A new bijective proof of Babson and Steingr\'{\i}msson's conjecture

TL;DR

This paper presents a novel bijective proof confirming the equidistribution of two permutation statistics, $stat$ and $maj$, resolving a conjecture by Babson and Steingrímsson with a combinatorial approach.

Contribution

The paper introduces a new bijective proof of Babson and Steingrímsson's conjecture, providing a combinatorial alternative to previous algebraic and computational proofs.

Findings

Confirmed that $(des,stat)$ and $(des,maj)$ are equidistributed on $S_n$.

Provided a new combinatorial bijection proof of the conjecture.

Enhanced understanding of permutation statistics and their distributions.

Abstract

Babson and Steingr\'{\i}msson introduced generalized permutation patterns and showed that most of the Mahonian statistics in the literature can be expressed by the combination of generalized pattern functions. Particularly, they defined a new Mahonian statistic in terms of generalized pattern functions, which is denoted $stat$. Given a permutation $\pi$, let $des(\pi)$ denote the descent number of $\pi$ and $maj(\pi)$ denote the major index of $\pi$. Babson and Steingr\'{\i}msson conjectured that $(des,stat)$ and $(des,maj)$ are equidistributed on $S_n$. Foata and Zeilberger settled this conjecture using q-enumeration, generating functions and Maple packages ROTA and PERCY. Later, Burstein provided a bijective proof of a refinement of this conjecture. In this paper, we give a new bijective proof of this conjecture.

Tables

Table 1. Table 2.1: The computation of the major index table of 13287546 13287546 13287546 .

$i$	${\pi }^{\left(i-1\right)}$	$m_i$
$8$	${}_4{}^{}1_5^{}{3}_3{2}_6{7}_2{5}_1{4}_7{6}_0$	$6$
$7$	${}_3{}^{}1_4^{}{3}_2{2}_5{5}_1{4}_6{6}_0$	$5$
$6$	${}_3{}^{}1_4^{}{3}_2{2}_5{5}_1{4}_0$	$0$
$5$	${}_2{}^{}1_3^{}{3}_1{2}_4{4}_0$	$4$
$4$	${}_2{}^{}1_3^{}{3}_1{2}_0$	$0$
$3$	${}_1{}^{}1_2^{}{2}_0$	$2$
$2$	${}_1{}^{}1_0^{}$	$0$

Table 2. Table 2.2: The computation of the stat table of 52718346 52718346 52718346 .

$i$	${\pi }^{\left(i-1\right)}$	$s_i$
$8$	${}_3{}^{}5_0^{}{2}_7{7}_1{1}_6{3}_5{4}_4{6}_2$	$6$
$7$	${}_3{}^{}5_0^{}{2}_1{1}_6{3}_5{4}_4{6}_2$	$1$
$6$	${}_3{}^{}5_0^{}{2}_1{1}_5{3}_4{4}_2$	$2$
$5$	${}_2{}^{}2_0^{}{1}_4{3}_3{4}_1$	$2$
$4$	${}_2{}^{}2_0^{}{1}_3{3}_1$	$1$
$3$	${}_2{}^{}2_0^{}{1}_1$	$1$
$2$	${}_1{}^{}1_0^{}$	$1$

Table 3. Table 2.3: The computation of ρ ​ ( 52718346 ) 𝜌 52718346 \rho(52718346) .

$i$	$s_i$	${\sigma }^{\left(i\right)}$
$5$	$2$	${}_3{}^{}5_2^{}{1}_4{2}_5{4}_1{3}_0$
$6$	$2$	${}_3{}^{}5_4^{}{6}_2{1}_5{2}_6{4}_1{3}_0$
$7$	$1$	${}_3{}^{}5_4^{}{6}_2{1}_5{2}_6{4}_7{7}_1{3}_0$
$8$	$6$	$56128473$

Equations

$$Des(\pi )=\{i:{\pi }_i>{\pi }_{i+1}\},$$

$$Asc(\pi )=\{i:{\pi }_i<{\pi }_{i+1}\}.$$

$$Inv(\pi )=\{(i,j):1\le i<j\le n,{\pi }_i>{\pi }_j\}.$$

$$maj(\pi )=\sum _{i\in Des(\pi )}i.$$

$$\sum _{\sigma \in Ob{j}_1}{q}^{s{t}_1(\sigma )}=\sum _{\sigma \in Ob{j}_2}{q}^{s{t}_2(\sigma )},$$

$$\sum _{\pi \in S_n}{q}^{inv(\pi )}=\sum _{\pi \in S_n}{q}^{maj(\pi )}=[n{]}_q!,$$

