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\title[On the combinatorics of the Schur function expansion of the  Rational Shuffle Conjecture]{Schur Function Expansions and the Rational Shuffle Conjecture}

\author{Dun Qiu\thanks{\href{mailto:duqiu@ucsd.edu}{duqiu@ucsd.edu}} \and Jeffrey Remmel\thanks{\href{mailto:remmel@math.ucsd.edu}{remmel@math.ucsd.edu}}}

\address{Department of Mathematics, University of California San Diego}

\received{\today}

\abstract{Gorsky and Negut introduced operators $Q_{m,n}$ on symmetric functions and conjectured that, in the case where $m$ and $n$ are relatively prime, the expansion of $Q_{m,n}(-1)^n$ in terms of the fundamental quasi-symmetric functions are given by polynomials introduced by Hikita. 
Later Bergeron, Garsia, Leven, and Xin extended and refined the conjectures of Gorsky and Negut to give a combinatorial interpretation of the coefficients that arise in expansion of $Q_{m,n}(-1)^n$ in terms of the fundamental 
quasi-symmetric functions for arbitrary $m$ and $n$ which 
we will call the rational shuffle conjecture. The rational shuffle conjecture was later proved by Mellit in 2016.  The main goal of this paper is 
to study the combinatorics of the coefficients that arise in the Schur function expansion of $Q_{m,n}(-1)^n$ in the case where $m$ or $n$ equals $3$.} 

\resume{Gorsky et Negut op\'erateurs introduit $Q_{m,n}$ sur les fonctions sym\'etriques et l'hypoth\`ese que, dans le cas o\`u $m$ et $n$ sont relativement premier, l'expansion de $Q_{m,n}(-1)^n$ en termes de la quasi-fondamental fonctions sym\'etriques sont donn\'ees par des polyn\^omes introduites par Hikita. Plus tard, Bergeron, Garsia, Leven, et Xin \'etendu et affin\'e les conjectures de Gorsky et Negut pour donner une interpr\'etation combinatoire des coefficients qui surgissent dans l'expansion de $Q_{m,n}(-1)^n$ en termes des fonctions sym\'etriques quasi arbitraire pour $m$ et $n$ que nous appellerons la conjecture rationnelle shuffle. Le rationnel shuffle conjecture a \'et\'e prouv\'ee par la suite de Mellit en 2016. L'objectif principal de cet article est d'\'etudier la combinatoire des coefficients qui surgissent dans l'expansion de la fonction Schur $Q_{m,n}(-1)^n$ dans le cas o\`u $m$ ou $n$ est \'egale \`a $3$.}

\keywords{Macdonald polynomials, parking functions, Dyck paths, Shuffle Conjecture}


\begin{document}

\maketitle

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

The rational shuffle conjecture as formulated by Gorsky and Negut \cite{GN} and Bergeron, Garsia, Leven, and Xin \cite{BGLX} gives a combinatorial interpretation of 
the coefficients that arise in the fundamental quasi-symmetric function expansion of certain operators on symmetric functions $Q_{m,n}$ applied to $(-1)^n$. 
This conjecture was proved by Mellit \cite{Mellit}. Leven gave a combinatorial proof of the Schur function expansion of $Q_{2,2n+1}(-1)$ and $Q_{2n+1,2}1$ in \cite{Em}, and the coefficient at $s_{1^n}$ in $\Qmn{m,n}(-1)^n$ is known as the rational $q,t$-Catalan number, computed by Gorsky and Mazin \cite{GM} for the case $n=3$ and studied by Lee, Li and Loehr \cite{LLL} for the case $n=4$. The coefficients at the hook-shaped Schur functions were discussed by Armstrong, Loehr and Warrington \cite{ALW}.
 
 \begingroup\emergencystretch=3em
In this paper, we explore the combinatorics of the Schur 
function expansion of 
$Q_{m,n}(-1)^n$ in the special case where either $m$ or $n$ is less than or equal to 3. In particular, we study the Schur function expansions of 
$Q_{2,2n}1$, $Q_{2n,2}1$, $Q_{3,n}(-1)^n$, and $Q_{m,3}(-1)$.  

\endgroup

To state our results, we must first recall the rational shuffle conjecture. 
This will require a series of definitions. 

Let $m$ and $n$ be positive integers.  An $(m, n)$-Dyck path is a lattice path from $(0,0)$ to $(m,n)$ which always remains weakly above the main diagonal $y = \frac{n}{m} x$. The cells that are cut through by the main diagonal will be called \textit{diagonal} cells. Here \fref{2}(a) gives an example of a $(5,7)$-Dyck path, and \fref{2}(b) gives an example of a $(4,6)$-Dyck path, where the diagonal cells are the light blue cells.

\begin{figure}[ht]
	\centering
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	\node at (4.5,-.7) {(a)};
	\path (11,1);
	\begin{customlegend}[legend cell align=left, %<= to align cells
	legend entries={ % <= in the following there are the entries
		area,
		pdinv,
		diagonal,
	},
	legend style={at={(10,6)},font=\footnotesize}] % <= to 
	\addlegendimage{area legend,orange!50!yellow!70!white,fill}
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	\node at (2.5,-.7) {(c)};
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	\caption{\fontsize{10.5}{0}\selectfont A $(5,7)$-Dyck path, a $(4,6)$-Dyck path, a $(5,7)$-parking function and its car ranks}
	\label{fig:2}
\end{figure}

\vspace{-2ex}
The number of full cells between an $(m, n)$-Dyck path $\Pi$ and the main diagonal is denoted $\area(\Pi)$. The collection of cells above a Dyck path $\Pi$ forms the Ferrers diagram (in English notation) of a partition $\lambda(\Pi)$. In the example of the rational Dyck path pictured 
in \fref{2}(a), $\lambda(\Pi)=(3,3,1,1)=\raisebox{-.2\height}{\begin{tikzpicture}[scale=0.1] \draw (3,0) grid (0,2) grid (1,-2);\end{tikzpicture}}$. Let $\chi(-)$ denotes the function that takes value $1$ if its argument is true, and $0$ otherwise. The path dinv of an $(m, n)$-Dyck path $\Pi$ is given by
{\small
	$$
	\pdinv(\Pi) = \sum_{c\in\lambda(\Pi)}\chi\left(\frac{arm(c)}{leg(c)+1}\leq\frac{m}{n}<\frac{arm(c)+1}{leg(c)}\right).
	$$
}

An $(m, n)$-\textit{parking function} $\PF$ is obtained by labeling the cells east of and adjacent to a north step of an $(m, n)$-Dyck path with the integers $1, \ldots,n$ in 
such a way that the numbers increase in each column as we read from bottom to top. We will refer to these labels as 
\textit{cars}. The underlying Dyck path is denoted as $\Pi(\PF)$. \fref{2}(c)  pictures a $(5,7)$-parking function based on the $(5,7)$-Dyck path pictured in \fref{2}(a).

