Width-$k$ Generalizations of Classical Permutation Statistics

TL;DR

This paper introduces width-$k$ generalizations of classical permutation statistics like descents and inversions, establishing new relationships and exploring their behavior in pattern avoidance contexts.

Contribution

It presents novel width-$k$ variants of classical permutation statistics and analyzes their properties, relationships, and behavior in pattern avoidance scenarios.

Findings

Width-$k$ statistics generalize classical permutation statistics.

New formulas and relationships are established for these statistics.

Behavior in pattern avoidance contexts is characterized.

Abstract

We introduce new natural generalizations of the classical descent and inversion statistics for permutations, called width-$k$ descents and width-$k$ inversions. These variations induce generalizations of the excedance and major statistics, providing a framework in which the most well-known equidistributivity results for classical statistics are paralleled. We explore additional relationships among the statistics providing specific formulas in certain special cases. Moreover, we explore the behavior of these width-$k$ statistics in the context of pattern avoidance.

Tables

Table 1. Table 1. The polynomials G n , k ​ ( q ) subscript 𝐺 𝑛 𝑘 𝑞 G_{n,k}(q) for n = 6 , 8 , 9 𝑛 6 8 9 n=6,8,9 and 1 ≤ k ≤ n 1 𝑘 𝑛 1\leq k\leq n .

Equations

$$\mathrm{des}\sigma$$

$$\mathrm{inv}\sigma$$

$$\mathrm{maj}\sigma$$

$$\mathrm{exc}\sigma$$

$$F_n^{\mathrm{st}}(q)=\sum _{\sigma \in {\mathfrak{S}}_n}{q}^{\mathrm{st}\sigma }.$$

$$\sum _{j\ge 0}(1+j{)}^n{q}^j=\frac{A_n(q)}{(1-q{)}^{n+1}}.$$

$${\mathrm{Des}}_k(\sigma )=\{i\in [n-k]\mid a_i>a_{i+k}\},$$

$${\mathrm{des}}_k(\sigma )=|{\mathrm{Des}}_k(\sigma )|.$$

$${\mathrm{Inv}}_k(\sigma )=\{(i,j)\in [n{]}^2\mid a_i>a_j\text{ and }j-k=mk,k\in \mathbb{Z}\}$$

$${\mathrm{inv}}_k(\sigma )=|{\mathrm{Inv}}_k(\sigma )|.$$

$${\mathrm{Des}}_{\{2,3\}}(\sigma )=\{1,4,5\}\text{ and }{\mathrm{Inv}}_{\{2,3\}}(\sigma )=\{(1,3),(1,7),(3,7),(4,7),(5,7)\}.$$

$${\mathrm{inv}}_K(\sigma )=\sum _{\varnothing ⊊K^{\prime }\subseteq K}(-1{)}^{|K^{\prime }|+1}{\mathrm{inv}}_{\mathrm{lcm}(K^{\prime })}(\sigma ),$$

$${\mathrm{inv}}_k(\sigma )=\sum _{j\ge 1}{\mathrm{des}}_{jk}(\sigma ).$$

$${\beta }_{n,k}^i(\sigma )=\{\begin{array}{ll}{a}_i{a}_{i+k}{a}_{i+2k}\dots a_{i+dk}& \text{ if }i\le r\\ a_i{a}_{i+k}{a}_{i+2k}\dots a_{i+(d-1)k}& \text{ if }r<i<k\end{array}$$

$$\varphi :{\mathfrak{S}}_n\to {\mathfrak{S}}_{d+1}^r\times {\mathfrak{S}}_d^{k-r}$$

$$M_{n,k}=\left(\genfrac{}{}{0ex}{}{n}{(d+1{)}^r,d^{k-r}}\right)$$

$$F_n^{{\mathrm{des}}_k}(q)$$

$$F_n^{{\mathrm{inv}}_k}(q)$$

$${\mathrm{des}}_k(\sigma )=\sum _{i=1}^k\mathrm{des}(\mathrm{std}{\beta }_{n,k}^i(\sigma )).$$

$$F_n^{{\mathrm{inv}}_k}(q)$$

$${\mathrm{Exc}}_K(\sigma )=\bigcup _{k\in K}\underset{i=1}{\overset{k}{⨄}}\{\{j\in [n-1]\mid \lceil j/k\rceil \in \mathrm{Exc}(\mathrm{std}{\beta }_{n,k}^i(\sigma ))\}\}$$

$${\mathrm{maj}}_K(\sigma )=\sum _{k\in K}\sum _{i\in {\mathrm{Des}}_k(\sigma )}\lceil \frac{i}{k}\rceil .$$

$$\sum _{i=1}^k\mathrm{maj}(\mathrm{std}{\beta }_{n,k}^i(\sigma ))\text{ and }{\mathrm{exc}}_K(\sigma )=\sum _{i=1}^k\mathrm{exc}(\mathrm{std}{\beta }_{n,k}^i(\sigma )).$$

