2.1 The scalar product and Foulkes’s theorem

Let $\left \langle \cdot ,\cdot \right \rangle \colon \Lambda \times \Lambda \rightarrow \mathbb {Q}$ denote the usual scalar product on symmetric functions defined by

$$\begin{align*}\left\langle m_{\lambda},h_{\mu}\right\rangle =\begin{cases} 1, & \text{if }\lambda=\mu\\ 0, & \text{otherwise,} \end{cases} \end{align*}$$

for all partitions $\lambda $ and $\mu $ and extending bilinearly, that is, by requiring that $\{m_{\lambda }\}$ and $\{h_{\mu }\}$ be dual bases. Then we have

$$\begin{align*}\left\langle p_{\lambda},p_{\mu}\right\rangle =\begin{cases} z_{\lambda}, & \text{if }\lambda=\mu\\ 0, & \text{otherwise,} \end{cases} \end{align*}$$

for all $\lambda $ and $\mu $ [Reference Stanley23, Proposition 7.9.3]. We extend the scalar product in the obvious way to symmetric functions involving other variables as well as symmetric functions of unbounded degree. Note that the scalar product is not always defined for the latter; for example, we have $\left \langle H(z),H\right \rangle =\sum _{n=0}^{\infty }z^{n}$ but $\left \langle H,H\right \rangle $ is undefined.

Given a composition L, let $r_{L}$ denote the skew Schur function of ribbon shape L. That is, for $L=(L_{1},L_{2},\dots ,L_{k})$ , we have

$$\begin{align*}r_{L}=\sum_{i_{1},\dots,i_{n}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}, \end{align*}$$

where the sum is over all $i_{1},\dots ,i_{n}$ satisfying

$$\begin{align*}\underset{L_{1}}{\underbrace{i_{1}\leq\cdots\leq i_{L_{1}}}}>\underset{L_{2}}{\underbrace{i_{L_{1}+1}\leq\cdots\leq i_{L_{1}+L_{2}}}}>\cdots>\underset{L_{k}}{\underbrace{i_{L_{1}+\cdots+L_{k-1}+1}\leq\cdots\leq i_{n}}.} \end{align*}$$

The next theorem, due to Foulkes [Reference Foulkes9] (see also [Reference Gessel12, Theorem 5] and [Reference Stanley24, Corollary 7.23.8]), will play a pivotal role in our approach to two-sided descent statistics.

Foulkes’s theorem is a special case of a more general theorem of Gessel on quasisymmetric generating functions, which we briefly describe below. For a composition $L=(L_{1},L_{2},\dots ,L_{k})$ , let , and recall that the fundamental quasisymmetric function $F_{L}$ is defined by

Moreover, given a set $\Pi $ of permutations, its quasisymmetric generating function $Q(\Pi )$ is defined by

The following is Corollary 4 of Gessel [Reference Gessel12].

Because $r_{M}$ is the quasisymmetric generating function for permutations whose inverse has descent composition M, Foulkes’s theorem follows from Theorem 2.2.

Gessel and Reutenauer [Reference Gessel and Reutenauer13] showed that, for any partition $\lambda $ , the quasisymmetric generating function for permutations with cycle type $\lambda $ is a symmetric function, and they used this fact along with Theorem 2.2 to study the joint distribution of $\operatorname {\mathrm {maj}}$ and $\operatorname {\mathrm {des}}$ over sets of permutations with restricted cycle structure, including cyclic permutations, involutions and derangements. The present authors later studied the distributions of $(\operatorname {\mathrm {pk}},\operatorname {\mathrm {des}})$ , $(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {des}})$ and $\operatorname {\mathrm {udr}}$ over permutations with restricted cycle structure [Reference Gessel and Zhuang15]. Our current work is a continuation of this line of research but for inverse descent classes.

2.2 Plethysm

Let A be a $\mathbb {Q}$ -algebra of formal power series (possibly containing $\Lambda $ ). Consider the operation $\Lambda \times A\rightarrow A$ , where the image of $(f,a)\in \Lambda \times A$ is denoted $f[a]$ , defined by the following two properties:

1. 1. For any $i\geq 1$ , $p_{i}[a]$ is the result of replacing each variable in a with its ith power.

2. 2. For any fixed $a\in A$ , the map $f\mapsto f[a]$ is a $\mathbb {Q}$ -algebra homomorphism from $\Lambda $ to A.

In other words, we have $p_{i}[f(x_{1},x_{2},\dots )]=f(x_{1}^{i},x_{2}^{i},\dots )$ for any $f\in \Lambda $ . If f contains variables other than the $x_{i}$ , then they are all raised to the ith power as well. For example, if q and t are variables, then $p_{i}[q^{2}tp_{m}]=q^{2i}t^{i}p_{im}$ . The map $(f,a)\mapsto f[a]$ is called plethysm. As with the scalar product, we extend plethysm in the obvious way to symmetric functions of unbounded degree with coefficients involving other variables; whenever we do so, we implicitly assume that any infinite sums involved converge as formal power series so that the plethysms are defined.

We will need several technical lemmas involving plethysm in order to evaluate the scalar products needed in our work. All of these lemmas are from [Reference Gessel and Zhuang15] by the present authors (or are easy consequences of results from [Reference Gessel and Zhuang15]).

A monic term is any monomial with coefficient 1.

Proof. By a special case of [Reference Gessel and Zhuang15, Lemma 2.5], we have

$$\begin{align*}f[\mathsf{m}]=\left\langle f,H(\mathsf{m})\right\rangle =\left\langle H(\mathsf{m}),f\right\rangle \end{align*}$$

and the result follows from the fact that $f\mapsto f[\mathsf {m}]$ is a $\mathbb {Q}$ -algebra homomorphism.

Proof. Given $f\in A$ and a variable $\alpha \in A$ , it is known [Reference Gessel and Zhuang15, Lemma 3.2] that

$$\begin{align*}f[k(1-\alpha)]=\left\langle f,E(-\alpha)^{k}H^{k}\right\rangle. \end{align*}$$

Then the map $f\mapsto \left \langle f,E(y)^{k}H^{k}\right \rangle $ is obtained by composing the map $f\mapsto f[k(1-\alpha )]$ with evaluation at $\alpha =-y$ , both of which are $\mathbb {Q}$ -algebra homomorphisms.

The two lemmas below are Lemmas 3.5 and 6.6 of [Reference Gessel and Zhuang15], respectively.

2.3 Symmetric function generating functions for descent statistics

As mentioned in the introduction, we will need generating functions for the ribbon skew Schur functions $r_{L}$ which keep track of the descent statistics that we are studying. Such generating functions were produced in previous work by the present authors and are stated in the next lemma. Part (a) is essentially a commutative version of [Reference Gessel and Zhuang14, Lemma 17], whereas (b)–(d) are given in [Reference Gessel and Zhuang15, Lemma 2.8].

Lemma 2.8. We have

(a) $$ \begin{align} \sum_{n=0}^{\infty}t^{n}\prod_{k=0}^{n}H(q^{k}x) & =\sum_{n=0}^{\infty}\frac{\sum_{L\vDash n}q^{\operatorname{\mathrm{maj}}(L)}t^{\operatorname{\mathrm{des}}(L)}r_{L}}{(1-t)(1-qt)\cdots(1-q^{n}t)}x^{n}, \end{align} $$

(b) $$ \begin{align} \frac{1}{1-tE(yx)H(x)}&=\frac{1}{1-t}+\\ &\quad\frac{1}{1+y}\sum_{n=1}^{\infty}\sum_{L\vDash n}\left(\frac{1+yt}{1-t}\right)^{n+1}\left(\frac{(1+y)^{2}t}{(y+t)(1+yt)}\right)^{\operatorname{\mathrm{pk}}(L)+1}\left(\frac{y+t}{1+yt}\right)^{\operatorname{\mathrm{des}}(L)+1}r_{L}x^{n},\nonumber \end{align} $$

(c) $$ \begin{align} \frac{H(x)}{1-tE(yx)H(x)}=\frac{1}{1-t}+ \sum_{n=1}^{\infty}\sum_{L\vDash n}\frac{(1+yt)^{n}}{(1-t)^{n+1}}\left(\frac{(1+y)^{2}t}{(y+t)(1+yt)}\right)^{\operatorname{\mathrm{lpk}}(L)}\left(\frac{y+t}{1+yt}\right)^{\operatorname{\mathrm{des}}(L)}r_{L}x^{n}, \end{align} $$

and

(d) $$ \begin{align} \frac{1+tH(x)}{1-t^{2}E(x)H(x)} & =\frac{1}{1-t}+\frac{1}{2(1-t)^{2}}\sum_{n=1}^{\infty}\sum_{L\vDash n}\frac{(1+t^{2})^{n}}{(1-t^{2})^{n-1}}\left(\frac{2t}{1+t^{2}}\right)^{\operatorname{\mathrm{udr}}(L)}r_{L}x^{n}. \end{align} $$

Some of these generating functions have nice power sum expansions that are expressible in terms of Eulerian polynomials and type B Eulerian polynomials. The nth type B Eulerian polynomial $B_{n}(t)$ encodes the distribution of the type B descent number over the nth hyperoctahedral group (see [Reference Zhuang30, Section 2.3] for definitions) but can also be defined by the formula

$$\begin{align*}\frac{B_{n}(t)}{(1-t)^{n+1}}=\sum_{k=0}^{\infty}(2k+1)^{n}t^{k} \end{align*}$$

analogous to Equation (1.3). Recall that $l(\lambda )$ is the number of parts of the partition $\lambda $ . We also define $o(\lambda )$ to be the number of odd parts of $\lambda $ , and use the notation $\sum _{\lambda \text { odd}}$ for a sum over partitions in which every part is odd.

The next lemma is [Reference Gessel and Zhuang15, Lemma 2.9].

Lemma 2.9. We have

(a) $$ \begin{align} \frac{1}{1-tE(yx)H(x)}=\sum_{\lambda}\frac{p_{\lambda}}{z_{\lambda}}\frac{A_{l(\lambda)}(t)}{(1-t)^{l(\lambda)+1}}x^{\left|\lambda\right|}\prod_{k=1}^{l(\lambda)}(1-(-y)^{\lambda_{k}}) \end{align} $$

where $\lambda _{1},\lambda _{2},\dots ,\lambda _{l(\lambda )}$ are the parts of $\lambda $ ,

(b) $$ \begin{align} \frac{H(x)}{1-tE(x)H(x)}=\sum_{\lambda}\frac{p_{\lambda}}{z_{\lambda}}\frac{B_{o(\lambda)}(t)}{(1-t)^{o(\lambda)+1}}x^{\left|\lambda\right|}, \end{align} $$

and

(c) $$ \begin{align} \frac{1+tH(x)}{1-tE(x)H(x)}=\sum_{\lambda\:\mathrm{odd}}\frac{p_{\lambda}}{z_{\lambda}}2^{l(\lambda)}\frac{A_{l(\lambda)}(t^{2})}{(1-t^{2})^{l(\lambda)+1}}x^{\left|\lambda\right|}+t\sum_{\lambda}\frac{p_{\lambda}}{z_{\lambda}}\frac{B_{o(\lambda)}(t^{2})}{(1-t^{2})^{o(\lambda)+1}}x^{\left|\lambda\right|}. \end{align} $$

2.4 Sums involving $z_\lambda $ and Stanley’s formula for doubly alternating permutations

The remainder of this section is not strictly about symmetric functions but relates to the constants $z_\lambda $ which appear in symmetric function theory, and a connection to an enumeration formula of Stanley obtained via symmetric function techniques.

To derive some of our formulas later on, we will need to evaluate several sums like

(2.1) $$ \begin{align} \sum_{\substack{\lambda\vdash n\\unicode{x0142}(\lambda) =k}}\frac{1}{z_\lambda}. \end{align} $$

The key to evaluating these sums is the well-known fact that $n!/z_\lambda $ is the number of permutations in $\mathfrak {S}_n$ of cycle type $\lambda $ ; see, for example, [Reference Stanley23, Proposition 1.3.2] and [Reference Stanley24, pp. 298–299]. Thus, Equation (2.1) is equal to $c(n,k)/n!$ , where as before, $c(n,k)$ is the unsigned Stirling number of the first kind, which counts permutations in $\mathfrak {S}_n$ with k cycles.

