This paper advances the enumeration of permutation classes avoiding certain patterns by deriving generating functions for most triples containing 1324, using combinatorial and analytic methods like generating trees and kernel method.
Contribution
It provides explicit generating functions for nearly all triples containing 1324, except one conjectured intractable case, enhancing understanding of Wilf classes.
Findings
01
Generated functions for most triples containing 1324
02
Used combinatorial and analytic methods such as generating trees and kernel method
03
Identified an intractable triple conjecture
Abstract
Recently, it has been determined that there are 242 Wilf classes of triples of 4-letter permutation patterns by showing that there are 32 non-singleton Wilf classes. Moreover, the generating function for each triple lying in a non-singleton Wilf class has been explicitly determined. In this paper, toward the goal of enumerating avoiders for the singleton Wilf classes, we obtain the generating function for all but one of the triples containing 1324. (The exceptional triple is conjectured to be intractable.) Our methods are both combinatorial and analytic, including generating trees, recurrence relations, and decompositions by left-right maxima. Sometimes this leads to an algebraic equation for the generating function, sometimes to a functional equation or a multi-index recurrence amenable to the kernel method.
Tables2
Table 1. Table 1. Small Wilf classes of three 4-letter patterns not counted by INSENC that include the pattern 1324 1324 1324
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Full text
Enumeration of small Wilf classes avoiding 1324 and two other 4-letter patterns
David Callan
Department of Statistics, University of Wisconsin, Madison, WI 53706
Recently, it has been determined that there are 242 Wilf
classes of triples of 4-letter permutation patterns by showing
that there are 32 non-singleton Wilf classes. Moreover,
the generating function for each triple lying in a non-singleton Wilf class has been explicitly determined.
In this paper, toward the goal of enumerating avoiders for the singleton Wilf classes, we obtain the generating function for all but one of the triples containing
1324. (The exceptional triple is conjectured to be intractable.) Our methods are both combinatorial and analytic, including generating trees,
recurrence relations, and decompositions by left-right maxima.
Sometimes this leads to an algebraic equation for the generating
function, sometimes to a functional equation or a multi-index
recurrence amenable to the kernel method.
In recent decades pattern avoidance has received a lot of attention. It has a
prehistory in the work of MacMahon [20] and Knuth [16], but the
current interest was sparked by a paper of Simion and Schmidt [24]. They thoroughly analyzed 3-letter patterns in permutations, including a bijection
between 123- and 132-avoiding permutations, thereby explaining the first
(nontrivial) instance of what is, in modern terminology, a Wilf class. Since then
the problem has been addressed on several other discrete structures, such as
compositions, k-ary words, and set partitions; see, e.g., the texts [14, 21]
and references contained therein.
Permutations avoiding a single 4-letter pattern have been well studied (see, e.g.,
[27, 28, 30, 1]), and the latter form 7 symmetry classes and 3 Wilf
classes. As for pairs of 4-letter patterns, there are 56 symmetry classes, for all
but 5 of which the avoiders have been enumerated [1]. Le [19]
established that these 56 symmetry classes form 38 distinct Wilf classes.
The (324)=2024 triples of 4-letter patterns split into 317 symmetry classes. It is known [11, 12] that the 317 symmetry classes split into 242 Wilf classes, 32 of which are large (a Wilf class is called large if it contains at least two symmetry classes, and small if it consists of a singleton symmetry class) and the large Wilf classes are all explicitly enumerated in [11, 12], where it is shown that each has an algebraic
generating function.
Our goal here is to enumerate (with one exception, see [2] and [26]) all the small Wilf classes that contain the pattern 1324.
Running the INSENC algorithm (regular insertion encoding, see [3, 29])
over all the 210 small Wilf classes
determines the generating function for 126 of them, as presented in the Appendix.
The remaining small classes that contain 1324 are listed in Table LABEL:tabgf1324
along with their generating functions, where the numbering follows that of Table 2
in the appendix to[5], based on lex order of counting sequences.
Section 2 contains some preliminary remarks, and Section 3
gives the proofs for all the results in Table LABEL:tabgf1324 not proved elsewhere.
2. Preliminaries
We say a permutation is standard if its support set is an initial segment of the positive integers, and for a permutation π whose support is any set of positive integers, St(π) denotes the standard permutation obtained by replacing the smallest entry of π by 1, the next smallest by 2, and so on.
Typically, for a a given triple T, we consider cases and analyze the structure of a T-avoider in each case to the point where we say that T-avoiders have such and such a form in that case. It is always to be understood that we are also asserting, without explicit mention, that a permutation of the specified form is a T-avoider, and this enables us to determine the various “contributions” to the generating function FT(x) for T-avoiders, yielding an equation for FT(x). The equation may be an explicit expression for FT(x) or an algebraic or functional equation. For all but one symmetry class, the generating function turns out to be algebraic of degree ≤4. For the exceptional class (Case 237, where {1432,1324,1243} and {4123,4231,4312} are representative triples), the generating function is conjectured not to satisfy any ADE (algebraic differential equation), see [2] and [25, Seq. A257562].
A permutation π expressed as π=i1π(1)i2π(2)⋯imπ(m) where i1<i2<⋯<im and
ij>max(π(j)) for 1≤j≤m is said to have mleft-right maxima (at i1,i2,…,im). Given
nonempty sets of numbers S and T, we will write S<T to mean
max(S)<min(T) (with the inequality vacuously holding if S or
T is empty). In this context, we will often denote singleton
sets simply by the element in question. Also, for a number k,
S−k means the set {s−k:s∈S}.
Throughout, C(x)=2x1−1−4x denotes the generating function for
the Catalan numbers Cn:=n+11(n2n)=(n2n)−(n−12n).
As is well known [16, 1], C(x) is the generating function for
(∣Sn(π)∣)n≥0 where π is any one of the six 3-letter
patterns. The identity C(x)=1−xC(x)1 or, equivalently,
xC(x)2=C(x)−1 is used to simplify some of our results.
Also throughout, L(x)=1−2x1−x denotes the generating
function for {213,231}-avoiders (resp. {213,123}-avoiders,
resp. {132,123}-avoiders), see [24], and K(x),K′(x) etc. are variously
used for other known generating functions.
3. Proofs
3.1. Case 49: {1324,2341,4123}
For this case, we need the following lemmas.
Lemma 1**.**
Let T={1324,2341,4123}. The generating function for the number of permutation (n−1)π′nπ′′∈Sn(T) is given by
[TABLE]
Proof.
Let us write an equation for H(x). Let π=(n−1)π′nπ′′∈Sn(T). If n=2 then we have a contribution of x2. So let us assume that n>2, so there are two cases, either n−2 belongs to π′ or to π′′.
•
n−2 belongs to π′: If π′′=∅ then we have a
contribution of x3(F{123,132}−1)=1−2xx3, see
[24]. So, we can assume that π′′=∅. If π′
has a letter between n−1 and n−2, then π can be written as
[TABLE]
which counted by
(1−x)3x5. Otherwise, π′ has no letter between
n−1 and n−2, which gives a contribution of
xH(x)−1−2xx3(1−x).
•
n−2 belongs to π′′: In this case, we have a
contribution of x(C(x)−1−xC(x))=x3C(x)3, where C(x) counts
the {123}-avoiders.
Hence, by adding all the contributions, we have
[TABLE]
which completes the proof.
∎
Lemma 2**.**
Let T={1324,2341,4123}. The generating function for the number of T-avoiders with exactly 2 left-right maxima is given by
Let us write an equation for G2(x). Let π=iπ′nπ′′∈Sn(T) be a permutation with exactly 2 left-right maxima. We
consider the following cases:
•
i=n−1: We have a contribution of H(x) as defined in
Lemma 1. So from now, we assume that π′′ contains
the letter n−1.
•
π′′=(n−1)π′′′: We have a contribution of xG2(x). So
we can assume that there is at least one letter between n and
n−1. Since π′′ avoids 1324 and 4123, we see that there
are exactly at most one letter between n and n−1 that greater
than i.
–
if there is a letter in nπ′′ between n and n−1 that
it is greater than i, then π can be written as
π=iπ′nkβ(n−1)⋯(k+1)α such that β<i and
α>i. By considering either β is empty or not, we
obtain the contributions 1−xx4L(x)2 and
(1−x)4x5, respectively. Recall L(x)=1−2x1−x
is the generating function for {213,231}-avoiders
(also for {213,123}-avoiders).
–
Thus, we can assume that π=π′nβ(n−1)α such
that α contains the subsequence (n−1)(n−2)⋯(i+1),
β<i and β is decreasing (since π avoids 4123,
and β is not empty). Suppose that β=ee′β′ then
e>e′>β′ and π′>e′. If π′ has a letter between e
and e′, then easy to see that the contribution is given by
(1−x)4x4+d, where d is the number of the letters
in π′′ that are greater than i. Otherwise, the contribution
is given by x3+dC(x)3+d, where d is the number of the
letters in π′′ that are greater than i. Therefore, we have a
contribution of
[TABLE]
which equals
[TABLE]
Hence, the various contributions give
[TABLE]
which completes the proof.
∎
Theorem 3**.**
Let T={1324,2341,4123}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and
G1(x)=xF{123}(x)=xC(x), see [16]. By Lemmas
1 and 2, we have that
[TABLE]
where
[TABLE]
Now, let us write an equation for G3(x). Let
π=i1π′i2π′′nπ′′′∈Sn(T) with exactly 3 left-right
maxima (i1,i2,n). Since π avoids 1324 and 2341, then
π′>π′′ and π′′′=βα such that
β>i2>α>i1. By considering the cases α,β
are empty or not, we obtain the contributions x3L(x)2,
x3(L(x)−1)L(x)/(1−x), x3(L(x)−1)L(x)/(1−x), and
x5/(1−x)4. Hence,
[TABLE]
.
By very similar techniques as in case G3(x), we obtain that
[TABLE]
Now, let us write an equation for Gm(x) and m≥5. Let
π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with
exactly m left-right maxima. Since π avoids T, we see that
π(s)=∅ for all s=3,4,…,m−1,
π(1)>π(2) and π(m)=βα such that
β>im−1>α>im−2. Thus, Gm(x)=xGm−1(x), for all m≥5.
Therefore,
[TABLE]
By substituting the expressions for G2(x),G3(x),G4(x) and simplifying, we obtain the stated generating function.
∎
3.2. Case 69: {1234,1324,3412}
A permutation π=π1π2⋯πn∈Sn determines n+1 positions, called sites, between its entries. The sites are denoted 1,2,…,n+1 left to right. In particular, site i is the space between πi−1 and πi, 2≤i≤n. Site i in π is said to be active (with respect to T) if, by inserting n+1 into π in site i, we get a permutation in Sn+1(T), otherwise inactive.
Say j is an ascent index for a permutation π=π1π2⋯πn of [n] if πj<πj+1, and then πj is an ascent bottom.
To construct the generating forest for T-avoiders, we first specify the labels. For n≥2, define the label of π∈Sn(T) to be (k,s), where k is the number of active sites in π and s is the number of active sites greater than the largest ascent index (LAI for short) with LAI taken to be 0 if there are no ascents, that is, if π is decreasing.
For instance, the active sites for π=12 are {1,2,3} and LAI =1, so the label for 12 is (3,2). Also, 12
has three children 312, 132 and 123 with active sites {1,3,4},{1,2,3} and {1,2,3}, respectively,
and LAI =2,1,2, respectively; hence labels (3,2),(3,2), and (3,1). Similarly, all 3 sites for 21 are active and and LAI =0, so its label is (3,3), and it has three children 321, 231 and 213 with active sites {1,2,3,4} in all three cases, and LAI =0,1,2, respectively; hence labels (4,4),(4,3), and (4,2).
An avoider π∈Sn(T) has a label (k,s) with k=s only if π is decreasing, in which case k=s=n+1.
Otherwise, 0≤s<k.
Proposition 4**.**
The roots for the generating forest T of Sn(T) are 12 and 21 with labels (3,2) and (3,3) respectively, and the succession rules for the labels of children,
in order of increasing insertion site, are given by
[TABLE]
As an example (bullets denote active sites, an underscore denotes last ascent bottom), the label of
\pi=\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,3\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,\underline{2}\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,6\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,5\,4\,1\in S_{6}(T) is (k,s)=(4,2); its children are
\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,7\,3\,\underline{2}\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,6\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,5\,4\,1,
\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,3\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,7\,\underline{2}\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,6\,5\,4\,1,
\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,3\,\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,\underline{2}\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,7\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,6\,5\,4\,1,
\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,3\,\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,2\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,\underline{6}\,\textrm{\raisebox{0.86108pt}{\tiny{\bullet}}}\,7\,5\,4\,1, in that order, with labels (3,2),(3,1),(4,2),(4,1) respectively.
The proof of Proposition 4 is based on induction by a routine checking of cases, and is left to the reader. We note a few properties of the active sites for π∈Sn(T). Site 1 is always active. If site n+1 is active, then so is site n. The active sites always form either an interval of integers (necessarily an initial segment of the positive integers) or a pair of intervals of integers. In the latter case, at least one of the intervals is of length 1. For example, 532416∈S6(T) has active sites {1,4,5,6} and
5342761∈S7(T) has active sites {1,2,5}.
Enumeration: Let Ak,s=Ak,s(t) be the generating function for the number of vertices labeled (k,s) in level n in the generating forest T, where the roots 12 and 21 are at level 2. Define
Substitute v=0 into (2) and into the derivative of (2) respect to v.
Solve the resulting system for the variables dvdA(u,v)∣v=0 and dvdA(1,v)∣v=0 to get
By a routine computer check, the solution of this equation is given by
[TABLE]
where
[TABLE]
Since A(1,1)=∑n≥2∣Sn(T)∣tn and so FT(t)=1+t+A(1,1), we get the following result.
Theorem 5**.**
Let T={1234,1324,3412}. Then
[TABLE]
3.3. Case 72: {1243,1324,3412}
First, we look at G2(x). For π=iπ′nπ′′ with 2 left-right maxima and d≥0
letters in π′′, let Hd(x) and Jd(x) denote the
generating functions in the repective cases i=n−1 and i<n−1. Note d≥1 in case i<n−1.
