---
author:
- 'O. Guibert and T. Mansour'
title: ' [restricted $132$-involutions and Chebyshev polynomials]{}'
---

[LABRI (UMR 5800), Université Bordeaux 1, 351 cours de la Libération, 33405 Talence Cedex, France]{}\
[guibert@labri.fr]{}, [toufik@labri.fr]{}

Abstract {#abstract .unnumbered}
========

We study generating functions for the number of involutions in $S_n$ avoiding (or containing once) $132$, and avoiding (or containing once) an arbitrary permutation $\tau$ on $k$ letters. In several interesting cases the generating function depends only on $k$ and is expressed via Chebyshev polynomials of the second kind. In particular, we establish that involutions avoiding both $132$ and $12\dots k$ have the same enumerative formula according to the length than involutions avoiding both $132$ and any [*double-wedge pattern*]{} possibly followed by fixed points of total length $k$. Many results are also shown with a combinatorial point of view.

Introduction
============

A permutation is a bijection from $[n]=\{1,2,\ldots,n\}$ to $[n]$. Let $S_n$ be the set of permutations of length $n$.

Let $\alpha\in S_n$ and $\tau\in S_k$ be two permutations. We say that $\alpha$ [*contains*]{} $\tau$ if there exists a subsequence $1\leq i_1<i_2<\dots<i_k\leq n$ such that $(\alpha_{i_1},\dots,\alpha_{i_k})$ is order-isomorphic to $\tau$; in such a context $\tau$ is usually called a [*pattern*]{}. We say that $\alpha$ [*avoids*]{} $\tau$, or is $\tau$-[*avoiding*]{}, if such a subsequence does not exist. The set of all $\tau$-avoiding permutations in $S_n$ is denoted $S_n(\tau)$. For an arbitrary finite collection of patterns $T$, we say that $\alpha$ avoids $T$ if $\alpha$ avoids any $\tau\in T$; the corresponding subset of $S_n$ is denoted $S_n(T)$.

While the case of permutations avoiding a single pattern has attracted much attention, the case of multiple pattern avoidance remains less investigated. In particular, it is natural, as the next step, to consider permutations avoiding pairs of patterns $\tau_1$, $\tau_2$. This problem was solved completely for $\tau_1,\tau_2\in S_3$ (see [@SS]), for $\tau_1\in S_3$ and $\tau_2\in S_4$ (see [@W]), and for $\tau_1,\tau_2\in S_4$ (see [@B1; @Km] and references therein). Several recent papers [@CW; @MV1; @Kr; @MV2] deal with the case $\tau_1\in S_3$, $\tau_2\in S_k$ for various pairs $\tau_1,\tau_2$. Another natural question is to study permutations avoiding $\tau_1$ and containing $\tau_2$ exactly $t$ times. Such a problem for certain $\tau_1,\tau_2\in S_3$ and $t=1$ was investigated in [@R], and for certain $\tau_1\in S_3$, $\tau_2\in S_k$ in [@RWZ; @MV1; @Kr]. The tools involved in these papers include continued fractions, Chebyshev polynomials, and Dyck words.

[*Chebyshev polynomials of the second kind*]{} (in what follows just Chebyshev polynomials) are defined by $$U_r(\cos\theta)=\frac{\sin(r+1)\theta}{\sin\theta}$$ for $r\geq0$. Evidently, $U_r(x)$ is a polynomial of degree $r$ in $x$ with integer coefficients. Chebyshev polynomials were invented for the needs of approximation theory, but are also widely used in various other branches of mathematics, including algebra, combinatorics, and number theory (see [@Ri]).

[*Dyck words*]{} are words $w$ of $\{x,\overline{x}\}^*$ verifying that $|w|_x = |w|_{\overline{x}}$ and that for all $w = w' w''$, $|w'|_x \geq |w'|_{\overline{x}}$. Dyck words of length $2n$ are enumerated by the $n$th Catalan number $C_n=\frac{1}{n+1}\binom{2n}{n}$ whose generating function is $C(x)=\frac{1-\sqrt{1-4x}}{2x}$.

We also consider words of $\{a,b^2\}^*$ of length $n$ enumerated by the $n$th Fibonacci number $F_n$ with $F_0=F_1=1$ and $F_{n}=F_{n-1}+F_{n-2}$ whose generating function is $F(x)=\frac{1}{1-x-x^2}$.

