\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amscd}
\usepackage{graphicx}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics}
\usepackage{latexsym}
\usepackage{epsf}
\usepackage{breakurl}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \underline{#1}}}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\begin{center}
\vskip 1cm{\LARGE\bf 
On 3- and 9-Regular Cubic Partitions
}
\vskip 1cm
\large
D. S. Gireesh and C. Shivashankar \\
Department of Mathematics\\
M. S. Ramaiah University of Applied Sciences\\
Bengaluru-560 058\\
Karnataka\\
India\\
\href{mailto:gireeshdap@gmail.com}{\tt gireeshdap@gmail.com}\\
\href{mailto:shankars224@gmail.com}{\tt shankars224@gmail.com}\\
\ \\
M. S. Mahadeva Naika \\
Department of Mathematics\\
Bengaluru Central University\\
Central College Campus\\ 
Bengaluru-560 001 \\
Karnataka\\
India\\
\href{mailto:msmnaika@rediffmail.com}{\tt msmnaika@rediffmail.com}
\end{center}

\vskip .2 in

\begin{abstract}
Let $a_3(n)$ and $a_9(n)$ be 3- and 9-regular cubic partitions of $n$. In this paper, we establish several infinite families of congruences modulo powers of 3. For example, for all non-negative integers $n$ and $\alpha$ we
find that
\[a_3\left (3^{2\alpha}n+\frac{3^{2\alpha}-1}{4}\right )\equiv 0 \pmod{3^{\alpha}}\] and
\[a_9\left (3^{\alpha+1}n+3^{\alpha+1}-1\right )\equiv 0 \pmod{3^{\alpha+1}}.\]
\end{abstract}

\section{Introduction}
\label{intro}
A partition of a positive integer $n$ is a non-increasing sequence of positive integers whose sum is $n$. Let $p(n)$ denote the number of partitions of $n$ and the generating function is
\[\sum\limits_{n\geq0}p(n)q^n=\frac1{f_1},\]
where, here and throughout the paper, we set
\[f_k=(q^k;q^k)_\infty=\prod\limits_{m=1}^{\infty}(1-q^{km}).\]
Chan \cite{HCC} studied the cubic partition function denoted by $a(n)$, whose generating function is
\[\sum\limits_{n\geq0}a(n)q^n=\frac1{f_1f_2}.\]
He found the generating function
\[\sum\limits_{n\geq0}a(3n+2)q^n=3\frac{f_3^3f_6^3}{f_1^4f_2^4}, \]
which readily implies that
\[a(3n+2)\equiv 0\pmod{3}.\]
Chan \cite{HCC1} established infinite family of congruence modulo powers of 3 for $a(n)$. For each $n, k\geq1$, he proved that
\begin{equation}
a(3^kn+c_k)\equiv0\pmod{3^{k+\delta(k)}},
\end{equation}
where $c_k$ is the reciprocal modulo $3^k$ of 8 and
\begin{displaymath}
\delta(k):= \begin{cases}
1, & \text{if $k$ is even;}\\
0, & \text{$k$ is odd.}
\end{cases}
\end{displaymath}

Zhao and Zhong \cite{ZZ} studied cubic partition pairs denoted by $b(n)$, the generating function satisfied by $b(n)$ is
\begin{equation}\label{deff122}
\sum\limits_{n\geq 0}b(n)q^n=\frac1{f_1^2f_2^2}.
\end{equation}
For each $n\geq 0$, they found Ramanujan's type congruences for $b(n)$, namely
\begin{align*}
b(5n+4)&\equiv 0\pmod{5},\\
b(7n+i)&\equiv 0\pmod{7},\\
b(9n+7)&\equiv 0\pmod{9},
\end{align*}
where $i\in\{2, 3, 4, 6\}$.

