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

0.12 MB·

by  Joanna N. Chen

A new bijective proof of Babson and Steingr´ımsson’s conjecture
Joanna N. Chen1, Shouxiao Li2
1College of Science
Tianjin University of Technology
Tianjin 300384, P.R. China
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1 2 College of Computer and Information Engineering
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2 Tianjin Agricultural University
n Tianjin 300384, P.R. China
a
J [email protected], [email protected].

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2
Abstract

]
O
C Babson and Steingr´ımsson introduced generalized permutation patterns and
. showed that most of the Mahonian statistics in the literature can be expressed
h
by the combination of generalized pattern functions. Particularly, they defined
t
a
a new Mahonian statistic in terms of generalized pattern functions, which is de-
m
noted stat. Given a permutation π, let des(π) denote the descent number of π
[
and maj(π) denote the major index of π. Babson and Steingr´ımsson conjectured

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that (des,stat) and (des,maj) are equidistributed on S . Foata and Zeilberger
v n
4 settled this conjecture using q-enumeration, generating functions and Maple pack-
4 ages ROTAandPERCY.Later, Bursteinprovidedabijectiveproofofarefinement
0
of this conjecture. In this paper, we give a new bijective proof of this conjecture.
8
0
.
1 Keywords: Euler-Mahonian, bijection, involution
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AMS Subject Classifications: 05A05, 05A15
1
:
v
i
X
1 Introduction
r
a
In this paper, we give a new bijective proof of a conjecture of Babson and Steingr´ımsson
[1] on Euler-Mahonian statistics.
Let S denote the set of all the permutations of [n] = {1,2,··· ,n}. Given a permu-
n
tation π = π π ···π ∈ S , a descent of π is a position i ∈ [n−1] such that π > π ,
1 2 n n i i+1
where π and π are called a descent top and a descent bottom, respectively. An ascent
i i+1
of π is a position i ∈ [n− 1] such that π < π , where π is called an ascent bottom
i i+1 i
and π is called an ascent top. The descent set and the ascent set of π are given by
i+1
Des(π) = {i: π > π },
i i+1
Asc(π) = {i: π < π }.
i i+1
1
The set of the inversions of π is
Inv(π) = {(i,j): 1 ≤ i < j ≤ n,π > π }.
i j
Let des(π), asc(π) and inv(π) be the descent number, the ascent number and the in-
version number of π, which are defined by des(π) = |Des(π)|, asc(π) = |Asc(π)| and
inv(π) = |Inv(π)|, respectively. The major index of π, denoted maj(π), is given by
maj(π) = i.
i∈Des(π)
X
Suppose that st is a statistic on the object Obj and st is a statistic on the object
1 1 2
Obj . If
2
qst1(σ) = qst2(σ),
σ∈XObj1 σ∈XObj2
we say that the statistic st over Obj is equidistributed with the statistic st over Obj .
1 1 2 2
A statistic on S is said to be Eulerian if it is equidistributed with the statistic des
n
on S . While a statistic on S is said to be Mahonian if it is equidistributed with the
n n
statistic inv on S . It is well-known that
n
qinv(π) = qmaj(π) = [n] !,
q
πX∈Sn πX∈Sn
where [n] = 1+q+···+qn−1 and [n] ! = [n] [n−1] ···[1] . Thus, the major index maj
q q q q q
is a Mahonian statistic. A pair of statistics on S is said to be Euler-Mahonian if it is
n
equidistributed with the joint distribution of the descent number and the major index.
In[1], BabsonandSteingr´ımsson introducedgeneralized permutationpatterns, where
they allow the requirement that two adjacent letters in a pattern must be adjacent in
the permutation. Let A be the alphabet {a,b,c,...} with the usual ordering. We write
patterns as words in 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 correspond-
ing letters in the permutation must be adjacent. Given a generalized pattern τ and a
permutation π, we say a subsequence of π is an occurrence (or instance) of τ in π if it
is order-isomorphic to τ and satisfies the above dash conditions. Let (τ)π denote the
number of occurrences of τ in π. Here, we see (τ) as a generalized pattern function. For
example, an occurrence of the generalized pattern b-ca in a permutation π = π π ···π
1 2 n
is a subsequence π π π such that i < j and π < π < π . For π = 4753162, we have
i j j+1 j+1 i j
(b-ca)π = 4.
