proved in | |||
Symmetric | Thm. 3.9 | ||
Thm. 3.10 | |||
|
Hankel | Thm. 3.13 | |
Lower triangular | Thm. 3.15 | ||
Thm. 3.15 | |||
Thm. 3.16 |
Citation: Weijie Liu, Liu Cong, Hongli Yuan, Jinshui Yang. The mechanism of kaolin clay flocculation by a cation-independent bioflocculant produced by Chryseobacterium daeguense W6[J]. AIMS Environmental Science, 2015, 2(2): 169-179. doi: 10.3934/environsci.2015.2.169
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Let
The notion of permutation pattern was introduced by Knuth [21,pp. 242-243] in 1968, but studied intensively and systematically for the first time by Simion and Schmidt [28] in 1985. Ever since then, it has become an active and prosperous research subject. The reader is referred to two book expositions, [5,Chapters 4 and 5] and [20], on this topic, as well as the numerous references therein. In the early 1980s, Herbert Wilf posed the problem of identifying equirestrictive sets of forbidden patterns. Let
In this paper, we will restrict ourselves to the case where
$ {\mathrm{DES}}(\pi): = \{i\in[n-1]:\pi(i) < \pi(i+1)\}, $ |
called the descent set of
$ {\mathsf{des}}(\pi): = |{\mathrm{DES}}(\pi)|\quad \text{and}\quad {\mathsf{iar}}(\pi): = \min({\mathrm{DES}}(\pi)\cup\{n\}), $ |
called the ${\boldsymbol des}$cent number and the ${\boldsymbol i}$nitial ${\boldsymbol a}$scending ${\boldsymbol r}$un of
$ A_n(t): = \sum\limits_{\pi\in{\mathfrak{S}}_n}t^{{\mathsf{des}}(\pi)}. $ |
Another statistic highlighted in our study is
$ {\mathsf{comp}}(\pi): = |\{i: \forall j\leq i, \: \pi(j)\leq i\}|. $ |
It is equal to the maximum number of components (see [1,7,8]) in an expression of
$ \sum\limits_{n\geq1}f(n)z^n = 1-\frac{1}{\sum_{n\geq0}n!z^n}. $ |
Thus, any statistic equidistributed with
For a (possibly set-valued) statistic
$ |{\mathfrak{S}}_n(P)^{{\mathsf{st}}}| = |{\mathfrak{S}}_n(Q)^{{\mathsf{st}}}|, $ |
meaning that for a fixed value of
Some highlights of our results will be outlined below. Before stating them, we need to recall some classical permutation statistics. For a permutation
$ LMAX(π):={π(i)∈[n]:π(j)<π(i),∀1≤j<i}andLMAXP(π):={i∈[n]:π(j)<π(i),∀1≤j<i}, $
|
the set of values and positions of the left-to-right maxima of
$ {\mathrm{DESB}}(\pi): = \{\pi(i+1)\in[n-1]: i\in{\mathrm{DES}}(\pi)\}, $ |
which is another set-valued extension of
The first one of our main results concerns a single pattern of length
Theorem 1.1. For every
(i) the two triples
(ii) the two quadruples
(iii) the quadruples
The result on the symmetry of
Next, Claesson, Kitaev and Steingrímsson [20,Thm 2.2.48] constructed a bijection between separable permutations of length
Corollary 1.2. The double Comtet statistics
We take the opportunity to announce the following refinement of Corollary 1.2, which appears in a separate article.
Theorem 1.3 ([16]). There exists an involution on
$ ∑π∈Sn(2413,3142)scomp(π)tiar(π)xLMAX(π)yDESB(π)=∑π∈Sn(2413,3142)siar(π)tcomp(π)xLMAX(π)yDESB(π), $
|
where
The proof of Theorem 1.1 provided in Sect. 3 is via two involutions on permutations that actually imply the even stronger symmetric phenomenon, namely the corresponding distribution matrices are Hankel; see Theorems 3.13 and 4.2. The proof of Theorem 1.3 is based on a combinatorial bijection on the so-called di-sk trees introduced in [17]. This bijection will also provide an alternative approach to Theorem 1.1(ii). The details will be given in [16].
Remark 1. Rubey's bijective proof of a slight modification (see Theorem 3.1) of Theorem 1.1(i) is via Dyck paths and the proof of Theorem 1.3 that will appear in [16] is based on di-sk trees. Our bijective and unified proof of Theorem 1.1(i)(ii), constructed directly on permutations, provides more insights into the symmetry of the double Comtet statistics, and therefore, it seems more likely to be extended to deal with such equidistributions over other classes of pattern-avoiding permutations.
Our third main result shows how
Recall that a polynomial in
$ \{t^k(1 + t)^{n-2k}\}_{0\leq k\leq n/2} $ |
with non-negative coefficients. Many polynomials arising from combinatorics and discrete geometry have been shown to be
$ A_n(t) = \sum\limits_{\pi\in{\mathfrak{S}}_n}t^{{\mathsf{des}}(\pi)} = \sum\limits_{k = 0}^{\lfloor\frac{n-1}{2}\rfloor}|\Gamma_{n, k}|t^k(1 + t)^{n-1-2k}, $ |
where
$ Sn(t):=∑π∈Sn(2413,3142)tdes(π)=⌊n−12⌋∑k=0|Γn,k(2413,3142)|tk(1+t)n−1−2k. $
|
(1) |
In a recent work [24] of Lin and Kim, they proved that
Theorem 1.4. For
$ ∑π∈Sn(2413,3142)tdes(π)xdd(π)yiar(π)=∑π∈Sn(2413,4213)tdes(π)xdd(π)yiar(π). $
|
(2) |
Theorem 1.4 refines Wang's equidistribution [30,Thm. 1.5] by the Comtet statistic
Besides the above three main results, we will also calculate the joint distribution of
proved in | |||
Symmetric | Thm. 3.9 | ||
Thm. 3.10 | |||
|
Hankel | Thm. 3.13 | |
Lower triangular | Thm. 3.15 | ||
Thm. 3.15 | |||
Thm. 3.16 |
proved in | |||
Hankel | Thm. 4.2 | ||
Thm. 4.2 | |||
Diagonal | Thm. 4.3 | ||
Thm. 4.4 | |||
Lower triangular | Thm. 4.6 | ||
Thm. 4.6 | |||
Upper triangular | Thm. 4.6 | ||
Lower triangular | Thm. 4.8 | ||
Thm. 4.8 | |||
Lower triangular | Thm. 4.9 | ||
No pattern | Thm. 4.10 | ||
Thm. 4.11 | |||
Thm. 4.11 | |||
Thm. 4.11 | |||
$ |
Ultimately zero | Thm. 4.11 |
The rest of this paper is organized as follows. In Section 2, we review some notation and terminology and prove two general lemmas concerning the direct sum operation of permutations. The classification of refined Wilf-equivalences for a single pattern of length
For a given permutation
● its reversal
● its complement
● its inverse
One thing we would like to point out, before we barge into classifying
Observation 1.
