The alternating run polynomials of permutations

TL;DR

This paper explores the properties of alternating run polynomials in permutations, generalizes related identities, and introduces semi-gamma-positivity, providing new combinatorial interpretations and connections between different permutation classes.

Contribution

It presents a generalization of the David-Barton identity, offers a combinatorial interpretation of q-alternating run polynomials, and introduces the concept of semi-gamma-positivity for these polynomials.

Findings

Generalized David-Barton identity relating alternating run and Eulerian polynomials

Combinatorial interpretation of q-alternating run polynomials using grammars

Proved semi-gamma-positivity of alternating run polynomials of dual Stirling permutations

Abstract

In this paper, we first consider a generalization of the David-Barton identity which relate the alternating run polynomials to Eulerian polynomials. By using context-free grammars, we then present a combinatorial interpretation of a family of q-alternating run polynomials. Furthermore, we introduce the definition of semi-gamma-positive polynomial and we show the semi-gamma-positivity of the alternating run polynomials of dual Stirling permutations. A connection between the up-down run polynomials of permutations and the alternating run polynomials of dual Stirling permutations is established.

Equations

$$R_{n,k}$$

$$T_{n,k}$$

$$R_{n+1,k}$$

$$T_{n+1,k}=k{T}_{n,k}+T_{n,k-1}+(n-k+2)T_{n,k-2},$$

$$R_n(x)={\left(\frac{1+x}{2}\right)}^{n-1}(1+w{)}^{n+1}{A}_n\left(\frac{1-w}{1+w}\right)$$

$$\pi (i_1)>\pi (i_2)<\pi (i_3)>\cdots \pi (i_k),$$

$$T_n(x)=\sum _{\pi \in {\mathfrak{S}}_n}{x}^{\mathrm{as}(\pi )}.$$

$$T_n(x)=\frac{1}{2}(1+x)R_n(x)$$

$$T(x,z)=:\sum _{n=0}^{\infty }{T}_n(x)\frac{z^n}{n!}=(1-x)\frac{1+\rho +2x{e}^{\rho z}+(1-\rho )e^{2\rho z}}{1+\rho -x^2+(1-\rho -x^2)e^{2\rho z}}.$$

$$\sum _{n=0}^{\infty }\frac{z^n}{n!}\sum _{k=0}^n{R}_{n+1,k}{x}^{n-k}=\frac{1-x}{1+x}{\left(\frac{\sqrt{1-x^2}+\sin (z\sqrt{1-x^2})}{x-\cos (z\sqrt{1-x^2})}\right)}^2.$$

$$f(x)=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }{\gamma }_k{x}^k(1+x{)}^{n-2k},$$

$$M_n(x)=\sum _{k=0}^{\lfloor (n+\delta )/2\rfloor }M(n,k)x^k(1+x{)}^{n+\delta -2k}$$

$$N_n(x)={\left(\frac{1+x}{2}\right)}^{n-\delta }(1+w{)}^{n+\delta }{M}_n\left(\frac{1-w}{1+w}\right)$$

$$N_n(x)=\sum _{k=0}^{\lfloor (n+\delta )/2\rfloor }\frac{1}{2^{k-2\delta }}M(n,k)x^k(1+x{)}^{n-\delta -k}.$$

$$1-w^2$$

$$\frac{1-w}{1+w}$$

$$1+\frac{1-w}{1+w}$$

$$N_n(x)$$

$$={\alpha }^{n-\delta }(1+w{)}^{n+\delta }\sum _{k}M(n,k)\frac{1}{(1+w{)}^{2k}}\frac{x^k}{{\alpha }^k}{\left(\frac{2}{1+w}\right)}^{n+\delta -2k}$$

$$=\sum _{k}M(n,k)x^k{\alpha }^{n-\delta -k}{2}^{n+\delta -2k}$$

$$=\sum _{k}M(n,k)x^k{\left(\frac{1+x}{2}\right)}^{n-\delta -k}{2}^{n+\delta -2k}$$

