} and \eqref{E456},
\begin{equation}
\PDO(96n+92) \equiv 0 \pmod{2^4} \label{E457}
\end{equation} and
\begin{equation}
\PDO(192n+56) \equiv 0 \pmod{2^4}. \label{E458}
\end{equation}
 From \eqref{E438}, \eqref{E446} and \eqref{E454},
\begin{equation}
\PDO(48n+32) \equiv \PDO(12n+8) \pmod{2^4}. \label{E459}
\end{equation}
Congruence \eqref{E434} follows from \eqref{E459} and mathematical
induction. Congruence \eqref{E435} follows from \eqref{E449},
\eqref{E457}, and \eqref{E434}. Similarly, congruence \eqref{E436}
follows from \eqref{E450}, \eqref{E458}, and \eqref{E434}.

By invoking \eqref{L12} in \eqref{E447} and extracting the terms containing even and odd powers of $q$,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+5) q^n  \equiv 8 \frac{f_1f_4^5}{f_8^2} \equiv 8 f_1f_4 \pmod{2^5} \label{E460}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+17) q^n  \equiv 16 \frac{f_1f_2^2f_8^2}{f_4} \equiv 16 f_1f_{16} \pmod{2^5}. \label{E461}
\end{equation}
Congruence \eqref{E437} follows from \eqref{E461}.
\end{proof}
\begin{proof}[Proofs of Theorem \ref{T5} and Theorem \ref{T7}]
 From \eqref{E438},
\begin{equation}
\frac{f_2^2f_3^4f_4^4}{f_1^8f_6^2}  \equiv \frac{f_3^4f_4^4}{f_2^2f_6^2} \pmod{2^3}. \label{E486}
\end{equation}
Using \eqref{E486}, we rewrite \eqref{E416} as
\begin{equation}
\sum_{n=0}^{\infty} \PDO(6n+3) q^n \equiv 4 \frac{f_3^4f_4^4}{f_2^2f_6^2} \pmod{2^5}. \label{E489}
\end{equation}
In view of \eqref{E1021} and \eqref{E1022},
\begin{equation}
\sum_{n=0}^{\infty} \PDO(6n+3) q^n \equiv 4 \psi^2(q^2) \phi^2(-q^3) \pmod{2^5}. \label{E492}
\end{equation}
Substituting \eqref{L41} in \eqref{E492} and extracting the terms involving $q^{3n+1}$,
we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(18n+9) q^n \equiv 4 q \psi^2(q^6) \phi^2(-q) \pmod{2^5}. \label{E493}
\end{equation}
Using \eqref{L42} in \eqref{E493} and extracting the terms involving $q^{3n+1}$,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(54n+27) q^n \equiv 4  \psi^2(q^2) \phi^2(-q^3) \pmod{2^5}. \label{E494}
\end{equation}
Congruence \eqref{E473} follows from \eqref{E492}, \eqref{E494},
and mathematical induction.

Applying \eqref{L13} in \eqref{E489} and using \eqref{E438},
\begin{align*}
\sum_{n=0}^{\infty} \PDO(6n+3) q^n &amp; \equiv 4 \frac{f_4^4f_{12}^{10}}{f_2^2f_6^4f_{24}^4} - 16q^3 \frac{f_4^4f_{24}^4}{f_2^2f_{12}^2}\\
&amp; \equiv 4 \frac{f_4^4f_{12}^2}{f_2^2f_6^4} + 16q^3 \frac{f_8^2f_{48}^2}{f_4f_{24}} \pmod{2^5}
\end{align*}
which implies that
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n+3) q^n \equiv 4 \frac{f_2^4f_6^2}{f_1^2f_3^4} \pmod{2^5} \label{E490}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n+9) q^n \equiv 16q \frac{f_4^2f_{24}^2}{f_2f_{12}} \pmod{2^5}. \label{E491}
\end{equation}
Congruence \eqref{E477} follows from \eqref{E491}.

 From \eqref{E491} and \eqref{E1022},
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+21) q^n \equiv 16 \psi(q) \psi(q^6) \pmod{2^5}. \label{E495}
\end{equation}
Invoking \eqref{L41} in \eqref{E495} and extracting the terms containing $q^{3n+2}$ and $q^{3n+1}$,
\begin{equation}
\PDO(72n+69) \equiv 0 \pmod{2^5} \label{E496}
\end{equation} for all $n \geq 0$ and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(72n+45) q^n \equiv 16 \psi(q^2) \psi(q^3) \pmod{2^5}. \label{E497}
\end{equation}
Using \eqref{L41} in \eqref{E497} and extracting the terms containing $q^{3n+1}$,
\begin{equation}
\PDO(216n+117) \equiv 0 \pmod{2^5}. \label{E498}
\end{equation}
Congruence \eqref{E478} follows from \eqref{E473} and \eqref{E498}.

