# The $q,t$-symmetry of the generalized $q,t$-Catalan number
$C_{(k_{1},k_{2},k_{3})}(q,t)$ and $C_{(k,k,k,k)}(q,t)$

Guoce Xin1 and Yingrui Zhang2,∗ 1,2School of Mathematical Sciences, Capital
Normal University, Beijing 100048, PR China<EMAIL_ADDRESS>&
<EMAIL_ADDRESS>
(Date: May 24, 2022)

###### Abstract.

We give two proofs of the $q,t$-symmetry of the generalized $q,t$-Catalan
number $C_{\vec{k}}(q,t)$ for $\vec{k}=(k_{1},k_{2},k_{3})$. One is by using
MacMahon’s partition analysis as we proposed; the other is a direct bijection.
We also prove $C_{(k,k,k,k)}(q,t)=C_{(k,k,k,k)}(t,q)$ by using MacMahon’s
partition analysis.

This work was partially supported by NSFC(12071311).

_Mathematic subject classification_ : Primary 05A19; Secondary 05E99.

_Keywords_ : $q,t$-Catalan numbers; $q,t$-symmetric; MacMahon’s partition
analysis.

##  1\. Introduction

In their study of the space $\mathcal{DH}_{n}$ of diagonal harmonics [2],
Garsia and Haiman introduced a $q,t$-analogue of the Catalan numbers, which
they called the $q,t$-Catalan sequence. There are several equivalent
characterizations of the (original) $q,t$-Catalan sequence, which includes two
combinatorial formulas: One is Haiman’s dinv-area $q,t$-Catalan sequence; the
other is Haglund’s area-bounce $q,t$-Catalan sequence. See [6, 8]. Further
information about $q,t$-Catalan sequence and related results can be found in
[3, 5, 7]. In [12], we introduced $q,t$-Catalan numbers $C_{\lambda}(q,t)$ of
type (a partition) $\lambda$ as an extension of Haglund and Haiman’s
combinatorial formula for ordinary $q,t$-Catalan numbers. We also investigated
the $q,t$-symmetry of $C_{\lambda}(q,t)$: The symmetry is easily proved when
the length of $\lambda$ is $\ell(\lambda)=2$; The symmetry may be proved by
using MacMahon’s partition analysis technique when $\ell(\lambda)=3$; No
symmetry holds in general when $\ell(\lambda)\geq 4$, but we conjecture the
$q,t$-symmetry of $C_{\lambda}(q,t)$ when $\lambda=((a+1)^{s},a^{n-s})$.

We find it better to define $C_{\vec{k}}(q,t)$ for any ordered partition,
i.e., vector $\vec{k}=(k_{1},\dots,k_{n})$ of positive integers, by

$\displaystyle
C_{\vec{k}}(q,t)=\sum_{D\in{\mathcal{D}}_{\vec{k}}}q^{\operatorname{\texttt{area}}(D)}t^{\operatorname{\texttt{bounce}}(D)},$

where the sum ranges over all $\vec{k}$-Dyck paths $D$, and
$\operatorname{\texttt{area}}(D)$ and $\operatorname{\texttt{bounce}}(D)$ are
two statistics of $D$.

In the special case when $k_{i}=k$ for all $i$, denote by
$k^{n}=(k,\dots,k)\in\mathbb{Z}^{n}$ and

$\displaystyle
C_{k^{n}}(q,t)=\sum_{D\in{\mathcal{D}}_{k^{n}}}q^{\operatorname{\texttt{area}}(D)}t^{\operatorname{\texttt{bounce}}(D)},$

where the sum ranges over all $k^{n}$-Dyck paths $D$.

Our main result is the following.

######  Theorem 1.

For $\vec{k}=(k_{1},k_{2},k_{3})$, the $q,t$-Catalan number $C_{\vec{k}}(q,t)$
is $q,t$-symmetric, i.e., $C_{\vec{k}}(q,t)=C_{\vec{k}}(t,q)$.

A direct corollary is the following, since

$C_{\lambda}(q,t)=\sum_{\lambda(\vec{k})=\lambda}C_{\vec{k}}(q,t),$

where $\lambda(\vec{k})$ is the partition obtained by arranging the entries of
$\vec{k}$ decreasingly.

######  Corollary 2.

For any partition $\lambda$ of length $3$, the $q,t$-Catalan number
$C_{\lambda}(q,t)$ is $q,t$-symmetric, i.e.,
$C_{\lambda}(q,t)=C_{\lambda}(t,q)$.

Note that one can define

$H_{\vec{k}}(q,t)=\sum_{D\in{\mathcal{D}}_{\vec{k}}}q^{\operatorname{\texttt{dinv}}(D)}t^{\operatorname{\texttt{area}}(D)},\quad\text{
and }H_{\lambda}=\sum_{\lambda(\vec{k})=\lambda}H_{\vec{k}}(q,t).$

We have $C_{\lambda}(q,t)=H_{\lambda}(q,t)$ by the fact that the sweep map
takes dinv to area, and area to bounce, but $C_{\vec{k}}(q,t)\neq
H_{\vec{k}}(q,t)$ because the sweep map takes a path in
${\mathcal{D}}_{\vec{k}}$ to a path in ${\mathcal{D}}_{\vec{k}^{\prime}}$ for
some $k^{\prime}$ with $\lambda(\vec{k})=\lambda(\vec{k}^{\prime})$. Indeed,
$H_{\vec{k}}(q,t)$ is not symmetric even when $\vec{k}$ is of length $3$. In
this sense, the bounce statistic is nicer than the dinv statistic.

We give two proofs of Theorem 1. One is by MacMahon’s partition analysis as we
proposed; the other is by a direct bijection. For
$\vec{k}=(k_{1},k_{2},k_{3})$, the statistics area and bounce have a simple
description. Dyck paths $D\in{\mathcal{D}}_{\vec{k}}$ are uniquely determined
by their red ranks $(r_{1}=0,r_{2},r_{3})$, which satisfy the conditions
$0\leq r_{2}\leq k_{1}$ and $0\leq r_{3}\leq r_{2}+k_{2}$. Let
$\lceil\alpha\rceil$ denote the least integer greater than or equal to
$\alpha$. The area of $D$ is simply
$\operatorname{\texttt{area}}(D)=r_{2}+r_{3}$, and the bounce of $D$ is given
by

$\displaystyle\operatorname{\texttt{bounce}}(D)=\left\\{\begin{array}[]{ll}2(k_{1}-r_{2})+r_{2}+k_{2}-r_{3}-\min(r_{2},k_{2}),&\text{
if }r_{2}+k_{2}-r_{3}\geq 2\min(r_{2},k_{2});\\\
2(k_{1}-r_{2})+\lceil{\frac{r_{2}+k_{2}-r_{3}}{2}}\rceil,&\text{otherwise.}\end{array}\right.$

The theorem can be proved by MacMahon’s partition analysis, which was
developed by MacMahon for solving problems in partition theory. The
fundamental ingredient is MacMahon’s Omega operator on the $\lambda$ variables
defined by

$\displaystyle\mathop{\Omega}_{\geq}\sum_{i_{1},i_{2},\dots
i_{n}\in\mathbb{Z}}A_{i_{1},\dots,i_{n}}\lambda_{1}^{i_{1}}\lambda_{2}^{i_{2}}\cdots\lambda_{n}^{i_{n}}$
$\displaystyle=\sum_{i_{1},i_{2},\dots i_{n}\geq 0}A_{i_{1},\dots,i_{n}},$
$\displaystyle\mathop{\Omega}_{\mu=}\sum_{i_{1},i_{2},\dots
i_{n}\in\mathbb{Z}}A_{i_{1},\dots,i_{n}}\mu_{1}^{i_{1}}\mu_{2}^{i_{2}}\cdots\mu_{n}^{i_{n}}$
$\displaystyle=A_{0,\dots,0}.$

In other words, the $\mathop{\Omega}_{\geq}$ operator extracts all terms with
nonnegative power in $\lambda_{1},\dots,\lambda_{n}$ and then set them to be
equal to $1$, and the $\mathop{\Omega}_{\mu=}$ operator extracts the term
independent of the $\mu$ variables. The to-be-eliminated variables
$\lambda_{i}$ or $\mu_{j}$ are usually clear from the context. Andrews et. al.
developed the Mathematica package Omega [1] to eliminate the $\lambda$
variables. We use Xin’s Ell2 Maple package [10] in our computation.

The paper is organized as follows. In this introduction, we have introduced
the basic concepts. In Section 2 we prove Theorem 1 by MacMahon’s partition
analysis. In Section 3, we give a bijective proof of Theorem 1. In Section 4
we prove $C_{k^{4}}(q,t)=C_{k^{4}}(t,q)$ by MacMahon’s partition analysis.

##  2\. An Algebraic Proof of Theorem 1

###  2.1. Basic Idea

For $\vec{k}=(k_{1},k_{2},k_{3})$, Dyck paths $D\in{\mathcal{D}}_{\vec{k}}$
are uniquely determined by their red ranks $(r_{1}=0,r_{2},r_{3})$, which
satisfy the conditions $0\leq r_{2}\leq k_{1}$ and $0\leq r_{3}\leq
r_{2}+k_{2}$.

The area of $D$ is simply given by
$\operatorname{\texttt{area}}(D)=r_{2}+r_{3}.$ We also have an explicit
formula of $\operatorname{\texttt{bounce}}(D)=b(r_{2},r_{3})$ as follows.