On the other hand, the statistic
Observation 2. The two mappings
Let
$ S(P)des,iar,comp(t,r,p;z):=∑n≥0∑π∈Sn(P)tdes(π)riar(π)pcomp(π)zn. $
|
(3) |
Most of the time we suppress the superindices
We also need the following operations on permutations.
Definition 2.1. For a word
There are two fundamental operations, called direct sum and skew sum, to construct a bigger permutation from two smaller ones. The direct sum
$ (\pi\oplus\sigma)_i = {πi,for i∈[1,k];σi−k+k,for i∈[k+1,k+l] $
|
and
$ (\pi\ominus\sigma)_i = {πi+l,for i∈[1,k];σi−k,for i∈[k+1,k+l]. $
|
For instance, we have
Proposition 2.2. A permutation is separable if and only if it can be built from the permutation
Definition 2.3. A nonempty permutation which is not the direct sum of two nonempty permutations is called indecomposable. Let
For instance,
Let
$ F_P(t_1, t_2, \ldots;z): = 1+\sum\limits_{n\geq1}z^n\sum\limits_{\pi\in{\mathfrak{S}}_n(P)}\prod_it_i^{{\mathsf{st}}_i(\pi)} $ |
and
$ I_P(t_1, t_2, \ldots;z): = \sum\limits_{n\geq1}z^n\sum\limits_{\pi\in \mathcal{I}_n(P)}\prod_it_i^{{\mathsf{st}}_i(\pi)}. $ |
We have the following general lemma regarding the direct sum decomposition of permutations, which is useful when considering the refinement of Wilf-equivalence by
Lemma 2.4. Let
1. We have the following functional equation:
$ FP(q)=11−qw+q(IP(t,t′)−w)(1−qIP(t,1))(1−qw), $
|
(4) |
where
$ FP(q):=FP(q,t1,…,t′1,…;z)andIP(t,t′):=IP(t1,…,t′1,…;z) $
|
are the generating functions with respect to
2. If
Proof. Note that if
$ \pi \text{ is $\sigma$-avoiding}\Longleftrightarrow \tau_i\text{ is $\sigma$-avoiding for each $i$}. $ |
Therefore, with respect to totally
$ FP(q)=1+∑k≥1qk(wk+k−1∑i=0wi(IP(t,t′)−w)IP(t,1)k−1−i)=11−qw+IP(t,t′)−wIP(t,1)−w(qIP(t,1)1−qIP(t,1)−qw1−qw), $
|
which becomes 4 after simplification.
In view of 4, the following three statements are equivalent:
(i)
(ii)
(iii)
For example, to see that (i) implies (ii), we first set
The following general lemma indicates that for a collection of indecomposable patterns, say
Lemma 2.5. Let
$ |{\mathfrak{S}}_n(P)^{{\mathsf{st}}', {\mathsf{comp}}, {\mathsf{st}}_1, {\mathsf{st}}_2, \ldots}| = |{\mathfrak{S}}_n(P)^{{\mathsf{comp}}, {\mathsf{st}}', {\mathsf{st}}_1, {\mathsf{st}}_2, \ldots}|. $ |
In particular, if
Proof. Let
$ FP(r,s)=11−rsw+r(IP(s)−sw)(1−rIP(1))(1−rsw), $
|
(5) |
where
$ \frac{1}{1-sw}+\frac{I_P(s)-sw}{(1-I_P(1))(1-sw)} = \frac{1}{1-sw}+\frac{sI_P(1)-sw}{(1-sI_P(1))(1-sw)}. $ |
Solving this equation gives
$ I_P(s)-sw = \frac{s(1-I_P(1))(I_P(1)-w)}{1-sI_P(1)}. $ |
Plugging this into 5 results in
$ FP(r,s)=1−rsw+(rsw+rs−r−s)IP(1)(1−rIP(1))(1−sIP(1))(1−rsw), $
|
(6) |
which is symmetric in
In this section, we deal with all patterns
$ \{213, 312, 321\}, \; \{132, 231\}, \; \text{and }\{123\}. $ |
While the
$ \{231, 312, 321\}, \; \{132, 213\}, \; \text{and }\{123\}. $ |
For the three patterns
Patterns
The pattern
$ {\mathsf{ldes}}(\pi): = \max(\{0\}\cup{\mathrm{DES}}(\pi)) $ |
be the position of the ${\boldsymbol l}$ast ${\boldsymbol des}$cent of
Theorem 3.1 (Rubey [27]). There exists an involution on
$ ∑π∈Sn(321)scomp(π)tn−ldes(π−1)xLMAXP(π)=∑π∈Sn(321)sn−ldes(π−1)tcomp(π)xLMAXP(π). $
|
(7) |
We explain here why Theorem 3.1 is equivalent to our Theorem 1.1 (i) up to the elementary transformation
$ n-{\mathsf{ldes}}(\pi^{\mathrm{rc}}) = {\mathsf{iar}}(\pi)\quad\text{and}\\ {\mathrm{LMAXP}}(\pi) = {\mathrm{LMIN}}(\pi^{-1}) = \overline{{\mathrm{LMAX}}((\pi^{-1})^{\mathrm{rc}})}, $ |
where
$ ∑π∈Sn(321)scomp(π)tn−ldes(π−1)xLMAXP(π)=∑(π−1)rc∈Sn(321)scomp((π−1)rc)tn−ldes(πrc)xLMAXP((π−1)rc)=∑(π−1)rc∈Sn(321)scomp(π)tiar(π)x¯LMAX(π)=∑π∈Sn(321)scomp(π)tiar(π)x¯LMAX(π). $
|
Therefore, equidistribution (7) is equivalent to Theorem 1.1 (i).