Apparently, for the first time the relation between restricted permutations and Chebyshev polynomials was discovered by Chow and West in [@CW], and later by Mansour and Vainshtein [@MV1; @MV2; @MV3; @MV4], Krattenthaler [@Kr]. These results related to a rational function $$R_k(x)=\frac{2t U_{k-1}(t)}{U_k(t)}, \qquad t=\frac{1}{2\sqrt{x}}$$ for all $k\geq 1$.\
An involution is a permutation such that its cycles are of length $1$ or $2$ that is $\alpha\in S_n$ is an involution if and only if $\alpha(\alpha_i)=i$ for all $i\in [n]$.

Some authors considered involutions with forbidden patterns.\
Regev in [@Regev] provided asymptotic formula for $12\cdots
k$-avoiding involutions of length $n$ and he also established that $1234$-avoiding involutions of length $n$ are enumerated by Motzkin numbers. Gessel [@Gessel] exhibited the enumeration of such $12\cdots k$-avoiding involutions of length $n$. Moreover, Gouyou–Beauchamps [@GouyouBYoung] obtained by an entirely bijective proof very nice exact formulas for the number of $12345$-avoiding and $123456$-avoiding involutions of length $n$.\
Gire [@GireThese] studied some permutations with forbidden subsequences and established a one-to-one correspondence between 1-2 trees having $n$ edges and permutations of length $n$ avoiding patterns $321$ and $231$, the latter being allowed in the case where it is itself a subsequence of the pattern $3142$. Guibert [@GuibertThese] also established bijections between 1-2 trees having $n$ edges and another set of permutations with forbidden subsequences and $3412$-avoiding involutions of length $n$ and $4321$-avoiding involutions of length $n$ (and so with $1234$-avoiding involutions of length $n$ by transposing the corresponding Young tableaux obtained by applying the Robinson-Schensted algorithm). He also shown [@GuibertThese] a bijection between vexillary involutions (that is $2143$-avoiding involutions) and $1243$-avoiding involutions. More recently, Guibert, Pergola and Pinzani [@GPP] established a one-to-one correspondence between 1-2 trees having $n$ edges and vexillary involutions of length $n$. So all these sets are enumerated by the $n$th Motzkin number $\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2i}C_{i}$. It remains a connected open problem: in [@GuibertThese] conjectures that $1432$-avoiding involutions of length $n$ are also enumerated by the $n$th Motzkin number.

In this paper we present a general approach to the study of involutions in $S_n$ avoiding $132$ (or containing $132$ exactly once), and avoiding (or containing exactly once) an arbitrary pattern $\tau\in
S_k$. As a consequence, we derive all the previously known results for this kind of problems, as well as many new results. Some results are also established by bijections as for an example a bijection between $132$-avoiding involutions and primitive Dyck words.

The paper is organized as follows. The case of involutions avoiding both $132$ and $\tau$ is treated in Section $2$. We present an explicit expression in terms of Chebyshev polynomials for several interesting cases. The case of involutions avoiding $132$ and containing $\tau$ exactly once is treated in section $3$. Here again we present an explicit expression in terms of Chebyshev polynomials for several interesting cases. Finally, the cases of involutions containing $132$ exactly once and either avoiding or containing exactly once an arbitrary pattern $\tau$ is treated in sections $4$ and $5$; respectively.

Avoiding $132$ and another pattern {#sec2}
==================================

Let $I_T(n)$ denote the number of involutions in $S_n(132)$ avoiding $T$, and let $I_T(x)=\sum_{n\geq 0}I_T(n)x^n$ be the corresponding generating function. The following proposition is the base of all the other results in this section, which holds immediately from definitions.

\[prom\] For any involution $\pi\in S_n(132)$ such that $\pi_j=n$ holds either,

1.  for $1\leq j\leq [n/2]$, $\pi=(\beta,n,\gamma,\delta,j)$, where

    -   $\beta$ is a $132$-avoiding permutation of the numbers $n-j+1,\dots,n-2,n-1$,

    -   $\delta$ is a $132$-avoiding permutation of the numbers $1,\dots,j-2,j-1$ such that $\delta\cdot\beta$ is the identity permutation of $S_{j-1}$,

    -   $\gamma$ is a $132$-avoiding involution of the numbers $j+1,j+2,\dots,n-j-1,n-j$;

2.  for $j=n$, $\pi=(\beta,n)$ where $\beta$ is an involution in $S_{n-1}(132)$.

As a corollary of Proposition \[prom\] we get the generating function for the number of involutions in $S_n(132)$ as follows.