Thus, with H(x):=∑d≥0Hd(x) and J(x):=∑d≥1Jd(x),
we have G2(x)=H(x)+J(x).
Lemma 6**.**
[TABLE]
Proof.
Clearly, H0(x)=x2K(x) where K(x)=F{132,3412}(x), so by [25, Seq. A001519] we have that H0(x)=1−3x+x2x2(1−2x). For d≥1, by considering the position of the letter n−2 in π′′ (either leftmost, rightmost, or in the middle), we obtain
[TABLE]
Note that the last equation also holds for d=0. Now sum over d≥0.
∎
By similar arguments, one can obtain the the following result for J(x).
Lemma 7**.**
We have
[TABLE]
∎
Theorem 8**.**
Let T={1243,1324,3412}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x). By Lemma 6 and Lemma 7, we have that
[TABLE]
Now, let us write an equation for Gm(x) with m≥3. Let π=i1π(1)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324,
π(s)<i1 for all s=1,2,…,m−1 and, since π avoids 1243, π(m)<i2.
If π(m−1)=∅ then we have a contribution of xGm−1(x).
Otherwise, since m≥3 and π avoids 3412, we see that π(m)<i1.
Moreover, since π avoids 3412 and 1324, we see that
π(1)>π(2)>⋯>π(m) and π(2)⋯π(m) is decreasing,
while π(m−1) is not empty, and π(1) avoids 132 and 3412.
Therefore, we have a contribution of xm+1K(x)/(1−x)m−1, where K(x) is given in the proof of Lemma 6. Hence,
[TABLE]
Summing over all m≥3 and using the expressions for G0(x) and G1(x), we obtain
[TABLE]
Solve for FT(x) using (4) to complete the proof.
∎
3.4. Case 75: {1243,1324,4231}
We make use of the following generating functions:
[TABLE]
denoted, respectively, L, A and B for short (all three follow from the main result in [23]).
Let Gm(x) be the generating function for T-avoiders π=i1π(1)⋯imπ(m) with m left-right maxima. Clearly, G0(x)=1 and G1(x)=xB. To get an explicit formula for Gm(x), we look at the cases m=2 and m≥3, and in the latter case we consider whether π(m) has a letter between i1 and i2 or not. We split the rather lengthy treatment into subsections.
3.4.1. Hm(x) and Hm′(x)
Define Hm(x) to be the generating function for permutations
[TABLE]
and define Hm′(x) to be the generating function for permutations
[TABLE]
Lemma 9**.**
For m≥1, Hm′(x)=xm+2Lm+1 and Hm(x) satisfies
[TABLE]
Proof.
First, we treat Hm′(x). Let π=(n−m)π(0)(n−m+1)π(1)⋯nπ(m)(n−m−1)∈Sn(T). In the case n=m+2, we have a contribution of xm+2. Otherwise, the letter n−m−2 belongs to π(j), where j=0,1,…,m. For each j=0,1,…,m−1, we have a contribution xj+1Hm−j′. For j=m, π(m) can be written as α(n−m−2)β where π(0)⋯π(m−1)α<β<n−m−2. The contribution for the case β=∅ is xHm′(x) and for the case β=∅ is xm+3(L−1). Hence,
[TABLE]
Setting m=1, we can solve for H1′(x) and then the result follows by induction on m.
Next, Hm(x). Let π=(n−m)π(0)(n−m+1)π(1)⋯nπ(m)∈Sn(T). In the case m=n−1, we have a contribution of xm+1. Otherwise, the letter n−m−1 belong to π(j) for some j∈{0,1,…,m}. If this j is <m, then π(0)=⋯=π(j−1)=∅,
and π(j) can be written as α(n−m−1)β where
α<βπ(j+1)⋯π(m). Then we have the contributions xj+1Hm−j, 1−xxm+3A and xm+2(L−1/(1−x)) for the cases α=∅, α is nonempty decreasing, and α contains a rise, respectively. If the letter n−m−1 belongs to π(m) then π(m) has the form α(n−m−1)β
and π(0)⋯π(m−1)α is both decreasing and <β. So we have the contributions Hm′(x), xm+2(1/(1−x)m+1−1)(A−1), and xm+2(B−1), for the cases β=∅, β=∅ and π(0)⋯π(m−1)α=∅, β=∅ and π(0)⋯π(m−1)α is not empty, respectively. Hence, Hm(x) satisfies the claimed relation,
∎
3.4.2. Dkm(x) and D′km(x)
Define
•
Dkm(x) to be the generating function for T-avoiders π=i1π(1)⋯imπ(m) such that i2=i1+k+1 and π(m) contains the subsequence (i1+k)(i1+k−1)⋯(i1+1), for all k≥1.
•
D′km(x) to be the generating function for T-avoiders π=i1π(1)⋯imπ(m) such that i2=i1+k+3 and π(m) contains the subsequence (i1+k+2)(i1+k+1)⋯(i1+3)(i1+1)(i2+2), for all k≥0.
Lemma 10**.**
For all k≥1,
[TABLE]
where
[TABLE]
Proof.
The proof is based on considering all possible positions of the next smallest element whose position we have not yet fixed. Let us write an equation for Dk2(x). Let π=iπ′nπ′′∈Sn(T) such that π′′ contains the subsequence (i+k)(i+k−1)⋯(i+1) and n=i+k+1. If i=1, then we have a contribution of xk+2. Otherwise, let i>1. If the letter i−1 belongs to π′ then it is the leftmost letter of π′, so we have a contribution of xDk2(x). If the letter i−1 belongs to π′′ and is on the left side of i+2, then there exists j∈[3,k−1] such that π=in(i+k)⋯(i+j)(i−1)⋯1(i+j−1)⋯(i+1), which gives a
contribution of 1−xxk+3, for all j=3,4,…,k−1. Let us denote the contribution of the case when the letter i−1 is between the letters i+2 and i+1 (resp. on right side of i+1) by Ek(x) (resp. Fk(x)). Then
[TABLE]
To write an equation for Ek(x),
let π=iπ′nπ(k)(i+k)⋯π(2)(i+2)π(1)(i+1)π(0)∈Sn(T) with n=i+k+1 and π(1)=α(i−1)β. Define Ek′(x) to be the contribution in case β=∅. If β=∅, then π=in(i+k)⋯(i+2)(i−1)β(i+1) which implies a contribution of xk+3(L−1). Thus
[TABLE]
By considering the possible positions of the letter i−2 in π=iπ′nπ(k)(i+k)⋯π(1)(i+1)∈Sn(T) with n=i+k+1 and π(1)=α(i−1), we obtain the equation
[TABLE]
To write an equation for Fk(x), let π=iπ′nπ(k)(i+k)⋯π(1)(i+1)π(0)∈Sn(T) with n=i+k+1 and π(0)=α(i−1)β.
Define Fk′(x) to be the contribution of the case β=∅.
When β=∅, by considering the possible positions of the letter i−2 in π, we obtain the contributions xk+3(B−1) when π′π(k)⋯π(1)α=∅, \frac{x^{k+4}}{1-x}\big{(}1+(k+1)\frac{x}{1-x}\big{)}(A-1) when π′ is
nonempty (and necessarily decreasing), and 1−x(k+1)xk+4(A−1) when π′= and π(k)⋯π(1)α=∅ (here either α=∅ and π(k)⋯π(1)=∅, or α=∅
and there exists j∈[1,k] such that π(j)=∅ and π(i)=∅ for all i=j). Thus,
[TABLE]
By considering the possible positions of the letter i−2 in π=iπ′nπ(k)(i+k)⋯π(1)(i+1)∈Sn(T) with n=i+k+1 and π(0)=α(i−1), we find
that
[TABLE]
which completes the proof.
∎
The next lemma, giving D′k2(x), can be established by similar arguments.
Lemma 11**.**
Let Pk(x) be the generating function for the number of permutations π=iπ′nπ(k)(i+k+2)⋯π(1)(i+3)π(0)∈Sn(T) such that n=i+k+3, π(0) contains the subsequence (i+1)(i+2). For all k≥0, D′k2(x)=1−x1Pk(x), where
[TABLE]
and the generating function Ek′′(x) for the number of permutations π=iπ′nπ(k)(i+k+2)⋯π(1)(i+3)π(0)∈Sn(T) such that n=i+k+3 and π(0)=α(i−1)β(i+1)(i+2) satisfies
Ek′′(x)=Ek′′′(x)+xk+5(L−1), where Ek′′′(x) is the generating function for the number of permutations π=iπ′nπ(k)(i+k+2)⋯π(1)(i+3)π(0)∈Sn(T) such that n=i+k+3 and π(0)=α(i−1)(i+1)(i+2) which satisfies satisfies
Ek′′′(x)=xk+5+xEk′′′(x)+kxk+6/(1−x)+xEk′′(x). ∎
Now, we are ready to write a formula for the generating function G2(x) for T-avoiders with 2 left-right maxima.
Proposition 12**.**
The generating function G2(x) is given by
[TABLE]
Proof.
Let π=iπ′nπ′′∈Sn(T) with 2 left-right maxima. If n=i+1, then the contribution is H1(x) by Lemma 9. If n>i, then the letters of π′′ that are greater than i form either a decreasing sequence (i+k)(i+k−1)⋯(i+1) or a decreasing sequence followed by an increasing sequence of at least two
terms (i+k)(i+k−1)⋯(i+1+s)(i+1)(i+2)⋯(i+s) with s≥2.
By Lemmas 10 and 11, we obtain the contributions Dk2(x) with k≥1 and D′k2(x) with k≥0, respectively. Thus,
[TABLE]
After working out explicit expressions for H1(x), ∑k≥1Dk2(x) and ∑k≥0D′k2(x) from Lemmas 9, 10 and 11, respectively, we complete the proof.
∎
Lemmas 10 and 11 suggest a method to study the generating functions Dkm(x),k≥1 and D′km(x),k≥0.
3.4.3. Formula for D1m(x)
Lemma 13**.**
Let m≥3. Then D1m(x)=xm+1Lm+(Sm+Sm′)/(1−x), where
[TABLE]
and
[TABLE]
Proof.
Let π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima such that π(m) has exactly one letter between i1 and i2. So, ij=i1+j for all j=2,3,…,m and π(m)=α(i1+1)β.
We denote the generating function for permutations π with β=∅ by E1m, and we denote the generating function for permutations π with β=∅ and i1−1∈π(j) by E1m,j, for j=2,3,…,m. Then, by considering the position of letter i−1 in π with β=∅, we obtain
[TABLE]
where
[TABLE]
with E1m,m=xE1m+xm+2(L−1). Hence, by induction on s, we have
[TABLE]
Thus, E1m=xm+1Lm.
Denote the generating function for permutations π with β=β′(i−1) by Sm, and the generating function for permutations π with β=β′(i−1)β′′ with β′′=∅ by Sm′. Clearly, D1m=E1m+(Sm+Sm′)/(1−x). By considering the position of the letter i−2 in the permutations counted by Sm, we obtain
[TABLE]
By considering the four possibilities where π(1)⋯π(m−1)α and β′ are empty or not, we complete the proof.
∎
3.4.4. A formula for D′0m
Lemma 14**.**
Let m≥3. Then D′0m(x)=1−xxm+2(L+x(L+L2+⋯+Lm−1)).
Proof.
As in the proof of Lemma 13, by considering the position of the letter i1−1 in permutations
π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima and π(m)=α(i1+1)β(i1+2), we obtain
[TABLE]
where D′0m,j is the generating function for the T-avoiders π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima and π(m)=α(i1+1)β(i1+2) such that the letter i1−1 belongs to π(j). Again, by considering the position of the letter i1−2, we have
Thus, by substituting the expression for D′0m,j into the equation for D′0m, we complete the proof.
∎
3.4.5. A formula for Dkm and D′km
In the next two lemmas we study the generating functions Dkm(x) with k≥2 and D′km with k≥1.
As before, by considering all possible positions of the next smallest element whose position we have not yet fixed, we obtain the following results.
Lemma 15**.**
Let m≥3 and k≥2. Define Skm,m−1 (resp. Skm,m, Ekm, E′km) to be the generating function for the T-avoiders π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima such that the letters between i1 and i2 in π(m) form a decreasing sequence j1j2⋯jk and the letter i1−1 lies between jk−1 and jk (resp. i1−1 lies on the right side
of jk, π(m) has no letter smaller than i1 on the right side of the letter j1, π(m) has at least one letter smaller than i1 on the right side of the letter j1). Then Dkm(x)=Ekm+E′km, where Ekm=(1−x)mxm+k,
[TABLE]
Lemma 16**.**
Let m≥3 and k≥1.
Define Ukm to be the generating function for the T-avoiders π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima such that the letters between i1 and i2 in π(m) form a subsequence (i2−1)⋯(i1+3)(i1−1)(i1+1)(i1+2) of k+1 terms. Then
[TABLE]
3.4.6. A formula for FT(x)
Now, we are ready to find an explicit formula for the generating function FT(x) by getting a formula for Gm(x).
Theorem 17**.**
Let T={1243,1324,4231}. Then
[TABLE]
Proof.
Fix m≥3. Let Gm′(x) be the generating function for the T-avoiders π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima and i2>i1+1. Let γ be the subsequence of letters of π(m) that are larger than i1 and smaller than i2. Since π avoids T, γ can be written as either
(i2−1)(i2−2)⋯(i1+1) or (i2−1)(i2−2)⋯(i1+s+1)(i1+1)(i1+2)⋯(i1+s) for s≥2. It follows that
[TABLE]
Thus,
[TABLE]
By Lemmas 9 and 13-16, we obtain after simplifying
[TABLE]
where p(x)=1−10x+49x2−149x3+296x4−403x5+408x6−322x7+187x8−74x9+18x10−2x11.
Since
[TABLE]
and we have determined each of the summands,
the result follows.
∎
3.5. Case 76: {3412,1324,2341}
Here we use the fact that
[TABLE]
denoted A for short (followed from the main result of see [23]).
Lemma 18**.**
The generating function G2(x) for T-avoiders with 2 left-right maxima is given by
[TABLE]
Proof.
Suppose π=iαnβ∈Sn(T) with 2 left-right maxima. If i=1, then π=1nβ
and where β avoids {213,2341,3412}, giving a contribution to G2(x) of x2A.
