On self-conjugate $(s, s+1,\ldots, s+k)$-core partitions

TL;DR

This paper proves that the number of self-conjugate (s, s+1, ..., s+k)-core partitions matches the count of symmetric (s, k)-Dyck paths, confirming a conjecture and expanding understanding of core partitions.

Contribution

It establishes a new combinatorial equivalence between self-conjugate core partitions and symmetric Dyck paths, confirming a previously conjectured relationship.

Findings

Self-conjugate (s, s+1, ..., s+k)-core partitions are equinumerous with symmetric (s, k)-Dyck paths.

Confirmed a conjecture by Cho, Huh, and Sohn.

Extended the combinatorial understanding of core partitions.

Abstract

Simultaneous core partitions have been widely studied since Anderson's work on the enumeration of $(s,t)$-core partitions. Amdeberhan and Leven showed that the number of $(s,s+1, \ldots, s+k)$-core partitions is equal to the number of $(s, k)$-Dyck paths. In this paper, we prove that self-conjugate $(s,s+1, \ldots, s+k)$-core partitions are equinumerous with symmetric $(s, k)$-Dyck paths, confirming a conjecture posed by Cho, Huh and Sohn.

Equations

$$\sum _{i\ge 0}\left(\genfrac{}{}{0ex}{}{\lfloor \frac{s}{2}\rfloor }{i}\right)\left(\genfrac{}{}{0ex}{}{i}{\lfloor \frac{i}{2}\rfloor }\right),$$

$$A_j=\{2i-1+2sj\mid jk+1\le i\le \lfloor \frac{s}{2}\rfloor \},$$

$$B_j=\{2i-1+2sj\mid jk+\lceil \frac{s+k}{2}\rceil +1\le i\le s\}.$$

$$P(s,k)=L(s,k)\cup R(s,k)$$

$$A_j^{\prime }=\{(i-\frac{k}{2}j,j)\mid jk+1\le i\le \lfloor \frac{s}{2}\rfloor ,i\in \mathcal{Z}\},$$

$$B_j^{\prime }=\{(s+1-i+\frac{k}{2}(j+1),-j-1)\mid jk+\lceil \frac{s+k}{2}\rceil +1\le i\le s,i\in \mathcal{Z}\}.$$

$$P^{\prime }(s,k)=L^{\prime }(s,k)\cup R^{\prime }(s,k)$$

$$\mathcal{Q}(m,k)=(\bigcup _{i=0}^{k-2}{\mathcal{F}\mathcal{B}}_0^{\prime }(m-i,2k))\cup {\mathcal{F}\mathcal{B}}_{-1}^{\prime }(m+1,2k).$$

$$P=H_3{U}_2{H}_1{U}_2{D}_2{D}_2{D}_2{H}_1{D}_2{H}_2{H}_1{U}_2{U}_2$$

$$Q=H_2{H}_1{U}_2{U}_2{U}_2{H}_1{U}_2{U}_2{U}_2{D}_2{H}_1{D}_2{H}_3$$

$$P=D_2{U}_2{H}_1{U}_2{D}_2{H}_2{D}_2{H}_1{D}_2{H}_3{U}_2{H}_1{H}_2$$

$$Q=U_2{H}_1{U}_2{D}_2{H}_2{U}_2{U}_2{H}_1{U}_2{H}_3{D}_2{H}_1{H}_2$$

$${\mathcal{Q}}^{\prime }(m,k)=(\bigcup _{i=0}^{k-1}{\mathcal{F}\mathcal{B}}_0^{\prime }(m-i,2k+1))\cup {\mathcal{F}\mathcal{B}}_{-1}^{\prime }(m+\frac{1}{2},2k+1).$$

$$J^{\ast }(P^{\prime }(2m+2,2k+1))=J(P^{\prime }(2m+1,2k+1)).$$

Full text

**On self-conjugate $(s,s+1,\ldots,s+k)$-core partitions **

Sherry H.F. Yan, Yao Yu, Hao Zhou

Department of Mathematics

Zhejiang Normal University

Jinhua 321004, P.R. China

huifangyan@hotmail.com

Abstract. Simultaneous core partitions have been widely studied since Anderson’s work on the enumeration of $(s,t)$-core partitions. Amdeberhan and Leven showed that the number of $(s,s+1,\ldots,s+k)$-core partitions is equal to the number of $(s,k)$-Dyck paths. In this paper, we prove that self-conjugate $(s,s+1,\ldots,s+k)$-core partitions are equinumerous with symmetric $(s,k)$-Dyck paths, confirming a conjecture posed by Cho, Huh and Sohn.

Keywords: core partition, $(s,k)$-Dyck path.

AMS Subject Classifications: 05A05, 05C30

1 Introduction

A partition $\lambda$ of a positive integer $n$ is defined to be a sequence of nonnegative integers $(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})$ such that $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{m}=n$ and $\lambda_{1}\geq\lambda_{2}\cdots\geq\lambda_{m}$. The Young diagram of $\lambda$ is defined to be an up- and left- justified array of $n$ boxes with $\lambda_{i}$ boxes in the $i$-th row. The *conjugate * of $\lambda$, denoted by $\lambda^{\prime}$, is the partition whose Young diagram is the reflection along the main diagonal of $\lambda$’s diagram, and $\lambda$ is said to be self-conjugate if $\lambda=\lambda^{\prime}$. The hook of each box $B$ in $\lambda$ consists of the box $B$ itself and boxes directly to the right and directly below $B$. For example, Figure 1 illustrates the Young diagram and the hook lengths of a self-conjugate partition $\lambda=(5,3,3,1,1)$.

For a positive integer $s$, a partition is said to be an $s$-core partition, or simply an $s$-core, if it contains no box whose hook length is a multiple of $s$. Furthermore, a partition is said to be $(t_{1},t_{2},\ldots,t_{m})$-core if it is simultaneously $t_{1}$-core, $t_{2}$-core, …, and $t_{m}$-core. Anderson [5] showed that the number of $(s,t)$-core partitions is the rational Catalan number ${1\over s+t}{s+t\choose s}$ when $s$ and $t$ are coprime to each other.

Simultaneous core partitions have attracted much attention since Anderson’s work [5]. Various results on the enumeration of such partitions are archived in [1, 4, 6, 7, 10, 12, 13, 14, 15, 17, 19, 20].

Ford, Mai and Sze [9] characterized the set of hook lengths of diagonal cells in self-conjugate $(s,t)$-core partitions, and they showed that the number of self-conjugate $(s,t)$-core partitions is given by ${\lfloor{s\over 2}\rfloor+\lfloor{t\over 2}\rfloor\choose\lfloor{s\over 2}\rfloor}$. Concerning the enumeration of $(s,s+1,s+2)$-core partitions, Amdeberhan [2] conjectured that the number of $(s,s+1,s+2)$-core partitions is equal to the number of Motzkin paths of length $s$. This conjecture was first confirmed by Yang, Zhong and Zhou [18]. Amdeberhan and Leven [3] further proved that the number of $(s,s+1,\ldots,s+k)$-core partitions is equal to the number of $(s,k)$-Dyck paths. The largest size of such partitions was derived by Xiong [16].