$b(r_{2},r_{3})=\left\\{\begin{array}[]{ll}2(k_{1}-r_{2})+r_{2}+k_{2}-r_{3}-\min(r_{2},k_{2}),&\text{
if }r_{2}+k_{2}-r_{3}\geq 2\min(r_{2},k_{2});\\\
2(k_{1}-r_{2})+\lceil{\frac{r_{2}+k_{2}-r_{3}}{2}}\rceil,&\text{otherwise.}\end{array}\right.$

This formula is piecewise linear in $r_{2}$ and $r_{3}$.

We prove Theorem 1 by MacMahon’s partition analysis technique, which applies
to sum over linear constraints. Indeed, we can construct the generating
function with respect to $k_{1},k_{2},k_{3},r_{2},r_{3}$:

$\displaystyle F(x_{1},x_{2},x_{3},y_{2},y_{3},q,t)$
$\displaystyle=\sum_{k_{1},k_{2},k_{3}\geq
0}x_{1}^{k_{1}}x_{2}^{k_{2}}x_{3}^{k_{3}}\sum_{D\in{\mathcal{D}}_{(k_{1},k_{2},k_{3})}}q^{\operatorname{\texttt{area}}(D)}t^{\operatorname{\texttt{bounce}}(D)}y_{2}^{r_{2}}y_{3}^{r_{3}}$
$\displaystyle=\sum_{k_{1},k_{2},k_{3}\geq
0}x_{1}^{k_{1}}x_{2}^{k_{2}}x_{3}^{k_{3}}\sum_{0\leq r_{2}\leq k_{1},\ 0\leq
r_{3}\leq
r_{2}+k_{2}}q^{r_{2}+r_{3}}t^{b(r_{2},r_{3})}y_{2}^{r_{2}}y_{3}^{r_{3}},$

where we allowed $k_{i}=0$ for $i=1,2,3$, which does not affect the
computation.

Then it is sufficient to prove the $q,t$ symmetry of

$F(x_{1},x_{2},x_{3},1,1,q,t)=\sum_{k_{1},k_{2},k_{3}\geq
0}x_{1}^{k_{1}}x_{2}^{k_{2}}x_{3}^{k_{3}}C_{(k_{1},k_{2},k_{3})}(q,t).$

We need to use the following Lemma to simplify our computation.

######  Lemma 3.

Let $A$ and $z$ be integers with $z\geq 2$. Write
$p=\lceil{\frac{A}{z}}\rceil$, so $0\leq pz-A\leq z-1$. Then

$\chi(zp\in\\{A+i|i=0\dots
z-1\\})=\mathop{\Omega}_{\mu=}\sum_{i=0}^{z-1}\mu^{A-zp+i}=\mathop{\Omega}_{\mu=}(\sum_{i=0}^{z-1}\mu^{i})\mu^{A-zp}$

where $\chi(true)=1$ and $\chi(false)=0$.

###### Proof.

It’s trivial.

###  2.2. Crude generating function

Due to the piecewise linearity of $b(r_{2},r_{3})$, we divide the generating
function $F(x_{1},x_{2},x_{3},y_{2},y_{3},q,t)$ into two parts, each having
two cases. It is convenient to write the generating function
$F(x_{1},x_{2},x_{3},y_{2},y_{3},q,t)=F=F_{1}+F_{2}=F_{11}+F_{12}+F_{21}+F_{22}$.

Part 1: $r_{2}>k_{2}$ (written as $r_{2}-k_{2}-1\geq 0$), which implies that
$\min(r_{2},k_{2})=k_{2}$. The corresponding generating function is $F_{1}$,
and we divide it into $F_{1}=F_{11}+F_{12}$ by considering the following two
cases:

Case 1: $r_{2}+k_{2}-r_{3}\geq 2k_{2}$, i.e., $r_{2}-r_{3}-k_{2}\geq 0$. Then,
we have

$b(r_{2},r_{3})=2(k_{1}-r_{2})+r_{2}+k_{2}-r_{3}-k_{2}=2k_{1}-r_{2}-r_{3}.$

So we have

$\displaystyle F_{11}=$
$\displaystyle\\!\\!\\!\\!\\!\sum_{k_{1},k_{2},k_{3}\geq
0}x_{1}^{k_{1}}x_{2}^{k_{2}}x_{3}^{k_{3}}\sum_{0\leq r_{2}\leq k_{1},\ 0\leq
r_{3}\leq r_{2}+k_{2},\atop r_{2}-k_{2}-1\geq 0,\ r_{2}-r_{3}-k_{2}\geq
0}q^{r_{2}+r_{3}}t^{2k_{1}-r_{2}-r_{3}}y_{2}^{r_{2}}y_{3}^{r_{3}}$
$\displaystyle=$
$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\sum_{k_{1},k_{2},k_{3},r_{2},r_{3}\geq
0}\\!\\!\mathop{\Omega}_{\geq}\left({x_{{1}}}^{k_{{1}}}{x_{{2}}}^{k_{{2}}}{x_{{3}}}^{k_{{3}}}{y_{{2}}}^{r_{{2}}}{y_{{3}}}^{r_{{3}}}{q}^{r_{{2}}+r_{{3}}}{t}^{2\,k_{{1}}-r_{{2}}-r_{{3}}}{\lambda_{{1}}}^{k_{{1}}-r_{{2}}}{\lambda_{{2}}}^{k_{{2}}+r_{{2}}-r_{{3}}}{\lambda_{{3}}}^{r_{{2}}-k_{{2}}-1}{\lambda_{{4}}}^{r_{{2}}-k_{{2}}-r_{{3}}}\right)$
$\displaystyle=$
$\displaystyle\mathop{\Omega}_{\geq}\frac{1}{{\lambda_{{3}}}\left(1-x_{{1}}{t}^{2}\lambda_{{1}}\right)\left(1-{\frac{x_{{2}}\lambda_{{2}}}{\lambda_{{3}}\lambda_{{4}}}}\right)\left(1-{\frac{\lambda_{{2}}\lambda_{{3}}\lambda_{{4}}qy_{{2}}}{t\lambda_{{1}}}}\right)\left(1-{\frac{y_{{3}}q}{\lambda_{{2}}\lambda_{{4}}t}}\right)(1-x_{3})};$

Case 2: $r_{2}+k_{2}-r_{3}<2k_{2}$, i.e., $k_{2}-r_{2}+r_{3}-1\geq 0$. Then,
we have

$\displaystyle
b(r_{2},r_{3})=2(k_{1}-r_{2})+\lceil{\frac{r_{2}+k_{2}-r_{3}}{2}}\rceil.$

Let $p=\lceil{\frac{r_{2}+k_{2}-r_{3}}{2}}\rceil$. Note that $p\geq 0$,
because the red ranks $(r_{1}=0,r_{2},r_{3})$ of a Dyck path
$D\in{\mathcal{D}}_{\vec{k}}$ satisfy $0\leq r_{3}\leq r_{2}+k_{2}$. By Lemma
3, we can write the generating function $F_{12}$ as follows.

$\displaystyle F_{12}=$
$\displaystyle\\!\\!\\!\\!\\!\sum_{k_{1},k_{2},k_{3}\geq
0,}x_{1}^{k_{1}}x_{2}^{k_{2}}x_{3}^{k_{3}}\sum_{0\leq r_{2}\leq k_{1},\ 0\leq
r_{3}\leq r_{2}+k_{2},\atop{k_{2}-r_{2}+r_{3}-1\geq 0,\ r_{2}-k_{2}-1\geq
0,\atop p\geq 0,\ 2p=r_{2}+k_{2}-r_{3}\text{ or
}2p=r_{2}+k_{2}-r_{3}+1}}q^{r_{2}+r_{3}}t^{2(k_{1}-r_{2})+p}y_{2}^{r_{2}}y_{3}^{r_{3}}$
$\displaystyle=$ $\displaystyle\sum_{k_{1},k_{2},k_{3},r_{2},r_{3},p\geq
0}\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}({x_{{1}}}^{k_{{1}}}{x_{{2}}}^{k_{{2}}}{x_{{3}}}^{k_{{3}}}{y_{{2}}}^{r_{{2}}}{y_{{3}}}^{r_{{3}}}{q}^{r_{{2}}+r_{{3}}}{t}^{2\,k_{{1}}-2\,r_{{2}}+p}\left(1+\mu\right){\mu}^{r_{{2}}+k_{{2}}-r_{{3}}-2\,p}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
{\lambda_{{1}}}^{k_{{1}}-r_{{2}}}{\lambda_{{2}}}^{k_{{2}}+r_{{2}}-r_{{3}}}{\lambda_{{3}}}^{r_{{2}}-k_{{2}}-1}{\lambda_{{4}}}^{k_{{2}}+r_{{3}}-r_{{2}}-1})$
$\displaystyle=$
$\displaystyle\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}\frac{1+\mu}{\lambda_{3}\lambda_{4}(1-x_{1}t^{2}\lambda_{1})\left(1-\frac{x_{2}\mu\lambda_{2}\lambda_{4}}{\lambda_{3}}\right)\left(1-\frac{y_{2}q\lambda_{2}\lambda_{3}\mu}{t^{2}\lambda_{1}\lambda_{4}}\right)\left(1-\frac{y_{3}q\lambda_{4}}{\lambda_{2}\mu}\right)\left(1-\frac{t}{\mu^{2}}\right)(1-x_{3})}.$

Part 2: $k_{2}\geq r_{2}$ (written as $k_{2}-r_{2}\geq 0$), which implies that
$\min(r_{2},k_{2})=r_{2}$. The corresponding generating function is $F_{2}$.
By a similar computation for Part 1, we divide $F_{2}$ into $F_{21}+F_{22}$,
whose formulas are given as follows.