In Corollary 3.7 we present a bijection, say
$ π=251634↦ω(π)=215634,butπ=251634↦˜ω(π)=215364. $
|
In view of Lemma 2.4 (2),
Theorem 3.2. For each
$ (LMAX,LMAXP,iar,comp)π=(LMAX,LMAXP,iar,comp)σ. $
|
(8) |
Sitting in the heart of our proof of Theorem 3.2, is a certain word composed of positive integers and a symbol
Definition 3.3. Given a nonempty set
$ w_{S, c}: = s_1\underbrace{\diamond\cdots\diamond}_{c_1}s_2\underbrace{\diamond\cdots\diamond}_{c_2}s_3\cdots s_k\underbrace{\diamond\cdots\diamond}_{c_k}. $ |
It is said to be an admissible word with respect to
$
i∑j=1cj≤si−i.
$
|
(*) |
Let
We also need to introduce the counterparts on
Theorem 3.4. There exist two bijections
$ (LMAX,LMAXP,iar,comp)π=(S,SP,ics,equ)wS,c, $
|
(9) |
$ (LMAX,LMAXP,iar,comp)σ=(T,SP,ics,equ)wT,d, $
|
(10) |
where
Proof. Since the constructions for the two bijections
$ S: = {\mathrm{LMAX}}(\pi) = \{\pi(i_1) = \pi(1), \pi(i_2), \ldots, \pi(i_k)\}. $ |
Let
It should be clear that
$ LMAX(ˆπ)=S=LMAX(ˆσ),LMAXP(ˆπ)=SP(wS,c)=LMAXP(ˆσ),iar(ˆπ)=ics(wS,c)=iar(ˆσ),comp(ˆπ)=equ(wS,c)=comp(ˆσ). $
|
Now set
$ \alpha^{-1}(\alpha(\pi)) = \pi, \quad \beta^{-1}(\beta(\sigma)) = \sigma, $ |
so
Proof of Theorem 3.2. Simply set
Remark 2. When composed with the complement map, our bijection
In view of (9), the pair
Definition 3.5. Given an admissible word
$ \sum\limits_{j = 1}^{i}c_j < s_i-i\le \sum\limits_{j = 1}^{i+1}c_j. $ |
For the previous example
Theorem 3.6. For
Proof. Take any
$ d_i = {ci+1if a−1≤i≤ℓ−1,si−i−∑ih=1chif i=ℓ,∑ih=1ch−∑i−1h=1dhif i=ℓ+1,ciotherwise. $
|
We denote
All it remains is to show that
$ c_i = {di−1if a≤i≤ℓ,0if i=a−1,di−1+diif i=ℓ+1,diotherwise. $
|
It is routine to check that
The following result is the restatement of Theorem 1.1 (i) and (ii).
Corollary 3.7. For
Proof. For each permutation
$ ({\mathrm{LMAX}}, {\mathsf{iar}}, {\mathsf{comp}})\:\pi = ({\mathrm{LMAX}}, {\mathsf{comp}}, {\mathsf{iar}})\:\rho. $ |
If
$ \rho = \alpha^{-1}(\psi^k(\alpha(\pi))). $ |
Combining Theorem 3.6 with (9), we verify that
$ LMAX(ρ)=LMAX(π),iar(ρ)=ics(ψk(α(π)))=ics(α(π))−k=iar(π)−k=comp(π),andcomp(ρ)=equ(ψk(α(π)))=equ(α(π))+k=comp(π)+k=iar(π), $
|
as desired. Now both
Finally, applying the bijection
For most of our calculations of the generating function
A square matrix is said to be Hankel if it has constant skew-diagonals. For the next theorem and Theorems 3.13 and 4.2, a key fact utilized by us is that
Lemma 3.8. Suppose
$ M(x,y)=xyx−y(N(x)−N(y)). $
|
(11) |
Proof. The Hankel condition enables us to group together terms along the same skew-diagonal. Noting that
$ M(x,y)=n∑i=1(mi1xiy+mi−12xi−2y2+⋯+m1ixyi)=n∑i=1mi1(xiy+⋯+xyi)=xyx−yn∑i=1mi1(xi−yi)=xyx−y(N(x)−N(y)), $
|
as desired.
Recall the Narayana polynomial
$ N:=N(t;z):=∑n≥0Nn(t)zn=1+(t−1)z−√1−2(t+1)z+(t−1)2z22tz. $
|
(12) |
Theorem 3.9. The generating function of the triple statistic
$ ˜S(312)des,iar,comp(t,r,p)=1−(r+p+tN)z+(rp+(r+p−1)tN)z2(1−rpz)(1−rz−tNz)(1−pz−tNz). $
|
(13) |
Proof. Conditioning on
$ S(t,r,p)=1+rpzS(t,r,p)+rpr−p(˜I(t,r)−˜I(t,p)), $
|
(14) |
where
$ ˜I(t,r):=∑n≥1zn∑π∈Sn(312)π(1)>1,comp(π)=1tdes(π)riar(π)=S(t,r,p)−1−rpzS(t,r,p)p|p=0=rz(˜S(t,r,0)−1). $
|
(15) |
Indeed, the first summand
Next, plugging (15) into (14) yields
$ (r−p)(1−rpz)˜S(t,r,p)=r˜S(t,r,0)−p˜S(t,p,0). $
|
(16) |
Setting
$ (r−p)(1−rpz)˜S(t,r,p)=(r−1)(1−rz)˜S(t,r,1)−(p−1)(1−pz)˜S(t,p,1). $
|
(17) |
It remains to calculate
●
●
Summing up these two cases and noting that
$ rz\tilde{\mathfrak{S}}(t, r, 1) = rz{\mathfrak{S}}(t, r, 1)+({\mathfrak{S}}(t, r, 1)-1)tzN. $ |
Solving for
$ \tilde{\mathfrak{S}}(t, r, 1) = \frac{1}{1-rz-tNz}. $ |
Plugging this back into (17), we establish (13) after simplification.