Substituting \eqref{L12} and \eqref{L14} in \eqref{E490},
and extracting the terms containing odd powers of $q$, we get
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(24n+15) q^n  &amp; \equiv 8 \frac{f_2^2f_6^{14}f_8^2}{f_1f_3^{12}f_4f_{12}^4} + 16q \frac{f_4^5f_6^2f_{12}^4}{f_1f_3^8f_8^2} \\ &amp; \equiv  8 \psi(q)\psi(q^4) + 16q \psi(q)\psi(q^{12}) \pmod{2^5}. \label{E501}
\end{align}
Employing \eqref{L41} in \eqref{E501} and extracting the terms involving $q^{3n+2}$,
we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(72n+63) q^n  \equiv 16 \psi(q^3)\psi(q^4) + 8 q \psi(q^3)\psi(q^{12}) \pmod{2^5}. \label{E502}
\end{equation}
Again, extracting the terms involving $q^{3n+2}$ in \eqref{E502}, we get
\begin{equation}
\PDO(216n+207) \equiv 0 \pmod{2^5}. \label{E503}
\end{equation}
Congruence \eqref{E475} follows from \eqref{E496}, \eqref{E503}, and \eqref{E473}.

Substituting \eqref{L13} and \eqref{L14} in \eqref{E415},
and extracting the terms containing even powers of $q$, we find
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n) q^n \equiv  \frac{f_2^{38}f_6^{10}}{f_1^{28}f_3^{4}f_4^{12}f_{12}^4} \equiv \frac{f_1^4f_2^6f_{12}^4}{f_3^4f_4^4f_6^6} \pmod{2^4}. \label{E540}
\end{equation}
Using \eqref{L13} and \eqref{L14} in \eqref{E540},
and extracting the terms containing even powers of $q$, we have
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n) q^n \equiv  \frac{f_1^4f_2^6f_6^{18}}{f_3^{20}f_4^4f_{12}^4} \equiv \frac{f_1^4f_2^6f_{12}^4}{f_3^4f_4^4f_6^6} \pmod{2^4}. \label{E541}
\end{equation}
 From \eqref{E540} and \eqref{E541},
\begin{equation}
\PDO(2(12n)) \equiv \PDO(12n) \pmod{2^4}. \label{E542}
\end{equation}
Congruence \eqref{E464} follows from \eqref{E542} and mathematical induction.