Case 1: $r_{2}+k_{2}-r_{3}\geq 2r_{2}$, i.e., $k_{2}-r_{2}-r_{3}\geq 0$. Then,
we have $b(r_{2},r_{3})=2k_{1}-2r_{2}+k_{2}-r_{3}$ and

$\displaystyle
F_{21}=\mathop{\Omega}_{\geq}\frac{1}{(1-x_{1}t^{2}\lambda_{1})(1-x_{2}t\lambda_{2}\lambda_{3}\lambda_{4})\left(1-\frac{y_{2}q\lambda}{t^{2}\lambda_{1}\lambda_{3}\lambda_{4}}\right)\left(1-\frac{y_{3}q}{t\lambda_{2}\lambda_{4}}\right)(1-x_{3})}.$

Case 2: $r_{2}+k_{2}-r_{3}<2r_{2}$, i.e., $r_{2}+r_{3}-k_{2}-1\geq 0$. Then,
we have

$\displaystyle
b(r_{2},r_{3})=2(k_{1}-r_{2})+\lceil{\frac{r_{2}+k_{2}-r_{3}}{2}}\rceil$

and

$\displaystyle
F_{22}=\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}\frac{1+\mu}{\lambda_{4}(1-x_{1}t^{2}\lambda_{1})\left(1-\frac{x_{2}\mu\lambda_{2}\lambda_{3}}{\lambda_{4}}\right)\left(1-\frac{y_{2}q\lambda_{2}\lambda_{4}\mu}{t^{2}\lambda_{1}\lambda_{3}}\right)\left(1-\frac{y_{3}q\lambda_{4}}{\lambda_{2}\mu}\right)\left(1-\frac{t}{\mu^{2}}\right)(1-x_{3})}.$

###  2.3. Obtain the generating function

By using the maple package Ell2, we obtain:

$\displaystyle F_{11}$
$\displaystyle=\frac{x_{{1}}y_{{2}}q\left({t+y_{{3}}q-{q}^{2}tx_{{1}}y_{{2}}y_{{3}}}\right)}{\left({1-x_{{1}}{t}^{2}}\right)\left({1-qy_{{2}}tx_{{1}}}\right)\left({1-{q}^{2}y_{{2}}y_{{3}}x_{{1}}}\right)\left({1-x_{{2}}qy_{{2}}x_{{1}}t}\right)(1-x_{3})},$
$\displaystyle F_{12}$
$\displaystyle=\frac{q^{4}y_{{3}}^{2}x_{{2}}y_{{2}}^{2}x_{{1}}^{2}\left({t+y_{{3}}q}\right)}{\left({1-x_{{1}}{t}^{2}}\right)\left({1-x_{{2}}qy_{{2}}x_{{1}}t}\right)\left({1-{q}^{2}y_{{2}}y_{{3}}x_{{1}}}\right)\left({1-{y_{{3}}}^{2}{q}^{3}x_{{2}}y_{{2}}x_{{1}}}\right)(1-x_{3})},$
$\displaystyle F_{21}$
$\displaystyle=\frac{1}{\left({1-tx_{{2}}}\right)\left({1-qx_{{2}}y_{{3}}}\right)\left({1-x_{{1}}{t}^{2}}\right)\left({1-x_{{2}}qy_{{2}}x_{{1}}t}\right)(1-x_{3})},$
$\displaystyle F_{22}$
$\displaystyle=\frac{y_{{3}}x_{{2}}y_{{2}}q^{2}x_{{1}}\left({t+y_{{3}}q}\right)}{\left({1-x_{{1}}{t}^{2}}\right)\left({1-qx_{{2}}y_{{3}}}\right)\left({1-x_{{2}}qy_{{2}}x_{{1}}t}\right)\left({1-{y_{{3}}}^{2}{q}^{3}x_{{2}}y_{{2}}x_{{1}}}\right)(1-x_{3})}.$

By adding the above four formulas and setting $y_{2}=y_{3}=1$, we obtain

$\displaystyle F(x_{1}$ $\displaystyle,x_{2},x_{3},1,1,q,t)$ $\displaystyle=$
$\displaystyle\frac{\left({1-q{t}^{2}x_{{1}}x_{{2}}}\right)\left({1-{q}^{2}tx_{{1}}x_{{2}}}\right)}{\left({1-qx_{{2}}}\right)\left({1-tx_{{2}}}\right)\left({1-qtx_{{1}}}\right)\left({1-{t}^{2}x_{{1}}}\right)\left({1-{q}^{2}x_{{1}}}\right)\left({1-qtx_{{1}}x_{{2}}}\right)(1-x_{3})}$
(1) $\displaystyle=$
$\displaystyle\frac{1}{\left(1-qtx_{1}\right)\left(1-qtx_{1}x_{2}\right)(1-x_{3})}\left(\frac{1}{1-{t}^{2}x_{{1}}}+{\frac{qx_{{2}}}{1-qx_{{2}}}}\right)\left(\frac{1}{1-{q}^{2}x_{{1}}}+{\frac{tx_{{2}}}{1-tx_{{2}}}}\right).$

The $q,t$-symmetry clearly follows from (1).

##  3\. A Bijective Proof of Theorem 1

In this section, we use the following notation to describe a Dyck path $D$ to
obtain a simpler formula of $\operatorname{\texttt{bounce}}(D)$. For
$\vec{k}=(a,c,e)$, A $\vec{k}$-Dyck path $D$ can be uniquely determined by the
parameters $b,d$ as follows. It starts with a red arrow $S^{a}$ (i.e. up step
$(1,a)$) followed by $b$ blue arrows $W$ (down step $(1,-1)$), then a red
arrow $S^{c}$ followed by $d$ blue arrows $W$, and a red arrow $S^{e}$
followed by $(a+c+e-b-d)$ blue arrows $W$. Clearly, the red ranks of $D$ are
$r_{1}=0,\ r_{2}=a-b$ and $r_{3}=a-b+c-d$. See [4, 11] for detailed concepts,
which are irrelevant here.

The formula of $\operatorname{\texttt{area}}(D)$ is as follows

$\operatorname{\texttt{area}}(D)=r_{2}+r_{3}=2a-2b+c-d.$

We recall that the formula of
$\operatorname{\texttt{bounce}}(D)=b(r_{2},r_{3})$, i.e.,

$b(r_{2},r_{3})=\left\\{\begin{array}[]{ll}2(k_{1}-r_{2})+r_{2}+k_{2}-r_{3}-\min(r_{2},k_{2}),&\text{
if }r_{2}+k_{2}-r_{3}\geq 2\min(r_{2},k_{2});\\\
2(k_{1}-r_{2})+\lceil{\frac{r_{2}+k_{2}-r_{3}}{2}}\rceil,&\text{otherwise.}\end{array}\right.$

Then, we have a simpler formula of $\operatorname{\texttt{bounce}}(D)=B(b,d)$
as follows:

$B(b,d)=\left\\{\begin{array}[]{ll}2b+d-\min(a-b,c),&\text{ if }d\geq
2\min(a-b,c);\\\
2b+\lceil{\frac{d}{2}}\rceil,&\text{otherwise.}\end{array}\right.$

We will give an involution (a bijection whose square is the identity map) on
${\mathcal{D}}_{\vec{k}}$ for $\vec{k}=(a,c,e)$ which interchanges
$\operatorname{\texttt{area}}$ and $\operatorname{\texttt{bounce}}$. If $a\leq
c$, we have $\min(a-b,c)=a-b$ and the formula of $B(b,d)$ becomes simpler. So
we divide the involution into two sub-sections: one for the case $a\leq c$ and
the other for the case $a>c$.

Now we need to give the following Lemma.

######  Lemma 4.

Giving two non-negative integers $c$ and $d$. If the following three
conditions hold

  1. (1)

$Y=\chi((c+\lceil{\frac{d}{2}}\rceil)~{}is~{}odd)$;

  2. (2)

$d^{\prime}=2\lfloor{\frac{d}{2}}\rfloor+Y$;

  3. (3)

$Y^{\prime}=\chi((c+\lceil{\frac{d^{\prime}}{2}}\rceil)~{}is~{}odd)$.

Then $Y^{\prime}=\chi(d~{}is~{}odd)$.

