%% if you are submitting an initial manuscript then you should have submission as an option here
%% if you are submitting a revised manuscript then you should have revision as an option here
%% otherwise options taken by the article class will be accepted
\documentclass[final]{FPSAC2022}
%% but DO NOT pass any options (or change anything else anywhere) which alters page size / layout / font size etc

%% note that the class file already loads {amsmath, amsthm, amssymb}

\newtheorem{thm}{Theorem}
\newtheorem{lem}{Lemma}


\usepackage[normalem]{ulem} 
\usepackage{color}

\renewcommand{\Im}{\operatorname{Im}}
\newcommand{\df}{\dfrac}
\newcommand{\Arg}{\operatorname{Arg}}
\newcommand{\mult}{\operatorname{mult}}
\newcommand{\even}{\operatorname{even}}
\newcommand{\Gal}{\operatorname{Gal}}
\newcommand{\Res}{\operatorname{Res}}
\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}
\newcommand{\s}{{\sigma}}
\newcommand{\Sig}{{\Sigma}}
\newcommand{\sss}{\sum^{\infty}_{\substack{
			c,d=-\infty\\(c,d)=1\\c,d\text{ odd }}}}
\renewcommand{\arctan}{\tan^{-1}}
\renewcommand{\a}{\alpha}
\renewcommand{\b}{\beta}
\newcommand{\e}{\epsilon}
\renewcommand{\d}{{\delta}}
\newcommand{\g}{\gamma}
\newcommand{\G}{\Gamma}
\renewcommand{\l}{\lambda}
\renewcommand{\L}{\Lambda}
\renewcommand{\k}{\kappa}
\renewcommand{\o}{{\omega}}

\renewcommand{\O}{{\Omega}}
\renewcommand{\r}{{\rho}}
\renewcommand{\t}{\tau}
\newcommand{\x}{\xi}
\newcommand{\z}{\zeta}
\newcommand{\CCal}{{\mathcal{C}}}
\newcommand{\ECal}{{\mathcal{E}}}
\newcommand{\HCal}{{\mathcal{H}}}
\newcommand{\JCal}{{\mathcal{J}}}
\newcommand{\ZBbb}{{\mathbf{Z}}}
\newcommand{\qu}{{\underline{q}}}
\newcommand{\thre}{{\sqrt{3}}}
\newcommand{\qq}{{\(\dfrac q{\qu}\)^2}}
\renewcommand{\(}{\left\(}
\renewcommand{\)}{\right\)}
\renewcommand{\[}{\left\[}
\renewcommand{\]}{\right\]}
\renewcommand{\i}{\infty}
\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}
\numberwithin{equation}{section}
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{refproof}{Proof}
\newtheorem{note}[theorem]{Note}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{case}[theorem]{Case}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{example}[]{Example}

\newcommand\mycom[2]{\genfrac{}{}{0pt}{}{#1}{#2}}
\numberwithin{equation}{section}
\newcommand{\abs}[1]{\lvert#1\rvert}
\newcommand{\blankbox}[2]{%
	\parbox{\columnwidth}{\centering
		\setlength{\fboxsep}{0pt}%
		\fbox{\raisebox{0pt}[#2]{\hspace{#1}}}%
	}%
}

\newenvironment{pf}
{\par\noindent{\it Proof of}\hskip 0.5em\ignorespaces}
{\hfill $\Box$\par\medskip}
\makeatletter
\def\proof{\@ifnextchar[{\@oproof}{\@nproof}}
\def\@oproof[#1][#2]{\trivlist\item[\hskip\labelsep\textit{#2 Proof of\
		#1.}~]\ignorespaces}
\def\@nproof{\trivlist\item[\hskip\labelsep\textit{Proof.}~]\ignorespaces}
%\smartqed
\def\endproof{\qed\endtrivlist}
\makeatother





%% define your title in the usual way
\title{Parity Biases in Partitions and Restricted Partitions}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
%\author[Optional shorter names]{Longer names me\thanks{\href{mailto:hello@world.c}{hello@world.c}. Longer names me was partially supported by Grant 2017.11.14.$\partial$\;supp.}\addressmark{1} \and You\addressmark{2}}

\author[K.~Banerjee, S.~Bhattacharjee, M.~G.~Dastidar, P.~J.~Mahanta, and M.~P.~Saikia]{
Koustav Banerjee\addressmark{1},
Sreerupa Bhattacharjee\addressmark{2},
Manosij Ghosh Dastidar\addressmark{3},
Pankaj Jyoti Mahanta\thanks{\href{mailto:pankaj@gonitsora.com}{pankaj@gonitsora.com}}\addressmark{4},
\and
Manjil P. Saikia\addressmark{5}
}

%% then use \addressmark to match authors to institutions here
\address{
\addressmark{1}Research Institute for Symbolic Computation, Johannes Kepler University, Linz, Austria\\
\addressmark{2}University of Lethbridge, Canada\\
\addressmark{3}Technische Universit{\"a}t Wien, Vienna, Austria\\
\addressmark{4}Gonit Sora, Dhalpur, Assam, India\\
\addressmark{5}Cardiff University, Cardiff, UK
}

%% put the date of submission here
\received{22 November 2021}

%% leave this blank until submitting a revised version
%\revised{}

%% put your English abstract here, or comment this out if you don't have one yet
%% please don't use custom commands in your abstract / resume, as these will be displayed online
%% likewise for citations -- please don't use \cite, and instead write out your citation as something like (author year)
\abstract{Let $p_{o}(n)$ (resp.\ $p_{e}(n)$) denote the number of partitions of $n$ with more odd parts (resp.\ even parts) than even parts (resp.\ odd parts). Recently, Kim, Kim, and Lovejoy proved that $p_{o}(n)>p_{e}(n)$ for all $n>2$ and conjectured that $d_{o}(n)>d_{e}(n)$ for all $n>19$, where $d_{o}(n)$ (resp.\ $d_{e}(n)$) denotes the number of partitions into distinct parts having more odd parts (resp.\ even parts) than even parts (resp.\ odd parts). In this paper we provide combinatorial proofs for both the result and the conjecture of Kim, Kim and Lovejoy. In addition, we discuss other results on partitions with restricted parts where our methods also work.}


%% put your keywords here, or comment this out if you don't have them yet
\keywords{Combinatorial inequalities, partitions, parity of parts, restricted partitions}

%% you can include your bibliography however you want, but using an external .bib file is STRONGLY RECOMMENDED and will make the editor's life much easier
%% regardless of how you do it, please use numerical citations, ie. [xx, yy] in the text

%% this sample uses biblatex, which (among other things) takes care of URLs in a more flexible way than bibtex
%% but you can use bibtex if you want
\usepackage[backend=bibtex]{biblatex}
\addbibresource{sample.bib}
%% note the \printbibliography command at the end of the file which goes with these biblatex commands

\begin{document}

\maketitle
%% note that you DO NOT have to put your abstract here -- it is generated by \maketitle and the \abstract and \resume commands above



\section{Introduction}\label{sec:intro}

In the theory of partitions, inequalities arising between two classes of partitions have a long tradition of study, for instance Alder's conjecture \cite{alder} and the Ehrenpreis problem \cite{gea3} are the most famous examples in this direction. In recent years there have been a number of results about partition inequalities. Very recently, Kim, Kim, and Lovejoy \cite{kkl} have given interesting inequalities which show biases in the parity of the partition functions. Proofs of such results employ a wide range of techniques, ranging from $q$-series methods, to combinatorial constructions and maps, to classical asymptotic analysis.