Now suppose i>1. If i−1 is the leftmost letter of α, we obtain a contribution of
xG2(x) by deleting i−1. If i−1∈α and is not the leftmost letter, then π
decomposes as π=iα′(i−1)α′′nβ′β′′,
where β′>i>i−1>β′′>α′>α′′ (to avoid 1324) and β′′ is decreasing
(to avoid 3412), as in Figure 1.
Consider four cases according as α′′,β′′ are empty or not:
α′′,β′′ both empty. Note that α′ is nonempty and avoids {132,2341,3412} and β′ avoids {213,2341,3412}. So the contribution is x3(A−1)A.
α′′=∅,β′′=∅. Here, β′ is decreasing (to avoid 2341),
while α′ is nonempty and avoids {132,2341,3412}. So the contribution is x4(A−1)/(1−x)2.
α′′=∅,β′′=∅. Here, α′ is decreasing (to avoid 2341)
and nonempty, α′′ is also decreasing (to avoid 3412) and nonempty, and β′ avoids {213,2341,3412}. So the contribution is x5A/(1−x)2.
α′′=∅,β′′=∅. Here, α′ and β′ are decreasing
(both to avoid 2341) and α′ is nonempty, α′′ is also decreasing (to avoid 3412) and nonempty, and β′′ is nonempty (and decreasing, as always). So the contribution is x6/(1−x)4.
Next, suppose i>1 and i−1∈β so that π=iαnβ′(i−1)β′′.
Then β′>i since j∈β′ with j<i makes inj(i−1) a 3412.
For d≥0, let Bd be the generating function for such permutations where β′ has d letters.
Lemma 19**.**
We have
[TABLE]
Proof.
First, let K be the generating function for T-avoiders i(i−2)(i−4)π′(i+1)(i−1)(i−3)π′′ with 2 left-right maxima such that i≥5 where n=i+1. Let us write an equation for K. When i=5 the contribution is x6.
Otherwise, we consider the position of i−5. If i−5 is either the leftmost letter of π′
or of π′′ then we have contribution of xK. Otherwise π′=α′(i−5)α′′ with α′=∅ such that i−5>π′′>α′>α′′ with α′′ and π′′ are decreasing. By consider whether α′′ is empty or not, we obtain the contribution x7(A−1+x2/(1−x)2)/(1−x). Thus
[TABLE]
which leads to
[TABLE]
Next, let K′ be the generating function for T-avoiders i(i−2)π′n(i−1)(i−3)π′′ with 2 left-right maxima such that n>i≥4. Let us write an equation for K′. When i=4 we have a contribution of x5A. If i−4 is the leftmost letter of π′ then we have a contribution K, and if i−4 is leftmost letter of π′′ then we have a contribution xK′. Otherwise π′=α′(i−4)α′′ with α′=∅ such that i−4>π′′>α′>α′′ with α′′ and π′′ decreasing. By considering whether α′′ is empty or not, we obtain the contribution x6(A−1+x2/(1−x)2)/(1−x). Thus K′=x5A+xK′+K+x6(A−1+x2/(1−x)2)/(1−x), which leads to
[TABLE]
Next, let K′′ be the generating function for T-avoiders i(i−2)π′n(i−1)π′′ with 2 left-right maxima such that n>i≥3. By using similar techniques as for finding formulas K and K′, we have
[TABLE]
which implies
[TABLE]
Lastly, let K′′′ be the generating function for T-avoiders iπ′n(i−1)j(i−2)π′′ with
2 left-right maxima such that n>j>i≥3.
By using similar techniques as for finding formulas K and K′, we obtain
[TABLE]
Now, we are ready to write an equation for the generating function B0 for T-avoiders
iπ′n(i−1)π′′ with 2 left-right maxima such that n>i≥2. Again, we decompose
the structure by looking at position of i−2. If i=2 then we have a contribution of x3A,
if i−2 is leftmost letter of π′ then we have a contribution of K′′, if i−2 belongs
to π′′ and is not the leftmost letter of it, we have a contribution
of x4A(A−1)+(1−x)2x6A+1−xx5(A−1)+(1−x)3x7, and
if i−2 belongs to π′′ we have a contribution of K′′′. Thus,
[TABLE]
which completes the proof.
∎
One can similarly show (details omitted) that
[TABLE]
and Bd=xBd−1 for all d≥2. Hence, the total contribution for the case i>1 and i−1∈β is given by B, where B−B1−B0=x(B−B0), which leads to
[TABLE]
Adding all the contributions,
[TABLE]
with solution for G2(x) as claimed.
∎
Theorem 20**.**
Let T={3412,1324,2341}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m left-right maxima. Clearly, G0(x)=1, G1(x)=xFT(x), and G2(x) is given above.
Now, let us write an equation for Gm(x) with m≥3. Let π=i1π(1)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324 and 2341, we can write π as
[TABLE]
where π(1)>π(2) and γ>im−1>β>im−2. By considering the four possibilities for whether π(2),β are empty of not, we obtain the contributions xmA2 (both empty), (1−x)2xm+1A (π(2) empty, β nonempty), (1−x)2xm+1A (π(2) nonempty, β empty) and (1−x)4xm+2 (both nonempty). Thus,
[TABLE]
Therefore,
[TABLE]
Substituting for G2(x) and solving for FT(x) completes the proof.
∎
3.6. Case 80: {1324,2341,3421}
In order to study this case, we need the following lemmas.
Lemma 21**.**
Let T={1324,2341,3421}. Let G2(x) be the generating function for the number of permutations π=iπ′nπ′′∈Sn(T) with exactly 2 left-right maxima. Then
[TABLE]
Proof.
In order to find a formula for the generating function G2(x), we refine it as follows. Let G2(x;d) be the generating function for the number of permutations π=iπ′nπ′′∈Sn(T) with exactly 2 left-right maxima and where π′′ has exactly d points smaller than i. Clearly,
[TABLE]
where K(x)=(1−x)2(1−2x)1−3x+3x2 (where we leave the proof to the reader). For d=1, our permutations can be written as i(i−1)⋯(i′+1)αn(n−1)⋯i′′i′β with α<i′ and i<β<i′′. By considering whether α and β are empty or not, one can show that
[TABLE]
For d≥2, our permutations can be written as
[TABLE]
such that all the letters that are greater than j1 in iα(1)⋯α(d) are decreasing and nβ(1)⋯β(d) is decreasing. We denote all the letters between j1 and j2 in π′ by δ. Now, let us write an equation for G2(x;d). If δ=β(2)=∅, then we have a contribution of xG2(x;d−1). If δ=∅, then π can be written as
[TABLE]
such that α(1)α(2)⋯α(d−1)γ′ is decreasing, jd+2−s>α(s)>jd+1−s for all s=1,2,…,d−1 with jd+1=i, j2>γ′=δ>j1 and j1>γ′′ where γ′′ avoids 132,2341,3421 and β(d+1) avoids 213,2341,3421. Thus, we have a contribution of (1−x)2dxd+3K(x)2. Otherwise, δ=∅ and β(2)=∅, so similarly, we have a contribution of (1−x)2d−1xd+3K(x)2. Therefore, for all d≥2,
[TABLE]
By summing over all d≥2, we obtain
[TABLE]
By using the values of G2(x;0) and G2(x;1), and then solving for G2(x), we complete the proof.
∎
Theorem 22**.**
Let T={1324,2341,3421}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x). Note that the generating function G2(x) is given in Lemma 21.
Let m=3. Each permutation of π∈Sn(T) with exactly 3 left-right maxima can be represented as either
•
π=i1π(1)(i1+1)nπ(3) with π(1)<i1 and π(3)>i1+1, where π(1) avoids 132,2341,3421 and π(3) avoids 213,2341,3421.
•
π=i1π(1)i2n(n−1)⋯(i2+1)(i1+1)(i1+2)⋯(i2−1), where i1+1<i2 and π(1) avoids 132,2341,3421.
•
π=i1(i1−1)⋯i′(i1+1)12⋯(i′−1)nπ(3), where i′>1 and π(3) avoids 213,2341,3421.
•
π=i1(i1−1)⋯i′i212⋯(i′−1)n(n−1)⋯(i2+1)(i1+1)(i1+2)⋯(i2−1), where i1+1<i2 and i′>1.
By finding the corresponding generating functions, we obtain
[TABLE]
Now, let us write an equation for Gm(x) with m≥4. Let π=i1π(1)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324 and 2341, we see that π(s)<i1 for all s=2,3,…,m−1 and π(m)>im−2. Hence, we have a contribution of xGm−1(x) (by removing the letter im−2). Thus, Gm(x)=xm−3G3(x) for all m≥3. Therefore,
[TABLE]
which, by substituting the values of Gj(x), j=0,1,2,3, we complete the proof.
∎
3.7. Case 84: {4231,1324,2341}
Lemma 23**.**
Let π=iαnβ∈Sn(T) with exactly 2 left-right maxima, i≥2 and i−1
in α. Then the generating function H(x) for such permutations is given by
[TABLE]
where G2(x) is the generating function for permutations in Sn(T) with exactly 2 left-right maxima and L(x)=1−2x1−x.
Proof.
Let us write an equation for H(x). If i−1 is the first letter of α, then the contribution is xG2(x). Otherwise, i−1 is the last letter of α (to avoid 1324 and 4231), and π can be written as π=iα′(i−1)nβ′β′′ such that ∅=α′<β′′<i<β′, where α′ avoids both 132 and 231 and β′′ avoids both 213 and 231. Thus, by considering whether β′=n(n−1)⋯(i+1) or not, we get contributions of x^{3}\big{(}L(x)-1\big{)}L(x)/(1-x) and x^{3}\big{(}L(x)-1\big{)}\big{(}L(x)-1/(1-x)\big{)}
(recall that F{132,231}=F{213,231}=L(x)). Hence,
[TABLE]
as required.
∎
Lemma 24**.**
Let π=iαnβ∈Sn(T) with exactly 2 left-right maxima, i≥2, and i−1 in β. Then the generating function H′(x) for such permutations is given by
[TABLE]
Proof.
In order to write an equation for H′(x), we define Ad(x) to be the generating function for permutations π=iαnβ′(i−1)β′′∈Sn(T) with exactly 2 left-right maxima, i≥2, and such that β′ has d letters that are greater than i (these d letters form a decreasing subsequence because π avoids 4231). We leave to the reader to show that
A0(x)=(1−x)(1−2x)3x3(1−3x+5x2−4x3), A1(x)=(1−x)(1−2x)4x4(1−3x+4x2−5x3+4x4), and for d≥2,
[TABLE]
Summing over all d≥0, we obtain
[TABLE]
as required.
∎
Lemma 25**.**
The generating function G2(x) for permutations in Sn(T) with exactly 2 left-right maxima is given by
[TABLE]
Proof.
Permutations in Sn(T) with exactly 2 left-right maxima whose first letter is 1 (and hence second letter is n) have the generating function x2L(x), where L(x)=F{231,213}(x)=1−2x1−x. Thus, by Lemmas 23 and 24, we obtain
[TABLE]
and solving for G2(x) completes the proof.
∎
Theorem 26**.**
Let T={4231,1324,2341}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xF{231,1324}(x)=x(1−2x)2(1−x)1−4x+5x2−x3) [23], and G2(x)
is given by Lemma 25. Now suppose m≥3 and
let π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324 and 2341, we see that π(3)⋯π(m−1)=∅, π(1)>π(2), and π(m)=αβ with α>im−1>β>im−2. By considering the four possibilities, π(2),β empty or not, we obtain the contributions xmL(x)2, \frac{x^{m}}{1-x}L(x)\big{(}L(x)-1\big{)},
\frac{x^{m}}{1-x}L(x)\big{(}L(x)-1\big{)} and \frac{x^{m}}{(1-x)^{2}}\big{(}L(x)-1\big{)}^{2}. Thus,
[TABLE]
Summing over m≥3, we obtain
[TABLE]
Using the expressions for G1(x),G2(x), we complete the proof.
∎
3.8. Case 86: {3412,2431,1324}
In this subsection, let A=1−3x+x21−2x denote the generating function
for F{132,3412}(x) and let B=(1−x)2(1−2x)1−3x+3x2 denote the
generating function for F{213,2431,3412}(x) (they can be derived from results in [23]).
Lemma 27**.**
Let H(x) be the generating function permutations in Sn(T) whose first letter is n−1. Then
[TABLE]
Proof.
Refine H(x) to Hd(x), the generating function for T-avoiders π=(n−1)π′nπ′′
where π′′ has d letters.
Since π avoids 3412, π′′ is a decreasing subsequence say j1j2⋯jd.
Clearly, H0(x)=x2A. If d≥2, then there is no letter between j1 and
jd in π′, otherwise π contains 2431. Thus Hd(x)=xd−1H1(x), for all d≥1.
Now, let us compute H1(x). If j1=1, then we have a contribution of x3A.
So, we can assume that j1>1. Define H1(x;e,d) to be the generating function
for T-avoiders π=(n−1)α1βnj1, where β has e letters smaller
than j1 and d letters greater than j1. Clearly, β is an increasing subsequence.
Now let us examine the following cases:
•
e=d=0. We have a contribution of xH1(x).
•
d=0 and e≥1. Here we see that α=α′α′′ with α′ a decreasing subsequence and α′′<e1, where e1 is the smallest letter in β. So we have a contribution of (1−x)e+1xe+4A.
•
e=0 and d≥1. Here we see that either α=α′α′′ with α′ decreasing and α′′<j1, or α=α′α′′(j1+1)α′′′ with α′ decreasing, j1>α′′>α′′′. So we have a contribution of x4(1−xA+(1−x)2x(A−1))(1−x)dxd.
•
e,d≥1. Here we see that α=α′α′′ with α′ decreasing
and α′′<e1, where e1 is the smallest letter in β.
So we have a contribution of
(1−x)d+e+1xd+e+6A.
Hence,
[TABLE]
which leads to
[TABLE]
Since H(x)=x2A(x)+∑d≥1Hd(x)=x2A(x)+1−x1H1(x), the result follows.
∎
Lemma 28**.**
The generating function for T-avoiders with 2 left-right maxima is given by
[TABLE]
The generating function for T-avoiders with 3 left-right maxima is given by
[TABLE]
Proof.