Substituting \eqref{L31} and \eqref{L14} in \eqref{E415}, and using \eqref{E438},
we get
\begin{align*}
\sum_{n=0}^{\infty} \PDO(6n) q^n   = &amp;\  \frac{f_4^{32}f_{12}^4}{f_2^{24}f_8^{10}f_{24}^2} + 16q^2 \frac{f_4^8f_8^6f_{12}^4}{f_2^{16}f_{24}^2}+ 4q^2 \frac{f_4^{26}f_6^2f_{24}^2}{f_2^{22}f_8^6f_{12}^2} \\
&amp;\ +64q^4 \frac{f_4^2f_6^2f_8^{10}f_{24}^2}{f_2^{14}f_{12}^2} + 32q^2 \frac{f_4^{17}f_6f_{12}}{f_2^{19}} + 8q
\frac{f_4^{20}f_{12}^4}{f_2^{20}f_8^2f_{24}^2}\\
&amp;\ +32q^3 \frac{f_4^{14}f_6^2f_8^2f_{24}^2}{f_2^{18}f_{12}^2} +4q\frac{f_4^{29}f_6f_{12}}{f_2^{23}f_8^8} + 64q^3 \frac{f_4^5f_6f_8^8f_{12}}{f_2^{15}}\\
\equiv &amp; \ \frac{f_2^8f_4^{16}f_{12}^4}{f_8^{10}f_{24}^2} + 16q^2 f_8^2f_{16}^2 + 4q^2 \frac{f_4^2f_6^2f_8^2f_{24}^2}{f_2^6f_{12}^2}\\ &amp; \ + 8qf_4^2f_8^2 + 4qf_2f_4f_6f_{12} \pmod{2^5},
\end{align*}
which yields
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n) q^n   \equiv  \frac{f_1^8f_2^{16}f_6^4}{f_4^{10}f_{12}^2} + 16q f_4^2f_8^2 + 4q \frac{f_2^2f_3^2f_4^2f_{12}^2}{f_1^6f_6^2} \pmod{2^5} \label{E507}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n+6) q^n \equiv  8f_2^2f_4^2 + 4f_1f_2f_3f_6 \pmod{2^5}. \label{E508}
\end{equation}
Substituting \eqref{L34} in \eqref{E508} and extracting
the terms containing even and odd powers of $q$, we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+6) q^n \equiv 4 \frac{f_1^2f_4^2f_{6}^4}{f_2^2f_{12}^2} + 8 f_1^2f_2^2 \pmod{2^5} \label{E510}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+18) q^n \equiv -4 \frac{f_2^4f_3^2f_{12}^2}{f_4^2f_6^2}  \pmod{2^5}. \label{E509}
\end{equation}
By \eqref{E438} and \eqref{E1022},
\begin{equation*}
\frac{f_2^4f_3^2f_{12}^2}{f_4^2f_6^2}  \equiv \psi^2(q^3) \pmod{2^2}
\end{equation*} and
\begin{equation*}
4\frac{f_1^2f_4^2f_6^4}{f_2^2f_{12}^2}+8f_1^2f_2^2  \equiv -4 \psi^2(q) \pmod{2^4}.
\end{equation*}
In view of above identities,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+6) q^n \equiv -4 \psi^2(q) \pmod{2^4} \label{E512}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(24n+18) q^n \equiv -4 \psi^2(q^3) \pmod{2^4}. \label{E511}
\end{equation}
It follows from \eqref{E511} that
\begin{equation}
\PDO(72n+42) \equiv 0 \pmod{2^4} \label{E514}
\end{equation} and
\begin{equation}
\PDO(72n+66) \equiv 0 \pmod{2^4} \label{E515}
\end{equation}
for all $n\geq0$ and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(72n+18) q^n \equiv -4 \psi^2(q) \pmod{2^4}. \label{E513}
\end{equation}
Employing \eqref{L11} in \eqref{E510} and \eqref{E509} and extracting the terms containing odd powers of $q$, we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+30) q^n \equiv -8 \frac{f_2^2f_3^4f_8^2}{f_1f_4f_6^2}
 + 16 \frac{f_1^3f_8^2}{f_4} \equiv 8 \psi(q) \psi(q^{4}) \pmod{2^5} \label{E517}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+42) q^n \equiv 8q  \frac{f_1^4f_6^2f_{24}^2}{f_2^2f_3f_{12}}
 \equiv 8q \psi(q^3) \psi(q^{12}) \pmod{2^5}. \label{E516}
\end{equation}
In view of \eqref{E516}, we have
\begin{equation}
\PDO(144n+42) \equiv 0 \pmod{2^5} \label{E519}
\end{equation} and
\begin{equation}
\PDO(144n+138) \equiv 0 \pmod{2^5} \label{E520}
\end{equation}
for all $n\geq0$ and
\begin{equation}
\sum\limits_{n=0}^{\infty} \PDO(144n+90) q^n \equiv 8 \psi(q)\psi(q^4) \pmod{2^5}. \label{E518}
\end{equation}
Substituting \eqref{L13}, \eqref{L14}, and \eqref{L31} in \eqref{E507},
and extracting the terms containing even and odd powers of $q$, we get
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(24n) q^n  \equiv &amp;  \frac{f_1^{12}f_2^{10}f_3^4}{f_4^8f_6^2}+ 16q\frac{f_1^{20}f_3^4f_4^8}{f_2^{14}f_6^2} + 16q\frac{f_2^8f_4^3f_6^4}{f_1^{13}f_3f_{12}}+ 8q \frac{f_2^{17}f_6f_{12}}{f_1^{16}f_4^3}\\
\equiv &amp; \frac{f_3^4f_4^8}{f_1^{20}f_2^6f_6^2}+16qf_4^6
+16q\frac{f_1^3f_4^3f_{12}}{f_3}+8qf_2f_4f_6f_{12} \pmod{2^5} \label{E522}
\end{align} and
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(24n+12) q^n &amp; \equiv  -8 \frac{f_1^{16}f_3^4}{f_2^2f_6^2}+ 4\frac{f_2^{20}f_6^4}{f_1^{17}f_3f_4^5f_{12}} + 16 f_2^2f_4^2\\
&amp; \equiv 4 \frac{f_2^4f_6^4}{f_1f_3f_4f_{12}}+8f_2^2f_4^2 \pmod{2^5}. \label{E521}
\end{align}
Employing \eqref{L33} in \eqref{E521} and extracting the terms involving even and odd powers of $q$, we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+12) q^n  \equiv 4 \frac{f_1^2f_4^2f_6^4}{f_2^2f_{12}^2} + 8f_1^2f_2^2 \pmod{2^5} \label{E523}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(48n+36) q^n  \equiv 4 \frac{f_2^4f_3^2f_{12}^2}{f_4^2f_6^2} \pmod{2^5}. \label{E524}
\end{equation}
In view of \eqref{E510}, \eqref{E509}, \eqref{E523}, and \eqref{E524}, we have
\begin{equation}
\PDO(2(24n+6)) \equiv \PDO(24n+6) \pmod{2^5} \label{E525}
\end{equation} and
\begin{equation}
\PDO(2(24n+18)) \equiv - \PDO(24n+18) \pmod{2^5}. \label{E526}
\end{equation}
Congruence \eqref{E467} follows from \eqref{E514}, \eqref{E526}, and
\eqref{E464}. Congruence \eqref{E468} follows from \eqref{E515},
\eqref{E526}, and \eqref{E464}.