Next we define $\dinv(\PF)$ and $\ides(\PF)$ for any parking function. 
We define the rank of a cell $(x,y)$ in the $(m,n)$-grid to be $\rank(x,y)=my-nx$, \fref{2}(d) shows the rank of the cars in \fref{2}(c). $\sigma(\PF)$, the word of $\PF$, is obtained by reading cars from highest to lowest ranks. In our example, $\sigma(\PF)=7563412$. We define 
$\ides(\PF)$ to be the descent set of $\sigma(\PF)^{-1}$, and we let 
$\tdinv(\PF) =\sum_{\mathrm{cars}\ i<j} \chi(\rank(i) < \rank(j) < \rank(i) + m).$
In \fref{2}(c), $\tdinv(\PF)=7$ since the pairs of cars contributing to $\tdinv$ are $(1,3)$, $(1,4)$, $(3,5)$, $(3,6)$, $(4,6)$, $(5,7)$ and $(6,7)$.

Our definition of $\dinv(\PF)$ will follow the formulation by Leven and Hicks \cite{Hicks}. Set $\frac{0}{0} = 0$ and $\frac{x}{0} = \infty$ for all $x\neq 0$, then
{\small
\[\dinvcorr(\Pi)=\sum_{c\in\lambda(\Pi)}\chi\left( \frac{arm(c)+1}{leg(c)+1}\leq\frac{m}{n}<\frac{arm(c)}{leg(c)} \right)-\sum_{c\in\lambda(\Pi)}\chi\left( \frac{arm(c)}{leg(c)}\leq\frac{m}{n}<\frac{arm(c)+1}{leg(c)+1} \right).\]
}
Then, for any  $(m, n)$-parking function $\PF$ with underlying Dyck path $\Pi$, we define 
	$$
	\dinv(\PF) = \tdinv(\PF)+\dinvcorr(\Pi).
	$$

There are alternative definitions of the statistic $\dinv$, see \cite{ALW, Hagbook}.

Given $S\subseteq \{1,2,\ldots,n-1\}$, let $F_{S}[X]$ denote the fundamental quasi-symmetric functions of Gessel \cite{gessel} associated to $S$, where $X=x_1+x_2+\cdots+x_n$. We define the \textit{Hikita polynomial} \cite{Hikita}
$\Hikita_{m,n}[X; q, t]$ where $m$ and $n$ are coprime by 
$$
\Hikita_{m,n}[X; q, t]=\sum_{\PF\in\PFc_{m,n}} t^{\area(\PF)}q^{\dinv(\PF)}F_{\ides(\PF)}[X].
$$

For any partition $\mu$ of $n$, let $\widetilde{H}_{\mu}$ be the modified Macdonald symmetric function \cite{Macbook} associated to $\mu$, and let $\nabla$ be the linear operator defined in terms of the
modified Macdonald symmetric functions $\widetilde{H}_{\mu }(X;q,t)$ by
$\nabla \widetilde{H}_{\mu } = t^{n(\mu )}q^{n(\mu ')} \widetilde{H}_{\mu }$, 
where $\mu '$ is its conjugate and
$n(\mu ) = \sum _{i}(i-1)\mu _{i}$. The classical shuffle conjecture of Haglund, Haiman, Loehr, Remmel, and Ulyanov \cite{HHLRU} is that $\nabla e_n=\Hikita_{n+1,n}[X; q, t]$
which was proved by Carlson and Mellit \cite{CM}.  

Gorsky and Negut \cite{GN} introduced operators $\Qmn{m,n}$ on symmetric 
functions and conjectured that the $\Qmn{m,n}(-1)^n=\Hikita_{m,n}[X; q, t]$ in the case where $m$ and $n$ are coprime. As shown in \cite{BGLX}, the $\Qmn{m,n}$ operators of the Gorsky-Negut conjecture can be defined in terms of the operators $D_k$ which were introduced in Bergeron and Garsia\cite{GB}. If $F[X]$ is a symmetric function and $M = (1-t)(1-q)$, then in plethystic notation, 
{\small
$$
D_k\ F[X]=F\left[X+\frac{M}{z}\right]\sum_{i\geq 0}  (-z)^i e_i[X]\bigg\vert_{z^k}.
$$
}
Then one can construct a family of symmetric function operators $\Qmn{a,b}$ for any pair of positive integers $(a, b)$ as follows. First for any $n\geq 0$, set $\Qmn{1,n} = D_n.$ Next, one can recursively define $\Qmn{m,n}$ for $m > 1$ as follows. Consider the $m\times n$ lattice with diagonal $y =\frac{n}{m}x$. Let $(a, b)$ be the lattice point which is closest to and below the diagonal. Set $(c, d) = (m -a, n -b)$. In such a case, we will write $\mathrm{Split}(m, n) = (a, b) + (c, d).$
Then we have the following recursive definition of the $\Qmn{m,n}$ operators:
	$$\Qmn{m,n} = \frac{1}{M}[\Qmn{c,d}, \Qmn{a,b}] = \frac{1}{M}(\Qmn{c,d} \Qmn{a,b}-\Qmn{a,b}\Qmn{c,d}).$$
\fref{3} gives an example of $\mathrm{Split}(3,4)$.
\begin{figure}[ht!]
	\centering	
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		\begin{tikzpicture}[scale=.8]
		\node[align=left,below] at (10,4) {$\mathrm{Split}(3,4) = (1,1)+(2,3) $,\\ \\$\Qmn{3,4} = \frac{1}{M}[\Qmn{2,3}, \Qmn{1,1}]=\frac{1}{M}[\Qmn{2,3},D_1] $\\${\color{white}\Qmn{3,4}} = \frac{1}{M}[\Qmn{2,3}, \Qmn{1,1}]=\frac{1}{M}[\frac{1}{M}[D_2, D_1],D_1] $\\${\color{white}\Qmn{3,4}} = \frac{1}{M^2} (D_2D_1D_1 - 2D_1D_2D_1 + D_1D_1D_2)$};
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		\draw[->,blue,very thick] (0,0) -- (1,1); \node[blue] at (1,.25) {$\Qmn{1,1}$};
		\draw[->,orange!80!red,thick] (1,1) -- (2,2); \node[orange!80!red] at (1.75,1.25) {$\Qmn{1,1}$};
		\draw[->,orange!80!red,thick] (2,2) -- (3,4); \node[orange!80!red] at (3,2.8) {$\Qmn{1,2}$};
		\end{tikzpicture}
	}
	\caption{The geometry of $\mathrm{Split}(3,4)$}
	\label{fig:3}
\end{figure}