$$maj=(a-cb)+(b-ca)+(c-ba)+(ba),$$

$$stat=(ac-b)+(ba-c)+(cb-a)+(ba).$$

$$adj(\pi )=|\{i:1\le i\le n\text{ and }{\pi }_i-{\pi }_{i+1}=1\}|,$$

$${}_8{1}_7{3}_6{2}_5{8}_4{7}_3{5}_2{4}_1{6}_0,$$

$${}_5{1}_6{3}_4{2}_7{8}_3{7}_2{5}_1{4}_8{6}_0.$$

$${\varphi }_{inv,n}:\{0,1,\dots ,n-1\}\times S_{n-1}\to S_n$$

$${\varphi }_{inv,9}(3,13287546)=132879546\text{and}{\varphi }_{maj,9}(3,13287546)=132897546.$$

$$inv({\varphi }_{inv,n}(i,\sigma ))=inv(\sigma )+i.$$

$$maj({\varphi }_{maj,n}(i,\sigma ))=maj(\sigma )+i.$$

$$(c_i,{\pi }^{(i-1)})={\varphi }_{inv,i}^{-1}({\pi }^{(i)}),$$

$$(m_i,{\pi }^{(i-1)})={\varphi }_{maj,i}^{-1}({\pi }^{(i)}).$$

$$E_n=\{w=w_1\cdots w_n|w_i\in [0,i-1],1\le i\le n\}.$$

$${}_5{1}_8{3}_0{2}_7{8}_1{7}_2{5}_3{4}_6{6}_4.$$

$$stat({\varphi }_{stat,n}(i,\sigma ))=stat(\sigma )+i.$$

$$(s_i,{\pi }^{(i-1)})={\varphi }_{stat,i}^{-1}({\pi }^{(i)}).$$

$$f_{i}+h_{i}=\left\{\begin{array}[]{ll}des(p),&\mbox{if $i$ is a descent of $p$ or $i=n$,}\\[3.0pt] n+des(p)+2,&\mbox{if $i$ is an ascent of $p$,}\\[6.0pt] 2des(p)+2,&\mbox{if $i=0$.}\end{array}\right.$$

$$\alpha^{(i+1)}=\phi_{maj,i+1}(l_{i+1},\alpha^{(i)})=\left\{\begin{array}[]{ll}\phi_{stat,i+1}(d-l_{i+1},\alpha^{(i)}),&\mbox{if $0\leq l_{i+1}\leq d$,}\\[3.0pt] \phi_{stat,i+1}(n+d+2-l_{i+1},\alpha^{(i)}),&\mbox{if $d+2\leq l_{i+1}\leq n$.}\end{array}\right.$$

$${\alpha }^{(i+1)}=$$

$$=$$

$$(ac-b){\pi }^{(k)}+(ba-c){\pi }^{(k)}+(cb-a){\pi }^{(k)}=(a-cb){\sigma }^{(k)}+(b-ca){\sigma }^{(k)}+(c-ba){\sigma }^{(k)}$$

$$(ac-b){\pi }^{(k)}+(ba-c){\pi }^{(k)}+(cb-a){\pi }^{(k)}$$

$$=$$

$$(\text{b}c-ba)p+(b-ca)p=(\text{b}c-b-\text{b}a)p+(b-ac)p.$$

$$A(p)$$

Full text

Keywords: Euler-Mahonian, bijection, involution

AMS Subject Classifications: 05A05, 05A15

1 Introduction

In this paper, we give a new bijective proof of a conjecture of Babson and Steingrímsson [1] on Euler-Mahonian statistics.

Let $S_{n}$ denote the set of all the permutations of $[n]=\{1,2,\cdots,n\}$. Given a permutation $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in S_{n}$, a descent of $\pi$ is a position $i\in[n-1]$ such that $\pi_{i}>\pi_{i+1}$, where $\pi_{i}$ and $\pi_{i+1}$ are called a descent top and a descent bottom, respectively. An ascent of $\pi$ is a position $i\in[n-1]$ such that $\pi_{i}<\pi_{i+1}$, where $\pi_{i}$ is called an ascent bottom and $\pi_{i+1}$ is called an ascent top. The descent set and the ascent set of $\pi$ are given by

[TABLE]

[TABLE]

The set of the inversions of $\pi$ is

[TABLE]

Let $des(\pi)$, $asc(\pi)$ and $inv(\pi)$ be the descent number, the ascent number and the inversion number of $\pi$, which are defined by $des(\pi)=|Des(\pi)|$, $asc(\pi)=|Asc(\pi)|$ and $inv(\pi)=|Inv(\pi)|$, respectively. The major index of $\pi$, denoted $maj(\pi)$, is given by

[TABLE]

Suppose that $st_{1}$ is a statistic on the object $Obj_{1}$ and $st_{2}$ is a statistic on the object $Obj_{2}$. If

[TABLE]

we say that the statistic $st_{1}$ over $Obj_{1}$ is equidistributed with the statistic $st_{2}$ over $Obj_{2}$.