Let $p_{o}(n)$ (resp.\ $p_{e}(n)$) denote the number of partitions of $n$ with more odd parts (resp.\ even parts) than even parts (resp.\ odd parts). Kim, Kim, and Lovejoy \cite{kkl} proved that $p_{o}(n)>p_{e}(n)$ for all $n>2$ and conjectured that $d_{o}(n)>d_{e}(n)$ for all $n>19$ where $d_{o}(n)$ (resp.\ $d_{e}(n)$) denotes the number of partitions into distinct parts having more odd parts (resp.\ even parts) than even parts (resp.\ odd parts). The primary goal of the present paper is to prove these two inequalities combinatorially.



We define a partition $\lambda$ of a non-negative integer $n$ to be an integer sequence $\lambda_{1} \geq \lambda_{2} \geq \dots \geq \lambda_{\ell} >0$. We say that $\lambda$ is a partition of $n$, denoted by $\lambda \vdash n$ and $\sum_{i=1}^{\ell} \lambda_{i} = n$. The set of partitions of $n$ is denoted by $P(n)$ and $|P(n)| = p(n)$. For $\lambda \vdash n$, we define $a(\lambda)$ to be the largest part of $\lambda$, $\ell(\lambda)$ to be the total number of parts of $\lambda$ and $\mult_{\lambda}(\lambda_{i}):=m_i$ to be the multiplicity of the part $\lambda_{i}$ in $\lambda$. We also use $\lambda= (\lambda_{1}^{m_{1}}\dots\lambda_{\ell}^{m_{\ell}})$ as an alternative notation for a partition. For $\lambda \vdash n$ with $\lambda = (\lambda_{1},\dots,\lambda_{\ell})$ and $\mu \vdash m$ with $\mu = (\mu_{1},\dots,\mu_{\ell^{'}})$, define the union $\lambda \cup \mu  \vdash m+n$ to be the partition with parts $\{\lambda_{i},\mu_{j}\}$ arranged in non-increasing order. For a partition $\lambda \vdash n$, we split $\lambda$ into $\lambda_{e}$ and $\lambda_{o}$, respectively into even and odd parts; \textit{i.e.}, $\lambda = \lambda_{e} \cup \lambda_{o}$. We denote by $\ell_{e}(\lambda)$ (resp.\ $\ell_{o}(\lambda)$) the number of even parts (resp.\ odd parts) of $\lambda$ and $\ell(\lambda) = \ell_{e}(\lambda)+\ell_{o}(\lambda)$.

The following sets of partitions are of interest in this paper.
\begin{definition}
	\begin{equation*}
		\begin{split}
			D(n) & := \{\lambda \in P(n):\mult_{\lambda}(\lambda_{i})=1  \ \text{for all} \  i\},\\
			P_{e}(n) & := \{\lambda \in P(n):  \ell_{e}(\lambda)> \ell_{o}(\lambda)\},\\
			P_{o}(n) & := \{\lambda \in P(n):  \ell_{o}(\lambda)> \ell_{e}(\lambda)\},\\
			D_{e}(n) & :=  P_{e}(n) \cap D(n), \quad \text{and}\\
			D_{o}(n) & := P_{o}(n) \cap D(n).
		\end{split}
	\end{equation*}
\end{definition}

\begin{definition}
	For all the sets defined above, their cardinalities will be denoted by lower case letters. For instance, $|P_e(n)|=p_e(n)$, $|D_e(n)|=d_e(n)$, and so on.
\end{definition}

Now, we state formally the main results proved in this paper.
\begin{theorem}[{\cite[Theorem 1]{kkl}}]\label{thm1} For all positive integers $n \neq 2$, we have
\begin{equation*}
	p_{o}(n)>p_{e}(n).
\end{equation*}
\end{theorem}

\begin{theorem}[Conjectured in~\cite{kkl}]\label{thm:distinct}
	For all positive integers $n > 19$, we have
	\begin{equation*}
	d_{o}(n)>d_{e}(n).
	\end{equation*}
\end{theorem}


Before we move on further, let us describe the fundamental principle behind the proofs. Let $X$ and $Y$ be two given sets, our goal is to prove that $|Y| > |X|$. We choose a subset $X_{0}\ (\varsubsetneq X)$ and define an injective map $f \colon X_{0} \rightarrow Y$. Then to prove $|Y| > |X|$, it is enough to find a suitable subset $Y_{0}\ \varsubsetneq Y\setminus f(X_{0})$ such that $|Y_{0}| > |X\setminus X_{0}|$. Throughout this paper, we follow the notation $x\mapsto y$ instead of writing $f(x)=y$ when the map $f$ is understood from the context.

The rest of the paper is organized as follows:In Section \ref{sec:proof-kkl} we give a combinatorial proof of the result of Kim, Kim and Lovejoy \cite{kkl}. In Section \ref{sec:distinct} we give a proof of the conjecture of Kim, Kim and Lovejoy \cite{kkl}. Finally in Section \ref{sec:rem} we present a very short discussion on further results that can be proved using our methods.

\section{Proof of Theorem \ref{thm1}}\label{sec:proof-kkl}
We begin by presenting the following lemma, used later in the proof of Theorem \ref{thm1}. We skip the proof, as it is routine and can be found in the full version of the paper \cite{full}.

\begin{lemma}\label{lem1}
	For all even positive integers $n$ with $n \geq 14$, we have
	\begin{equation*}
		\sum_{k=1}^{\frac{n-6}{2}}\Bigl\lfloor\frac{n-2k-2}{4}\Bigr\rfloor > 1 + \sum_{k=1}^{\lfloor \frac{n-2}{6}\rfloor}\Bigl\lfloor \frac{n-6k+2}{4}\Bigr\rfloor + \sum_{k=1}^{\lfloor \frac{n-6}{6}\rfloor}\Bigl\lfloor \frac{n-6k-2}{4}\Bigr\rfloor.
	\end{equation*}
	For all odd positive integers $n$ with $n \geq 11$, we have
	\begin{equation*}
	\sum_{k=1}^{\frac{n-5}{2}}\Bigl\lfloor\frac{n-2k-1}{4}\Bigr\rfloor > 1 + \Bigl\lfloor \frac{n-4}{4}\Bigr\rfloor+\sum_{k=1}^{\lfloor \frac{n-5}{6}\rfloor}\Bigl\lfloor \frac{n-6k-1}{4}\Bigr\rfloor + \sum_{k=1}^{\lfloor \frac{n-9}{6}\rfloor}\Bigl\lfloor \frac{n-6k-5}{4}\Bigr\rfloor.
	\end{equation*}
\end{lemma}

Let
\begin{equation*} 
\begin{split}
G^{0}_{e}(n) & := \{\lambda \in P_{e}(n):  \ell_{e}(\lambda)- \ell_{o}(\lambda)=1 \ \text{and} \ a(\lambda) \ \equiv 0 \ ( \text{mod} \ 2) \},\\
\overline{G^{0}_{e}}(n) & := \{\lambda \in G^{0}_{e}(n) : \lambda_{3} \geq 3\},\\
G^{1}_{e}(n) & := \{\lambda \in P_{e}(n):  \ell_{e}(\lambda)- \ell_{o}(\lambda)=1 \ \text{and} \ a(\lambda) \ \equiv 1 \ ( \text{mod} \ 2) \},\\
G^{2}_{e}(n) & := \{\lambda \in P_{e}(n): \ell_{e}(\lambda)- \ell_{o}(\lambda) \geq 2 \},\\
\text{and} \ \ G_{e}(n) & := G^{1}_{e}(n) \ \cup \ G^{2}_{e}(n).
\end{split}
\end{equation*}