Recall that
$ C:=S(321)des,iar,comp(t,1,1)=1−√1−4tz2+4z2−4z2z(tz−z+1), $
|
(18) |
which is the generating function of the descent polynomials on
Theorem 3.10. The generating function of the triple statistic
$ ˜S(321)des,iar,comp(t,r,p)=(rpz−rz+tz)C2−(rpz+p−1)C+p(1−rpz)(1−rzC)(p+C−pC). $
|
(19) |
Proof. Recently, Fu, Han and Lin [15,Lemma 4.5] generalized (18) to
$ H:=S(321)des,iar,comp(t,r,1)=1−rzC+trz2C2(1−rz)(1−rzC). $
|
For convenience, let
$ S(321)des,iar,comp(t,r,p)=11−rpz+p(I(r)−rz)(1−pI(1))(1−rpz)=1−p(I(1)−I(r))−rpz(1−pI(1))(1−rpz). $
|
(20) |
It follows that
$ I(1) = 1-1/C\quad\text{and}\quad I(r)-I(1) = (H/C-1)(1-rz). $ |
Substituting these back to (20) yields (19).
Pattern
Now we move onto the class of
Proposition 3.11. For any permutation
1. For
2. When read from left to right, the values of the left-to-right maxima of
3. The first
4. Provided
The next theorem strengthens Theorem 1.1 (iii).
Theorem 3.12. For all positive integers
$ ∑π∈Sn(132)xLMAX(π)yLMIN(π)riar(π)pcomp(π)=∑π∈Sn(132)xLMAX(π)yLMIN(π)rcomp(π)piar(π). $
|
(21) |
In particular, we have
$ ∑π∈Sn(132)tdes(π)riar(π)pcomp(π)=∑π∈Sn(132)tdes(π)rcomp(π)piar(π). $
|
(22) |
Proof. We begin by noting that for each
i.
ii.
iii.
In terms of the two operations deletion and insertion that we introduce in Definition 2.1, we let
$ σ=φ(π):=insn,n(delπ(1)(π)), $
|
with
$ π=φ−1(σ):=insσ(1),1(deln(σ)) $
|
being the inverse map. We illustrate this definition by giving an example, where the letters affected by this map have been overlined.
$ π=ˉ5ˉ6ˉ734ˉ82ˉ9¯101¯11σ=ˉ5ˉ634ˉ72ˉ8ˉ91¯10¯11 $
|
Applying Proposition 3.11, it is rountine to verify i, ii, and iii, and we leave the details to the reader. Items ii and iii ensure that
$ ({\mathrm{LMAX}}, {\mathrm{LMIN}}, {\mathsf{iar}}, {\mathsf{comp}})\; \tau = ({\mathrm{LMAX}}, {\mathrm{LMIN}}, {\mathsf{comp}}, {\mathsf{iar}})\; \pi. $ |
Pairing permutations in this way leads to (21).
Finally, by Proposition 3.11 (1) we have
Theorem 3.13. We have
$ ˜S(132)des,iar,comp(t,r,p)=11−rpz+(1−z)(N−1)t(1−rz)(1−pz)(1−z−(N−1)tz). $
|
(23) |
Proof. The proof is analogous to that of Theorem 3.9. Noting that
$ S(t,r,p)=11−rpz+rpr−p∑n≥2zn∑π∈In(132)tdes(π)(riar(π)−piar(π))=1+(rpz)21−rpz+rpr−p∑n≥1zn∑π∈In(132)tdes(π)(riar(π)−piar(π)). $
|
Converting to
$ ˜S(t,r,p)=rpz1−rpz+r˜S(t,r,0)−p˜S(t,p,0)r−p. $
|
(24) |
Plugging in
$ \tilde{\mathfrak{S}}(t, r, 1) = \frac{rz}{1-rz}+\frac{r\tilde{\mathfrak{S}}(t, r, 0)-\tilde{\mathfrak{S}}(t, 1, 0)}{r-1}. $ |
Now solve for
$ ˜S(t,r,p)=rpz1−rpz+(r−1)(˜S(t,r,1)−rz1−rz)−(p−1)(˜S(t,p,1)−pz1−pz)r−p. $
|
(25) |
Next, we decompose each
$ \tilde{\mathfrak{S}}(t, r, 1) = \frac{(1-z)(1+t(N-1))}{(1-rz)(1-z-tz(N-1))}. $ |
We plug this back into (25) and simplify to arrive at (23).
We deal with the three remaining classes, namely,
Patterns
Theorem 3.14. For every
$ (des,iar,comp)π=(des,comp,iar)σ. $
|
In particular, the matrices
Proof. We define
$ \sigma: = \theta(\pi): = {\operatorname{ins}}_{\pi(1), 1}(\theta(\nu)\oplus\theta(\mu)). $ |
The following facts can be readily verified.
So we see
The equidistribution between
Theorem 3.15. We have
$ ˜S(213)des,iar,comp(t,r,p)=(1−rz)(tN−t+1)(1−rpz)(1−rz(tN−t+1))and $
|
(26) |
$ ˜S(231)des,iar,comp(t,r,p)=(1−pz)(tN−t+1)(1−rpz)(1−pz(tN−t+1)). $
|
(27) |
Proof. We begin with the calculation of
●
●
●
Summing up these three cases and noting that
$ S(t,r,p)=1+rpz+trpz(N−1)+rpz(S(t,r,p)−1)+trpz(S(t,r,1)−1)(N−1). $
|
Now we plug in
Decomposing each
Pattern
For
$ ˜S(123)des,iar,comp(t,r,p)=A(t,p)+rB(t,p), $
|
where
$ pzA(t,p):=∑n≥1zn∑π∈Sn(123),iar(π)=1tdes(π)pcomp(π)=pz+tpz2+(tp+t2p+tp2)z3+⋯,pzB(t,p):=∑n≥2zn∑π∈Sn(123),iar(π)=2tdes(π)pcomp(π)=p2z2+(tp+tp2)z3+⋯. $
|
By (18), the generating function for the descent polynomials on
$ C∗:=S(123)des,iar,comp(t,1,1)−1=S(321)des,iar,comp(t−1,1,1;tz)−1t=−1+2tz+2tz2−2t2z2+√1−4tz−4tz2+4t2z22t2z(tz−z−1). $
|
(28) |
Theorem 3.16. We have
$ A(t,p)=(p−1)tz2(1−tz)2+(1−tz)C∗(1−tz+z)z, $
|
(29) |
$ B(t,p)=(p−1)z1−tz+C∗1−tz+z. $
|
(30) |
Thus,
$ \tilde{\mathfrak{S}}(123)^{{\mathsf{des}}, {\mathsf{iar}}, {\mathsf{comp}}}(t, r, p) = \frac{(1-p)z(trz-tz-r)}{(1-tz)^2}+\frac{(1+rz-tz) C^*}{z(1+z-tz)}. $ |
Proof. For
$ {A(t,p)=tpz2(1−tz)2+C∗z−B(t,1)−tz2(1−tz)2,B(t,p)=pz+z(A(t,1)−1)+tzB(t,p). $
|
Solving this system of equations gives rise to (29) and (30).