A statistic on $S_{n}$ is said to be Eulerian if it is equidistributed with the statistic $des$ on $S_{n}$. While a statistic on $S_{n}$ is said to be Mahonian if it is equidistributed with the statistic $inv$ on $S_{n}$. It is well-known that

[TABLE]

where $[n]_{q}=1+q+\cdots+q^{n-1}$ and $[n]_{q}!=[n]_{q}[n-1]_{q}\cdots[1]_{q}$. Thus, the major index $maj$ is a Mahonian statistic. A pair of statistics on $S_{n}$ is said to be Euler-Mahonian if it is equidistributed with the joint distribution of the descent number and the major index.

In [1], Babson and Steingrímsson introduced generalized permutation patterns, where they allow the requirement that two adjacent letters in a pattern must be adjacent in the permutation. Let $\mathcal{A}$ be the alphabet $\{a,b,c,\ldots\}$ with the usual ordering. We write patterns as words in $\mathcal{A}$, where two adjacent letters may or may not be separated by a dash. Two adjacent letters without a dash in a pattern indicates that the corresponding letters in the permutation must be adjacent. Given a generalized pattern $\tau$ and a permutation $\pi$, we say a subsequence of $\pi$ is an occurrence (or instance) of $\tau$ in $\pi$ if it is order-isomorphic to $\tau$ and satisfies the above dash conditions. Let $(\tau)\pi$ denote the number of occurrences of $\tau$ in $\pi$. Here, we see $(\tau)$ as a generalized pattern function. For example, an occurrence of the generalized pattern $b$-$ca$ in a permutation $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$ is a subsequence $\pi_{i}\pi_{j}\pi_{j+1}$ such that $i<j$ and $\pi_{j+1}<\pi_{i}<\pi_{j}$. For $\pi=4753162$, we have $(b$-$ca)\pi=4$.

Further, Babson and Steingrímsson [1] showed that almost all of the Mahonian permutation statistics in the literature can be written as linear combinations of generalized patterns. We list some of them below.

[TABLE]

[TABLE]

They conjectured that the statistic $(des,stat)$ is Euler-Mahonian.

Conjecture 1.1

The distribution of the bistatistic $(des,stat)$ is equal to that of $(des,maj)$.

In 2001, D. Foata and D. Zeilberger [4] gave a proof of this conjecture using q-enumeration and generating functions and an almost completely automated proof via Maple packages ROTA and PERCY.

Given a permutation $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$, let $F(\pi)=\pi_{1}$ be the first letter of $\pi$ and

[TABLE]

where $\pi_{n+1}=0$. Burstein [2] provided a bijective proof of the following refinement of Conjecture 1.1 as follows.

Theorem 1.2

Statistics $(adj,des,F,maj,stat)$ and $(adj,des,F,stat,maj)$ are equidistributed over $S_{n}$ for all $n$.

In this paper, we will give a new bijective proof of Conjecture 1.1, which does not preserve the statistic $adj$.

2 A new bijective proof of Conjecture 1.1

In this section, we recall a particular bijection $\varphi$ on $S_{n}$ that maps the inversion number to the major index, which is due to Carlitz [3] and stated more clearly in [6] and [7]. Based on this, we give an analogous bijection which proves Conjecture 1.1.

To give a description of $\varphi$, we first recall two labeling schemes for permutations. It involves accounting for the effects of inserting a new largest element into a permutation, and so it is known as the insertion method.

Given a permutation $\sigma=\sigma_{1}\sigma_{2}\cdots\sigma_{n-1}$ in $S_{n-1}$. To obtain a permutation $\pi\in S_{n}$, we can insert $n$ in $n$ spaces, namely, immediately before $\sigma_{1}$ or immediately after $\sigma_{i}$ for $1\leq i\leq n$. In order to keep track of how the insertion of $n$ affects the inversion number and the major index, we may define two labelings of the $n$ inserting spaces.

The inv-labeling of $\sigma$ is given by numbering the spaces from right to left with $0,1,\ldots,n-1$. The maj-labeling of $\sigma$ is obtained by labeling the space after $\sigma_{n-1}$ with [math], labeling the descents from right to left with $1,2,\ldots,\rm{des(\sigma)}$ and labeling the remaining spaces from left to right with $des(\sigma)+1,\ldots,n$. As an example, let $\sigma=13287546$, the inv-labeling of $\sigma$ is

[TABLE]

while the maj-labeling of $\sigma$ is given by

[TABLE]

For $n\geq 2$, we define the map

[TABLE]

by setting $\phi_{inv,n}(i,\sigma)$ to be the permutation obtained by inserting $n$ in the space labeled $i$ in the inv-labeling of $\sigma$. By changing the inv-labeling to maj-labeling, we obtain the map $\phi_{maj,n}$. As an example,

[TABLE]

For maps $\phi_{inv,n}$ and $\phi_{maj,n}$, we have the following two lemmas.