Lin \cite{Lin} studied the cubic partition pairs and established  the following congruences modulo 27:
\begin{align}
b(27n+16)&\equiv 0\pmod{27},\label{LC1}\\
b(27n+25)&\equiv 0\pmod{27},\label{LC2}\\
b(81n+61)&\equiv 0\pmod{27}.\label{LC3}
\end{align}
Also, he proposed the following conjectures:
\begin{conjecture} For each $n\geq 0$,
\begin{equation}\label{8161}
b(81n+61)\equiv 0\pmod{81}.
\end{equation}
\end{conjecture}

\begin{conjecture}
\begin{align}
\sum\limits_{n\geq 0}b(81n+7)q^n&\equiv 9\frac{f_2f_3^2}{f_6}\pmod{81},\\
\sum\limits_{n\geq 0}b(81n+34)q^n&\equiv 36\frac{f_1f_6^2}{f_3}\pmod{81}.
\end{align}
\end{conjecture}
The above conjectures were proved by Gireesh and Naika \cite{GM}, Chern  \cite{Chern}, and Lin et al.\ \cite{LWX}.

Motivated by the above results, in this paper, we study 3- and 9-regular cubic partitions, which are defined as follow:

\begin{itemize}
\item
Let $a_3(n)$ denote the number of 3-regular cubic partitions of $n$, whose generating function is
\begin{equation}\label{defa3}
\sum\limits_{n\geq 0}a_3(n)q^n=\frac{f_3f_6}{f_1f_2}.
\end{equation}

\item
Let $a_9(n)$ denote the number of 9-regular cubic partitions of $n$, whose generating function is
\begin{equation}\label{defa9}
\sum\limits_{n\geq0}a_9(n)q^n=\frac{f_9f_{18}}{f_1f_2}.
\end{equation}
\end{itemize}

We shall show that
\begin{equation}\label{3p31}
\sum\limits_{n\geq 0}a_3(3n+2)q^n=3\frac{f_3^3f_6^3}{f_1^3f_2^3}
\end{equation}
and
\begin{equation}\label{3a32}
\sum\limits_{n\geq 0}a_9(3n+2)q^n=3\frac{f_3^4f_6^4}{f_1^4f_2^4}.
\end{equation}
These are analogous to Ramanujan's most beautiful identities \cite[pp.~239, 243]{RAMAU}
\begin{equation}\label{Rama54}
\sum\limits_{n\geq 0}p(5n+4)q^n=5\frac{f_5^5}{f_1^6}
\end{equation}
and
\begin{equation}\label{Rama75}
\sum\limits_{n\geq 0}p(7n+5)q^n=7\frac{f_7^3}{f_1^4}+49q\frac{f_7^7}{f_1^8}.
\end{equation}
We also obtain infinite families of congruences modulo powers of 3 for $a_3(n)$ and $a_9(n)$, which are stated in the following theorems:
\begin{theorem}\label{th1}
For each $n,\alpha \geq0$,
\begin{align}
a_3\left (3^{2\alpha}n+\frac{3^{2\alpha}-1}{4}\right )&\equiv 0 \pmod{3^{\alpha}},\label{3ac1}\\
a_3\left (3^{2\alpha+1}n+\frac{3^{2\alpha+2}-1}{4}\right )&\equiv 0 \pmod{3^{\alpha+1}},\label{3ac2}\\
a_3\left (3^{2\alpha+2}n+\frac{7\times 3^{2\alpha+1}-1}{4}\right )&\equiv 0 \pmod{3^{\alpha+2}},\label{3ac3}\\
a_3\left (3^{2\alpha+2}n+\frac{11\times 3^{2\alpha+1}-1}{4}\right )&\equiv 0 \pmod{3^{\alpha+2}}.\label{3ac4}
\end{align}
\end{theorem}

\begin{theorem}\label{th2}
For each $n,\alpha \geq0$,
\begin{equation}\label{9ac}
a_9\left (3^{\alpha+1}n+3^{\alpha+1}-1\right )\equiv 0 \pmod{3^{\alpha+1}}.
\end{equation}
\end{theorem}

The results \eqref{3ac1}--\eqref{9ac} are  analogous to Ramanujan's congruences modulo powers of 5 \cite{Hir}, for $n,\alpha\geq0$,
\begin{equation}
p\left (5^{2\alpha+1}n+\frac{19\times 5^{2\alpha+1}+1}{24}\right )\equiv 0 \pmod{5^{2\alpha+1}}
\end{equation}
and
\begin{equation}
p\left (5^{2\alpha+2}n+\frac{23\times 5^{2\alpha+2}+1}{24}\right )\equiv 0 \pmod{5^{2\alpha+2}}.
\end{equation}