The following corollary can be proved combinatorially from analyzing the designated
Corollary 3.17. For
$ ∑n≥2zn∑π∈S∗n(123)riar(π)pcomp(π)=r2p2z21−z+(r+p)rpz3(1−z)2+(z3+2z4)rp(1−z)2(1−2z). $
|
(31) |
In particular, the distribution of
Proof. To calculate the generating function in (31), we need to extract the coefficients of
$ \frac{rpt(1-tz) C^*}{1-tz+z}, $ |
then take partial derivative
In this section, we let
The Wilf-classification of pairs of length
$ {(132,213),(132,312),(213,231),(231,312),(231,321)},{(132,231)},{(213,312)},{(312,321)},{(123,132)},{(123,213)}; $
|
the class enumerated by
$ \{(132, 321)\}, \, \, \{(123, 231)\}, \, \, \{(213, 321)\}, \, \, \{(123, 312)\}; $ |
and the terminating (i.e., enumerated by
$ {(132,231),(132,312),(213,231),(213,312)},{(132,213)}{(123,132),(123,213)},{(231,312),(231,321),(312,321)} $
|
and the class enumerated by
$ \{(132, 321), (213, 321)\}, \, \, \{(123, 231), (123, 312)\}. $ |
All the above refined Wilf-equivalences can be easily proven by setting
For
Pattern pairs
First note that if the pattern
Theorem 4.1. For all positive integers
$ ∑π∈Sn(P)xLMAX(π)yLMIN(π)riar(π)pcomp(π)=∑π∈Sn(P)xLMAX(π)yLMIN(π)rcomp(π)piar(π). $
|
In particular, we have
$ ∑π∈Sn(P)tdes(π)riar(π)pcomp(π)=∑π∈Sn(P)tdes(π)rcomp(π)piar(π). $
|
This symmetry can also be seen directly from the following generating functions.
Theorem 4.2. We have
$ ˜S(132,312)des,iar,comp(t,r,p)=11−rpz+(1−z)tz(1−rz)(1−pz)(1−z−tz), $
|
(32) |
$ ˜S(132,321)des,iar,comp(t,r,p)=11−rpz+tz(1−rz)(1−pz)(1−z). $
|
(33) |
Proof. The proof is quite analogous to that of Theorem 3.13. First for (32), Theorem 4.1 tells us that
$ ˜S(t,r,p)=rpz1−rpz+(r−1)(˜S(t,r,1)−rz1−rz)−(p−1)(˜S(t,p,1)−pz1−pz)r−p. $
|
(34) |
Next, note that all
●
●
Summing up all cases gives us
$ S(t,r,p)=11−rpz+pz(S(t,r,p)−11−rpz)+trpz2(1−tz)(1−rz)+tpz21−tz(S(t,r,1)−11−rz). $
|
Set
$ \tilde{\mathfrak{S}}(132, 312)^{{\mathsf{des}}, {\mathsf{iar}}, {\mathsf{comp}}}(t, r, 1) = \frac{1-z}{(1-rz)(1-z-tz)}, $ |
then plug this back into (34) and simplify, we get (32). The proof of (33) is simpler noting that for
Pattern pair
The first thing to notice is that for every
$ S(t,r,p)=S(t,rp,1),andS(t,r,p)=1+rpzS(t,r,p)+trpz(S(t,1,1)−1). $
|
Solving these two functional equations gives us
Theorem 4.3.
$ \tilde{\mathfrak{S}}(213, 231)^{{\mathsf{des}}, {\mathsf{iar}}, {\mathsf{comp}}}(t, r, p) = \frac{1-z}{(1-rpz)(1-z-tz)}. $ |
Pattern pair
For every permutation
●
●
●
●
Summing up all cases above gives rise to
Theorem 4.4.
$ ˜S(123,312)des,iar,comp(t,r,p)=1+rpz1−tz+(r+p)tz2(1−tz)2+t2z3(1−tz)3. $
|
For the remaining choices of
Pattern pairs
Recall the two bijections,
●
●
Then the following theorem is a quick corollary of Theorems 3.2 and 3.14.
Theorem 4.5. For each
Next, we compute the generating functions for these three pairs.
Theorem 4.6. We have
$ ˜S(213,312)des,iar,comp(t,r,p)=1−rz(1−rpz)(1−(r+t)z), $
|
(35) |
$ ˜S(231,312)des,iar,comp(t,r,p)=1−pz(1−rpz)(1−(p+t)z), $
|
(36) |
$ ˜S(231,321)des,iar,comp(t,r,p)=1−(1+p−t)z+(1−t)pz2(1−rpz)(1−(p+1)z+(1−t)pz2). $
|
(37) |
Proof. In view of Theorem 4.5, (35) follows from (36) by switching variables
Pattern pairs
For the same reason that the bijection
Theorem 4.7. For each
Next, note that each permutation
$ {\mathfrak{S}}(t, r, p) = \frac{1}{1-rpz}+pz({\mathfrak{S}}(t, r, p)-\frac{1}{1-rpz})+trpz({\mathfrak{S}}(t, 1, 1)-1). $ |
Solving this and applying Theorem 4.7, we can deduce the following theorem.
Theorem 4.8. We have
$ ˜S(132,213)des,iar,comp(t,r,p)=11−rpz+tz(1−rz)(1−z−tz),˜S(132,231)des,iar,comp(t,r,p)=11−rpz+tz(1−pz)(1−z−tz). $
|
Pattern pair
Note that each permutation
Theorem 4.9. We have
$ ˜S(213,321)des,iar,comp(t,r,p)=11−rpz+tz(1−z)(1−rz)(1−rpz). $
|
Pattern pair
Noting that both
$ FP(q)=11−qw+q(IP(t,t′)−w)(1−qIP(t,1))(1−qw) $
|
from Lemma 2.4 (1) to reduce the calculation to
$ I312,321(t,r):=∑n≥1zn∑π∈Sn(312,321)comp(π)=1tdes(π)riar(π). $
|
Now any indecomposable
$ I_{312, 321}(t, r) = rz+\frac{trz^2}{1-rz}, $ |
with which we can deduce
Theorem 4.10.