Lemma 2.1

For $\sigma\in S_{n-1}$ and $i\in\{0,1,\ldots,n-1\}$, we have

[TABLE]

Lemma 2.2

For $\sigma\in S_{n-1}$ and $i\in\{0,1,\ldots,n-1\}$, we have

[TABLE]

Lemma 2.1 is easy to verified. For a detailed proof of Lemma 2.2, see [5].

Given a permutation $\pi\in S_{n}$, let $\pi^{(i)}$ be the restriction of $\pi$ to the letters $1,2,\ldots,i$ for $1\leq i\leq n$ . For $2\leq i\leq n$, let

[TABLE]

[TABLE]

By setting $c_{1}=0$ and $m_{1}=0$, we obtain two sequences $c_{1}c_{2}\cdots c_{n}$ and $m_{1}m_{2}\cdots m_{n}$ in $E_{n}$, where

[TABLE]

Define $\gamma(\pi)=c_{1}c_{2}\cdots c_{n}$ and $\mu(\pi)=m_{1}m_{2}\cdots m_{n}$. It is not hard to check that both $\gamma$ and $\mu$ are bijections. The sequence $c_{1}c_{2}\cdots c_{n}$ is called the inversion table of $\pi$, while $m_{1}m_{2}\cdots m_{n}$ is called the major index table of $\pi$. Moreover, we have $\sum_{i=1}^{n}c_{i}=inv(\pi)$ and $\sum_{i=1}^{n}m_{i}=maj(\pi).$ As an example, let $\pi=13287546$, then $\pi^{(8)}=\pi$ and $\mu(\pi)=00204056$ as computed in Table 2.1.

Now, we can define the bijection $\varphi$ that maps the inversion number to the major index by letting $\varphi=\mu^{-1}\gamma$. Clearly, $\varphi$ is a bijection which proves that statistics $inv$ and $maj$ are equidistributed over $S_{n}$.

In the following of this section, we will construct a bijection $\rho$ to prove Conjecture 1.1, which is, to some extent, an analogue of the above bijection $\varphi$.

First, we define a stat-labeling of $\sigma\in S_{n}$. Label the descents of $\sigma$ and the space after $\sigma_{n}$ by $0,1,\ldots,des(\sigma)$ from left to right. Label the space before $\sigma_{0}$ by $des(\sigma)+1$. The ascents of $\sigma$ are labeled from right to left by $des(\sigma)+2,\ldots,n$. As an example, for $\sigma=13287546$, we have the stat-labeling of $\sigma$ as follows

[TABLE]

Based on the stat-labeling, we define the map $\phi_{stat,n}$ for $n\geq 2$. For a permutation $\sigma=\sigma_{1}\sigma_{2}\ldots\sigma_{n-1}$ and $0\leq i\leq n-1$, define $\phi_{stat,n}(i,\sigma)$ to be the permutation obtained by inserting $n$ in the position labeled $i$ in the stat-labeling of $\sigma$. For instance, $\phi_{stat,9}(7,13287546)=132987546$.

It should be noted that unlike the properties of $\phi_{inv,n}$ and $\phi_{maj,n}$ stated in Lemma 2.1 and Lemma 2.2, we deduce that $\phi_{stat,n}(i,\sigma))\neq stat(\sigma)+i$ for $i=des(\sigma)+1$. For the map $\phi_{stat,n}$, we have the following property.

Lemma 2.3

For $\sigma\in S_{n-1}$ and $i\in\{0,1,\ldots,des(\sigma),des(\sigma)+2,\ldots,n-1\}$, we have

[TABLE]

*Proof.  *First, we recall that $stat=(ac$-$b)+(ba$-$c)+(cb$-$a)+(ba)$. To prove this lemma, we have to consider the changes of the statistic $stat$ brought by inserting $n$ into $\sigma$. Assume that $\sigma=\sigma_{1}\sigma_{2}\cdots\sigma_{n-1}$, there are three cases for us to consider.

• •

Case 1 $\colon$ $n$ is inserted into the space after $\sigma_{n-1}$.

Write $\pi=\sigma n$. Clearly, the insertion of $n$ does not bring new $ac$-$b$, $cb$-$a$ and $ba$ patterns. While $n$ can form new $ba$-$c$ patterns of $\pi$ with the $ba$ patterns of $\sigma$. It follows that $stat(\pi)-stat(\sigma)=(ba)\sigma=des(\sigma)$. Notice that the label of the space after $\sigma_{n-1}$ is $des(\sigma)$. Hence, in this case the lemma holds.

• •

Case 2 $\colon$ $n$ is inserted into a descent.

Suppose that $\tau$ is the permutation obtained by inserting $n$ to the position $i$ and $\sigma_{i}>\sigma_{i+1}$. Moreover, let this descent be the $k$-th descent from left to right. We claim that $stat(\tau)-stat(\sigma)=k-1$.