Further, Babson and Steingr´ımsson [1] showed that almost all of the Mahonian per-
mutation statistics in the literature can be written as linear combinations of generalized
patterns. We list some of them below.
maj = (a−cb)+(b−ca)+(c−ba)+(ba),
2
stat = (ac−b)+(ba−c)+(cb−a)+(ba).
They conjectured that the statistic (des,stat) is Euler-Mahonian.
Conjecture 1.1 Thedistribution of thebistatistic (des,stat) is equal to thatof (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 π = π π ···π , let F(π) = π be the first letter of π and
1 2 n 1
adj(π) = |{i: 1 ≤ i ≤ n and π −π = 1}|,
i i+1
where π = 0. Burstein [2] provided a bijective proof of the following refinement of
n+1
Conjecture 1.1 as follows.
Theorem 1.2 Statistics (adj,des,F,maj,stat) and (adj,des,F,stat,maj) are equidis-
tributed over S for all n.
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 ϕ on S that maps the inversion number
n
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 ϕ, 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 σ = σ σ ···σ in S . To obtain a permutation π ∈ S ,
1 2 n−1 n−1 n
we can insert n in n spaces, namely, immediately before σ or immediately after σ for
1 i
1 ≤ i ≤ 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 σ is given by numbering the spaces from right to left with
0,1,...,n−1. The maj-labeling of σ is obtained by labeling the space after σ with 0,
n−1
labeling the descents from right to left with 1,2,...,des(σ) and labeling the remaining
spaces from left to right with des(σ)+1,...,n. As an example, let σ = 13287546, the
inv-labeling of σ is
1 3 2 8 7 5 4 6 ,
8 7 6 5 4 3 2 1 0
3
while the maj-labeling of σ is given by
1 3 2 8 7 5 4 6 .
5 6 4 7 3 2 1 8 0
For n ≥ 2, we define the map
φ : {0,1,...,n−1}×S → S
inv,n n−1 n
by setting φ (i,σ) to be the permutation obtained by inserting n in the space labeled
inv,n
i in the inv-labeling of σ. By changing the inv-labeling to maj-labeling, we obtain the
map φ . As an example,
maj,n
φ (3,13287546) = 132879546 and φ (3,13287546) = 132897546.
inv,9 maj,9
For maps φ and φ , we have the following two lemmas.
inv,n maj,n
Lemma 2.1 For σ ∈ S and i ∈ {0,1,...,n−1}, we have
n−1
inv(φ (i,σ)) = inv(σ)+i.
inv,n
Lemma 2.2 For σ ∈ S and i ∈ {0,1,...,n−1}, we have
n−1
maj(φ (i,σ)) = maj(σ)+i.
maj,n
Lemma 2.1 is easy to verified. For a detailed proof of Lemma 2.2, see [5].
Given a permutation π ∈ S , let π(i) be the restriction of π to the letters 1,2,...,i
n
for 1 ≤ i ≤ n . For 2 ≤ i ≤ n, let
(c ,π(i−1)) = φ−1 (π(i)),
i inv,i
(m ,π(i−1)) = φ−1 (π(i)).
i maj,i
By setting c = 0 and m = 0, we obtain two sequences c c ···c and m m ···m in
1 1 1 2 n 1 2 n
E , where
n
E = {w = w ···w |w ∈ [0,i−1],1 ≤ i ≤ n}.
n 1 n i
Define γ(π) = c c ···c and µ(π) = m m ···m . It is not hard to check that both γ
1 2 n 1 2 n
and µ are bijections. The sequence c c ···c is called the inversion table of π, while
1 2 n
m m ···m is called the major index table of π. Moreover, we have n c = inv(π)
1 2 n i=1 i
and n m = maj(π). As an example, let π = 13287546, then π(8) = π and µ(π) =
i=1 i P
00204056 as computed in Table 2.1.
P
Now, we can define the bijection ϕ that maps the inversion number to the major
index by letting ϕ = µ−1γ. Clearly, ϕ is a bijection which proves that statistics inv and
maj are equidistributed over S .
n
4
i π(i−1) m
i
8 1 3 2 7 5 4 6 6
4 5 3 6 2 1 7 0
7 1 3 2 5 4 6 5
3 4 2 5 1 6 0
6 1 3 2 5 4 0
3 4 2 5 1 0
5 1 3 2 4 4
2 3 1 4 0
4 1 3 2 0
2 3 1 0
3 1 2 2
1 2 0
2 1 0
1 0
Table 2.1: The computation of the major index table of 13287546.