We split the set $G_{e}(n)$ into the parity of length of partition as $G_{e}(n) = G_{e,0}(n) \ \cup \ G_{e,1}(n)$ with $G_{e,i}(n) = \{\lambda \in G_{e}(n) : \ell(\lambda) \equiv i \ ( \text{mod} \ 2)\}$ $(i=0,1)$ and let $\overline{G_{e}}(n) := G_{e,0}(n) \ \cup \ G_{e,1}(n) \ \cup \overline{G^{0}_{e}}(n)$. Therefore,
\begin{equation}\label{eq0}
P_{e}(n) \setminus \overline{G_{e}}(n) = \{\lambda \in G^{0}_{e}(n) : 0 \leq \lambda_{3} \leq 2 \}.
\end{equation}

We construct a map $f \colon \overline{G_{e}}(n) \rightarrow P_{o}(n)$ by defining maps $f|_{G_{e,0}(n)} = f_{1}$, $f|_{G_{e,1}(n)} = f_{2}$ and $f|_{\overline{G^{0}_{e}}(n)} = f_{3}$ such that $\{f_{i}\}_{1 \leq i \leq 3}$ are injective with the following properties:  $ f_{1}(G_{e,0}(n)) \ \cap \ f_{2}(G_{e,1}(n))=f_{1}(G_{e,0}(n)) \ \cap \ f_{3}(\overline{G^{0}_{e}}(n)) =  f_{2}(G_{e,1}(n)) \ \cap \ f_{3}(\overline{G^{0}_{e}}(n)) = \emptyset$, so as to conclude that the map $f$ is injective. Then we will choose a subset $\overline{P_{o}}(n) \ \varsubsetneq P_{o}(n) \setminus f(\overline{G_{e}}(n))$ with $|\overline{P_{o}}(n)|>|P_{e}(n) \setminus \overline{G_{e}}(n)|$.

Let $\lambda \in G_{e,0}(n)$ with $\lambda_{e} = (\lambda_{e_{1}},\dots,\lambda_{e_{k}})$  and $\lambda_{o} = (\lambda_{o_{1}},\dots,\lambda_{o_{m}})$ where $k+m = \ell(\lambda)$. Since $\lambda \in G_{e,0}(n)$, $\ell(\lambda) =2r$ for some $r \in \mathbb{Z}_{>0}$ and $k >r$ because, $k-m \geq 1 \ \text{implies} \ 2k \geq k+m+1 = 2r+1$. 

We define $f_{1} \colon G_{e,0}(n) \rightarrow P_{o}(n)$ by $f_{1}(\lambda) := \mu$ with $$\mu_{e} = ((\lambda_{o_{1}}+1),\dots,(\lambda_{o_{m}}+1))$$ and $$\mu_{o} = ((\lambda_{e_{1}}+1),\dots,(\lambda_{e_{k-r}}+1),(\lambda_{e_{k-r+1}}-1),\dots,(\lambda_{e_{k}}-1)).$$ Here we note that $\mu \in P(n)$ and $f_{1}$ reverses the parity of parts; \textit{i.e.}, for $\lambda$ with $k$ even and $m$ odd parts, we get $f_{1}(\lambda) = \mu$ with $k$ odd and $m$ even parts and $\mu \in P_{o}(n)$. Suppose for $\lambda^{'} \neq \lambda^{''} (\in G_{e,0}(n))$ with $\ell(\lambda^{'}) = \ell(\lambda^{''})$, we have $\mu^{'} = f_{1}(\lambda^{'}) = f_{1}(\lambda^{''}) = \mu^{''}$. Then $\ell_{e}(\lambda) = \ell_{e}(\lambda^{''})$ and so, $\ell_{o}(\lambda) = \ell_{o}(\lambda^{''})$. Now, since $\lambda^{'}$ and $\lambda^{''}$ are distinct, by the definition of $f_{1}$ we have at least a tuple $(i,j) \in \mathbb{Z}_{>0} \times \mathbb{Z}_{>0}$ such that $\mu^{'}_{i} \neq \mu^{''}_{j}$. Next, we consider the case when $\lambda^{'} \neq \lambda^{''} (\in G_{e,0}(n))$ with $\ell(\lambda^{'}) \neq \ell(\lambda^{''})$ and it is immediate that $\ell(\mu^{'}) \neq \ell(\mu^{''})$ and therefore, $\mu^{'} \neq \mu^{''}$. So, $f_{1}$ is an injective map.

For $\lambda \in G_{e,1}(n)$ with $\lambda_{e} = (\lambda_{e_{1}},\dots,\lambda_{e_{k}})$  and $\lambda_{o} = (\lambda_{o_{1}},\dots,\lambda_{o_{m}})$ where $k+m = \ell(\lambda) = 2r+1$ for some $r \in \mathbb{Z}_{> 0}$, we note that, $k > r$ and $k-m \geq 3$. But $k-m =1$ holds only when $a(\lambda)$ is odd, because $k = m+1$ and $a(\lambda)$ is even implies that $\lambda \in G^{0}_{e}(n)$. We exclude the condition $k = m+2$ as it contradicts that $k+m = 2r+1$.

 We define $f_{2} \colon G_{e,1}(n) \rightarrow P_{o}(n)$ with $f_{2}(\lambda) := \mu$, where
 \begin{equation}\label{map1}
 \mu_{e} =
  \begin{cases}
 ((\lambda_{o_{1}}+1),\dots,(\lambda_{o_{m}}+1)) \cup (\lambda_{e_{1}}+2) &\quad \text{if $a(\lambda)$ is even},\\
 ((\lambda_{o_{2}}+1),\dots,(\lambda_{o_{m}}+1)) &\quad \text{if $a(\lambda)$ is odd},\\
 \end{cases}
 \end{equation}
and
  \begin{equation}\label{map2}
 \mu_{o} =
 \begin{cases}
((\lambda_{e_{2}}+1),\dots,(\lambda_{e_{k-r-1}}+1),(\lambda_{e_{k-r}}-1),\dots,(\lambda_{e_{k}}-1))  & \!\text{if $a(\lambda)$ is even},\\
((\lambda_{e_{1}}+1),\dots,(\lambda_{e_{k-r-1}}+1),(\lambda_{e_{k-r}}-1),\dots,(\lambda_{e_{k}}-1)) \cup (\lambda_{o_{1}}+2) &\!\text{if $a(\lambda)$ is odd}.\\
 \end{cases}
 \end{equation}
For $a(\lambda)$ even, $\ell_{o}(\mu)-\ell_{e}(\mu) = k-1-(m+1) =k-m-2 \geq 1$ and for $a(\lambda)$ odd, $\ell_{o}(\mu)-\ell_{e}(\mu) = k+1-(m-1) =k-m+2 \geq 3.$ Hence, $\mu \in P_{o}(n)$ and by a similar argument as given above, one can show that $f_{2}$ is injective.

Next, for $\lambda \in \overline{G^{0}_{e}}(n)$ with $\lambda = (\lambda_{1},\dots,\lambda_{\ell})$, we define $f_{3} \colon \overline{G^{0}_{e}}(n) \rightarrow P_{o}(n)$ by  $$f_{3}(\lambda) = \mu = ((\lambda_{1}+1),\lambda_{4},\dots,\lambda_{\ell}) \cup ((\lambda_{2}-2),(\lambda_{3}-2)) \cup (2,1).$$ Independent of whether $\lambda_{2}$ and $\lambda_{3}$ are odd or even, we can observe that $\ell_{o}(\mu)-\ell_{e}(\mu) = 1$ and $a(\mu) = \lambda_{1}+1$ is odd. By definition, $f_{3}$ is an injective map. 