The rational shuffle conjecture of Gorsky and Negut in the case where $m$ and 
$n$ are relatively prime which was proved by Mellit \cite{Mellit} is the following. 
\begin{theorem}[Mellit] If $m$ and $n$ are coprime positive integers, then 
$\Qmn{m,n}(-1)^n=\Hikita_{m,n}[X; q, t]$. 
\end{theorem}

When $m$ and $n$ are coprime and $k \geq 1$, we defined the return of 
a $(km,kn)$-parking function $\PF$, $ret(\PF)$ to be the \textit{smallest} positive integer $i$ such that the supporting path of $\PF$ goes through the point $(im, in)$. Then Garsia, Leven, Wallach, and Xin \cite{GLWX} conjectured the following theorem 
which was also proved by Mellit \cite{Mellit}. 
\begin{theorem}[Mellit]\label{theorem:GLWX}
	For all pairs of coprime positive integers $(m,n)$ and any $k\in\mathbb{Z}^+$, we have
	$$
	\Qmn{km,kn}(-1)^{kn}=\sum_{\PF\in\PFc_{km,kn}} [ret(\PF)]_{\frac{1}{t}}t^{\area(\PF)}q^{\dinv(\PF)}F_{\ides(\PF)}[X].
	$$
\end{theorem}

A more important goal is to find the Schur function expansion of $\nabla e_n$ since that would allows us to find the character generating function 
of the ring of diagonal invariants, see \cite{Haiman}. More generally, we would like to find a combinatorial interpretation of of the coefficients that arise in the Schur function expansion of $\Qmn{m,n}(-1)^n$. The main goal of this paper is to find such Schur function expansions in the case where $m$ or $n$ equals 3. The Schur function expansion of $\Qmn{m,n}(-1)^n$ in the case where $m$ and $n$ are coprime and either $m$ or $n$ equals 2 was given by Leven \cite{Em}. That is, let $\qtn{n}=\frac{q^n-t^n}{q-t}=q^{n-1}+q^{n-2}t+\cdots+t^{n-1}$ be the 
$q,t$-analogue of $n$, then Leven proved that for any $k \geq 0$,
\begin{align*}
\Qmn{2k+1,2}1 &= \Hikita_{2k+1,2}[X; q, t]=\qtn{k}s_2+\qtn{k+1}s_{1,1},\text{ and}\\
\Qmn{2,2k+1}(-1) &=\Hikita_{2,2k+1}[X; q, t]=\textstyle \sum_{r=0}^{k}\qtn{k+1-r}s_{2^r1^{2k+1-2r}}.
\end{align*}
Using the results of \cref{theorem:GLWX}, we can prove the following.  
\begin{theorem}
$\Qmn{2k,2}1=\Hikita_{2k,2}[X; q, t]=(\qtn{k}+\qtn{k-1})s_2+(\qtn{k+1}+\qtn{k})s_{1,1}$ and\\\hspace*{23.5mm}$\Qmn{2,2k}1=\Hikita_{2,2k}[X; q, t]=\sum_{r=0}^{k}(\qtn{k+1-r}+\qtn{k-r})s_{2^r1^{2k+1-2r}}.$
\end{theorem}
The main goal of this paper is to study the combinatorics of 
the Schur function expansion of $\Qmn{m,n} (-1)^n$ in the case where either $m$ or $n=3$. 
Given that Mellit has proved the rational shuffle conjecture, we can 
find the Schur function expansion in one of two ways. That is, we can 
use the properties of the $\Qmn{m,n}$ to find the Schur function expansion 
of $\Qmn{m,n} (-1)^n$ which we will refer to as working on the {\em symmetric 
function side} of the rational shuffle conjecture. Second, one could 
start with the Hikita polynomial $H_{m,n}[X;q,t]$ and expand 
that polynomial into Schur functions which we will call working 
on the {\em combinatorial side} of the rational shuffle conjecture. 
In the case where $n=3$, we can prove the following theorem by either 
working on the symmetric function side 
or the combinatorial side of the rational shuffle conjecture.  
\begin{theorem}\label{theorem:1}
For any $k\geq 0$,
\vspace{-1ex}
{\small
\begin{align*}
\Qmn{3k+1,3}(-1) & \!\begin{multlined}[t] = \Hikita_{3k+1,3}[X; q, t] = \left(\textstyle \sum_{i=0}^{k-1}(qt)^{k-1-i}\qtn{3i+1}\right)s_3\\
+\left(\textstyle \sum_{i=0}^{k-1}(qt)^{k-1-i}(\qtn{3i+2}+\qtn{3i+3})\right)s_{2,1}+\left(\textstyle \sum_{i=0}^{k}(qt)^{k-i}\qtn{3i+1}\right)s_{1^3},\end{multlined}\\
\Qmn{3k+2,3}(-1) &\!\begin{multlined}[t] = \Hikita_{3k+2,3}[X; q, t] = \left(\textstyle\sum_{i=0}^{k-1}(qt)^{k-1-i}\qtn{3i+2}\right)s_3\\
+\left(\textstyle \sum_{i=0}^{k}(qt)^{k-1-i}(\qtn{3i}+\qtn{3i+1})\right)s_{2,1}+\left(\textstyle \sum_{i=0}^{k}(qt)^{k-i}\qtn{3i+2}\right)s_{1^3},\end{multlined}\\
\Qmn{3k,3}(-1) & \!\begin{multlined}[t] = \Hikita_{3k,3}[X; q, t] = \left(\textstyle \sum_{i=0}^{k-1}(qt)^{k-1-i}\left(\qtn{3i-1}+\qtn{3i}+\qtn{3i+1}\right)\right)s_3\\
\begin{multlined}[t]+\left((qt)^{k+1}\left(\qtn{3}+2\qtn{2}+\qtn{1}\right)\right.\\
+\left.\textstyle \sum_{i=1}^{k-1}(qt)^{k-1-i}\left(\qtn{3i}+2\qtn{3i+1}+2\qtn{3i+2}+\qtn{3i+3}\right)\right)s_{2,1}\end{multlined}\\
+\left(\textstyle\sum_{i=0}^{k}(qt)^{k-i}(\qtn{3i-1}+\qtn{3i}+\qtn{3i+1})\right)s_{1^3}.\end{multlined}
\end{align*}
}
\end{theorem}