Now, let $d(n,k)$ be the number of permutations in $\mathfrak {S}_n$ with k cycles, all of odd length; let $e(n,k)$ be the number of permutations in $\mathfrak {S}_n$ with k odd cycles (and any number of even cycles); and let $f(n,k,m)$ be the number of permutations in $\mathfrak {S}_n$ with k odd cycles and m cycles in total. Note that $d(n,k)$ , $e(n,k)$ and $f(n,k,m)$ are 0 if $n-k$ is odd. Then by the same reasoning as above, we have the additional sum evaluations in the next lemma.

Lemma 2.10. We have

(a) $$ \begin{align} &\sum_{\substack{\lambda\vdash n\\unicode{x0142}(\lambda) =k}}\frac{1}{z_\lambda}=\frac{c(n,k)}{n!}, \end{align} $$

(b) $$ \begin{align} &\sum_{\substack{\lambda\vdash n\\ \mathrm{odd}\\ l(\lambda)=k }} \frac{1}{z_{\lambda}}= \frac{d(n,k)}{n!}, \end{align} $$

(c) $$ \begin{align} &\sum_{\substack{\lambda\vdash n\\ o(\lambda)=k }} \frac{1}{z_{\lambda}}=\frac{e(n,k)}{n!}, \end{align} $$

and

(d) $$ \begin{align} &\sum_{\substack{\lambda\vdash n\\ o(\lambda)=k\\ l(\lambda)=m }} \frac{1}{z_{\lambda}}=\frac{f(n,k,m)}{n!}. \end{align} $$

The numbers $c(n,k)$ , $d(n,k)$ , $e(n,k)$ and $f(n,k,m)$ all have simple exponential generating functions. Let

Proposition 2.11. We have

(a) $$ \begin{align} &\sum_{n=0}^\infty \sum_{k=0}^n c(n,k) v^k \frac{u^n}{n!} = e^{-v\log(1-u)}=\frac{1}{(1-u)^v}, \end{align} $$

(b) $$ \begin{align} &\sum_{n=0}^\infty \sum_{k=0}^n d(n,k) v^k \frac{u^n}{n!} = e^{vL(u)}=\left(\frac{1+u}{1-u}\right)^{v/2}, \end{align} $$

(c) $$ \begin{align} &\sum_{n=0}^\infty\sum_{k=0}^n e(n,k) v^k \frac{u^n}{n!} =\frac{e^{vL(u)}}{\sqrt{1-u^2}}=\frac{(1+u)^{(v-1)/2}}{(1-u)^{(v+1)/2}}, \end{align} $$

and

(d) $$ \begin{align} &\sum_{n=0}^\infty\sum_{k,m=0}^n f(n,k,m) v^k w^m \frac{u^n}{n!} =\frac{e^{vwL(u)}}{(1-u^2)^{w/2}} =\frac{(1+u)^{(v-1)w/2}}{(1-u)^{(v+1)w/2}}. \end{align} $$

Proof. We prove only (c); the proofs of the other formulas are similar (and (a) is well known). For (c) we want to count permutations in which odd cycles are weighted v and even cycles are weighted 1. So by the exponential formula for permutations [Reference Stanley24, Corollary 5.1.9], we have

$$ \begin{align*} \sum_{n=0}^\infty \sum_{k=0}^n e(n,k) \frac{u^n}{n!} v^k &=\exp\biggl(v\sum_{m=1}^\infty \frac{u^{2m-1}}{2m-1} +\sum_{m=1}^\infty \frac{u^{2m}}{2m}\biggr)\\ &=\exp\Bigl(\frac{v}{2}\bigl(-\log(1-u)+\log(1+u)\bigr)+\frac{1}{2}\bigl(-\log(1-u) -\log(1+u)\bigr)\Bigr)\\ &=\frac{(1+u)^{(v-1)/2}}{(1-u)^{(v+1)/2}}.\\[-47pt] \end{align*} $$

A permutation $\pi $ is called alternating if $\pi (1)>\pi (2)<\pi (3)>\pi (4)<\cdots $ , and it is well known that alternating permutations in $\mathfrak {S}_n$ are counted by the nth Euler number $E_{n}$ , whose exponential generating function is $\sum _{n=0}^\infty E_n x^n/n! = \sec x + \tan x$ . The series $L(u)$ defined above makes an appearance in Stanley’s work on doubly alternating permutations: alternating permutations whose inverses are alternating. Let $\tilde E_n$ denote the number of doubly alternating permutations in $\mathfrak {S}_n$ . Stanley showed that the ordinary generating function for doubly alternating permutations is

(2.2) $$ \begin{align} \sum_{n=0}^{\infty} \tilde E_n u^n = \frac{1}{\sqrt{1-u^{2}}}\sum_{r=0}^{\infty}E_{2r}^{2}\frac{L(u)^{2r}}{(2r)!}+\sum_{r=0}^{\infty}E_{2r+1}^{2}\frac{L(u)^{2r+1}}{(2r+1)!} \end{align} $$

[Reference Stanley21, Theorem 3.1]. Stanley’s proof of Equation (2.2) uses Foulkes’s theorem, but Equation (2.2) can also be recovered from our results, as shall be demonstrated at the end of Sections 3.2 and 4.3.

We note that Stanley’s formula (2.2) can be expressed in terms of the numbers $d(n,k)$ and $e(n,k)$ . By Proposition 2.11, we have

$$ \begin{gather*} \frac{1}{\sqrt{1-u^2}}\frac{L(u)^{2r}}{(2r)!} = \sum_{n=0}^\infty e(n, 2r) \frac{u^n}{n!} \qquad\text{and}\qquad \frac{L(u)^{2r+1}}{(2r+1)!} = \sum_{d=0}^\infty d(n, 2r+1) \frac{u^n}{n!}. \end{gather*} $$

Thus, Equation (2.2) is equivalent to the statement that

$$ \begin{align*} \tilde E_n = \begin{cases} \displaystyle\frac{1}{n!}\sum_r e(n,2r)E_{2r}^2 = \frac{1}{n!}\sum_{k=0}^n e(n,k) E_{k}^2,& \text{ if }n\text{ is even,}\\ \displaystyle\frac{1}{n!}\sum_r d(n, 2r+1)E_{2r+1}^2 = \frac{1}{n!}\sum_{k=0}^n d(n,k) E_{k}^2,& \text{ if }n\text{ is odd.} \end{cases} \end{align*} $$

3.1 Peaks, descents and their inverses

Define the polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ipk}},\operatorname {\mathrm {des}},\operatorname {\mathrm {ides}})}(y,z,s,t)$ by

which encodes the desired two-sided distribution.

Theorem 3.1. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{\left(\frac{(1+ys)(1+zt)}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(\frac{(1+y)^{2}s}{(y+s)(1+ys)},\frac{(1+z)^{2}t}{(z+t)(1+zt)},\frac{y+s}{1+ys},\frac{z+t}{1+zt}\right)}{(1+y)(1+z)}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{(1+yx)(1+zx)}{(1-yzx)(1-x)}\right)^{ij}s^{i}t^{j}\nonumber \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} &\frac{\left(\frac{(1+ys)(1+zt)}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(\frac{(1+y)^{2}s}{(y+s)(1+ys)},\frac{(1+z)^{2}t}{(z+t)(1+zt)},\frac{y+s}{1+ys},\frac{z+t}{1+zt}\right)}{(1+y)(1+z)}\\ &\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{\lambda\vdash n}\frac{\prod_{i=1}^{l(\lambda)}(1-(-y)^{\lambda_{i}})(1-(-z)^{\lambda_{i}})}{z_{\lambda}}\frac{A_{l(\lambda)}(s)A_{l(\lambda)}(t)}{(1-s)^{l(\lambda)+1}(1-t)^{l(\lambda)+1}}.\nonumber \end{align} $$

Proof. To prove this theorem, we shall compute $\left \langle (1-sE(yx)H(x))^{-1},(1-tE(z)H)^{-1}\right \rangle $ in three different ways. First, from Lemma 2.8 (b) we have

(3.1) $$ \begin{align} & \left\langle \frac{1}{1-sE(yx)H(x)},\frac{1}{1-tE(z)H}\right\rangle \notag \\ & \qquad\qquad\qquad\qquad=\frac{1}{(1-s)(1-t)}+\sum_{m,n=1}^{\infty}\frac{\left(\frac{1+ys}{1-s}\right)^{m+1}\left(\frac{1+zt}{1-t}\right)^{n+1}\sum_{L\vDash m,\,M\vDash n}N_{L,M}\left\langle r_{L},r_{M}\right\rangle }{(1+y)(1+z)}x^{m}, \end{align} $$

where

Upon applying Foulkes’s theorem (Theorem 2.1), Equation (3.1) simplifies to

(3.2) $$ \begin{align} & \left\langle \frac{1}{1-sE(yx)H(x)},\frac{1}{1-tE(z)H}\right\rangle \nonumber \\ & \quad=\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{\left(\frac{(1+ys)(1+zt)}{(1-s)(1-t)}\right)^{n+1}\sum_{\pi\in\mathfrak{S}_{n}}N_{\operatorname{\mathrm{Comp}}(\pi),\operatorname{\mathrm{Comp}}(\pi^{-1})}}{(1+y)(1+z)}x^{n}\nonumber \\ & \quad=\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{\left(\frac{(1+ys)(1+zt)}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(\frac{(1+y)^{2}s}{(y+s)(1+ys)},\frac{(1+z)^{2}t}{(z+t)(1+zt)},\frac{y+s}{1+ys},\frac{z+t}{1+zt}\right)}{(1+y)(1+z)}x^{n}. \end{align} $$

Second, we have

(3.3) $$ \begin{align} \left\langle \frac{1}{1-sE(yx)H(x)},\frac{1}{1-tE(z)H}\right\rangle & =\left\langle \sum_{i=0}^{\infty}E(yx)^{i}H(x)^{i}s^{i},\sum_{j=0}^{\infty}E(z)^{j}H^{j}t^{j}\right\rangle \nonumber \\ & =\sum_{i,j=0}^{\infty}\left\langle E(yx)^{i}H(x)^{i},E(z)^{j}H^{j}\right\rangle s^{i}t^{j}\nonumber \\ & =\sum_{i,j=0}^{\infty}\left\langle E(yx),E(z)^{j}H^{j}\right\rangle ^{i}\left\langle H(x),E(z)^{j}H^{j}\right\rangle ^{i}s^{i}t^{j}\nonumber \\ & =\sum_{i,j=0}^{\infty}\frac{(1+yx)^{ij}}{(1-yzx)^{ij}}\frac{(1+zx)^{ij}}{(1-x)^{ij}}s^{i}t^{j}, \end{align} $$

where the last two steps are obtained using Lemmas 2.4 and 2.5, respectively. Finally, from Lemma 2.9 (a) we have

(3.4) $$ \begin{align} & \left\langle \frac{1}{1-sE(yx)H(x)},\frac{1}{1-tE(z)H}\right\rangle \nonumber \\ & \qquad\qquad=\sum_{\lambda,\mu}\frac{\prod_{i=1}^{l(\lambda)}(1-(-y)^{\lambda_{i}})\prod_{j=1}^{l(\mu)}(1-(-z)^{\mu_{i}})}{z_{\lambda}z_{\mu}}\frac{A_{l(\lambda)}(s)A_{l(\mu)}(t)}{(1-s)^{l(\lambda)+1}(1-t)^{l(\mu)+1}}\left\langle p_{\lambda},p_{\mu}\right\rangle x^{\left|\lambda\right|}\nonumber \\ & \qquad\qquad=\sum_{\lambda}\frac{\prod_{i=1}^{l(\lambda)}(1-(-y)^{\lambda_{i}})(1-(-z)^{\lambda_{i}})}{z_{\lambda}}\frac{A_{l(\lambda)}(s)A_{l(\lambda)}(t)}{(1-s)^{l(\lambda)+1}(1-t)^{l(\lambda)+1}}x^{\left|\lambda\right|}. \end{align} $$

Combining Equation (3.2) with Equation (3.3) yields part (a), whereas combining Equation (3.2) with Equation (3.4) and then extracting coefficients of $x^{n}$ yields part (b).

