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\vskip 1cm{\LARGE\bf 
A Note  on the \\
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Andrews-Ericksson-Petrov-Romick Bijection\\
\vskip .1in
for MacMahon's Partition Theorem
}
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\large
Beaullah Mugwangwavari and Darlison Nyirenda \\
School of Mathematics\\
University of the Witwatersrand \\
P. O. Wits 2050\\
Johannesburg\\
South Africa \\
\href{mailto:712040@students.wits.ac.za}{\tt 712040@students.wits.ac.za}\\
\href{mailto:darlison.nyirenda@wits.ac.za}{\tt darlison.nyirenda@wits.ac.za} \\
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\begin{abstract}
 Andrews' generalization of MacMahon's partition theorem states that the number of partitions of $n$ in which odd multiplicities are at least $2r + 1$  is equal to the number of partitions in which odd parts are congruent to $\modd{2r + 1} {4r + 2}$. In this note, we give a bijective proof of this generalization. Our result naturally extends the bijection of Andrews, Ericksson, Petrov, and Romik for MacMahon's partition theorem.
\end{abstract}



\section{Introduction}\label{introd}
Let $n \in \mathbb{Z}_{\geq 1}$. A \textit{partition} of $n$ is a sequence of positive integers $(\lambda_1, \lambda_2, \ldots)$  where $\lambda_1 \geq \lambda_2 \geq \cdots $ and  $\sum\limits_{j \geq 1} \lambda_j = n$. 
The components $\lambda_i$ are called \textit{parts} of a partition. If further restrictions are imposed on the parts, one gets some interesting partition functions. One such example is the number of partitions of $n$ into distinct parts. This is interesting because it relates to the number of partitions of $n$ into odd parts, a case of Euler's partition identity \cite{and2}.  A partition identity relates two partition functions. One of the partition identities of interest in this paper is Theorem \ref{MacMahon}, which is due to MacMahon.
\begin{theorem}[MacMahon, \cite{PAM}] \label{MacMahon}
The number of partitions of $n$ wherein no part appears with multiplicity one is equal to the number of partitions of $n$ where parts are even or congruent to $\modd{3} {6}$.
\end{theorem}
Andrews, Ericksson, Petrov, and Romik \cite{gepr} gave the first bijective proof of Theorem \ref{MacMahon} which we describe in the sequel. 

Let $C_{n}$ denote the set of partitions of $n$ wherein no part appears with multiplicity one and let $D_{n}$ denote the set of partitions of $n$ wherein parts are even or congruent to $\modd{3} {6}$. The bijection goes as follows:
let $ \lambda = (l^{h_{l}}, (l-1)^{h_{l-1}}, \ldots, 3^{h_{3}}, 2^{h_{2}}, 1^{h_{1}}) \in C_{n}$ where $h_{i}$ is the multiplicity of $i$. Since no part appears exactly once, it follows that $h_{i} \in \{0, 2, 3, 4,\ldots \}$. Uniquely decompose $ h_{i}$ as $$h_{i} = k_{i} + g_{i},\,\,\text{where} \,\,\, k_{i} \in \{0, 3\},\,\, g_{i} \in \{0, 2, 4, 6, 8,\ldots \}.$$ 
For $j\geq 1$, define $d_j$ as  follows:
\begin{align*}
d_{6t + 1} &amp; = d_{6t + 5} = 0, \\
d_{6t + 2} &amp; = \frac{1}{2}g_{3t + 1}, \\
d_{6t + 4} &amp; = \frac{1}{2}g_{3t + 2}, \\
d_{6t + 3} &amp; = \frac{1}{3}k_{2t + 1} + g_{6t + 3}, \\
d_{6t + 6} &amp; = \frac{1}{3}k_{2t + 2} + g_{6t + 6},
\end{align*} 
where $t = 0,1,2,\ldots$

The partition $(f^{d_f},(f-1)^{d_{f-1}},\ldots, 2^{d_2}, 1^{d_1})$ is in $D_n$.
We shall call the above mapping  the 
{\it Andrews-Ericksson-Petrov-Romik bijection}.

Andrews later gave a generalization for Theorem \ref{MacMahon} which we recall below.
\begin{theorem}[Andrews, \cite{NoteA}] \label{andthm}
Let $A_{r}(n)$ denote the set of partitions of $n$ wherein parts appearing an odd number of times actually appear at least $2r + 1$ times and let $B_{r}(n)$ denote the set of partitions of $n$ wherein odd parts are congruent to $\modd{2r + 1} {4r + 2}$. Then $$ \vert A_{r}(n)\vert = \vert B_{r}(n)\vert.$$
\end{theorem}
However, there is no bijective proof that naturally extends  Andrews-Ericksson-Petrov-Romik bijection to prove Theorem \ref{andthm}.  For this, you may consult \cite{JSSF} and the references therein. Our aim in this short note is to supply a bijection for Theorem \ref{andthm}, which naturally generalizes Andrews-Ericksson-Petrov-Romik bijection and is different from the ones given in the literature. 
\section{The bijection}
We now describe our bijection for Theorem \ref{andthm}.
We need to establish the one-to-one correspondence between the sets $A_{r}(n)$ and $B_{r}(n)$. 