We treat G2(x) and leave the similar derivation of G3(x) to the reader. Let π=iπ′nπ′′∈Sn(T) with 2 left-right maxima. By Lemma 27, the contribution for the case i=n−1 is H(x). So let us assume that i<n−1, that is, π′′ contains the letter i+1. Since π avoids 2431, we can write π as π=iα′nα′′β′ where α′α′′<i<β′, β′=∅ and α′′ is decreasing. If α′′=∅, then we have contribution of x2A(B−1). If α′′ has exactly one letter then we have a contribution of \frac{x^{3}}{1-x}A(B-1)+x^{3}\big{(}A-1/(1-x)\big{)}(B-1) corresponding to the cases α′ decreasing or not. If α′′ has at least two letters, then β′>i>α′>α′′, so the contribution is x4A(B−1)/(1−x).
By adding all the contributions, we complete the proof.
∎
Theorem 29**.**
Let T={3412,2431,1324}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1,G1(x)=xFT(x), and G2(x),G3(x) are given in Lemma 28. Now let us write an equation for Gm(x) with m≥4. Suppose π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324 we see that π(j)<i1 for all j=1,2,…,m−1. With α=π(m−1) and
β the subsequence of all letters between im−2 and im−1 in π(m), we have contributions as follows:
•
If α=β=∅, then we have xGm−1(x);
•
If α=∅ and β=∅, then we have xm+1A/(1−x)m−2;
•
If α=∅ and β=∅, then we have xm+1AB/(1−x)m−1;
•
If α=∅ and β=∅, then we have xm+2A/(1−x)m−1.
Thus, Gm(x)=xGm−1(x)+xm+1A/(1−x)m−2+xm+1AB/(1−x)m−1+xm+2A/(1−x)m−1, for all m≥4. Summing over m≥4, we obtain
[TABLE]
The result follows by substituting the expressions for G2(x), G3(x), A and B, and solving for FT(x).
∎
3.9. Case 88: {3412,3421,1324}
For this case we outline a proof based on a labelled generating forest and also give a proof based on left-right maxima.
3.9.1. Labelled generating forest
The table in the following lemma both recursively defines labels and gives valid succession rules. The j-th entry on the right hand side in the rules gives the label when n+1 is inserted into the j-th active site left to right. A label ki,1≤i≤5, always indicates k active sites. The proof is omitted.
Lemma 30**.**
The generating forest F is given by
[TABLE]
∎
Theorem 31**.**
Let T={3412,3421,1324}. Then
[TABLE]
Proof.
Let ak(x), bk(x), ck(x), dk(x) and ek(x) be the generating functions for the number of permutations in the nth level of the generating forest F with label k1, k2, k3, k4 and k5, respectively.
By Lemma 30, we have
[TABLE]
with a3(x)=x2+x∑j≥3aj(x), c2(x)=xc2(x)+x∑j≥3dj(x), c3(x)=x2+xc3(x)+x∑j≥4(bj(x)+cj(x)), d3(x)=xc2(x)+xc3(x)+xb4(x) and e1(x)=x∑j≥1ej(x).
Now let A(x,v)=∑k≥3ak(x)vk, B(x,v)=∑k≥3bk(x)vk, C(x,v)=∑k≥3ck(x)vk, D(x,v)=∑k≥2dk(x)vk and E(x,v)=∑k≥2ek(x)vk.
By the above recurrences, we see that A(x,v)−(x2+xA(x,1))v3=xvA(x,v). Thus, by substituting v=1, we obtain A(x,1)=1−2xx2, which implies A(x,v)=(1−xv)(1−2x)x2v3(1−x).
Thus, by the equation for bk(x), we have B(x,v)=(1−xv)(1−2x)x3v4(1−x).
By the equations for ck(x),dk(x),ek(x), we have
[TABLE]
with c2(x)=1−xxD(x,1), c3(x)=x2+xB(x,1)+xC(x,1)−1−xx2D(x,1), d3(x)=(1−x)2x2D(x,1) and e1(x)=xE(x,1)v. Note that b4(x)=1−2xx3(1−x) (from B(x,v)).
Substituting v=1 in the first two equations, and solving for C(x,1) and D(x,1), we obtain
[TABLE]
which implies
[TABLE]
We solve the equation for E(x,v) by using the kernel method (see, e.g., [15] for an exposition), taking v=1−x1. Using the expressions for A(x,v), B(x,v), C(x,v) and D(x,v), this gives
[TABLE]
Since FT(x)=1+x+A(x,1)+B(x,1)+C(x,1)+D(x,1)+E(x,1), the result follows.
∎
3.10. Case 93: {1324,2413,3421}
In order to study this case, we need the following lemmas.
Lemma 32**.**
Let T={1324,2413,3421}. Let Hm(x) be the generating function for the number of T-avoiders of the form π=(n+1−m)π(1)(n+2−m)π(2)⋯nπ(m) such that π(n+1)∈Sn+1(T). Then
[TABLE]
where K(x)=F{132,3421}(x)=1+(1−x)(1−2x)2x(1−3x+3x2) (see [25, Seq. A005183]) and L(x)=1−2x1−x.
Proof.
Let us write a formula for Hm(x). Let π=(n+1−m)π(1)(n+2−m)π(2)⋯nπ(m) such that π(n+1)∈Sn+1(T). Since π avoids 1324, we see that π(1)>⋯>π(m). If π(2)=⋯=π(m)=∅, then π(1) avoids 132 and 3412, which gives a contribution of xmK(x). Otherwise, since π avoids 3421, there exists a unique j such that π(j)=∅ with 2≤j≤m. Moreover, π(1) avoids 132,231. Thus, we have a contribution of 1−xxm+1L(x). Hence, Hm(x)=xmK(x)+1−x(m−1)xm+1L(x), as required.
∎
Lemma 33**.**
Let T={1324,2413,3421}. Let Hm′(x) be the generating function for the number of permutations π=(n+1−m)π(1)(n+2−m)π(2)⋯nπ(m)∈Sn(T). Then
[TABLE]
with H2′(x)=(1−2x)3x2(1−4x+5x2).
Proof.
We leave the formula for H2′(x) to the reader. Now we write a formula for Hm′(x) with m≥3. Let π=(n+1−m)π(1)(n+2−m)π(2)⋯nπ(m)∈Sn(T). If π(m−1)=∅, then we have a contribution of xHm−1′(x). Thus, we can assume that π(m−1)=∅. Since π avoids 1324 and 2413, we have that π(2)=⋯=π(m−2)=∅, π(1)>π(m)>π(m−1) and π(m−1)π(m) is increasing. Note π(1) avoids 132 and 231. Thus, we have a contribution of (1−x)2xm+1L(x). Hence,
Hm′(x)=xHm−1′(x)+(1−x)(1−2x)xm+1, as required.
∎
Similar consideration yield the following result.
Lemma 34**.**
Let T={1324,2413,3421}. Let Hm′′(x) be the generating function for the number of permutations π=(n+1−m)π(1)(n+2−m)π(2)⋯nπ(m)∈Sn(T) such that π(m) has a letter smaller than n+1−m. Then
[TABLE]
with H2′′(x)=(1−x)(1−2x)3x3(1−4x+6x2−2x3).
Now we are ready to state the formula for FT(x).
Theorem 35**.**
Let T={1324,2413,3421}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
Now let us write an equation for Gm(x) with m≥2. Let π=i1π(1)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324 and 2413, we see that π(s)<i1 for all s=1,2,…,m−1 and π(m)=β(m)⋯β(1) such that is−1<β(s)<is for all s=1,2,…,m (with i0=0). We consider cases:
•
β(2)=⋯β(j−1)=∅ and β(j)=∅ with j=2,3,…,m−1. Since π avoids T, we see that π(s)=∅ for all s=j,j+1,…,m−1 and β(s)=∅ for all s=j,m, where β(m) avoids 132 and 231, and β(j) is nonempty increasing. Thus, by Lemma 32, we have a contribution of 1−xxm+2−jHj−1(x)L(x).
•
β(2)=⋯β(m−1)=∅ and β(m)=∅. If β(1)=∅, then we have a contribution of xHm−1(x)(K(x)−1), see Lemma 32. Otherwise, we have a contribution of Hm′′(x)(L(x)−1), see Lemma 34.
•
β(2)=⋯β(m)=∅. For this case, we have a contribution of Hm′(x), see Lemma 33.
By summing over all contributions, we obtain
[TABLE]
for all m≥2. Hence, by summing over all m≥2 and using the initial conditions G0(x) and G1(x), we have
Hence, by solving for FT(x), we complete the proof.
∎
3.11. Case 99: {1324,3142,4231}
Throughout this case, we abbreviate
F{132,4231}(x) by K(x). Recall K(x)=1+(1−x)(1−2x)2x(1−3x+3x2) [25, Seq. A005183]. To find FT(x), we use the following lemmas.
Lemma 36**.**
Let T={1324,3142,4231}. Define Jm(x) to be the generating function for the number of permutations π=i1i2⋯imπ′∈Sn(T) with exactly m left-right maxima, all occurring at the start, such that i1>1. Set J(x)=∑m≥2Jm(x). Then
J2(x)=(1−x)2(1−2x)2x3(1−2x+3x2−x3) and
[TABLE]
Proof.
Let us write an equation for Jm(x) with m≥3. Let π=i1i2⋯imπ′∈Sn(T) with exactly m left-right maxima. If im−1=n−1, then we have a contribution of xJm−1(x). Thus, we can assume that im−1<n−1. If π′=(n−1)π′′, then we have a contribution of xJm(x), otherwise, since π avoids T (note that 1 belongs to π′), we can write π as
π′=i′(i′−1)⋯(im−1+1)αβ, such that α<i1, i′<β<n and n−1 belongs to β. Note that β avoids 213 and 231, and α avoids 132 and 231. Thus, we have a contribution of
xm(1+x)(L(x)−1)2.
Hence,
[TABLE]
By very similar techniques, we find that J2(x)=x(K(x)−1)+xJ2(x)+x2(1+x)(L(x)−1)2. Thus,
J2(x)=(1−x)2(1−2x)2x3(1−2x+3x2−x3) and the displayed recurrence for Jm then readily yields the stated expression for J(x).
∎
Lemma 37**.**
Let T={1324,3142,4231}. Define Jm′(x) to be the generating function for the number of permutations π=i1i2⋯imπ′∈Sn(T) with exactly m left-right maxima such that i1=1 (and im=n). Then
[TABLE]
Proof.
Let us write an equation for Jm′(x) with m≥3. Let π=i1i2⋯imπ′∈Sn(T) with exactly m left-right maxima such that i1=1. By considering whether i2=2 or not (we leave the details to the reader), we obtain the contributions
xJm−1′(x) and (1−x)m−1xm+1L(x). Hence,
[TABLE]
with J2′(x)=x2L(x). The result follows by solving this recurrence.
∎
Theorem 38**.**
Let T={1324,3142,4231}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m left-right maxima. Clearly, G0(x)=1 and G1(x)=xF{132,4231}(x)=xK(x).
Now, let us write an equation for Gm(x) with m≥2. Let π=i1π(1)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324, we see that π(s)<i1 for all s=2,3,…,m−1. We consider the following cases:
•
π(m) has a letter smaller than i1. Since π avoids 4231, we see that π(1)⋯π(m−1) is decreasing. Assume that π(j)=∅ and π(j+1)=⋯π(m−1)=∅. Since π avoids 3142, we have that
π(m) has no letters between i1 and ij. Thus, we have a contribution of
1−x1Jm(x) when j=1, and (1−x)jxjJm+1−j(x) when j=2,3,…,m−1, where Jd(x) is defined in Lemma 36. Hence,
the total contribution for this case is given by
[TABLE]
•
π(m)>i1 and π(2)=⋯π(m−1)=∅. Then we have a contribution of L(x)Jm′(x), where Jm′(x) is given in Lemma 37.
•
π(m)>i1 and there exists j such that π(j)=∅ and π(j+1)=⋯π(m−1)=∅, where j=2,3,…,m−1. As before, we obtain the contributions (1−x)j−1xj−1(L(x)−1)Jm+1−j′(x) when j≤m−2, and (1−x)m−2xmL(x)(L(x)−1) when j=m−1. Hence, the total contribution from this case is
[TABLE]
Hence,
[TABLE]
Summing over m≥2 and using the expressions for G0(x) and G1(x), we obtain
[TABLE]
where J(x)=∑d≥2Jd(x), J2(x) and J′(x)=∑d≥3Jd′(x) are given in Lemmas 36 and 37. After several algebraic operations, we complete the proof.
∎
3.12. Case 132: {1324,2341,2413}
Theorem 39**.**
Let T={1324,2341,2413}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
Let us write an equation for G2(x). Let π=iπ′nπ′′∈Sn(T) with exactly 2 left-right maxima. The contribution of the case π′′>i is x2K(x)2, where
K(x)=1−3x+x21−2x is the generating function for {132,2341}-
(or {213,2341}-) avoiders (for example, see [23]). If there is a letter in π′′ smaller than i,
we can write π as π=iπ′n(n−1)⋯(i+1)π′′ such that π′′ is not empty.
Thus, we have a contribution of \frac{x}{1-x}\big{(}F_{T}(x)-1-xK(x)\big{)},
where 1−xx2K(x) counts the permutations of the form iπ′n(n−1)⋯(i+1).
Hence, G_{2}(x)=x^{2}K(x)^{2}+\frac{x}{1-x}\big{(}F_{T}(x)-1-xK(x)\big{)}.
Next, let us write an equation for G3(x). Let π=i1π′i2π′′i3π′′′∈Sn(T) with exactly 3 left-right maxima. Since π avoids 1324 and 2413, we can write π=i1π′i2π′′i3αβ such that
π′′<π′<i1<β<i2<α<i3. By considering the four cases where π′′,β
are empty or not (if both are nonempty we get an occurrence of 2413, so we
actually have three cases),
we obtain
[TABLE]
Finally, let us write an equation for Gm(x) with m≥4. Let π=i1π(1)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids T, we see that i1>π(1)>π(2), π(j)=∅ for j=3,4,…,m−1, and π(m)=αβ such that im>α>im−1>β. Again, by considering the four cases where
π(2),β are empty or not, we obtain that
[TABLE]
By summing for m≥4 and using the expressions for G0(x),G1(x),G2(x) and G3(x), we obtain
[TABLE]
which, by solving for FT(x), completes the proof.