A ballot $(s,k)$-path of height $n$ is a lattice path from $(0,0)$ to $(s,n)$ using up steps $U_{k}=({k\over 2},1)$, down steps $D_{k}=({k\over 2},-1)$ and horizontal steps $H_{\ell}=(\ell,0)$ for some positive integer $\ell$ with $1\leq\ell<k$, and never lying below the $x$-axis. A ballot $(s,k)$-path of height [math] is said to be an (s,k)-Dyck path. Notice that an (s,2)-Dyck path is just a Motzkin path of length $s$. We say that an $(s,k)$-Dyck path is symmetric if its reflection about the line $x={s\over 2}$ is itself. For example, a symmetric $(10,4)$-Dyck path $P=U_{2}H_{2}H_{2}H_{2}D_{2}$ is shown in Figure 2.

In this paper, we are mainly concerned with the enumeration of self-conjugate $(s,s+1,\ldots,s+k)$-core partitions, confirming the following conjecture posed by Cho, Huh and Sohn [8].

Conjecture 1.1

For given positive integers $s$ and $k$, the number of self-conjugate $(s,s+1,\ldots,s+k)$-core partitions is equal to the number of symmetric $(s,k)$-Dyck paths.

The case when $k=2$ has been confirmed by Cho, Huh and Sohn [8]. To be more precise, they showed that the number of self-conjugate $(s,s+1,s+2)$-core partitions is given by

[TABLE]

which counts the number of symmetric Motzkin paths of length $s$.

2 Proof of Conjecture 1.1

In this section, we aim to confirm Conjecture 1.1. We begin with some definitions and notations.

Following the poset terminology given by Stanley [11], for two elements $x$ and $y$ in a poset $(P,\prec)$, we say that $y$ covers $x$ if $x\prec y$ and there exists no element $z\in P$ such that $x\prec z\prec y$. An element $x$ in $P$ is said to be of rank $s$ if it covers an element of rank $s-1$. Note that the elements of rank [math] in $P$ are just the minimal elements. The Hasse diagram of a finite poset $P$ is a graph whose vertices are the elements of $P$, whose edges are the cover relations, and such that if $y$ covers $x$ then there is an edge connecting $x$ and $y$ and $y$ is placed above $x$. An order ideal of $P$ is a subset $I$ such that if any $y\in I$ and $x\prec y$ in $P$, then $x\in I$.

For a partition $\lambda$, let $MD(\lambda)$ denote the set of main diagonal hook lengths. It is apparent that $MD(\lambda)$ is a set of distinct odd integers when $\lambda$ is self-conjugate. In the following theorem, Ford, Mai and Sze [9] provided a characterization of the set $MD(\lambda)$ of any self-conjugate $t$-core partition $\lambda$.

Theorem 2.1

( [9]) * Let $\lambda$ be a self-conjugate partition. Then $\lambda$ is $t$-core if and only if $MD(\lambda)$ verifies the following properties.*

• (i)

$MD(\lambda)$* is a set of distinct odd integers;*

• (ii)

If $h\in MD(\lambda)$ with $h>2t$, then $h-2t\in MD(\lambda)$;

• (iii)

If $h_{1},h_{2}\in MD(\lambda)$, then $h_{1}+h_{2}$ is not a multiple of $2t$.

Definition 2.1

Define $L(s,k)=\bigcup_{j\geq 0}A_{j},$ and $R(s,k)=\bigcup_{j\geq 0}B_{j},$ where

[TABLE]

and

[TABLE]

Lemma 2.2

Let $x$ be a positive odd integer with $x>2s$. Then $x$ is contained in $L(s,k)$ (resp. $R(s,k)$) if and only if $x-2t\in L(s,k)$ (resp. $x-2t\in R(s,k)$ ) for all $s\leq t\leq s+k$.

*Proof.  *The necessity follows directly from the definition of $L(s,k)$ (resp. $R(s,k)$). Now we proceed to prove the sufficiency part of the theorem. Let $x=2i-1+2js$ with $1\leq i\leq s$ and $j\geq 1$. Assume that $x-2t\in L(s,k)$ for all $s\leq t\leq s+k$. We proceed to show that $x\in L(s,k)$. We claim that $i>k$. Since $x-2s-2k\in L(s,k)$, we have either $i>k$ or $j\geq 2$. Notice that $x-2s=2i-1+2(j-1)s$. Then $x-2s\in L(s,k)$ implies that $\lfloor{s\over 2}\rfloor\geq i\geq k(j-1)+1$. If $j\geq 2$, then we also have $i>k$. Then $x-2s-2k=2(i-k)-1+2(j-1)s\in L(s,k)$ implies that $i-k\geq k(j-1)+1$, which is equivalent to $i\geq kj+1$. On the other hand, $x-2s=2i-1+2(j-1)s\in L(s,k)$ implies that $i\leq\lfloor{s\over 2}\rfloor$. Hence we have $x\in L(s,k)$.

Assume that $x-2t\in R(s,k)$ for all $s\leq t\leq s+k$. We proceed to verify that $x\in R(s,k)$ by considering the following two cases.

If $j\geq 2$, then $x-2s=2i-1+2(j-1)s\in R(s,k)$ implies that $i-k(j-1)=i-kj+k\geq\lceil{s+k\over 2}\rceil+1$, which leads to $i-k\geq\lceil{s+k\over 2}\rceil+1$. Then $x-2s-2k=2(i-k)-1+2(j-1)s\in R(s,k)$ implies that $i-kj=i-k-k(j-1)\geq\lceil{s+k\over 2}\rceil+1$. Hence, we have $x\in R(s,k)$.

If $j=1$, then $x-2s-2k=2i-1-2k\in R(s,k)$ implies that $i-k\geq\lceil{s+k\over 2}\rceil+1$, which is equivalent to $i\geq k+\lceil{s+k\over 2}\rceil+1$. Thus, we have $x\in R(s,k)$. This completes the proof.

Definition 2.2

Let

[TABLE]

with the following poset structure: if $x,y\in P(s,k)$, then $y$ covers $x$ if and only if $y=x+2s+2t$ for some $0\leq t\leq k$.

Lemma 2.2 tells us that $P(s,k)$ can be viewed as the disjoint union of the posets $L(s,k)$ and $R(s,k)$. In the Hasse diagram of $L(s,k)$ (resp. $R(s,k)$), each element $x$ of rank $\ell$ covers exactly $k+1$ elements $x-2s,x-2s-2$, $\ldots,x-2s-2k$ of rank $\ell-1$ for all $\ell\geq 1$. For example, the Hasse diagrams of $P(20,3)$ and $P(20,4)$ are illustrated in Figure 3 and 4, respectively.

An order ideal $I$ of $P(s,k)$ is said to be *nice * if $I$ does not contain two elements $h_{1},h_{2}$ such that $h_{1}+h_{2}\in\{2s,2s+2,\ldots,2s+2k\}$.

In the following theorem, we establish a correspondence between self-conjugate $(s,s+1,\ldots,s+k)$-core partitions and the nice order ideals of the poset $P(s,k)$.