$${\mathrm{exc}}_{\{2,3\}}(\sigma )=|\{1,3,1,2\}|=4$$

$${\mathrm{maj}}_{\{2,3\}}(\sigma )=\lceil \frac{1}{2}\rceil +\lceil \frac{5}{2}\rceil +\lceil \frac{4}{3}\rceil =6.$$

$$G_{n,k}(q)=\sum _{\sigma \in {\mathfrak{S}}_n}{q}^{{\mathrm{des}}_k(\sigma )-{\mathrm{des}}_{n-k}(\sigma )}.$$

$$F_n^{\mathrm{st}}(\Pi ;q)=\sum _{\sigma \in {\mathrm{Av}}_n(\Pi )}{q}^{\mathrm{st}\sigma },$$

$$C_n=\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$$

$${\pi }^r=a_m{a}_{m-1}\dots a_1\text{ and }{\pi }^c=(m+1-a_1)(m+1-a_2)\dots (m+1-a_m).$$

$${\Pi }^r=\{{\pi }^c\mid \pi \in \Pi \}\text{ and }{\Pi }^c=\{{\pi }^r\mid \pi \in \Pi \}$$

$$T_n(\Pi ;t_1,\dots ,t_{n-1})=\sum _{\sigma \in {\mathrm{Av}}_n(\Pi )}\prod _{k=1}^{n-1}{t}_k^{{\mathrm{des}}_k(\sigma )}.$$

$$T_n({\Pi }^{\prime };t_1,\dots ,t_{n-1})=t_1^{n-1}{t}_2^{n-2}\cdots t_{n-1}{T}_n(\Pi ;t_1^{-1},\dots ,t_{n-1}^{-1}).$$

Full text

Department of Mathematics, Michigan State University, East Lansing, MI 48824-1027, USA. E-mail: davisr@math.msu.edu

1. Introduction

Let $\mathfrak{S}_{n}$ denote the set of permutations $\sigma=a_{1}\cdots a_{n}$ of $[n]$, and let $\mathfrak{S}=\mathfrak{S}_{1}\cup\mathfrak{S}_{2}\cup\dots$. A function $\operatorname{st}:\mathfrak{S}_{n}\to\mathbb{N}$ is called a statistic, and the systematic study of permutation statistics is generally accepted to have begun with MacMahon [6]. In particular, four of the most well-known statistics are the descent, inversion, major, and excedance statistics, defined respectively by

[TABLE]

where $\operatorname{Des}\sigma=\{i\in[n-1]\ |\ a_{i}>a_{i+1}\}$.

Given any statistic $\operatorname{st}$, one may form the generating function

[TABLE]

A famous result due to MacMahon [6] states that $F_{n}^{\operatorname{des}}(q)=F_{n}^{\operatorname{exc}}(q)$, and that both equal are to the Eulerian polynomial $A_{n}(q)$. The Eulerian polynomials themselves may be defined via the identity

[TABLE]

Another well-known result, due to MacMahon [5] and Rodrigues [7], states $F_{n}^{\operatorname{inv}}(q)=F_{n}^{\operatorname{maj}}(q)=[n]_{q}!$ Foata famously provided combinatorial proofs of the equidistributivity of each; see [4] for a treatment of these. Thus any statistic $\operatorname{st}$ for which $F_{n}^{\operatorname{st}}(q)=A_{n}(q)$ is called Eulerian, and if $F_{n}^{\operatorname{st}}(q)=[n]_{q}!$ then $\operatorname{st}$ is called Mahonian. These four statistics have many generalizations; in this article, we discuss new variations, induced from a simple generalization of $\operatorname{des}$.

For each of the following definitions, we assume $n\in\mathbb{Z}_{>0}$, $k\in[n-1]$, $\emptyset\neq K\subseteq[n-1]$, and $\sigma=a_{1}a_{2}\ldots a_{n}\in\mathfrak{S}_{n}$. We define a width-$k$ descent of $\sigma$ to be an index $i\subseteq[n-k]$ for which $a_{i}>a_{i+k}$. Thus the width-$1$ descents are the usual descents of a permutation. Denote the set of all width-$k$ descents of $\sigma$ by

[TABLE]

and set

[TABLE]

If one is interested in descents of $\sigma$ of various widths, first let $K\subseteq[n-1]$ denote the set of widths under consideration. Now, we define $\operatorname{Des}_{K}(\sigma)$ to be the multiset $\bigcup_{k\in K}\operatorname{Des}_{k}(\sigma)$, and $\operatorname{des}_{K}(\sigma)=|\operatorname{Des}_{K}(\sigma)|$.