$ ˜S(312,321)des,iar,comp(t,r,p)=11−rpz+(1−z)tz(1−rpz)(1−rz)(1−(1+p)z+(1−t)pz2). $
|
Pattern pairs
These four pattern sets all contain the pattern
Theorem 4.11. We have
$ ˜S(123,132)des,iar,comp(t,r,p)=1+rpz+tpz21−tz+tz(1+z−tz)(1+(r−t)z+(1−r)tz2)(1−tz)((1−tz)2−tz2),˜S(123,213)des,iar,comp(t,r,p)=1+rpz1−tz+tz(1−tz+rz)(1−tz+z)(1−tz)((1−tz)2−tz2),˜S(123,231)des,iar,comp(t,r,p)=1+rpz1−tz+(1+p−tpz)tz2(1−tz)3,˜S(123,321)des,iar,comp(t,r,p)=1+(t+rp)z+(1+r)(1+p)tz2+(2r+t+pt)tz3. $
|
This section aims to characterize the pattern pair
$ [1001],[210110001],[7310331011100001],[28124101211410443101111000001],[1215218510524617510181712410554310111110000001]. $
|
The integer sequence formed by the entries in the upper-left corner of
$ 1, 1, 2, 7, 28, 121, 550, 2591, \ldots. $ |
This sequence matches A010683 in the OEIS [25], a sequence that counts, among many combinatorial objects, dissections of a convex polygon with
The first result in this section is a consequence of Theorems 1.3 and 1.4.
Corollary 5.1. For
$ ∑π∈Sn(2413,3142)tdes(π)xcomp(π)yiar(π)=∑π∈Sn(2413,4213)tdes(π)xcomp(π)yiar(π). $
|
(38) |
Consequently,
$ ∑π∈Sn(2413,4213)tdes(π)xcomp(π)yiar(π)=∑π∈Sn(2413,4213)tdes(π)xiar(π)ycomp(π) $
|
(39) |
Proof. Since the patterns
The identity (39) follows directly from (38) and Theorem 1.3.
Remark 3. In view of Corollary 5.1, one may wonder that if (2) can be further refined by
Lin and Kim [24,Theorem 5.1] showed that, among all permutation classes avoiding two patterns of length 4, the three classes below are the only nontrivial classes which are
Theorem 5.2 ([24]). We have the refined Wilf-equivalences:
$ (2413, 3142)\sim_{{\mathsf{des}}}(2413, 4213)\sim_{{\mathrm{DES}}}(2314, 3214)\sim_{{\mathrm{DES}}}(3412, 4312). $ |
It should be noted that
Theorem 5.3. We have
$ (2413, 3142)\sim_{({\mathsf{des}}, {\mathsf{iar}}, {\mathsf{comp}})}(2413, 4213)\sim_{({\mathsf{des}}, {\mathsf{iar}}, {\mathsf{comp}})}(3412, 4312). $ |
Consequently,
$ ∑π∈Sn(3412,4312)tdes(π)xcomp(π)yiar(π)=∑π∈Sn(3412,4312)tdes(π)xiar(π)ycomp(π). $
|
(40) |
In order to prove Theorem 5.3, we need a set-valued version of Lemma 2.4. For an integer
$ {\mathrm{ST}}(\pi) = \bigcup\limits_{i = 1}^k c_i + {\mathrm{ST}}(\tau_i), $ |
where
Lemma 5.4. Let
Proof. Since
Obviously, the assertion is true for
Proof of Theorem 5.3. The refined Wilf-equivalence
$ (2413, 4213)\sim_{({\mathrm{DES}}, {\mathsf{comp}})}(3412, 4312) $ |
is a direct consequence of Theorem 5.2 and Lemma 5.4, as the set-valued statistic
Next we compute the generating function
$ \{(2413, 3142), (2413, 4213), (3412, 4312)\}. $ |
Theorem 5.5. Let
$ ˜S(S)(t,r,p)=(1/z+1−r−p)S(t)+(1−r)(1−p)S(t)2(1−rpz)(1+(1−p)S(t))(1+(1−r)S(t)), $
|
(41) |
where
$ S(t)=z+(1+t)zS(t)+tzS(t)2+tS(t)3. $
|
(42) |
Proof. The functional equation (42) for the generating function
$ S(S)(t,r,p;z)=1−rpz+(rpz+rp−r−p)IS(1−rIS)(1−pIS)(1−rpz), $
|
(43) |
where
$ I_{{\mathcal{S}}} = \frac{S(t)}{1+S(t)}. $ |
Substitute this into (43) and simplify, we get (41).
Aided by the computer program, we make the following conjecture, whose validity will complete the characterization of pattern pairs of length
Conjecture 1. Let
$ (1324,2134),(1324,3124),(1423,4123),(1432,4132),(2134,2314),(2314,3124)(2431,4231),(2431,3241),(3241,3421),(3421,4231),(3421,4321). $
|
Moreover, if
Remark 4. In view of Lemma 2.4, the second assertion for the
In the rest of this section, we aim to confirm Conjecture 1 for the pattern pair
Theorem 5.6. We have the refined Wilf equivalence
$ (2413, 4213)\sim_{({\mathrm{LMAXP}}, {\mathsf{comp}})}(2431, 4231). $ |
In particular,
In view of Lemma 5.4, to prove Theorem 5.6, it is sufficient to prove the refined Wilf-equivalence
For
$ {\mathrm{AVA}}(\pi): = \{k\in[n]: {\operatorname{ins}}_k(\pi)\in{\mathfrak{S}}_{n}(2431, 4231)\} = \{k_1 > k_2 > \cdots\}. $ |
Clearly, if
We have the following growth rule for
Lemma 5.7. Suppose
$ {\mathrm{AVA}}({\operatorname{ins}}_j(\pi)) = {[j,n+1],if m≤j≤n−1;[m,n+1],if j=n. $
|
Proof. For
The definition of
Lemma 5.8. (Lin and Kim [24,Lemma 5.3]). Suppose
$ {\mathrm{AVA}}(\pi) = \{n = k_1 > k_2 > \cdots > k_m = 1\}. $ |
Then, for
$ {\mathrm{AVA}}({\operatorname{ins}}_{k_j}(\pi)) = \{n+1\geq k_j+1 > k_j > k_{j+1} > \cdots > k_m = 1\}. $ |
We are ready to prove Theorem 5.6 by constructing the generating trees for both classes.