We now show that the images of $\{f_{i}\}_{1 \leq i \leq 3}$ are mutually disjoint by considering the following cases
\begin{enumerate}
	\item  By definition of the maps given above, $f_{1}(G_{e,0}(n)) \varsubsetneq P^{0}_{o}(n)$ where $$P^{0}_{o}(n) := \{\mu \in P_{o}(n) : \ell(\mu) \ \equiv 0 \ ( \text{mod} \ 2) \}$$ and $f_{2}(G_{e,1}(n)) \varsubsetneq P^{1}_{o}(n)$ with $$P^{1}_{o}(n) := \{\mu \in P_{o}(n) : \ell(\mu) \ \equiv 1 \ ( \text{mod} \ 2) \}.$$ So, $f_{1}(G_{e,0}(n)) \ \cap \ f_{2}(G_{e,1}(n)) = \emptyset$.
	\item For $\lambda \in G_{e,0}(n)$ with $\ell_{e}(\lambda)-\ell_{o}(\lambda) \geq 2$, we have $f_{1}(\lambda) = \mu \in P_{o}(n)$ with $\ell_{o}(\mu)-\ell_{e}(\mu) \geq 2$ and for $\lambda \in G_{e,0}(n)$ with $\ell_{e}(\lambda)-\ell_{o}(\lambda) = 1$, $f_{1}(\lambda) = \mu \in P_{o}(n)$ with $\ell_{o}(\mu)-\ell_{e}(\mu) = 1$ but then $a(\mu) = \lambda_{1}+1$ is even. Considering $\lambda \in \overline{G^{0}_{e}}(n)$, $f_{3}(\lambda) = \mu \in P_{o}(n)$ with $a(\mu) = \mu_{1}$ is odd and $\ell_{o}(\mu)-\ell_{e}(\mu) = 1$. Therefore, $f_{1}( G_{e,0}(n)) \ \cap \ f_{3}(\overline{G^{0}_{e}}(n)) = \emptyset$.
	\item Let us consider $\lambda \in G_{e,1}(n)$ with $a(\lambda)$ even and $\ell_{e}(\lambda)-\ell_{o}(\lambda)=3$. Then $f_{2}(\lambda) = \mu \in P_{o}(n)$ with $a(\mu)$ even and $\ell_{o}(\mu)-\ell_{e}(\mu)=1$. For $\ell_{e}(\lambda)-\ell_{o}(\lambda) \geq 4$, we have $f_{2}(\lambda) = \mu$ with $\ell_{o}(\mu)-\ell_{e}(\mu) \geq 2$. Moreover, if $a(\lambda)$ is odd then it is immediate that $\ell_{o}(\mu)-\ell_{e}(\mu) \geq 3$ and consequently, $f_{2}( G_{e,1}(n)) \ \cap \ f_{3}(\overline{G^{0}_{e}}(n)) = \emptyset$. So, the map $f \colon \overline{G_{e}}(n) \rightarrow P_{o}(n)$ is injective.
\end{enumerate}
For $\mu \in P_{o}(n)$ with its odd component $\mu_{o} =(\mu_{o_{1}},\dots,\mu_{o_{s}})$, we define
\begin{equation*}
\overline{P_{o}}(n) := \{\mu \in P_{o}(n) : \ell_{e}(\mu) =2 \ \text{and} \ \mu_{o_{i}}=1 \ \text{for all} \ 1 \leq i \leq s \}.
\end{equation*}
By the definition of $f$, it is clear that $\overline{P_{o}}(n) \varsubsetneq P_{o}(n) \setminus f(\overline{G_{e}}(n))$. Now, it remains to show that $|\overline{P_{o}}(n)| > |P_{e}(n) \setminus \overline{G_{e}}(n)|$.

For $n$ even and for $\lambda \in \overline{P_{o}}(n)$, we have $\ell_{o}(\lambda) = 2k+2$ for some $k \in \mathbb{Z}_{>0}$. Here we observe that
\begin{equation}\label{card1}
|\{\lambda \in P(n) : \lambda_{1}+\lambda_{2}=n; \lambda_{1}, \lambda_{2} \ \text{both even}                     \}|=\Bigl\lfloor\frac{n}{4}\Bigr\rfloor
\end{equation}
and 
\begin{equation}\label{card2}
|\{\lambda \in P(n) : \lambda_{1}+\lambda_{2}=n; \lambda_{1} \ \text{even}, \lambda_{2} \ \text{odd and} \ \lambda_{2} \in \mathbb{Z}_{\geq 3}                     \}|=\Bigl\lfloor\frac{n-3}{4}\Bigr\rfloor.
\end{equation}
Since, $\lambda \in \overline{P_{o}}(n)$ with $n$ an even positive integer and $\ell_{o}(\lambda) = 2k+2$, for each $k \in \mathbb{Z}_{>0}$, then by \eqref{card1}, 
\begin{equation}\label{eq1}
|\{\lambda \in \overline{P_{o}}(n) : \lambda_{1}+\lambda_{2}+(2k+2) \times 1=n; \lambda_{1}, \lambda_{2} \ \text{both even}                     \}|=\Bigl\lfloor\frac{n-2k-2}{4}\Bigr\rfloor
\end{equation}
and $1\leq k \leq \dfrac{n-6}{2}$ because $k$ maximizes only when both $\lambda_{1}$ and $\lambda_{2}$ minimizes; \textit{i.e.}, only the instance $2+2+(2k+2)\times1=n$ which implies $k = \dfrac{n-6}{2}$. Therefore we have,
\begin{equation}\label{eq2}
|\overline{P_{o}}(n)| = \sum_{k=1}^{\frac{n-6}{2}}\Bigl\lfloor\frac{n-2k-2}{4}\Bigr\rfloor.
\end{equation}

Similarly, for $n$ odd, we have $\ell_{o}(\lambda) = 2k+1$ with $1 \leq k \leq \dfrac{n-5}{2}$ and 
\begin{equation}\label{eq3}
|\{\lambda \in \overline{P_{o}}(n) : \lambda_{1}+\lambda_{2}+(2k+1) \times 1=n; \lambda_{1}, \lambda_{2} \ \text{even}                     \}|=\Bigl\lfloor\frac{n-2k-1}{4}\Bigr\rfloor.
\end{equation}
Consequently, 
\begin{equation}\label{eq4}
|\overline{P_{o}}(n)| = \sum_{k=1}^{\frac{n-5}{2}}\Bigl\lfloor\frac{n-2k-1}{4}\Bigr\rfloor.
\end{equation}