$$=\sum _k\frac{1}{2^{k-2\delta }}M(n,k)x^k(1+x{)}^{n-\delta -k},$$

$$A_n(x)=\sum _{k=1}^{\lfloor (n+1)/2\rfloor }a(n,k)x^k(1+x{)}^{n+1-2k}$$

$$a(n,k)=ka(n-1,k)+(2n-4k+4)a(n-1,k-1),$$

$$R_n(x)=\sum _{k=1}^{\lfloor (n+1)/2\rfloor }\frac{1}{2^{k-2}}a(n,k)x^k(1+x{)}^{n-1-k}.$$

$$B_n(x)=\sum _{\pi \in B_n}{x}^{{\mathrm{des}}_B(\pi )}.$$

$$B_n(x)=\sum _{k=0}^{\lfloor n/2\rfloor }b(n,k)x^k(1+x{)}^{n-2k},$$

$$b(n,k)=(1+2k)b(n-1,k)+4(n-2k+1)b(n-1,k-1),$$

$$b_n(x)=\sum _{k=0}^{\lfloor n/2\rfloor }\frac{1}{2^k}b(n,k)x^k(1+x{)}^{n-k}.$$

$$b_n(x)={\left(\frac{1+x}{2}\right)}^n(1+w{)}^n{B}_n\left(\frac{1-w}{1+w}\right).$$

$$b_{n+1}(x)=(1+x+2n{x}^2)b_n(x)+2x(1-x^2)b_n^{\prime }(x),$$

$$c_{n+1}(x)=(2n{x}^2+3x-1)c_n(x)+2x(1-x^2)c_n^{\prime }(x).$$

$$c_n(x)=\sum _{\pi \in {\hat{B}}_n}{x}^{\mathrm{altrun}(\pi )}.$$

$$D^n(a)$$

Full text

Keywords: Alternating runs; Eulerian polynomials; Semi-$\gamma$-positivity; Stirling permutations

1. Introduction

The enumeration of permutations by number of alternating runs was first studied by André [1]. Knuth [19, Section 5.1.3] has discussed this topic in connection to sorting and searching. Over the past few decades, the study of alternating runs of permutations was initiated by David and Barton [12, 157-162].

Let $\mathfrak{S}_{n}$ denote the symmetric group of all permutations of $[n]=\{1,2,\ldots,n\}$. Let $\pi=\pi(1)\pi(2)\cdots\pi(n)\in\mathfrak{S}_{n}$. An alternating run of $\pi$ is a maximal consecutive subsequence that is increasing or decreasing (see [1, 22]). An up-down run of $\pi$ is an alternating run of $\pi$ endowed with a 0 in the front (see [13, 22]). Let ${\rm altrun\,}(\pi)$ (resp. ${\rm udrun\,}(\pi)$) be the number of alternating runs (resp. up-down runs) of $\pi$. For example, if $\pi=324156$, then ${\rm altrun\,}(\pi)=4,~{}{\rm udrun\,}(\pi)=5$. We define

[TABLE]

It is well known that these numbers satisfy the following recurrence relations

[TABLE]

[TABLE]

with the initial conditions $R_{1,0}=1$ and $R_{1,k}=0$ for $k\geq 1$, $T_{0,0}=1$ and $T_{0,k}=0$ for $k\geq 1$ (see [1, 13]). The alternating run polynomial and up-down run polynomial are respectively defined by $R_{n}(x)=\sum_{k=0}^{n-1}R_{n,k}x^{k}$ and $T_{n}(x)=\sum_{k=0}^{n}T_{n,k}x^{k}$.