In the following of this section, we will construct a bijection ρ to prove Conjecture
1.1, which is, to some extent, an analogue of the above bijection ϕ.
First, we define a stat-labeling of σ ∈ S . Label the descents of σ and the space
n
after σ by 0,1,...,des(σ) from left to right. Label the space before σ by des(σ)+1.
n 0
The ascents of σ are labeled from right to left by des(σ)+2,...,n. As an example, for
σ = 13287546, we have the stat-labeling of σ as follows
1 3 2 8 7 5 4 6 .
5 8 0 7 1 2 3 6 4
Based on the stat-labeling, we define the map φ for n ≥ 2. For a permutation
stat,n
σ = σ σ ...σ and 0 ≤ i ≤ n − 1, define φ (i,σ) to be the permutation ob-
1 2 n−1 stat,n
tained by inserting n in the position labeled i in the stat-labeling of σ. For instance,
φ (7,13287546) = 132987546.
stat,9
It should be noted that unlike the properties of φ and φ stated in Lemma
inv,n maj,n
2.1 and Lemma 2.2, we deduce that φ (i,σ)) 6= stat(σ) +i for i = des(σ)+ 1. For
stat,n
the map φ , we have the following property.
stat,n
Lemma 2.3 For σ ∈ S and i ∈ {0,1,...,des(σ),des(σ)+2,...,n−1}, we have
n−1
stat(φ (i,σ)) = stat(σ)+i.
stat,n
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 σ. Assume
that σ = σ σ ···σ , there are three cases for us to consider.
1 2 n−1
• Case 1 : n is inserted into the space after σ .
n−1
Write π = σn. 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 π with the ba patterns of σ. It
follows that stat(π)−stat(σ) = (ba)σ = des(σ). Notice that the label of the space
after σ is des(σ). Hence, in this case the lemma holds.
n−1
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• Case 2 : n is inserted into a descent.
Suppose that τ is the permutation obtained by inserting n to the position i and
σ > σ . Moreover, let this descent be the k-th descent from left to right. We
i i+1
claim that stat(τ)−stat(σ) = k −1.
The insertion of n forms some new ac-b patterns, and the number of these new
patterns is |{j, j > i and σ > σ }|. Moreover, n forms k − 1 new ba-c patterns
j i
with the former k − 1 descents, while it destroys the ba-c patterns of σ by the
number |{j, j > i and σ > σ }|. It is easy to verify the functions (cb-a) and (ba)
j i
do not change. Hence, we conclude that stat(τ) −stat(σ) = 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 : n is inserted into an ascent.
Supposethatpisthepermutationobtainedbyinserting nintothepositioni, where
σ < σ . Moreover, we assume that there are k descents to the left of position i.
i i+1
We claim that stat(p)−stat(σ) = k +n−i+1.
Now we proceed to prove this claim. The insertion of n to σ brings new ac-b
patterns, the number of which is |{j, j ≥ i + 1 and σ ≥ σ }|. Moreover,
j i+1
the insertion of n brings k new ba-c patterns with the former k descents. Also,
nσ σ , where l > i + 1 and σ < σ , forms a new cb-a pattern of p. Notice
i+1 l l i+1
that (ba)p −(ba)σ = 1. By combining all above, we see that stat(τ) −stat(σ) =
k+n−i+1. The claim is verified. By the stat-labeling of σ, 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 π ∈ S ,
n
for 2 ≤ i ≤ n, let
(s ,π(i−1)) = φ−1 (π(i)).
i stat,i
Set s = 0 and ν(π) = s s ···s . It is easily checked that ν is a bijection from S to
1 1 2 n n
E . As an example, ν(52718346) = 01112216, which is computed in Table 2.2.
n
i π(i−1) s
i
8 5 2 7 1 3 4 6 6
3 0 7 1 6 5 4 2
7 5 2 1 3 4 6 1
3 0 1 6 5 4 2
6 5 2 1 3 4 2
3 0 1 5 4 2
5 2 1 3 4 2
2 0 4 3 1
4 2 1 3 1
2 0 3 1
3 2 1 1
2 0 1
2 1 1
1 0
Table 2.2: The computation of the stat table of 52718346.