[(see [@SS Prop. 5])]{} Let $C(t)$ be the generating function for the Catalan numbers; then $$I_\varnothing(x)=\frac{1}{1-x-x^2C(x^2)}.$$

Proposition \[prom\] yields for all $n\geq 1$, $$I_\varnothing(n)=\sum_{j=1}^{[n/2]} C_{j-1}I_\varnothing(n-2j)+I_\varnothing(n-1),$$ where $C_{j-1}$ is the $(j-1)$th Catalan number. Besides $I_\emptyset(0)=1$, therefore in terms of generating function we get that $$I_\varnothing(x)=1+x^2C(x^2)I_\varnothing(x)+xI_\varnothing(x).$$

We can also prove this result by a bijective point of view.

Let $P_{x,\overline{x}} = \{ w \in \{x,\overline{x}\}^* :
\mbox{ for all } w = w' w'', |w'|_x \geq |w'|_{\overline{x}} \}$ be the language of primitive Dyck words.\
The number of such words of $P_{x,\overline{x}}$ of length $n$ is the central binomial coefficient $\binom{n}{\lfloor\frac{n}{2}\rfloor}$ with $n \geq 0$. Indeed, any primitive Dyck word $w$ of $P_{x,\overline{x}}$ can be uniquely written as $w_0 x w_1 x \ldots x w_p$ where $w_i$ is a Dyck word (that is $w_i \in P_{x,\overline{x}}$ and $|w_i|_x =
|w_i|_{\overline{x}}$) for all $0 \leq i \leq p$, but $w$ can also be uniquely written as $w_0 \overline{x} w_1 \overline{x} \dots
\overline{x} w_{\lceil \frac{p}{2} \rceil - 1} \overline{x}
w_{\lceil \frac{p}{2} \rceil} x w_{\lceil \frac{p}{2} \rceil + 1}
x \ldots x w_p$. So primitive Dyck words $w$ of $P_{x,\overline{x}}$ of length $n$ are in bijection with bilateral words of $\{ w \in \{x,\overline{x}\}^* : |w|_x =
|w|_{\overline{x}} \mbox{ or } |w|_x = |w|_{\overline{x}} - 1 \}$ of length $n$ trivially enumerated by $\binom{n}{\lfloor\frac{n}{2}\rfloor}$.

\[Phi\] There is a bijection $\Phi$ between involutions in $S_n(132)$ and primitive Dyck words of $P_{x,\overline{x}}$ of length $n$. Moreover, the number of fixed points of the involution corresponds to the difference between the number of letters $x$ and $\overline{x}$ into the primitive Dyck word.

Let $\pi$ be a $132$-avoiding involution on $[n]$ having $p$ fixed points. According to Proposition \[prom\] we have $\pi=\pi'\pi''x\pi'''$ with $|\pi'|=\frac{n-p}{2}$ ($\pi'$ has no fixed points and constitutes cycles with $\pi''$ or $\pi'''$), $\pi''$ does not contain fixed point and $\pi(x)=x$ ($x$ is the first fixed point). We obtain two $132$-avoiding involutions on $[n+1]$ from $\pi$: the first one is given by inserting a fixed point between $\pi'$ and $\pi''$, and the second one (iff $\pi$ has at least one fixed point) is given by modifying the first fixed point $x$ by a cycle starting between $\pi'$ and $\pi''$. All $132$-avoiding involutions can be obtained (and only once) by applying this rule, starting from the empty involution of length $0$.\
Let $w$ be a primitive Dyck word of $P_{x,\overline{x}}$ of length $n$ such that $|w|_x-|w|_{\overline{x}}=p$. So we have $w = w_0 x
w_1 x \ldots x w_p$ where $w_i$ are Dyck words for all $0 \leq i
\leq p$. We obtain two primitive Dyck words of length $n+1$ from $w$: $x w$ and $x w_0 \overline{x} w_1 x \cdots x w_p$ (iff $p>0$). All primitive Dyck words can be obtained (and only once) by applying this rule, starting from the empty word of length $0$.

Clearly, these two generating trees for the $132$-avoiding involutions and the primitive Dyck words can be characterized by the following succession system:\
$\left\{\begin{array}{lcll}
    \multicolumn{4}{l}{(0)} \\
    (0) & \leadsto & (1) & \\
    (p) & \leadsto & (p+1) , (p-1) & \mbox{if } p \geq 1
    \end{array}\right.$

Figure \[fig-ag\] shows the bijection $\Phi$ between $132$-avoiding involutions and primitive Dyck words (and the labels of the succession system which characterizes them) for the first values.