A descent of $\pi\in\mathfrak{S}_{n}$ is an index $i\in[n-1]$ such that $\pi(i)>\pi(i+1)$. Denote by ${\rm des\,}(\pi)$ the number of descents of $\pi$. The classical Eulerian polynomial is defined by $A_{n}(x)=\sum_{\pi\in\mathfrak{S}_{n}}x^{{\rm des\,}(\pi)+1}$. By solving a differential equation, David and Barton [12, 157-162] established the identity:

[TABLE]

for $n\geq 2$, where $w=\sqrt{\frac{1-x}{1+x}}$. Using (2), Bóna proved that the polynomial $R_{n}(x)$ has only real zeros (see [4]). Moreover, one can prove that $R_{n}(x)$ has the zero $x=-1$ with the multiplicity $\lfloor\frac{n}{2}\rfloor-1$ by using (2), which can also be obtained based on the recurrence relation of $R_{n}(x)$ (see [25]). Motivated by (2), Zhuang [31] proved several identities expressing polynomials counting permutations by various descent statistics in terms of Eulerian polynomials.

Let us now recall another combinatorial interpretation of $T_{n}(x)$. An alternating subsequence of $\pi$ is a subsequence $\pi(i_{1})\cdots\pi(i_{k})$ satisfying

[TABLE]

where $i_{1}<i_{2}<\cdots<i_{k}$ (see [28]). Denote by ${\rm as\,}(\pi)$ the number of terms of the longest alternating subsequence of $\pi$. By definition, we see that ${\rm as\,}(\pi)={\rm udrun\,}(\pi)$. Thus

[TABLE]

There has been much recent work related to the numbers $R_{n,k}$ and $T_{n,k}$. In [3], Bóna and Ehrenborg proved that $R_{n,k}^{2}\geq R_{n,k-1}R_{n,k+1}$. Subsequently, Bóna [4, Section 1.3.2] noted that

[TABLE]

for $n\geq 2$. Set $\rho=\sqrt{1-x^{2}}$. Stanley [28, Theorem 2.3] showed that

[TABLE]

By using (3) and (4), Stanley [28] obtained explicit formulas of $T_{n,k}$ and $R_{n,k}$. Canfield and Wilf [6] presented an asymptotic formula for $R_{n,k}$. In [21], another explicit formula of $R_{n,k}$ was obtained by combining the derivative polynomials of tangent function and the following generating function obtained by Carlitz [7]:

[TABLE]

In [22], several convolution formulas of the polynomials $R_{n}(x)$ and $T_{n}(x)$ are obtained by using Chen’s grammars. By generalizing a reciprocity formula of Gessel, Zhuang [30] obtained generating function for permutation statistics that are expressible in terms of alternating runs. Very recently, Josuat-Vergès and Pang [18] showed that alternating runs can be used to define subalgebras of Solomon’s descent algebra.

In this paper, we continue the work initiated by David and Barton [12]. In Section 2, we consider a generalization of (2). In Section 3, we present a combinatorial interpretation of a family of $q$-alternating run polynomials by using Chen’s grammars. In Section 4, we show the semi-$\gamma$-positivity of the alternating run polynomials of dual Stirling permutations.

2. The David-Barton type identity

Let $f(x)=\sum_{i=0}^{n}f_{i}x^{i}$ be a symmetric polynomial, i.e., $f_{i}=f_{n-i}$ for any $0\leq i\leq n$. Then $f(x)$ can be expanded uniquely as

[TABLE]

and it is said to be $\gamma$-positive if $\gamma_{k}\geq 0$ for $0\leq k\leq\lfloor\frac{n}{2}\rfloor$ (see [15]). The $\gamma$-positivity provides an approach to study symmetric and unimodal polynomials and has been extensively studied (see [2, 5, 10, 20] for instance).

The first main result of our paper is the following, which shows that the David-Barton type identities often occur in combinatorics and geometry.

Theorem 1**.**

Let

[TABLE]

be a symmetric polynomial, where $\delta$ is a fixed integer. Set $w=\sqrt{\frac{1-x}{1+x}}$. Then

[TABLE]

if and only if

[TABLE]

Proof.

Set $\alpha=\frac{1+x}{2}$. Note that

[TABLE]

It follows from (5) that

[TABLE]

and vice versa. This completes the proof. ∎

The reader is referred to [2] for a survey of some recent results on $\gamma$-positivity. For any $\gamma$-positive polynomial $M_{n}(x)$, we can define an associated polynomial $N_{n}(x)$ by using (6). And then we get a David-Barton type identity (5). As illustrations, in the rest of this section, we shall present two examples.