The formulas in Theorem 3.1 and others appearing later in this paper can be ‘inverted’ upon making appropriate substitutions. For example, Theorem 3.1 (a) can be rewritten as

$$ \begin{align*} &\frac{1}{(1-\alpha)(1-\beta)}+\frac{1}{(1+u)(1+v)}\sum_{n=1}^{\infty}\left(\frac{(1+u\alpha)(1+v\beta)}{(1-\alpha)(1-\beta)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}(y,z,s,t)x^{n}\\ &\quad=\sum_{i,j=0}^{\infty}\left(\frac{(1+ux)(1+vx)}{(1-uvx)(1-x)}\right)^{ij}\alpha^{i}\beta^{j}, \end{align*} $$

where

$$ \begin{align*} u&=\frac{1+s^{2}-2ys-(1-s)\sqrt{(1+s)^{2}-4ys}}{2(1-y)s}, & \alpha&=\frac{(1+s)^{2}-2ys-(1+s)\sqrt{(1+s)^{2}-4ys}}{2ys},\\ v&=\frac{1+t^{2}-2zt-(1-t)\sqrt{(1+t)^{2}-4zt}}{2(1-z)t}, \quad\text{and} & \beta&=\frac{(1+t)^{2}-2zt-(1+t)\sqrt{(1+t)^{2}-4zt}}{2zt}. \end{align*} $$

3.2 Peaks and inverse peaks

We will now consider specializations of the polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ipk}},\operatorname {\mathrm {des}},\operatorname {\mathrm {ides}})}(y,z,s,t)$ that give the distributions of the (mixed) two-sided statistics $(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ipk}})$ , $(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})$ and $(\operatorname {\mathrm {des}},\operatorname {\mathrm {ides}})$ . Let us begin with the two-sided distribution of $\operatorname {\mathrm {pk}}$ , which is encoded by

Theorem 3.2. We have

(a) $$ \begin{align} \frac{1}{(1-s)(1-t)}+\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{(1+s)(1+t)}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right)x^{n} =\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}s^{i}t^{j} \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \quad\left(\frac{(1+s)(1+t)}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right) =4\sum_{i,j=0}^{\infty}\sum_{k=0}^{n}{2ij \choose k}{2ij+n-k-1 \choose n-k}s^{i}t^{j} \end{align} $$

and

(c) $$ \begin{align} \quad\left(\frac{(1+s)(1+t)}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right) =\frac{1}{n!}\sum_{k=0}^n 4^{k+1}d(n,k)\frac{A_{k}(s)A_{k}(t)}{(1-s)^{k+1}(1-t)^{k+1}}\quad \end{align} $$

with $d(n,k)$ as defined in Section 2.4.

Proof. Parts (a) and (c) are obtained from evaluating Theorem 3.1 (a) and (b), respectively, at $y=z=1$ , with the sum on $\lambda $ in (c) evaluated by Lemma 2.10 (b). From the identities

$$\begin{align*}(1+x)^{k}=\sum_{n=0}^{k}{k \choose n}x^{n}\quad\text{and}\quad(1-x)^{-k}=\sum_{n=0}^{\infty}{k+n-1 \choose n}x^{n}, \end{align*}$$

we obtain

(3.5) $$ \begin{align} \left(\frac{1+x}{1-x}\right)^{2ij}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}{2ij \choose k}{2ij+n-k-1 \choose n-k}x^{n}. \end{align} $$

Substituting Equation (3.5) into (a) and extracting coefficients of $x^{n}$ yields (b).

The first several polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ipk}})}(s,t)$ are displayed in Table 1. The coefficients of $s^{k}t$ —equivalently, the coefficients of $st^{k}$ by symmetry—are characterized by the following proposition.

Table 1

Joint distribution of $\operatorname {\mathrm {pk}}$ and $\operatorname {\mathrm {ipk}}$ over $\mathfrak {S}_{n}$

Proposition 3.3 was first stated as a corollary of a more general result of Troyka and Zhuang [Reference Troyka and Zhuang26], namely, that for any composition L of $n\geq 1$ , there is exactly one permutation in $\pi \in \mathfrak {S}_{n}$ with descent composition L such that $\operatorname {\mathrm {ipk}}(\pi )=0$ . However, it is also possible to prove Proposition 3.3 directly from Theorem 3.2.

Next, we obtain a surprisingly simple formula expressing the two-sided peak polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ipk}})}(s,t)$ in terms of products of the ordinary peak polynomials

giving the distribution of the peak number over $\mathfrak {S}_{n}$ .

Theorem 3.4. For all $n\geq 1$ , we have

$$ \begin{align*} P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}(s,t) &= \frac{1}{n!}\sum_{k=0}^n d(n,k) \left((1-s)(1-t)\right)^{\frac{n-k}{2}}P_{k}^{\operatorname{\mathrm{pk}}}(s)P_{k}^{\operatorname{\mathrm{pk}}}(t). \end{align*} $$

Since $d(n,k)=0$ when $n-k$ is odd, the above formula does not involve any square roots.

Proof. It is known [Reference Stembridge25] that

(3.6) $$ \begin{align} A_{n}(t)=\left(\frac{1+t}{2}\right)^{n+1}P_{n}^{\operatorname{\mathrm{pk}}}\left(\frac{4t}{(1+t)^{2}}\right) \end{align} $$

for $n\geq 1$ . Combining Equation (3.6) with Theorem 3.2 (c) and then replacing the variables s and t with u and v, respectively, yields

$$ \begin{align*} P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}\left(\frac{4u}{(1+u)^{2}},\frac{4v}{(1+v)^{2}}\right) =\frac{1}{n!}\sum_{k=0}^n d(n,k)\left(\frac{(1-u)(1-v)}{(1+u)(1+v)}\right)^{n-k}P_{k}^{\operatorname{\mathrm{pk}}}\left(\frac{4u}{(1+u)^{2}}\right)P_{k}^{\operatorname{\mathrm{pk}}}\left(\frac{4v}{(1+v)^{2}}\right). \end{align*} $$

Setting $s=4u/(1+u)^{2}$ and $t=4v/(1+v)^{2}$ , we obtain

(3.7) $$ \begin{align} P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}(s,t) =\frac{1}{n!}\sum_{k=0}^n d(n,k) \left(\frac{(1-u)(1-v)}{(1+u)(1+v)}\right)^{n-k}P_{k}^{\operatorname{\mathrm{pk}}}(s)P_{k}^{\operatorname{\mathrm{pk}}}(t). \end{align} $$

It can be verified that $u=2s^{-1}(1-\sqrt {1-s})-1$ and $v=2t^{-1}(1-\sqrt {1-t})-1$ , which lead to

(3.8) $$ \begin{align} \frac{(1-u)(1-v)}{(1+u)(1+v)}=\sqrt{(1-s)(1-t)}. \end{align} $$

Substituting Equation (3.8) into Equation (3.7) completes the proof.

The substitution used in the proof of Theorem 3.4 allows us to invert the formulas in Theorem 3.2 and others involving the same quadratic transformation. For example, Theorem 3.2 (b) is equivalent to

$$ \begin{align*} \begin{aligned} \frac{P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}(s,t)}{(1-s)^{\frac{n+1}{2}}(1-t)^{\frac{n+1}{2}}} & =4\sum_{i,j=0}^{\infty}\sum_{k=0}^{n}{2ij \choose k}{2ij+n-k-1 \choose n-k}u^{i}v^{j}, \end{aligned} \end{align*} $$

where, as before, $u=2s^{-1}(1-\sqrt {1-s})-1$ and $v=2t^{-1}(1-\sqrt {1-t})-1$ . In fact, we can express u and v in terms of the Catalan generating function

as $u=(s/4)C(s/4)^{2}$ and $v=(t/4)C(t/4)^{2}$ .

If we multiply the formula of Theorem 3.4 by $u^n$ and sum over n using Proposition 2.11 (b), we obtain the following generating function for the two-sided peak polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ipk}})}(s,t)$ .

Theorem 3.5. We have

$$ \begin{align*} \sum_{n=0}^\infty P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}(s,t) u^n = \sum_{k=0}^\infty \frac{1}{k!} \left(\frac{L(\sqrt{(1-s)(1-t)}u)}{\sqrt{(1-s)(1-t)}}\right)^k P_{k}^{\operatorname{\mathrm{pk}}}(s)P_{k}^{\operatorname{\mathrm{pk}}}(t), \end{align*} $$

with $L(u)$ as defined in Section 2.4.

We can derive the odd part of Stanley’s generating function (2.2) for doubly alternating permutations from Theorem 3.5. While an alternating permutation $\pi $ satisfies $\pi (1)>\pi (2)<\pi (3)>\pi (4)<\cdots $ , we say that $\pi $ is reverse alternating if $\pi (1)<\pi (2)>\pi (3)<\pi (4)>\cdots $ . It is evident by symmetry that reverse alternating permutations are also counted by the Euler numbers $E_n$ . As shown by Stanley [Reference Stanley21], the number of doubly alternating permutations in $\mathfrak {S}_n$ is equal to the number $\tilde {E}_n$ of reverse alternating permutations in $\mathfrak {S}_n$ whose inverses are reverse alternating.

It is readily verified that a permutation $\pi $ in $\mathfrak {S}_n$ has at most $(n-1)/2$ peaks and has exactly $(n-1)/2$ peaks if and only if n is odd and $\pi $ is reverse alternating. Thus, we have

$$\begin{align*}\lim_{s\rightarrow0}P_{k}^{\operatorname{\mathrm{pk}}}(s^{-2})s^{k+1}=\lim_{s\rightarrow0}\sum_{\pi\in\mathfrak{S}_{k}}s^{k-2\operatorname{\mathrm{pk}}(\pi)-1}=\begin{cases} 0, & \text{if }k\text{ is even,}\\ E_{k}, & \text{if }k\text{ is odd.} \end{cases} \end{align*}$$

Similarly, we have

$$\begin{align*}\lim_{s\rightarrow0}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}})}(s^{-2},t^{-2})s^{n+1}t^{n+1}=\begin{cases} 0, & \text{if }n\text{ is even,}\\ \tilde{E}_{n}, & \text{if }n\text{ is odd.} \end{cases} \end{align*}$$

So, to obtain Stanley’s formula from Theorem 3.5, we first replace s with $s^{-2}$ , t with $t^{-2}$ and u with $stu$ ; then we multiply by $st$ and take the limit as $s,t\to 0$ . The substitution takes

$$\begin{align*}\left(\frac{L(\sqrt{(1-s)(1-t)}u)}{\sqrt{(1-s)(1-t)}}\right)^k P_{k}^{\operatorname{\mathrm{pk}}}(s)P_{k}^{\operatorname{\mathrm{pk}}}(t) \end{align*}$$

to

$$ \begin{align*} \left(\frac{L(\sqrt{(s^{2}-1)(t^{2}-1)}u)}{\sqrt{(1-s^{-2})(1-t^{-2})}}\right)^{k}P_{k}^{\operatorname{\mathrm{pk}}}(s^{-2})P_{k}^{\operatorname{\mathrm{pk}}}(t^{-2}) =\left(\frac{L(\sqrt{(s^{2}-1)(t^{2}-1)}u)}{\sqrt{(s^{2}-1)(t^{2}-1)}}\right)^{k}s^{k}P_{k}^{\operatorname{\mathrm{pk}}}(s^{-2})t^{k}P_{k}^{\operatorname{\mathrm{pk}}}(t^{-2}); \end{align*} $$

multiplying by $st$ and taking $s,t\to 0$ gives $L(u)^k E_k^2$ for k odd and $0$ for k even, as desired.

We will later get the even part of Stanley’s generating function (2.2) in a similar way from Theorem 4.4, which is the analogue of Theorem 3.5 for left peaks.

Before proceeding, we note that every formula like Theorem 3.4 has an analogous generating function like Theorem 3.5. We will only write out these generating functions for $(\operatorname {\mathrm {pk}}, \operatorname {\mathrm {ipk}})$ and $(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ilpk}})$ – which are used in our rederivation of Stanley’s formula – as well as for $(\operatorname {\mathrm {des}},\operatorname {\mathrm {ides}})$ .

3.3 Peaks and inverse descents

Next, define the polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ by

We omit the proof of the next theorem as it is similar to that of Theorem 3.2, with the main difference being that we specialize at $y=1$ and $z=0$ (as opposed to $y=z=1$ ).