The insertion of $n$ forms some new $ac$-$b$ patterns, and the number of these new patterns is $|\{j,~{}j>i~{}\text{and}~{}\sigma_{j}>\sigma_{i}\}|$. Moreover, $n$ forms $k-1$ new $ba$-$c$ patterns with the former $k-1$ descents, while it destroys the $ba$-$c$ patterns of $\sigma$ by the number $|\{j,~{}j>i~{}\text{and}~{}\sigma_{j}>\sigma_{i}\}|$. It is easy to verify the functions $(cb$-$a)$ and $(ba)$ do not change. Hence, we conclude that $stat(\tau)-stat(\sigma)=k-1$. The claim is verified. Notice that the label of the $k$-th descent from left to right is also $k-1$. It follows that the lemma holds for this case.

• •

Case 3 $\colon$ $n$ is inserted into an ascent.

Suppose that $p$ is the permutation obtained by inserting $n$ into the position $i$, where $\sigma_{i}<\sigma_{i+1}$. Moreover, we assume that there are $k$ descents to the left of position $i$. We claim that $stat(p)-stat(\sigma)=k+n-i+1$.

Now we proceed to prove this claim. The insertion of $n$ to $\sigma$ brings new $ac$-$b$ patterns, the number of which is $|\{j,~{}j\geq i+1~{}\text{and}~{}\sigma_{j}\geq\sigma_{i+1}\}|$. Moreover, the insertion of $n$ brings $k$ new $ba$-$c$ patterns with the former $k$ descents. Also, $n\sigma_{i+1}\sigma_{l}$, where $l>i+1$ and $\sigma_{l}<\sigma_{i+1}$, forms a new $cb$-$a$ pattern of $p$. Notice that $(ba)p-(ba)\sigma=1$. By combining all above, we see that $stat(\tau)-stat(\sigma)=k+n-i+1$. The claim is verified. By the stat-labeling of $\sigma$, we see that the label of this position is also $k+n-i+1$. Hence, in this case the lemma holds.

By combining the three cases above, we compete the proof.

Based on the stat-labeling, we define the stat table of a permutation. Given $\pi\in S_{n}$, for $2\leq i\leq n$, let

[TABLE]

Set $s_{1}=0$ and $\nu(\pi)=s_{1}s_{2}\cdots s_{n}$. It is easily checked that $\nu$ is a bijection from $S_{n}$ to $E_{n}$. As an example, $\nu(52718346)=01112216$, which is computed in Table 2.2.

Now we can give the definition of the map $\rho$ which proves Conjecture 1.1. Given $\pi\in S_{n}$, let $\sigma=\rho(\pi)$, where $\sigma$ can be constructed as follows. Assume that $F(\pi)=k$ and $\pi^{(k)}=kp_{2}\cdots p_{k}$. Then, let $\sigma^{(k)}=k(k-p_{k})(k-p_{k-1})\cdots(k-p_{2})$. Assume that $s=s_{1}s_{2}\cdots s_{n}=\nu(\pi)$. For $k+1\leq i\leq n$, let $\sigma^{(i)}=\phi_{maj,i}(s_{i},\sigma^{(i-1)})$. Clearly, $\sigma=\sigma^{(n)}$ can be constructed by the procedure above. As an example, we compute $\rho(\pi)$, where $\pi=52718346$. It is straightforward to see that $k=5$ and $\pi^{(5)}=52134$.

By the computation in Table 2.3, we see that $\rho(52718346)=56128473$. Notice that in this example, the descent number and the first letter of both of the preimage and image of $\rho$ are the same. In fact, these properties always holds.

Lemma 2.4

For $\pi\in S_{n}$, we have $F(\pi)=F(\rho(\pi))$ and $des(\pi)=des(\rho(\pi))$.

*Proof.  *Suppose that $\sigma=\rho(\pi)$ and $k=F(\pi)$. We proceed to show that $F(\sigma^{(i)})=F(\pi^{(i)})=k$ and $des(\sigma^{(i)})=des(\pi^{(i)})$ for $k\leq i\leq n$ by induction. Let $\pi^{(k)}=kp_{2}\cdots p_{k}$, then we have $\sigma^{(k)}=k(k-p_{k})(k-p_{k-1})\cdots(k-p_{2})$. It is routine to check that $F(\sigma^{(k)})=F(\pi^{(k)})=k$ and $des(\sigma^{(k)})=des(\pi^{(k)})$, hence, we omit the details here.

Now assume that $des(\sigma^{(l)})=des(\pi^{(l)})=d$ and $F(\sigma^{(l)})=F(\pi^{(l)})=k$ for $k\leq l\leq n-1$. We proceed to show that $des(\sigma^{(l+1)})=des(\pi^{(l+1)})$ and $F(\sigma^{(l+1)})=F(\pi^{(l+1)})=k$.