For example, Foata and Schützenberger [14] discovered that

[TABLE]

for $n\geq 1$, where the numbers $a(n,k)$ satisfy the recurrence relation

[TABLE]

with the initial conditions $a(1,1)=1$ and $a(1,k)=0$ for $k\neq 1$ (see [10, 26] for instance). By using the David-Barton identity (2) and Theorem 1, we immediately get the following result.

Proposition 2**.**

For $n\geq 2$, we have

[TABLE]

Let $\pm[n]=\{\pm 1,\pm 2,\ldots,\pm n\}$. Let $B_{n}$ be the hyperoctahedral group of rank $n$. Elements of $B_{n}$ are signed permutations of $\pm[n]$ with the property that $\pi(-i)=-\pi(i)$ for all $i\in[n]$. In the sequel, we always assume that signed permutations in $B_{n}$ are prepended by 0. That is, we identify a signed permutation $\pi=\pi(1)\cdots\pi(n)$ with the word $\pi(0)\pi(1)\cdots\pi(n)$, where $\pi(0)=0$. A type $B$ descent is an index $i\in\{0,1,\ldots,n-1\}$ such that $\pi(i)>\pi(i+1)$. Let ${\rm des\,}^{B}(\pi)$ be the number of type $B$ descents of $\pi$. The type $B$ Eulerian polynomials are defined by

[TABLE]

It is well known that

[TABLE]

where the numbers $b(n,k)$ satisfy the recurrence relation

[TABLE]

with the initial conditions $b(1,0)=1$ and $b(1,k)=0$ for $k\neq 0$ (see [2, 10, 26]).

Define

[TABLE]

Then by Theorem 1, we get the following result.

Proposition 3**.**

For $n\geq 1$, we have

[TABLE]

Combining (7) and (8), we see that the polynomials $b_{n}(x)$ satisfy the recurrence relation

[TABLE]

with the initial conditions $b_{0}(x)=1,~{}b_{1}(x)=1+x$. For $n\geq 1$, we define $b_{n}(x)=\frac{1+x}{x}c_{n}(x)$. It follows from (9) that the polynomials $c_{n}(x)$ satisfy the recurrence relation

[TABLE]

Let $\widehat{B}_{n}=\{\pi\in B_{n}\mid\pi(1)>0\}$. There is a combinatorial interpretation of $c_{n}(x)$ (see [11, 29]):

[TABLE]

3. The $q$-alternating runs polynomials

For an alphabet $A$, let $\mathbb{Q}[[A]]$ be the rational commutative ring of formal power series in monomials formed from letters in $A$. A Chen’s grammar (which is known as context-free grammar) over $A$ is a function $G:A\rightarrow\mathbb{Q}[[A]]$ that replaces a letter in $A$ by an element of $\mathbb{Q}[[A]]$, see [8, 9, 24] for details. The formal derivative $D:=D_{G}$ is a linear operator defined with respect to a context-free grammar $G$. Following [9], a grammatical labeling is an assignment of the underlying elements of a combinatorial structure with variables, which is consistent with the substitution rules of a grammar.

Let us now recall two results on context-free grammars.

Proposition 4** ([22, Theorem 6]).**

If $G=\{a\rightarrow ab,~{}b\rightarrow bc,~{}c\rightarrow b^{2}\}$, then

[TABLE]

Proposition 5** ([22, Theorem 9]).**

If $G=\{a\rightarrow 2ab,~{}b\rightarrow bc,~{}c\rightarrow b^{2}\}$, then

[TABLE]

Combining Leibniz’s formula and Proposition 4, we see that

[TABLE]

Motivated by Propositions 4 and  5, it is natural to consider the grammar

[TABLE]

Note that $D_{G_{1}}(a)=qab,~{}D_{G_{1}}^{2}(a)=a(q^{2}b^{2}+qbc)$. By induction, it is easy to verify that

[TABLE]

It follows from (10) that

[TABLE]

which leads to the recurrence relation

[TABLE]

The $q$-alternating run polynomials are defined by

[TABLE]

In particular, $R_{n}(x;1)=T_{n}(x)$, $R_{n}(x;2)=R_{n+1}(x).$ The first few $R_{n}(x;q)$ are given as follows:

[TABLE]

We define

[TABLE]

Proposition 6**.**

We have $R(x,z;q)=T^{q}(x,z)$, where $T(x,z)$ is given by (4). Therefore,

[TABLE]

Moreover, we have $R_{n}(x;-q)=R_{n}(-x;q)$ and $R_{n}(-x;-q)=R_{n}(x;q)$.