Substituting \eqref{L14}, \eqref{L22}, and \eqref{L31} in \eqref{E522},
and extracting even and odd powers of $q$, we get
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(48n) q^n  &amp; \equiv  \frac{f_2^{72}f_6^4}{f_1^{72}f_4^{18}f_{12}^2} +4 q \frac{f_2^{66}f_3^2f_{12}^2}{f_1^{70}f_4^{14}f_6^2} -48q \frac{f_1^2f_2^2f_6^4}{f_3^2}\\
&amp;\equiv \frac{f_2^8f_6^4}{f_1^8f_4^2f_{12}^2} +4q \frac{f_2^2f_3^2f_4^2f_{12}^2}{f_1^6f_6^2} + 16q f_2f_4f_6f_{12} \pmod{2^5} \label{E527}
\end{align} and
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(48n+24) q^n  &amp; \equiv 4 \frac{f_2^{69}f_3f_6}{f_1^{71}f_4^{16}}+16\frac{f_2^{60}f_6^{4}}{f_1^{68}f_4^{10}f_{12}^2} + 8 f_1f_2f_3f_6\\
 &amp;\equiv 12 f_1f_2f_3f_6 + 16f_2^2f_4^2 \pmod{2^5}. \label{E528}
\end{align}
Substituting \eqref{L34} in \eqref{E528} and extracting the terms containing even and odd powers of $q$, we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(96n+24) q^n  \equiv -4 \frac{f_1^2f_4^2f_6^4}{f_2^2f_{12}^2} + 16f_1^2f_2^2 \pmod{2^5} \label{E530}
\end{equation} and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(96n+72) q^n  \equiv -12 \frac{f_2^4f_3^2f_{12}^2}{f_4^2f_6^4} \pmod{2^5}. \label{E529}
\end{equation}
From \eqref{E509} and \eqref{E529}, we have
\begin{equation}
\PDO(2^2(24n+18)) \equiv 3 \PDO(24n+18) \pmod{2^5}. \label{E531}
\end{equation}
Employing \eqref{L11} in \eqref{E530} and extracting the terms containing odd powers of $q$,
we get
\begin{equation}
\sum_{n=0}^{\infty} \PDO(192n+120) q^n  \equiv 8 \frac{f_2^2f_3^4f_8^2}{f_1f_4f_6^2} \equiv  8\psi(q) \psi(q^4) \pmod{2^5}. \label{E532}
\end{equation}
Substituting \eqref{L14} and \eqref{L31} in \eqref{E527},
and extracting the terms containing even and odd power of $q$, we have
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(96n) q^n
&amp;\equiv  \frac{f_2^{26}f_3^4}{f_1^{20}f_4^8f_6^2} + 16q \frac{f_2^2f_3^4f_4^8}{f_1^{12}f_6^2} +16q\frac{f_2^8f_4^3f_6^4}{f_1^{13}f_3f_{12}}+8q \frac{f_2^{17}f_6f_{12}}{f_1^{16}f_4^3}\\
&amp;\equiv  \frac{f_3^4f_4^8}{f_1^{20}f_2^6f_6^2}+16qf_4^6+ 16q\frac{f_1^3f_4^3f_{12}}{f_3} + 8qf_2f_4f_6f_{12} \pmod{2^5} \label{E534}
\end{align} and
\begin{align}
\nonumber \sum_{n=0}^{\infty} \PDO(96n+48) q^n  &amp; \equiv 8 \frac{f_3^4f_2^{14}}{f_1^{16}f_6^2} + 4 \frac{f_2^{20}f_6^4}{f_1^{17}f_3f_4^5f_{12}} +16f_1f_2f_3f_6\\ &amp;\equiv  8f_2^2f_4^2+4 \frac{f_2^4f_6^4}{f_1f_3f_4f_{12}}+16f_1f_2f_3f_6 \pmod{2^5}. \label{E533}
\end{align}
By \eqref{E534} and \eqref{E522}
\begin{equation}
\PDO(4(24n)) \equiv \PDO(24n) \pmod{2^5}. \label{E535}
\end{equation}
Congruence \eqref{E472} follows from \eqref{E535} and mathematical induction.