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Now we can give the definition of the map ρ which proves Conjecture 1.1. Given
π ∈ S , let σ = ρ(π), where σ can be constructed as follows. Assume that F(π) = k
n
and π(k) = kp ···p . Then, let σ(k) = k(k − p )(k − p )···(k − p ). Assume that
2 k k k−1 2
s = s s ···s = ν(π). For k +1 ≤ i ≤ n, let σ(i) = φ (s ,σ(i−1)). Clearly, σ = σ(n)
1 2 n maj,i i
can be constructed by the procedure above. As an example, we compute ρ(π), where
π = 52718346. It is straightforward to see that k = 5 and π(5) = 52134. By the
i s σ(i)
i
5 2 5 1 2 4 3
3 2 4 5 1 0
6 2 5 6 1 2 4 3
3 4 2 5 6 1 0
7 1 5 6 1 2 4 7 3
3 4 2 5 6 7 1 0
8 6 56128473
Table 2.3: The computation of ρ(52718346).
computation in Table 2.3, we see that ρ(52718346) = 56128473. Notice that in this
example, the descent number and the first letter of both of the preimage and image of ρ
are the same. In fact, these properties always holds.
Lemma 2.4 For π ∈ S , we have F(π) = F(ρ(π)) and des(π) = des(ρ(π)).
n
Proof. Suppose that σ = ρ(π) and k = F(π). We proceed to show that F(σ(i)) =
F(π(i)) = k and des(σ(i)) = des(π(i)) for k ≤ i ≤ n by induction. Let π(k) = kp ···p ,
2 k
then we have σ(k) = k(k−p )(k−p )···(k−p ). It is routine to check that F(σ(k)) =
k k−1 2
F(π(k)) = k and des(σ(k)) = des(π(k)), hence, we omit the details here.
Now assume that des(σ(l)) = des(π(l)) = d and F(σ(l)) = F(π(l)) = k for k ≤ l ≤
n−1. We proceed to show that des(σ(l+1)) = des(π(l+1)) and F(σ(l+1)) = F(π(l+1)) = k.
Let s = s s ···s = ν(π). By the constructions of ν and ρ, we may see that
1 2 n
π(l+1) = φ (s ,π(l)) and σ(l+1) = φ (s ,σ(l)). Notice that both in the maj-
stat,l+1 l+1 maj,l+1 l+1
labeling of σ(l) and the stat-labeling of π(l), the descents and the space after the last
letter are labeled by {0,1,...,d}, the space before the first element is labeled by d+1,
and the ascents are labeled by {d+2,...,l}. Since F(π(l)) = k = F(π), it is easy to see
that s 6= d+1. It follows that F(σ(l+1)) = F(π(l+1)) = k.
l+1
Ifs < d+1, thenl+1isinsertedintothedescents orthespaceafterthelastelement
l+1
of π(l) and σ(l). Hence, we deduce that des(σ(l+1)) = des(π(l+1)) = d. If s > d+1, then
l+1
l + 1 is inserted into the ascents of π(l) and σ(l). Hence, we deduce that des(σ(l+1)) =
des(π(l+1)) = d+1. Combining the two cases above, we have des(σ(l+1)) = des(π(l+1)).
Notice that σ(n) = σ and π(n) = π, we complete the proof.
Base on the construction of ρ and Lemma 2.4, we have the following theorem.
Theorem 2.5 The map ρ is an involution on S .
n
7
Proof. Given a permutation π ∈ S , it suffices for us to show that ρ2(π) = π. That is,
n
writing σ = ρ(π), we need to show that ρ(σ) = π.