Theorem 2.3

Let $\lambda$ be a self-conjugate partition. Then $\lambda$ is $(s,s+1,\ldots,s+k)$-core if and only if $MD(\lambda)$ is a nice order ideal of the poset $P(s,k)$.

*Proof.  *Assume that $\lambda$ is a self-conjugate $(s,s+1,\ldots,s+k)$-core partition, we proceed to show that $MD(\lambda)$ is a nice order ideal of $P(s,k)$. By (ii) and (iii) of Theorem 2.1, it suffices to show that $MD(\lambda)$ is a subset of $P(s,k)$. Suppose to the contrary, there exists an element $x$ of $MD(\lambda)$ such that $x$ is not contained in $P(s,k)$. Choose $x$ to the smallest such element. Examining the definition of the poset $P(s,k)$, we have $x>2s$. We claim that $x>2s+2k$. If not, suppose that $x=2s+2t+1$ for some $0\leq t\leq k-1$. Then by (ii) of Theorem 2.1, $MD(\lambda)$ contains the element $y=1$. Then there are two elements $x=2s+2t+1$ and $y=1$ in $MD(\lambda)$ such that $x+y=2s+2t+2$, contradicting (iii). By (ii) of Theorem 2.1 and the selection of $x$, we have $x-2t\in P(s,k)$ for all $s\leq t\leq s+k$. From Lemma 2.2, it follows that $x\in P(s,k)$, which contradicts the selection of $x$. Hence we have reached the conclusion that $MD(\lambda)$ is a subset of $P(s,k)$.

Conversely, given a nice order ideal $I$ of $P(s,k)$, we wish to show that $I=MD(\lambda)$ for some self-conjugate $(s,s+1,\ldots,s+k)$-core partition $\lambda$. It suffices to prove that $I$ verifies properties (i)- (iii) of Theorem 2.1. It is easy to check that $I$ verifies properties (i) and (ii). We proceed to show that $I$ also has property (iii). Let $h_{1},h_{2}\in I$. For any $1\leq t\leq k$, we have $h_{1}=2p-1+2(s+t)q$ and $h_{2}=2p^{\prime}-1+2(s+t)q^{\prime}$ where $1\leq p,p^{\prime}\leq s+t$, and $q,q^{\prime}\geq 0$. Let $x=2p-1$ and $y=2p^{\prime}-1$. Since $I$ is a nice order ideal of $P(s,k)$, the elements $x$ and $y$ are contained in $I$ and $x+y\neq 2(s+t)$. This implies that $h_{1}+h_{2}$ is not a multiple of $2(s+t)$ for all $0\leq t\leq k$. Thus we have concluded that $I$ verifies (iii). This completes the proof.

Definition 2.3

Define $L^{\prime}(s,k)=\bigcup_{j\geq 0}A^{\prime}_{j},$ and $R^{\prime}(s,k)=\bigcup_{j\geq 0}B^{\prime}_{j},$ where

[TABLE]

and

[TABLE]

Definition 2.4

Let

[TABLE]

with the following poset structure:

• •

If $(a,b),(a^{\prime},b^{\prime})\in L^{\prime}(s,k)$, then $(a,b)$ covers $(a^{\prime},b^{\prime})$ if and only if $|a-a^{\prime}|\leq{k\over 2}$ and $b=b^{\prime}+1$;

• •

If $(a,b),(a^{\prime},b^{\prime})\in R^{\prime}(s,k)$, then $(a,b)$ covers $(a^{\prime},b^{\prime})$ if and only if $|a-a^{\prime}|\leq{k\over 2}$ and $b=b^{\prime}-1$.

For example, the Hasse diagrams of $P^{\prime}(20,3)$ and $P^{\prime}(20,4)$ are shown in Figure 5 and Figure 6, respectively.

Now we present a map $\chi:P(s,k)\rightarrow P^{\prime}(s,k)$ as follows.

• •

If $x=2i-1+2sj$ with $jk+1\leq i\leq\lfloor{s\over 2}\rfloor$ and $j\geq 0$, set $\chi(x)=(i-{k\over 2}j,j)$;

• •

If $x=2i-1+2sj$ with $jk+\lceil{s+k\over 2}\rceil+1\leq i\leq s$ and $j\geq 0$, set $\chi(x)=(s+1-i+{k\over 2}(j+1),-j-1)$.

It is easily seen that the map $\chi:P(s,k)\rightarrow P^{\prime}(s,k)$ is an isomorphism of posets. Furthermore, $\chi$ maps a nice order ideal of $P(s,k)$ to an order ideal of $P^{\prime}(s,k)$ containing no elements $(a,0)$ and $(a^{\prime},-1)$ such that $|a-a^{\prime}|\leq{k\over 2}$.

Theorem 2.4

For given positive integers $s$ and $k$, the nice order ideals of $P(s,k)$ are in one-to-one correspondence with order ideals of $P^{\prime}(s,k)$ containing no elements $(a,0)$ and $(a^{\prime},-1)$ such that $|a-a^{\prime}|\leq{k\over 2}$

Denote by $J(P^{\prime}(s,k))$ the set of order ideals of $P^{\prime}(s,k)$ containing no elements $(a,0)$ and $(a^{\prime},-1)$ such that $|a-a^{\prime}|\leq{k\over 2}$.

2.1 On self-conjugate $(s,s+1,\ldots,s+2k)$-core partitions

It is routine to check that the poset $P^{\prime}(2m,2k)$ is identical with the poset $P^{\prime}(2m+1,2k)$. Relying on Theorems 2.3 and 2.4, we are led to the following result.

Theorem 2.5

For positive integers $m$ and $k$, self-conjugate $(2m,2m+1,\ldots,2m+k)$-core partitions are equinumerous with self-conjugate $(2m+1,2m+2,\ldots,2m+k+1)$-core partitions.

Combining Theorems 2.3 and 2.5, we will focus on the enumeration of order ideals $I\in J(P^{\prime}(2m,2k))$ for any positive integers $m$ and $k$. We begin with some definitions and notations.

If a ballot $(s,k)$-path is allowed to go below the $x$-axis, then it is called a free ballot $(s,k)$-path. Denote by $\mathcal{FB}_{n}(s,k)$ the set of all free ballot $(s,k)$-paths of height $n$. Let $\mathcal{SD}(s,k)$ denote the set of all symmetric $(s,k)$-Dyck paths.

Throughout this paper we identify a (free) ballot $(s,k)$-path with a word of $U_{k\over 2}$’s, $D_{k\over 2}$’s and $H_{\ell}$’s . Let $p$ be a step running from the point $(x_{1},y_{1})$ to the point $(x_{2},y_{2})$. Then we say that the point $(x_{1},y_{1})$ is its *starting

• point, and the point $(x_{2},y_{2})$ is its ending point. The starting point and the *ending * point of a path are defined analogously.

Let $\mathcal{FB}^{\prime}_{n}(m,k)$ be the set of free ballot $(m,k)$-paths of height $n$ which start with either a horizontal step or a down step. Let

[TABLE]

We describe a map $\varphi:J(P^{\prime}(2m,2k))\rightarrow\mathcal{Q}(m,k)$ as follows. Let $I\in J(P^{\prime}(2m,2k))$. First we color the point $(a,b)$ with $a,b\in\mathcal{Z}$ and $0\leq a\leq m+1$ in the plane by black and white according to the following rules.