By substituting \eqref{L33} and \eqref{L34} in \eqref{E533}, and
extracting the terms containing even and odd powers of $q$,
\begin{equation}
\sum_{n=0}^{\infty} \PDO(192n+48) q^n  \equiv  8f_1^2f_2^2+ 20 \frac{f_1^2f_4^2f_6^4}{f_2^2f_{12}^2} \pmod{2^5} \label{E536}
\end{equation}
and
\begin{equation}
\sum_{n=0}^{\infty} \PDO(192n+144) q^n  \equiv  - 12 \frac{f_2^4f_3^2f_{12}^2}{f_4^2f_6^2} \pmod{2^5}. \label{E537}
\end{equation}
 From \eqref{E510} and \eqref{E536},
\begin{equation}
\PDO(2^3(24n+6)) \equiv -3 \PDO(24n+6) \pmod{2^5}. \label{E538}
\end{equation}
In view of \eqref{E509} and \eqref{E537},
\begin{equation}
\PDO(2^3(24n+18)) \equiv 3 \PDO(24n+18) \pmod{2^5}. \label{E539}
\end{equation}
Congruence \eqref{E474} follows from \eqref{E496}, \eqref{E520},
\eqref{E526}, \eqref{E531}, \eqref{E539}, and \eqref{E472}. Congruence
\eqref{E479} follows from \eqref{E519}, \eqref{E526}, \eqref{E531},
\eqref{E539}, and \eqref{E472}.