In the case where $m=3$, we have a conjectured formula but we have not 
been able to prove it. Nevertheless, by working on the 
combinatorial side of the rational shuffle conjecture we can prove 
many remarkable facts about the Schur function expansion of 
$Q_{3,n}(-1)^n$. Let $[s_\lambda]_{m,n}$ be the coefficient of $s_{\lambda}$ in the polynomial $\Qmn{m,n}(-1)$ and $\Hikita_{m,n}[X; q, t]$, then we can combinatorially prove the following 
facts about $[s_\lambda]_{m,3}$, $[s_\lambda]_{3,n}$ and $[s_\lambda]_{m,n}$.
\begin{theorem}\label{theorem:2}
{\bf (a)} $[s_{1^3}]_{m-3,3}=[s_{3}]_{m,3}$,
{\bf (b)}  $[s_{3^{a+1}2^b1^c}]_{3,n}=[s_{3^a2^b1^c}]_{3,n-3}$,
{\bf (c)}  $[s_{1^3}]_{n,3}=[s_{1^n}]_{3,n}$,
{\bf (d)}  $[s_{21}]_{n,3}=[s_{21^{n-2}}]_{3,n}$, and 
{\bf (e)}  $[s_{2^a1^b}]_{3,n}=[s_{2^b1^a}]_{3,3(a+b)-n}$.
\end{theorem}
Next we state a general theorem and two conjectures.   
\begin{theorem}\label{theorem:3}
For all $m,n > 0$ and $\lambda'\vdash (n-am)$, {\bf (a)} $[s_{1^n}]_{m-n,n}=[s_{n}]_{m,n}$,
{\bf (b)} $[s_{m^a\lambda'}]_{m,n}=[s_{\lambda'}]_{m,n-am}$, {\bf (c)} $[s_{1^n}]_{m,n}=[s_{1^m}]_{n,m}$, and {\bf (d)} $[s_{k1^{n-k}}]_{m,n}=[s_{k1^{m-k}}]_{n,m}$.
\end{theorem}
Note that \tref{3}$(d)$ is a result about the hook-shaped Schur functions.

\begin{conjecture}
	Let $a<b$, then
	$[s_{2^a1^b}]_{3,n}=\sum_{i=0}^{a}\qtn{b+i}+(qt)[s_{2^a1^{b-3}}]_{3,n-3}.$
\end{conjecture}
\begin{conjecture}
	$[s_{(m-1)^{\alpha_{m-1}}(m-2)^{\alpha_{m-2}}\cdots1^{\alpha_{1}}}]_{m,n}=[s_{(m-1)^{\alpha_{1}}(m-2)^{\alpha_{2}}\cdots1^{\alpha_{m-1}}}]_{m,(m\sum_{i=1}^{m-1}\alpha_i-n)}.$
\end{conjecture}\addtocounter{conjecture}{-2}

The outline of this paper is as follows. First, in \cref{sec:schur_basis_expansion}, we shall  
outline the proof of \cref{theorem:1} from both the symmetric function 
side and the combinatorial side. The combinatorics involved in proofs 
from the combinatorial side leads to very interesting combinatorial proofs, 
but due to lack of space, we can only briefly describe some of the ideas involved. 
Nevertheless, our combinatorial proofs allow us to prove many general facts 
of the Schur function expansion of $\Qmn{m,n}$ in general which are not at all obvious 
from the symmetric function side.  
We will exhibit such ideas in \cref{sec:schur_basis_expansion,sec:combinatorial_results,sec:combinatorial_results_again}.
 
In \cref{sec:combinatorial_results}, we shall briefly discuss 
the proof of \cref{theorem:2}.  We shall see that 
the results of \cref{theorem:2} show that the problem of computing 
the Schur function expansion of $\Qmn{3,n}(-1)^n$ can be reduced to the problem 
finding the coefficients of Schur functions of the form $s_{2^a,1^b}$ 
in $\Qmn{3,n}(-1)^n$. We will state a conjecture for such coefficients at the end 
of \cref{sec:combinatorial_results}.  In \cref{sec:combinatorial_results_again}, we extend some combinatorial results to  $\Qmn{m,n}(-1)^n$ case.

\vspace{-2ex}
\section{Schur Basis Expansion of $(m,3)$ Case}\label{sec:schur_basis_expansion}


\subsection{Symmetric Function Side --- $\Qmn{m,3}$(-1)}
We need the following lemma from \cite{BGLX} to prove the symmetric function side of the theorem.

\vspace{-1ex}
\begin{lemma}\label{lemma:1}
	For any positive $m,n$,
	$
	\nabla \Qmn{m,n}\nabla^{-1}=\Qmn{m+n,n}.
	$
\end{lemma}
\vspace{-1ex}
This allows us to prove the formula by induction:
\vspace{-1ex}
\begin{align*}
\Qmn{m+3,3}(-1) &= \nabla \Qmn{m,3} \nabla^{-1}(-1) = \nabla \Qmn{m,3} (-1)= \nabla ([s_3]_{m,3}s_3+[s_{21}]_{m,3}s_{21}+[s_{1^3}]_{m,3}s_{1^3})\\
&= [s_3]_{m,3}\nabla s_3+[s_{21}]_{m,3}\nabla s_{21}+[s_{1^3}]_{m,3}\nabla s_{1^3}.
\end{align*}	
\vspace{-1ex}
One can directly calculate that 
\begin{align*}
	\nabla s_3 &= (qt)^2s_{21}+(qt)^2\qtn{2}s_{1^3},\qquad
	\nabla s_{21}=- (qt)\qtn{2}s_{21}-(qt)\qtn{3}s_{1^3},\\
	\nabla s_{1^3}&= s_3+(\qtn{2}+\qtn{3})s_{21}+(qt+\qtn{4})s_{1^3},
\end{align*}
and the base cases of \tref{1} are $\Qmn{1,3}(-1)$, $\Qmn{2,3}(-1)$ and $\Qmn{3,3}(-1)$, which can be verified by breaking the $\Qmn{m,3}$ operators into $D_k$ operators and applying symmetric function manipulations. With the base cases verified, one prove \tref{1} by inducting on $m$. 