Proof.

By rewriting (12) in terms of generating function $R(x,z;q)$, we obtain

[TABLE]

It is routine to check that the generating function $T^{q}(x,z)$ satisfies (14). Also, this generating function gives $T^{q}(0,z)=T^{q}(x,0)=1$. Hence $R(x,z;q)=T^{q}(x,z)$. It is routine to check that

[TABLE]

which leads to the desired result. ∎

We say that $\pi\in\mathfrak{S}_{n}$ is a circular permutation if it has only one cycle. Let $A=\{x_{1},x_{2},\ldots,x_{k}\}$ be a finite set of positive integers, and let ${\mathcal{C}}_{A}$ be the set of all circular permutations of $A$. We will write a permutation $w\in{\mathcal{C}}_{A}$ by using its canonical presentation $w=y_{1}y_{2}\cdots y_{k}$, where $y_{1}=\min A,y_{i}=w^{i-1}(y_{1})$ for $2\leq i\leq k$ and $y_{1}=w^{k}(y_{1})$. A cycle peak (resp. cycle double ascent, cycle double descent) of $w$ is an entry $y_{i}$, $2\leq i\leq k$, such that $y_{i-1}<y_{i}>y_{i+1}$ (resp. $y_{i-1}<y_{i}<y_{i+1}$, $y_{i-1}>y_{i}>y_{i+1}$), where we set $y_{k+1}=\infty$. Let ${\rm cpk\,}(w)$ (resp. ${\rm cdasc\,}(w)$, ${\rm cddes\,}(w)$, ${\rm cyc\,}(w)$) be the number of cycle peaks (resp. cycle double ascents, cycle double descents, cycles) of $w$.

Definition 7**.**

A cycle run of a circular permutation $w$ is an alternating run of $w$ endowed with a $\infty$ in the end. Let ${\rm crun\,}(w)$ be the number of cycle runs of $w$.

It is clear that ${\rm crun\,}(w)=2{\rm cpk\,}(w)+1$. In the following discussion we always write $\pi\in\mathfrak{S}_{n}$ in standard cycle decomposition: $\pi=w_{1}\cdots w_{k}$, where the cycles are written in increasing order of their smallest entry and each of these cycles is expressed in canonical presentation. We define

[TABLE]

In particular, ${\rm crun\,}((1)(2)\cdots(n))=\sum_{i=1}^{n}{\rm crun\,}(i)=\sum_{i=1}^{n}{\rm altrun\,}(i\infty)=n$. We can now present the second main result.

Theorem 8**.**

For $n\geq 1$, we have

[TABLE]

Proof.

For $\pi\in\mathfrak{S}_{n}$, we first put a $\infty$ in the end of each cycle. We then introduce a grammatical labeling of $\pi$ as follows:

• ($L_{1}$)

Put a subscript label $q$ at the end of each cycle of $\pi$;

• ($L_{2}$)

Put a superscript label $a$ at the end of $\pi$;

• ($L_{3}$)

Put a superscript label $b$ before each $\infty$;

• ($L_{4}$)

If $\pi(i)$ is a cycle peak, then put a superscript label $b$ before $\pi(i)$ and a superscript label $b$ right after $\pi$;

• ($L_{5}$)

If $\pi(i)$ is a cycle double ascents, then put the superscript label $c$ before $\pi(i)$;

• ($L_{6}$)

If $\pi(i)$ is a cycle double descents, then put the superscript label $c$ right after $\pi(i)$.