=590.0pt

\[nb132ptfix\] The number of $132$-avoiding involutions of length $n$ is $\binom{n}{\lfloor \frac{n}{2} \rfloor}$. Moreover, the number of $132$-avoiding involutions of length $n$ having $p$ fixed points with $0 \leq p \leq n$ (and $p$ is odd iff $n$ is odd) is the ballot number $\binom{n}{\frac{n+p}{2}} -
\binom{n}{\frac{n+p}{2}+1}$.

Indeed, the number of primitive Dyck words $w$ of $P_{x,\overline{x}}$ according to the length and $|w|_x-|w|_{\overline{x}}$ is given by the ballot number (or Delannoy number [@Errera] or distribution $\alpha$ of the Catalan number [@Kre]).

In particular, the number of $132$-avoiding involutions of length $2n$ without fixed points is $C_n$ the $n$th Catalan number. The following theorem is the base of all the other results in this section.

\[thg\] Let $T$ set of patterns, $T'=\{(\tau,|\tau|+1) : \tau\in T\}$, and let $S_T(x)$ be the generating function for the number of $T$-avoiding permutations in $S_n(132)$. Then $$I_{T'}(x)=\frac{1}{1-x^2S_T(x^2)}+\frac{x}{1-x^2S_T(x^2)}I_T(x).$$

Proposition \[prom\] with definitions of $T'$ yields for $n\geq
1$, $$I_{T'}(n)=I_T(n-1)+\sum_{j=1}^{[n/2]} s_T(j-1)I_{T'}(n-2j),$$ where $s_T(j-1)$ is the number of permutations in $S_{j-1}(132,T)$. Hence, in terms of generating functions we have $$I_{T'}(x)-1=x\cdot I_T(x)+x^2 S_T(x^2)\cdot\frac{I_{T'}(x)+I_{T'}(-x)}{2}+x^2S_T(x^2)\cdot\frac{I_{T'}(x)-I_{T'}(-x)}{2},$$ so the theorem holds.

Avoiding $132$ and $12\dots k$ {#subsec21}
------------------------------

\[ex1\] [(see [@SS])]{} Let us find $I_{123}(x)$; let $T'=\{123\}$ and $T=\{12\}$, so Theorem \[thg\] gives $$I_{123}(x)=\frac{1}{1-x^2S_{12}(x^2)}+\frac{x}{1-x^2S_{12}(x^2)}I_{12}(x),$$ where by definitions $S_{12}(x)=I_{12}(x)=\frac{1}{1-x}$, hence $$I_{123}(x)=\frac{1+x}{1-2x^2},$$ which means the number of involutions $I_{123}(n)$ is given by $2^{[n/2]}$ for all $n\geq 0$.

Similarly, $I_{1234}(x)=\frac{1}{1-x-x^2}$, so the number of involutions $I_{1234}(n)$ is given by $F_n$ the $n$th Fibonacci number.

The case of varying $k$ is more interesting. As an extension of Example \[ex1\] let us consider the case $T=\{12\dots k\}$.

\[th12k\] For all $k\geq 1$, $$I_{12\dots k}(x)=\frac{1}{x\cdot U_k\left( \frac{1}{2x} \right)}
          \sum_{j=0}^{k-1} U_j\left( \frac{1}{2x} \right).$$

Immediately, the theorem holds for $k=1$. Let $k\geq 2$; Theorem \[thg\] gives $$I_{12\dots k}(x)=\frac{1}{1-x^2S_{12\dots (k-1)}(x^2)}+\frac{x}{1-x^2S_{12\dots (k-1)}(x^2)}I_{12\dots (k-1)}(x).$$ On the other hand, the generating function for the sequence $S_n(132,12\dots(k-1))$ is given by $R_{k-1}(x)$ (see [@CW Th. 1]) with $R_k(x)=\frac{1}{1-xR_{k-1}(x)}$ (see [@MV1]) we get that $$I_{12\dots k}(x)=R_k(x^2)+xR_k(x^2)I_{12\dots (k-1)}(x).$$ Besides $I_{1}(x)=R_0(x)=1$, hence by use induction on $k$ and definitions of $R_k(x)$ the theorem holds.

We consider now a combinatorial point of view for this result.