Let F(π) = k, then by Lemma 2.4, we know that k = F(π) = F(σ). Write π(k) =
kp ···p , then we have σ(k) = k(k−p )···(k−p ). Assume that ν(π) = s = s s ···s
2 k k 2 1 2 n
and µ(σ) = m m ···m . By the construction of ρ, we have σ(i) = φ (s ,σ(i−1)) for
1 2 n maj,i i
k +1 ≤ i ≤ n. It follows that m = s for k +1 ≤ i ≤ n.
i i
Suppose that α = ρ(σ), in the following, we proceed to show that α(i) = π(i) for
k ≤ i ≤ n by induction. By the definition of ρ, we have α(k) = kp ···p = π(k). Assume
2 k
that α(i) = π(i) holds for k ≤ i ≤ n−1, we aim to show that α(i+1) = π(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 ∈ S , assume the maj-labeling of p is f f ···f , while the
n 0 1 n
stat-labeling of p is h h ···h . Then it is easily checked that
0 1 n
des(p), if i is a descent of p or i = n,
f +h = n+des(p)+2, if i is an ascent of p, (2.1)
i i 

 2des(p)+2, if i = 0.

Let ν(σ) = l l ···l and d = des(α(i)). Then, we have
1 2 n
φ (d−l ,α(i)), if 0 ≤ l ≤ d,
α(i+1) = φ (l ,α(i)) = stat,i+1 i+1 i+1
maj,i+1 i+1 φ (n+d+2−l ,α(i)), if d+2 ≤ l ≤ n.
( stat,i+1 i+1 i+1
By the proof of Lemma 2.4, we see that des(σ(i)) = des(α(i)) = d. Recall that ν(σ) =
l l ···l and µ(σ) = m m ···m . Hence, it follows from (2.4) that
1 2 n 1 2 n
α(i+1) = φ (m ,α(i))
stat,i+1 i+1
= φ (s ,π(i))
stat,i+1 i+1
= π(i+1).
Notice that α = α(n) and π = π(n). Hence, we have π = ρ(σ), namely, ρ2(π) = π. This
completes the proof.
Indeed, the involution ρ also preserves some addtional statistics, which is stated in
the following proposition.
Proposition 2.6 For any π = π π ···π ∈ S , we have maj(ρ(π)) = stat(π) and
1 2 n n
stat(ρ(π)) = maj(π).
Proof. Write σ = ρ(π). Let k = F(π) and π(k) = kp ···p . Then we have σ(k) =
2 k
k(k − p )···(k − p ). In the following, we will show that maj(σ(i)) = stat(π(i)) for
k 2
k ≤ i ≤ n by induction.
8
First, we show that maj(σ(k)) = stat(π(k)). By the proof of Lemma 2.4, we have
des(π(k)) = des(σ(k)). Hence (ba)π(k) = (ba)σ(k). It suffices for us to show that
(ac−b)π(k)+(ba−c)π(k)+(cb−a)π(k) = (a−cb)σ(k)+(b−ca)σ(k)+(c−ba)σ(k) (2.2)
Given a pattern p = p p ···p , by putting a line under p (resp. p ), we mean that
1 2 k 1 k
an instance of p must begin(resp. end) with the leftmost(resp. rightmost) letter of the
permutation. By putting a dot under p , we mean that an instance of p must not begin
1
with the leftmost letter. For instance, (cb-a) is a function which maps a permutation,
¯
say τ = τ τ ···τ , to |{i,τ τ τ forms a 321 pattern of τ}|.
1 2 n 1 2 i
By the definition of ρ, we know that
(ac−b)π(k) +(ba−c)π(k) +(cb−a)π(k)
=(ac−b)π(k) +(ba−c)π(k) +(cb−a)π(k) +(c.b−a)π(k)
¯
=(b−ac)σ(k) +(a−cb)σ(k) +(c−b−a)σ(k) +(c. −ba)σ(k)
¯ ¯
Hence, to prove (2.2), we need to prove that for any permutation p ∈ S with p = k,
k 1
(c−ba)p+(b−ca)p = (c−b−a)p+(b−ac)p. (2.3)
¯ ¯ ¯
We define sets A(p),C(p) and multisets B(p),D(p) as follows.
A(p) = {p : p p p forms a 321 pattern of p},
i 1 i i+1
B(p) = {p : p p p forms a 231 pattern of p},
i i j j+1
C(p) = {p : p p p forms a 321 pattern of p},
i 1 i m
D(p) = {p : p p p forms a 213 pattern of p}.
i i j j+1
To prove (2.3), it is enough to show that A∪B = C ∪D, where the union operator is a
multiset union. First, we show that C ∪D ⊆ A∪B.