Proof of Theorem 5.6. Label each
$ ΩSch={(2)(k)⇝(k+1),(k+1),(k),(k−1),…,(3). $
|
(44) |
This means that the initial permutation
We can construct a generating tree (an infinite rooted and labeled tree) for
$ {\mathrm{LMAXP}}(\pi) = \{i: p_i \text{ is a star node}\}. $ |
For instance, the second
It can be readily checked that Lemma 5.8 gives the same rewriting rule
The main purpose of this section is to prove Theorem 1.4. We begin with the motivation that leads to the discovery of Theorem 1.4.
Recall that a sequence
● constructed a bijection
$ Sn(t)=∑π∈Sn(2413,3142)tdes(π)=∑e∈In(021)tasc(e)=∑π∈Sn(2413,4213)tdes(π), $
|
(45) |
where
● proved combinatorially via the so-called modified Foata–Strehl action that
$ ∑π∈Sn(2413,4213)tdes(π)=⌊n−12⌋∑k=0|Γn,k(2413,4213)|tk(1+t)n−1−2k. $
|
(46) |
Recall that
$ |Γn,k(2413,3142)|=|Γn,k(2413,4213)| $
|
(47) |
for all
$ ∑π∈Sn(2413,3142)tdes(π)xdd(π)=∑π∈Sn(2413,4213)tdes(π)xdd(π), $
|
(48) |
where
Theorem 1.4 is a refinement of Wang's equidistribution (48) by the Comtet statistic
As we will see, some easy combinatorial arguments on
For each inversion sequence
$ In,k:=|{e∈In(021):izero(e)=k}|=|{π∈Sn(2413,4213):iar(π)=k}|. $
|
(49) |
Thus,
$ I_{n, k} = |\{\pi\in{\mathfrak{S}}_n(2413, 3142):{\mathsf{iar}}(\pi) = k\}| $ |
by Theorem 1.4. We have the following recurrence relation for
Theorem 6.1. We have
$ In,1=n−1∑k=12k−1In−1,kfor n≥2, $
|
(50) |
$ In,i=In−1,i−1+n−1∑k=i2k−iIn−1,kfor 2≤i≤n. $
|
(51) |
Proof. Let
● each of the
● but all zeros after the first positive entry of
Recursion (51) follows from similar reasoning.
We will prove Theorem 1.4 by showing that the generating functions for both sides of (2) satisfy the same algebraic functional equation. We begin with the calculation of the generating function for the right-hand side of (2):
$ G(t,x,y;z):=∑n≥1zn∑π∈Sn(2413,4213)tdes(π)xdd(π)yiar(π)=yz+(y2+txy)z2+(y3+2txy2+2ty+t2x2y)z3+⋯. $
|
For any
$ G(t, x, y;z) = \sum\limits_{n\geq1}z^n\sum\limits_{e\in \mathfrak{I}_n(021)}t^{{\mathsf{asc}}(e)}x^{{\mathsf{da}}(e)}y^{{\mathsf{izero}}(e)}. $ |
Lemma 6.2. We have the algebraic functional equation for
$ y3z+(txy2z+3y3z−2y2z−y2)G+c2G2+c3G3=0, $
|
(52) |
where
$ c2:=2txy2z−2txyz+3y3z+tyz−4y2z−2y2+yz+2yandc3:=txy2z−2txyz+y3z+txz+tyz−2y2z−tz−y2+yz+t+2y−1. $
|
Proof. Let
$ {\tilde {\mathfrak{I}}}_n^{(0)}(021): = \{(e, \phi)\in{\tilde {\mathfrak{I}}}_n(021): \phi(1) = 0\}\quad \text{and} \quad {\tilde {\mathfrak{I}}}_n^{(1)}(021): = {\tilde {\mathfrak{I}}}_n(021)\setminus{\tilde {\mathfrak{I}}}_n^{(0)}(021). $ |
For each
$ {\mathsf{asc}}({\mathfrak{a}}): = {\mathsf{asc}}(e)+|\{i\in[r-1]: \phi(i) < \phi(i+1)\}| $ |
and
$ {\mathsf{da}}({\mathfrak{a}}): = {\mathsf{da}}(e)+\chi(\phi(r-1) < \phi(r)); $ |
otherwise,
$ {\mathsf{asc}}({\mathfrak{a}}): = {\mathsf{asc}}(e)+|\{i\in[r-1]: \phi(i) < \phi(i+1)\}|+1 $ |
and
$ {\mathsf{da}}({\mathfrak{a}}): = {\mathsf{da}}(e)+\chi(\phi(r-1) < \phi(r))+\chi(e_2 = 1). $ |
The reason of defining these two statistics in this way will become transparent when we decompose
$ ˜G0(t,x;z):=∑n≥1zn∑a∈˜I(0)n(021)tasc(a)xda(a), and˜G1(t,x;z):=∑n≥1zn∑a∈˜I(1)n(021)tasc(a)xda(a). $
|
For convenience, we use the convention that
Each
●
● and
This decomposition is reversible and satisfies
$ asc(e)=asc(ˆe)+asc(a),da(e)=da(ˆe)+da(a),andizero(e)=izero(ˆe)+1. $
|
Turning the above decomposition into generating function yields
$ G=yz(1+G)(1+˜G1). $
|
(53) |
Similar decomposition as above for
$ {˜G0=z(1+˜G0+˜G1)(1+˜G1),˜G1=z(tx+t˜G0+tz(1−x)(1+˜G1)+˜G1)(1+˜G1). $
|
Eliminating
$ ˉG1=1+(txz−2z)ˉG1+(tz2−2txz2+z2+2z)ˉG21+(txz3−tz3+tz2−z2)ˉG31. $
|
On the other hand, solving (53) gives
Next, we continue to compute the generating function for the left-hand side of (2):
$ S=S(t,x,y;z):=∑n≥1zn∑π∈Sn(2413,3142)tdes(π)xdd(π)yiar(π)=yz+(y2+txy)z2+(y3+2txy2+2ty+t2x2y)z3+⋯. $
|
This will be accomplished by applying Stankova's block decomposition [29] (see also [23]) of separable permutations that we now recall.