Let $\pi$ be a $132$-avoiding involution. Clearly, if $\pi$ avoids $12\dots k$ then $\pi$ has less than $k$ fixed points. Moreover, if $\pi$ of length $n$ having less than $k$ fixed points is obtained from an $132$-avoiding involution $\sigma$ of length less than $n$ having $k$ fixed points (and take $\sigma$ as big as possible) by applying the rules described for bijection $\Phi$ given by Theorem \[Phi\], then $\pi$ contains a subsequence $1 2
\ldots k$ because the first fixed points of $\sigma$ become cycles into $\pi$ such that the beginning of these cycles and the last remaining fixed points of $\sigma$ into $\pi$ constitute a subsequence of type $1 2 \ldots k$. So the succession system $(*)$\
$\left\{\begin{array}{lcll}
    (0) \\
    (0) & \leadsto & (1) \\
    (p) & \leadsto & (p+1) , (p-1) & 1 \leq p \leq k-2 \\
    (k-1) & \leadsto & (k-2)
    \end{array}\right.$\
characterizes the generating tree of the involutions avoiding both $132$ and $1 2 \ldots k$.

It is easy to see that for $k$ odd, the number of involutions of length $2m$ avoiding both $132$ and $12\dots k$ is the twice of the number of involutions of length $2m-1$ avoiding both $132$ and $12\dots k$.

Moreover, the reader can note that the set of labels of this succession system is finite and so the corresponding generating function is rational. More precisely, we immediately deduce from the previous succession system that the number of involutions of length $n$ avoiding both $132$ and $12\dots k$ and having $p$ fixed points is given by the $(p+1)$th component of the vector given by $V_k . M_k^n$ where $V_k = ( \ 1 \ 0 \ 0 \ \dots \ 0 \ )$ is a vector of $k$ elements and $M_k = \left(
\begin{array}{ccccccccc}
    0 & 1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
    1 & 0 & 1 & 0 & 0 & \cdots & 0 & 0 & 0 \\
    0 & 1 & 0 & 1 & 0 & \cdots & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 1 & \cdots & 0 & 0 & 0 \\
      &   &   &   &   & \ddots &   &   &   \\
    0 & 0 & 0 & 0 & 0 & \cdots & 1 & 0 & 1 \\
    0 & 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 0 \\
\end{array} \right)$ is a $k \times k$ matrix.\
In another way, we can see that as an automaton where the states are $0, 1, \dots, k-1$ and the transitions are arrows from $i$ to $i+1$ for $0 \leq i < k-1$ and from $i$ to $i-1$ for $0<i\leq
k-1$.

The bijection $\Phi$ establishes a one-to-one correspondence between involutions of length $n$ avoiding both $132$ and $12\dots
k$ and having $p$ fixed points, and primitive Dyck words $w = w_0
x w_1 x \ldots x w_p$ of $P_{x,\overline{x}}$ of length $n$ such that $w_i$ is a Dyck word of height less than $k-p+i$ (that is $w_i \in P_{x,\overline{x}}$ and $|w_i|_x = |w_i|_{\overline{x}}$ and for all $w_i = w' w''$, $|w'|_x-|w'|_{\overline{x}}<k-p+i$) for all $0 \leq i \leq p$.\
In particular involutions of length $2n$ avoiding both $132$ and $12\dots k$ without fixed points are in bijection by $\Phi$ with Dyck words of length $2n$ of height less than $k$.

These Dyck words of bounded height was considered by Kreweras [@Kre] and Viennot [@V]. In particular, Dyck words of length $2n$ of height less than $1$, $2$, $3$, $4$, $5$ are respectively enumerated by $0$, $1$, $2^{n-1}$, $F_{n-2}$, $\frac{3^{n-1}+1}{2}$ for all $n \geq 1$.\
We provide some simple bijections for special cases $k=3,4,5$ (related to Example \[ex1\]) by generating some well known words in the same way as involutions avoiding both $132$ and $1 2 \ldots k$.

Fist of all, we consider the case $k=3$ and the words of $\{a,b\}^*$ or $a\{a,b\}^*$ enumerated by the powers of $2$ we can generate from the empty word labeled $(0)$ by the rules:\
$\left\{\begin{array}{lcl}
    w (0) & \leadsto & a w (1) \\
    a w (1) & \leadsto & a w (2) , b w (0) \\
    w (2) & \leadsto & a w (1)
    \end{array}\right.$\
such that the words labeled $(0)$ start by $b$ whereas the words labeled $(1)$ or $(2)$ start by $a$.\
So, words of $\{a,b\}^n$ (respectively $a\{a,b\}^n$) are in bijection with involutions avoiding both $132$ and $123$ of length $2n$ (respectively $2n+1$) enumerated by $2^n$ (respectively $2^n$).