Now for $n$ even we will show that
\begin{equation}\label{eq5}
|P_{e}(n) \setminus \overline{G_{e}}(n)| = 1 + \sum_{k=1}^{\lfloor \frac{n-2}{6}\rfloor}\Bigl\lfloor \frac{n-6k+2}{4}\Bigr\rfloor + \sum_{k=1}^{\lfloor \frac{n-6}{6}\rfloor}\Bigl\lfloor \frac{n-6k-2}{4}\Bigr\rfloor.
\end{equation}
We interpret the set $P_{e}(n) \setminus \overline{G_{e}}(n)$ as a disjoint union of its three proper subsets given by\\
$P_{e}(n) \setminus \overline{G_{e}}(n) = A_{1} \cup A_{2} \cup A_{3}$ where,
\begin{equation*}
A_{1} = \{\lambda \in P_{e}(n) \setminus \overline{G_{e}}(n): 0 \leq \lambda_{3} \leq 1\}, \quad A_{2} = \bigcup_{k \geq 1} A_{2,k}, \quad \text{and} \quad A_{3} = \bigcup_{k \geq 1} A_{3,k},
\end{equation*}
with
\begin{multline*}
A_{2,k} = \{\lambda=(\lambda_1,\lambda_2,2,\dots,2,1,\dots,1)\vdash n: \lambda_1\ \text{and}\ \lambda_2\ \text{even}, \\  \mult_{\lambda}(2)=2k-1, \mult_{\lambda}(1)=2k \},
\end{multline*}
and
\begin{multline*}
A_{3,k} = \{\lambda=(\lambda_1,\lambda_2,2,\dots,2,1,\dots,1)\vdash n: \lambda_1\ \text{even},\lambda_2\ \text{odd},\\   \mult_{\lambda}(2)=2k, \mult_{\lambda}(1)=2k-1\}.
\end{multline*}

Next, we explicitly describe the sets and derive their cardinality by separating into three cases.

\textit{Case 1(E):} We observe that $|A_{1}|=1$ since we have only one possibility $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},0,0)$. We reject the other three possibilities; \textit{i.e.}, $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},0,1)$ as $\lambda_{2} \geq \lambda_{3}$, $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},1,0)$ as $n$ even and $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},1,1)$ as $\lambda \in P_{e}(n) \setminus \overline{G_{e}}(n)$. Next, we look at the subset of $A_1$, say $A_{1,\geq 2}:=\{\lambda\in A_1: \lambda_2 \geq 2\}$ and note that $A_{1,\geq 2}=\emptyset$. This is because for $\lambda \in A_{1,\geq 2}$, there are altogether four possibilities for $\lambda_3 \in \{0,1\}$.

For $\lambda_3=0$, the choice $(\lambda_1,\lambda_2,\lambda_3)=(\lambda_{1},\lambda_{2},0)$ and $\lambda_2$ is even is impossible as $\lambda \in P_{e}(n) \setminus \overline{G_{e}}(n)$ and if $\lambda_2$ is odd, is again an impossible option since $n$ is even. Whereas for $\lambda_3=1$, the choice $(\lambda_1,\lambda_2,\lambda_3)=(\lambda_{1},\lambda_{2},1)$ and $\lambda_2$ is even is impossible as $n$ is even and if $\lambda_2$ is odd, again an impossible option since $\lambda \in P_{e}(n) \setminus \overline{G_{e}}(n)$.

\textit{Case 2(E):} By \eqref{card1},
\begin{equation}\label{eq6}
|A_{2,k}|=\Bigl \lfloor\frac{n-6k+2}{4}\Bigr\rfloor
\end{equation}
and $1\leq k\leq\lfloor\frac{n-2}{6}\rfloor$ because $k$ maximizes only when both $\lambda_{1} \ \text{and} \ \lambda_{2}$ 
minimizes; \textit{i.e.}, the instance $2+2+(2k-1)\times 2+(2k)\times 1=n$ which implies $k \leq \lfloor\frac{n-2}{6}\rfloor$. By \eqref{eq6},
\begin{equation}\label{req1}
A_{2}=\bigcup_{k=1}^{\lfloor\frac{n-2}{6}\rfloor} A_{2,k} \ \text{and} \ |A_{2}|=\sum_{k=1}^{\lfloor\frac{n-2}{6}\rfloor} \Bigl \lfloor\frac{n-6k+2}{4}\Bigr\rfloor.
\end{equation} 

\textit{Case 3(E):} From \eqref{card2}, it follows that
\begin{equation}\label{eq7}
|A_{3,k}|=\Bigl \lfloor\frac{(n-6k+1)-3}{4}\Bigr\rfloor=\Bigl \lfloor\frac{n-6k-2}{4}\Bigr\rfloor
\end{equation}
and $1\leq k\leq\lfloor\frac{n-6}{6}\rfloor$ because $k$ maximizes only when both $\lambda_{1} \ \text{and} \ \lambda_{2}$ 
minimizes; \textit{i.e.}, the instance $4+3+(2k)\times 2+(2k-1)\times 1=n$ which implies $k \leq \lfloor\frac{n-6}{6}\rfloor$. By \eqref{eq7},
\begin{equation}\label{req2}
A_{3}=\bigcup_{k=1}^{\lfloor\frac{n-6}{6}\rfloor} A_{3,k} \ \text{and} \ |A_{3}|=\sum_{k=1}^{\lfloor\frac{n-6}{6}\rfloor} \Bigl \lfloor\frac{n-6k-2}{4}\Bigr\rfloor.
\end{equation}
By \textit{Case 1(E)}, \eqref{req1} and \eqref{req2} we have \eqref{eq5}.

For all odd integers, $n$ with $n\geq 9$, we will show that 
\begin{equation}\label{eq8}
|P_{e}(n) \setminus \overline{G_{e}}(n)| = 1+ \Bigl\lfloor \frac{n-4}{4}\Bigr\rfloor+ \sum_{k=1}^{\lfloor \frac{n-5}{6}\rfloor}\Bigl\lfloor \frac{n-6k-1}{4}\Bigr\rfloor + \sum_{k=1}^{\lfloor \frac{n-9}{6}\rfloor}\Bigl\lfloor \frac{n-6k-5}{4}\Bigr\rfloor.
\end{equation}
Similarly as before, we write $P_{e}(n) \setminus \overline{G_{e}}(n)$ as a disjoint union of its four proper subsets given by
$P_{e}(n) \setminus \overline{G_{e}}(n) =B_{0} \cup B_{1} \cup B_{2} \cup B_{3}$ where,
\begin{equation*}
\begin{split}
B_{0} &= \{\lambda=(\lambda_1,\lambda_2,1) \in P_{e}(n) \setminus \overline{G_{e}}(n):\lambda_{2} \geq 4 \}\\
B_{1} & = \{\lambda \in P_{e}(n) \setminus \overline{G_{e}}(n): 0 \leq \lambda_{2} \leq 2 \ \text{and} \ 0 \leq \lambda_{3} \leq 1 \}\\
B_{2} & = \bigcup_{k \geq 1} B_{2,k}, \quad \text{and} \quad B_{3}  = \bigcup_{k \geq 1} B_{3,k},
\end{split}
\end{equation*}
with
\begin{multline*}
B_{2,k}  = \{\lambda=(\lambda_1,\lambda_2,2,\dots,2,1,\dots,1)\vdash n: \lambda_1\ \text{and}\ \lambda_2\ \text{even},  \\ \mult_{\lambda}(2)=2k, \mult_{\lambda}(1)=2k+1 \},
\end{multline*}
\begin{multline*}
B_{3,k}  = \{\lambda=(\lambda_1,\lambda_2,2,\dots,2,1,\dots,1)\vdash n: \lambda_1\ \text{even},\lambda_2\ \text{odd},  \\ \mult_{\lambda}(2)=2k+1, \mult_{\lambda}(1)=2k\}.
\end{multline*}

\textit{Case 1(O):} For $\lambda= (\lambda_1,\lambda_2,1) \in B_0$ and $n$ odd, it follows that both $\lambda_1$ and $\lambda_2$ are even. Therefore minimal choice for $n$ is $9$, otherwise $\lambda_1 \geq \lambda_2 \geq 4$ with the constraint that both $\lambda_1$ and $\lambda_{2}$ even would be an impossibility. Moreover, we can observe that 
\begin{equation}\label{new0}
|B_0| =
\Bigl\lfloor \frac{n-4}{4}\Bigr\rfloor.
\end{equation}

\textit{Case 2(O):} We observe that $|B_{1}|=1$ since we have only one possibility $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},2,1)$. We reject the other three possibilities; \textit{i.e.}, $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},0,1)$ as $\lambda_{2} \geq \lambda_{3}$, $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},0,0)$ and $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},2,0)$ as $n$ odd, $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},1,0)$ and $(\lambda_{1},\lambda_{2},\lambda_{3})=(\lambda_{1},1,1)$ as $\lambda \in P_{e}(n) \setminus \overline{G_{e}}(n)$.