\subsection{Combinatorial Side --- $\Hikita_{m,3}[X; q, t]$}

Hikita \cite{Hikita} in 2012 proved that the Hikita polynomials $\Hikita_{m,n}[X;q,t]$ are symmetric (in $X$) for any coprime $m,n$.  

A weak composition of $n$ is a sequence of non-negative integers summing up to $n$. Suppose that  
$\gamma = (\gamma_1, \ldots, \gamma_n)$ is a weak composition of 
$n$ into $n$ parts. We let $X =(x_1, \ldots, x_n)$, and $\Delta_{\gamma}(X) =  \det||x_i^{\gamma_j+n-j}|| =  \sum_{\sg  \in S_n} sgn(\sg) (x_{\sg(1)}^{\gamma_1+n-1} \cdots 
x_{\sg(n)}^{\gamma_n+n-n})$. \\
Let $\Delta(X) = \det||x_i^{n-j}||$  
be the Vandermonde determinant, then the Schur function 
$s_{\gamma}(X)$ 
associated to $\gamma$ is defined to be 
$s_{\gamma}(X) = \frac{\Delta_{\gamma}(X)}{\Delta(X)}$. 
It is well known that for any such weak composition $\gamma$,  
either $s_{\gamma}(X) =0$ or there is a 
partition $\lambda$ of $n$ such that 
$s_{\gamma}(X) = \pm s_{\lambda}(X)$.
In fact, there is a well-known straightening relation 
which allows on to prove that fact. Namely, 
if $\gamma_{i+1}  > 0$, then $s_{(\gamma_1, \ldots, \gamma_i, \gamma_{i+1}, \ldots, \gamma_n)}(X) = - s_{(\gamma_1, \ldots, \gamma_{i+1}-1, \gamma_{i}+1, \ldots, \gamma_n)}(X)$.


Suppose $\alpha = (\alpha_1, \ldots, \alpha_k)$ is 
a composition of $n$ with $k$ parts. We associate 
a subset $S(\alpha)$ of $\{1, \ldots, n-1\}$ with $\alpha$ 
by setting $S(\alpha) = \{\alpha_1, \alpha_1+\alpha_2, \ldots , \alpha_1 + \cdots + \alpha_{k-1}\}$. 
We let $\tilde{\alpha}$ be the weak composition 
of $n$ with $n$ parts by adding a sequence of $n-k$ 0's at the 
end of $\alpha$.  For example, if 
$\alpha =(2,3,2,1)$, then $S(\alpha) =\{2,5,7\}$ and 
$\tilde{\alpha} = (2,3,2,1,0,0,0,0)$. 

In a remarkable and important paper, Egge, Loehr and Warrington  
\cite{ELW} gave a combinatorial description of how 
to start with the quasi-symmetric function expansion of a homogeneous 
symmetric function of degree $n$, 
$P(X) = \sum_{\alpha \Vdash n} a_{\alpha} F_\alpha(X)$
and transform it into an expansion in terms of Schur functions 
$P(X) = \sum_{\lambda \vdash n} b_{\lambda} s_{\lambda}(X).$
The following theorem due to Garsia and Remmel \cite{GR} is implicit 
in the work of \cite{ELW}, but is not explicitly stated and it allows 
one to find the Schur function expansion by using the straightening laws. 
\begin{theorem}\label{main}
Suppose that $P(X)= \sum_{\alpha \Vdash n} a_{\alpha} F_\alpha(X)$ is a symmetric function which is homogeneous of degree $n$, then 
$ 
P(X) = \sum_{\alpha \Vdash n} 
a_{\alpha} s_{\tilde{\alpha}}(X).$
\end{theorem}


Let $\pides(\sigma)$ be the composition set of $\ides(\sigma)$, then \tref{2} and the 
straightening action allow us to transform $\Hikita_{m,n}[X;q,t]$ into Schur function expansion that
$$
\Hikita_{m,n}[X; q, t]=\sum_{\PF\in\PFc_{m,n}} t^{\area(\PF)}
q^{\dinv(\PF)} F_{\ides(\PF)} [X]=\sum_{\PF\in\PFc_{m,n}} t^{\area(\PF)}
q^{\dinv(\PF)} s_{\pides(\PF)}.
$$

Any parking function $\PF\in\PFc_{m,3}$ has $3$ rows, thus has only $3$ cars: $1,2,3$, so the word $\sigma(\PF)$ can be any permutation $\sigma\in\mathcal{S}_3$. \taref{1} shows the $s_{\pides}$ contribution of the $6$ permutations in $\mathcal{S}_3$.

\begin{table}[ht!]
	\centering
	\begin{tabular}{|c|C{1.4cm}|C{1.4cm}|C{1.4cm}|C{1.4cm}|C{1.4cm}|C{1.4cm}|}
		\hline
		$\sigma\in\mathcal{S}_3$ & $123$ & $132$ & $213$ & $231$ & $312$ & $321$\\\hline
		$s_{\pides}$ & $s_{3}$ & $s_{21}$ & $s_{12}=0$ & $s_{21}$ & $s_{12}=0$ & $s_{1^3}$\\\hline
	\end{tabular}
	\caption{$s_{\pides}$ contribution of the permutations in $\mathcal{S}_3$}
	\label{table:1}
\end{table}


By our notation, $\Hikita_{m,3}[X; q, t]=[s_{3}]_{m,3}s_3+[s_{21}]_{m,3}s_{21}+[s_{1^3}]_{m,3}s_{1^3}$. For an example of how one can work out the combinatorial side 
of rational shuffle conjecture in the case where $n=3$, we shall briefly describe 
how one can use \cref{main} to compute $[s_{3}]_{3k+1,3}$. 

From \taref{1}, we see that only parking functions in $\PFc_{3k+1,3}$ with a word $123$ contribute to the coefficient of $s_{3}$. We also notice that the $3$ cars should be in different columns, otherwise there are cars $i<j$ with $\rank(i)<\rank(j)$, contradicting that the 
word of the parking function is $123$. Thus we have one $\PF\in\PFc_{3k+1,3}$ with word $123$ on each $(3k+1,3)$ Dyck path which has no two consecutive north steps. 