Theorem 3.6. We have

(a) $$ \begin{align} \qquad\frac{1}{(1-s)(1-t)}+\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1+s}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ides}})}\left(\frac{4s}{(1+s)^{2}},t\right) =\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{ij}s^{i}t^{j}\qquad \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \qquad\left(\frac{1+s}{(1-s)(1-t)}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ides}})}\left(\frac{4s}{(1+s)^{2}},t\right) & =2\sum_{i,j=0}^{\infty}\sum_{k=0}^{n}{ij \choose k}{ij+n-k-1 \choose n-k}s^{i}t^{j} \end{align} $$

and

(c) $$ \begin{align} \quad\, \left(\frac{1+t}{(1-s)(1-t)}\right)^{\! n+1}P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ides}})}\left(\frac{4s}{(1+s)^{2}},t\right) &=\frac{1}{n!}\sum_{k=0}^n 2^{k+1} d(n,k) \frac{A_{k}(s)A_{k}(t)}{(1-s)^{k+1}(1-t)^{k+1}}. \end{align} $$

We display the first several polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ in Table 2, collecting terms with the same power of s in order to display the symmetry in their coefficients. This symmetry follows from the fact that the reverse of a permutation $\pi \in \mathfrak {S}_{n}$ satisfies $\operatorname {\mathrm {ides}}(\pi ^{r})=n-1-\operatorname {\mathrm {ides}}(\pi )$ [Reference Zhuang31, Proposition 2.6 (c)] but has the same number of peaks as $\pi $ .

Table 2

Joint distribution of $\operatorname {\mathrm {pk}}$ and $\operatorname {\mathrm {ides}}$ over $\mathfrak {S}_{n}$

Furthermore, observe that the coefficients of $t^{k}s$ in $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ are binomial coefficients; this is a consequence of the following proposition, which is Corollary 9 of [Reference Troyka and Zhuang26].

In addition to being symmetric, the coefficients of $s^{k}$ in $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ seem to be unimodal polynomials in t, which we know holds for $k=0$ in light of Proposition 3.7. In the last section of this paper, we will state the conjecture that the polynomials $[s^{k}]\,P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ are in fact $\gamma $ -positive, a property which implies unimodality and symmetry.

The next formula expresses $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ in terms of the products $P_{k}^{\operatorname {\mathrm {pk}}}(s)A_{k}(t)$ . The proof is very similar to that of Theorem 3.4 and so it is omitted.

Theorem 3.8. For all $n\geq 1$ , we have

$$ \begin{align*} P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ides}})}(s,t) & =\frac{1}{n!}\sum_{k=0}^n d(n,k)\left((1-s)^{1/2}(1-t)\right)^{n-k}P_{k}^{\operatorname{\mathrm{pk}}}(s)A_{k}(t). \end{align*} $$

3.4 Descents and inverse descents

To conclude this section, we state the analogous results for the two-sided Eulerian polynomials

$$ \begin{align*} A_{n}(s,t) & =P_{n}^{(\operatorname{\mathrm{pk}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}(1,1,s,t)=\sum_{\pi\in\mathfrak{S}_{n}}s^{\operatorname{\mathrm{des}}(\pi)+1}t^{\operatorname{\mathrm{ides}}(\pi)+1}. \end{align*} $$

Recall that parts (a) and (b) were originally due to Carlitz, Roselle and Scoville [Reference Carlitz, Roselle and Scoville5], whereas (c) and (d) are new.

Theorem 3.9. We have

(a) $$ \begin{align} \qquad\sum_{n=0}^{\infty}\frac{A_{n}(s,t)}{(1-s)^{n+1}(1-t)^{n+1}}x^{n} & =\sum_{i,j=0}^{\infty}\frac{s^{i}t^{j}}{(1-x)^{ij}} \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \qquad\frac{A_{n}(s,t)}{(1-s)^{n+1}(1-t)^{n+1}} & =\sum_{i,j=0}^{\infty}{ij+n-1 \choose n}s^{i}t^{j} \end{align} $$

and

(c) $$ \begin{align} \qquad A_{n}(s,t) =\frac{1}{n!}\sum_{k=0}^n c(n,k) \left((1-s)(1-t)\right)^{n-k}A_{k}(s)A_{k}(t). \end{align} $$

Moreover, we have

(d) $$ \begin{align} \qquad \sum_{n=0}^\infty A_n(s,t) u^n = \sum_{k=0}^\infty \frac{1}{k!}\left(\log\frac{1}{1-(1-s)(1-t)u}\right)^k \frac{A_k(s)A_k(t)}{(1-s)^k(1-t)^k}. \end{align} $$

Parts (a)–(c) are proven similarly to Theorems 3.2, 3.4 and 3.6 except that we evaluate Theorem 3.1 at $y=z=0$ . Part (c) can also be derived directly from (b):

$$ \begin{align*} \frac{A_{n}(s,t)}{(1-s)^{n+1}(1-t)^{n+1}} &=\sum_{i,j=0}^{\infty}{ij+n-1 \choose n}s^{i}t^{j}\\ &=\sum_{i,j=0}^{\infty}\frac{1}{n!}\sum_{k=0}^n c(n,k) (ij)^k s^{i}t^{j}\\ &=\frac{1}{n!}\sum_{k=0}^n c(n,k)\sum_{i=0}^\infty i^k s^i \sum_{j=0}^\infty j^k t^j\\ &=\frac{1}{n!}\sum_{k=0}^n c(n,k)\frac{A_k(s)}{(1-t)^{k+1}}\frac{A_k(t)}{(1-t)^{k+1}}. \end{align*} $$

Part (d) is obtained from (c) using Proposition 2.11 (a).

4.1 Left peaks, descents and their inverses

In this section, we will examine (mixed) two-sided distributions involving the left peak number $\operatorname {\mathrm {lpk}}$ . Let us first derive a generating function formula for the polynomials

giving the joint distribution of $\operatorname {\mathrm {lpk}}$ , $\operatorname {\mathrm {ilpk}}$ , $\operatorname {\mathrm {des}}$ and $\operatorname {\mathrm {ides}}$ over $\mathfrak {S}_{n}$ .

Theorem 4.1. We have

$$ \begin{align*} &\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{(1+ys)^{n}(1+zt)^{n}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(\frac{(1+y)^{2}s}{(y+s)(1+ys)},\frac{(1+z)^{2}t}{(z+t)(1+zt)},\frac{y+s}{1+ys},\frac{z+t}{1+zt}\right)}{(1-s)^{n+1}(1-t)^{n+1}}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\frac{(1+yx)^{i(j+1)}(1+zx)^{(i+1)j}}{(1-yzx)^{ij}(1-x)^{(i+1)(j+1)}}s^{i}t^{j}. \end{align*} $$

Proof. From Lemma 2.8 (c), we have

(4.1) $$ \begin{align} & \left\langle \frac{H(x)}{1-sE(yx)H(x)},\frac{H}{1-tE(z)H}\right\rangle \nonumber \\ & \qquad\qquad\qquad=\frac{1}{(1-s)(1-t)}+\sum_{m,n=1}^{\infty}\sum_{\substack{L\vDash m\\ M\vDash n } }\frac{(1+ys)^{m}(1+zt)^{n}}{(1-s)^{m+1}(1-t)^{n+1}}\acute{N}_{L,M}\left\langle r_{L},r_{M}\right\rangle x^{m}, \end{align} $$

where

By Foulkes’s theorem (Theorem 2.1), Equation (4.1) simplifies to

(4.2) $$ \begin{align} & \left\langle \frac{1}{1-sE(yx)H(x)},\frac{1}{1-tE(z)H}\right\rangle \nonumber \\ & \quad=\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{(1+ys)^{n}(1+zt)^{n}\sum_{\pi\in\mathfrak{S}_{n}}\acute{N}_{\operatorname{\mathrm{Comp}}(\pi),\operatorname{\mathrm{Comp}}(\pi^{-1})}}{(1-s)^{n+1}(1-t)^{n+1}}x^{n}\nonumber \\ & \quad=\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{(1+ys)^{n}(1+zt)^{n}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(\frac{(1+y)^{2}s}{(y+s)(1+ys)},\frac{(1+z)^{2}t}{(z+t)(1+zt)},\frac{y+s}{1+ys},\frac{z+t}{1+zt}\right)}{(1-s)^{n+1}(1-t)^{n+1}}x^{n}. \end{align} $$

We now compute the same scalar product in a different way. We have

(4.3) $$ \begin{align} \left\langle \frac{H(x)}{1-sE(yx)H(x)},\frac{H}{1-tE(z)H}\right\rangle &=\left\langle \sum_{i=0}^{\infty}E(yx)^{i}H(x)^{i+1}s^{i},\sum_{j=0}^{\infty}E(z)^{j}H^{j+1}t^{j}\right\rangle \nonumber \\ &=\sum_{i,j=0}^{\infty}\left\langle E(yx)^{i}H(x)^{i+1},E(z)^{j}H^{j+1}\right\rangle s^{i}t^{j}\nonumber \\ &=\sum_{i,j=0}^{\infty}\left\langle (E(yx)^{i}H(x)^{i+1})[X+1],E(z)^{j}H^{j}\right\rangle s^{i}t^{j}\nonumber \\ &=\sum_{i,j=0}^{\infty}\left\langle (E(yx)[X+1])^{i}(H(x)[X+1])^{i+1},E(z)^{j}H^{j}\right\rangle s^{i}t^{j}\nonumber \\ &=\sum_{i,j=0}^{\infty}\left\langle \frac{(1+yx)^{i}}{(1-x)^{i+1}}E(yx)^{i}H(x)^{i+1},E(z)^{j}H^{j}\right\rangle s^{i}t^{j}\nonumber \\ &=\sum_{i,j=0}^{\infty}\frac{(1+yx)^{i}}{(1-x)^{i+1}}\left\langle E(yx),E(z)^{j}H^{j}\right\rangle ^{i}\left\langle H(x),E(z)^{j}H^{j}\right\rangle ^{i+1}s^{i}t^{j}\nonumber \\ &=\sum_{i,j=0}^{\infty}\frac{(1+yx)^{i(j+1)}}{(1-yzx)^{ij}}\frac{(1+zx)^{(i+1)j}}{(1-x)^{(i+1)(j+1)}}s^{i}t^{j}; \end{align} $$

here, we are using Lemma 2.6 for the third equality, and in the last three steps, we apply Lemmas 2.7, 2.4 and 2.5, respectively. Combining (4.2) and (4.3) completes the proof.

4.2 Left peaks, inverse peaks, descents and inverse descents

Next, let us consider the joint distribution of $\operatorname {\mathrm {lpk}}$ , $\operatorname {\mathrm {ipk}}$ , $\operatorname {\mathrm {des}}$ and $\operatorname {\mathrm {ides}}$ over $\mathfrak {S}_{n}$ . Define

Theorem 4.2. We have

$$ \begin{align*} &\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{(1+ys)^{n}(1+zt)^{n+1}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(\frac{(1+y)^{2}s}{(y+s)(1+ys)},\frac{(1+z)^{2}t}{(z+t)(1+zt)},\frac{y+s}{1+ys},\frac{z+t}{1+zt}\right)}{(1+z)(1-s)^{n+1}(1-t)^{n+1}}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+yx}{1-yzx}\right)^{ij}\left(\frac{1+zx}{1-x}\right)^{(i+1)j}s^{i}t^{j}. \end{align*} $$

Proof. The two sides of the above equation are obtained from evaluating the scalar product

$$\begin{align*}\left\langle \frac{H}{1-sE(yx)H(x)},\frac{1}{1-tE(z)H}\right\rangle \end{align*}$$

in two different ways; the proof is similar to that of Theorem 3.1 and so we omit the details.

4.3 Left peaks and inverse left peaks

We now consider specializations of $P_{n}^{(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ilpk}},\operatorname {\mathrm {des}},\operatorname {\mathrm {ides}})}(y,z,s,t)$ and $P_{n}^{(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ipk}},\operatorname {\mathrm {des}},\operatorname {\mathrm {ides}})}(y,z,s,t)$ , beginning with the two-sided left peak polynomials

Theorem 4.3. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{((1+s)(1+t))^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right)x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\frac{(1+x)^{2ij+i+j}}{(1-x)^{2ij+i+j+1}}s^{i}t^{j}\nonumber \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \:\frac{((1+s)(1+t))^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right) =\sum_{i,j=0}^{\infty}\sum_{k=0}^{n}{2ij+i+j \choose k}{2ij+i+j+n-k \choose n-k}s^{i}t^{j} \end{align} $$

and

(c) $$ \begin{align} \quad\,\,\,\frac{((1+s)(1+t))^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right) &=\sum_{k=0}^n e(n,k) \frac{B_{k}(s)B_{k}(t)}{(1-s)^{k+1}(1-t)^{k+1}} \end{align} $$

with $e(n,k)$ as defined in Section 2.4.