∎
3.13. Case 150: {1324,3421,3241}
In this section, define L=F{213,231}(x)=1−2x1−x (see [24]). We also have, by the simple decomposition π′nπ′′ of 132-avoiders,
[TABLE]
and
[TABLE]
.
Lemma 40**.**
The generating function for T-avoiders of the form
•
(d+1)nπ′(d+2)* with n−3≥d≥1 is given by*
[TABLE]
•
(d+1)nπ′* with n−2≥d≥1 is given by*
[TABLE]
Moreover,
[TABLE]
Proof.
Let us write an equation for Ed(x). Let π=(d+1)nπ′(d+2)∈Sn(T). If n=d+3, we have a contribution of xd+3, so let us assume that n≥d+4 and consider 3 cases:
•
If d+3 on left side of letter 1, π can be written as (d+1)n(n−1)⋯n′(d+3)π′(d+2) with π′<n′, so we have 1−xxEd(x).
•
If d+3 between the letters j and j+1, j=1,2,…,d−1, then π can be written as
[TABLE]
such that the subsequence nα(1)⋯α(j+1) is decreasing and greater than π′ and π′>d+3, where π′ avoids {231,213}. So, we have (1−x)jxd+4L(x).
•
If d+3 between the letters d and d+2, then π can be written as
[TABLE]
such that the subsequence nα(1)⋯α(d+1) is decreasing and greater than π′ and π′>d+3, where π′ avoids {231,213}. So, we have (1−x)d+1xd+4L(x).
Thus, Ed(x)=xd+3+1−xxEd(x)+∑j=2d+1(1−x)jxd+4L, as claimed.
Similarly, we can write an equation for Dd(x) and obtain
[TABLE]
where xd+2, 1−xxDd(x), (1−x)j+1xd+3L and 1−xEd(x)+(1−x)d+1xd+3(B−1/(1−x)) are the respective contributions for the cases n=d+2, the letter d+2 on the left side of the letter 1, the letter d+2 between the letters j and j+1 with j=1,2,…,d−1, and the letter d+2 on the right side of the letter d.
By summing over d≥1, we complete the proof.
∎
Lemma 41**.**
For m≥3, the generating function for T-avoiders π=i1π(1)⋯imπ(m) with m left-right maxima and π(1)=∅ is given by
[TABLE]
Proof.
Let π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima and π(1)=∅. By the definitions we see that π(2)⋯π(m−1)=∅.
We have the following contributions to Gm,1(x):
•
If π(m)>im−1, then we have xm(A−1)B.
•
If π(m) has a letter smaller than i1, then π(1)<π(m)=π′(k+1)(k+2)⋯(i1−1)(i1+1)⋯(i2−1), π′>im−1, π′ avoids {213,231} and π(1) avoids {132,3421,3241}. This leads to a contribution of (1−x)2xm+1(A−1)L.
•
If π(m)>i1 and π(m) has a letter between ij and ij+1, where j=1,2,…,m−2, then π(1)<i1<π(m), where π(m)=π′(ij+1)⋯(ij+1−1) with π′ a {213,231}-avoider, and π(1) avoids {132,3421,3241}. This leads to a contribution of 1−xxm+1(A−1)L.
Adding these contributions yields the result.
∎
Using similar arguments, we obtain the following result.
Lemma 42**.**
The generating function for T-avoiders π=i1π(1)⋯imπ(m) with m left-right maxima and π(1)=⋯=π(s−1)=∅ and π(s)=∅, 2≤s≤m−2, is given by
[TABLE]
Lemma 43**.**
Let T={1324,3421,3241}. Then the generating function for T-avoiders π=i1π(1)⋯imπ(m) with m left-right maxima and π(1)=⋯=π(m−2)=∅
and π(m−1)=∅ is given by
[TABLE]
where
[TABLE]
Proof.
To write an equation for Gm,m−1(x), suppose π=i1π(1)⋯imπ(m) with m left-right maxima and π(1)=⋯=π(m−2)=∅ and π(m−1)=∅. Let α be the subsequence of π(m) consisting of letters smaller than π(1), and let β be the subsequence of π(m) consisting of letters between i1 and im−1. Note that π(m)<αβ. Now let us consider the following four cases:
•
α=β=∅: Here π(m−1)<i1 and π(m)>im−1. So we have a contribution Nm.
•
α=∅ and β=∅: Here π(m−1)<i1, and there exits 1≤j≤m−2 such that π(m)=π′(ij+1)⋯(ij+1−1) and π′ avoids {213,231}. So we have a contribution Nm′.
•
α=∅ and β=∅. In this case π(m) can be written as
[TABLE]
where β(s)>β(s−1)>⋯>β(0)>im−1 and π(m−1)<i1−s. Let Mm be the contribution of this case. By considering whether β(s) is a decreasing sequence (including the empty case) or contains an ascent, we obtain the contributions 1−xx(Nm+Mm) and xm+2(L−1/(1−x))/(1−x)3. Thus,
[TABLE]
with solution for Mm as above.
•
α=∅ and β=∅. In this case, there exists j, 1≤j≤m−2 such that π(m) can be written as
[TABLE]
where β(s)>β(s−1)>⋯>β(0)>im−1 and π(m−1)<i1−s. Let Mm′ be the contribution of this case. By considering whether β(0) is empty or not, we obtain the contributions x(Nm′+Mm′) and (m−2)xm+3(L−1)/((1−x)2(1−2x)). Thus,
[TABLE]
with solution for Mm′ as above.
This completes the proof.
∎
Theorem 44**.**
Let T={1324,3421,3241}. Then
[TABLE]
Proof.
First, we study the generating function Gm′(x) for T-avoiders π with m
left-right maxima such that π1=1. Clearly, Gm′(x)/x is the generating function
for the {213,3421,3241}-avoiders with m−1 left-right maxima.
It is not hard to see that
[TABLE]
Now let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
Let us find the generating function G2(x). Suppose π=iπ′nπ′′∈Sn(T) with 2 left-right maxima. If π′′>i, then we have a contribution of x2(A−1)B. Otherwise, we can write π as
[TABLE]
where β(s)>β(s−1)>⋯>β(0)>i and π′<i−s. Let us denote the contribution of this case by K. By considering whether β(s) is decreasing or contains a rise, we obtain the contributions of 1−xx(K+x2(A−1)B) and (1−x)2x3(A−1)(L−1−x1). Thus,
[TABLE]
and G2(x)−G2′(x)=x2(A−1)B+K.
Now let us find an explicit formula for Gm(x). By Lemmas 40–43, we see that
[TABLE]
where Gm,m(x) counted by xm−2(Dm(x)+(m−2)Em(x)/(1−x)). Thus,
[TABLE]
By substituting the expressions for Gm(x)−Gm′(x) and solving for FT(x), we complete the proof.
∎
3.14. Case 151: {1324,1342,3421}
Here, we focus on the number of left-right maxima and begin with the case where n is the second letter. Let Jd(x) be the generating for permutations of the form (d+1)nα∈Sn(T).
Lemma 45**.**
J0(x)=x2L(x)=1−2xx2(1−x)* and for all d≥1,*
[TABLE]
Proof.
Clearly, J0(x)=x2F{213,231}(x)=x2L(x). Thus, we assume that d≥1 and let us write an equation for Jd(x).
If d=n−2 then we have a contribution of xd+2. Otherwise, d≤n−3, and consider the position of the letter d+2. Note that the letters 1,2,…,d occur in that order since π starts (d+1)n and avoids 3421.
•
The letter d+2 appears on the left side of the letter 1. In this case, (d+1)nα=(d+1)n(n−1)⋯(n′+1)(d+2)α′, which leads to a contribution of 1−xxJd(x).
•
The letter d+2 appears between the letters p and p+1, where 1≤p≤d−1. Then, (d+1)nα=(d+1)nα′(d+2)(p+1)(p+2)⋯d, where all the letters in α′ which are >d+2 (resp. <p+1) are decreasing (resp. increasing). Thus, we have a contribution of
[TABLE]
•
The letter d+2 appear on the right side of d. In this case, π has the form (d+1)nα′(d+2)β where all the letters in α′ greater than d+2 form a decreasing subsequence γ with γ>β>d+2, and β avoids {213,231}. Thus, we have a contribution of
[TABLE]
Hence, by adding all the contributions, we have
[TABLE]
and the stated expression for Jd(x) follows.
∎
Now set J(x)=∑d≥0Jd(x), the generating function for T-avoiders whose largest letter occurs in second position.
Corollary 46**.**
[TABLE]
∎
Lemma 47**.**
Let G2(x) be the generating function for permutations in Sn(T) with exactly two left-right maxima. Then
[TABLE]
Proof.
Define G2(x;d) to be the generating function for permutations in iπ′nπ′′∈Sn(T) with two left-right maxima such that π′′ contains d letters smaller than i. If d=0 so that π has the form iαnβ with α<i<β,
then α avoids {132,3421} and β avoids {213,231}. Hence
G2(x;0)=x2A(x)L(x),
where A(x)=1+(1−x)2(1−2x)2x(1−3x+3x2) is the generating function for Sn({132,3421}) [23].
From now on, we assume that d≥1. The letters j1,…,jd in π′′ that are
less than i are increasing, and so π can be decomposed as
[TABLE]
where i>α(1)>jd>α(2)>⋯>α(d)>j1>α(d+1).
Now, we treat the following cases:
•
α(j)=∅ for all j=2,3,…,d+1. In this case, we have a contribution of L(x)Jd(x), see Lemma 45.
•
α(s)=∅ and α(j)=∅ for all j=s+1,s+2,…,d+1, where s=2,3,…,d+1. In this case, α(j) is a decreasing sequence for j=1,2,…,s−1, α(s) avoids {132,231}, and π′′=β(1)j1⋯β(d)jdγ, where β=β(1)⋯β(d) is a decreasing subsequence such that β>γ and γ avoids {213,231}. So, we have a contribution of
[TABLE]
for s=2,3,…,d, and
[TABLE]
for s=d+1.
•
α(d+1)=∅ and α(s) contains a rise (a rise of π is an index i such that πi<πi+1) with s=1,2,…,d. In this case, α(1)⋯α(s−1) is a decreasing sequence, α(s+1)⋯α(d) is empty, α(d) is increasing and π′′ decomposes as in the previous bullet. So the contribution is
[TABLE]
Hence, for d≥1,
[TABLE]
Summing over d≥0 and using Lemma 45, we obtain the stated formula for G2(x).
∎
Lemma 48**.**
Let G3(x) be the generating for the number of permutations in Sn(T) with exactly three left-right maxima. Then
[TABLE]
Proof.
We consider permutations π=i1π(1)i2π(2)i3π(3)∈Sn(T) with exactly three left-right maxima.
Since π avoids 1324 and 1342, we see that π(1)>π(2)β where β is all the letters in π(3) smaller than i1. If π(2)=∅ and β=∅ then we have a contribution of x3A(x)L(x). If π(2)=∅ and β=∅, then we have a contribution of x^{3}L(x)\big{(}J(x)-x^{2}L(x)\big{)}. If π(2)=∅, we see that π(1)>π(2)β, π(1) avoids {132,231}, π(2)β is an increasing sequence. Suppose β=j1j2⋯jd, then π(3)=β(1)j1⋯β(d)jdβ(d+1), where β(1)⋯β(d) is a decreasing sequence
which is greater than β(d+1), and β(d+1) avoids {213,231}. Thus, we have a contribution of x3L(x)1−xxL(x)1−x/(1−x)1. Hence,
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x), and G2(x) and G3(x) are given by Lemmas 47 and 48, respectively. Now let m≥4 and let us write an equation for Gm(x).
Suppose π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. If π(2) is empty then we have a contribution of xGm−1(x).
Otherwise, π(j)=∅ for all j=3,4,…,m−1, π(m)>im−1 and
π(1)>π(2), where π(m) avoids {213,231}, π(1) avoids
{132,231}, and π(2) is an increasing sequence. Recall that
F{213,231}(x)=F{132,231}(x)=1−2x1−x [24]. Thus, we have
[TABLE]
for all m≥4. Summing over m≥4, we obtain
[TABLE]
Solving for FT(x) using the expressions above for G1(x),G2(x),G3(x), we complete the proof.
∎
3.15. Case 153: {4231,1324,1342}
Here, we use the same techniques as in Case 84.
Lemma 50**.**
Let m≥3. The generating function for permutations π=π1π2⋯πn∈Sn(T) with exactly m left-right maxima and π1=1 is given by 1−2xxm(1−x).
Proof.
Such permutations can be written as 12⋯(m−1)nπ′ where π′ avoids 213 and 231, and the result follows.
∎
Lemma 51**.**
Let m≥3. The generating function Gm(x) for permutations π=π1π2⋯πn∈Sn(T) with exactly m left-right maxima satisfies the recurrence
[TABLE]
Proof.
To write an equation for Gm(x), let π=i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima. Since π avoids 1324 and 1342 we see that π(2)⋯π(m) has no letter between i1 and im−1. If π(2)⋯π(m) has no letter
smaller than π(1), the contribution is xmF{132,231}(x)F{213,231}(x)=xmL2(x). Otherwise, π(1) is a decreasing sequence, and then we have a contribution of
The next two lemmas give the generating function for permutations in Sn(T) with exactly 2 left-right maxima. We leave the proof to the diligent reader (very similar to Case 84).
Lemma 52**.**
Let π=iαnβ∈Sn(T) with exactly 2 left-right maxima and i≥2.
•
If i−1 is the left most letter of α, then the generating function for such permutations π is given by xG2(x), where G2(x) is the generating function for permutations in Sn(T) with exactly 2 left-right maxima.
•
If i−1 is in α but not the first letter in α, then the generating function for such permutations π is given by x^{3}L(x)\big{(}(L(x)-1\big{)}.