Let $\lambda(\PF)=\{\lambda_1,\lambda_2\}$ be the partition associated with the Dyck path $\Pi(\PF)$, then we can count both $\area$ and $\dinv$ from $\lambda(\PF)$. We have $\area(\PF)=3k-\lambda_1-\lambda_2$, and we can also write the formula for $\dinv$:
{\small
\[\dinv(\PF) = \begin{cases}
\lambda_1-2 & \textrm{if } \lambda_2\geq1, 1\leq\lambda_1-\lambda_2\leq k,\textrm{ and } \lambda_1\leq k,\\
2\lambda_1-k-3 & \textrm{if } \lambda_2\geq1, 1\leq\lambda_1-\lambda_2\leq k,\textrm{ and } \lambda_1\geq k+1,\\
2\lambda_2+k-2 & \textrm{if } \lambda_2\geq1 \textrm{ and } \lambda_1-\lambda_2\geq k+1.
\end{cases}\]
}
\vspace{-3ex}
\begin{figure}[ht!]
	\centering	
	\scalebox{.7}{\begin{tikzpicture}[scale=0.5]
	\fillshade{1/1,2/1,3/1,3/2,4/2,5/2,5/3,6/3,7/3};
	\PFmnu{0,0}{7}{3}{0/3,2/2,3/1,0/0};
	\draw[fill=green!30!white] (0,2) rectangle (3,3);
	\draw[fill=green!30!white] (0,1) rectangle (2,2);
	\node at (1.5,2.5) {$\lambda_1$};
	\node at (1,1.5) {$\lambda_2$};
	\node at (-1.5,2) {$\lambda(\PF)$};
	\end{tikzpicture}}
	\caption{Example: a $\PF\in\PFc_{7,3}$ with word $123$}
	\label{fig:4}\vspace{-2ex}
\end{figure}

For $[s_{3}]_{3k+1,3}=\sum_{i=0}^{k-1}(qt)^{k-1-i}\qtn{3i+1}$, we construct each term $(qt)^{k-1-i}\qtn{3i+1}$ as a sequence of parking functions. For each $i$, we have $3$ branches of partitions (parking functions) to obtain $(qt)^{k-1-i}\qtn{3i+1}$:
\vspace{-1ex}
\begin{align*}
	\Lambda_1 &= \{(k+i+1,k),(k+i,k-1),\ldots,(k+1,k-i)\},\\
	\Lambda_2 &= \{(2k,i),(2k-1,i-1),\ldots,(2k+1-i,1)\},\\
	\Lambda_3 &= \{(k+1,i+1),(k,i+1),\ldots,(i+2,i+1)\}.
\end{align*}\vspace{-3ex}

\begin{figure}[h]
	\centering	
	\scalebox{.8}{
		\begin{tikzpicture}[scale=0.42]
		\partitiontwo{0}{0}{11}{k+i+1}{6}{k};
		\partitiontwo{0}{-4}{10}{k+i}{5}{k-1};
		\partitiontwo{0}{-8}{9}{k+i+1-r}{4}{k-r};
		\partitiontwo{0}{-12}{8}{k+2}{3}{k-i+1};
		\partitiontwo{0}{-16}{7}{k+1}{2}{k-i};
		
		\partitiontwo{13}{0}{12}{2k}{4}{i};
		\partitiontwo{13}{-4}{11}{2k-1}{3}{i-1};
		\partitiontwo{13}{-8}{10}{2k-r}{2}{i-r};
		\partitiontwo{13}{-12}{9}{2k+1-i}{1}{1};
		
		\partitiontwo{8}{-16}{6}{k}{2}{k-i};
		\partitiontwo{15}{-16}{5}{k-1}{2}{k-i};
		\partitiontwo{22}{-16}{3}{k-i+1}{2}{k-i};
		
		\arrowx{9.5,-1.5}{12.5,-1.5}
		\arrowx{9.5,-5.5}{12.5,-5.5}
		\arrowx{9.5,-9}{12.5,-9}
		\arrowx{9,-13}{12,-13}
		
		\arrowx{12.5,-2}{10,-3.5}
		\arrowx{12.5,-6}{11.5,-6.5}
		\arrowx{10,-7.5}{9,-8}
		\dotdot{11,-6.5}{11,-7}{11,-7.5};
		\arrowx{12.5,-9.5}{11,-10}
		\arrowx{9.5,-11}{8,-11.5}
		\dotdot{10.25,-10.5}{10.25,-11}{10.25,-11.5};
		\arrowx{12.5,-13.5}{7,-15.5}
		
		\arrowx{4.5,-17.5}{7.5,-17.5}
		\arrowx{11.5,-17.5}{14.5,-17.5}
		\arrowx{18,-17.5}{20,-17.5}
		\dotdot{20.5,-17.5}{21,-17.5}{21.5,-17.5}
		
		\fille{0}{-1}{q^{3i}\cdot(qt)^{k-1-i}}
		\fille{0}{-5}{q^{3i-2}t^2\cdot(qt)^{k-1-i}}
		\fille{0}{-9}{q^{3i-2r}t^{2r}\cdot(qt)^{k-1-i}}
		\fille{0}{-13}{q^{i+2}t^{2i-2}\cdot(qt)^{k-1-i}}
		\fille{0}{-17}{q^{i}t^{2i}\cdot(qt)^{k-1-i}}
		
		\fillw{25}{-1}{q^{3i-1}t\cdot(qt)^{k-1-i}}
		\fillw{24}{-5}{q^{3i-3}t^3\cdot(qt)^{k-1-i}}
		\fillw{23}{-9}{q^{3i-2r-1}t^{2r+1}\cdot(qt)^{k-1-i}}
		\fillw{22}{-13}{q^{i+1}t^{2i-1}\cdot(qt)^{k-1-i}}
		
		\fillls{12}{-15.5}{q^{i-1}t^{2i+1}\cdot(qt)^{k-1-i}}
		\fillls{19}{-15.5}{q^{i-2}t^{2i+2}\cdot(qt)^{k-1-i}}
		\fillw{25}{-17}{t^{3i}\cdot(qt)^{k-1-i}}
		\draw [thick, blue,decorate,decoration={brace,amplitude=5pt,mirror},xshift=0.4pt,yshift=-0.4pt](-7.5,0) -- (-7.5,-17.5) node[blue,midway,xshift=-.5cm] {$\Lambda_1$};
		\draw [thick, blue,decorate,decoration={brace,amplitude=5pt,mirror},xshift=0.4pt,yshift=-0.4pt](32,-14) -- (32,0) node[blue,midway,xshift=.5cm] {$\Lambda_2$};
		\draw [thick, blue,decorate,decoration={brace,amplitude=5pt,mirror},xshift=0.4pt,yshift=-0.4pt](-7.5,-18.5) -- (30,-18.5) node[blue,midway,yshift=-.5cm] {$\Lambda_3$};
		\end{tikzpicture}
	}
	\caption{The construction of $(qt)^{k-1-i}\qtn{3i+1}$}
	\label{fig:5}
\end{figure}