Now, define a width-$k$ inversion of $\sigma$ to be a pair $(i,j)$ for which $a_{i}>a_{j}$ and $j-i=mk$ for some positive integer $m$. Denote the set of width-$k$ inversions by

[TABLE]

and set

[TABLE]

Again, one may be interested in width-$k$ inversions for multiple values of $k$, so for $K\subseteq[n-1]$ define $\operatorname{Inv}_{K}(\sigma)=\bigcup_{k\in K}\operatorname{Inv}_{k}(\sigma)$. Additionally, let $\operatorname{inv}_{K}(\sigma)=|\operatorname{Inv}_{K}(\sigma)|$.

Example 1.1**.**

If $\sigma=4136572$, then

[TABLE]

Thus, $\operatorname{des}_{\{2,3\}}(\sigma)=3$, $\operatorname{inv}_{\{2,3\}}(\sigma)=5$.

As we will see in the next section, the above definitions motivate us to generalize $\operatorname{exc}$ and $\operatorname{maj}$ in such a way that well-known known relationships among $\operatorname{des},\operatorname{inv},\operatorname{maj},\operatorname{exc}$ are paralleled. However, it is not very convenient to work with $\operatorname{exc}_{K}$ and $\operatorname{maj}_{K}$ directly. So, this paper will focus much more on $\operatorname{des}_{K}$ and $\operatorname{inv}_{K}$.

The main focus of this paper is to explore these new statistics and their relationships among each other. While they are well-behaved for special cases of $K\subseteq[n-1]$, formulas for more general cases have been more elusive. Section 3 continues this exploration by considering the same statistics for classes of permutations avoiding a variety of patterns.

2. Main Results

We begin with simple expressions for $\operatorname{des}_{K}(\sigma)$ and $\operatorname{inv}_{K}(\sigma)$ for a fixed $\sigma\in\mathfrak{S}_{n}$. First, we point out that if $K\subseteq[n-1]$, $j\in K$ and $ij\in K$ for some positive integer $i$, then for any $\sigma\in\mathfrak{S}_{n}$, $\operatorname{inv}_{K}(\sigma)=\operatorname{inv}_{K\setminus\{ij\}}(\sigma)$. This occurs because for any $k\in[n-1]$, $\operatorname{inv}_{k}(\sigma)$ counts all descents whose widths are multiples of $k$. Thus $ij$ is already accounted for in $\operatorname{inv}_{K\setminus\{j\}}(\sigma)$. This follows quickly from the definitions of $\operatorname{inv}_{k}$ and $\operatorname{des}_{k}$.

Proposition 2.1**.**

For any nonempty $K\subseteq[n-1]$,

[TABLE]

where we set $\operatorname{inv}_{\operatorname{lcm}(K^{\prime})}(\sigma)=0$ if $\operatorname{lcm}(K^{\prime})\geq n$.

Proof.

We first consider the case where $K=\{k\}$. The elements of $\operatorname{Inv}_{k}(\sigma)$ are pairs of the form $(i,i+jk)$ for some positive integer $j$. Such an element exists if and only if there is a width-$(jk)$ descent of $\sigma$ at $i$. Thus, $\operatorname{inv}_{k}(\sigma)$ simply counts the number of width-$jk$ descents of $\sigma$ for all possible $j$. This leads to the equality

[TABLE]

The formula for general $K$ then follows from inclusion-exclusion: by adding the number of all width-$k$ inversions for all $k\in K$, we are twice counting any instances of a width-$\operatorname{lcm}(k_{1},k_{2})$ inversion since such an inversion is also of widths $k_{1}$ and $k_{2}$. If $K$ contains three distinct elements $k_{1},k_{2},k_{3}$, then $\operatorname{lcm}(k_{1},k_{2},k_{3})$ would have been added three times (once for each $\operatorname{inv}_{k_{i}}(\sigma)$) and subtracted three times (once for each $\operatorname{inv}_{\{k_{i},k_{j}\}}(\sigma)$, $i<j$), so it must be added again for the sum. Extending this argument to range over larger subsets of $K$ results in the claimed formula. ∎

Example 2.2**.**

Let us return to $\sigma=4136572$. We saw from Example 1.1 that $\operatorname{inv}_{\{2,3\}}(\sigma)=5$, where four inversions have width $2$ and two have width $3$. But since the inversion $(1,7)$ has width both $2$ and $3$, it must also have width $\operatorname{lcm}(2,3)=6$. So, it contributes two summands of $1$ and one summand of $-1$.