\textit{Case 3(O):} By \eqref{card1},
\begin{equation}\label{eq9}
|B_{2,k}|=\Bigl \lfloor\frac{n-6k-1}{4}\Bigr\rfloor
\end{equation}
and $1\leq k\leq\lfloor\frac{n-5}{6}\rfloor$ because $k$ maximizes only when both $\lambda_{1} \ \text{and} \ \lambda_{2}$ 
minimizes; \textit{i.e.}, the instance $2+2+(2k)\times 2+(2k+1)\times 1=n$ which implies $k \leq \lfloor\frac{n-5}{6}\rfloor$. By \eqref{eq9},
\begin{equation}\label{req3}
B_{2}=\bigcup_{k=1}^{\lfloor\frac{n-5}{6}\rfloor} B_{2,k} \ \text{and} \ |B_{2}|=\sum_{k=1}^{\lfloor\frac{n-5}{6}\rfloor} \Bigl \lfloor\frac{n-6k-1}{4}\Bigr\rfloor.
\end{equation}

\textit{Case 4(O):} From \eqref{card2}, it follows that
\begin{equation}\label{eq10}
|B_{3,k}|=\Bigl \lfloor\frac{(n-6k-2)-3}{4}\Bigr\rfloor=\Bigl \lfloor\frac{n-6k-5}{4}\Bigr\rfloor
\end{equation}
and $1\leq k\leq\lfloor\frac{n-9}{6}\rfloor$ because $k$ maximizes only when both $\lambda_{1} \ \text{and} \ \lambda_{2}$ 
minimizes; \textit{i.e.}, the instance $4+3+(2k+1)\times 2+(2k)\times 1=n$ which implies $k \leq \lfloor\frac{n-9}{6}\rfloor$. By \eqref{eq10},
\begin{equation}\label{req4}
B_{3}=\bigcup_{k=1}^{\lfloor\frac{n-9}{6}\rfloor} B_{3,k} \ \text{and} \ |B_{3}|=\sum_{k=1}^{\lfloor\frac{n-9}{6}\rfloor} \Bigl \lfloor\frac{n-6k-5}{4}\Bigr\rfloor.
\end{equation}
By \textit{Case 2(O)}, \eqref{new0}, \eqref{req3} and \eqref{req4}, we have \eqref{eq8}.

Therefore, by Lemma \ref{lem1}, $|\overline{P_{o}}(n)| > |P_{e}(n) \setminus \overline{G_{e}}(n)|$ for all $n \geq \mathbb{Z}_{\geq 14} \cup \{11,13\}$. To conclude the proof, it remains to check $n \in \{1,3,4,5,6,7,8,9,10,12\}$ which we did by numerically checking in \Mathematica.

\section{Proof of Theorem \ref{thm:distinct}}\label{sec:distinct}


Following the definitions in Section \ref{sec:proof-kkl}, set
\begin{equation*}
\begin{split}
H^0_e(n) &:= G^0_e(n) \cap D(n),\\
\overline{H^{0}_{e}}(n) &:= \{\lambda \in H^0_e(n): \ell_o(\lambda)>1\},\\
\text{and}\ \ H_e(n)&:= G_e(n) \cap D(n).
\end{split}
\end{equation*}
We split $H_e(n)$ into $H_{e,0}(n)=G_{e,0}(n) \cap D(n)$ and $H_{e,1}(n)=G_{e,1}(n) \cap D(n)$. Similarly, define the map $f \colon H_e(n) \rightarrow D_o(n)$ by $f|_{H_{e,0}(n)}=f_1$ and $f|_{H_{e,1}(n)}=f_2$. Since $H_e(n) \subsetneq G_e(n)$, we conclude that the map $f$ is injective by \eqref{map1} and \eqref{map2}.  

Now we show that $d_o(n) - |f(H_e(n))|>d_e(n) - |H_e(n)|$ for all $n>31$. The subset $D_o(n) \setminus f(H_e(n))$ contains different classes of partitions, one of which is 
$$\overline{D_{o}}(n):= \{\lambda \in D_o(n) \setminus f(H_e(n)) : \ell_o(\lambda)-\ell_e(\lambda)=1\ \text{and}\ a(\lambda) \equiv 1  \pmod 2\}.$$ We note that $\overline{D_{o}}(n)$ may contain other classes of partitions depending on whether $n$ is even or odd.

For a partition $\lambda \in \overline{H^{0}_{e}}(n)$, we split $\lambda$ into its even component $\lambda_e=(\lambda_{e_{1}},\lambda_{e_{2}},\dots,\lambda_{e_{m+1}})$ and odd component $\lambda_o=(\lambda_{o_{1}},\lambda_{o_{2}},\dots,\lambda_{o_{m}})$ for some $m \in \mathbb{Z}_{\geq 2}$. Now, me make a transformation of $\lambda$ into $\lambda^\ast$ with
\begin{equation*}
\lambda^\ast = (\lambda_{e_{1}}+\lambda_{o_{1}},\lambda_{e_{2}}+\lambda_{o_{2}}) \cup (\lambda_{e_{3}},\lambda_{e_{4}},\dots,\lambda_{e_{m+1}}) \cup (\lambda_{o_{3}},\lambda_{o_{4}},\dots,\lambda_{o_{m}}) \in \overline{D_{o}}(n).
\end{equation*} 
We observe that two partitions, say $\lambda, \overline{\lambda} \in \overline{H^{0}_{e}}(n)$, where
\begin{equation*}
\begin{split}
\lambda &= (\lambda_{e_{1}},\lambda_{e_{2}},\dots,\lambda_{e_{m+1}}) \cup (\lambda_{o_{1}},\lambda_{o_{2}},\dots,\lambda_{o_{m}})\\
\text{and}\ \ \overline{\lambda} &= (\overline{\lambda}_{e_1},\overline{\lambda}_{e_2},\dots,\overline{\lambda}_{e_{m+1}}) \cup (\overline{\lambda}_{o_1},\overline{\lambda}_{o_2},\dots,\overline{\lambda}_{o_m}),
\end{split}
\end{equation*}
transform to a same partition, say $\mu \in \overline{D_{o}}(n)$ if and only if
\begin{equation*}
\lambda_{e_{1}}-\overline{\lambda}_{e_{1}} = \overline{\lambda}_{o_{1}}-\lambda_{o_{1}} \equiv 0 \pmod 2\ \ \text{or}/ \text{and}\ \ \lambda_{e_{2}}-\overline{\lambda}_{e_{2}} = \overline{\lambda}_{o_{2}}-\lambda_{o_{2}} \equiv 0  \pmod 2.
\end{equation*}
If those cases arise we subtract some multiple of $2$ from the greatest part of the resultant partition and add the multiple of $2$ to the other even parts which are present in the partition, and continue this process till we have a repetition among the parts of the partition. This process is injective by its definition, and we denote it by $g$. For example, consider partitions $\lambda=(12,10,6,2)\cup(7,3,1)$ and $\overline{\lambda}=(10,8,6,2)\cup(9,5,1)$ in $\overline{H^{0}_{e}}(41)$, then both $\lambda$ and $\overline{\lambda}$ maps to the same partition $\mu=(19,13,6,2,1) \in \overline{D_{o}}(41)$. Consequently, by the process $g$, finally $\lambda\mapsto (19,13,6,2,1)$, whereas $\overline{\lambda} \mapsto (17,13,8,2,1)$. As a trivial remark, $\overline{H^{0}_{e}}(n) = \emptyset$ for all positive even integers $n \leq 14$, since $6+4+2+3+1=16$ is the least possible option. 