###### Proof.

We have

$c+\lceil{\frac{d^{\prime}}{2}}\rceil=c+\lfloor{\frac{d}{2}}\rfloor+Y$

and

$c+\lfloor{\frac{d}{2}}\rfloor+Y=\left\\{\begin{array}[]{ll}c+\lfloor{\frac{d}{2}}\rfloor+1=c+\lceil{\frac{d}{2}}\rceil,&\hbox{
if }c+\lceil{\frac{d}{2}}\rceil\text{is odd and }d\text{ is odd};\\\
c+\lfloor{\frac{d}{2}}\rfloor+1=c+\lceil{\frac{d}{2}}\rceil+1,&\hbox{ if
}c+\lceil{\frac{d}{2}}\rceil\text{is odd and }d\text{ is even};\\\
c+\lfloor{\frac{d}{2}}\rfloor=c+\lceil{\frac{d}{2}}\rceil-1,&\hbox{ if
}c+\lceil{\frac{d}{2}}\rceil\text{is even and }d\text{ is odd};\\\
c+\lfloor{\frac{d}{2}}\rfloor=c+\lceil{\frac{d}{2}}\rceil,&\hbox{ if
}c+\lceil{\frac{d}{2}}\rceil\text{is even and }d\text{ is
even}.\end{array}\right.$

So,

$c+\lceil{\frac{d^{\prime}}{2}}\rceil\left\\{\begin{array}[]{ll}is~{}odd,&\hbox{
if }d\text{ is odd};\\\ is~{}even,&\hbox{ if }d\text{ is
even}.\end{array}\right.$

We get

$Y^{\prime}=\chi(d~{}is~{}odd).$

###  3.1. $\vec{k}=(a,c,e)$ with $a\leq c$

The formula of $B(b,d)$ is

$B(b,d)=\left\\{\begin{array}[]{ll}3b+d-a,&\text{ if }2(a-b)\leq d;\\\
2b+\lceil{\frac{d}{2}}\rceil,&\text{ if }2(a-b)>d.\end{array}\right.$

Here, we give a map $\varphi:\overline{D}\mapsto\varphi(\overline{D})$ where
$\varphi(\overline{D})$ is determined by its two values
$(b^{\prime},d^{\prime})$, i.e., $\varphi(b,d)=(b^{\prime},d^{\prime})$ which
$(b^{\prime},d^{\prime})$ is defined as follows:

$(b^{\prime},d^{\prime})=\left\\{\begin{array}[]{ll}(b,3a-5b+c-d),&\text{ if
}2(a-b)\leq d\text{ and }3b+d-a\leq c;\\\
(\frac{a-b+c-d-X}{2},2a-2b+X),&\text{ if }2(a-b)\leq d\text{ and }3b+d-a>c;\\\
(a-\lfloor{\frac{d}{2}}\rfloor,2d-2b+c-3\lceil{\frac{d}{2}}\rceil)&\text{ if
}2(a-b)>d\text{ and }2b+\lceil{\frac{d}{2}}\rceil\leq c;\\\
(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}},2\lfloor{\frac{d}{2}}\rfloor+Y),&\text{
if }2(a-b)>d\text{ and }2b+\lceil{\frac{d}{2}}\rceil>c.\end{array}\right.$

where $X$ is $\chi((a-b+c-d)~{}is~{}odd)$ and $Y$ is
$\chi((c+\lceil{\frac{d}{2}}\rceil)~{}is~{}odd)$.

According to the formulas of $B(b,d)$ and $(b^{\prime},d^{\prime})$, we define
the classification situation as follows:

L1:

$2(a-b)\leq d$,

L2:

$2(a-b)>d$;

and

L11:

$2(a-b)\leq d\text{ and }3b+d-a\leq c$,

L12:

$2(a-b)\leq d\text{ and }3b+d-a>c$,

L21:

$2(a-b)>d\text{ and }2b+\lceil{\frac{d}{2}}\rceil\leq c$,

L22:

$2(a-b)>d\text{ and }2b+\lceil{\frac{d}{2}}\rceil>c$.

######  Proposition 5.

the map $\varphi$ is an involution on ${\mathcal{D}}_{\vec{k}}$ for
$\vec{k}=(a,c,e)$ with $a\leq c$ and it interchanges
$\operatorname{\texttt{area}}$ and $\operatorname{\texttt{bounce}}$.

We prove Proposition 5 in two steps as follows:

  1. step 1:

    1. (a)

$D=\varphi(\overline{D})$ is a Dyck path, i.e., $D\in{\mathcal{D}}_{\vec{k}}$,
satisfying $a-b^{\prime}\geq 0$ and $a-b^{\prime}+c-d^{\prime}\geq 0$;

    2. (b)

the map $\varphi$ is an involution, i.e.,
$\varphi(b^{\prime},d^{\prime})=(b^{\prime\prime},d^{\prime\prime})=(b,d)$.
Indeed, the map $\varphi$ is also an involution when restricted to condition
L11, and the same situation holds for L22; It exchanges conditions L12 and
L21. See Figure 1.

  2. step 2:

The map $\varphi$ interchanges $\operatorname{\texttt{area}}$ and
$\operatorname{\texttt{bounce}}$, i.e.,
$\operatorname{\texttt{area}}(D)=\operatorname{\texttt{bounce}}(\overline{D})$
and
$\operatorname{\texttt{bounce}}(D)=\operatorname{\texttt{area}}(\overline{D})$.

Figure 1.  The involution $\varphi$ and its image.

###### Proof.

Recall that $a\leq c$, $a-b\geq 0$ and $a-b+c-d\geq 0$.

  1. step 1:

Case 11: When condition L11 holds, we have
$(b^{\prime},d^{\prime})=(b,3a-5b+c-d)$.

We can obtain

    1. (a)

$\displaystyle a-b^{\prime}$ $\displaystyle=a-b\geq 0;$ $\displaystyle
a-b^{\prime}+c-d^{\prime}$ $\displaystyle=a-b+c-(3a-5b+c-d)=d-2(a-b)+2b\geq
2b\geq 0.$

    2. (b)

It is easy to check that the condition L11 for $(b^{\prime},d^{\prime})$
holds:

$2(a-b^{\prime})-d^{\prime}=2(a-b)-(3a-5b+c-d)=3b+d-a-c\leq 0$

and

$3b^{\prime}+d^{\prime}-a=3b+(3a-5b+c-d)-a=2(a-b)-d+c\leq c.$

So,

$\displaystyle(b^{\prime\prime},d^{\prime\prime})=(b^{\prime},3a-5b^{\prime}+c-d^{\prime})=(b,3a-5b+c-(3a-5b+c-d))=(b,d).$

Case 12: When condition L12 holds, we have
$(b^{\prime},d^{\prime})=({\frac{a-b+c-d-X}{2}},2a-2b+X).$

We can obtain

    1. (a)

$\displaystyle a-b^{\prime}$
$\displaystyle=a-{\frac{a-b+c-d-X}{2}}=\frac{a+b-c+d+X}{2};$ $\displaystyle
a-b^{\prime}+c-d^{\prime}$ $\displaystyle=a-{\frac{a-b+c-d-X}{2}}+c-(2a-2b+X)$
$\displaystyle=\frac{-3a+5b+c+d-X}{2}.$

Recall that condition L12: $2(a-b)\leq d$ and $3b+d-a>c$. Then $2b\geq 2a-d$,
$3b>a+c-d$ and $a+b-c+d>2a-2b\geq 0$. We have $5b>3a+c-2d$. Thus,

$\displaystyle a-b^{\prime}=\frac{a+b-c+d+X}{2}>0;$ $\displaystyle
a-b^{\prime}+c-d^{\prime}=\frac{-3a+5b+c+d-X}{2}>\frac{2c-d-X}{2}.$

By $a\leq c$ and $d\leq a+c\leq 2c$, we have $2c-d-X<0$ only if $2c=d$ and
$X=1$. But when $2c=d$, we have $a=c$, $b=0$ and $a-b+c-d=0$, i.e., $X=0$.
This contradicts with $X=1$. So, $a-b^{\prime}+c-d^{\prime}>0$.

    2. (b)

It is easy to check that the condition L21 for $(b^{\prime},d^{\prime})$
holds:

$\displaystyle 2(a-b^{\prime})-d^{\prime}$
$\displaystyle=2a-(a-b+c-d-X)-(2a-2b+X)$
$\displaystyle=2a+c-(a\\!-\\!b\\!-\\!d)-2a+2b$ $\displaystyle=3b+d-a-c>0$

and

$\displaystyle 2b^{\prime}+\lceil{\frac{d^{\prime}}{2}}\rceil$
$\displaystyle=a-b+c-d-X+a-b+X$ $\displaystyle=2a-2b-d+c\leq c.$

So,

$\displaystyle(b^{\prime\prime},d^{\prime\prime})$
$\displaystyle=(a-\lfloor{\frac{d^{\prime}}{2}}\rfloor,2d^{\prime}-2b^{\prime}+c-3\lceil{\frac{d^{\prime}}{2}}\rceil)$
$\displaystyle=(a-(a-b),4a-4b+2X-(a-b+c-d-X)+c-3(a-b+X)$
$\displaystyle=(b,d).$

Case 21: When condition L21 holds, we have

$(b^{\prime},d^{\prime})=(a-\lfloor{\frac{d}{2}}\rfloor,2d-2b+c-3\lceil{\frac{d}{2}}\rceil).$

We can obtain

    1. (a)

$\displaystyle a-b^{\prime}$
$\displaystyle=a-(a-\lfloor{\frac{d}{2}}\rfloor)=\lfloor{\frac{d}{2}}\rfloor\geq
0;$ $\displaystyle a-b^{\prime}+c-d^{\prime}$
$\displaystyle=\lfloor{\frac{d}{2}}\rfloor+c-(2d-2b+c-3\lceil{\frac{d}{2}}\rceil)=2b-d+2\lceil{\frac{d}{2}}\rceil\geq
0.$

    2. (b)

It is easy to check that the condition L12 for $(b^{\prime},d^{\prime})$
holds:

$\displaystyle 2(a-b^{\prime})-d^{\prime}$
$\displaystyle=2a-2b^{\prime}-d^{\prime}$
$\displaystyle=2a-2(a-\lfloor{\frac{d}{2}}\rfloor)-(2d-2b+c-3\lceil{\frac{d}{2}}\rceil))$
$\displaystyle=2b+\lceil{\frac{d}{2}}\rceil-c\leq 0$

and

$\displaystyle 3b^{\prime}+d^{\prime}-a$
$\displaystyle=3(a-\lfloor{\frac{d}{2}}\rfloor)+(2d-2b+c-3\lceil{\frac{d}{2}}\rceil)-a$
$\displaystyle=2(a-b)-d+c>c.$