Proof. Evaluating Theorem 4.1 at $y=z=1$ yields part (a), whereas (b) follows from (a) and

$$\begin{align*}\frac{(1+x)^{2ij+i+j}}{(1-x)^{2ij+i+j+1}}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}{2ij+i+j \choose k}{2ij+i+j+n-k \choose n-k}x^{n}. \end{align*}$$

To prove (c), first observe that Lemma 2.9 (b) implies

$$ \begin{align*} \begin{aligned} \left\langle \frac{H(x)}{1-sE(x)H(x)},\frac{H}{1-tEH}\right\rangle & =\sum_{\lambda,\mu}\frac{1}{z_{\lambda}z_{\mu}}\frac{B_{o(\lambda)}(s)B_{o(\mu)}(t)}{(1-s)^{o(\lambda)+1}(1-t)^{o(\mu)+1}}\left\langle p_{\lambda},p_{\mu}\right\rangle x^{\left|\lambda\right|}\\ & =\sum_{\lambda}\frac{1}{z_{\lambda}}\frac{B_{o(\lambda)}(s)B_{o(\lambda)}(t)}{(1-s)^{o(\lambda)+1}(1-t)^{o(\lambda)+1}}x^{\left|\lambda\right|}, \end{aligned} \end{align*} $$

but this same scalar product is also given by

$$ \begin{align*} \left\langle \frac{H(x)}{1-sE(x)H(x)},\frac{H}{1-tEH}\right\rangle =\frac{1}{(1-s)(1-t)}\sum_{n=1}^{\infty}\frac{((1+s)(1+t))^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right)x^{n}, \end{align*} $$

which is obtained from evaluating Equation (4.2) at $y=z=1$ . Equating these two expressions, extracting coefficients of $x^{n}$ , and applying Lemma 2.10 (c) completes the proof.

See Table 3 for the first several polynomials $P_{n}^{(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ilpk}})}(s,t)$ .

Table 3

Joint distribution of $\operatorname {\mathrm {lpk}}$ and $\operatorname {\mathrm {ilpk}}$ over $\mathfrak {S}_{n}$

Next, we express these two-sided left peak polynomials as a sum involving products of the ordinary left peak polynomials

Theorem 4.4. For all $n\geq 1$ , we have

$$ \begin{align*} P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}})}(s,t) & =\frac{1}{n!}\sum_{k=0}^n e(n,k)\left((1-s)(1-t)\right)^{\frac{n-k}{2}}P_{k}^{\operatorname{\mathrm{lpk}}}(s)P_{k}^{\operatorname{\mathrm{lpk}}}(t). \end{align*} $$

Proof. The proof follows in the same way as the proof of Theorem 3.4 except that we use the formula

(4.4) $$ \begin{align} B_{n}(t)=(1+t)^{n}P_{n}^{\operatorname{\mathrm{lpk}}}\left(\frac{4t}{(1+t)^{2}}\right) \end{align} $$

[Reference Petersen18, Proposition 4.15] in place of (3.6). The details are omitted.

From Theorem 4.4 and Proposition 2.11 (c), we can obtain a generating function for the two-sided left peak polynomials, analogous to Theorem 3.5 for the two-sided peak polynomials.

Theorem 4.5. We have

$$ \begin{align*} \sum_{n=0}^\infty P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}})}(s,t) u^n = \frac{1}{\sqrt{1-(1-s)(1-t)u^2}}\sum_{k=0}^\infty \frac{1}{k!} \left(\frac{L(\sqrt{(1-s)(1-t)}u)}{\sqrt{(1-s)(1-t)}}\right)^{\! k} P_{k}^{\operatorname{\mathrm{lpk}}}(s)P_{k}^{\operatorname{\mathrm{lpk}}}(t). \end{align*} $$

Just as we derived from Theorem 3.5 the odd part of Stanley’s formula (2.2) for doubly alternating permutations, we can derive the even part of (2.2) from Theorem 4.5. First, we note that a permutation $\pi $ in $\mathfrak {S}_n$ has at most $n/2$ left peaks and has exactly $n/2$ left peaks if and only if n is even and $\pi $ is alternating. Then we have

$$\begin{align*}\lim_{s\rightarrow0}P_{k}^{\operatorname{\mathrm{lpk}}}(s^{-2})s^{k}=\lim_{s\rightarrow0}\sum_{\pi\in\mathfrak{S}_{k}}s^{k-2\operatorname{\mathrm{lpk}}(\pi)}=\begin{cases} 0, & \text{if }n\text{ is odd,}\\ E_{k}, & \text{if }n\text{ is even}, \end{cases} \end{align*}$$

and similarly

$$\begin{align*}\lim_{s\rightarrow0}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ilpk}})}(s^{-2},t^{-2})s^{n}t^{n}=\begin{cases} 0, & \text{if }n\text{ is odd,}\\ \tilde{E}_{n}, & \text{if }n\text{ is even.} \end{cases} \end{align*}$$

Therefore, to obtain the even part of Stanley’s formula, we take Theorem 4.5 and replace s with $s^{-2}$ , t with $t^{-2}$ and u with $stu$ ; then we take the limit as $s,t\to 0$ , similarly to the computation for the odd part.

4.4 Left peaks and inverse peaks

We proceed to give analogous formulas for the polynomials

encoding the joint distribution of $\operatorname {\mathrm {lpk}}$ and $\operatorname {\mathrm {ipk}}$ over $\mathfrak {S}_n$ .

Theorem 4.6. We have

(a) $$ \begin{align} \frac{1}{(1-s)(1-t)}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{(1+s)^{n}(1+t)^{n+1}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ipk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right)x^{n} =\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij+j}s^{i}t^{j} \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \frac{(1+s)^{n}(1+t)^{n+1}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ipk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right) =2\sum_{i,j=0}^{\infty}\sum_{k=0}^{n}{2ij+j \choose k}{2ij+j+n-k-1 \choose n-k}s^{i}t^{j} \end{align} $$

and

(c) $$ \begin{align} \quad\frac{(1+s)^{n}(1+t)^{n+1}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ipk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right) =\frac{1}{n!}\sum_{k=0}^n 2^{k+1} d(n,k) \frac{B_{k}(s)A_{k}(t)}{(1-s)^{k+1}(1-t)^{k+1}}. \end{align} $$

Proof. The proof of parts (a) and (b) is the same as that of Theorem 4.3 (a) and (b), except that we specialize Theorem 4.2 as opposed to Theorem 4.1. To prove (c), notice that from Lemma 2.9 (a)–(b) we have

(4.5) $$ \begin{align} \left\langle \frac{1}{1-sE(x)H(x)},\frac{H}{1-tEH}\right\rangle & =\sum_{\lambda\text{ odd}}\sum_{\mu}\frac{2^{l(\lambda)}}{z_{\lambda}z_{\mu}}\frac{A_{l(\lambda)}(s)B_{o(\mu)}(t)}{(1-s)^{l(\lambda)+1}(1-t)^{o(\mu)+1}}\left\langle p_{\lambda},p_{\mu}\right\rangle x^{\left|\mu\right|}\nonumber \\ & =\sum_{\lambda\text{ odd}}\frac{2^{l(\lambda)}}{z_{\lambda}}\frac{A_{l(\lambda)}(s)B_{l(\lambda)}(t)}{(1-s)^{l(\lambda)+1}(1-t)^{l(\lambda)+1}}x^{\left|\lambda\right|}, \end{align} $$

and the same scalar product can be shown to be equal to

(4.6) $$ \begin{align} &\left\langle \frac{1}{1-sE(x)H(x)},\frac{H}{1-tEH}\right\rangle =\frac{1}{(1-s)(1-t)}\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\frac{1}{2}\sum_{n=1}^{\infty}\frac{(1+s)^{n+1}(1+t)^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ipk}})}\left(\frac{4s}{(1+s)^{2}},\frac{4t}{(1+t)^{2}}\right)x^{n} \end{align} $$

by using Lemma 2.8 (b)–(c). Equating Equations (4.5) and (4.6), extracting coefficients of $x^{n}$ , and then applying Lemma 2.10 (b) completes the proof of (c).

Table 4 displays the first several polynomials $P_{n}^{(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ipk}})}(s,t)$ . The following proposition, which is Corollary 11 of [Reference Troyka and Zhuang26], characterizes the coefficients of $s^{k}t$ in $P_{n}^{(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ipk}})}(s,t)$ .

Table 4

Joint distribution of $\operatorname {\mathrm {lpk}}$ and $\operatorname {\mathrm {ipk}}$ over $\mathfrak {S}_{n}$

The next formula expresses the polynomial $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ilpk}})}(s,t)$ as a sum involving products of the peak and left peak polynomials. The proof follows in the same way as that of Theorem 3.4, except that we begin by substituting both Equations (3.6) and (4.4) into Theorem 4.6 (c).

Theorem 4.8. For all $n\geq 1$ , we have

$$ \begin{align*} P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ipk}})}(s,t) & =\frac{1}{n!}\sum_{k=0}^n d(n,k) \left((1-s)(1-t)\right)^{\frac{n-k}{2}}P_{k}^{\operatorname{\mathrm{lpk}}}(s)P_{k}^{\operatorname{\mathrm{pk}}}(t). \end{align*} $$

4.5 Left peaks and inverse descents

Finally, define

We omit the proofs of the following theorems as they are very similar to the proofs of analogous results presented earlier.

Theorem 4.9. We have

(a) $$ \begin{align} \frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{(1+s)^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ides}})}\left(\frac{4s}{(1+s)^{2}},t\right)x^{n} =\sum_{i,j=0}^{\infty}\frac{(1+x)^{i(j+1)}}{(1-x)^{(i+1)(j+1)}}s^{i}t^{j}, \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \qquad\frac{(1+s)^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ides}})}\left(\frac{4s}{(1+s)^{2}},t\right) =\sum_{i,j=0}^{\infty}\sum_{k=0}^{n}{i(j+1) \choose k}{ij+i+j+n-k \choose n-k}s^{i}t^{j}\qquad \end{align} $$

and

(c) $$ \begin{align} \quad\,\frac{(1+s)^{n}}{((1-s)(1-t))^{n+1}}P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ides}})}\left(\frac{4s}{(1+s)^{2}},t\right) & =\frac{1}{n!}\sum_{k,m=0}^n f(n,k,m)\frac{B_{k}(s)A_{m}(t)}{(1-s)^{k+1}(1-t)^{m+1}} \end{align} $$

with $f(n,k,m)$ as defined in Section 2.4.

Theorem 4.10. For all $n\geq 1$ , we have

$$ \begin{align*} P_{n}^{(\operatorname{\mathrm{lpk}},\operatorname{\mathrm{ides}})}(s,t) & =\frac{1}{n!}\sum_{k,m=0}^n f(n,k,m) (1-s)^{\frac{n-k}{2}}(1-t)^{n-m}P_{k}^{\operatorname{\mathrm{lpk}}}(s)A_{m}(t). \end{align*} $$

See Table 5 for the first several polynomials $P_{n}^{(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ides}})}(s,t)$ .

Table 5

Joint distribution of $\operatorname {\mathrm {lpk}}$ and $\operatorname {\mathrm {ides}}$ over $\mathfrak {S}_{n}$

5.1 Up-down runs and inverse up-down runs

Consider the two-sided up-down run polynomials

Theorem 5.1. We have

$$ \begin{align*} &\frac{1}{(1-s)(1-t)}+\frac{1}{4(1-s)^{2}(1-t)^{2}}\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}(1+t^{2})^{n}}{(1-s^{2})^{n-1}(1-t^{2})^{n-1}}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{iudr}})}\left(\frac{2s}{1+s^{2}},\frac{2t}{1+t^{2}}\right)x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}\left(1+s\left(\frac{1+x}{1-x}\right)^{j}+t\left(\frac{1+x}{1-x}\right)^{i}+st\frac{(1+x)^{i+j}}{(1-x)^{i+j+1}}\right)s^{2i}t^{2j}. \end{align*} $$

The proof of Theorem 5.1 follows the same structure as our proofs from Sections 3–4: We compute an appropriate scalar product in multiple ways and set them equal to each other. We shall provide the full proof here but will omit the proofs for all remaining results in this section due to their similarity with what has been presented earlier.