Let $(l^{h_{l}}, (l-1)^{h_{l-1}}, \ldots, 3^{h_{3}}, 2^{h_{2}}, 1^{h_{1}}) \in A_{r}(n)$. In this notation, $h_{i}$ is the multiplicity of $i$. Note that $h_i$ can be zero for some $i$. Clearly, $h_{i} \in \{0, 2, 4, \ldots, 2r, 2r + 1, 2r + 2, \hdots \}$. 

 Write $h_i$ as 
$$ h_{i} = k_{i} + g_{i}\,\,\,\text{where}\,\, k_{i} \in \{0, 2r + 1\}\,\,\,\text{and}\,\, g_{i} \in \{0, 2, 4, 6, 8, 10,\ldots \}$$
This decomposition is unique since $k_i$ and $g_i$ can be made explicit, i.e.,
$$ k_{i} = (2r + 1)\left( \frac{1 - (-1)^{h_{i}}}{2} \right)\,\,\,\text{and}\,\,\, g_{i} = h_{i} - (2r + 1)\left( \frac{1 - (-1)^{h_{i}}}{2} \right).$$
For $j\geq 1$, define $d_j$ as follows:
\begin{align*}
d_{(4r + 2)t + 2j - 1} &amp; = 0  \,\,\, \text{for} \,\, j\in\{1,2, \ldots, 2r + 1\}\setminus \{r + 1\}, \\
d_{(4r + 2)t + 2j} &amp; = \frac{1}{2}g_{(2r + 1)t + j} \,\,\, \text{for} \,\, j\in\{1,2, \ldots, 2r\}, \\
d_{(4r + 2)t + (2r + 1)j} &amp; = \frac{1}{2r + 1}k_{2t + j} + g_{(4r + 2)t + (2r + 1)j} \,\,\, \text{for} \,\, j \in \{1, 2\},  
\end{align*}
where $t = 0,1,2, \ldots$.

The image is thus given by 
$$ (\ldots, f^{d_f}, (f-1)^{d_{f-1}},\ldots, 2^{d_2}, 1^{d_1}).$$
We claim that  $(\ldots, f^{d_f}, (f-1)^{d_{f-1}},\ldots, 2^{d_2}, 1^{d_1}) \in B_{r}(n)$. In order to show that $$ (l^{h_{l}}, (l-1)^{h_{l-1}}, \ldots, 3^{h_{3}}, 2^{h_{2}}, 1^{h_{1}}) \mapsto (\ldots, f^{d_f}, (f-1)^{d_{f-1}},\ldots, 2^{d_2}, 1^{d_1})$$ defines a bijection from $A_{r}(n)$ onto $B_{r}(n)$, it suffices to show that 
$\sum\limits_{i \geq 1}id_i = n$.