•
If i−1 is a letter in β such that there are exactly d letters between n and i−1 in π that are greater than i. We denote the generating function for the number of such permutations by Ad(x). Then
A0(x)=(1−x)(1−2x)3x3(1−3x+5x2−4x2), A1(x)−xA0(x)=(1−2x)5x5(2−5x+4x2),
and for all d≥2,
[TABLE]
Lemma 53**.**
Let G2(x) be the generating function for permutations in Sn(T) with exactly 2 left-right maxima. Then
[TABLE]
Proof.
With the definitions introduced in Lemma 52, we have
where x2L(x) counts the permutations 1nπ′∈Sn(T),
and the result follows.
∎
Theorem 54**.**
Let T={4231,1324,1342}. Then
[TABLE]
Proof.
Note that the number of permutations in Sn(T) with leftmost letter is n is given by G1(x)=xF{231,1324}(x)=xF{132,4231}(x)=1−xx(1+x(L(x)−1)L(x)), where L(x)=1−2x1−x, see [23]. Thus, by Lemma 51 and Lemma 53, we obtain
[TABLE]
which, by solving for FT(x), completes the proof.
∎
3.16. Case 156: T={1324,2341,2431}
Theorem 55**.**
Let T={1324,2341,2431}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
Let us write an equation for G2(x). Let π=iπ′nπ′′∈Sn(T) with exactly 2 left-right maxima. The contributions for the cases n−2≥i=1 and i=n−1≥1 are x2(M(x)−1) and x\big{(}F_{T}(x)-1\big{)} respectively, where M(x)=1−2x−x21−x−x2 is the generating function for {213,2341,2431}-avoiders (for example, see [23]). Denote the contribution for the case 2≤i≤n−2 by H(x). Then G_{2}(x)=x^{2}\big{(}M(x)-1\big{)}+x\big{(}F_{T}(x)-1\big{)}+H(x). To find a formula for H(x) we consider the position of i−1 in π, which leads to either π=iα(i−1)β′nβ′′β′′′ with β′′′>iα>β′β′′ or π=iα′nα′′(i−1)β′β′′ with β′′>i>α′α′′>β′. By examining the four possibilities, α,β′β′′ either empty or not, in the first case and examining the four possibilities for α′α′′,β′ in the second case, we obtain that
[TABLE]
where K(x)=1−3x+x21−2x is the generating function for {132,2341}-avoiders. Thus,
[TABLE]
Next, let us write an equation for G3(x). Let π=i1π′i2π′′i3π′′′∈Sn(T)
with exactly 3 left-right maxima. Since π avoids T, we can
write π=i1π′i2π′′i3βα where π′′<π′<i1<β<i2<α<i3.
By considering the four cases where π′′,β are empty or not, we obtain
[TABLE]
Finally, let us write an equation for Gm(x) with m≥4. Let π=i1π(1)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids T, we see that i1>π(1)>π(2), π(j)=∅ for j=3,4,…,m−1, and π(m) has the form βα where im>α>im−1>β. Thus, π avoids T if and only if the permutation that is obtained from π by removing im−2 avoids T. Hence Gm(x)=xGm−1(x)=⋯=xm−3G3(x), for m≥3.
By summing over m≥4 and using the expressions for G0(x),G1(x),G2(x) and G3(x), we obtain
[TABLE]
which, by solving for FT(x), completes the proof.
∎
3.17. Case 158: {1324,1342,3412}
Here, it is convenient to consider J(x), the generating function for T-avoiders whose maximal letter occurs in second position, and its refinement to Jd(x), the generating function for T-avoiders whose maximal letter occurs in second position and whose first letter is d, that is,
permutations of the form dnπ′′∈Sn(T).
Lemma 56**.**
With the preceding notation,
[TABLE]
Proof.
Suppose π∈Sn(T) has the form dnπ′′. Since π avoids 3412, we see that π′′ contains the subsequence (d−1)(d−2)⋯1. Therefore, if d=n−1, the contribution is xd+1 and otherwise, by considering whether n−1 occurs before or after the string (d−1)(d−2)⋯1, we find
[TABLE]
for all d≥1.
Solving this recurrence successively for d=1 and d=2 gives the first two stated expressions.
Also, summing over all d≥1 yields
[TABLE]
from which the expression for J(x) follows.
∎
Lemma 57**.**
Let G2(x) be the generating function for the number of T-avoiders
with exactly 2 left-right maxima. Then
G2(x)=(1−x)2(1−2x)(1−3x)(1−3x+x2)x2(1−7x+19x2−23x3+9x4).
Proof.
Again, we refine G2(x) by defining G2(x;d) to be the generating function for the number of permutations iπ′nπ′′∈Sn(T) with exactly 2 left-right maxima where π′′ has d letters smaller than i. Since π avoids 3412, we see that π′′ is decreasing. Now, let us write an equation for G2(x;d). Clearly, G2(x;0)=x2F{132,3412}(x)F{213,231}(x)=1−2xx2(1−x)K(x), where F{132,3412}(x)=K(x)=1−3x+x21−2x (see [25, Seq A001519]) and F{213,231}(x)=1−2x1−x (see [24]).
For d=1, let j denote the letter in π′′ smaller than i. Thus i>2 and π can be written, according as j>1 or j=1, as either
(1) π=i(i−1)⋯(j+1)α(1)n(n−1)⋯(i′+1)i′jβ(1) where
1∈α(1) and j>α(1), i′>β(1) and β(1) avoids 213 and 231, and α(1) avoids 132 and 3412, or (2) π=iαnβ where α consists of the letters 2,3,…,i−1 in some order
and α avoids 132 and 3412.
Hence,
[TABLE]
where J2(x) is defined above.
Now, let d≥2. Since π avoids 1324 and 1342, we can express π as
[TABLE]
where i>α(d+1)>j1>α(d)>j2>⋯>α(2)>jd>α(1). We consider three cases:
•
α(s)=∅ and α(s+1)=⋯=α(d+1)=∅ with s=3,4,…,d+1. In this case α(1),…,α(s−1) are decreasing and
α(s) avoids 132 and 3412, while β(2)=⋯=β(d+1)=∅ and β(1) is decreasing. Hence, we have a contribution of \frac{x^{d+2}}{(1-x)^{s}}\big{(}K(x)-1\big{)}.
•
α(3)=⋯=α(d+1)=∅ and α(2)=∅. In this case, α(1) and β(1) are decreasing, and β(1)>β(2)>⋯>β(d+1). Thus, we have a contribution of \frac{x}{(1-x)^{2}}\big{(}K(x)-1\big{)}J_{d}(x).
•
α(2)=⋯=α(d+1)=∅. In this case, we have a contribution of K(x)Jd+1(x).
Hence, for all d≥2,
[TABLE]
Summing over d≥0 and using Lemma 56, we obtain the stated expression for
G2(x).
∎
Theorem 58**.**
Let T={1324,1342,3412}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x), and G2(x) is given by Lemma 57.
Now, let us write an equation for Gm(x) with m≥3. A T-avoider i1π(1)⋯imπ(m) with m≥3 left-right maxima
i1,i2,…,im has the restricted form shown in Figure 2,
where i1,i2,…,im−1 are increasing consecutive integers, A is a list of zero or more decreasing consecutive integers, π(m) is composed of B1 and B2, and regions not marked are empty. Furthermore, π(1) avoids {132,3412} and, of course, i1imπ(m)
avoids T. Conversely, every permutation of this form with m≥3 satisfying the latter two conditions is a T-avoider. Hence we get contributions as follows: K(x) from π(1); J(x) from i1imπ(m); xm−2 from i2,…,im−1; and 1/(1−x)m−2
from A.
So Gm(x)=xm−2K(x)J(x)/(1−x)m−2. By summing over m≥3, we obtain
[TABLE]
Now solve for FT(x) using Lemma 57 to complete the proof.
∎
3.18. Case 180: {1342,2314,4231}
Theorem 59**.**
Let T={1342,2314,4231}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xF{231}(x)=xC(x) (see [16]).
Now let m≥3 and let us write equation for Gm(x).
Let π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 2314 and 1342, we see that π(j)>ij−1 for all j=2,3,…,m−1. If π(m) has a letter smaller than any letter in π(1), then ijπ(j)=ij(ij−1)⋯(ij−1+1) for all j=1,2,…,m−1, and π(m)=im(im−1)⋯(im−1+1)β such that β is non empty permutation in Si0(231). Hence, we get a contribution of (1−x)mxm(C(x)−1). Otherwise, π(j)>ij−1 for all j=2,3,…,m, which gives a contribution of xmCm(x). Hence,
[TABLE]
for all m≥3.
Now, let us focus on the case m=2. First, let H be the generating function for permutations iπ′nπ′′∈Sn(T) with exactly 2 left-right maxima and containing the subsequence n(n−1)⋯(i+1). Let us write an equation for H. If i=1, then we have a contribution of x2/(1−x). Otherwise, n−1≥i≥2 and consider the position of i−1.
If (i−1) is the first letter in π′ then we have a contribution of xH.
If (i−1)∈π′ but is not the first letter in π′, then π must have the form
iα(i−1)βn(n−1)⋯(i+1) with α<β<i−1, α=∅,
and α,β both 231-avoiders, which implies a contribution of
x^{3}\big{(}C(x)-1\big{)}C(x)/(1-x). Otherwise, i−1∈π′′ and then π can be decomposed
as in Figure 3.
where ↓ indicates a region of decreasing entries. Since St(iAnB) (B is spread over two regions) is of the type counted by H and D contributes C(x), we get a contribution of x/(1−x)C(x)H.
Hence,
[TABLE]
which leads to
[TABLE]
Now let us write an equation for G2(x). Let π=iπ′nπ′′ with exactly 2 left-right maxima. If i=1, then we have a contribution of x2C(x). Otherwise, n−1≥i≥2 and again consider the position of i−1. If π′=(i−1)π′′′, then we have a contribution of xG2(x). If π′=α(i−1)β such that α is not empty then π=iα(i−1)βnπ′′ with α<β<i−1<π′′<n, which gives a contribution of x^{3}\big{(}C(x)-1\big{)}C(x)^{2}. Thus, we can assume that i−1 belongs to π′′. In this case, π can be written as π=iπ′nα′(i−1)α′′α′′′ such that iπ′nα′ has 2 left-right maxima and contains the subsequence n(n−1)⋯(j+1), each letter in α′′ is greater than each letter smaller than i in π′α′, and i<α′′′<j+1, where α′′,α′′′ avoids 231 and iπ′nα′ avoids T. Thus, we have a contribution of xHC(x)2. Hence,
[TABLE]
which implies
[TABLE]
Summing over m≥0, we obtain
[TABLE]
and this expression simplifies to the stated form.
∎
3.19. Case 184: {1324,2431,3241}
Theorem 60**.**
Let T={1324,2431,3241}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
Let us write an equation for Gm(x) with m≥2. Suppose π∈Sn(T) has exactly m left-right maxima. Since π avoids 3241 and 2431, we can express π as
[TABLE]
where
π(1)<⋯<π(m)<i1<ρ(1)<i2<ρ(2)<⋯<im−1<ρ(m−1)<im. Since π avoids 1324 we see that at most one element of
L:={π(1),…,π(m−1)} is nonempty and at most one element of
R:={ρ(1),…,ρ(m−1)} is nonempty. Thus, we have the following cases:
•
if L and R are both lists of empty words, then π avoids T if and only if π(m) avoids T, so we have a contribution xmFT(x).
•
if, say, π(s)∈L is nonempty and R is a list of empty words, then π avoids T if and only if π(s) avoids 132 and 3241, and π(m) avoids 213 and 2431. Thus, we have a contribution of x^{m}K(x)\big{(}K(x)-1\big{)}, where K(x)=1−3x+x21−2x
is the generating function for {132,3241}-avoiders [25, Seq. A001519]. By symmetry,
K(x) is also the generating function for
{213,2431}-avoiders since R∘C∘I({213,2431})={132,3241}), where R= reverse, C= complement, and I= inverse on permutations.
•
if ρ(t)∈R is nonempty and L is a list of empty words, then, similarly, we have the same contribution of x^{m}K(x)\big{(}K(x)-1\big{)}.
•
lastly, suppose π(s)∈L and ρ(t)∈R are nonempty. If t≥s+1, then
π avoids T if and only if π(m)=∅, π(s) avoids 132 and 3241,
and ρ(t) avoids 213 and 2431, which gives a contribution of x^{m}\big{(}K(x)-1\big{)}^{2}.
Otherwise, 1≤t≤s, and we have the same conditions except π(m) must be increasing rather than empty, which leads to a contribution of \frac{x^{m}}{1-x}\big{(}K(x)-1\big{)}^{2}.
By adding all the contributions, we find that for all m≥2,
[TABLE]
Summing over m≥2, we obtain
[TABLE]
and solving for FT(x) completes the proof.
∎
3.20. Case 187: {1324,2314,2431}
Theorem 61**.**
Let T={1324,2314,2431}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
Let us write an equation for G2(x). Suppose π∈Sn(T) has 2 left-right maxima so that
π=iπ′nπ′′. If i=n−1, then the contribution is given by x(FT(x)−1).
Otherwise, we denote the contribution by H. So G2(x)=x(FT(x)−1)+H. For H, π can be written as π=iπ′nαβ with 1≤i≤n−2 and π′α<i<β and hence
β=∅. Note that β avoids both 213 and 2431 and π′α
avoids both 231 and 132. By considering whether π′α is empty or not and whether
i−1 belongs to π′ or to α, we find that
[TABLE]
where L=1−3x+x21−2x is the generating function for {213,2431}-avoiders
[25, Sequence A001519] and K=1−2x1−x is the generating function for {132,231}-avoiders [24].
Hence, H=(1−3x+x2)(1−2x)2x3(1−x)(1−2x+2x2), which implies
[TABLE]
Now, let us write an equation for Gm(x) with m≥3. Suppose π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324 and 2314, we see that π(2)=⋯=π(m−1)=∅. We consider the following two cases:
•
π(m)<im−1. In this case, by removing the letter n, we obtain a bijection between such permutations of length n and T-avoiders with m−1 left-right maxima of length n−1. Therefore, we have a contribution of xGm−1(x).
•
π(m) contains a letter between im−1 and im=n. In this case, since π avoids 1324 and 2314, we see that π(m)>im−1. So π avoids T if and only if π(m) avoids 213 and 2431 and π(1) avoids 231 and 132. So the contribution
for this case is xmK(L−1).