• •

If $b\geq 0$, then the point $(a,b)$ is colored by black if and only if $(a,b)\in I$;

• •

If $b<0$, then the point $(a,b)$ is colored by black if and only if $(a,b)\notin I$.

Now we recursively define a sequence $P_{0},P_{1},\ldots P_{n}$ of points where $P_{0}=(0,0)$. Assume that $P_{i-1}$ has been already determined. Let $P_{i-1}=(a_{i-1},b_{i-1})$. We proceed to demonstrate how to get $P_{i}$. There are two cases.

Case 1. There exists a black point $(a_{i-1}+\ell,b_{i-1})$ with $1\leq\ell\leq 2k-1$. If the point $(a_{i-1}+k,b_{i-1}+1)$ is black, then set $P_{i}=(a_{i-1}+k,b_{i-1}+1)$. Otherwise, we choose $\ell$ to the smallest such integer and set $P_{i}=(a_{i-1}+\ell,b_{i-1})$.

Case 2. There does not exist a black point $(a_{i-1}+\ell,b_{i-1})$ with $1\leq\ell\leq 2k-1$. If the point $(a_{i-1}+k,b_{i-1}-1)$ is black, set $P_{i}=(a_{i-1}+k,b_{i-1}-1)$. Otherwise, set $P_{i}=P_{i-1}$ and $n=i-1$.

Let $\varphi(I)$ be the resulting path with lattice points $P_{0},P_{1},\ldots,P_{n}$. For example, let $I=\{(1,0),(2,0),(3,0),(4,0),(5,0),(3,1),(8,-1),(10,-1)\}\in J(P^{\prime}(20,4))$. By applying the map $\varphi$, we get a path $\varphi(I)$ as shown in Figure 7.

Lemma 2.6

For any order ideal $I\in J(P^{\prime}(2m,2k))$, the resulting path $\varphi(I)$ verifies the following properties.

• (i)

The points lying above the path $\varphi(I)$ are white;

• (ii)

The points lying below the path $\varphi(I)$ are black.

*Proof.  *Here we only verify (i). By similar arguments, one can prove (ii). We follow notations in the definition of $\varphi$. Let $P^{(\ell)}$ be the partial path of $\varphi(I)$ whose lattice points are $P_{0},P_{1},\ldots,P_{\ell}$. We prove by induction on $\ell$. Clearly the assertion holds for $\ell=0$. Assume that the points lying above the path $P^{(\ell-1)}$ are white. Let $P_{\ell-1}=(a,b)$. Now we proceed to show that the assertion also holds for $P^{(\ell)}$. We have three cases.

Case 1. $P_{\ell}=(a+k,b+1)$. When $y>b+1$ and $a<x\leq a+k$, the point $(x,y)$ must be white. If not, by the cover relation of $I$, the point $(a,y-1)$ is also black, which contradicts the induction hypothesis. It remains to show that the point $(x,b+1)$ is white for any $a<x<a+k$. If not, Let $c$ be the smallest integer such that the point $(c,b+1)$ is black and $a<c<a+k$. By the cover relation of $I$, it follows that the point $(c-k,b)$ is black. By the induction hypothesis, we must have $P_{\ell-2}=(a-k,b+1)$. Then by the construction of the path, we must have $P_{\ell-1}=(c,b+1)$, which contradicts the selection of $P_{\ell-1}$.

Case 2. $P_{\ell}=(a+i,b)$ for some $1\leq i\leq 2k-1$. We claim that the point $(x,b+1)$ is white for any $a<x\leq a+i$. If not, Let $c$ be the smallest integer such that the point $(c,b+1)$ is black and $a<c\leq a+i$. According to the selection of $P_{\ell}$, we have $c\neq a+k$. By the cover relation of $I$, the point $(c-k,b)$ is black. Hence, by the induction hypothesis, we must have $P_{\ell-2}=(a-k,b+1)$. Then by the construction of the path, we must have $P_{\ell-1}=(c,b+1)$, which contradicts the selection of $P_{\ell-1}$. Hence, by the cover relation of $I$, the point $(x,y)$ must be white for any $y>b+1$ and $a<x\leq a+i$.

Case 3. $P_{\ell}=(a+k,b-1)$. Due to the selection of the lattice point $P_{\ell}$, the point $(x,b)$ is white for any $a<x\leq a+k$. Then by the cover relation of $I$, the point $(x,y)$ must be white for any $y>b+1$ and $a<x\leq a+k$. This completes the proof.

Lemma 2.7

The map $\varphi$ is well defined, that is, for any $I\in J(P^{\prime}(2m,2k))$ we have $\varphi(I)\in\mathcal{Q}(m,k)$.

*Proof.  *Let $\varphi(I)$ be path with lattice points $P_{0},P_{1},\ldots,P_{n}$. It is apparent that the resulting path is a lattice path consisting of up steps $U_{k}$, down steps $D_{k}$ and horizontal steps $H_{\ell}$ for some $1\leq\ell\leq 2k-1$. Moreover, since $(k,1)$ is not contained in $P^{\prime}(2m,2k)$, $\varphi(I)$ must start with either a down step or a horizontal step. In order to show that $\varphi$ is well defined, it remains to verify that we have either $P_{n}=(m-i,0)$ for some $0\leq i\leq k-2$ or $P_{n}=(m+1,-1)$. Notice that $(k,-1)$ is not contained in $P^{\prime}(m,k)$. Thus the point $(k,-1)$ is always colored by black. This ensures that we have $n>0$.

Suppose that $P_{n}=(a,b)$. According to the construction of the path, the point $P_{n}$ is black. If $b>0$, then we have $(a,b)\in I$ and $(a+k,b-1)\in I$ by the cover relation of $I$. It follows that the point $(a+k,b-1)$ is black and the path can not terminate at the point $(a,b)$. So we have $b\leq 0$.

We claim that $P_{n}\neq(m-i,0)$ for any $i\geq k-1$. If not, we have $(m-i,0)\in I$, which implies that $(m-i+k,-1)\notin I$. It follows that the point $(m-i+k,-1)$ is black and the path can not terminate at the point $(a,b)$.

Suppose that $b<0$ and $a\leq m+1-k$. Since the point $(a,b)$ is the ending point of $\varphi(I)$, the point $(a+k,b-1)$ must be white. This implies that we have $(a+k,b-1)\in I$. Recall that $(a+k,b-1)$ covers $(a,b)$. Therefore, we have $(a,b)\in I$ and the point $(a,b)$ is white, a contradiction.

Suppose that $b<0$ and $m+2-k\leq a\leq m$. Since $m+1-a\leq k-1$ and $\varphi(I)$ terminates at $(a,b)$, the point $(m+1,b)$ must be white. By the cover relation of $I$, we have $(m+1,-1)\in I$, a contradiction.

We claim that $P_{n}\neq(m+1,b)$ for any $b<-1$. If not, by (i) of Lemma 2.6, the point $(m+1,-1)$ is white, yielding a contradiction with the fact that the point $(m+1,-1)$ is black.