We come now to a function which helps demonstrate the interactions among the width-$k$ statistics. Let $n$ and $k$ be positive integers for which $n=dk+r$ for some $d,r\in\mathbb{Z}$ with $0\leq r<k$. To each $\sigma=a_{1}\dots a_{n}\in\mathfrak{S}_{n}$ we may then associate the set of disjoint substrings $\beta_{n,k}(\sigma)=\{\beta_{n,k}^{1}(\sigma),\dots,\beta_{n,k}^{k}(\sigma)\}$ where

[TABLE]

Now, define

[TABLE]

by setting $\phi(\sigma)=(\operatorname{std}\beta_{n,k}^{1}(\sigma),\dots,\operatorname{std}\beta_{n,k}^{k}(\sigma))$, where $\operatorname{std}$ is the standardization map, that is, the permutation obtained by replacing the smallest element of $\sigma$ with $1$, the second-smallest element with $2$, etc. Note in particular that each $\operatorname{std}\beta_{n,k}^{i}(\sigma)$ is a permutation of $[d+1]$ or $[d]$.

The first of the identities in the following proposition was originally established in [8], though with slightly different notation. We provide an alternate proof and extend their result to width-$k$ inversions.

Theorem 2.3**.**

Let $n$ and $k$ be positive integers such that $n=dk+r$, where $0\leq r<k$, and let $A_{i}(q)$ denote the $i^{th}$ Eulerian polynomial. Also let $M_{n,k}$ denote the multinomial coefficient

[TABLE]

where $i^{j}$ indicates $i$ repeated $j$ times. We then have the identities

[TABLE]

Proof.

Let $k\in[n-1]$ and consider $\phi$ defined above. Note that $\phi$ is an $M_{n,k}$-to-one function since, given $(\sigma_{1},\dots,\sigma_{k})\in\mathfrak{S}_{d+1}^{r}\times\mathfrak{S}_{d}^{k-r}$, there are $M_{n,k}$ ways to partition $[n]$ into the subsequences $\beta_{n,k}^{i}(\sigma)$ such that $\operatorname{std}\beta^{i}_{n,k}(\sigma)=\sigma_{i}$ for all $i$. Also note that

[TABLE]

Thus,

[TABLE]

This proves the first identity.

The second identity follows completely analogously, with the main modification being that an element of $\operatorname{Inv}_{k}(\sigma)$ corresponds to a usual inversion in some unique $\operatorname{std}\beta_{n,k}^{j}(\sigma)$. ∎

Given $\operatorname{des}_{K}$ and $\operatorname{inv}_{K}$, one must wonder what the corresponding generalizations of $\operatorname{exc}$ and $\operatorname{maj}$ are whose relationships with $\operatorname{des}_{K}$ and $\operatorname{inv}_{K}$ parallel that of the classical statistics. To do this, we define the multiset

[TABLE]

and set $\operatorname{exc}_{K}(\sigma)=|\operatorname{Exc}_{K}(\sigma)|$, and also set

[TABLE]

We could equivalently state that

[TABLE]

These are exactly the definitions needed in order to obtain identities that parallel $F_{n}^{\operatorname{des}}(q)=F_{n}^{\operatorname{exc}}(q)$ and $F_{n}^{\operatorname{inv}}(q)=F_{n}^{\operatorname{maj}}(q)$, as we will soon see.

One important distinct to make between $\operatorname{exc}$ and $\operatorname{exc}_{K}$ is the following. If $\sigma=a_{1}a_{2}\dots a_{n}$ and $\tau=b_{1}b_{2}\dots b_{n}$, then even if $a_{i}=b_{i}$ for some $i$, one cannot say $i\in\operatorname{Exc}_{K}(\sigma)$ if and only if $i\in\operatorname{Exc}_{K}(\tau)$. For example, if $\sigma=4136572$, then $1\in\operatorname{Exc}_{2}(\sigma)$, but if $\tau=4153627$ then $1\notin\operatorname{Exc}_{2}(\tau)$.

Example 2.4**.**

Again let $\sigma=4136572$. We then have

[TABLE]

and

[TABLE]

By constructing a nearly identical argument as in the proof Theorem 2.3, and using the facts that $F_{n}^{\operatorname{des}}(q)=F_{n}^{\operatorname{exc}}(q)$ and $F_{n}^{\operatorname{inv}}(q)=F_{n}^{\operatorname{maj}}(q)$, we have the following corollary.

Corollary 2.5**.**

The identities $F_{n}^{\operatorname{des}_{k}}(q)=F_{n}^{\operatorname{exc}_{k}}(q)$ and $F_{n}^{\operatorname{inv}_{k}}(q)=F_{n}^{\operatorname{maj}_{k}}(q)$ hold.

Now that we have established the analogous parallels between the four classical statistics and their width-$k$ counterparts, we wish to explore what other structure is present. A simple proposition relates $\operatorname{des}_{k}$ and $\operatorname{inv}_{k}$ when $k$ is large.

Corollary 2.6**.**

For all $k\geq n/2$, $F_{n}^{\operatorname{des}_{k}}(q)=F_{n}^{\operatorname{inv}_{k}}(q)$.

Proof.