Thus, Gm(x)=xGm−1(x)+xmK(L−1).
Summing over m≥3, we obtain
[TABLE]
Hence, using the expressions for G2(x), G1(x) and G0(x), we obtain
[TABLE]
and solving for FT(x) completes the proof.
∎
3.21. Case 193: {1324,2431,3142}
Observe that each pattern here contains 132. So if a permutation avoids 132, then it certainly avoids T.
Theorem 62**.**
Let T={1324,2431,3142}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
Now let us write an equation for Gm(x) with m≥2. Let π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. We see that
i1>π(1)>π(2)>⋯>π(m−1) (to avoid 1324) and π(m) can be written as α(1)α(2)⋯α(m) such that π(m−1)>α(1)
and i1<α(2) (to avoid 3142) and
α(2)<i2<α(3)<⋯<im−1<α(m)<im (to avoid 2431).
Furthermore, π(j) avoids 132 for all j=1,2,…,m−1 (or im is the 4 of a 1324),
and at most one of α(2),…,α(m) is nonempty (to avoid 1324).
We consider two cases according as α(2)=⋯=α(m)=∅ or not:
•
α(2)=⋯=α(m)=∅. Here, α(1) only needs to avoid T
and we have a contribution of xmC(x)m−1FT(x)
•
There is a unique j in the interval [2,m] such that α(j)=∅.
Here, α(1) must avoid 132 (or α(j) contains the 4 of 1324), and
α(j) must avoid both 213 ≈324 (or i1 is the 1 of a 1324) and 2431.
Also, π(j)=⋯=π(m)=∅ (to avoid 3142). So π(1),…,π(j−1),α(1) each contribute C(x) and we get a contribution of
x^{m}C(x)^{j}\big{(}K(x)-1\big{)} where K(x)=1−3x+x21−2x is the generating function
for {213,2431} avoiders [25, Seq. A001519].
Thus,
[TABLE]
Summing over m≥2 and using the expressions for G1(x) and G0(x), we obtain
[TABLE]
Solving for FT(x) gives the stated generating function after simplification.
∎
3.22. Case 195: {1324,2341,1243}
In this subsection, let A=1−3x+x21−2x and B=(1−x)(1−x−x2)1−x+x3 denote the generating functions for F{213,2341}(x) and F{132,213,2341}(x) (they can be derived from results in [23]). The first few lemmas refer to permutations with exactly 2 left-right maxima.
Lemma 63**.**
The generating function Jm(x) for permutations of the form
[TABLE]
with 2 left-right maxima satisfies
[TABLE]
Moreover, if J(x,w)=∑m≥1Jm(x)wm−1, then
[TABLE]
Proof.
Let us write an equation for Jm(x). Let π=(n−m−1)α(m+1)nα(m−1)(n−1)⋯α(0)(n−m)∈Sn(T). If n=m+2 then we have a contribution of xm+2. Otherwise, we can consider the position of the letter n−m−2. If the letter n−m−2 belongs to α(s) with s=1,2,…,m, then there is no letter smaller than n−m−2 on its left side (otherwise, π contains 1243), so we have a contribution of xm+1−sJs. So, we can assume that the letter n−m−2 belongs to α(0), that is, α(0)=γ(n−m−2)γ′. Since π avoids 1324, we have α(m)⋯α(1)γ>γ′. In the case γ′=∅ then we have a contribution of xJm(x). Otherwise, since π avoids 2341, we have that α(m)⋯α(1)γ is decreasing and γ′ avoids {132,2341}, which gives a contribution of (1−x)m+1xm+3(A−1). Hence, by adding all contributions, we obtain
[TABLE]
Multiplying by wm−1 and summing over m≥1, we complete the proof.
∎
Lemma 64**.**
The generating function Km(x) for T-avoiders of the form
[TABLE]
with 2 left-right maxima satisfies K1(x)=J1(x)+K2(x) and for all m≥2,
Let us write an equation for Km(x). Clearly, K1(x)=J1(x)+K2(x). Let m≥2, and let π=iπ′nα(1)(i+m)⋯α(m)(i+1)∈Sn(T) with 2 left-right maxima. If i=n−m−1 then we have a contribution of Jm(x) (see Lemma 63). Otherwise, since π avoids 1243 and 1324, the letter i+m+1 belongs to either α(1) or α(2). The former case gives contribution of Km+1(x). The latter case gives a contribution of (1−x)(1−2x)(1−x−x2)xm+2, where the proof details are left to the reader. Thus, for all m≥2,
The generating function M(x) for permutations of the form iπ′nπ′′(i+1)π′′′(i+2)∈Sn(T) with 2 left-right maxima is given by
[TABLE]
Proof.
Let us write an equation for M(x). By Lemma 64, the case π′′′=∅ gives a contribution of xK1(x). Otherwise, π′π′′>π′′′ and the subsequence of π′π′′ consisting of letters smaller than i is decreasing, and the subsequence of π′′ consisting of letters greater than i is also decreasing, and π′′′ is a nonempty permutation that avoids {132,2341}. Hence, we have a contribution of (1−x)(1−2x)x4(A−1). By adding all the contributions, we complete the proof.
∎
Lemma 66**.**
The generating function N(x) for permutations of the form iπ′nπ′′(i+2)π′′′(i+3)(i+1)∈Sn(T) with 2 left-right maxima is given by
[TABLE]
Proof.
Let us write an equation for N(x). If π′′′=∅, then, as in the previous lemma, the subsequence of π′π′′ consisting of letters smaller than i is decreasing, and the subsequence of π′′ consisting of letters greater than i is also decreasing. Thus, we have a contribution of (1−x)(1−2x)x6. If π′′′=∅, we get a contribution of (1−x)(1−2x)(1−x−x2)x5(x3+(1−x)2) (proof omitted). Adding the two contributions, we complete the proof.
∎
Lemma 67**.**
Define the following generating functions for T-avoiders with 2 left-right maxima:
•
Bm(x)* for permutations of the form iπ′nπ′′∈Sn(T) such that π′′ contains the subsequence (i+m)(i+m−1)⋯(i+1);*
•
Hm(x)* for permutations of the form*
[TABLE]
•
Em(x)* for permutations of the form*
[TABLE]
such that the letter (n−m−2)
belongs to α(1);
•
Dm(x)* for permutations of the form*
[TABLE]
such that the letter (n−m−2)
belongs to α(0).
Then for all m≥1,
[TABLE]
with t=1/(1−x), where M(x),N(x),Jm(x) are defined in the three preceding lemmas.
Proof.
(a) To write an equation for B1(x), suppose π=iπ′nπ′′∈Sn(T) with 2 left-right maxima. If n=i+1, then we have H1(x). Otherwise, the letter i+2 appears either on the left side of i+1, which leads to a contribution of B2(x), or on right side of i+1. In the latter case π can be written as
π=iπ′nγ(i+1)γ′(i+2)(i+3)⋯(i+s) with s≥2, which gives a contribution of 1−x1M(x), see Lemma 65. Thus, B1(x)=H1(x)+B2(x)+1−x1M(x).
Let us write an equation for Bm(x) with m≥2. As in the case B1(x), we obtain
(b) The recurrence for the generating function Hm(x) can be obtained by using very similar techniques as in Lemma 63 using the definitions of Em(x) and Dm(x).
(c) By mapping each permutation (n−m−1)α(m+1)nα(m)(n−1)⋯α(1)(n−m)α(0) to
(n−m−2)α(m+1)nα(m)(n−1)⋯α(1)(n−m−1)α(0), we obtain the required relation.
(d) Let us write an equation for Em(x). Let
[TABLE]
with 2 left-right maxima and α(1)=γ(n−m−2)γ′ where the letter n−m−2 belongs to α(1). The contribution of the case
α′=α(m+1)⋯α(2)γ=∅ is xm+2(FT(x)−1). So we can assume that α′=∅. If γ′=∅ then by considering whether
γ′ is decreasing or not, we get contributions of 1−xxm+4(tm+1−1)B and xm+1(tm+1−1)(A−1/(1−x)), respectively. Thus, we can assume that α′=∅ and β=∅, and by considering either α(0) is empty or not, we obtain the contributions x(Jm(x)−xm+2) and xm+3(tm+1−1)(B−1), respectively. Sum all contributions to complete the proof.
∎
Define H(x,w)=∑m≥1Hm(x)wm−1 and E(x,w)=∑m≥1Em(x)wm−1. Thus, this recurrence can be written as
[TABLE]
which implies
[TABLE]
This type of functional equation can be solved systematically using
the kernel method (see, e.g., [15] for an exposition) by taking w=C(x). We find
[TABLE]
Lemma 67(d) and some mathematical programming yields
[TABLE]
which leads to
[TABLE]
Now we are ready to give the main result of this subsection.
Theorem 68**.**
Let T={1324,2341,1243}. Then FT(x)=
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
First, we treat G2(x). Each π=iπ′nπ′′∈Sn(T) with 2 left-right maxima can be decomposed as either (n−1)π′nπ′′ or iπ′nπ′′ with 1≤i<n−2 with respective contributions to G2(x) of x(FT(x)−1) and B1(x). From Lemma 67(a) we have
[TABLE]
which implies
[TABLE]
where M(x) and N(x) are given in Lemmas 65 and 66, respectively, and
[TABLE]
where
[TABLE]
and H(x,0) and E(x,C(x)) are given in (5) and (6).
Now, let us write a formula for G3(x). Let π=i1π′i2π′′i3π′′′∈Sn(T) with 3 left-right maxima. then i1<π′′′<i2 and π′>π′′. By considering whether π′,π′′ are empty or not, we obtain the contributions x3A (π′=π′′=∅), x3(A−1)/(1−x) (π′=∅,π′′=∅), x3B/(1−x)2 (π′=∅,π′′=∅) and [math]. Thus,
[TABLE]
Now, let us write a formula for Gm(x) with m≥4. Let i1π(1)⋯imπ(m)∈Sn(T) with m left-right maxima. Then π(3)=⋯=π(m)=∅ and π(1)>π(2). By considering whether π(2) is empty or not, we have
[TABLE]
Summing over m≥0, we obtain
[TABLE]
Substituting the formula for G2(x), we get
[TABLE]
where M(x) and N(x) are given in Lemmas 65 and 66, respectively, and
H(x,0) and E(x,C(x)) are given in (5) and (6), respectively.
Solving for FT(x) and simplifying the result, we complete the proof.
∎
3.23. Case 210: {1243,1324,2431}
Theorem 69**.**
Let T={1243,1324,2431}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
For Gm(x) with m≥2, suppose π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324, π(s)<i1 for all s=1,2,…,m−1. Since π avoids 1243, π(m)<i2.
First, suppose π(m) has a letter greater than i1 and consider two cases:
•
all the letters between i1 and i2 in π(m) are increasing.
Here, π(1)⋯π(m−1) is decreasing (π avoids 1243), and π(m)
can be decomposed as (π avoids 2431) α(i1+1)⋯(i2−1) and,
writing π(1)⋯π(m−1) as jsjs−1⋯j1, α
can be further decomposed as α(0)α(1)⋯α(s) with
α(0)<j1<α(1)<⋯<js<α(s)<i1, where α(0) avoids 132
and α(i) is an increasing subword for all i=1,2,…,s.
In short, π has the schematic form shown in the Figure 4 where ↓ denotes decreasing, ↑ denotes increasing, and blank regions are empty.
If A is empty, then so is C and, in any case, the position of im is determined by AC.
From Figure 4, π can be written as i1A1imBCD,
where A1 consists of A and the left-right maxima i2,…,im−1,
and π is determined by St(A1C), St(B), and St(D). The latter two contribute C(x) and 1−xx respectively. Now AC is a {132,231}-avoider and St(A1C) can be viewed as a {132,231}-avoider decorated with m−2 dots (for
i2,…,im−1) placed in the spaces between its entries but not after the first ascent. For example, the 4 boxes shown accept dots in □5□2□1□34. The generating function for such constructs with M dots, each dot counting as a letter, can be shown to be xML(x)M+1 where L(x)=1−2x1−x is the generating function for {132,231}-avoiders.
Thus, with M=m−2, the contribution of this case is 1−xxm+1L(x)m−1C(x).
•
the letters between i1 and i2 in π(m) do not form an increasing sequence, where the sequence π(1)⋯π(m−1)α is decreasing, and π(m)=αβ with α<i1<β<i2. Thus, we have a contribution of \frac{x^{m}}{(1-x)^{m}}\big{(}L(x)-\frac{1}{1-x}\big{)}.
Hence, for all m≥2,
[TABLE]
where Hm(x) is the generating function for those permutations π∈Sn(T) with π(m)<i1 (in other words i1=n+1−m).
Now let us write an equation for Hm(x).
Let π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima such that ij=n+j−m for all j. If π(1)=∅ then we have a contribution of xHm−1(x). Otherwise, if we assume that π(1) has exactly s left-right maxima, it is not hard to see that those permutations are bijection with the permutations with exactly m+s−1 left-right maxima. Thus,
[TABLE]
with H1(x)=G1(x) (by definitions).
Define G(x;t)=∑m≥0Gm(x)tm and H(x;t)=∑m≥1Hm(x)tm. Note that G(x;1)=FT(x). Hence, the above recurrences can be written as
[TABLE]
Hence,
[TABLE]
This equation can be solved by the kernel method (see, e.g., [15] for an exposition) taking t=C(x) and leads, after simplification, to our theorem.
∎
3.24. Case 211: {1234,1324,2341}
All three patterns contain 123 and so T-avoiders consist of 123-avoiders together with T-avoiders that contain a 123. The former are counted by C(x). To count the latter, let abc at positions i,j,k be the leftmost 123 pattern (smallest i, then smallest j, then smallest k). This 123 pattern divides the permutation diagram into rectangular regions as shown
in Figure 5 where the bullets represent the leftmost 123.
Shaded regions are empty for the indicated reason, where (min) refers to the minimal, that is, leftmost property of the 123 pattern, and unshaded regions are labeled
α,α′ etc. A down arrow (↓) indicates necessarily decreasing entries, again for the indicated reason.