\section{Preliminaries}
Define
$$\zeta=\frac{f_1f_2}{qf_9f_{18}}$$
and
$$T=\frac{f_3^4f_6^4}{q^3f_9^4f_{18}^4}.$$
Let $H$ be the ``huffing'' operator defined by
\[H\left (\sum a_nq^n\right )=\sum a_{3n}q^{3n}.\]
From Chan \cite[(11)--(19)]{HCC1}, for each $i\geq1$, we have
\begin{equation}
H\left (\frac1{\zeta^i}\right )=\sum_{j=1}^i\frac{m_{i,j}}{T^j},
\end{equation}
where $m_{i,j}$'s are defined in the following matrix.

The $m_{i,j}$ form a matrix $M$, the first nine rows of which are
\begin{equation}\label{matrix}
M=\left (\begin{matrix} 3&0&0&0&0&0&0&\cdots\\
2&3^3&0&0&0&0&0&\cdots\\
1&3^3&3^5&0&0&0&0&\cdots\\
0&2\cdot 3^2&2^2\cdot 3^4&3^7&0&0&0&\cdots\\
0&5&2\cdot 3^3\cdot 5&3^6\cdot 5&3^9&0&0&\cdots\\
0&1&2\cdot 3^2\cdot 7 &3^6\cdot 5&2\cdot 3^9&3^{11}&0&\cdots\\
0&0&2\cdot 3\cdot 7&2^2\cdot 3^4 \cdot 7&3^8\cdot 7&3^{10}\cdot 7&3^{13}&\cdots\\
0&0&2^3&2\cdot 3^3 \cdot 19&2^4\cdot3^7&2^2\cdot 3^9\cdot 7&2^3\cdot 3^{12}&\cdots\\
0&0&1&2^2\cdot 3^4&3^9&3^9\cdot 5^2&2^2\cdot 3^{13}&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots & \ddots
\end{matrix}\right )\\
\end{equation}
and for $i\geq4$, $m_{i,1}=0$, and for $j\geq2$, 
\begin{equation}\label{mij}
m_{i,j}=9m_{i-1,j-1}+3m_{i-2,j-1}+m_{i-3,j-1}.
\end{equation}
In fact $m_{4i-3,j}=0$ for $j\leq i-1$, so we can write
\begin{equation}\label{Hz1}
H\left (\frac1{\zeta^{4i-3}}\right )=\sum_{j=i}^{4i-3}\frac{m_{4i-3,j}}{T^j}
=\sum_{j=1}^{3i-2}\frac{m_{4i-3,i+j-1}}{T^{i+j-1}}
=\sum_{j=1}^{3i-2}\frac{a_{i,j}}{T^{i+j-1}},
\end{equation}
where 
\begin{equation}\label{aij}
a_{i,j}=m_{4i-3,i+j-1}.
\end{equation}

Similarly, $m_{4i-1,j}=0$ if $j\leq i-1$, so we can write
\begin{equation}\label{Hz2}
H\left (\frac1{\zeta^{4i-1}}\right )
=\sum_{j=i}^{4i-1}\frac{m_{4i-1,j}}{T^j}\\
=\sum_{j=1}^{3i}\frac{m_{4i-1,i+j-1}}{T^{i+j-1}}
=\sum_{j=1}^{3i}\frac{b_{i,j}}{T^{i+j-1}},
\end{equation}
where \begin{equation}\label{bij}
b_{i,j}=m_{4i-1,i+j-1}.
\end{equation}

And $m_{4i,j}=0$ if $j\leq i$, so we can write
\begin{equation}\label{Hz3}
H\left (\frac1{\zeta^{4i}}\right )
=\sum_{j=1+i}^{4i}\frac{m_{4i,j}}{T^j}\\
=\sum_{j=1}^{3i}\frac{m_{4i,i+j}}{T^{i+j}}
=\sum_{j=1}^{3i}\frac{c_{i,j}}{T^{i+j}},
\end{equation}
where 
\begin{equation}\label{cij}
c_{i,j}=m_{4i,i+j}.
\end{equation}