So,

$(b^{\prime\prime},d^{\prime\prime})=({\frac{a-b^{\prime}+c-d^{\prime}-X^{\prime}}{2}},2a-2b^{\prime}+X^{\prime})$

where $X^{\prime}$ is $\chi(a-b^{\prime}+c-d^{\prime}~{}is~{}odd)$ which is
equal to $\chi(d~{}is~{}odd)$ by the formula
$a-b^{\prime}+c-d^{\prime}=2b-d+2\lceil{\frac{d}{2}}\rceil$ from $(a)$. Then
we have

$\displaystyle(b^{\prime\prime},d^{\prime\prime})$
$\displaystyle=(b+\lceil{\frac{d}{2}}\rceil-{\frac{d}{2}}-{\frac{X^{\prime}}{2}},2a-2(a-\lfloor{\frac{d}{2}}\rfloor)+X^{\prime})$
$\displaystyle=(b+\lceil{\frac{d}{2}}\rceil-{\frac{d+X^{\prime}}{2}},2\lfloor{\frac{d}{2}}\rfloor+X^{\prime})$
$\displaystyle=(b,d).$

Case 22: When condition L22 holds, we have

$(b^{\prime},d^{\prime})=(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}},2\lfloor{\frac{d}{2}}\rfloor+Y).$

We can obtain

    1. (a)

$\displaystyle a-b^{\prime}$
$\displaystyle=a-(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}})={\frac{2b+d-c+\lfloor{\frac{d}{2}}\rfloor+Y}{2}};$
$\displaystyle a-b^{\prime}+c-d^{\prime}$
$\displaystyle={\frac{2b+2d-c-\lceil{\frac{d}{2}}\rceil+Y}{2}}+c-(2\lfloor{\frac{d}{2}}\rfloor+Y)$
$\displaystyle={\frac{2b+2d+c-\lceil{\frac{d}{2}}\rceil+Y-4\lfloor{\frac{d}{2}}\rfloor-2Y}{2}}$
$\displaystyle={\frac{2b+2d+c-\lceil{\frac{d}{2}}\rceil-4\lfloor{\frac{d}{2}}\rfloor-Y}{2}}$
$\displaystyle={\frac{2b+d+c-3\lfloor{\frac{d}{2}}\rfloor-Y}{2}}.$

Recall that condition L22: $2(a-b)>d$ and $2b+\lceil{\frac{d}{2}}\rceil>c$. We
are in the case $a\leq c$. Then $2b+\lceil{\frac{d}{2}}\rceil>c\geq
a>{\frac{d}{2}}+b$ and $2b+d-c+\lfloor{\frac{d}{2}}\rfloor+Y>0$.

We know

$a-b^{\prime}>0$

and
$2b+d+c-3\lfloor{\frac{d}{2}}\rfloor-Y>3b+{\frac{3d}{2}}-3\lfloor{\frac{d}{2}}\rfloor-Y>0$,
i.e.,

$a-b^{\prime}+c-d^{\prime}>0.$

    2. (b)

It is easy to check that the condition L22 for $(b^{\prime},d^{\prime})$
holds:

$\displaystyle 2(a-b^{\prime})-d^{\prime}$
$\displaystyle=2a-2(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}})-(2\lfloor{\frac{d}{2}}\rfloor+Y)$
$\displaystyle=2b+2d-c-\lceil{\frac{d}{2}}\rceil+Y-2\lfloor{\frac{d}{2}}\rfloor-Y$
$\displaystyle=2b+\lceil{\frac{d}{2}}\rceil-c>0$

and

$\displaystyle 2b^{\prime}+\lceil{\frac{d^{\prime}}{2}}\rceil$
$\displaystyle=2(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}})+\lfloor{\frac{d}{2}}\rfloor+Y$
$\displaystyle=2a-2b-2d+c+\lceil{\frac{d}{2}}\rceil-Y+\lfloor{\frac{d}{2}}\rfloor+Y$
$\displaystyle=2a-2b-d+c>c.$

So,

$\displaystyle(b^{\prime\prime},d^{\prime\prime})$
$\displaystyle=(a-b^{\prime}-d^{\prime}+{\frac{c+\lceil{\frac{d^{\prime}}{2}}\rceil-Y^{\prime}}{2}},2\lfloor{\frac{d^{\prime}}{2}}\rfloor+Y^{\prime})$
$\displaystyle=(a-(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}})-(2\lfloor{\frac{d}{2}}\rfloor+Y)+{\frac{c+\lfloor{\frac{d}{2}}\rfloor+Y-Y^{\prime}}{2}},2\lfloor{\frac{d}{2}}\rfloor+Y^{\prime})$
$\displaystyle=(b+{\frac{d-2\lfloor{\frac{d}{2}}\rfloor-Y^{\prime}}{2}},2\lfloor{\frac{d}{2}}\rfloor+Y^{\prime})$

where $Y^{\prime}$ is
$\chi((c+\lceil{\frac{d^{\prime}}{2}}\rceil)~{}is~{}odd)$.

Recall that $Y$ is $\chi((c+\lceil{\frac{d}{2}}\rceil)~{}is~{}odd)$ and
$d^{\prime}=2\lfloor{\frac{d}{2}}\rfloor+Y$. So, by Lemma 4, we have

$(b^{\prime\prime},d^{\prime\prime})=(b+{\frac{d-2\lfloor{\frac{d}{2}}\rfloor-Y^{\prime}}{2}},2\lfloor{\frac{d}{2}}\rfloor+Y^{\prime})=(b,d).$

  2. step 2:

Case 11: When condition L11 holds, we have
$\operatorname{\texttt{area}}(\overline{D})=2a-2b+c-d$ and
$\operatorname{\texttt{bounce}}(\overline{D})=3b+d-a$. We calculate

$\displaystyle\operatorname{\texttt{area}}(D)$
$\displaystyle=2a-2b^{\prime}+c-d^{\prime}=2a-2b+c-(3a-5b+c-d)$
$\displaystyle=3b+d-a=\operatorname{\texttt{bounce}}(\overline{D});$
$\displaystyle\operatorname{\texttt{bounce}}(D)$
$\displaystyle=3b^{\prime}+d^{\prime}-a=3b+(3a-5b+c-d)-a$
$\displaystyle=2a-2b+c-d=\operatorname{\texttt{area}}(\overline{D}).$

Case 12: When condition L12 holds, we have
$\operatorname{\texttt{area}}(\overline{D})=2a-2b+c-d$ and
$\operatorname{\texttt{bounce}}(\overline{D})=3b+d-a$. We calculate

$\displaystyle\operatorname{\texttt{area}}(D)$
$\displaystyle=2a-2b^{\prime}+c-d^{\prime}=2a-(a-b+c-d-X)+c-(2a-2b+X)$
$\displaystyle=2a+c-(a\\!-\\!b\\!+\\!c\\!-\\!d)-2a+2b$
$\displaystyle=3b+d-a=\operatorname{\texttt{bounce}}(\overline{D});$
$\displaystyle\operatorname{\texttt{bounce}}(D)$
$\displaystyle=2b^{\prime}+\lceil{\frac{d^{\prime}}{2}}\rceil=a-b+c-d-X+a-b+X$
$\displaystyle=2a-2b+c-d=\operatorname{\texttt{area}}(\overline{D}).$

Case 21: When condition L21 holds, we have
$\operatorname{\texttt{area}}(\overline{D})=2a-2b+c-d$ and
$\operatorname{\texttt{bounce}}(\overline{D})=2b+\lceil{\frac{d}{2}}\rceil$.
We calculate

$\displaystyle\operatorname{\texttt{area}}(D)$
$\displaystyle=2a-2b^{\prime}+c-d^{\prime}$
$\displaystyle=2a-2(a-\lfloor{\frac{d}{2}}\rfloor)+c-(2d-2b+c-3\lceil{\frac{d}{2}}\rceil))$
$\displaystyle=2b+\lceil{\frac{d}{2}}\rceil=\operatorname{\texttt{bounce}}(\overline{D});$
$\displaystyle\operatorname{\texttt{bounce}}(D)$
$\displaystyle=3b^{\prime}+d^{\prime}-a$
$\displaystyle=3(a-\lfloor{\frac{d}{2}}\rfloor)+(2d-2b+c-3\lceil{\frac{d}{2}}\rceil)-a$
$\displaystyle=2a-2b+c-d=\operatorname{\texttt{area}}(\overline{D}).$

Case 22: When condition L12 holds, we have
$\operatorname{\texttt{area}}(\overline{D})=2a-2b+c-d$ and
$\operatorname{\texttt{bounce}}(\overline{D})=2b+\lceil{\frac{d}{2}}\rceil$.
We calculate

$\displaystyle\operatorname{\texttt{area}}(D)$
$\displaystyle=2a-2b^{\prime}+c-d^{\prime}$
$\displaystyle=2a-2(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}})+c-(2\lfloor{\frac{d}{2}}\rfloor+Y)$
$\displaystyle=2b+2d-c-\lceil{\frac{d}{2}}\rceil+Y+c-2\lfloor{\frac{d}{2}}\rfloor-Y$
$\displaystyle=2b+\lceil{\frac{d}{2}}\rceil=\operatorname{\texttt{bounce}}(\overline{D});$
$\displaystyle\operatorname{\texttt{bounce}}(D)$
$\displaystyle=2b^{\prime}+\lceil{\frac{d^{\prime}}{2}}\rceil$
$\displaystyle=2(a-b-d+{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}})+\lfloor{\frac{d}{2}}\rfloor+Y$
$\displaystyle=2a-2b-2d+c+\lceil{\frac{d}{2}}\rceil-Y+\lfloor{\frac{d}{2}}\rfloor+Y$
$\displaystyle=2a-2b+c-d=\operatorname{\texttt{area}}(\overline{D}).$