Thus
\begin{align*}
\sum\limits_{i \geq 1}id_i &amp; =  \sum\limits_{j = 1}^{r}\sum\limits_{t = 0}^{\infty}((4r + 2)t + 2j - 1)d_{(4r + 2)t + 2j - 1} + \sum\limits_{j = r + 2}^{2r + 1}\sum\limits_{t = 0}^{\infty}((4r + 2)t + 2j - 1)d_{(4r + 2)t + 2j - 1}  \\
                          &amp; \qquad + \, \sum\limits_{j = 1}^{2r}\sum\limits_{t = 0}^{\infty}((4r + 2)t + 2j)d_{(4r + 2)t + 2j} +                  \sum\limits_{j = 1}^{2}\sum\limits_{t = 0}^{\infty}((4r + 2)t + (2r + 1)j)d_{(4r + 2)t + (2r + 1)j} \\
                           &amp; =   \sum\limits_{j = 1}^{2r}\sum\limits_{t = 0}^{\infty}((2r + 1)t + j)g_{(2r + 1)t + j} +  \sum\limits_{j = 1}^{2}\sum\limits_{t = 0}^{\infty}((4r + 2)t + (2r + 1)j)g_{(4r + 2)t + (2r + 1)j}                \\
                           &amp; \qquad + \sum\limits_{j = 1}^{2}\sum\limits_{t = 0}^{\infty}(2t + j)k_{2t + j}.
\end{align*}
Using the notation $$ S(a,b) =  \sum\limits_{t = 0}^{\infty}(at + b)g_{at + b}\,\,\,\,\,\text{and}\,\,\,\,\,T(a,b) =  \sum\limits_{t = 0}^{\infty}(at + b)k_{at + b},$$
observe that 
\begin{align*}
\sum\limits_{t = 0}^{\infty}((2r + 1)t + j)g_{(2r + 1)t + j} &amp; =  \sum\limits_{\ell = 0, 2r + 1}\sum\limits_{t = 0}^{\infty}((4r + 2)t + j + \ell )g_{(4r + 2)t + j + \ell}\\   
&amp; = S(4r + 2, j)  + S(4r + 2, 2r + 1)
\end{align*}
and 
\begin{align*}
\sum\limits_{t = 0}^{\infty}(2t + j)k_{2t + j}  &amp; = \sum\limits_{t = 0}^{\infty} \sum\limits_{\ell = 0}^{2r} ((4r + 2)t + j + 2\ell)k_{(4r + 2)t + j + 2\ell} \\
&amp; = \sum\limits_{\ell = 0}^{2r} T(4r + 2, j + 2\ell).
\end{align*}
Hence
\begin{align*}
\sum\limits_{i \geq 1}id_i &amp; = \sum\limits_{j = 1}^{2r} S(4r + 2, j)  +  \sum\limits_{j = 1}^{2r}S(4r + 2, j + 2r + 1) +  \sum\limits_{j = 1}^{2}S(4r + 2, (2r + 1)j)   \\
&amp; \qquad    + \sum\limits_{\ell = 0}^{2r}T(4r + 2, 1 + 2\ell) +  \sum\limits_{\ell = 0}^{2r} T(4r + 2, 2 + 2\ell) \\
&amp; = \sum\limits_{j = 1}^{2r} (S(4r + 2, j)  + S(4r + 2, j + 2r + 1) + T(4r + 2, 1 + 2j) + T(4r + 2, 2 + 2j) )  \\
&amp; \qquad + \sum\limits_{j = 1}^{2}S(4r + 2, (2r + 1)j) + T(4r + 2, 1) + T(4r + 2, 2) \\
&amp; = \sum\limits_{j = 1}^{2r + 1} (S(4r + 2, j)  + S(4r + 2, j + 2r + 1)) + \sum\limits_{j = 0}^{2r}(T(4r + 2, 1 + 2j) + T(4r + 2, 2 + 2j)) \\
&amp; = \sum\limits_{j = 1}^{2r + 1} (S(4r + 2, j)  + S(4r + 2, j + 2r + 1)) + \sum\limits_{j = 1}^{2r + 1}(T(4r + 2, 2j - 1) + T(4r + 2, 2j)) \\
&amp; = \sum\limits_{j = 1}^{4r + 2} S(4r + 2, j) + \sum\limits_{j = 1}^{4r + 2}T(4r + 2, j)
\end{align*}
\begin{align*}
&amp; = \sum\limits_{j = 1}^{4r + 2} (S(4r + 2, j) + T(4r + 2, j))\\ 
&amp; = \sum\limits_{j = 1}^{4r + 2} \sum\limits_{t = 0}^{\infty} ((4r + 2)t + j)( g_{(4r + 2)t + j} + k_{(4r + 2)t + j}) \\
&amp; = \sum\limits_{t = 0}^{\infty} \sum\limits_{j = 1}^{4r + 2} ((4r + 2)t + j)h_{(4r + 2)t + j}  \\
&amp; = \sum\limits_{i = 1}^{\infty}ih_{i} \\
&amp; = n.
\end{align*}
\begin{remark}
Setting $r = 1$ in the bijection yields the bijection given by Andrews et al.~\cite{gepr}.
\end{remark}
For example, consider $n = 15$ and $r = 2$. Table \ref{tab1} demonstrates the correspondence.
The inverse of our map is described in the following section.
\begin{table}[H]
	\begin{center}
		\begin{tabular}{ccc}
			$A_{2}(15)$ &amp; $\stackrel{}{\longrightarrow}$&amp; $B_{2}(15)$\\ \hline
			(2,2,2,2,2,1,1,1,1,1) &amp; $\mapsto$ &amp; (10,5) \\
			(4,4,1,1,1,1,1,1,1) &amp; $\mapsto$ &amp; (8,5,2)\\
			(3,3,2,2,1,1,1,1,1) &amp; $\mapsto$ &amp; (6,5,4)\\
			(3,3,1,1,1,1,1,1,1,1,1) &amp; $\mapsto$ &amp; (6,5,2,2)\\
			(2,2,2,2,1,1,1,1,1,1,1) &amp; $\mapsto$ &amp; (5,4,4,2)\\
			(2,2,1,1,1,1,1,1,1,1,1,1,1) &amp; $\mapsto$ &amp; (5,4,2,2,2)\\ 
			(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1) &amp; $\mapsto$ &amp; (5,2,2,2,2,2)	\\ \hline
		\end{tabular}
		\caption{The map $A_{r}(n)\rightarrow B_{r}(n)$ for $r = 2, n=15$.}\label{tab1}
	\end{center}
\end{table}

\section{The inverse mapping}
We now describe the inverse to our bijection.