Proof. First, Lemma 2.8 (d) along with Foulkes’s theorem (Theorem 2.1) leads to

(5.1) $$ \begin{align} & \left\langle \frac{1+sH(x)}{1-s^{2}E(x)H(x)},\frac{1+tH}{1-t^{2}EH}\right\rangle =\frac{1}{(1-s)(1-t)}\nonumber \\ & \qquad\qquad\qquad+\frac{1}{4(1-s)^{2}(1-t)^{2}}\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}(1+t^{2})^{n}}{(1-s^{2})^{n-1}(1-t^{2})^{n-1}}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{iudr}})}\left(\frac{2s}{1+s^{2}},\frac{2t}{1+t^{2}}\right)x^{n}. \end{align} $$

Next, using Lemmas 2.4 and 2.5, we obtain

(5.2) $$ \begin{align} & \left\langle \frac{1+sH(x)}{1-s^{2}E(x)H(x)},\frac{1}{1-t^{2}EH}\right\rangle =\sum_{i,j=0}^{\infty}\left\langle (1+sH(x))E(x)^{i}H(x)^{i},E^{j}H^{j}\right\rangle s^{2i}t^{2j}\nonumber \\ & \qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\left\langle E(x)^{i}H(x)^{i},E^{j}H^{j}\right\rangle +s\left\langle E(x)^{i}H(x)^{i+1},E^{j}H^{j}\right\rangle \right)s^{2i}t^{2j}\nonumber \\& \qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\left\langle E(x),E^{j}H^{j}\right\rangle ^{i}\left\langle H(x),E^{j}H^{j}\right\rangle ^{i}+s\left\langle E(x),E^{j}H^{j}\right\rangle ^{i}\left\langle H(x),E^{j}H^{j}\right\rangle ^{i+1}\right)s^{2i}t^{2j}\nonumber \\ & \qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\left(\frac{1+x}{1-x}\right)^{2ij}+s\left(\frac{1+x}{1-x}\right)^{2ij+j}\right)s^{2i}t^{2j}\nonumber \\ & \qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}\left(1+s\left(\frac{1+x}{1-x}\right)^{j}\right)s^{2i}t^{2j}. \end{align} $$

Similarly, using Lemmas 2.4–2.7, we obtain

(5.3) $$ \begin{align} & \left\langle \frac{1+sH(x)}{1-s^{2}E(x)H(x)},\frac{tH}{1-t^{2}EH}\right\rangle =\sum_{i,j=0}^{\infty}\left\langle (1+sH(x))E(x)^{i}H(x)^{i},E^{j}H^{j+1}\right\rangle s^{2i}t^{2j+1}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left\langle ((1+sH(x))E(x)^{i}H(x)^{i})[X+1],E^{j}H^{j}\right\rangle s^{2i}t^{2j+1}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left\langle (E(x)^{i}H(x)^{i}+sE(x)^{i}H(x)^{i+1})[X+1],E^{j}H^{j}\right\rangle s^{2i}t^{2j+1}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left\langle (E(x)^{i}H(x)^{i}+sE(x)^{i}H(x)^{i+1})[X+1],E^{j}H^{j}\right\rangle s^{2i}t^{2j+1}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left\langle \left(\frac{1+x}{1-x}\right)^{i}E(x)^{i}H(x)^{i}+s\frac{(1+x)^{i}}{(1-x)^{i+1}}E(x)^{i}H(x)^{i+1},E^{j}H^{j}\right\rangle s^{2i}t^{2j+1}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\Bigg(\left(\frac{1+x}{1-x}\right)^{i}\left\langle E(x),E^{j}H^{j}\right\rangle ^{i}\left\langle H(x),E^{j}H^{j}\right\rangle ^{i}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad\quad +s\frac{(1+x)^{i}}{(1-x)^{i+1}}\left\langle E(x),E^{j}H^{j}\right\rangle ^{i}\left\langle H(x),E^{j}H^{j}\right\rangle ^{i+1}\Bigg)s^{2i}t^{2j+1}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\left(\frac{1+x}{1-x}\right)^{2ij+i}+s\frac{(1+x)^{2ij+i+j}}{(1-x)^{2ij+i+j+1}}\right)s^{2i}t^{2j+1}\nonumber \\ & \qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}\left(t\left(\frac{1+x}{1-x}\right)^{i}+st\frac{(1+x)^{i+j}}{(1-x)^{i+j+1}}\right)s^{2i}t^{2j}. \end{align} $$

Summing Equations (5.2) and (5.3) yields

$$ \begin{align*} \left\langle \frac{1+sH(x)}{1-s^{2}E(x)H(x)},\frac{1+tH}{1-t^{2}EH}\right\rangle = \sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}\left(1+s\left(\frac{1+x}{1-x}\right)^{j}+t\left(\frac{1+x}{1-x}\right)^{i}+st\frac{(1+x)^{i+j}}{(1-x)^{i+j+1}}\right)s^{2i}t^{2j}; \end{align*} $$

comparing this with Equation (5.1) completes the proof.

The first several polynomials $P_{n}^{(\operatorname {\mathrm {udr}},\operatorname {\mathrm {iudr}})}(s,t)$ are given in Table 6. We note that the permutations in $\mathfrak {S}_{n}$ with n up-down runs are precisely the alternating permutations in $\mathfrak {S}_{n}$ , so the coefficient of $s^{n}t^{n}$ of $P_{n}^{(\operatorname {\mathrm {udr}},\operatorname {\mathrm {iudr}})}(s,t)$ is the number of doubly alternating permutations in $\mathfrak {S}_{n}$ . In fact, Stanley’s formula (2.2) for doubly alternating permutations can be derived from Theorem 5.1, although it is simpler to work with our formulas for $(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ipk}})$ and $(\operatorname {\mathrm {lpk}},\operatorname {\mathrm {ilpk}})$ .

Table 6

Joint distribution of $\operatorname {\mathrm {udr}}$ and $\operatorname {\mathrm {iudr}}$ over $\mathfrak {S}_{n}$

To invert Theorem 5.1 and other formulas in this section, we will need to solve $s=2u/(1+u^{2})$ for u, and the solution is given by $u=s^{-1}(1-\sqrt {1-s^{2}})=(s/2)C(s^{2}/4)$ , where $C(x)$ is the Catalan number generating function. Hence, Theorem 5.1 can be written as

$$ \begin{align*} &\frac{1}{(1-u)(1-v)}+\frac{1}{4(1-u)^{2}(1-v)^{2}}\sum_{n=1}^{\infty}\frac{(1+u^{2})^{n}(1+v^{2})^{n}}{(1-u^{2})^{n-1}(1-v^{2})^{n-1}}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{iudr}})}(s,t)x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}\left(1+u\left(\frac{1+x}{1-x}\right)^{j}+v\left(\frac{1+x}{1-x}\right)^{i}+uv\frac{(1+x)^{i+j}}{(1-x)^{i+j+1}}\right)u^{2i}v^{2j}, \end{align*} $$

where $u=s^{-1}(1-\sqrt {1-s^{2}})=(s/2)C(s^{2}/4)$ and $v=t^{-1}(1-\sqrt {1-t^{2}})=(t/2)C(t^{2}/4)$ .

5.2 Up-down runs, inverse peaks and inverse descents

Next, let us study the statistic $(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ipk}},\operatorname {\mathrm {ides}})$ and its specializations $(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ipk}})$ and $(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ides}})$ . Define

Theorem 5.2. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}}{(1-s^{2})^{n-1}}\left(\frac{1+yt}{1-t}\right)^{n+1}\frac{P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{ides}})}\left(\frac{2s}{1+s^{2}},\frac{(1+y)^{2}t}{(y+t)(1+yt)},\frac{y+t}{1+yt}\right)}{2(1-s)^{2}(1+y)}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{(1+x)(1+yx)}{(1-yx)(1-x)}\right)^{ij}\left(1+s\left(\frac{1+yx}{1-x}\right)^{j}\right)s^{2i}t^{j}\nonumber \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \qquad & \frac{(1+s^{2})^{n}}{(1-s^{2})^{n-1}}\left(\frac{1+yt}{1-t}\right)^{n+1}\frac{P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{ides}})}\left(\frac{2s}{1+s^{2}},\frac{(1+y)^{2}t}{(y+t)(1+yt)},\frac{y+t}{1+yt}\right)}{2(1-s)^{2}(1+y)}\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{\substack{\lambda\vdash n\\ \mathrm{odd} } }\frac{2^{l(\lambda)}}{z_{\lambda}}\frac{A_{l(\lambda)}(s^{2})A_{l(\lambda)}(t)}{(1-s^{2})^{l(\lambda)+1}(1-t)^{l(\lambda)+1}}\prod_{k=1}^{l(\lambda)}(1-(-y)^{\lambda_{k}})\nonumber\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad+s\sum_{\lambda\vdash n}\frac{1}{z_{\lambda}}\frac{B_{o(\lambda)}(s^{2})A_{l(\lambda)}(t)}{(1-s^{2})^{o(\lambda)+1}(1-t)^{l(\lambda)+1}}\prod_{k=1}^{l(\lambda)}(1-(-y)^{\lambda_{k}}).\nonumber \end{align} $$

As before, we set $y=1$ to specialize to $(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ipk}})$ , immediately arriving at parts (a) and (b) of the following theorem. Part (c) is proven similarly to Theorem 3.4, except that we also use the formula

$$ \begin{align*} 2^{n}A_{n}(t^{2})+tB_{n}(t^{2})=\frac{(1+t)^{2}(1+t^{2})^{n}}{2}P_{n}^{\operatorname{\mathrm{udr}}}\left(\frac{2t}{1+t^{2}}\right) \end{align*} $$

[Reference Gessel and Zhuang15, Section 6.3], where

Theorem 5.3. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\frac{1}{4(1-s)^{2}}\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}}{(1-s^{2})^{n-1}}\left(\frac{1+t}{1-t}\right)^{n+1}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ipk}})}\left(\frac{2s}{1+s^{2}},\frac{4t}{(1+t)^{2}}\right)x^{n}\\ &\quad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}\left(1+s\left(\frac{1+x}{1-x}\right)^{j}\right)s^{2i}t^{j}\nonumber \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} \frac{1}{4(1-s)^{2}}\frac{(1+s^{2})^{n}}{(1-s^{2})^{n-1}}\left(\frac{1+t}{1-t}\right)^{n+1}&P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ipk}})}\left(\frac{2s}{1+s^{2}},\frac{4t}{(1+t)^{2}}\right)\\ &\quad=\frac{1}{n!}\sum_{k=0}^n 2^{k}d(n,k)\frac{(2^{k}A_{k}(s^{2})+sb:{k}(s^{2}))A_{k}(t)}{(1-s^{2})^{k+1}(1-t)^{k+1}}\nonumber \end{align} $$

and

(c) $$ \begin{align} P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ipk}})}(s,t) & =\frac{1}{n!}\sum_{k=0}^n d(n,k)\left((1-s^{2})(1-t)\right)^{\frac{n-k}{2}}P_{k}^{\operatorname{\mathrm{udr}}}(s)P_{k}^{\operatorname{\mathrm{pk}}}(t). \end{align} $$

The first several polynomials $P_{n}^{(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ipk}})}(s,t)$ are displayed in Table 7. The coefficients of $s^{k}t$ in $P_{n}^{(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ipk}})}(s,t)$ are binomial coefficients, just like the coefficients of $st^{k}$ in $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ from Table 2.

Table 7

Joint distribution of $\operatorname {\mathrm {udr}}$ and $\operatorname {\mathrm {ipk}}$ over $\mathfrak {S}_{n}$

Unlike the analogous results given earlier, Proposition 5.4 was not proven in [Reference Troyka and Zhuang26], so we shall supply a proof here.

Setting $y=0$ in Theorem 5.2 yields the analogous result for $(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ides}})$ .