We can write \eqref{Hz1} as
\begin{equation}
H\left (\left (q\frac{f_9f_{18}}{f_1f_2}\right )^{4i-3}\right )
=\sum_{j=1}^{3i-2}a_{i,j}\left (q^3\frac{f_9^4f_{18}^4}{f_3^4f_6^4}\right )^{i+j-1},
\end{equation}
which can be rearranged to
\begin{equation}
H\left (q^{i-3}\left(\frac{f_3f_6}{f_1f_2}\right )^{4i-3}\right )
=\sum_{j=1}^{3i-2}a_{i,j}q^{3j-3}\left (\frac{f_9f_{18}}{f_3f_6}\right )^{4j-1}.
\end{equation}

The equation \eqref{Hz2} can be written as
\begin{equation}
H\left (\left (q\frac{f_9f_{18}}{f_1f_2}\right )^{4i-1}\right )
=\sum_{j=1}^{3i}b_{i,j}\left (q^3\frac{f_9^4f_{18}^4}{f_3^4f_6^4}\right )^{i+j-1},
\end{equation}
which implies that
\begin{equation}
H\left (q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right )^{4i-1}\right )
=\sum_{j=1}^{3i}b_{i,j}q^{3j-3}\left (\frac{f_9f_{18}}{f_3f_6}\right )^{4j-3}.
\end{equation}

Similarly \eqref{Hz3} is
\begin{equation}
H\left (\left (q\frac{f_9f_{18}}{f_1f_2}\right )^{4i}\right )
=\sum_{j=1}^{3i}c_{i,j}\left (q^3\frac{f_9^4f_{18}^4}{f_3^4f_6^4}\right )^{i+j},
\end{equation}
and this can be rearranged to
\begin{equation}
H\left (q^i\left(\frac{f_3f_6}{f_1f_2}\right )^{4i}\right )
=\sum_{j=1}^{3i}c_{i,j}q^{3j}\left (\frac{f_9f_{18}}{f_3f_6}\right )^{4j}.
\end{equation}

\section{Generating functions}
In this section, we deduce generating functions that are useful in proving our main results.
\begin{theorem}
For each  $\alpha\geq 0$,
\begin{equation}\label{3p33}
\sum_{n\geq0}a_3\left(3^{2\alpha}n+\frac{3^{2\alpha}-1}{4}\right )q^n
=\sum_{i\geq1}x_{2\alpha,i}q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i-3}
\end{equation}
and
\begin{equation}\label{3p34}
\sum_{n\geq0}a_3\left (3^{2\alpha+1}n+\frac{3^{2\alpha+2}-1}{4}\right )q^n
=\sum_{i\geq1}x_{2\alpha+1,i}q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i-1},
\end{equation}
where the coefficient vectors ${\mathbf{x}}_{\alpha}=(x_{\alpha,1},x_{\alpha,2},\ \ldots\ )$
are given by
\begin{equation}\label{x0}
\mathbf{x}_0=(x_{0,1},x_{0,2},x_{0,3},\ \ldots\ )=(1,0,0,\ \ldots\ ),
\end{equation}
and
\begin{align}
{\mathbf{x}}_{\alpha+1}&={\mathbf{x}}_{\alpha}A\ \ \text{if}\ \alpha\ \text{is even},\\
{\mathbf{x}}_{\alpha+1}&={\mathbf{x}}_{\alpha}B\ \ \text{if}\ \alpha\ \text{is odd},
\end{align}
where $A=(a_{i,j})_{i,j\geq1}$ and $B=(b_{i,j})_{i,j\geq1}$.
\end{theorem}
\begin{proof}
The identity \eqref{defa3} is the $\alpha=0$ case of \eqref{3p33}.