Since $k\geq n/2$, the sets $\beta_{n,k}^{i}(\sigma)$ contain at most two elements. So, width-$k$ descents and width-$k$ inversions of $\sigma$ are identical. ∎

We now show that interesting behavior occurs when considering the function

[TABLE]

According to computational data, the following conjecture holds for all $n\leq 9$ and $1\leq k<n$ for which $\gcd(k,n)=1$.

Conjecture 2.7**.**

If $\gcd(k,n)=1$, then $G_{n,k}(q)=nq^{1-k}A_{n-1}(q)$.

Several illustrative polynomials are given in Table 1. This does not hold more generally, and it would be interesting to determine if a general formula exists.

Question 2.8**.**

For which values of $n,k$ does there exist a closed formula for $G_{n,k}(q)$, and what is the formula?

3. Pattern Avoidance

We say that a permutation $\sigma\in\mathfrak{S}_{n}$ contains the pattern $\pi\in\mathfrak{S}_{m}$ if there exists a subsequence $\sigma^{\prime}$ of $\sigma$ such that $\operatorname{std}(\sigma^{\prime})=\pi$. If no such subsequence exists, then we say that $\sigma$ avoids the pattern $\pi$. If $\Pi\subseteq\mathfrak{S}$, then we say $\sigma$ avoids $\Pi$ if $\sigma$ avoids every element of $\Pi$. The set of all permutations of $\mathfrak{S}_{n}$ avoiding $\Pi$ is denoted $\operatorname{Av}_{n}(\Pi)$. In a mild abuse of notation, if $\Pi=\{\pi\}$, we will write $\operatorname{Av}_{n}(\pi)$.

In this section, we consider the functions

[TABLE]

which specializes to $F_{n}^{\operatorname{st}}(q)$ if $\Pi=\emptyset$. In most instances, $F_{n}^{\operatorname{des}_{k}}$ will be the main focus, but $F_{n}^{\operatorname{des}_{K}}$ and $F_{n}^{\operatorname{inv}_{k}}$ will also make appearances.

An important concept within pattern avoidance is that of Wilf equivalence. Two sets $\Pi,\Pi^{\prime}\subset\mathfrak{S}$ are said to be Wilf equivalent if $|\operatorname{Av}_{n}(\Pi)|=|\operatorname{Av}_{n}(\Pi^{\prime})|$ for all $n$. In this case, we write $\Pi\equiv\Pi^{\prime}$ to denote this Wilf equivalence, which is indeed an equivalence relation. For example, it is known [3, 6] that whenever $\pi,\pi^{\prime}\in\mathfrak{S}_{3}$, then $|\operatorname{Av}_{n}(\pi)|=|\operatorname{Av}_{n}(\pi^{\prime})|=C_{n}$, where

[TABLE]

is the $n^{th}$ Catalan number.

Proving whether $\Pi\equiv\Pi^{\prime}$ is often quite difficult, and their Wilf equivalence does not imply that $F_{n}^{\operatorname{st}}(\Pi;q)=F_{n}^{\operatorname{st}}(\Pi^{\prime};q)$. However, in some instances, the problems of establishing these identities have straightforward solutions by applying basic transformations on the elements of the avoidance classes. Given $\pi=a_{1}\dots a_{n}\in\mathfrak{S}_{m}$, let $\pi^{r}$ denote its reversal and $\pi^{c}$ denote its complement, respectively defined by

[TABLE]

Similarly, given $\Pi\subseteq\mathfrak{S}$, we let

[TABLE]

be the reversal and complement of $\Pi$, respectively.

Our results of this section begin with a multivariate generalization of $F_{n}^{\operatorname{st}}(\Pi;q)$, and with showing how its specializations describe relationships among $\Pi,\Pi^{r}$, and $\Pi^{c}$.

Definition 3.1**.**

Fix $\Pi\subseteq\mathfrak{S}$. Define

[TABLE]

Several conclusions quickly follow. This function specializes to $F_{n}^{\operatorname{des}_{K}}(\Pi;q)$ by setting $t_{i}=q$ for $i\in K$ and $t_{i}=1$ for all $i\notin K$. We can also recover $F_{n}^{\operatorname{inv}_{K}}(\Pi;q)$ from $T_{n}$ by setting $t_{i}=q$ whenever $i\in[n-1]\cap k\mathbb{Z}$ for some $k\in K$, and setting $t_{i}=1$ otherwise. Additionally, when $\Pi=\{\pi\}$, a nice duality appears, providing a mild generalization of Lemma 2.1 from [1].

Proposition 3.2**.**

For any $\Pi\subseteq\mathfrak{S}$, let $\Pi^{\prime}$ denote either $\Pi^{r}$ or $\Pi^{c}$. We then have

[TABLE]

Consequently,

[TABLE]

Proof.

It is enough to prove the claim when the set of patterns being avoided is $\{\pi\}$ for some $\pi\in\mathfrak{S}_{m}$, since the full result follows by applying the argument to all elements of $\Pi$ simultaneously.