###  3.2. $\vec{k}=(a,c,e)$ with $a>c$

The formula of $B(b,d)$ is

$B(b,d)=\left\\{\begin{array}[]{ll}2b-c+d,&\text{ if }a-b>c\text{ and }2c\leq
d;\\\ 3b+d-a,&\text{ if }a-b\leq c\text{ and }2(a-b)\leq d;\\\
2b+\lceil{\frac{d}{2}}\rceil,&\text{ if }\min(2(a-b),2c)>d.\end{array}\right.$

Here, we give a map $\psi:D\mapsto\psi(D)\in{\mathcal{D}}_{\vec{k}}$, where
$\psi(D)$ is determined by its two values $(b^{\prime},d^{\prime})$, i.e.,
$\psi(b,d)=(b^{\prime},d^{\prime})$ which $(b^{\prime},d^{\prime})$ is defined
as follows:

$\displaystyle(b^{\prime},d^{\prime})$
$\displaystyle=\left\\{\begin{array}[]{ll}(a-b+c-d,d),&\text{ if }a-b>c\text{
and }2c\leq d\\\ &\text{ except }b=0\text{ and }d=2c;\\\ (a-c,d),&\text{ if
}b=0\text{ and }d=2c;\\\ (0,d),&\text{ if }b=a-c\text{ and }d=2(a-b);\\\
({\frac{a-b+c-d-X}{2}},2a-2b+X),&\text{ if }a-b\leq c\text{ and }2(a-b)\leq
d\\\ &\text{ except }b=a-c\text{ and }d=2(a-b);\\\
(a-\lfloor{\frac{d}{2}}\rfloor,2d-2b+c-3\lceil{\frac{d}{2}}\rceil),&\text{ if
}\min(2(a-b),2c)>d\\\ &\text{ and }2b+\lceil{\frac{d}{2}}\rceil\leq c;\\\
(a-b-d-{\frac{c+\lceil{\frac{d}{2}}\rceil-Y}{2}},2\lfloor{\frac{d}{2}}\rfloor+Y),&\text{
if }\min(2(a-b),2c)>d\\\ &\text{ and
}2b+\lceil{\frac{d}{2}}\rceil>c.\end{array}\right.$

where $X$ is $\chi((a-b+c-d)~{}is~{}odd)$ and $Y$ is
$\chi((c+\lceil{\frac{d}{2}}\rceil)~{}is~{}odd)$.

According to the formulas of $B(b,d)$ and $(b^{\prime},d^{\prime})$, we define
the classification situation as follows:

G1:

$a-b>c$ and $2c\leq d$,

G2:

$a-b\leq c$ and $2(a-b)\leq d$,

G3:

$\text{min}(2(a-b),2c)>d$;

and

G11:

$a-b>c$ and $2c\leq d$ except $b=0$ and $2c=d$(G1/G12),

G12:

$b=0$ and $2c=d$,

G21:

$a-b=c$ and $2(a-b)=d$,

G22:

$a-b\leq c$ and $2(a-b)\leq d$ except $a-b=c$ and $2(a-b)=d$(G2/G21),

G31:

$\text{min}(2(a-b),2c)>d\text{ and }2b+\lceil{\frac{d}{2}}\rceil\leq c$,

G32:

$\text{min}(2(a-b),2c)>d\text{ and }2b+\lceil{\frac{d}{2}}\rceil>c$.

######  Proposition 6.

The map $\psi$ is an involution on ${\mathcal{D}}_{\vec{k}}$ for
$\vec{k}=(a,c,e)$ with $a>c$ and it interchanges
$\operatorname{\texttt{area}}$ and $\operatorname{\texttt{bounce}}$.

Figure 2. The involution $\psi$ and its image.

###### Proof.

This proof is similar to the case $a\leq c$, but we note that condition G12
contains only one Dyck path, so does condition G21. They exchange with each
other under the map $\psi$.

We only outline the two steps of the proof as follows and omit the details.

  1. step 1:

    1. (a)

For a given Dyck path $\overline{D}\in{\mathcal{D}}_{\vec{k}}$,
$D=\psi(\overline{D})$ is also in ${\mathcal{D}}_{\vec{k}}$, i.e., satisfying
$a-b^{\prime}\geq 0$ and $a-b^{\prime}+c-d^{\prime}\geq 0$;

    2. (b)

The map $\psi$ is an involution, i.e.,
$\psi(b^{\prime},d^{\prime})=(b^{\prime\prime},d^{\prime\prime})=(b,d)$.
Indeed, the map $\psi$ is also an involution when restricted to condition G11,
and the same situation holds for G32; It exchanges the two Dyck paths under
the conditions G12 and G21; It exchanges conditions G22 and G31. See Figure 2.

  2. step 2:

The map $\psi$ interchanges $\operatorname{\texttt{area}}$ and
$\operatorname{\texttt{bounce}}$, i.e.,
$\operatorname{\texttt{area}}(D)=\operatorname{\texttt{bounce}}(\overline{D})$
and
$\operatorname{\texttt{bounce}}(D)=\operatorname{\texttt{area}}(\overline{D})$.

##  4\. An Algebraic Proof of $C_{k^{4}}(q,t)=C_{k^{4}}(t,q)$

In this section, we use the same idea in section 2 to prove
$C_{k^{4}}(q,t)=C_{k^{4}}(t,q)$. A $k^{4}$-Dyck path $D$ can be uniquely
determined by the parameters $a,b,c$ as follows. It starts with a red arrow
$S^{k}$ (i.e. up step $(1,k)$) followed by $a$ blue arrows $W$ (down step
$(1,-1)$), then a red arrow $S^{k}$ followed by $b$ blue arrows $W$, and a red
arrow $S^{k}$ followed by $c$ blue arrows $W$, finally a red arrow $S^{k}$
followed by $(4k-a-b-c)$ blue arrows $W$. The parameters $a,b,c$ satisfy the
conditions $0\leq a\leq k$ , $0\leq b\leq 2k-a$ and $0\leq c\leq 3k-a-b$. The
red ranks $(r_{1}=0,r_{2},r_{3},r_{4})$ of $D$ can be written by the
parameters $a,b,c$ as follows.

$r_{2}=k-a,\ r_{3}=2k-a-b,\ r_{4}=3k-a-b-c.$

Then, the area of $D$ is simply
$\operatorname{\texttt{area}}(D)=r_{2}+r_{3}+r_{4}=6k-3a-2b-c$ and the bounce
of $D$ is given in [9] by

$\displaystyle\operatorname{\texttt{bounce}}(D)=$ (12)
$\displaystyle\left\\{\begin{array}[]{ll}6a+3b+c-4k,&\text{ if }b\geq
2k-2a\text{ and }c\geq 4k-2a-2b;\\\ 5a+2b+\lceil{\frac{c}{2}}\rceil-2k,&\text{
if }b\geq 2k-2a\text{ and }c<4k-2a-2b;\\\ 4a+2b+c-2k,&\text{ if
}b<2k-2a,b\text{ is even and }c\geq 3k-a-\frac{3b}{2};\\\
2a+\frac{b}{2}+k+\lceil{\frac{3a+\frac{3b}{2}+c-3k}{2}}\rceil,&\text{ if
}b<2k-2a,b\text{ is even and }\\\ &\ \ \ 3k-3a-\frac{3b}{2}\leq
c<3k-a-\frac{3b}{2};\\\ 3a+b+\lceil{\frac{c}{3}}\rceil,&\text{ if
}b<2k-2a,b\text{ is even and }c<3k-3a-3s;\\\ 4a+2b+c-2k+1,&\text{ if
}b<2k-2a,b\text{ is odd and }c\geq 3k-a-\frac{3(b+1)}{2}+1;\\\
2a+\frac{b+1}{2}+k&\\\ \ \ \ \ \ \ \ \
+\lceil{\frac{3a+\frac{3(b+1)}{2}+c-3k-1}{2}}\rceil,&\text{ if
}b<2k-2a,b\text{ is odd and }\\\ &\ \ \ 3k-3a-\frac{3(b+1)}{2}+1\leq
c<3k-a-\frac{3(b+1)}{2}+1;\\\ 3a+b+1+\lceil{\frac{c-1}{3}}\rceil,&\text{ if
}b<2k-2a,b\text{ is odd and }c<3k-3a-\frac{3(b+1)}{2}+1.\end{array}\right.$

We can construct the generating function with respect to $k,a,b,c$:

$\displaystyle H(x,y_{2},y_{3},y_{4},q,t)$ $\displaystyle=\sum_{k\geq
0}x^{k}\sum_{D\in{\mathcal{D}}_{K}}q^{\operatorname{\texttt{area}}(D)}t^{\operatorname{\texttt{bounce}}(D)}y_{2}^{a}y_{3}^{b}y_{4}^{c}$
$\displaystyle=\sum_{k\geq 0}x^{k}\sum_{0\leq a\leq k,\ 0\leq b\leq 2k-a,\
0\leq c\leq
3k-a-b}q^{6k-3a-2b-c}t^{\operatorname{\texttt{bounce}}(D)}y_{2}^{a}y_{3}^{b}y_{4}^{c},$

where we allowed $k=0$, which does not affect the computation.

Then it is sufficient to prove the $q,t$ symmetry of

$H(x,1,1,1,q,t)=\sum_{k\geq 0}x^{k}C_{k^{4}}(q,t).$

###  4.1. Crude generating function

Due to the piecewise linearity of $\operatorname{\texttt{bounce}}(D)$ in
formula (4), we divide the generating function $H(x,y_{2},y_{3},y_{4},q,t)$
into three parts as follows.