Suppose \eqref{3p33} holds for some $\alpha\geq 0$. Then
\begin{equation}
\sum_{n\geq0}a_3\left(3^{2\alpha}n+\frac{3^{2\alpha}-1}{4}\right )q^n
=\sum_{i\geq1}x_{2\alpha,i}q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i-3},
\end{equation}
which is equivalent to
\begin{equation}\label{3p39}
\sum_{n\geq0}a_3\left(3^{2\alpha}n+\frac{3^{2\alpha}-1}{4}\right )q^{n-2}
=\sum_{i\geq1}x_{2\alpha,i}q^{i-3}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i-3}.
\end{equation}
Applying the operator $H$ to \eqref{3p39}, we find that
\begin{align*}
\sum_{n\geq0}a_3\left(3^{2\alpha}(3n+2)+\frac{3^{2\alpha}-1}{4}\right )q^{3n}
&=\sum_{i\geq1}x_{2\alpha,i}H\left(q^{i-3}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i-3}\right)\\
&=\sum_{i\geq1}x_{2\alpha,i}\sum\limits_{j=1}^{3i-2}a_{i,j}q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j-1}\\
&=\sum\limits_{j\geq1}\left(\sum_{i\geq1}x_{2\alpha,i}a_{i,j}\right)q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j-1}\\
&=\sum\limits_{j\geq1}x_{2\alpha+1,j}q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j-1},
\end{align*}
which implies equation \eqref{3p34}.

Now suppose \eqref{3p34} holds for some $\alpha\geq0$. Then
\begin{equation}\label{3p310}
\sum_{n\geq0}a_3\left (3^{2\alpha+1}n+\frac{3^{2\alpha+2}-1}{4}\right )q^n
=\sum_{i\geq1}x_{2\alpha+1,i}q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i-1}.
\end{equation}
Applying the operator $H$ to \eqref{3p310}, we find that
\begin{align*}
\sum_{n\geq0}a_3\left (3^{2\alpha+1}(3n)+\frac{3^{2\alpha+2}-1}{4}\right )q^{3n}
&=\sum_{i\geq1}x_{2\alpha+1,i}H\left(q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i-1}\right)\\
&=\sum_{i\geq1}x_{2\alpha+1,i}\sum_{j=1}^{3i}b_{i,j}q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j-3}\\
&=\sum_{j\geq1}\left(\sum_{i\geq1}x_{2\alpha+1,i}b_{i,j}\right)q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j-3}\\
&=\sum_{j\geq1}x_{2\alpha+2,j}q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j-3}.
\end{align*}
After simplification, we obtain \eqref{3p33} with $\alpha+1$ in place of $\alpha$. This completes the proof of \eqref{3p33} and \eqref{3p34} by induction.
\end{proof}

\begin{theorem}For each $\alpha\geq0$,
\begin{equation}\label{3a9}
\sum_{n\geq0}a_9\left(3^{\alpha+1}n+3^{\alpha+1}-1\right )q^n
=\sum_{i\geq1}y_{\alpha,i}q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i}
\end{equation}
where the coefficient vectors ${\mathbf{Y}}_{\alpha}=(y_{\alpha,1},y_{\alpha,2},\ \ldots\ )$
are given by
\begin{equation}\label{y0}
\mathbf{Y}_0=(y_{0,1},y_{0,2},y_{0,3},\ \ldots\ )=(3,0,0,\ \ldots\ ),
\end{equation}
and
\begin{equation}
{\mathbf{Y}}_{\alpha+1}={\mathbf{Y}}_{\alpha}C,
\end{equation}
where $C=(c_{i,j})_{i,j\geq1}$.
\end{theorem}
\begin{proof}
We prove this by induction on $\alpha$. The identity \eqref{3a32} is the $\alpha=0$ case of \eqref{3a9}.