So far, we have reached the conclusion that we have either $P_{n}=(m-i,0)$ for some $0\leq i\leq k-2$ or $P_{n}=(m+1,-1)$. This implies that the map $\varphi$ is well defined, completing the proof.

Lemma 2.8

For any order ideal $I\in J(P^{\prime}(2m,2k))$, the resulting path $\varphi(I)$ verifies the following properties.

• (i)

The points on the path $\varphi(I)$ are black if and only if they are ending points of steps;

• (ii)

For $b<0$, the point $(a,b)$ to the right of the ending point of the path $\varphi(I)$ is black;

• (iii)

For $b\geq 0$, the point $(a,b)$ to the right of the ending point of the path $\varphi(I)$ is white.

*Proof.  *Point (i) follows directly from the construction of $\varphi(I)$.

Recall that the ending point of $\varphi(I)$ is either the point $(m-i,0)$ for some $0\leq i\leq k-2$ or the point $(m+1,-1)$. Clearly, it suffices to prove $\varphi(I)$ verifies properties (ii) and (iii) for the former case. Since $\varphi(I)$ terminates at $(m-i,0)$, the point $(a,0)$ is white for any $m-i<a\leq m+1$. Hence by the cover relation of $I$, the point $(a,b)$ is white for any $m-i<a\leq m+1$ and $b\geq 0$. Moreover, the point $(a,-1)$ is black for any $m-i<a\leq m+1$. This implies that the point $(a,b)$ is black for any $m-i<a\leq m+1$ and $b\leq-1$. Thus, $\varphi(I)$ has properties (ii) and (iii). This completes the proof.

Theorem 2.9

The map $\varphi$ is a bijection between $J(P^{\prime}(2m,2k))$ and $\mathcal{Q}(m,k)$.

*Proof.  *First we give the inverse map $\varphi^{\prime}$ from the set $\mathcal{Q}(m,k)$ to the set $J(P^{\prime}(2m,2k))$. Let $P\in\mathcal{Q}(m,k)$. First we color the point $(a,b)$ with $a,b\in\mathcal{Z}$ and $0\leq a\leq m+1$ in the plane by black and white according to the following rules.

• •

If the point $(a,b)$ lies below the path, we color it by black;

• •

If the point $(a,b)$ lies above the path, we color it by white;

• •

For a point $(a,b)$ on the path, we color it by black if and only if it is an ending point of a step;

• •

For the point (a,b) to the right to the endpoint of $P$, we color it by white when $b\geq 0$, and color it by black when $b<0$.

For $b<0$, let $(a,b)\in\varphi^{\prime}(P)$ if and only if the point $(a,b)$ is white. For $b\geq 0$, let $(a,b)\in\varphi^{\prime}(P)$ if and only if the point $(a,b)$ is black. It is easily seen that $\varphi^{\prime}(P)\in J(P^{\prime}(2m,2k))$. Lemmas 2.6 and 2.8 guarantee that the maps $\varphi$ and $\varphi^{\prime}$ are inverses of each other, and thus the map $\varphi$ is a bijection. This completes the proof.

In the following, we aim to establish a bijection between $\mathcal{Q}(m,k)$ and $\mathcal{SD}(2m,2k)$.

Let $\mathcal{B}^{e}(s,k)$ and $\mathcal{B}^{o}(s,k)$ denote the set of ballot $(s,k)$-paths of even height and the set of ballot $(s,k)$-paths of odd height, respectively. The *complement * of the path $P$, denoted by $P^{c}$, is obtained from $P$ by interchanging the up steps and the down steps. The *reverse complement * of the path $P$, denoted by $P^{{}^{rc}}$, is obtained from $P$ by encoding the steps of $P$ from right to left and interchanging the up steps and the down steps. For example, the complement of the path $P=H_{1}U_{3}D_{3}D_{3}U_{3}D_{3}H_{2}$ is given by $H_{1}D_{3}U_{3}U_{3}D_{3}U_{3}H_{2}$, while the reverse complement of the path $P$ is given by $H_{2}U_{3}D_{3}U_{3}U_{3}D_{3}H_{1}$.

Theorem 2.10

There is a bijection $\alpha$ between the set $\mathcal{FB}_{0}(s,k)$ and the set $\mathcal{B}^{e}(s,k)$.

*Proof.  *Let $P\in\mathcal{FB}_{0}(s,k)$. Suppose that $a$ is the lowest point of $P$. If there are more than one such lowest point, we choose $a$ to be the leftmost one. Now we proceed to construct a path $\alpha(P)$. Clearly, $P$ can be uniquely decomposed as $P=P^{(1)}P^{(2)}$, where $P^{(1)}$ is the subpath of $P$ which goes from the starting point of $P$ to the point $a$, and $P^{(2)}$ is the remaining subpath of $P$. Set $\alpha(P)=P^{(2)}(P^{(1)})^{rc}$. It is apparent that the resulting path is a ballot $(s,k)$-path of even height.

In order to show that the map $\alpha$ is a bijection, we describe a map $\alpha^{\prime}:\mathcal{B}^{e}(s,k)\rightarrow\mathcal{FB}_{0}(s,k)$. Let $Q$ be a ballot $(s,k)$-path of height $2\ell$ for some $\ell\geq 0$. Now we proceed to demonstrate how to get a path $\alpha^{\prime}(Q)$. Let $b$ the rightmost lattice point of $Q$ which is on the line $y=\ell$. Then $Q$ can be uniquely decomposed as $Q=Q^{(1)}Q^{(2)}$, where $Q^{(1)}$ is the subpath of $Q$ which goes from the starting point of $Q$ to the point $b$, and $Q^{(2)}$ is the remaining subpath of $Q$. Set $\alpha^{\prime}(Q)=(Q^{(2)})^{rc}Q^{(1)}$. One can easily verify that the resulting path $\alpha^{\prime}(Q)\in\mathcal{FB}_{0}(s,k)$. Moreover, the point $b$ becomes the leftmost lowest point of the resulting path. This ensures that the maps $\alpha$ and $\alpha^{\prime}$ are inverses of each other. Thus the map $\alpha$ is a bijection, completing the proof.

Example 2.1

The decomposition of a path

[TABLE]

is shown in Figure 8. By applying the map $\alpha$, its corresponding path is illustrated in Figure 9.

Example 2.2

The decomposition of a path

[TABLE]

is shown in Figure 9. By applying the map $\alpha^{\prime}$, its corresponding path is illustrated in Figure 8.

Theorem 2.11

There is a bijection $\beta$ between the set $\mathcal{FB}_{-1}(s,k)$ and the set $\mathcal{B}^{o}(s,k)$.

*Proof.  *Let $P\in\mathcal{FB}_{-1}(s,k)$. Now we proceed to construct a path $\beta(P)$ as follows. Suppose that $a$ is the rightmost lattice point of $P$ which is on the $x$-axis. Then $P$ can be uniquely decomposed as $P=P^{(1)}P^{(2)}$, where $P^{(1)}$ is the subpath of $P$ which goes from the starting point of $P$ to the point $a$, and $P^{(2)}$ is the remaining subpath of $P$. Set $\beta(P)=\alpha(P^{(1)})(P^{(2)})^{c}$. It is routine to check that the resulting path $\beta(P)$ is a ballot $(s,k)$-path of odd height.