Theorem 5.5. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\frac{1}{2(1-s)^{2}}\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}}{(1-s^{2})^{n-1}(1-t)^{n+1}}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ides}})}\left(\frac{2s}{1+s^{2}},t\right)x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{ij}\left(1+\frac{s}{(1-x)^{j}}\right)s^{2i}t^{j}\nonumber \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} &\frac{(1+s^{2})^{n}}{2(1-s)^{2}(1-s^{2})^{n-1}(1-t)^{n+1}}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ides}})}\left(\frac{2s}{1+s^{2}},t\right)x^{n}\\ &\qquad\quad\qquad=\frac{1}{n!}\sum_{k=0}^n 2^k d(n,k)\frac{A_{k}(s^{2})A_{k}(t)}{(1-s^{2})^{k+1}(1-t)^{k+1}} +\frac{s}{n!}\sum_{k,m=0}^n f(n,k,m)\frac{B_{k}(s^{2})A_{m}(t)}{(1-s^{2})^{k+1}(1-t)^{m+1}}. \nonumber \end{align} $$

Table 8 contains the first several polynomials $P_{n}^{(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ides}})}(s,t)$ .

Table 8

Joint distribution of $\operatorname {\mathrm {udr}}$ and $\operatorname {\mathrm {ides}}$ over $\mathfrak {S}_{n}$

5.3 Up-down runs, inverse left peaks and inverse descents

Define

Theorem 5.6. We have

$$ \begin{align*} &\frac{1}{(1-s)(1-t)}+\frac{1}{2(1-s)^{2}}\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}(1+yt)^{n}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ilpk}},\operatorname{\mathrm{ides}})}\left(\frac{2s}{1+s^{2}},\frac{(1+y)^{2}t}{(y+t)(1+yt)},\frac{y+t}{1+yt}\right)}{(1-s^{2})^{n-1}(1-t)^{n+1}}x^{n}\\[7pt] &\qquad\qquad\qquad\qquad\qquad\quad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{i(j+1)}\left(\frac{1+yx}{1-yx}\right)^{ij}\left(1+s\frac{(1+yx)^{j}}{(1-x)^{j+1}}\right)s^{2i}t^{j}. \end{align*} $$

Setting $y=1$ in Theorem 5.6 yields part (a) of the following theorem. Part (b) is obtained by computing the scalar product

$$\begin{align*}\left\langle \frac{1+sH(x)}{1-sE(x)H(x)},\frac{H}{1-tEH}\right\rangle \end{align*}$$

using Lemmas 2.8 (c)–(d) and 2.9 (b)–(c). Part (c) is obtained from (b) in a way similar to the proof of Theorem 4.4, making use of Equation (4.4).

Theorem 5.7. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\frac{1}{2(1-s)^{2}}\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}(1+t)^{n}}{(1-s^{2})^{n-1}(1-t)^{n+1}}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ilpk}})}\left(\frac{2s}{1+s^{2}},\frac{4t}{(1+t)^{2}}\right)x^{n}\\[7pt] &\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{i(2j+1)}\left(1+s\frac{(1+x)^{j}}{(1-x)^{j+1}}\right)s^{2i}t^{j}\nonumber \end{align} $$

and, for all $n\geq 1$ , we have

(b) $$ \begin{align} &\frac{(1+s^{2})^{n}(1+t)^{n}}{2(1-s)^{2}(1-s^{2})^{n-1}(1-t)^{n+1}} P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ilpk}})}\left(\frac{2s}{1+s^{2}},\frac{4t}{(1+t)^{2}}\right)\\[7pt] &\qquad\qquad\quad\qquad=\frac{1}{n!}\sum_{k=0}^n 2^k d(n,k) \frac{A_{k}(s^{2})B_{k}(t)}{(1-s^{2})^{k+1}(1-t)^{k+1}} +\frac{s}{n!}\sum_{k=0}^n e(n,k) \frac{B_{k}(s^{2})B_{k}(t)}{(1-s^{2})^{k+1}(1-t)^{k+1}}\nonumber \end{align} $$

and

(c) $$ \begin{align} &(1+s)^{2}(1+s^{2})^{n}P_{n}^{(\operatorname{\mathrm{udr}},\operatorname{\mathrm{ilpk}})}\left(\frac{2s}{1+s^{2}},t\right)\\ & \quad\qquad =\frac{1}{n!}\sum_{k=0}^n 2^{k+1} d(n,k) \left((1-s^{2})(1-t)^{1/2}\right)^{n-k}A_{k}(s^{2})P_{k}^{\operatorname{\mathrm{lpk}}}(t)\nonumber\\ & \qquad\qquad\quad +\frac{2s}{n!}\sum_{k=0}^n e(n,k)\left((1-s^{2})(1-t)^{1/2}\right)^{n-k}B_{k}(s^{2})P_{k}^{\operatorname{\mathrm{lpk}}}(t).\nonumber \end{align} $$

See Table 9 for the first several polynomials $P_{n}^{(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ilpk}})}(s,t)$ .

Table 9

Joint distribution of $\operatorname {\mathrm {udr}}$ and $\operatorname {\mathrm {ilpk}}$ over $\mathfrak {S}_{n}$

5.4 Biruns

Recall that a birun is a maximal monotone consecutive subsequence, whereas an up-down run is either a birun or an initial descent. In [Reference Gessel and Zhuang15], the authors gave the formula

$$ \begin{align*} \frac{2+tH(x)+tE(x)}{1-t^{2}E(x)H(x)} & =\frac{2}{1-t}+\frac{2t}{(1-t)^{2}}xh_{1}+\frac{(1+t)^{3}}{2(1-t)}\sum_{n=2}^{\infty}\sum_{L\vDash n}\frac{(1+t^{2})^{n-1}}{(1-t^{2})^{n}}\left(\frac{2t}{1+t^{2}}\right)^{\operatorname{\mathrm{br}}(L)}x^{n}r_{L} \end{align*} $$

(cf. Lemma 2.8), which allows us to produce formulas for (mixed) two-sided distributions involving the number of biruns. For example, computing the scalar product

(5.4) $$ \begin{align} \left\langle \frac{2+sH(x)+sE(x)}{1-s^{2}E(x)H(x)},\frac{2+tH+tE}{1-t^{2}EH}\right\rangle \end{align} $$

would yield a formula for the two-sided distribution of $\operatorname {\mathrm {br}}$ , but we chose not to derive this formula as it would be complicated to write down. Looking back at the formula in Theorem 5.1 for the two-sided distribution of $\operatorname {\mathrm {udr}}$ , we see that the right-hand side is a summation whose summands are each a sum involving four terms, which is because the numerators in the scalar product

$$\begin{align*}\left\langle \frac{1+sH(x)}{1-s^{2}E(x)H(x)},\frac{1+tH}{1-t^{2}EH}\right\rangle \end{align*}$$

that we sought to compute has two terms each. The numerators in the scalar product (5.4) contain three terms each, which will lead to nine terms in the formula for $(\operatorname {\mathrm {br}},\operatorname {\mathrm {ibr}})$ as opposed to four.

On the other hand, formulas for the polynomials

have fewer such terms, and so we present them below.

Theorem 5.8. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\frac{2stx}{(1-s)^{2}(1-t)^{2}}+\frac{(1+s)^{3}}{8(1-s)}\sum_{n=2}^{\infty}\frac{(1+s^{2})^{n-1}(1+t)^{n+1}P_{n}^{(\operatorname{\mathrm{br}},\operatorname{\mathrm{ipk}})}\left(\frac{2s}{1+s^{2}},\frac{4t}{(1+t)^{2}}\right)}{(1-s^{2})^{n}(1-t)^{n+1}}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{2ij}\left(1+s\left(\frac{1+x}{1-x}\right)^{j}\right)s^{2i}t^{j}\nonumber \end{align} $$

and

(b) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\frac{stx}{(1-s)^{2}(1-t)^{2}}+\frac{(1+s)^{3}}{4(1-s)}\sum_{n=2}^{\infty}\frac{(1+s^{2})^{n-1}P_{n}^{(\operatorname{\mathrm{br}},\operatorname{\mathrm{ides}})}\left(\frac{2s}{1+s^{2}},t\right)}{(1-s^{2})^{n}(1-t)^{n+1}}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\frac{1}{2}\sum_{i,j=0}^{\infty}\left(\frac{1+x}{1-x}\right)^{ij}\left(2+\frac{s}{(1-x)^{j}}+s(1+x)^{j}\right)s^{2i}t^{j}.\nonumber \end{align} $$

The first several polynomials $P_{n}^{(\operatorname {\mathrm {br}},\operatorname {\mathrm {ipk}})}(s,t)$ and $P_{n}^{(\operatorname {\mathrm {br}},\operatorname {\mathrm {ides}})}(s,t)$ are displayed in Tables 10–11. Notice that the coefficients of $s^{k}t$ in $P_{n}^{(\operatorname {\mathrm {br}},\operatorname {\mathrm {ipk}})}(t)$ are twice the coefficients of $s^{k}t$ in $P_{n-1}^{(\operatorname {\mathrm {udr}},\operatorname {\mathrm {ipk}})}(t)$ ; we shall give a simple proof of this fact.

Table 10

Joint distribution of $\operatorname {\mathrm {br}}$ and $\operatorname {\mathrm {ipk}}$ over $\mathfrak {S}_{n}$

Table 11

Joint distribution of $\operatorname {\mathrm {br}}$ and $\operatorname {\mathrm {ides}}$ over $\mathfrak {S}_{n}$

In Table 11, we first collect powers of s in displaying the $P_{n}^{(\operatorname {\mathrm {br}},\operatorname {\mathrm {ides}})}(s,t)$ to showcase the symmetry present in the coefficients of $s^{k}$ . This symmetry can be explained in the same way as for the polynomials $P_{n}^{(\operatorname {\mathrm {pk}},\operatorname {\mathrm {ides}})}(s,t)$ : Upon taking reverses, we have that $\operatorname {\mathrm {ides}}(\pi ^{r})=n-1-\operatorname {\mathrm {ides}}(\pi )$ but $\operatorname {\mathrm {br}}(\pi ^{r})=\operatorname {\mathrm {br}}(\pi )$ . Note that the coefficients of $s^{k}$ for $k\geq 2$ also seem to be unimodal, but in general they are not $\gamma $ -positive.

6.1 A rederivation of the Garsia–Gessel formula

We now return full circle by showing how our approach can be used to rederive Garsia and Gessel’s formula (1.4) for the polynomials $A_{n}(s,t,q,r)=\sum _{\pi \in \mathfrak {S}_{n}}s^{\operatorname {\mathrm {des}}(\pi )}t^{\operatorname {\mathrm {ides}}(\pi )}q^{\operatorname {\mathrm {maj}}(\pi )}r^{\operatorname {\mathrm {imaj}}(\pi )}$ .

Proof of the Garsia–Gessel formula.