Suppose \eqref{3a9} holds for some $\alpha\geq 0$. Then
\begin{equation}
\sum_{n\geq0}a_9\left(3^{\alpha+1}n+3^{\alpha+1}-1\right )q^n
=\sum_{i\geq1}y_{\alpha,i}q^{i-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4i},
\end{equation}
which is equivalent to
\begin{equation}\label{3p9}
\sum_{n\geq0}a_9\left(3^{\alpha+1}n+3^{\alpha+1}-1\right )q^{n-2}
=q^{-3}\sum_{i\geq1}y_{\alpha,i}q^i\left(\frac{f_3f_6}{f_1f_2}\right)^{4i}.
\end{equation}
Applying the operator $H$ to \eqref{3p9}, we find that
\begin{align*}
\sum_{n\geq0}a_9\left(3^{\alpha+1}(3n+2)+3^{\alpha+1}-1\right)q^{3n}
&=q^{-3}\sum_{i\geq1}y_{\alpha,i}H\left(q^i\left(\frac{f_3f_6}{f_1f_2}\right)^{4i}\right)\\
&=\sum_{i\geq1}y_{\alpha,i}\sum\limits_{j=1}^{3i}c_{i,j}q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j}\\
&=\sum\limits_{j\geq1}\left(\sum_{i\geq1}y_{\alpha,i}c_{i,j}\right)q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j}\\
&=\sum\limits_{j\geq1}y_{\alpha+1,j}q^{3j-3}\left(\frac{f_9f_{18}}{f_3f_6}\right)^{4j},
\end{align*}
which implies that
\begin{equation}
\sum_{n\geq0}a_9\left(3^{\alpha+2}n+3^{\alpha+2}-1\right )q^n=\sum\limits_{j\geq1}y_{\alpha+1,j}q^{j-1}\left(\frac{f_3f_6}{f_1f_2}\right)^{4j},
\end{equation}
which is \eqref{3a9} with $\alpha+1$ for $\alpha$ .
\end{proof}

\section{Congruences}
Let $\nu(N)$ be the largest power of 3 that divides $N$. Note that $\nu(0)=+\infty$.

\begin{proof}[Proof of Theorem \ref{th1}]
It follows from \eqref{matrix} and \eqref{mij} that
\begin{equation}\label{numij}
\nu(m_{i.j})\geq 3j-i-1,
\end{equation}
and from \eqref{aij}, \eqref{bij} and \eqref{numij},
\begin{equation}\label{nuaij}
\nu(a_{i.j})\geq 3(i+j-1)-(4i-3)-1=3j-i-1
\end{equation}
and
\begin{equation}\label{nubij}
\nu(b_{i.j})\geq 3(i+j-1)-(4i-1)-1=3j-i-3.
\end{equation}
It is not hard to show that
\begin{equation}\label{nux1}
\nu(x_{2\alpha,j})\geq \alpha+3j-4
\end{equation}
and
\begin{equation}\label{nux2}
\nu(x_{2\alpha+1,j})\geq \alpha+1+3(j-1).
\end{equation}
The identity \eqref{nux1} is true for $\alpha=0$, by \eqref{x0}.

Suppose \eqref{nux1} is true for some $\alpha\geq0$. Then
\begin{align*}
\nu(x_{2\alpha+1,j})&\geq \underset{i\geq 1}{\text{min}}\left(\nu(x_{2\alpha,i})+\nu(a_{i,j})\right)\\
&=\nu(x_{2\alpha,1})+\nu(a_{1,j})\\
&\geq \alpha+3j-2\\
&\geq \alpha+1+3(j-1),
\end{align*}
which is \eqref{nux2}.

Now suppose \eqref{nux2} is true for all $\alpha\geq0$. Then
\begin{align*}
\nu(x_{2\alpha+2,j})&\geq \underset{i\geq 1}{\text{min}}\left(\nu(x_{2\alpha+1,i})+\nu(b_{i,j})\right)\\
&=\nu(x_{2\alpha+1,1})+\nu(b_{1,j})\\
&\geq \alpha+1+3j-4,
\end{align*}
which is \eqref{nux1} with $\alpha+1$ in place of $\alpha$. This completes the proof of \eqref{nux1} and \eqref{nux2}  by induction.

The congruence \eqref{3ac1} follows from \eqref{3p33} together with \eqref{nux1}, and the congruence \eqref{3ac2} follows from \eqref{3p34} together with \eqref{nux2}.