Depending on the parity of $n$, it remains to analyze the left over set
\begin{equation}\label{leftoverset}
\widetilde{H^0_e}(n):= \{\lambda \in H^0_e(n) : \ell_{e}(\lambda)-\ell_{o}(\lambda)=1, \ell_{o}(\lambda)\leq 1\ \text{and}\ a(\lambda) \equiv 0 (\text{mod}\ 2)\},
\end{equation}
which is unmapped yet (after applying the map $f$ and $g$).

For $n$ to be an even positive integer, we observe that $\widetilde{H^0_e}(n)$ consists of only one partition $(n)$. An even integer $n$ can be expressed as a sum of two consecutive odd integers if and only if $n$ is divisible by 4. If $n$ is divisible by 4, then for some definite odd integer $\lambda_{o_1}$ we get $(\lambda_{o_1},\lambda_{o_1}-2)\in D_o(n)$, which is not mapped yet. So we map $(n)$ to $(\lambda_{o_1},\lambda_{o_1}-2)$. If $n$ is not divisible by 4, then for some definite odd integer $\lambda_{o_1}$ we get $(\lambda_{o_1},\lambda_{o_1}-2,2)\in D_o(n)$, which is not mapped yet. So in this case we map $(n)$ to $(\lambda_{o_1},\lambda_{o_1}-2,2)$. Therefore, by some elementary observations we get that the theorem is true for all even integer $n>6$.

Let $n$ be odd. We rewrite \eqref{leftoverset} as
\begin{equation*}
\widetilde{H^0_e}(n):= \{\lambda \in H^0_e(n) : \ell_{e}(\lambda)=2, \ell_{o}(\lambda)= 1\ \text{and}\ a(\lambda) \equiv 0 (\text{mod}\ 2)\}.
\end{equation*}
Write a partition $\lambda \in \widetilde{H^0_e}(n)$ into its even component $\lambda_e=(\lambda_{e_1},\lambda_{e_{2}})$ and odd component $\lambda_o=(\lambda_{o_1})$. We split $\widetilde{H^0_e}(n)$ into following three classes:
\begin{enumerate}
	\item $\widetilde{H}^0_{e,1}(n):= \{\lambda \in \widetilde{H^0_e}(n): \lambda_{e_2}=2\}$,
	\item $\widetilde{H}^0_{e,2}(n):= \{\lambda \in \widetilde{H^0_e}(n): \lambda_{e_2}\geq 6\}$, and
	\item $\widetilde{H}^0_{e,3}(n):= \{\lambda \in \widetilde{H^0_e}(n): \lambda_{e_2}= 4\}$.
\end{enumerate}
Now we consider the following three classes of partitions from the set of partitions, say $\widetilde{D_o}(n) \subsetneq D_o(n)$ which have no preimage yet:
\begin{enumerate}
	\item $\widetilde{D}_{o,1}(n):= \{\pi \in \widetilde{D_o}(n) : \ell(\pi)=4\ \text{and}\ \pi_{o_1}-\pi_{o_2}=2 \}$,
	\item $\widetilde{D}_{o,2}(n):= \{\pi \in \widetilde{D_o}(n) : \ell_o(\pi)=3\ \text{and}\ \pi_{o_1}-\pi_{o_2}=2 \}$, and
	\item $\widetilde{D}_{o,3}(n):= \{\pi \in \widetilde{D_o}(n) : \ell_o(\pi)-\ell_{e}(\pi)=1, a(\pi)\equiv 0 (\text{mod}\ 2)\ \text{and}\ \pi_{e_1}-\pi_{o_1}=1\ \text{or}\ 3 \}$.
\end{enumerate}
Now we construct an injective map from $\widetilde{H}^0_{e,1}(n)$ to $\widetilde{D}_{o,1}(n)$. Let $\lambda=(\lambda_{e_1},2)\cup (\lambda_{o_1}) \in \widetilde{H}^0_{e,1}(n)$. Define a transformation $S$ such that $S(\lambda)=(\lambda_{e_1})\cup(\lambda_{o_1}+1,1)$. Now define $S^\ast$ such that
\begin{equation*}
S^\ast(S(\lambda))=
\begin{cases}
S(\lambda) & \quad \text{if $\lambda_{e_1} \equiv 0 (\text{mod}\ 4)$},\\
(\lambda_{e_1}-2)\cup (\lambda_{o_1}+1,3)& \quad \text{if $\lambda_{e_1} \equiv 2 (\text{mod}\ 4)$.}
\end{cases}	
\end{equation*}
Now define $S^{\ast\ast}$ such that $S^{\ast\ast}(S^\ast(S(\lambda)))=(\lambda_{o_2},\lambda_{o_2}-2)\cup(\lambda_{o_3})\cup(\lambda_{o_1}+1)$, where $\lambda_{o_2}+\lambda_{o_2}-2=\lambda_{e_1} \text{~or~} \lambda_{e_1}-2$, and $\lambda_{o_3}=1 \text{~or~} 3$ accordingly. For example: $(24,5,2)$ maps to $(13,11,6,1)$ and $(22,7,2)$ maps to $(11,9,8,3)$. This process is injective.