First consider when $\pi^{\prime}=\pi^{c}$ and $\sigma\in\operatorname{Av}_{n}(\pi)$. Because $\sigma\in\operatorname{Av}_{n}(\pi)$ if and only if $\sigma^{c}\in\operatorname{Av}_{n}(\pi^{c})$, we have that for each $k$, $i\in\operatorname{Des}_{k}(\sigma)$ if and only if $i\notin\operatorname{Des}_{k}(\sigma^{c})$. This implies $\operatorname{Des}_{k}(\sigma^{c})=[n-k]\setminus\operatorname{Des}_{k}(\sigma)$, hence $\operatorname{des}_{k}(\sigma^{c})=n-k-\operatorname{des}_{k}(\sigma)$. So,

[TABLE]

Proving that the result holds for $\pi^{\prime}=\pi^{r}$ follows similarly.

The second identity in the proposition statement holds by applying the first identity twice: first for $\Pi^{r}$ and then for $\Pi^{c}$. ∎

The above identities significantly reduce the amount of work needed to study $F_{n}^{\operatorname{des}_{K}}(\Pi;q)$ for all $\Pi\subseteq\mathfrak{S}_{n}$. For the remainder of this paper, we systematically approach $\Pi$ for $|\Pi|\leq 2$.

3.1. Avoiding singletons

By Proposition 3.2, we immediately get

[TABLE]

and

[TABLE]

So, studying $F_{n}^{\operatorname{des}_{k}}(\pi;q)$ for $\pi\in\mathfrak{S}_{3}$ reduces to studying the function for a choice of one pattern from $\{123,321\}$ and one from the remaining patterns. For some choices of $\Pi$, the permutations in $\operatorname{Av}_{n}(\Pi)$ are especially highly structured, which leads to similar arguments throughout the rest of this paper.

We begin with $\pi=312$. Notice that if $a_{1}\dots a_{n}\in\operatorname{Av}_{n}(312)$ and $a_{i}=1$, then $\operatorname{std}(a_{1}\dots a_{i-1})\in\operatorname{Av}_{i-1}(312)$, $\operatorname{std}(a_{i+1}\dots a_{n})\in\operatorname{Av}_{n-i}(312)$, and

[TABLE]

Proposition 3.3**.**

For all $n$,

[TABLE]

where $C_{i}$ is the $i^{th}$ Catalan number.

Proof.

First consider when $\sigma=a_{1}\dots a_{n}\in\operatorname{Av}_{n}(312)$ and $a_{i}=1$ for some $i\leq k$. By the discussion preceding this proposition, $j\notin\operatorname{Des}_{k}(\sigma)$ for any $j\leq i$. So, none of the $C_{i-1}$ possible permutations that make up $\operatorname{std}(a_{1}\dots a_{i-1})$ contribute to $\operatorname{des}_{k}(\sigma)$. The only contributions to $\operatorname{des}_{k}(\sigma)$ come from $\operatorname{std}(a_{i+1}\dots a_{n})\in\operatorname{Av}_{n-i}(312;q)$. The overall contribution to $F_{n}^{\operatorname{des}_{k}}(312;q)$ is the first summand of the identity.

Now suppose $a_{i}=1$ for some $i>k$. Each choice of $a_{1}\dots a_{i}$, contributes to $\operatorname{des}_{k}(\sigma)$ as usual, but there will be an additional width-$k$ descent produced at $i-k$. The elements $a_{i+1}\dots a_{n}$ contribute to $\operatorname{des}_{k}(\sigma)$ as before. The overall contribution to $F_{n}^{\operatorname{des}_{k}}(312;q)$ is the first summand of the identity. Since we have considered all possible indices $i$ for which we could have $a_{i}=1$, we add the two cases together and are done. ∎

Note that when we set $q=1$, the above recursion specializes to the well-known recursion for Catalan numbers $C_{n+1}=\sum_{i=0}^{n}C_{i}C_{n-i}$.

We can use the previous proposition in conjunction with Proposition 3.2 to produce formulae for $F_{n}^{\operatorname{des}_{k}}(\Pi;q)$ and $F_{n}^{\operatorname{inv}_{k}}(\Pi;q)$ whenever $\Pi$ is a single element of $\mathfrak{S}_{3}$ other than $123$ or $321$. The only nontrivial work required, then, is to compute the degrees of the two polynomials.

Corollary 3.4**.**

For all $n$,

[TABLE]

Proof.