From \eqref{3p34} and \eqref{nux2}, we have
\begin{equation}\label{3c1}
\sum\limits_{n\geq 0}a_3\left(3^{2\alpha+1}n+\frac{3^{2\alpha+2}-1}{4}\right)q^n\equiv 3^{\alpha+1}\frac{f_3^3f_6^3}{f_1^3f_2^3}\pmod{3^{\alpha+4}}.
\end{equation}
By the binomial theorem, it is easy to see that
\begin{equation}\label{fp3}
f_1^3\equiv f_3\pmod{3}.
\end{equation}
In view of \eqref{fp3}, the congruence \eqref{3c1} can be expressed as
\begin{align}
\sum\limits_{n\geq 0}a_3\left(3^{2\alpha+1}n+\frac{3^{2\alpha+2}-1}{4}\right)q^n\equiv 3^{\alpha+1}\frac{f_9f_{18}}{f_3f_6}\pmod{3^{\alpha+2}}.\label{3c31}
\end{align}
Equating the coefficients of $q^{3n+1}$ and $q^{3n+2}$ in \eqref{3c31}, we obtain \eqref{3ac3} and \eqref{3ac4}, respectively.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th2}]
It follows from \eqref{cij} and \eqref{numij} that
\begin{equation}\label{nucij}
\nu(c_{i.j})\geq 3(i+j)-4i-1=3j-i-1.
\end{equation}
It is not hard to show that
\begin{equation}\label{nuy1}
\nu(y_{\alpha,j})\geq \alpha+1+3(j-1).
\end{equation}
The identity \eqref{nuy1} is true for $\alpha=0$, by \eqref{y0}.

Suppose \eqref{nuy1} is true for some $\alpha\geq0$. Then
\begin{align*}
\nu(y_{\alpha+1,j})&\geq \underset{i\geq 1}{\text{min}}\left(\nu(y_{\alpha,i})+\nu(c_{i,j})\right)\\
&=\nu(y_{\alpha,1})+\nu(c_{1,j})\\
&\geq \alpha+1+3j-2\\
&\geq \alpha+2+3(j-1),
\end{align*}
which is \eqref{nuy1} with $\alpha+1$ for $\alpha$.

The congruence \eqref{9ac} follows from \eqref{3a9} together with \eqref{nuy1}.
\end{proof}	

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

\begin{thebibliography}{9}
\bibitem{HCC}
H.-C. Chan, Ramanujan's cubic continued fraction and a generalization of his `most beautiful identity', \textit{ Int. J. Number Theory} {\bf6} (2010), 673--680.

\bibitem{HCC1}
 H.-C. Chan, Ramanujan's cubic continued fraction and Ramanujan type congruences for a ceratin partition function, \textit{ Int. J. Number Theory} \textbf{  6} (2010), 819--834.

\bibitem{Chern}
S. Chern, Arithmetic Properties for Cubic Partition Pairs Modulo Powers of 3, \textit{ Acta. Math. Sin.-English Ser.} \textbf{33} (2017), 1504--1512.

\bibitem{GM}
D. S. Gireesh and M. S. Mahadeva Naika, General family of congruences modulo large powers of 3 for cubic partition pairs, \textit{New Zealand J. Math.} \textbf{47} (2017), 43--56.

\bibitem{Hir}
M. D. Hirschhorn and D. C. Hunt, A simple proof of the Ramanujan conjecture for powers of 5, \textit{ J. Reine Angew. Math.} \textbf{326} (1981), 1--17.

\bibitem{Lin}
 B.~L.~S. Lin, Congruences modulo 27 for cubic partition pairs, \textit{J. Number Theory} \textbf{171} (2017), 31--42.

\bibitem{LWX}
B.~L.~S. Lin,  L. Wang, and  E. X. W. Xia, Congruences for cubic partition pairs modulo powers of 3, \textit{ Ramanujan J.} \textbf{46} (2018), 563--578.

\bibitem{RAMAU}
S. Ramanujan, \textit{The Lost Notebook and Other Unpublished Papers},
Narosa, New Delhi, 1998.

\bibitem{ZZ}
H. Zhao and Z. Zhong, Ramanujan type congruences for a partition function,\textit{ Electron. J. Combin.} \textbf{18} (2011), \#P58.
\end{thebibliography}



\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: } 
partition, 3-regular cubic partition,
9-regular cubic partition, congruence.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A335602},
\seqnum{A335604}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 27 2019.
Revised versions received  June 4 2020; June 17 2020; June 18 2020.
Published in {\it Journal of Integer Sequences}, June 24 2020.

\bigskip
\hrule
\bigskip

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


\end{document}