We seek to compute the the scalar product

$$\begin{align*}\left\langle \sum_{i=0}^{\infty}s^{i}\prod_{k=0}^{i}H(q^{k}x),\sum_{j=0}^{\infty}t^{j}\prod_{l=0}^{j}H(r^{l})\right\rangle \end{align*}$$

in two different ways. First, from Lemma 2.8 (a) we have

(6.1) $$ \begin{align} & \left\langle \sum_{i=0}^{\infty}s^{i}\prod_{k=0}^{i}H(q^{k}x),\sum_{j=0}^{\infty}t^{j}\prod_{l=0}^{j}H(r^{l})\right\rangle \nonumber \\ & \qquad\qquad\qquad\qquad=\sum_{m,n=0}^{\infty}\frac{\sum_{L\vDash m,M\vDash n}s^{\operatorname{\mathrm{des}}(L)}t^{\operatorname{\mathrm{des}}(M)}q^{\operatorname{\mathrm{maj}}(L)}r^{\operatorname{\mathrm{maj}}(M)}\left\langle r_{L},r_{M}\right\rangle }{(1-s)(1-qs)\cdots(1-q^{m}s)(1-t)(1-rt)\cdots(1-r^{n}t)}x^{m}\nonumber \\ & \qquad\qquad\qquad\qquad=\sum_{n=0}^{\infty}\frac{\sum_{\pi\in\mathfrak{S}_{n}}s^{\operatorname{\mathrm{des}}(\pi)}t^{\operatorname{\mathrm{ides}}(\pi)}q^{\operatorname{\mathrm{maj}}(\pi)}r^{\operatorname{\mathrm{imaj}}(\pi)}}{(1-s)(1-qs)\cdots(1-q^{n}s)(1-t)(1-rt)\cdots(1-r^{n}t)}x^{n}\nonumber \\ & \qquad\qquad\qquad\qquad=\sum_{n=0}^{\infty}\frac{A_{n}(s,t,q,r)}{(1-s)(1-qs)\cdots(1-q^{n}s)(1-t)(1-rt)\cdots(1-r^{n}t)}x^{n}, \end{align} $$

where, as usual, we apply Foulkes’s theorem (Theorem 2.1) to calculate the scalar product $\left \langle r_{L},r_{M}\right \rangle $ . Second, observe that

$$ \begin{align*} \left\langle \sum_{i=0}^{\infty}s^{i}\prod_{k=0}^{i}H(q^{k}x),\sum_{j=0}^{\infty}t^{j}\prod_{l=0}^{j}H(r^{l})\right\rangle & =\sum_{i,j=0}^{\infty}s^{i}t^{i}\left\langle \prod_{k=0}^{i}H(q^{k}x),\prod_{l=0}^{j}H(r^{l})\right\rangle \\ & =\sum_{i,j=0}^{\infty}s^{i}t^{i}\prod_{k=0}^{i}\prod_{l=0}^{j}\left\langle H(q^{k}x),H(r^{l})\right\rangle \end{align*} $$

by Lemma 2.3. To evaluate $\left \langle H(q^{k}x),H(r^{l})\right \rangle $ , let us recall that $h_{n}=\sum _{\lambda \vdash n}m_{\lambda }$ . Then

$$ \begin{align*} \left\langle H(q^{k}x),H(r^{l})\right\rangle & =\left\langle \sum_{i=0}^{\infty}h_{i}q^{ki}x^{i},\sum_{j=0}^{\infty}h_{j}r^{lj}\right\rangle \\ & =\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{\lambda\vdash i}\left\langle m_{\lambda},h_{j}\right\rangle q^{ki}x^{i}r^{lj}\\ & =\sum_{j=0}^{\infty}q^{kj}x^{j}r^{lj}\\ & =\frac{1}{1-xq^{k}r^{l}}, \end{align*} $$

whence

(6.2) $$ \begin{align} \left\langle \sum_{i=0}^{\infty}s^{i}\prod_{k=0}^{i}H(q^{k}x),\sum_{j=0}^{\infty}t^{j}\prod_{l=0}^{j}H(r^{l})\right\rangle =\sum_{i,j=0}^{\infty}s^{i}t^{i}\prod_{k=0}^{i}\prod_{l=0}^{j}\frac{1}{1-xq^{k}r^{l}}. \end{align} $$

Combining Equations (6.1) and (6.2) yields Garsia and Gessel’s formula (1.4).

6.2 Major index and other statistics

Lemma 2.8 (a) also enables us to derive formulas for mixed two-sided distributions involving the major index. Define

We leave the proofs of the following formulas to the interested reader.

Theorem 6.1. We have

(a) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\frac{1}{1+y}\sum_{n=1}^{\infty}\left(\frac{1+yt}{1-t}\right)^{n+1}\frac{P_{n}^{(\operatorname{\mathrm{maj}},\operatorname{\mathrm{ipk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(q,\frac{(1+y)^{2}t}{(y+t)(1+yt)},s,\frac{y+t}{1+yt}\right)}{(1-s)(1-qs)\cdots(1-q^{n}s)}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}s^{i}t^{j}\prod_{k=0}^{i}\left(\frac{1+q^{k}yx}{1-q^{k}x}\right)^{j},\nonumber \end{align} $$

(b) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\sum_{n=1}^{\infty}\frac{(1+t)^{n}}{(1-t)^{n+1}}\frac{P_{n}^{(\operatorname{\mathrm{maj}},\operatorname{\mathrm{ilpk}},\operatorname{\mathrm{des}},\operatorname{\mathrm{ides}})}\left(q,\frac{(1+y)^{2}t}{(y+t)(1+yt)},s,\frac{y+t}{1+yt}\right)}{(1-s)(1-qs)\cdots(1-q^{n}s)}x^{n}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}s^{i}t^{j}\prod_{k=0}^{i}\frac{(1+q^{k}yx)^{j}}{(1-q^{k}x)^{j+1}}\nonumber \end{align} $$

and

(c) $$ \begin{align} &\frac{1}{(1-s)(1-t)}+\frac{1}{2(1-s)^{2}}\sum_{n=1}^{\infty}\frac{(1+s^{2})^{n}}{(1-s^{2})^{n-1}}\frac{P_{n}^{(\operatorname{\mathrm{maj}},\operatorname{\mathrm{iudr}},\operatorname{\mathrm{des}})}\left(q,\frac{2s}{1+s^{2}},t\right)}{(1-t)(1-qt)\cdots(1-q^{n}t)}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum_{i,j=0}^{\infty}s^{2i}t^{j}\left(1+s\prod_{k=0}^{j}\frac{1}{1-q^{k}x}\right)\prod_{k=0}^{j}\left(\frac{1+q^{k}x}{1-q^{k}x}\right)^{i}.\nonumber \end{align} $$

One may obtain formulas for $(\operatorname {\mathrm {maj}},\operatorname {\mathrm {ipk}})$ , $(\operatorname {\mathrm {maj}},\operatorname {\mathrm {ilpk}})$ and $(\operatorname {\mathrm {maj}},\operatorname {\mathrm {iudr}})$ by specializing Theorem 6.1 appropriately. For example, setting $y=1$ in Theorem 6.1 (a), multiplying both sides by $1-s$ , and then taking the limit of both sides as $s\rightarrow 1$ yields

$$\begin{align*}\frac{1}{1-t}+\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1+t}{1-t}\right)^{n+1}\frac{P_{n}^{(\operatorname{\mathrm{maj}},\operatorname{\mathrm{ipk}})}\left(q,\frac{4t}{(1+t)^{2}}\right)}{(1-q)(1-q^{2})\cdots(1-q^{n})}x^{n}=\sum_{j=0}^{\infty}t^{j}\prod_{k=0}^{\infty}\left(\frac{1+q^{k}x}{1-q^{k}x}\right)^{j}, \end{align*}$$

where .

7.1 Real-rootedness

A univariate polynomial is called real-rooted if it has only real roots. We conjecture that the distributions of $\operatorname {\mathrm {ipk}}$ and $\operatorname {\mathrm {ilpk}}$ over permutations in $\mathfrak {S}_{n}$ with any fixed value of $\operatorname {\mathrm {pk}}$ , $\operatorname {\mathrm {lpk}}$ , and $\operatorname {\mathrm {udr}}$ —as well as that of $\operatorname {\mathrm {ipk}}$ upon fixing $\operatorname {\mathrm {br}}$ —are all encoded by real-rooted polynomials. This conjecture has been empirically verified for all $n\leq 50$ ; our formulas from Sections 3–6 played a crucial role in formulating and gathering supporting evidence for this conjecture as they have allowed us to efficiently compute the polynomials in question.

Conjecture 7.1 would imply that all of these polynomials are unimodal and log-concave and can be used to show that the distributions of $\operatorname {\mathrm {ipk}}$ and $\operatorname {\mathrm {ilpk}}$ over permutations in $\mathfrak {S}_{n}$ with a fixed value k for the relevant statistics each converge to a normal distribution as $n\rightarrow \infty $ .

It is worth noting that the peak and left peak polynomials

$$\begin{align*}P_{n}^{\operatorname{\mathrm{pk}}}(t)=\sum_{\pi\in\mathfrak{S}_{n}}t^{\operatorname{\mathrm{pk}}(\pi)+1}=\sum_{\pi\in\mathfrak{S}_{n}}t^{\operatorname{\mathrm{ipk}}(\pi)+1}\quad\text{and}\quad P_{n}^{\operatorname{\mathrm{lpk}}}(t)=\sum_{\pi\in\mathfrak{S}_{n}}t^{\operatorname{\mathrm{lpk}}(\pi)}=\sum_{\pi\in\mathfrak{S}_{n}}t^{\operatorname{\mathrm{ilpk}}(\pi)} \end{align*}$$

are known to be real-rooted (see, e.g., [Reference Petersen18, Reference Stembridge25]), so Conjecture 7.1 would also have the consequence that both of these ‘real-rooted distributions’ can be partitioned into real-rooted distributions based on the value of another statistic.

7.2 Gamma-positivity

Real-rootedness provides a powerful method of proving unimodality results in combinatorics, but when the polynomials in question are known to be symmetric,Footnote ³ an alternate avenue to unimodality is $\gamma $ -positivity. Any symmetric polynomial $f(t)\in \mathbb {R}[t]$ with center of symmetry $n/2$ can be written uniquely as a linear combination of the polynomials $\{t^{j}(1+t)^{n-2j}\}_{0\leq j\leq \left \lfloor n/2\right \rfloor }$ —referred to as the gamma basis—and $f(t)$ is called $\gamma $ -positive if its coefficients in the gamma basis are nonnegative. The $\gamma $ -positivity of a polynomial directly implies its unimodality, and $\gamma $ -positivity has appeared in many contexts within combinatorics and geometry; see [Reference Athanasiadis1] for a detailed survey.

The prototypical example of a family of $\gamma $ -positive polynomials are the Eulerian polynomials $A_{n}(t)$ , as established by Foata and Schützenberger [Reference Foata and Schützenberger8], and there is also a sizable literature on the $\gamma $ -positivity of ‘Eulerian distributions’ (polynomials encoding the distribution of the descent number $\operatorname {\mathrm {des}}$ ) over various restricted subsets of $\mathfrak {S}_{n}$ . For example, the Eulerian distribution over linear extensions of sign-graded posets [Reference Brändén2], r-stack-sortable permutations [Reference Brändén3], separable permutations [Reference Fu, Lin and Zeng10] and involutions [Reference Wang28] are all known to be $\gamma $ -positive. The two-sided Eulerian distribution $(\operatorname {\mathrm {des}},\operatorname {\mathrm {ides}})$ is also known to satisfy a refined $\gamma $ -positivity property which was conjectured by Gessel and later proved by Lin [Reference Lin16].

Define

to be the polynomial encoding the distribution of $\operatorname {\mathrm {ides}}$ over permutations in $\mathfrak {S}_{n}$ with k peaks, or equivalently, the distribution of $\operatorname {\mathrm {des}}$ (i.e., the Eulerian distribution) over permutations in $\mathfrak {S}_{n}$ whose inverses have k peaks. We conjecture that these polynomials are $\gamma $ -positive as well.

In fact, we shall give a stronger conjecture which refines by ‘pinnacle sets’. Given a permutation $\pi \in \mathfrak {S}_{n}$ , the pinnacle set $\operatorname {\mathrm {Pin}}(\pi )$ of $\pi $ is defined by

In other words, $\operatorname {\mathrm {Pin}}(\pi )$ contains all of the values $\pi (k)$ at which k is a peak of $\pi $ . Given $n\geq 1$ and $S\subseteq [n]$ , define the polynomial $\hat {A}_{n,S}^{\operatorname {\mathrm {ides}}}(t)$ by

this gives the distribution of the inverse descent number over permutations in $\mathfrak {S}_{n}$ with a fixed pinnacle set S or, equivalently, the Eulerian distribution over permutations in $\mathfrak {S}_{n}$ with ‘inverse pinnacle set’ S.

Note that

$$\begin{align*}\hat{A}_{n,k}^{\operatorname{\mathrm{ides}}}(t)=\sum_{\substack{S\subseteq[n]\\ \left|S\right|=k } }\hat{A}_{n,S}^{\operatorname{\mathrm{ides}}}(t). \end{align*}$$

Because the sum of $\gamma $ -positive polynomials with the same center of symmetry is again $\gamma $ -positive, a positive resolution to Conjecture 7.3 would imply Conjecture 7.2. However, we do not have nearly as much empirical evidence to support Conjecture 7.3. Since we do not have a formula for the polynomials $\hat {A}_{n,k}^{\operatorname {\mathrm {ides}}}(t)$ , we were only able to verify Conjecture 7.3 up to $n=10$ , whereas Conjecture 7.2 has been verified for all $n\leq 80$ with the assistance of Theorem 3.6 (a).