The degrees of the above polynomials are given by identifying a permutation in $\operatorname{Av}_{n}(312)$ with the most possible descents. This is satisfied by $n(n-1)\dots 21\in\operatorname{Av}_{n}(312)$ which has $n-k$ descents of width $k$. Determining the number of width-$k$ inversions in this permutation is done similarly. ∎

Although $321$ is Wilf equivalent to $312$, it is not so obvious how to construct a recurrence relation for $F_{n}^{\operatorname{des}_{k}}(321;q)$. To help describe the elements of $\operatorname{Av}_{n}(321)$, first let $\sigma=a_{1}\dots a_{n}\in\mathfrak{S}$ and call $a_{i}$ a left-right maximum if $a_{i}>a_{j}$ for all $j<i$. Thus $\sigma\in\operatorname{Av}_{n}(321)$ if and only if its set of non-left-right maxima form an increasing subsequence of $\sigma$. Indeed, if the non-left-right maxima did contain a descent, then there would be some left-right maximum preceding both elements, which violates the condition that $\sigma$ avoid $321$. Despite such a description, using it to reveal $F_{n}^{\operatorname{des}_{k}}(321;q)$ has thus far been unsuccessful. So, we leave the following as an open question.

Question 3.5**.**

Is there a closed formula or simple recursion for $F_{n}^{\operatorname{des}_{k}}(321;q)$ (equivalently, for $F_{n}^{\operatorname{des}_{k}}(123;q)$)?

3.2. Avoiding doubletons

At this point, we begin studying the functions $F_{n}^{\operatorname{des}_{k}}(\Pi;q)$ when avoiding doubletons from $\mathfrak{S}_{3}$. Recall that, by the Erdős-Szekeres theorem [2], there are no permutations in $\mathfrak{S}_{n}$ for $n\geq 5$ that avoid both $123$ and $321$. Thus, we will not consider this pair. Additionally, the functions $F_{n}^{\operatorname{inv}_{k}}(\Pi;q)$ are quite unwieldy, so we will not consider these either.

For our first nontrivial example, we begin with $\{123,132\}$. Permutations $a_{1}\dots a_{n}\in\operatorname{Av}_{n}(123,132)$ have the following structure: for any $i=1,\dots,n$, if $a_{i}=n$, then the substring $a_{1}a_{2}\dots a_{i-1}$ is decreasing and consists of the elements from the interval $[n-i+1,n-1]$. Additionally, the substring $a_{i+1}\dots a_{n}$ is an element of $\operatorname{Av}_{n-i}(123,132)$. This structure makes it easy to show that there are $2^{n-1}$ elements of $\operatorname{Av}_{n}(123,132)$ [9].

Proposition 3.6**.**

For all $n$ and $1\leq k\leq n-1$,

[TABLE]

Proof.

Let $\sigma=a_{1}\dots a_{n}\in\operatorname{Av}_{n}(123,132;q)$. If $a_{i}=n$ for $i\leq k$, then it is clear from the preceding description of the elements in $\operatorname{Av}_{n}(123,132)$ that all of $1,\dots,\min(i,n-k)$ are elements of $\operatorname{Des}_{k}(\sigma)$. If $j>n-k$ for some $j$, then $j\notin\operatorname{Des}_{k}(\sigma)$ since $a_{j+k}$ does not exist. The remaining elements of $\operatorname{Des}_{k}(\sigma)$ arise as a width-$k$ descent of $a_{i+1}\dots a_{n}$, which accounts for the factor of $F_{n-i}^{\operatorname{des}_{k}}(123,132;q)$ in the first summand.

Now, if $k+1\leq n-k$ and $a_{i}=n$ for $i=k+1,\dots,n-k$, then every element of $1,\ldots,i$ except $i-k$ is an element of $\operatorname{Des}_{k}(\sigma)$. If $k+1>n-k$, then the second summand is empty and nothing is lost by continuing to the case of $a_{i}=n$ for $i\geq\max(k+1,n-k+1)$. Again, the remaining elements of $\operatorname{Des}_{k}(\sigma)$ are the width-$k$ descents of $a_{i+1}\dots a_{n}$, hence the additional factor of $F_{n-i}^{\operatorname{des}_{k}}(123,132;q)$. This accounts for the second summand.

Finally, let $m=\max(k+1,n-k+1)$. If $a_{i}=n$ for $i\geq m$, then all of $1,\ldots,n-k$ are descents except $i-k$, and these are the only possible width-$k$ descents. In particular, none of $i+1,i+2,\dots,n$ can be the index for a width-$k$ descent. This leads to the sum

[TABLE]

which accounts for the third summand. ∎

Next, we consider $\{123,312\}$. We proceed similarly as before but with some minor differences, reflecting the new structure we encounter. If $\sigma=a_{1}\dots a_{n}\in\operatorname{Av}_{n}(123,312)$ and $a_{i}=1$ for some $i<n$, then $\sigma$ is of the form

[TABLE]

since neither subsequence $a_{1}\dots a_{i-1}$ and $a_{i+1}\dots a_{n}$ may contain an ascent. If $a_{n}=1$, though, then $\operatorname{std}(a_{1}\dots a_{n-1})\in\operatorname{Av}_{n-1}(123,312)$.