The weight of $\pi$ is the product of its labels. When $n=1,2$, we have

[TABLE]

Then the weight of $(1^{b})_{q}^{a}$ is given by $D_{G_{1}}(a)$, and the sum of weights of the elements in $\mathfrak{S}_{2}$ is given by $D_{G_{1}}^{2}(a)$. Hence the result holds for $n=1,2$. Let

[TABLE]

Suppose we get all labeled permutations in $r_{n-1}(i,j)$, where $n\geq 3$. Let $\pi^{\prime}$ be obtained from $\pi\in r_{n-1}(i,j)$ by inserting the entry $n$. We distinguish the following four cases:

• ($c_{1}$)

If we insert $n$ as a new cycle, then $\pi^{\prime}\in r_{n-1}(i+1,j+1)$. This case corresponds to the substitution rule $a\rightarrow qab$.

• ($c_{2}$)

If we insert $n$ before a $\infty$, then $\pi^{\prime}\in r_{n-1}(i,j)$. This case corresponds to the substitution rule $b\rightarrow bc$;

• ($c_{3}$)

If we insert $n$ before or right after a cycle peak, then $\pi^{\prime}\in r_{n-1}(i,j)$. This case corresponds to the substitution rule $b\rightarrow bc$;

• ($c_{4}$)

If we insert $n$ before a cycle double ascents or right after a cycle double descents, then $\pi^{\prime}\in r_{n-1}(i+2,j)$. This case corresponds to the substitution rule $c\rightarrow b^{2}$.

In each case, the insertion of $n$ corresponds to one substitution rule in the grammar (10). It is easy to check that the action of $D_{G_{1}}$ on elements of $\mathfrak{S}_{n-1}$ generates all elements of $\mathfrak{S}_{n}$. Using (11) and by induction, we present a constructive proof of (15). This completes the proof. ∎

We define

[TABLE]

[TABLE]

By using the principle of inclusion-exclusion, it is routine to verify that

[TABLE]

Hence

[TABLE]

A permutation $\pi\in\mathfrak{S}_{n}$ is a derangement if $\pi(i)\neq i$ for any $i\in[n]$. Let $\mathcal{D}_{n}$ denote the set of derangements in $\mathfrak{S}_{n}$. Then

[TABLE]

Proposition 9**.**

Set $d_{n}(x)=R_{n}(x,0;1)$. Then the polynomials $d_{n}(x)$ satisfy the recurrence

[TABLE]

with the initial conditions $d_{0}(x)=1,~{}d_{1}(x)=0$. In particular, $d_{n}(-1)=-(n-1)$ for $n\geq 1$.

Proof.

Let $d(x,z)=\sum_{n=0}^{\infty}d_{n}(x)\frac{z^{n}}{n!}$. It follow from (16) that

[TABLE]

By rewriting (1) in terms of generating function $T(x,z)$, we obtain

[TABLE]

Hence

[TABLE]

which yields the desired recurrence relation. ∎

Let $d_{n}(x)=\sum_{k=0}^{n}d_{n,k}x^{k}$. By using (18), it is not hard to verify that

[TABLE]

4. Semi-$\gamma$-positive polynomials

Let $g(x)=\sum_{i=0}^{2n}g_{i}x^{i}$ be a symmetric polynomial. Note that

[TABLE]

Hence $g(x)$ can be expanded as

[TABLE]

It is clear that if $\gamma_{i}\geq 0$ for all $0\leq i\leq n$, then $\lambda_{k}\geq 0$ for all $0\leq k\leq n$. Furthermore, we have

[TABLE]

Similarly, if $h(x)=\sum_{i=0}^{2n+1}h_{i}x^{i}$ a symmetric polynomial, then we have

[TABLE]

Hence $h(x)$ can be expanded as

[TABLE]

Definition 10**.**

If $f(x)=(1+x)^{\nu}\sum_{k=0}^{n}\lambda_{k}x^{k}(1+x^{2})^{n-k}$ and $\lambda_{k}\geq 0$ for all $0\leq k\leq n$, then we say that $f(x)$ is semi-$\gamma$-positive, where $\nu=0$ or $\nu=1$.