In order to show that the map $\beta$ is a bijection, we describe a map $\beta^{\prime}:\mathcal{B}^{o}(s,k)\rightarrow\mathcal{FB}_{-1}(s,k)$. Let $Q$ be a ballot $(s,k)$-path of odd height $2\ell+1$ for some nonnegative integer $\ell$ . Now we proceed to demonstrate how to get a path $\beta^{\prime}(Q)$. Let $b$ the rightmost lattice point of $P$ which is on the line $y=2\ell$. Then $Q$ can be uniquely decomposed as $Q=Q^{(1)}Q^{(2)}$, where $Q^{(1)}$ is the subpath of $Q$ which goes from the starting point of $Q$ to the point $b$, and $Q^{(2)}$ is the remaining subpath of $Q$. Set $\beta^{\prime}(Q)=\alpha^{-1}(Q^{(1)})(Q^{(2)})^{c}$. One can easily verify that the resulting path $\beta^{\prime}(Q)\in\mathcal{FB}_{-1}(s,k)$. Moreover, the ending point of $\alpha^{-1}(Q^{(1)})$ becomes the rightmost lattice point which lies on the $x$-axis. This ensures that the maps $\beta$ and $\beta^{\prime}$ are inverses of each other. Thus the map $\beta$ is a bijection, completing the proof.

Example 2.3

The decomposition of a path

[TABLE]

is shown in Figure 10. By applying the map $\beta$, its corresponding path is illustrated in Figure 11.

Example 2.4

The decomposition of a path

[TABLE]

is shown in Figure 11. By applying the map $\beta^{\prime}$, its corresponding path is illustrated in Figure 10.

With abuse of notation, we use $H_{0}$ to denote the empty step. Now we are in the position to establish a bijection between the set $\mathcal{Q}(m,k)$ and the set $\mathcal{SD}(2m,2k)$.

Theorem 2.12

There is a bijection between the set $\mathcal{Q}(m,k)$ and the set $\mathcal{SD}(2m,2k)$.

*Proof.  *We first describe a map $\delta:\mathcal{Q}(m,k)\rightarrow\mathcal{SD}(2m,2k)$. Let $P$ be a path in $\mathcal{Q}(m,k)$, and let $Q$ be the path obtained form $P$ by removing its first step. We proceed to construct a path $\delta(P)$ as follows.

• (i)

If $P\in\mathcal{FB}^{\prime}_{0}(m-i,2k)$ for some $0\leq i\leq k-2$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq\ell\leq 2k-1$, set $\delta(P)=\alpha(Q)H_{\ell}H_{2i}H_{\ell}(\alpha(Q))^{rc}$.

• (ii)

If $P\in\mathcal{FB}^{\prime}_{0}(m-i,2k)$ for some $0\leq i\leq k-2$ and $P$ starts with a down step $D_{k}$, set $\delta(P)=\beta(Q^{c})D_{k}H_{2i}U_{k}(\beta(Q^{c}))^{rc}$.

• (iii)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+1,2k)$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq\ell\leq k$, set $\delta(P)=\beta(Q)H_{2\ell-2}(\beta(Q))^{rc}$.

• (iv)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+1,2k)$ and $P$ starts with a horizontal step $H_{\ell}$ for some $k+1\leq\ell\leq 2k-1$, set $\delta(P)=\beta(Q)U_{k}H_{2(\ell-k-1)}D_{k}(\beta(Q))^{rc}$.

• (v)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+1,2k)$ and $P$ starts with a down step $D_{k}$, set $\delta(P)=\alpha(Q)H_{2k-2}(\alpha(Q))^{rc}$.

It is easy to check that $\delta(P)\in SD(2m,2k)$, and thus the map $\delta$ is well defined.

In order to show the map $\delta$ is a bijection, we describe a map $\delta^{\prime}:\mathcal{SD}(2m,2k)\rightarrow\mathcal{Q}(m,k)$. Let $P$ be a path in $\mathcal{SD}(2m,2k)$, we proceed to construct a path $\delta^{\prime}(P)$ as follows.

• ( i*′* )

If $P=QH_{\ell}H_{2i}H_{\ell}Q^{rc}$ with $1\leq\ell\leq 2k-1$ and $0\leq i\leq k-2$ and $Q$ is of even height, set $\delta^{\prime}(P)=H_{\ell}\alpha^{-1}(Q)$.

• ( ii*′* )

If $P=QD_{k}H_{2i}U_{k}Q^{rc}$ with $0\leq i\leq k-2$ and $Q$ is of odd height, set $\delta^{\prime}(P)=D_{k}(\beta^{-1}(Q))^{c}$.

• ( iii*′* )

If $P=QH_{2i}Q^{rc}$ with $0\leq i\leq k-1$ and $Q$ is of odd height, set $\delta^{\prime}(P)=H_{i+1}\beta^{-1}(Q)$.

• ( iv*′* )

If $P=QU_{k}H_{2i}D_{k}Q^{rc}$ with $0\leq i\leq k-2$ and $Q$ is of odd height, set $\delta^{\prime}(P)=H_{k+1+i}\beta^{-1}(Q)$.

• ( v*′* )

If $P=QH_{2k-2}Q^{rc}$ and $Q$ is of even height, set $\delta^{\prime}(P)=D_{k}\alpha^{-1}(Q)$.

It is not difficult to check that the resulting path $\delta^{\prime}(P)\in\mathcal{Q}(m,k)$. This implies that the map $\delta$ is well defined. One can easily verify that ( i*′* ), ( ii*′* ), ( iii*′* ), ( iv*′* ) and ( v*′* ) respectively reverse the procedures of ( i ), ( ii ), ( iii ), ( iv ) and ( v ). This implies that the maps $\delta$ and $\delta^{\prime}$ are inverses of each other. Hence, the map $\delta$ is a bijection, completing the proof.

Lemma 2.13

There is a bijection between $\mathcal{SD}(2m,2k)$ and $\mathcal{SD}(2m+1,2k)$.

*Proof.  *First we discribe a map $\xi:\mathcal{SD}(2m,2k)\rightarrow\mathcal{SD}(2m+1,2k)$. For any path $P\in\mathcal{SD}(2m,2k)$, it can be uniquely decomposed as $P=QH_{2\ell}Q^{rc}$ for some $0\leq\ell<{k}$. Let $\xi(P)=QH_{2\ell+1}Q^{rc}$. It is easily seen that the map $\xi$ is reversible, and thus is a bijection. This completes the proof.

Theorems 2.3, 2.4, 2.5, 2.9 and 2.12 together with Lemma 2.13 lead to the following result.

Theorem 2.14

For given positive integers $s$ and $k$, the number of self-conjugate $(s,s+1,\ldots,s+2k)$-core partitions is equal to the number of symmetric $(s,2k)$-Dyck paths.

2.2 On self-conjugate $(s,s+1,\ldots,s+2k+1)$-core partitions

Let

[TABLE]

In the plane, a point $(x,y)$ is said to be visible if either $x=a+{1\over 2}$ for some $a\in\mathcal{Z}$ and $y$ is an odd integer, or $x=a$ for some $a\in\mathcal{Z}$ and $y$ is an even integer.