\end{proof}
\begin{proof}[Proof of Theorem \ref{T6}]
Using \eqref{E438} and \eqref{E1022} in \eqref{E490},
\begin{equation}
\sum_{n=0}^{\infty} \PDO(12n+3) q^n \equiv 4 \psi^2(q) \pmod{2^4}. \label{E499}
\end{equation}
 From \eqref{E499}, \eqref{E512}, \eqref{E525} and \eqref{E464},
\begin{equation}
\sum_{n=0}^{\infty}\PDO(2^{\alpha}(24n+6)) q^n \equiv  -4 \psi^2(q) \pmod{2^4} \label{E470}
\end{equation}
 From \eqref{E513}, \eqref{E525} and \eqref{E464},
\begin{equation}
\sum_{n=0}^{\infty}\PDO(2^{\alpha}(72n+18)) q^n \equiv \begin{cases}
-4 \psi^2(q) \pmod{2^4}, &amp; \text{if } \alpha =0; \\
 4 \psi^2(q) \pmod{2^4}, &amp; \text{if } \alpha \neq 0. \end{cases} \label{E465}
\end{equation}
Combining \eqref{E499}, \eqref{E470}, and \eqref{E465},
\begin{equation}
\sum\limits_{n=0}^{\infty} \PDO(2^{\alpha}r(12n+3)) q^n \equiv  \begin{cases}
(-1)^{r+1}4  \sum\limits_{k,m=0}^{\infty} q^{k(k+1)/2+m(m+1)/2} \pmod{2^4}, &amp; \textrm{if } \alpha =0; \\
(-1)^{r} 4 \sum\limits_{k,m=0}^{\infty} q^{k(k+1)/2+m(m+1)/2} \pmod{2^4}, &amp; \textrm{if } \alpha \neq 0. \end{cases}\label{E574}
\end{equation}
Theorem \ref{T6} follows from \eqref{E574}.
\end{proof}
\begin{proof} [Proof of Theorem \ref{T8}]
 From \eqref{E517}, \eqref{E525}, \eqref{E532}, \eqref{E538} and \eqref{E472}, we have
\begin{equation}
\sum_{n=0}^{\infty} \PDO(2^{\alpha} (48n+30)) q^n \equiv 8 \psi(q) \psi(q^4) \pmod{2^5}. \label{E482}
\end{equation}
 From \eqref{E518}, \eqref{E526}, \eqref{E531}, \eqref{E539} and \eqref{E472}, we have
\begin{equation}
\sum_{n=0}^{\infty} \PDO(2^{\alpha} (144n+90)) q^n \equiv  \begin{cases}
8 \psi(q) \psi(q^4) \pmod{2^5}, &amp; \text{if } \alpha = 0; \\
-8 \psi(q) \psi(q^4)\pmod{2^5}, &amp;  \text{if } \alpha \neq 0. \end{cases} \label{E483}
\end{equation}
Theorem \ref{T8} follows from \eqref{E482} and \eqref{E483}.
\end{proof}
\begin{proof}[Proofs of Theorems \ref{T10}, \ref{T11}, \ref{T12} and \ref{T13}]
From \eqref{E455},
\begin{equation}
\sum\limits_{n=0}^{\infty} \PDO (96n+44) q^n \equiv 8 f_2 \frac{f_6^2}{f_3} \pmod{2^4}. \label{E485}
\end{equation}
Replacing $n$ by $8n+3$ in \eqref{E434}, we see that for all $\alpha, n \geq 0$,
\begin{equation}
\PDO(4^{\alpha}(96n+44)) \equiv \PDO(96n+44) \pmod{2^4}. \label{E487}
\end{equation}
In view of \eqref{E1022}, \eqref{E562}, \eqref{E485} and \eqref{E487},
\begin{equation}
\sum\limits_{n=0}^{\infty} \PDO (4^{\alpha}(24n+11) q^n \equiv 8 f_2 \psi(q^3) \pmod{2^4}. \label{E488}
\end{equation}
Theorem \ref{T10} follows from  \eqref{E488}. From \eqref{E456},
\begin{equation}
\sum\limits_{n=0}^{\infty} \PDO (192n+152) q^n \equiv 8 f_1 \frac{f_{12}^2}{f_6} \pmod{2^4}. \label{E566}
\end{equation}
Replacing $n$ by $16n+12$ in \eqref{E434}, we see that for all $\alpha, n \geq 0$,
\begin{equation}
\PDO(4^{\alpha}(192n+152)) \equiv \PDO(192n+152) \pmod{2^4}. \label{E567}
\end{equation}
In view of \eqref{E1022}, \eqref{E565}, \eqref{E566} and \eqref{E567},
\begin{equation}
\sum\limits_{n=0}^{\infty} \PDO (4^{\alpha}(48n+38) q^n \equiv 8 f_1 \psi(q^6) \pmod{2^4}. \label{E568}
\end{equation}
Theorem \ref{T11} follows from \eqref{E568}. Theorem \ref{T12} follows from \eqref{E102} and \eqref{E460}. Theorem \ref{T13} follows from \eqref{E102} and \eqref{E461}.
\end{proof}

The proofs of Corollaries \ref{C6}, \ref{C8}, \ref{C10}, \ref{C11},
\ref{C12} and \ref{C13} are similar to the proof of Corollary
\ref{C34}; hence we omit the details.

\begin{proof}[Proof of Theorem \ref{T0}]
Replacing $n$ by $2n+1$ in \eqref{E429},
\begin{equation}
\PDO(2^{\alpha+2}3^{\beta}(72n+66)) \equiv 0 \pmod{9}. \label{E552}
\end{equation}
Congruence \eqref{E543} follows readily from \eqref{E551} and \eqref{E552}.
Replacing $n$ by $4n+3$ in \eqref{E429},
\begin{equation}
\PDO(2^{\alpha+2}3^{\beta}(144n+138)) \equiv 0 \pmod{9}. \label{E558}
\end{equation}
Congruence \eqref{E544} follows readily from \eqref{E555} and \eqref{E558}.
\end{proof}


\section{Acknowledgments}
The authors would like to thank the anonymous referee for his/her
valuable comments and suggestions. The second author was supported by CSIR
Senior Research Fellowship (No.\ 09/039(0111)/2014-EMR-I).

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\end{thebibliography}

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


\noindent \emph{Keywords: }
partition with designated summand, congruence, theta function.

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\noindent (Concerned with sequences
\seqnum{A077285} and
\seqnum{A102186}.)

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\vspace*{+.1in}
\noindent
Received October 25 2016;
revised version received January 13 2017; January 23 2017. 
Published in {\it Journal of Integer Sequences}, January 28 2017.

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