It should be noted that a semi-$\gamma$-positive polynomial is not always $\gamma$-positive. From the above discussion it follows that we have the following result.

Proposition 11**.**

If $f(x)=(1+x)^{\nu}\left(f_{1}(x^{2})+xf_{2}(x^{2})\right)$ is a semi-$\gamma$-positive polynomial, then both $f_{1}(x)$ and $f_{2}(x)$ are $\gamma$-positive.

In the following, we shall show the semi-$\gamma$-positivity of the alternating run polynomials of dual Stirling permutations. Following [16], a Stirling permutation of order $n$ is a permutation of the multiset $\{1,1,\ldots,n,n\}$ such that for each $i$, $1\leq i\leq n$, all entries between the two occurrences of $i$ are larger than $i$. There has been much recent work on Stirling permutations, see [17, 24] and references therein.

Denote by $\mathcal{Q}_{n}$ the set of Stirling permutations of order $n$. Let $\sigma=\sigma_{1}\sigma_{2}\cdots\sigma_{2n}\in\mathcal{Q}_{n}$. Let $\Phi$ be the injection which maps each first occurrence of entry $j$ in $\sigma$ to $2j$ and the second $j$ to $2j-1$, where $j\in[n]$. For example, $\Phi(221331)=432651$. Let $\Phi(\mathcal{Q}_{n})=\{\pi\mid\sigma\in\mathcal{Q}_{n},\Phi(\sigma)=\pi\}$ be the set of dual Stirling permutations of order $n$. Clearly, $\Phi(\mathcal{Q}_{n})$ is a subset of $\mathfrak{S}_{2n}$. For $\pi\in\Phi(\mathcal{Q}_{n})$, the entry $2j$ is to the left of $2j-1$, and all entries in $\pi$ between $2j$ and $2j-1$ are larger than $2j$, where $1\leq j\leq n$. Noted that $\pi\in\Phi(\mathcal{Q}_{n})$ always ends with a descending run. The alternating runs polynomials of dual Stirling permutations are defined by

[TABLE]

According to [23], the numbers $F_{n,k}$ satisfy the recurrence relation

[TABLE]

with the initial conditions $F_{0,0}=1,~{}F_{1,1}=1$ and $F_{n,0}=0$ for $n\geq 1$. It follows from (19) that

[TABLE]

The first few $F_{n}(x)$ are given as follows:

[TABLE]

Let

[TABLE]

By induction, it is to verify that

[TABLE]

Lemma 12** ([23]).**

If

[TABLE]

then we have

[TABLE]

We now recall another combinatorial interpretation of $F_{n}(x)$. An occurrence of an ascent-plateau of $\sigma\in\mathcal{Q}_{n}$ is an index $i$ such that $\sigma_{i-1}<\sigma_{i}=\sigma_{i+1}$, where $i\in\{2,3,\ldots,2n-1\}$. An occurrence of a left ascent-plateau is an index $i$ such that $\sigma_{i-1}<\sigma_{i}=\sigma_{i+1}$, where $i\in\{1,2,\ldots,2n-1\}$ and $\sigma_{0}=0$. Let ${\rm ap\,}(\sigma)$ and ${\rm la\,}(\sigma)$ be the numbers of ascent-plateaus and left ascent-plateaus of $\sigma$, respectively. The number of flag ascent-plateaus of $\sigma$ is defined by

[TABLE]

Clearly, ${\rm fap\,}(\sigma)={\rm ap\,}(\sigma)+{\rm la\,}(\sigma)$. Following [24, Section 3], we have

[TABLE]

Thus,

[TABLE]

In fact, it is easy to verify that ${\rm fap\,}(\sigma)={\rm altrun\,}(\Phi(\sigma))$ for any $\sigma\in\mathcal{Q}_{n}$.

Proposition 13**.**

For $n\geq 1$, we have

[TABLE]

where the numbers $\gamma_{n,k}$ satisfy the recurrence relation

[TABLE]

with the initial conditions $\gamma_{1,1}=1$ and $\gamma_{1,k}=0$ for $k\neq 1$. In particular,

[TABLE]

Proof.