We describe a map $\psi:J(P^{\prime}(2m,2k+1))\rightarrow\mathcal{Q^{\prime}}(m,k)$ as follows. Let $I\in J(P^{\prime}(2m,2k+1))$. For the visible point $(a,b)$ in the plane where $b\in\mathcal{Z}$ and $0\leq a\leq m+{1\over 2}$, we color it by black and white according to the following rules.

• •

If $b\geq 0$ , then the point $(a,b)$ is colored by black if and only if $(a,b)\in I$;

• •

If $b<0$, then the point $(a,b)$ is colored by black if and only if $(a,b)\notin I$.

Now we recursively define a sequence $P_{0},P_{1},\ldots P_{n}$ of points where $P_{0}=(0,0)$. Assume that $P_{i-1}$ has been already determined. Let $P_{i-1}=(a_{i-1},b_{i-1})$. We proceed to get $P_{i}$ by the following procedure.

Case 1. There exists a black point $(a_{i-1}+\ell,b_{i-1})$ with $1\leq\ell\leq 2k$. If the point $(a_{i-1}+k+{1\over 2},b_{i-1}+1)$ is black, then set $P_{i}=(a_{i-1}+k+{1\over 2},b_{i-1}+1)$. Otherwise, choose $\ell$ to the smallest such integer and set $P_{i}=(a_{i-1}+\ell,b_{i-1})$.

Case 2. There does not exist a black point $(a_{i-1}+\ell,b_{i-1})$ with $1\leq\ell\leq 2k$. If the point $(a_{i-1}+k+{1\over 2},b_{i-1}-1)$ is black, set $P_{i}=(a_{i-1}+k+{1\over 2},b_{i-1}-1)$. Otherwise, set $P_{i}=P_{i-1}$ and $n=i-1$.

Let $\psi(I)$ be the resulting path with lattice points $P_{0},P_{1},\ldots,P_{n}$. For example, let $I=\{(1,0),(6,-2),(7.5,-1),(6.5,-1),(5.5,-1),(4.5,-1)\}\in J(P^{\prime}(20,3))$. By applying the map $\psi$, we get a path $\psi(I)$ as shown in Figure 12.

By the same reasoning as in the proof of Lemmas 2.6, 2.7, 2.8 and Theorem 2.9, we can deduce the following results.

Lemma 2.15

The map $\psi$ is well defined, that is, for any $I\in J(P^{\prime}(2m,2k+1))$ we have $\psi(I)\in\mathcal{Q}^{\prime}(m,k)$.

Lemma 2.16

For any $I\in J(P^{\prime}(2m,2k+1))$, the resulting path $\psi(I)$ verifies the following properties.

• (i)

The visible points on the path $\psi(I)$ are black if and only if they are ending points of steps ;

• (ii)

For $b<0$, the visible point $(a,b)$ to the right of the ending point of the path $\psi(I)$ is black ;

• (iii)

For $b\geq 0$, the visible point $(a,b)$ to the right of the ending point of the path $\psi(I)$ is white;

• (iv)

The visible points lying above the path $\psi(I)$ are white;

• (v)

The visible points lying below the path $\psi(I)$ are black;

Theorem 2.17

The map $\psi$ is a bijection between $J(P^{\prime}(2m,2k+1))$ and $\mathcal{Q}^{\prime}(m,k)$.

In the following theorem, we establish a bijection between the set $\mathcal{Q}^{\prime}(m,k)$ and the set $\mathcal{SD}(2m,2k+1)$.

Theorem 2.18

There is a bijection between the set $\mathcal{Q}^{\prime}(m,k)$ and the set $\mathcal{SD}(2m,2k+1)$.

*Proof.  *We first describe a map $\gamma:\mathcal{Q}^{\prime}(m,k)\rightarrow\mathcal{SD}(2m,2k+1)$. Let $P$ be a path in $\mathcal{Q}^{\prime}(m,k)$ and let $Q$ be the path obtained from $P$ by removing its first step. Now we proceed to demonstrate the procedure to get a path $\gamma(P)$.

• (i)

If $P\in\mathcal{FB}^{\prime}_{0}(m-i,2k+1))$ for some $0\leq i\leq k-1$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq\ell\leq 2k$, set $\gamma(P)=\alpha(Q)H_{\ell}H_{2i}H_{\ell}(\alpha(Q))^{rc}$.

• (ii)

If $P\in\mathcal{FB}^{\prime}_{0}(m-i,2k+1))$ for some $0\leq i\leq k-1$ and $P$ starts with a down step $D_{k+{1\over 2}}$, set $\gamma(P)=\beta(Q^{c})D_{k+{1\over 2}}H_{2i}U_{k+{1\over 2}}(\beta(Q^{c}))^{rc}$.

• (iii)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+{1\over 2},2k+1))$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq\ell\leq k$, set $\gamma(P)=\beta(Q)H_{2\ell-1}(\beta(Q))^{rc}$.

• (iv)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+{1\over 2},2k+1))$ and $P$ starts with a horizontal step $H_{\ell}$ for some $k+1\leq\ell\leq 2k$, set $\gamma(P)=\beta(Q)U_{k+{1\over 2}}H_{2(\ell-k-1)}D_{k+{1\over 2}}(\beta(Q))^{rc}$.

• (v)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+{1\over 2},2k+1))$ and $P$ starts with a down step $D_{k+{1\over 2}}$, set $\gamma(P)=\alpha(Q)H_{2k}(\alpha(Q))^{rc}$.

One can easily check that $\gamma(P)\in\mathcal{SD}(2m,2k+1)$, and thus the map $\gamma$ is well defined.

Next we describe a map $\gamma^{\prime}:\mathcal{SD}(2m,2k+1)\rightarrow\mathcal{Q}^{\prime}(m,k)$. Let $P$ be a path in $\mathcal{SD}(2m,2k+1)$, we proceed to construct a path $\gamma^{\prime}(P)$ by the following procedure.

• ( i*′* )

If $P=QH_{\ell}H_{2i}H_{\ell}Q^{rc}$ with $1\leq\ell\leq 2k$ and $0\leq i\leq k-1$ and $Q$ is of even height, set $\gamma^{\prime}(P)=H_{\ell}\alpha^{-1}(Q)$.

• ( ii*′* )

If $P=QD_{k+{1\over 2}}H_{2i}U_{k+{1\over 2}}Q^{rc}$ with $0\leq i\leq k-1$ and $Q$ is of odd height, set $\gamma^{\prime}(P)=D_{k+{1\over 2}}(\beta^{-1}(Q))^{c}$.

• ( iii*′* )

If $P=QH_{2i-1}Q^{rc}$ with $1\leq i\leq k$ and $Q$ is of odd height, set $\gamma^{\prime}(P)=H_{i}\beta^{-1}(Q)$.

• ( iv*′* )

If $P=QU_{k+{1\over 2}}H_{2i}D_{k+{1\over 2}}Q^{rc}$ with $0\leq i\leq k-1$ and $Q$ is of odd height, set $\gamma^{\prime}(P)=H_{k+1+i}\beta^{-1}(Q)$.