Let p = a, then we know that C(p) = {a+ 1,a+ 2,...,k − 1}. If p ∈ C(p) is a
k i
descent top, it is easy to see that p ∈ A(p). If p ∈ C(p) is an ascent bottom, we claim
i i
that p ∈ B(p). This claim will be proved together with case 4 in the following.
i
Given an element p in D(p), if the multiplicity of p is x, there exists a set
i i
{j ,j +1,j ,j +1,...,j ,j +1},
1 1 2 2 x x
which is ordered by increasing order, satisfying that
p p p , p p p , ..., p p p
i j1 j1+1 i j2 j2+1 i jx jx+1
are instances of 213 pattern. We claim that there exists j + 1 ≤ r < j such that
1 1 2
p p p forms a 231 pattern.
i r1 r1+1
Choose the smallest g such that j +1 ≤ g < j and g is a descent. If p < p ,
1 1 1 2 1 g1+1 i
then p p p formsa 231 pattern, the claim is verified. Otherwise, we seek the smallest
i g1 g1+1
9
g + 1 ≤ g < j such that g is a descent. If p < p , the claim is verified. If not,
1 2 2 2 g2+1 i
we repeat the above process. Since p > p , the process must be terminated. Hence,
i j2
the claim is verified. By a similar means, we deduce that there exists j +1 ≤ r < j
l l l+1
where 1 ≤ l ≤ x−1 such that p p p forms a 231 pattern. The claim is verified.
i rl rl+1
To analyze the element p in D(p), we consider four cases.
i
• p is a descent top and 1 < p ≤ a−1.
i i
Supposethatthemultiplicityofp ofthistypeinD(p)isx. Bytheabovestatement,
i
weseethatp p p formsa231pattern, wherej +1 ≤ r < j and1 ≤ l ≤ x−1.
i rl rl+1 l l l+1
Thus, we deduce that there are x−1 p s in B(p). Notice that there is one p left
i i
in B(p). Clearly, we can set this p to be the element of A(p) consisting of p , p
i 1 i
and p .
i+1
• p is an ascent bottom and 1 ≤ p ≤ a−1.
i i
Suppose that the multiplicity of p of this type in D(p) is x. Similarly, we know
i
that p p p forms a 231 pattern, where j + 1 ≤ r < j and 1 ≤ l ≤ x − 1.
i rl rl+1 l l l+1
Since p < p , we have j > i+1. Based on this, it can be easily seen that there
i i+1 1
exists r such that i+1 ≤ r < j and p p p forms a 231 pattern. Hence, we
0 0 1 i r0 r0+1
deduce that in this case there are x p s in B(p).
i
• p is a descent top and a+1 ≤ p ≤ k −1.
i i
Suppose that the multiplicity of p of this type in D(p) is x. Similarly, we deduce
i
that p p p forms a 231 pattern, where j + 1 ≤ r < j and 1 ≤ l ≤ x − 1.
i rl rl+1 l l l+1
What’s more, it follows from p > a that there exists j ≤ r < m such that p p a
i x x i rx
forms a 231 pattern. Hence, we deduce that p in B(p) and its multiplicity is x.
i
• p is an ascent bottom and a+1 ≤ p ≤ k −1.
i i
Suppose that the multiplicity of p of this case in D(p) is x. Notice that p is also
i i
an element of C(p) with multiplicity equals 1. Hence, in this case, we have to prove
that there are x+1 p s in B(p).
i
Similarlywiththeabovecases, wededucethatp p p formsa231pattern, where
i rl rl+1
j +1 ≤ r < j and 1 ≤ l ≤ x−1. Since p > a, there exists j ≤ r < m such
l l l+1 i x x
that p p a forms a 231 pattern. By p < p , we have j > i+1. Based on this, it
i rx i i+1 1
can be easy seen that there exists r such that i+1 ≤ r < j and p p p forms
0 0 1 i r0 r0+1
a 231 pattern. Hence, we deduce that p in B(p) and its multiplicity is x+1.
i
Combining all above, we deduce that C ∪D ⊆ A∪B. By a similar analysis, we can
prove that A∪B ⊆ C ∪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∪B = C ∪D. This proves that maj(σ(k)) = stat(π(k)).
Now assume that maj(σ(i)) = stat(π(i)) for k ≤ i ≤ n−1, we proceed to show that
maj(σ(i+1)) = stat(π(i+1)). Write ν(π) = s s ···s . Then, by Lemma 2.2, Lemma 2.3
1 2 n
10