Part 1:

$b\geq 2k-2a$;

Part 2:

$b<2k-2a$ and $b$ is even;

Part 3:

$b<2k-2a$ and $b$ is odd.

There are two cases for Part 1 and 3 cases for Part 2 and Part 3 each, so it
is convenient to write the generating function
$H(x,y_{2},y_{3},y_{4},q,t)=H=H_{1}+H_{2}+H_{3}=H_{11}+H_{12}+H_{21}+H_{22}+H_{23}+H_{31}+H_{32}+H_{33}$.

Part 1: $b\geq 2k-2a$, i.e., $2a+b-2k\geq 0$. By a similar computation for
section 2, the formulas $H_{11}$ and $H_{12}$ are given as follows.

Case 1: $c\geq 4k-2a-2b$, i.e., $2a+2b+c-4k\geq 0$. Then, we have

$\operatorname{\texttt{bounce}}(D)=6a+3b+c-4k.$

and

$\displaystyle H_{11}=$ $\displaystyle\sum_{k\geq 0}x^{k}\sum_{0\leq a\leq k,\
0\leq b\leq 2k-a,\ 0\leq c\leq 3k-a-b,\atop 2a+b-2k\geq 0,\ 2a+2b+c-4k\geq
0}q^{6k-3a-2b-c}t^{6a+3b+c-4k}y_{2}^{a}y_{3}^{b}y_{4}^{c}$ $\displaystyle=$
$\displaystyle\sum_{k,a,b,c\geq
0}\mathop{\Omega}_{\geq}\left(x^{k}q^{6k-3a-2b-c}t^{6a+3b+c-4k}y_{2}^{a}y_{3}^{b}y_{4}^{c}\lambda_{1}^{k-a}\lambda_{2}^{2k-a-b}\lambda_{3}^{3k-a-
b-c}\lambda_{4}^{2a+b-2k}\lambda_{5}^{2a+2b+c-4k}\right)$ $\displaystyle=$
$\displaystyle\mathop{\Omega}_{\geq}\frac{1}{(1-\frac{xq^{6}\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}}{t^{4}\lambda_{4}^{2}\lambda_{5}^{4}})(1-\frac{t^{6}y_{2}\lambda_{4}^{2}\lambda_{5}^{2}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}})(1-\frac{t^{3}y_{3}\lambda_{4}\lambda_{5}^{2}}{q^{2}\lambda_{2}\lambda_{3}})(1-\frac{ty_{4}\lambda_{5}}{q\lambda_{3}})}.$

Case 2: $c<4k-2a-2b$, i.e., $4k-2a-2b-c-1\geq 0$. Then, we have

$\displaystyle\operatorname{\texttt{bounce}}(D)=5a+2b+\lceil\frac{c}{2}\rceil-2k,$

and

$\displaystyle
H_{12}=\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}\frac{1+\mu}{\lambda_{5}(1-\frac{xq^{6}\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}\lambda_{5}^{4}}{t^{2}\lambda_{4}^{2}})(1-\frac{t^{5}y_{2}\lambda_{4}^{2}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{5}^{2}})(1-\frac{t^{2}y_{3}\lambda_{4}}{q^{2}\lambda_{2}\lambda_{3}\lambda_{5}^{2}})(1-\frac{y_{4}\mu}{q\lambda_{3}\lambda_{5}})(1-\frac{t}{\mu^{2}})}.$

Part 2: $b<2k-2a$ (i.e.,$2k-2a-b-1\geq 0$) and $b$ is even. Let $b=2s$, then
$2k-2a-2s-1\geq 0$ and
$\operatorname{\texttt{area}}(D)=6k-3a-2b-c=6k-3a-4s-c$.

Case 1: $c\geq 3k-a-\frac{3b}{2}$, i.e., $a+3s+c-3k\geq 0$. Then, we have

$\operatorname{\texttt{bounce}}(D)=4a+2b+c-2k=4a+4s+c-2k.$

We can write the generating function $H_{21}$ as follows.

$\displaystyle H_{21}=\sum_{k\geq 0}x^{k}\sum_{0\leq a\leq k,\ 0\leq 2s\leq
2k-a,\ 0\leq c\leq 3k-a-2s,\atop 2k-2a-2s-1\geq 0,\ a+3s+c-3k\geq
0}q^{6k-3a-4s-c}t^{4a+4s+c-2k}y_{2}^{a}y_{3}^{2s}y_{4}^{c}$ $\displaystyle=$
$\displaystyle\sum_{k,a,s,c\geq
0}\mathop{\Omega}_{\geq}\left(x^{k}q^{6k-3a-4s-c}t^{4a+4s+c-2k}y_{2}^{a}y_{3}^{2s}y_{4}^{c}\lambda_{1}^{k-a}\lambda_{2}^{2k-a-2s}\lambda_{3}^{3k-a-2s-c}\lambda_{4}^{2k-2a-2s-1}\lambda_{5}^{a+3s+c-3k}\right)$
$\displaystyle=$
$\displaystyle\mathop{\Omega}_{\geq}\frac{1}{\lambda_{4}(1-\frac{xq^{6}\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}\lambda_{4}^{2}}{t^{2}\lambda_{5}^{3}})(1-\frac{t^{4}y_{2}\lambda_{5}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}^{2}})(1-\frac{t^{4}y_{3}^{2}\lambda_{5}^{3}}{q^{4}\lambda_{2}^{2}\lambda_{3}^{2}\lambda_{4}^{2}})(1-\frac{ty_{4}\lambda_{5}}{q\lambda_{3}})}.$

Case 2: $3k-3a-\frac{3b}{2}\leq c<3k-a-\frac{3b}{2}$, i.e., $3a+3s+c-3k\geq 0$
and $3k-a-3s-c-1\geq 0$. Then, we have

$\operatorname{\texttt{bounce}}(D)=2a+\frac{b}{2}+k+\lceil{\frac{3a+\frac{3b}{2}+c-3k}{2}}\rceil=2a+s+k+\lceil{\frac{3a+3s+c-3k}{2}}\rceil.$

Let $p=\lceil{\frac{3a+3s+c-3k}{2}}\rceil$. By Lemma 3, we can write the
generating function $H_{22}$ as follows.

$\displaystyle H_{22}$
$\displaystyle=\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}\frac{1+\mu}{\lambda_{4}\lambda_{6}(1-\frac{xq^{6}t\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}\lambda_{4}^{2}\lambda_{6}^{3}}{\mu^{3}\lambda_{5}^{3}})(1-\frac{t^{2}y_{2}\mu^{3}\lambda_{5}^{3}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}^{2}\lambda_{6}})(1-\frac{ty_{3}^{2}\mu^{3}\lambda_{5}^{3}}{q^{4}\lambda_{2}^{2}\lambda_{3}^{2}\lambda_{4}^{2}\lambda_{6}^{3}})(1-\frac{y_{4}\mu\lambda_{5}}{q\lambda_{3}\lambda_{6}})(1-\frac{t}{\mu^{2}})};$

Case 3: $c<3k-3a-\frac{3b}{2}$, i.e., $3k-3a-3s-c-1\geq 0$. Then, we have

$\operatorname{\texttt{bounce}}(D)=3a+b+\lceil{\frac{c}{3}}\rceil=3a+2s+\lceil{\frac{c}{3}}\rceil.$

Let $p=\lceil{\frac{c}{3}}\rceil$. By Lemma 3, we can write the generating
function $H_{23}$ as follows.

$\displaystyle H_{23}$
$\displaystyle=\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}\frac{1+\mu+\mu^{2}}{\lambda_{4}\lambda_{5}(1-xq^{6}\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}\lambda_{4}^{2}\lambda_{5}^{3})(1-\frac{t^{3}y_{2}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}^{2}\lambda_{5}^{3}})(1-\frac{t^{2}y_{3}^{2}}{q^{4}\lambda_{2}^{2}\lambda_{3}^{2}\lambda_{4}^{2}\lambda_{5}^{3}})(1-\frac{y_{4}\mu}{q\lambda_{3}\lambda_{5}})(1-\frac{t}{\mu^{3}})};$

Part 3: $b<2k-2a$ (i.e.,$2k-2a-b-1\geq 0$) and $b$ is odd. Let $b=2s+1$, then
$2k-2a-2s-2\geq 0$ and
$\operatorname{\texttt{area}}(D)=6k-3a-2b-c=6k-3a-4s-c-2$. By a similar
computation for Part 2, the formulas $H_{31},H_{32},H_{33}$ are given as
follows.