• ( v*′* )

If $P=QH_{2k}Q^{rc}$ and $Q$ is of even height, set $\gamma^{\prime}(P)=D_{k+{1\over 2}}\alpha^{-1}(Q)$.

It is not difficult to check that the resulting path $\gamma^{\prime}(P)\in\mathcal{Q}^{\prime}(m,k)$. This implies that the map $\gamma$ is well defined. Furthermore, one can easily verify that ( i*′* ), ( ii*′* ), ( iii*′* ), ( iv*′* ) and ( v*′* ) respectively reverse the procedures of ( i ), ( ii ), ( iii ), ( iv ) and ( v ). This implies that the maps $\gamma$ and $\gamma^{\prime}$ are inverses of each other. Hence, the map $\gamma$ is a bijection, completing the proof.

Combining Theorems 2.3, 2.4, 2.17 and 2.18, we are led to the following result.

Theorem 2.19

For given positive integer $m$ and nonnegative integer $k$, the number of self-conjugate $(2m,2m+1,\ldots,2m+2k+1)$-core partitions is equal to the number of symmetric $(2m,2k+1)$-Dyck paths.

Let $J^{*}(P^{\prime}(2m+2,2k+1))$ denote the order ideals $I\in J(P^{\prime}(2m+2,2k+1))$ such that $I$ does not contain the element $(m+1,0)$. It is easily seen that

[TABLE]

Let $\mathcal{Q}^{*}(m,k)=\mathcal{Q}^{\prime}(m,k)\setminus\mathcal{FB}^{\prime}_{0}(m,2k+1)$. The construction of the map $\psi$ together with Lemma 2.16 leads to the following result.

Theorem 2.20

The map $\psi$ induces a bijection between the set $J^{*}(P^{\prime}(2m+2,2k+1))$ and the set $\mathcal{Q}^{*}(m+1,k)$.

In the following theorem, we aim to establish a bijection between the set $\mathcal{Q}^{*}(m+1,k)$ and the set $\mathcal{SD}(2m+1,2k+1)$.

Theorem 2.21

There is a bijection between the set $\mathcal{Q}^{*}(m+1,k)$ and the set $\mathcal{SD}(2m+1,2k+1)$.

*Proof.  *We first describe a map $\eta:\mathcal{Q}^{*}(m+1,k)\rightarrow\mathcal{SD}(2m+1,2k+1)$. Let $P$ be a path in $\mathcal{Q}^{*}(m+1,k)$ and let $Q$ be the path obtained from $P$ by removing its first step. Now we proceed to demonstrate the procedure to get a path $\eta(P)$.

• (i)

If $P\in\mathcal{FB}^{\prime}_{0}(m+1-i,2k+1))$ for some $1\leq i\leq k-1$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq\ell\leq 2k$, set $\eta(P)=\alpha(Q)H_{\ell}H_{2i-1}H_{\ell}(\alpha(Q))^{rc}$.

• (ii)

If $P\in\mathcal{FB}^{\prime}_{0}(m+1-i,2k+1))$ for some $1\leq i\leq k-1$ and $P$ starts with a down step $D_{k+{1\over 2}}$, set $\eta(P)=\beta(Q^{c})D_{k+{1\over 2}}H_{2i-1}U_{k+{1\over 2}}(\beta(Q^{c}))^{rc}$.

• (iii)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+1+{1\over 2},2k+1))$ and $P$ starts with a horizontal step $H_{\ell}$ for some $1\leq\ell\leq k+1$, set $\eta(P)=\beta(Q)H_{2\ell-2}(\beta(Q))^{rc}$.

• (iv)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+1+{1\over 2},2k+1))$ and $P$ starts with a horizontal step $H_{\ell}$ for some $k+2\leq\ell\leq 2k$, set $\eta(P)=\beta(Q)U_{k+{1\over 2}}H_{2(\ell-k-1)-1}D_{k+{1\over 2}}(\beta(Q))^{rc}$.

• (v)

If $P\in\mathcal{FB}^{\prime}_{-1}(m+1+{1\over 2},2k+1))$ and $P$ starts with a down step $D_{k+{1\over 2}}$, set $\eta(P)=\alpha(Q)H_{2k-1}(\alpha(Q))^{rc}$.

It is easy to verify that $\eta(P)\in\mathcal{SD}(2m+1,2k+1)$, and thus the map $\eta$ is well defined.

Next we describe a map $\eta^{\prime}:\mathcal{SD}(2m+1,2k+1)\rightarrow\mathcal{Q}^{*}(m+1,k)$. Let $P$ be a path in $\mathcal{SD}(2m+1,2k+1)$, we proceed to construct a path $\eta^{\prime}(P)$ by the following procedure.

• ( i*′* )

If $P=QH_{\ell}H_{2i-1}H_{\ell}Q^{rc}$ with $1\leq\ell\leq 2k$ and $1\leq i\leq k-1$ and $Q$ is of even height, set $\eta^{\prime}(P)=H_{\ell}\alpha^{-1}(Q)$.

• ( ii*′* )

If $P=QD_{k+{1\over 2}}H_{2i-1}U_{k+{1\over 2}}Q^{rc}$ with $1\leq i\leq k-1$ and $Q$ is of odd height, set $\eta^{\prime}(P)=D_{k+{1\over 2}}(\beta^{-1}(Q))^{c}$.

• ( iii*′* )

If $P=QH_{2i}Q^{rc}$ with $0\leq i\leq k$ and $Q$ is of odd height, set $\eta^{\prime}(P)=H_{i+1}\beta^{-1}(Q)$.

• ( iv*′* )

If $P=QU_{k+{1\over 2}}H_{2i-1}D_{k+{1\over 2}}Q^{rc}$ with $1\leq i\leq k-1$ and $Q$ is of odd height, set $\eta^{\prime}(P)=H_{k+1+i}\beta^{-1}(Q)$.

• ( v*′* )

If $P=QH_{2k-1}Q^{rc}$ and $Q$ is of even height, set $\eta^{\prime}(P)=D_{k+{1\over 2}}\alpha^{-1}(Q)$.

It is routine to check that the resulting path $\eta^{\prime}(P)\in\mathcal{Q}^{*}(m+1,k)$. Therefore, the map $\eta^{\prime}$ is well defined. Furthermore, one can easily verify that ( i*′* ), ( ii*′* ), ( iii*′* ), ( iv*′* ) and ( v*′* ) respectively reverse the procedures of ( i ), ( ii ), ( iii ), ( iv ) and ( v ). This implies that the maps $\eta$ and $\eta^{\prime}$ are inverses of each other. Hence, the map $\eta$ is a bijection, completing the proof.

Theorems 2.3, 2.4, 2.20 and 2.21 together with Equation (2.1) yield the following result.

Theorem 2.22

For given positive integer $m$ and nonnegative integer $k$, the number of self-conjugate $(2m+1,2m+2,\ldots,2m+2k+2)$-core partitions is equal to the number of symmetric $(2m+1,2k+1)$-Dyck paths.

Combining Theorems 2.14, 2.19 and 2.22, we complete the proof of Conjecture 1.1.

Acknowledgments. This work was supported by the National Natural Science Foundation of China (11671366).