We first consider a change of the grammar (20). Set $a=yz$ and $b=y+z$. Then we have $D(x)=xa,~{}D(a)=a(b^{2}-2a),~{}D(b)=ab$. If

[TABLE]

then by induction, we see that there exist integers $\gamma_{n,k}$ such that

[TABLE]

Note that

[TABLE]

By comparing the coefficients of $a^{k}b^{2n-2k+2}$, we immediately get (22). Moreover, it is clear that $\gamma_{n,0}=0$ for $n\geq 1$. By using (23), upon taking $a=yz$ and $b=y+z$, we get

[TABLE]

Then comparing (24) with (21), we see that $F_{n}(x)=\sum_{k=1}^{n}\gamma_{n,k}x^{k}(1+x)^{2n-2k}$ for $n\geq 1$. By using (22), we obtain

[TABLE]

which yields the desired explicit formula. ∎

For $n\geq 1$, let $\gamma_{n}(x)=\sum_{k=1}^{n}\gamma_{n,k}x^{k}$. It follows from (22) that

[TABLE]

The first few $\gamma_{n}(x)$ are $\gamma_{0}(x)=1,~{}\gamma_{1}(x)=x,~{}\gamma_{2}(x)=x-x^{2},~{}\gamma_{3}(x)=x-x^{2}+3x^{3}$. From Proposition 13, we see that for any positive even integer $n$, the polynomial $F_{n}(x)$ is not $\gamma$-positive.

We can now present the third main result of this paper.

Theorem 14**.**

The polynomial $F_{n}(x)$ is semi-$\gamma$-positive. More precisely, we have

[TABLE]

where the numbers $f_{n,k}$ satisfy the recurrence relation

[TABLE]

with the initial conditions $f_{0,0}=1$ and $f_{n,0}=0$ for $n\geq 1$. Let $f_{n}(x)=\sum_{k=0}^{n}f_{n,k}x^{k}$. Then

[TABLE]

where $T(x,z)$ is given by (4).

Proof.

We first consider the grammar (20). Note that

[TABLE]

Set $u=yz$ and $v=y^{2}+z^{2}$. Then we have $D(x)=xu,~{}D(u)=uv$ and $D(v)=4u^{2}$. If

[TABLE]

then by induction we see that there exist nonnegative integers $f_{n,k}$ such that

[TABLE]

Note that

[TABLE]

By comparing the coefficients of $u^{k}v^{n+1-k}$, we get (25). Moreover, it follows from (27) that $f_{0,0}=1$ and $f_{n,0}=0$ for $n\geq 1$. By using (28), upon taking $u=yz$ and $v=y^{2}+z^{2}$, we get

[TABLE]

By comparing (29) with (21), we get

[TABLE]

We now consider a change of the grammar (10). Set $q=\frac{1}{2},~{}a=x,~{}b=2u,~{}c=v$. Then

[TABLE]

which are the substitution rules in the grammar (27). Hence it follows from (13) that

[TABLE]

which leads to $f(x,z)=R(2x,z;1/2)=\sqrt{T(2x,z)}$. This completes the proof. ∎

Combining (26) and (30), we immediately get the following result.

Corollary 15**.**

We have

[TABLE]

It would be interesting to present a combinatorial interpretation of Corollary 15. By using (26), it is not hard to verify that

[TABLE]

It should be noted that the numbers $f_{n,n}$ appear as A012259 in [27].

5. Concluding remarks

This paper gives a survey of some results related to alternating runs of permutations. We present a method to construct David-Barton type identities, and based on the survey [2], one can derive several David-Barton type identities. Moreover, we introduce the definition of semi-$\gamma$-positive polynomial. The $\gamma$-positivity of a polynomial $f(x)$ is a sufficient (not necessary) condition for the semi-$\gamma$-positivity of $f(x)$. In particular, we show that the alternating run polynomials of dual Stirling permutations are semi-$\gamma$-positive.