Lemma 6.3 (Stankova [29]). A permutation
(i)
$ A_1, A_2, \ldots, A_k, n, B_1, B_2, \ldots, B_l, \quad (|k-l|\leq1), $ |
where
(ii) The elements in any block form a permutation that avoids both
See Fig. 2 for a transparent illustration of this lemma. Condition (ii) is clear, while condition (i) is equivalent to saying that
For convenience, we need to introduce two variants of the double descents. Let
$ {\mathsf{dd}}_0(\pi): = |\{i\in[n]: \pi(i-1) > \pi(i) > \pi(i+1)\}|, $ |
where
$ {\mathsf{dd}}_{\infty}(\pi): = |\{i\in[n]: \pi(i-1) > \pi(i) > \pi(i+1)\}|, $ |
where
$ L = L(t, x, y;z): = \sum\limits_{n\geq1}z^n\sum\limits_{\pi\in{\mathfrak{S}}_n(2413, 3142)}t^{{\mathsf{des}}(\pi)}x^{{\mathsf{dd}}_0(\pi)}y^{{\mathsf{iar}}(\pi)} = yz+(ty+y^2)z^2+\cdots $ |
and
$ R = R(t, x;z): = \sum\limits_{n\geq1}z^n\sum\limits_{\pi\in{\mathfrak{S}}_n(2413, 3142)}t^{{\mathsf{des}}(\pi)}x^{{\mathsf{dd}}_{\infty}(\pi)} = xz+(1+tx^2)z^2+\cdots. $ |
Set
Lemma 6.4. Let
$ {L1=S1(1+tS1)1+txS1,R=S1(S1+x)1+S1,S1=tS31+tzS21+(z+txz)S1+z,(1−z)˜L=tS1B(1+B)1−tRB+tzS1˜L(2+L1+B+tR−tRL1B)(1−tRB)(1−tRL1),S=B(1+tR)1−tRB+z˜L(1+tR)2(1−tRB)(1−tRL1). $
|
(54) |
Proof. The first three equations of (54) were proved by Wang [30]. We begin with the proof of the fifth equation in (54) by writing
$ A_1, A_2, \ldots, A_k, n, B_1, B_2, \ldots, B_l, \quad (|k-l|\leq1), $ |
where
$ 2yzB^j(tR)^j+2\sum\limits_{i = 1}^jzB^{i-1}\tilde L L_1^{j-i}(tR)^j. $ |
$ yzB^{j+1}(tR)^j+\sum\limits_{i = 1}^{j+1}zB^{i-1}\tilde L L_1^{j+1-i}(tR)^j. $ |
$ yzB^{j}(tR)^{j+1}+\sum\limits_{i = 1}^{j}zB^{i-1}\tilde L L_1^{j-i}(tR)^{j+1}. $ |
Summing over all the above cases gives the fifth equation of (54). The fourth equation of (54) is obtained by writing
We are ready to verify Theorem 1.4.
Proof of Theorem 1.4. We aim to verify that
$ y^3z+(txy^2z+3y^3z-2y^2z-y^2)S+c_2S^2+c_3S^3, $ |
where
In this paper, we launch a systematic study of the Wilf-equivalence refined by two permutation statistics, namely
In view of Lemmas 2.5 and 5.4, we pose the following open problem about a set-valued extension of Lemma 2.5 for further investigation.
Problem. Let
$|{\mathfrak{S}}_n(P)^{{\mathrm{ST}}, {\mathsf{iar}}}| = |{\mathfrak{S}}_n(P)^{{\mathrm{ST}}, {\mathsf{comp}}}|\Longleftrightarrow|{\mathfrak{S}}_n(P)^{{\mathrm{ST}}, {\mathsf{iar}}, {\mathsf{comp}}}| = |{\mathfrak{S}}_n(P)^{{\mathrm{ST}}, {\mathsf{comp}}, {\mathsf{iar}}}|$? |
In particular, we suspect that the equivalence above holds when
Conjecture 2. Let
$ |Sn(P)LMAX,iar|=|Sn(P)LMAX,comp|⟺|Sn(P)LMAX,iar,comp|=|Sn(P)LMAX,comp,iar|. $
|
It is our hope, that the results presented and conceived (see also Conjecture 1) here, would attract more people to work on Wilf-equivalences refined by Comtet statistics, or to unearth and study new Comtet statistics in general.
The authors thank the anonymous referees for their valuable comments and suggestions. Part of this work was initiated at the Research Center for Mathematics and Interdisciplinary Sciences at Shandong University. The authors would like to thank the center for the excellent working condition and the hospitality extended during their stay.
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proved in | |||
Symmetric | Thm. 3.9 | ||
Thm. 3.10 | |||
|
Hankel | Thm. 3.13 | |
Lower triangular | Thm. 3.15 | ||
Thm. 3.15 | |||
Thm. 3.16 |
proved in | |||
Hankel | Thm. 4.2 | ||
Thm. 4.2 | |||
Diagonal | Thm. 4.3 | ||
Thm. 4.4 | |||
Lower triangular | Thm. 4.6 | ||
Thm. 4.6 | |||
Upper triangular | Thm. 4.6 | ||
Lower triangular | Thm. 4.8 | ||
Thm. 4.8 | |||
Lower triangular | Thm. 4.9 | ||
No pattern | Thm. 4.10 | ||
Thm. 4.11 | |||
Thm. 4.11 | |||
Thm. 4.11 | |||
$ |
Ultimately zero | Thm. 4.11 |
proved in | |||
Symmetric | Thm. 3.9 | ||
Thm. 3.10 | |||
|
Hankel | Thm. 3.13 | |
Lower triangular | Thm. 3.15 | ||
Thm. 3.15 | |||
Thm. 3.16 |
proved in | |||
Hankel | Thm. 4.2 | ||
Thm. 4.2 | |||
Diagonal | Thm. 4.3 | ||
Thm. 4.4 | |||
Lower triangular | Thm. 4.6 | ||
Thm. 4.6 | |||
Upper triangular | Thm. 4.6 | ||
Lower triangular | Thm. 4.8 | ||
Thm. 4.8 | |||
Lower triangular | Thm. 4.9 | ||
No pattern | Thm. 4.10 | ||
Thm. 4.11 | |||
Thm. 4.11 | |||
Thm. 4.11 | |||
$ |
Ultimately zero | Thm. 4.11 |