$\displaystyle H_{31}$
$\displaystyle=\mathop{\Omega}_{\geq}\frac{t^{3}y_{3}\lambda_{5}^{2}}{q^{2}\lambda_{2}\lambda_{3}\lambda_{4}^{2}(1-\frac{xq^{6}\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}\lambda_{4}^{2}}{t^{2}\lambda_{5}^{3}})(1-\frac{t^{4}y_{2}\lambda_{5}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}^{2}})(1-\frac{t^{4}y_{3}^{2}\lambda_{5}^{3}}{q^{4}\lambda_{2}^{2}\lambda_{3}^{2}\lambda_{4}^{2}})(1-\frac{ty_{4}\lambda_{5}}{q\lambda_{3}})};$
$\displaystyle H_{32}$
$\displaystyle=\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}\frac{t(1+\mu)\mu^{2}y_{3}\lambda_{5}^{2}}{q^{2}\lambda_{2}\lambda_{3}\lambda_{4}^{2}\lambda_{6}^{3}(1-\frac{xq^{6}t\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}\lambda_{4}^{2}\lambda_{6}^{3}}{\mu^{3}\lambda_{5}^{3}})(1-\frac{t^{2}\mu^{3}y_{2}\lambda_{5}^{3}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}^{2}\lambda_{6}})}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \
\times\frac{1}{(1-\frac{t\mu^{3}y_{3}^{2}\lambda_{5}^{3}}{q^{4}\lambda_{2}^{2}\lambda_{3}^{2}\lambda_{4}^{2}\lambda_{6}^{3}})(1-\frac{\mu
y_{4}\lambda_{5}}{q\lambda_{3}\lambda_{6}})(1-\frac{t}{u^{2}})};$
$\displaystyle H_{33}$
$\displaystyle=\mathop{\Omega}_{\geq}\mathop{\Omega}_{\mu=}\frac{t^{2}(1+\mu+\mu^{2})y_{3}}{q^{2}\mu\lambda_{2}\lambda_{3}\lambda_{4}^{2}\lambda_{5}^{3}(1-xq^{6}\lambda_{1}\lambda_{2}^{2}\lambda_{3}^{3}\lambda_{4}^{2}\lambda_{5}^{3})(1-\frac{t^{3}y_{2}}{q^{3}\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}^{2}\lambda_{5}^{3}})}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \
\times\frac{1}{(1-\frac{t^{2}y_{3}^{2}}{q^{4}\lambda_{2}^{2}\lambda_{3}^{2}\lambda_{4}^{2}\lambda_{5}^{3}})(1-\frac{\mu
y_{4}}{q\lambda_{3}\lambda_{5}})(1-\frac{t}{\mu^{3}})}.$

###  4.2. Obtain the generating function

By using the maple package Ell2, we obtain:

$\displaystyle H_{11}$
$\displaystyle=\frac{1-{q}^{2}{t}^{8}{x}^{2}y_{{2}}{y_{{3}}}^{3}y_{{4}}}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-q{t}^{3}x{y_{{3}}}^{2}y_{{4}}}\right)\left({1-q{t}^{4}xy_{{2}}{y_{{4}}}^{2}}\right)\left({1-q{t}^{5}xy_{{2}}y_{{3}}}\right)\left({1-{t}^{6}xy_{{2}}y_{{3}}y_{{4}}}\right)},$
$\displaystyle H_{12}$
$\displaystyle=\frac{y_{{2}}q^{2}t^{3}x\left({q+ty_{{4}}}\right)}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-{q}^{3}{t}^{3}xy_{{2}}}\right)\left({1-q{t}^{5}xy_{{2}}y_{{3}}}\right)\left({1-q{t}^{4}xy_{{2}}{y_{{4}}}^{2}}\right)},$
$\displaystyle H_{21}$
$\displaystyle=\frac{q^{3}txy_{{4}}^{3}}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-{q}^{3}tx{y_{{4}}}^{3}}\right)\left({1-q{t}^{3}x{y_{{3}}}^{2}y_{{4}}}\right)\left({1-q{t}^{4}xy_{{2}}{y_{{4}}}^{2}}\right)},$
$\displaystyle H_{22}$
$\displaystyle=\frac{y_{{2}}y_{{4}}^{3}x^{2}t^{4}q^{5}\left({q+ty_{{4}}}\right)}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-{q}^{3}tx{y_{{4}}}^{3}}\right)\left({1-{q}^{3}{t}^{3}xy_{{2}}}\right)\left({1-q{t}^{4}xy_{{2}}{y_{{4}}}^{2}}\right)},$
$\displaystyle H_{23}$
$\displaystyle=\frac{q^{4}x\left({{q}^{2}+t{y_{{4}}}^{2}+qty_{{4}}}\right)}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-{q}^{6}x}\right)\left({1-{q}^{3}tx{y_{{4}}}^{3}}\right)\left({1-{q}^{3}{t}^{3}xy_{{2}}}\right)},$
$\displaystyle H_{31}$
$\displaystyle=\frac{q^{2}t^{2}xy_{{4}}y_{{3}}\left({q+ty_{{4}}-{q}^{2}{t}^{3}x{y_{{3}}}^{2}y_{{4}}}\right)}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-{q}^{3}tx{y_{{4}}}^{3}}\right)\left({1-q{t}^{3}x{y_{{3}}}^{2}y_{{4}}}\right)\left({1-q{t}^{4}xy_{{2}}{y_{{4}}}^{2}}\right)},$
$\displaystyle H_{32}$
$\displaystyle=\frac{y_{{2}}y_{{3}}y_{{4}}x^{2}t^{5}q^{5}\left({q+ty_{{4}}}\right)}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-{q}^{3}tx{y_{{4}}}^{3}}\right)\left({1-{q}^{3}{t}^{3}xy_{{2}}}\right)\left({1-q{t}^{4}xy_{{2}}{y_{{4}}}^{2}}\right)},$
$\displaystyle H_{33}$
$\displaystyle=\frac{y_{{3}}t^{2}xq^{4}\left({1+{q}^{5}xy_{{4}}+{q}^{4}tx{y_{{4}}}^{2}}\right)}{\left({1-{q}^{2}{t}^{2}x{y_{{3}}}^{2}}\right)\left({1-{q}^{6}x}\right)\left({1-{q}^{3}tx{y_{{4}}}^{3}}\right)\left({1-{q}^{3}{t}^{3}xy_{{2}}}\right)}.$

By adding the above eight formulas and setting $y_{2}=y_{3}=y_{4}=1$, we
obtain

(13) $\displaystyle
H(x,1,1,1,q,t)=\frac{N}{\left({1-{q}^{3}tx}\right)\left({1-{q}^{2}{t}^{2}x}\right)\left({1-q{t}^{3}x}\right)\left({1-{q}^{6}x}\right)\left({1-{t}^{6}x}\right)}$

where

$\displaystyle N=1$
$\displaystyle+\left({q}^{5}t+{q}^{4}{t}^{2}+{q}^{3}{t}^{3}+{q}^{2}{t}^{4}+q{t}^{5}+{q}^{4}t+{q}^{3}{t}^{2}+{q}^{2}{t}^{3}+q{t}^{4}\right)x$
$\displaystyle+\left({q}^{6}{t}^{5}+{q}^{5}{t}^{6}-{q}^{7}{t}^{3}-{q}^{6}{t}^{4}-{q}^{5}{t}^{5}-{q}^{4}{t}^{6}-{t}^{7}{q}^{3}-{q}^{5}{t}^{4}-{q}^{4}{t}^{5}\right){x}^{2}$
$\displaystyle-\left({q}^{8}{t}^{8}+{q}^{9}{t}^{6}+{q}^{8}{t}^{7}+{q}^{7}{t}^{8}+{q}^{6}{t}^{9}\right){x}^{3}.$

Then $C_{k^{4}}(q,t)=C_{k^{4}}(t,q)$ clearly follows from (13).

##  5\. Concluding remark

In this paper we give two proofs of the $q,t$-symmetry of the generalized
$q,t$-Catalan number $C_{\vec{k}}(q,t)$ for $\vec{k}$ of length $3$.

For $\vec{k}$ of length $n\geq 4$, we conjectured the $q,t$-symmetry of
$C_{\lambda}(q,t)$ when $\lambda=((a+1)^{s},a^{n-s})$. MacMahon’s partition
analysis technique may be used to attack the conjecture provided that we can
find a piecewise linear formula of the bounce statistic. Such a formula seems
DO exist, but becomes complicated even for $n=4$.

## References

  * [1] G. E. Andrews, P. Paule, and A. Riese, MacMahon’s partition analysis: the Omega package, European J. Combin. 22(2001).
  * [2] A. Garsia and M. Haiman, A remarkable $q,t$-Catalan sequence and $q$-Lagrange inversion, J. Algebraic Combinatorics 5 (1996), 191–244.
  * [3] A. Garsia and J. Haglund. ”A proof of the q, t-Catalan positivity conjecture”. Discrete Math. 256 (2002), pp. 677–717.
  * [4] A. Garsia and G. Xin, Dinv and Area, Electron. J. Combin., 24 (1) (2017), P1.64.
  * [5] E. Gorsky and M. Mazin, Compactified Jacobians and $q,t$-Catalan Numbers I, J. Combin. Theory Ser. A, 120 (2013), 49–63.
  * [6] J. Haglund,  Conjectured Statistics for the $q,t$-Catalan numbers, Advances in Mathematics 175 (2003), 319–334.
  * [7] J. Haglund, The $q,t$-Catalan numbers and the space of diagonal harmonics, with an appendix on the combinatorics of Macdonald polynomials, AMS University Lecture Series, 2008.
  * [8] Nicholas A. Loehr. Conjectured Statistics for the Higher $q,t$-Catalan Sequences[J]. Electronic Journal of Combinatorics, 2005, 12(1):318–344.
  * [9] M. Niu. The $q,t$-symmetry of the generalized $q,t$-Catalan number $C_{\lambda=(k^{s},(k-1)^{n-s})}(q,t)$ when $n=4$, M.S. Thesis. Capital Normal University, 2022. (Written in Chinese)
  * [10] G. Xin, A fast algorithm for MacMahon’s partition analysis, Electron. J. Combin. 11 (2004),R58, 20 pp. (electronic).
  * [11] G. Xin and Y. Zhang, On the Sweep Map for $\vec{k}$-Dyck Paths, Electron. J. Combin., 26 (3) (2019), P3.63.
  * [12] G. Xin and Y. Zhang, Dinv, Area and Bounce for $\vec{k}$-Dyck Paths, arXiv:2011.04927.