Let $s=s_{1}s_{2}\cdots s_{n}=\nu(\pi)$. By the constructions of $\nu$ and $\rho$, we may see that $\pi^{(l+1)}=\phi_{stat,l+1}(s_{l+1},\pi^{(l)})$ and $\sigma^{(l+1)}=\phi_{maj,l+1}(s_{l+1},\sigma^{(l)})$. Notice that both in the maj-labeling of $\sigma^{(l)}$ and the stat-labeling of $\pi^{(l)}$, the descents and the space after the last letter are labeled by $\{0,1,\ldots,d\}$, the space before the first element is labeled by $d+1$, and the ascents are labeled by $\{d+2,\ldots,l\}$. Since $F(\pi^{(l)})=k=F(\pi)$, it is easy to see that $s_{l+1}\neq d+1$. It follows that $F(\sigma^{(l+1)})=F(\pi^{(l+1)})=k$.

If $s_{l+1}<d+1$, then $l+1$ is inserted into the descents or the space after the last element of $\pi^{(l)}$ and $\sigma^{(l)}$. Hence, we deduce that $des(\sigma^{(l+1)})=des(\pi^{(l+1)})=d$. If $s_{l+1}>d+1$, then $l+1$ is inserted into the ascents of $\pi^{(l)}$ and $\sigma^{(l)}$. Hence, we deduce that $des(\sigma^{(l+1)})=des(\pi^{(l+1)})=d+1$. Combining the two cases above, we have $des(\sigma^{(l+1)})=des(\pi^{(l+1)})$. Notice that $\sigma^{(n)}=\sigma$ and $\pi^{(n)}=\pi$, we complete the proof.

Base on the construction of $\rho$ and Lemma 2.4, we have the following theorem.

Theorem 2.5

The map $\rho$ is an involution on $S_{n}$.

*Proof.  *Given a permutation $\pi\in S_{n}$, it suffices for us to show that $\rho^{2}(\pi)=\pi$. That is, writing $\sigma=\rho(\pi)$, we need to show that $\rho(\sigma)=\pi$.

Let $F(\pi)=k$, then by Lemma 2.4, we know that $k=F(\pi)=F(\sigma)$. Write $\pi^{(k)}=kp_{2}\cdots p_{k}$, then we have $\sigma^{(k)}=k(k-p_{k})\cdots(k-p_{2})$. Assume that $\nu(\pi)=s=s_{1}s_{2}\cdots s_{n}$ and $\mu(\sigma)=m_{1}m_{2}\cdots m_{n}$. By the construction of $\rho$, we have $\sigma^{(i)}=\phi_{maj,i}(s_{i},\sigma^{(i-1)})$ for $k+1\leq i\leq n$. It follows that $m_{i}=s_{i}$ for $k+1\leq i\leq n$.

Suppose that $\alpha=\rho(\sigma)$, in the following, we proceed to show that $\alpha^{(i)}=\pi^{(i)}$ for $k\leq i\leq n$ by induction. By the definition of $\rho$, we have $\alpha^{(k)}=kp_{2}\cdots p_{k}=\pi^{(k)}$. Assume that $\alpha^{(i)}=\pi^{(i)}$ holds for $k\leq i\leq n-1$, we aim to show that $\alpha^{(i+1)}=\pi^{(i+1)}$. To achieve this, we need to mention the following property of the maj-labeling and the stat-labeling of a single permutation.

For a permutation $p\in S_{n}$, assume the maj-labeling of $p$ is $f_{0}f_{1}\cdots f_{n}$, while the stat-labeling of $p$ is $h_{0}h_{1}\cdots h_{n}$. Then it is easily checked that

[TABLE]

Let $\nu(\sigma)=l_{1}l_{2}\cdots l_{n}$ and $d=des(\alpha^{(i)})$. Then, we have

[TABLE]

By the proof of Lemma 2.4, we see that $des(\sigma^{(i)})=des(\alpha^{(i)})=d$. Recall that $\nu(\sigma)=l_{1}l_{2}\cdots l_{n}$ and $\mu(\sigma)=m_{1}m_{2}\cdots m_{n}$. Hence, it follows from (2.4) that

[TABLE]

Notice that $\alpha=\alpha^{(n)}$ and $\pi=\pi^{(n)}$. Hence, we have $\pi=\rho(\sigma)$, namely, $\rho^{2}(\pi)=\pi$. This completes the proof.

Indeed, the involution $\rho$ also preserves some addtional statistics, which is stated in the following proposition.

Proposition 2.6

For any $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in S_{n}$, we have $maj(\rho(\pi))=stat(\pi)$ and $stat(\rho(\pi))=maj(\pi)$.