The branch $\Lambda_1$ contains $\lambda$'s such that $\lambda_1-\lambda_2=i+1\leq k$ with $\lambda_2\geq i+1$, 
the branch $\Lambda_2$ contains all $\lambda$'s such that $\lambda_1-\lambda_2=2k-i>k$, and
the branch $\Lambda_3$ contains $\lambda$'s such that $\lambda_2=i+1$ and $\lambda_1-\lambda_2\leq k-i$. Notice that $|\Lambda_1|=|\Lambda_2|+1$, and the last partition of $\Lambda_1$ is the same as the first partition in $\Lambda_3$. So as shown in \fref{5}, the {construction} begins with \textit{alternatively} taking partitions from $\Lambda_1$ and $\Lambda_2$, {ending with} the {last partition of $\Lambda_1$}. Then {continue} the chain by taking partitions in $\Lambda_3$ and end the chain with the last partition $(k-i+1,k-i)$ in $\Lambda_3$. \fref{6} shows the combinatorics of 
$[s_{3}]_{13,3}$.
\begin{figure}[ht!]	\centering
\includegraphics[width=\textwidth]{13,3ex.png}
	\caption{The combinatorics of 
$[s_{3}]_{13,3}=\qtn{10}+(qt)\qtn{7}+(qt)^2\qtn{4}+(qt)^3\qtn{1}$.}
	\label{fig:6}\vspace{-2ex}
\end{figure}

We can combinatorially prove  $[s_{21}]_{3k+1,3}=\sum_{i=0}^{k-1}(qt)^{k-1-i}(\qtn{3i+2}+\qtn{3i+3})$ in a similar way. In this case, we have $2$ possible diagonal words: $132$ and $312$, we can obtain $6$ branches of parking functions for each $i$ from $0$ to $k-1$ based on the diagonal words and the shape of their partitions. $3$ of the branches contribute $(qt)^{k-1-i}\qtn{3i+2}$ and the rest contribute $(qt)^{k-1-i}\qtn{3i+2}$ to the coefficient of $s_{21}$.


To prove that $[s_{1^3}]_{3k+1,3}=\sum_{i=0}^{k}(qt)^{k-i}\qtn{3i+1}=[s_3]_{3k+4,3}$, 
we can give a combinatorial proof of a more general result, namely, 
$[s_{1^3}]_{m,3}=[s_3]_{m+3,3}$. 

Note that a parking function with $\pides$ can be straightened to $1^3$ must have word $321$. One $\PF\in\PFc_{m,3}$ with word $321$ is correspond with an $(m,3)$ Dyck path. As shown is \fref{7}, we can obtain a $\PF\in\PFc_{m+3,3}$ with word $123$ by pushing a staircase into a $\PF\in\PFc_{m,3}$ with word $321$. The fillings of the cars are fixed by the $\pides$ respectively.
\begin{figure}[ht!]
	\centering
	\scalebox{.7}{
	\begin{tikzpicture}[scale=0.5]
	\fillshadea{0/2,0/3,-1/3,7/2,7/3,8/3}
	\draw[help lines] (-2,3) grid (0,2) grid (-1,1);
	\fillshade{1/1,2/1,2/2,3/2,3/3,4/3,7/1,8/1,9/1,9/2,10/2,11/2,11/3,12/3,13/3};
	\PFmnu{0,0}{4}{3}{0/1,0/2,1/3,0/0};
	\arrowx{4.2,1.5}{5.8,1.5};
	\PFmnu{6,0}{7}{3}{0/3,1/2,3/1,0/0};
	\end{tikzpicture}}
	\caption{Bijection between $\PFc_{m,3}$ with word $321$ and $\PFc_{m+3,3}$ with word $123$}
	\label{fig:7}
\end{figure}

\section{Combinatorial Results about $[s_{\lambda}]_{3,n}$}\label{sec:combinatorial_results}

In $(3,n)$ case, we have $n$ cars, i.e. the word of a $(3,n)$ parking function is a permutation of $[n]$. First, we can prove the following.
\begin{enumerate}
\vspace{-1ex}
\item Let $i<j$ be two cars in the parking function. If $i$ appears to the left of $j$ in the diagonal word, then the cars $i,j$ must be in different columns.
\vspace{-1ex}
\item The parts in the composition set $\pides(\PF)$ of a parking function $\PF\in\PFc_{m,n}$ are less than or equal to $m$.
\end{enumerate}
\vspace{-1ex}
Thus, in the $(3,n)$ case, $[s_{\lambda}]_{3,n} \neq 0$ implies that 
$\lambda\vdash n$ must be of the form $3^a2^b1^c$; $[s_{\lambda}]_{m,n} \neq 0$ only if the partition $\lambda$ only has parts of size less than or equal to $m$.