Given that $(\ldots, f^{d_f}, (f-1)^{d_{f-1}},\ldots, 2^{d_2}, 1^{d_1}) \in B_{r}(n)$. Define $g_i$ and $k_i$ as
$$  g_{(2r + 1)t + j} = 2d_{(4r + 2)t + 2j}, j = 1,2, \ldots 2r, t = 0,1,2,\ldots$$

For the remaining cases, we have, for $t = 0,1,2,\ldots$,
$$
g_{(4r + 2)t + (2r + 1)j} = 
\begin{cases}
d_{(4r + 2)t + (2r + 1)j}, &amp; \text{if $d_{(4r + 2)t + (2r + 1)j} \equiv \modd{0} {2},\ j = 1,2$;} \\
d_{(4r + 2)t + (2r + 1)j} - 1, &amp; \text{if $d_{(4r + 2)t + (2r + 1)j} \equiv \modd{1} {2},\ j = 1,2$,}
\end{cases}
$$ 
and
$$
k_{2t + j} = 
\begin{cases}
0, &amp; \text{if $d_{(4r + 2)t + (2r + 1)j} \equiv \modd{0} {2},\ j = 1,2$;} \\
2r + 1, &amp; \text{if $d_{(4r + 2)t + (2r + 1)j} \equiv \modd{1} {2},\ j= 1,2$.}
\end{cases}
$$ 
Then the partition $(\ldots, 3^{g_3 + k_3}, 2^{g_2 + k_2}, 1^{g_1 + k_1})$ is in $A_{r}(n)$. 

For example, Table \ref{tab2} demonstrates the inverse mapping when $n = 15$. 
\begin{table}[htp]
	\begin{center}
		\begin{tabular}{ccc}
			$B_{2}(15)$ &amp; $\stackrel{}{\longrightarrow}$&amp; $A_{2}(15)$\\ \hline
			(10,5) &amp; $\mapsto$ &amp; (2,2,2,2,2,1,1,1,1,1)\\
			(8,5,2) &amp; $\mapsto$ &amp; (4,4,1,1,1,1,1,1,1)\\
			(6,5,4) &amp; $\mapsto$ &amp; (3,3,2,2,1,1,1,1,1)\\
			(6,5,2,2) &amp; $\mapsto$ &amp; (3,3,1,1,1,1,1,1,1,1,1)\\
			(5,4,4,2) &amp; $\mapsto$ &amp; (2,2,2,2,1,1,1,1,1,1,1) \\
			(5,4,2,2,2) &amp; $\mapsto$ &amp; (2,2,1,1,1,1,1,1,1,1,1,1,1)\\
			(5,2,2,2,2,2) &amp; $\mapsto$ &amp; (1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)\\ \hline
		\end{tabular}
		\caption{The inverse map $B_{r}(n)\rightarrow A_{r}(n)$ for $n=15$, $r = 2$.}\label{tab2}
	\end{center}
\end{table} 

\begin{thebibliography}{9}
\bibitem{NoteA}G. E. Andrews, A generalization of a partition theorem of MacMahon,  {\it J. Combin. Theory} \textbf{3} (1967), 100--101.
\bibitem{gepr} G. E. Andrews, H. Eriksson, F. Petrov, and D. Romik, Integrals, partitions and MacMahon's theorem, {\it J. Combin. Theory Ser. A} \textbf{114} (2007), 545--554. 
\bibitem{JSSF}S. Fu and J. A. Sellers, Bijective proofs of partition identities of MacMahon, Andrews, and Subbarao, {\it Electron. J. Combin.} \textbf{21} (2014), P2.41.
\bibitem{PAM} P. A. MacMahon, { \it Combinatory Analysis}, Vol.~2, Cambridge University Press, 1916.
\bibitem{and2} D. Nyirenda, A note on a finite version of Euler's partition identity, \textit{Australas. J. Combin.} \textbf{71} (2018), 241--245.
\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11P81; Secondary 11P83, 05A15. 

\noindent \emph{Keywords: } partition, bijection.

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\vspace*{+.1in}
\noindent
Received June 27 2020;
revised versions received  August 25 2020; April 25 2021.
Published in {\it Journal of Integer Sequences}, April 25 2021.

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