Proposition 3.7**.**

For all $n$ and $1\leq k<n$,

[TABLE]

Proof.

Suppose $a_{i}=1$ for some $i\leq k$. Using the description of elements in $\operatorname{Av}_{n}(123,312)$, we know there are $n-k-i$ width-$k$ descents. Since there is only one such permutation for each $i$, we simply add all of the $q^{n-k-i}$, so long as $n-k-i\geq 0$. If this inequality does not hold, then there are no width-$k$ descents in this range. This accounts for the first summand.

The second summand is computed very similarly to that of the second summand in Proposition 3.6. The final summand is a direct result of noting that when $a_{n}=1$, then $(a_{1}-1)\dots(a_{n-1}-1)$ may be any element of $\operatorname{Av}_{n-1}(123,312;q)$, which accounts for the factor $F_{n-1}^{\operatorname{des}_{k}}(123,312;q)$. For each of these choices, we know $n-k\in\operatorname{Des}_{k}(\sigma)$, hence the factor of $q$. Adding the sums results in the identity claimed. ∎

Now we consider $\{132,231\}$. Note that if $a_{1}\dots a_{n}\in\operatorname{Av}_{n}(132,231)$, then $a_{i}\neq n$ for any $1<i<n$. If $a_{1}=n$, then $\operatorname{std}(a_{2}\dots a_{n})\in\operatorname{Av}_{n-1}(132,231)$, and similarly if $a_{n}=n$. Once again, this allows us to quickly compute that $|\operatorname{Av}_{n}(132,231)|=2^{n-1}$.

Proposition 3.8**.**

Let $K=\{k_{1},\ldots,k_{l}\}$ be a nonempty subset of $[n-1]$ whose elements are listed in increasing order. We then have

[TABLE]

where $k_{0}=1$ and $k_{l+1}=n$.

Proof.

It follows from the description of elements in $\operatorname{Av}_{n}(132,231)$ that there are two summands in a recurrence for $F_{n}^{\operatorname{des}_{K}}(132,231;q)$: one corresponding to $a_{1}=n$ and one corresponding to $a_{n}=n$. When $a_{1}=n$, then there are $|K|=l$ copies of $1\in\operatorname{Des}_{K}(\sigma)$; if $a_{n}=n$, then $a_{n}$ makes no contribution to $\operatorname{Des}_{K}(\sigma)$. Thus, by deleting $n$ from $\sigma$, we get the recurrence

[TABLE]

Repeating this, a factor of $1+q^{l}$ appears until we get to

[TABLE]

At this point, note that

[TABLE]

Repeating the previous argument ends up with the identity claimed. ∎

Next, consider $\{132,213\}$. If $\sigma=a_{1}\dots a_{n}\in\operatorname{Av}_{n}(132,213)$, then if $a_{i}=n$ for any $i$, then $a_{1}\dots a_{i-1}$ must be increasing in order to avoid $213$. Moreover,

[TABLE]

in order to avoid $132$, and $\operatorname{std}(a_{i+1}\dots a_{n})\in\operatorname{Av}_{n-i}(132,213)$.

Proposition 3.9**.**

For all $n$ and $1\leq k<n$,

[TABLE]

Proof.

The proof of this is entirely analogous to the proof of Proposition 3.6. ∎

Next, we consider $\{132,312\}$. For each $i=1,\ldots,n-1$, either $a_{i+1}=\max\{a_{1},\ldots,a_{i}\}+1$ or $a_{i+1}=\min\{a_{1},\ldots,a_{i}\}-1$. This doubleton often results in especially pleasant formulas, and our results are no exception.

Proposition 3.10**.**

Let $K=\{k_{1},\ldots,k_{l}\}$ be a nonempty subset of $[n-1]$ whose elements are listed in increasing order. We then have

[TABLE]

where $k_{0}=1$ and $k_{l+1}=n$.

Proof.

From the description of elements $\sigma=a_{1}\dots a_{n}\in\operatorname{Av}_{n}(132,312)$, we know that either $a_{n}=1$ or $a_{n}=n$. In the former case, $n-k\in\operatorname{Des}_{K}(\sigma)$ for each $k\in K$, and in the latter case, $n-k\notin\operatorname{Des}_{K}(\sigma)$ for each $k\in K$. This leads to the recurrence

[TABLE]

Following an analogous argument as in the proof of Proposition 3.8 obtains the result. ∎

From the general formula given above, we can quickly determine $F_{n}^{\operatorname{inv}_{k}}(132,312;q)$.

Corollary 3.11**.**

For fixed $n,k$, write $n=dk+r$ for unique nonnegative integers $d,r$ such that $0\leq r<k$. We then have

[TABLE]

Proof.

This is a direct consequence of the general formulas from Propositions 3.8 and 3.10, and recalling that

[TABLE]

∎

Acknowledgement

The author thanks Bruce Sagan for his many helpful comments.