Our next objective is to embed the set $\widetilde{H}^0_{e,2}(n)$ into a subset of $\widetilde{D_o}(n)$ which is not mapped till now. Define a transformation $U$ such that $U(\lambda)=((\lambda_{e_1-3},3)\cup (\lambda_{o_{1}}))\cup(\lambda_{e_{2}},2)$ for $\lambda \in \widetilde{H}^0_{e,2}(n)$. Associated with $U$, let us define $U^\ast$ in such a way that
\begin{equation*}
U^\ast(U(\lambda))=
\begin{cases}
U(\lambda) &\text{if}~ \lambda_{o_1}\neq3,  \lambda_{e_1}-3\neq \lambda_{o_1},\\
(\lambda_{e_1}-3,5,1)\cup(\lambda_{e_2}-2,2) &\text{if}~ \lambda_{o_1}=3,\\
\bigl(\text{this transformation is impossible for}\ n \leq 17\bigr)\\
\bigl(\text{\textit{e.g.}}\ (10,6,3)\mapsto (7,5,4,2,1)\bigr)\\
((\lambda_{e_1}-3,\lambda_{o_1}-2)\cup (5))\cup(\lambda_{e_2}-2,2) &\text{if}~ \lambda_{e_1}-3=\lambda_{o_1},\\
\bigl(\text{this transformation is impossible for}\ n \leq 25\bigr) \\
\bigl(\text{\textit{e.g.}}\ (12,9,6)\mapsto (9,7,5,4,2)\bigr)\\
((\lambda_{e_1}-3,\lambda_{o_1}-4)\cup(5))\cup(\lambda_{e_2}-2,4) &\text{if}~ \lambda_{e_1}-1=\lambda_{o_1}, \lambda_{e_2}\neq6,\\
\bigl(\text{this transformation is impossible for}\ n \leq 29\bigr)\\
\bigl(\text{e.g.}\ (12,11,8)\mapsto (9,7,6,5,4)\bigr)\\
((\lambda_{e_1}-3,\lambda_{o_1}-4)\cup(3))\cup(6,4) &\text{if}~ \lambda_{e_1}-1=\lambda_{o_1}, \lambda_{e_2}=6.\\
\bigl(\text{this transformation is impossible for}\ n \leq 23 \bigr)\\
\bigl(\text{\textit{e.g.}}\ (10,9,6)\mapsto (7,6,5,4,3)\bigr)
\end{cases}
\end{equation*}
Denote the resulting transformation $U^\ast U$ by $\widetilde{U}$. Note that $\ell(\widetilde{U}(\lambda))=5$, where the resulting partitions; \textit{i.e.}, images under $\widetilde{U}$, contains parts from $\{3,5\}$ and its smallest even part from $\{2,4\}$. For a partition $\lambda \in H^0_e(n)$ with $\ell(\lambda)=7$, $\ell(g(\lambda))=5$. Now, assume $\widetilde{U}(\lambda)=g(\mu)$ for some partition $\mu$ with $\ell(\mu)=7$. In the map $g$, after applying the first transformation, the other transformations are applied on the even parts only (if it is necessary). If $\widetilde{U}(\lambda)=g(\mu)$, then we remove one of $2$, $3$, $4$, or $5$ (which may exist in $g(\mu)$) from $g(\mu)$ by applying a similar transformation.

Now we compare the number of partitions in $\widetilde{H}^0_{e,3}(n)$ with $\widetilde{D}_{o,2}(n)$ and $\widetilde{D}_{o,3}(n)$,$$\bigl|\widetilde{H}^0_{e,3}(n)\bigr|=\left\lfloor\frac{n-3}{4}\right\rfloor.$$ Let $\pi=(\pi_{o_1},\pi_{o_1}-2,\pi_{o_3}) \in \widetilde{D}_{o,2}(n)$, and $\pi_{o_1}=2k+1$. Then the least possible value of $k$ is given by $(2k+1)+(2k-1)\geq2\{n-(2k+1)-(2k-1)\}+6,$ which implies $k=\big\lceil\frac{n+3}{6}\big\rceil$. So the total number of partitions in $\widetilde{D}_{o,2}(n)$ is at least $\big\lceil\frac{n-4k}{4}\big\rceil$, which is equal to $\big\lceil\frac{n}{4}-\big\lceil\frac{n+3}{6}\big\rceil\big\rceil$. Any odd integer $n$ is of the form $12\ell+r$, where $r=1,3,5,7,9,~\text{or}~11$. Calculating for all the six cases, we get the total number of partitions in this class is
\begin{equation*}
=\begin{cases}
\left\lfloor\frac{n}{12}\right\rfloor & \text{if } n\neq12\ell+9,\\
\left\lfloor\frac{n}{12}\right\rfloor+1 & \text{if } n=12\ell+9.
\end{cases}
\end{equation*}
Now, $\left\lfloor\frac{n-3}{4}\right\rfloor-\begin{cases}
\left\lfloor\frac{n}{12}\right\rfloor & \text{if } n\neq12\ell+9\\
\left\lfloor\frac{n}{12}\right\rfloor+1 & \text{if } n=12\ell+9
\end{cases}=\left\lfloor\frac{n}{6}\right\rfloor-1,\left\lfloor\frac{n}{6}\right\rfloor, ~\text{or}~ \left\lfloor\frac{n}{6}\right\rfloor+1.$

It remains to estimate a lower bound of $\bigl|\widetilde{D}_{o,3}(n)\bigr|$. Let $\pi=(\pi_{o_1},\pi_{o_2},\pi_{o_3})\cup(\pi_{e_1},\pi_{e_2}) \in \widetilde{D}_{o,3}(n)$. The greatest odd part $\pi_{o_1}$ is the largest possible value if $\pi$ contains $3$, $1$ and $2$. So the largest possible value of $\pi_{o_1}$ is $\dfrac{n-9}{2}$ or $\dfrac{n-7}{2}$ if $n\equiv3 \pmod 4$ or $n\equiv1 \pmod 4$ respectively. The smallest possible value of $\pi_{o_1}$ is greater than $\left\lfloor\frac{n}{5}\right\rfloor$. When $\pi_{o_1}$ is not the largest or the smallest possible value, then for each possible value of $\pi_{o_1}$, we get at least 6 partitions. So total number of partitions is greater than
\begin{equation*}
 6\times\frac{1}{2}\times\bigg\{\frac{n-9}{2}-1-\bigg(\bigg\lfloor\frac{n}{5}\bigg\rfloor+1+1\bigg)\bigg\}\geq3\times\bigg\{\frac{n-9}{2}-\frac{n}{5}-3\bigg\} =\frac{9(n-25)}{10}.
\end{equation*}
Now, $\left\lfloor\frac{9(n-25)}{10}\right\rfloor > \left\lfloor\frac{n}{6}\right\rfloor+1$ for all odd positive integers $n>31$.

This proves the theorem for all $n>31$. We can verify the result numerically for $19<n\leq31$.


\section{Other Results}\label{sec:rem}

Our method can be amended to prove other results where biases in parity are found for restricted partitions. If $q_{o}(n)$ (resp.\ $q_{e}(n)$) denotes the number of partitions of $n$ with more odd parts (resp.\ even parts) than even parts (resp.\ odd parts) where the smallest part is at least $2$, then we have the following theorem.

\begin{theorem}\label{thm:reverse}
	For all positive integers $n>7$, we have
	\begin{equation*}
		q_o(n)<q_e(n).
	\end{equation*}
\end{theorem}

 These parity biases seems to also occur for more restricted partition functions. For a nonempty set $S \subsetneq \mathbb{Z}_{\geq 0}$, define $P_e^S(n):=\{\lambda \in P_e(n): \lambda_i \notin S\}$ and $P_o^S(n):=\{\lambda \in P_o(n): \lambda_i \notin S\}$. Consequently, denote the number of partitions in $P_e^S(n)$ (resp.\ $P_o^S(n)$) by $p_e^S(n)$ (resp.\ $p_o^S(n)$).
The above definition leads us to the following results that describes not only the parity of parts but also its arithmetic by putting a constraint on its support.

\begin{theorem}\label{thm:missingset1}
	For all $n\geq1$ we have $$p_o^{\{2\}}(n)>p_e^{\{2\}}(n).$$
\end{theorem}


\begin{theorem}\label{thm:missingset2}
	If $S=\{1,2\}$, then for all integers $n>8$, we have
	$$p^{S}_o(n)>p^{S}_e(n).$$
\end{theorem}


\acknowledgements{The first author would like to acknowledge that the research was funded by the Austrian Science Fund (FWF): W1214-N15, project DK6. The last author is partially supported by the Leverhulme Trust Research Project Grant RPG-2019-083 and thanks Prof.\ Michael Schlosser (Vienna) for suggesting to look at parity biases in partitions with restrictions on the parts.}

%% if you use biblatex then this generates the bibliography
%% if you use some other method then remove this and do it your own way
\printbibliography



\end{document}

.