Also, αα′ avoids 123 (or the middle bullet ends a 2341), while β lies to the left of β′ and γ>γ′
(both to avoid 1324). We consider 4 cases according as β′,γ′ are empty or not:
•
β′,γ′ both empty. Here, St(αaα′c) is a a 123-avoider τ of length ≥2 with last entry >1. Say τ has length n, with 1 in position i<n and last entry ℓ≥2. Now β serves to “decorate” τ with 0 or more dots inserted arbitrarily in the ℓ−1 spaces between 1 and ℓ. Similarly, γ amounts to 0 or more dots inserted in the n−i spaces between i and n. Thus, the contribution of this case is
[TABLE]
where C(n,i,ℓ) is the number of 123-avoiders of length n with 1 in position i and last entry ℓ.
It is known that C(n,i,ℓ)=n+i−ℓn−2−i+ℓ(n−1n+i−ℓ), a generalized Catalan number, and the displayed sum evaluates to the compact expression x3C(x)5.
•
β′ empty, γ′ nonempty. Here, γ′ avoids 123 and 132 and so the contribution is x3C(x)5(L−1)=x3C(x)51−2xx where L=1−2x1−x is the generating function for {123,132}-avoiders.
•
β′ nonempty, γ′ empty. Here, β′ avoids 123 and 213 and so the contribution is x3C(x)5(L−1)=x3C(x)51−2xx since L=1−2x1−x is also the generating function for {123,213}-avoiders.
•
β′,γ′ both nonempty. In case γ′ is decreasing, the only restriction on β′ is to avoid 123 and 213 (since γ′=∅). So γ′ contributes 1−xx and β′ contributes L−1. In case γ′ is not decreasing, β′is decreasing (to avoid 1234) and γ′ contributes L−1−x1 while β′ contributes 1−xx. So, β′,γ′ together contribute \frac{x}{1-x}(L-1)+\big{(}L-\frac{1}{1-x}\big{)}\frac{x}{1-x}=\frac{x^{2}}{(1-x)^{2}(1-2x)} for an overall contribution of x3C(x)5(1−x)2(1−2x)x2
Hence, summing over 123-avoiders and the 4 cases,
[TABLE]
Simplifying this expression, we have established
Theorem 70**.**
Let T={1234,1324,2341}. Then
[TABLE]
3.25. Case 212: {1324,2413,2431}
Note that each pattern contains 132.
Theorem 71**.**
Let T={1324,2413,2431}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
For m=2,π has the form iαnβ with α<i, and β cannot contain both
letters >i and <i (2413 and 2431). If β has no letter >i, then i=n−1 and deleting it gives a contribution of x\big{(}F_{T}(x)-1\big{)}. Otherwise, β>i and α avoids 132
(due to n) and β avoids both 213 (due to i) and 2431 and is nonempty,
contributing x^{2}C(x)\big{(}K(x)-1\big{)}, where K(x)=1−3x+x21−2x
is the generating function for {213,2431}-avoiders [25, Seq. A001519].
So G_{2}(x)=x\big{(}F_{T}(x)-1\big{)}+x^{2}C(x)\big{(}K(x)-1\big{)}.
Now let us write equation for Gm(x) for m≥3.
Suppose π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324, we have π(s)<i1 for all s=1,2,…,m−1. Since π avoids 2413 and 2431, either π(2) is empty or π(m) has no letter between i1 and i2 (or both). Thus, we have three cases
•
π(2)=∅ and π(m) has no letter between i1 and i2. Here, i2 and its position is determined by the rest of the avoider, and we have a contribution of xGm−1(x).
•
π(2)=∅ and π(m) has a letter between i1 and i2.
Here, π(3)=⋯=π(m−1)=∅ (or they would contain the 1 of a 2413),
and since π avoids both 2413 and 2431, i1<π(m).
Additionally, π(m) is not empty and avoids both 213 (or i1 is the 1 of a 1324)
and 2431, and π(1) avoids 132 (or im is the 4 of a 1324).
Hence, the contribution is x^{m}C(x)\big{(}K(x)-1\big{)}.
•
π(2)=∅ and π(m) has no letter between i1 and i2.
Since π avoids 2413, we see that i1>π(1)>π(s) for all s=2,3,…,m−1.
Also, π(1) avoids 132 and i2=i1+1. So π can be recovered from St(i1π(1))
and St(i2π(2)⋯imπ(m)), giving respective contributions of
xC(x) and Gm−1(x)−Gm−1′(x) where Gm−1′(x) counts the π(2) empty case, that is, Gm′(x) is the generating function for T-avoiders with m left-right maxima in which the first letter is smaller than the second.
Thus, the contribution is xC(x)\big{(}G_{m-1}(x)-G^{\prime}_{m-1}(x)\big{)}.
Hence, for all m≥3,
[TABLE]
For G2′(x), a T-avoider has the form π=inπ′. If π′ is empty,
the contribution is x2. So suppose π′=∅.
If i=n−1 then we have x2(FT(x)−1), but if i<n−1, then, since π avoids 2413 and 2431
we see that i=1 and π=1nπ′ with π′=∅ avoiding 213 and 2431, giving a contribution of x2(K(x)−1). So
[TABLE]
For Gm′(x) with m≥3, using the same three cases as above, we find by similar arguments
the recurrence
for m≥3, with G2,G1,G0 as above.
Summing over m≥3, we obtain
[TABLE]
Solving for FT(x) completes the proof.
∎
3.26. Case 213: {2431,1324,1342}
Theorem 72**.**
Let T={2431,1324,1342}. Then
[TABLE]
Proof.
Let Gm(x) be the generating function for T-avoiders with m
left-right maxima. Clearly, G0(x)=1 and G1(x)=xFT(x).
First, we find a formula for G2(x). Since π
avoids 2431, we can write a T-avoider π with 2 left-right maxima as π=iαnβγ where αβ<i<γ, and γ avoids both 132 and 231 (or i is the 1 of a 1324 or 1342). If γ=∅, then π=n−1αnβ and, by deleting n−1, the contribution is x\big{(}F_{T}(x)-1\big{)}. Now suppose γ=∅
and let H denote the contribution of such permutations to G2(x).
If αβ=∅, then π=1nγ with γ=∅ and the contribution is
x2(J−1) where J=1−2x1−x is the generating function for {213,231}-avoiders. Henceforth, αβ=∅, and so i>1 and consider 3 cases:
i−1∈α. Here, π=iα′i−1α′′nβγ where α′>α′′β(1324) and α′ avoids 132. So the contribution is xC(x)H.
i−1 is the last letter of β. Deleting i−1 gives a one-size-smaller avoider and the
contribution is xH.
i−1∈β but is not the last letter of β. Here α=∅ (2431) and π=inβ′i−1β′′γ where β′′=∅, β′>β′′ (1324) and β′,β′′ each avoid 132. So, with C(x),C(x)−1,J−1 respectively from β′,β′′,γ, the contribution
is x^{3}C(x)\big{(}C(x)-1\big{)}(J-1).
Thus, H=x^{2}(J-1)+xC(x)H+xH+x^{3}C(x)\big{(}C(x)-1\big{)}(J-1) and so
[TABLE]
which leads to
[TABLE]
Now let m≥3 and let us write an equation for Gm(x).
Let π=i1π(1)i2π(2)⋯imπ(m)∈Sn(T) with exactly m left-right maxima. Since π avoids 1324 and 1342, we can express π as
[TABLE]
such that i1>π(1)>⋯>π(m−2)>αβ and im−1<γ. Note that π avoids T if and only if π(j) avoids 132 for j=1,2,…,m−2 and the standard form of i1αnβγ is a T-avoider with two left-right maxima. Hence,
[TABLE]
By summing for m≥2 and using the expressions for G1(x) and G0(x), we obtain
[TABLE]
with solution as stated.
∎
For the next two cases we set
aT(n)=∣Sn(T)∣ and let aT(n;j1,j2,...,js) denote the number of permutations in Sn(T) whose first s letters are j1j2...js.
3.27. Case 231: {1324,1342,2341}
Set b(n;j)=aT(n;j,j+1).
Lemma 73**.**
For 1≤j≤n−2,
[TABLE]
and aT(n;n)=aT(n;n−1)=aT(n;n−2)=aT(n−1).
Proof.
The initial conditions aT(n;n)=aT(n;n−1)=aT(n−1) easily follow from the definitions. For 1≤j≤n−2, we have
[TABLE]
So assume 1≤j≤n−2 and let π=jiπ′ be a member of Sn(T). We consider several cases on i. Since π avoids 1324 and 1342, we have that either 1≤i≤j+1 or i=n. If
i=n, then π avoids T if and only if jπ′ avoids T, so aT(n;j,n)=aT(n−1,j). So,
[TABLE]
Let 1≤i≤j−1. If π avoids T then iπ′ avoids T. On other hand, if π contains 1324 (resp. 1342), then iπ′ contains 1324 (resp. 1324). Also, if π contains 2341 where j does not occur in the corresponding occurrence, then iπ′ contains 2341. Thus, we assume that π contains jabc with c<j<a<b. If c<i, then iπ′ contains 2341, otherwise iπ′ contains 1342. Therefore, iπ′ avoids T if and only if π avoids T, which implies aT(n−1;j,i)=aT(n;i). Hence,
[TABLE]
as required.
∎
Define AT(n;v)=∑j=1naT(n;j)vj−1 and B(n;v)=∑j=1n−2b(n;j)vj−1. Then Lemma 73 can be written as
[TABLE]
with AT(0;v)=AT(1;v)=1.
Define AT(x,v)=∑n≥0AT(n;v)xn and B(x,v)=∑n≥3B(n;v)xn. Then, the above recurrence can be written as
[TABLE]
Lemma 74**.**
[TABLE]
Proof.
Let π=j(j+1)π′∈Sn(T) with 1≤j≤n−2. Since π avoids 2341, we can express π as π=j(j+1)αβ with α<j<j+1<β. Note that π avoids T if and only if α avoids both 132,2341 and β avoids both 213,231. By [25, Seq. A001519] we have that F{132,2341}(x)=1−3x+x21−2x, and by [24] we have that F{213,231}(x)=1−2x1−x, yielding
[TABLE]
which leads to B(x,v)=(1−3xv+x2v2)(1−2x)x3(1−2xv), as required.
∎
To solve the preceding functional equation, we apply the kernel method (see, e.g., [15] for an exposition) and take v=21+1−4x=1/C(x). Then
[TABLE]
After simplification, this gives the following result.
Theorem 75**.**
Let T={1324,1342,2341}. Then
[TABLE]
3.28. Case 241: {1324,1243,1234}
Set b(n;j)=aT(n;j,n−1).
Lemma 76**.**
For 1≤j≤n−2,
[TABLE]
and aT(n;n)=aT(n;n−1)=aT(n;n−2)=aT(n−1),
b(n;n)=b(n;n−2)=aT(n−2) and b(n;n−1)=0.
Proof.
The initial conditions aT(n;n)=aT(n;n−1)=aT(n;n−2)=aT(n−1),
b(n;n)=b(n;n−2)=aT(n−2) and b(n;n−1)=0 easily follow from the definitions. For 1≤j≤n−2, we have
[TABLE]
So assume 1≤j≤n−3 and let π=jiπ′ be a member of Sn(T). We consider several cases for i. If i=n, then aT(n;j,n)=aT(n−1;j). If 1≤i<j then jiπ′ (respectively, jnπ′) avoids T if and only if iπ′ (respectively, jπ′) avoids T, so aT(n;j,i)=aT(n−1;i) for all i=1,2,…,j−1 and aT(n;j,n)=aT(n−1;j). Since π avoids 1234 and 1243, we see that either i<j or j>n−2, and aT(n;j,n)=aT(n−1;j). Thus,
[TABLE]
which completes the proof of the first recurrence relation.
For the second relation, by similar reasons, we have
[TABLE]
Clearly, aT(n;j,n−1,n)=aT(n−2,j). The permutation j(n−1)(n−2)π′′ contains 1324, so aT(n;j,n−1,n−2)=0. The permutation j(n−1)iπ′′ with j+1≤i≤n−3 contains either 1234 or 1243, so aT(n;j,n−1,i)=0.
Thus,
[TABLE]
Let j(n−1)iπ′′∈Sn. If i(n−1)π′′ contains a pattern in T then j(n−1)iπ′′ contains the same pattern in T. Now, suppose i(n−1)π′′ avoids T, so if π=j(n−1)iπ′′ contains 1234 or 1243, then i(n−1)π′′ contains 1234 or 1243, which implies that π avoids 1234 and 1243. If π contains 1324, then we can assume that 1324 occurs in
π as a subsequence jabc with j<b<a<c, otherwise i(n−1)π′ contains 1324. Also, we can assume that b=n−1 which gives c=n, otherwise iabc occurs in i(n−1)π′′. Since j≤n−3, π has an element d such that either j<d<b or b<d<n−1. Thus, i(n−1)π′′ contains either idba or ibda or ibad, that is i(n−1)π′′ does not avoid T. Therefore, π avoids T. Hence,
[TABLE]
for all i=1,2,…,j−1, which implies
[TABLE]
This completes the proof.
∎
Define AT(n;v)=∑j=1naT(n;j)vj−1 and B(n;v)=∑j=1nb(n;j)vj−1. Then Lemma 76 can be written as
[TABLE]
with AT(0;v)=AT(1;v)=1, AT(2;v)=1+v, B(0;v)=B(1;v)=0 and B(2;v)=v.
Define AT(x,v)=∑n≥0AT(n;v)xn and B(x,v)=∑n≥0B(n;v)xn. Then, the above recurrence can be written as
[TABLE]
By multiplying (12) by v2(1−v)2(1−1−vxv) and (11) by v2(1−v)2, then adding the results, we obtain
[TABLE]
where K(x,v)=xv(1−v2)+x2(1−3v+v2)−v2(1−v)2.
To solve the preceding functional equation, we apply the kernel method (see, e.g., [15] for an exposition) and take
[TABLE]
which satisfies K(x;v+)=K(x,v−)=0. Hence, since FT(x)=AT(x,1), we obtain the following result.
Theorem 77**.**
Let T={1324,1243,1234}.
Then FT(x) is given by
[TABLE]
Acknowledgement
We thank Richard Mather for pointing out some inaccuracies in the previous version.
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