*Proof.  *Write $\sigma=\rho(\pi)$. Let $k=F(\pi)$ and $\pi^{(k)}=kp_{2}\cdots p_{k}$. Then we have $\sigma^{(k)}=k(k-p_{k})\cdots(k-p_{2})$. In the following, we will show that $maj(\sigma^{(i)})=stat(\pi^{(i)})$ for $k\leq i\leq n$ by induction.

First, we show that $maj(\sigma^{(k)})=stat(\pi^{(k)})$. By the proof of Lemma 2.4, we have $des(\pi^{(k)})=des(\sigma^{(k)}).$ Hence $(ba)\pi^{(k)}=(ba)\sigma^{(k)}$. It suffices for us to show that

[TABLE]

Given a pattern $p=p_{1}p_{2}\cdots p_{k}$, by putting a line under $p_{1}$(resp. $p_{k}$), we mean that an instance of $p$ must begin(resp. end) with the leftmost(resp. rightmost) letter of the permutation. By putting a dot under $p_{1}$, we mean that an instance of $p$ must not begin with the leftmost letter. For instance, $(\b{c}b$-$a)$ is a function which maps a permutation, say $\tau=\tau_{1}\tau_{2}\cdots\tau_{n}$, to $|\{i,\tau_{1}\tau_{2}\tau_{i}~{}\text{forms a 321 pattern of}~{}\tau\}|.$

By the definition of $\rho$, we know that

[TABLE]

Hence, to prove (2.2), we need to prove that for any permutation $p\in S_{k}$ with $p_{1}=k$,

[TABLE]

We define sets $A(p),C(p)$ and multisets $B(p),D(p)$ as follows.

[TABLE]

To prove (2.3), it is enough to show that $A\cup B=C\cup D$, where the union operator is a multiset union. First, we show that $C\cup D\subseteq A\cup B$.

Let $p_{k}=a$, then we know that $C(p)=\{a+1,a+2,\ldots,k-1\}$. If $p_{i}\in C(p)$ is a descent top, it is easy to see that $p_{i}\in A(p)$. If $p_{i}\in C(p)$ is an ascent bottom, we claim that $p_{i}\in B(p)$. This claim will be proved together with case 4 in the following.

Given an element $p_{i}$ in $D(p)$, if the multiplicity of $p_{i}$ is $x$, there exists a set

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which is ordered by increasing order, satisfying that

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are instances of $213$ pattern. We claim that there exists $j_{1}+1\leq r_{1}<j_{2}$ such that $p_{i}p_{r_{1}}p_{r_{1}+1}$ forms a $231$ pattern.

Choose the smallest $g_{1}$ such that $j_{1}+1\leq g_{1}<j_{2}$ and $g_{1}$ is a descent. If $p_{g_{1}+1}<p_{i}$, then $p_{i}p_{g_{1}}p_{g_{1}+1}$ forms a $231$ pattern, the claim is verified. Otherwise, we seek the smallest $g_{1}+1\leq g_{2}<j_{2}$ such that $g_{2}$ is a descent. If $p_{g_{2}+1}<p_{i}$, the claim is verified. If not, we repeat the above process. Since $p_{i}>p_{j_{2}}$, the process must be terminated. Hence, the claim is verified. By a similar means, we deduce that there exists $j_{l}+1\leq r_{l}<j_{l+1}$ where $1\leq l\leq x-1$ such that $p_{i}p_{r_{l}}p_{r_{l}+1}$ forms a $231$ pattern. The claim is verified.

To analyze the element $p_{i}$ in $D(p)$, we consider four cases.

• •

$p_{i}$ is a descent top and $1<p_{i}\leq a-1$.

Suppose that the multiplicity of $p_{i}$ of this type in $D(p)$ is $x$. By the above statement, we see that $p_{i}p_{r_{l}}p_{r_{l}+1}$ forms a $231$ pattern, where $j_{l}+1\leq r_{l}<j_{l+1}$ and $1\leq l\leq x-1$. Thus, we deduce that there are $x-1$ $p_{i}$s in $B(p)$. Notice that there is one $p_{i}$ left in $B(p)$. Clearly, we can set this $p_{i}$ to be the element of $A(p)$ consisting of $p_{1}$, $p_{i}$ and $p_{i+1}$.

• •

$p_{i}$ is an ascent bottom and $1\leq p_{i}\leq a-1$.

Suppose that the multiplicity of $p_{i}$ of this type in $D(p)$ is $x$. Similarly, we know that $p_{i}p_{r_{l}}p_{r_{l}+1}$ forms a $231$ pattern, where $j_{l}+1\leq r_{l}<j_{l+1}$ and $1\leq l\leq x-1$. Since $p_{i}<p_{i+1}$, we have $j_{1}>i+1$. Based on this, it can be easily seen that there exists $r_{0}$ such that $i+1\leq r_{0}<j_{1}$ and $p_{i}p_{r_{0}}p_{r_{0}+1}$ forms a $231$ pattern. Hence, we deduce that in this case there are $x$ $p_{i}$s in $B(p)$.