We can give bijective proofs of the following results. In each case, we 
shall briefly describe the idea how the bijection works. 
\vspace{-1ex}
\begin{result}$[s_{3^{a+1}2^b1^c}]_{3,n}=[s_{3^a2^b1^c}]_{3,n-3}$.
\end{result}
\vspace{-1ex}
$\PF\in\PFc_{3,n}$ with $\pides\ 3^{a+1}2^b1^c$ must have the cars $1,2,3$ placed at the bottom of $3$ columns in a rank decreasing way. The bijection is that we can remove these $3$ cars and delete the corresponding north steps of the Dyck path to obtain a parking function of size $(3,n-3)$ with cars labeled from $4$ to $n$; then we subtract $3$ from the labels to obtain a $\PF'\in\PFc_{3,n-3}$ with $\pides\ 3^{a}2^b1^c$, as \fref{8}$(a)$ shows. Clearly, the $\area$ is unchanged since the diagonal cells are still the previous diagonal cells, and the $\dinv$ statistic is also not changed since this manipulation keeps the rank of all the unremoved cars. This tells us that we only need to consider the coefficients $[s_{2^a1^b}]_{3,n}$.
\vspace{-1ex}
\begin{result}
	$[s_{1^3}]_{n,3}=[s_{1^n}]_{3,n}$.
\end{result}
\vspace{-1ex}
The bijection for this identity is that we can transpose the path of $\PF\in\PFc_{n,3}$ and fill the word $(n,n-1,\ldots,1)$ to get $\PF'$. It's easy to verify that we get $\PF'\in\PFc_{3,n}$ with same $\area$ and $\dinv$. \fref{8}$(b)$ shows an example of this bijection.
\begin{figure}[ht!]
	\centering
	\scalebox{.7}{\begin{tikzpicture}[scale=0.5]
		\fillshade{1/0,1/1,1/2,2/2,2/3,2/4,3/4,3/5,3/6,6/1,6/2,7/2,7/3,8/3,8/4,11/1,11/2,12/2,12/3,13/3,13/4};
		\PFmnu{0,-1}{3}{7}{0/3,0/5,0/6,1/2,1/4,1/7,2/1,0/0};
		\draw[orange, very thick] (0,-1)--(0,0)--(1,0)--(1,3)--(2,3)--(2,6)--(3,6);
		\arrowx{3.2,2}{4.8,2};
		\PFmnu{5,0}{3}{4}{0/5,0/6,1/4,1/7,0/0};
		\arrowx{8.2,2}{9.8,2};
		\PFmnu{10,0}{3}{4}{0/2,0/3,1/1,1/4,0/0};
		\node at (6.5,-1) {$(a)$};
		\path (16,0);
		\end{tikzpicture}}
	\scalebox{.7}{\begin{tikzpicture}[scale=0.5]
		\fillshade{1/1,2/1,2/2,3/2,3/3,4/3,7/1,7/2,8/2,8/3,9/3,9/4};
		\PFmnu{0,0}{4}{3}{0/1,0/2,1/3,0/0};
		\arrowx{4.2,1.5}{5.8,1.5};
		\PFmnu{6,0}{3}{4}{0/1,0/2,0/4,1/3,0/0};
		\node at (4.5,-1) {$(b)$};
		\end{tikzpicture}}
	\caption{Bijection $[s_{3^{a+1}2^b1^c}]_{3,n}\longrightarrow[s_{3^a2^b1^c}]_{3,n-3}$ and bijection $[s_{1^3}]_{n,3}\longrightarrow[s_{1^n}]_{3,n}$}
	\label{fig:8}\vspace{-2ex}	
\end{figure}
\vspace{-1ex}
\begin{result}
$[s_{21}]_{n,3}=[s_{21^{n-2}}]_{3,n}$.
\end{result}
\vspace{-1ex}
The bijective proof of this result is similar to the previous result. 
That is, one transposes the path and and labels the path to produce $\pides\ 21^{n-2}$. If there are only $2$ peaks in the Dyck path, then the filling of cars in both $(n,3)$ and $(3,n)$ cases are unique. Otherwise, in any rational $(n,3)$-Dyck path with $3$ peaks, there are $2$ kinds of words: $132$ and $312$ having $\pides\ 21$ in $(n,3)$ case, which means that there are $2$ choices to locate the car $1$ in the $(3,n)$ case. The words that have $\pides\ 21^{n-2}$ must be of the form $(n,n-1,\ldots,i+1,1,i,i-1,\ldots,2)$, so the car $1$ can be placed at the bottom of either the second column or the third column. We are able to match the $2$ possible positions of car $1$ in both $(n,3)$ and $(3,n)$ cases by the $\dinv$ statistic, thus prove the result.
\vspace{-1ex}
\begin{result}
$[s_{2^a1^b}]_{3,n}=[s_{2^b1^a}]_{3,3(a+b)-n}$.
\end{result}
\vspace{-1ex}
We have found the straightening action in parking functions combinatorially from $\pides \{\cdots1,3\cdots\}\textnormal{ to }\pides \{\cdots2,2\cdots\}$, which is an \textit{involution} whose fixed points are the coefficients of $[s_{2^a1^b}]_{3,n}$. Further we have found a bijection between the fixed parking functions with $\pides\ 2^a1^b$ and the fixed parking functions with $\pides\ 2^b1^a$, mapping the $2$ cars (or $1$ car) causing part $2$ (or $1$) in $\pides\ 2^a1^b$ to $1$ car (or $2$ cars) causing part $1$ (or $2$) in $\pides\ 2^b1^a$.


The four results above prove \tref{2}. Finally, we conjecture a recursive formula for $[s_{2^a1^b}]_{3,n}$.
\begin{conjecture}
Let $a<b$, then
$[s_{2^a1^b}]_{3,n}=\sum_{i=0}^{a}\qtn{b+i}+(qt)[s_{2^a1^{b-3}}]_{3,n-3}.$
\end{conjecture}
We verified this formula by Maple for $n<27$. If this conjecture is true, then we have solved the Schur function expansion in the $(3,n)$ case.

\vspace{-2ex}
\section{Combinatorial Results about $[s_{\lambda}]_{3,n}$}\label{sec:combinatorial_results_again}
Using the ideas of the previous section, we can give bijective proofs of the 
following general results. \addtocounter{theorem}{-2}
\begin{theorem}
For all $m,n > 0$ and $\lambda'\vdash (n-am)$,
	{\bf (a)} $[s_{1^n}]_{m-n,n}=[s_{n}]_{m,n}$, 
	{\bf (b)} $[s_{m^a\lambda'}]_{m,n}=[s_{\lambda'}]_{m,n-am}$,
	{\bf (c)} $[s_{1^n}]_{m,n}=[s_{1^m}]_{n,m}$, and 
	{\bf (d)} $[s_{k1^{n-k}}]_{m,n}=[s_{k1^{m-k}}]_{n,m}$.
\end{theorem}
We haven't completely understood how to use straightening to compute 
the coefficients of $s_{\lambda}$ for general $(m,n)$ case, but 
computations in Maple have led us to conjecture the following.
\begin{conjecture}
	$[s_{(m-1)^{\alpha_{m-1}}(m-2)^{\alpha_{m-2}}\cdots1^{\alpha_{1}}}]_{m,n}=[s_{(m-1)^{\alpha_{1}}(m-2)^{\alpha_{2}}\cdots1^{\alpha_{m-1}}}]_{m,(m\sum_{i=1}^{m-1}\alpha_i-n)}.$
\end{conjecture}

\vspace{-3ex}
\printbibliography


\end{document}