• •

$p_{i}$ is a descent top and $a+1\leq p_{i}\leq k-1$.

Suppose that the multiplicity of $p_{i}$ of this type in $D(p)$ is $x$. Similarly, we deduce that $p_{i}p_{r_{l}}p_{r_{l}+1}$ forms a $231$ pattern, where $j_{l}+1\leq r_{l}<j_{l+1}$ and $1\leq l\leq x-1$. What’s more, it follows from $p_{i}>a$ that there exists $j_{x}\leq r_{x}<m$ such that $p_{i}p_{r_{x}}a$ forms a $231$ pattern. Hence, we deduce that $p_{i}$ in $B(p)$ and its multiplicity is $x$.

• •

$p_{i}$ is an ascent bottom and $a+1\leq p_{i}\leq k-1$.

Suppose that the multiplicity of $p_{i}$ of this case in $D(p)$ is $x$. Notice that $p_{i}$ is also an element of $C(p)$ with multiplicity equals $1$. Hence, in this case, we have to prove that there are $x+1$ $p_{i}$s in $B(p)$.

Similarly with the above cases, we deduce that $p_{i}p_{r_{l}}p_{r_{l}+1}$ forms a $231$ pattern, where $j_{l}+1\leq r_{l}<j_{l+1}$ and $1\leq l\leq x-1$. Since $p_{i}>a$, there exists $j_{x}\leq r_{x}<m$ such that $p_{i}p_{r_{x}}a$ forms a $231$ pattern. By $p_{i}<p_{i+1}$, we have $j_{1}>i+1$. Based on this, it can be easy seen that there exists $r_{0}$ such that $i+1\leq r_{0}<j_{1}$ and $p_{i}p_{r_{0}}p_{r_{0}+1}$ forms a $231$ pattern. Hence, we deduce that $p_{i}$ in $B(p)$ and its multiplicity is $x+1$.

Combining all above, we deduce that $C\cup D\subseteq A\cup B$. By a similar analysis, we can prove that $A\cup B\subseteq C\cup D$. We omit it here. As an example, if $p=978452613$, we have $A(p)=\{5,6,8\}$, $B(p)=\{2,4,4,5,7\}$, $C(p)=\{4,5,6,7,8\}$ and $D(p)=\{2,4,5\}$. It can be verified that $A\cup B=C\cup D$. This proves that $maj(\sigma^{(k)})=stat(\pi^{(k)})$.

Now assume that $maj(\sigma^{(i)})=stat(\pi^{(i)})$ for $k\leq i\leq n-1$, we proceed to show that $maj(\sigma^{(i+1)})=stat(\pi^{(i+1)})$. Write $\nu(\pi)=s_{1}s_{2}\cdots s_{n}$. Then, by Lemma 2.2, Lemma 2.3 and the construction of $\rho$, we have

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This proves $maj(\sigma^{(i)})=stat(\pi^{(i)})$ for $k\leq i\leq n$. Notice that $\pi=\pi^{(n)}$ and $\sigma=\sigma^{(n)}$. We deduce that $maj(\sigma)=maj(\rho(\pi))=stat(\pi)$. By Theorem 2.5, we see that $\rho$ is an involution. This implies that $stat(\rho(\pi))=maj(\pi)$. This completes the proof.

Combining Lemma 2.4, Theorem 2.5 and Proposition 2.6, we give a proof of Conjecture 1.1.

It should be mentioned that Burstein [2] provided a direct bijective proof of a refinement of Conjecture 1.1. The bijection $\chi$ is given as follows. Given a permutation $\pi\in S_{n}$ with $F(\pi)=k$. Let $\pi^{\prime}=\chi(\pi)$ with $\pi^{\prime}(1)=k$ and

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In addition to preserving the statistics $des$ and $F$, the bijection $\chi$ also preserves the statistic $adj$, while our bijection does not. As an example, set $\pi=543617982$, then $\rho(\pi)=\sigma=539784621$ and $\chi(\pi)=\pi^{\prime}=537684921$. It can be checked that $adj(\pi)=adj(\pi^{\prime})$, while $adj(\pi)\neq adj(\sigma)$. Moreover, it is easily seen that for $\pi\in S_{n}$ with $F(\pi)=n$, we have $\rho(\pi)=\chi(\pi)=\sigma$. We note that in this case both Burstein and us have to prove $maj(\sigma)=stat(\pi)$. Different form our proof in Proposition 2.6, Burstein gave the following two relations, which implies that $maj(\sigma)=stat(\pi)$.

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Acknowledgments. We wish to thank the anonymous referees for their valuable comments and suggestions.