Next we consider the case $k=4$ and the words of $\{a,b^2\}^\ast$ enumerated by the Fibonacci numbers we can generate from the empty word labeled $(0)$ by the rules:\
$\left\{\begin{array}{lcl}
    w (0) & \leadsto & a w (1) \\
    a w (1) & \leadsto & a a w (2) , b^2 w (0) \\
    a w (2) & \leadsto & b^2 w (3) , a a w (1) \\
    w (3) & \leadsto & a w (2)
    \end{array}\right.$\
such that the words labeled $(0)$ or $(3)$ start by $b^2$ whereas the words labeled $(1)$ or $(2)$ start by $a$.\
So, words of $\{a,b^2\}^\ast$ of length $n$ are in bijection with involutions in $S_n(132,1234)$ enumerated by $F_n$ the $n$th Fibonacci number.

Now we consider the case $k=5$ and the words of $\{a,b,c\}^\ast a$ or $\{a,b,c\}^\ast a \cup b \{a,b,c\}^\ast a$ enumerated by the powers of $3$ we can generate from the empty word labeled $(0)$ by the rules:\
$\left\{\begin{array}{lcl}
    w (0) & \leadsto & a w (1) \\
    w (1) & \leadsto & b w (2) , w (0) \\
    w = b w' = b b^* c w (2) & \leadsto & w (3) , a w' (1) \\
    w = b w' = b b^* a w (2) & \leadsto & c w' (3) , w (1) \\
    w (3) & \leadsto & w (4) , b w (2) \\
    w (4) & \leadsto & c w (3)
    \end{array}\right.$\
such that the words labeled $(0)$ or $(1)$ start by $b^\ast a$, the words labeled $(3)$ or $(4)$ start by $b^\ast c$, and the words labeled $(2)$ start by $b$ (and have one letter more than words labeled $(0)$ or $(4)$ at the same level).\
So, words of $\{a,b,c\}^n a$ (respectively $\{a,b,c\}^n a \cup b \{a,b,c\}^n a$) are in bijection with involutions avoiding both $132$ and $12345$ of length $2n+1$ (respectively $2n+2$) enumerated by $3^n$ (respectively $2.3^n$).

Figure \[fig-forbid\] (an output of the software `forbid` [@GuibertThese]) shows the first values for the number of involutions avoiding both $132$ and $12\dots k$ for $3 \leq k \leq
5$ according to the number of fixed points.

Involutions $\pi \in S_n(132,123)$ according to $|\{\pi(x)=x\}|$ for $1 \leq n \leq 15$

             =1 =2 =2 =4 =4 =8 =8 =16 =16 =32 =32 =64 =64 =128 =128
          2:  -  1  0  2  0  4  0   8   0  16   0  32   0   64    0
          1:  1  0  2  0  4  0  8   0  16   0  32   0  64    0  128
          0:  0  1  0  2  0  4  0   8   0  16   0  32   0   64    0
             1: 2: 3: 4: 5: 6: 7:  8:  9: 10: 11: 12: 13:  14:  15: [n]

Involutions $\pi \in S_n(132,1234)$ according to $|\{\pi(x)=x\}|$ for $1 \leq n \leq 15$

             =1 =2 =3 =5 =8 =13 =21 =34 =55 =89 =144 =233 =377 =610 =987
          3:  -  -  1  0  3   0   8   0  21   0   55    0  144    0  377
          2:  -  1  0  3  0   8   0  21   0  55    0  144    0  377    0
          1:  1  0  2  0  5   0  13   0  34   0   89    0  233    0  610
          0:  0  1  0  2  0   5   0  13   0  34    0   89    0  233    0
             1: 2: 3: 4: 5:  6:  7:  8:  9: 10:  11:  12:  13:  14:  15: [n]

Involutions $\pi \in S_n(132,12345)$ according to $|\{\pi(x)=x\}|$ for $1 \leq n \leq 15$

             =1 =2 =3 =6 =9 =18 =27 =54 =81 =162 =243 =486 =729 =1458 =2187
          4:  -  -  -  1  0   4   0  13   0   40    0  121    0   364     0
          3:  -  -  1  0  4   0  13   0  40    0  121    0  364     0  1093
          2:  -  1  0  3  0   9   0  27   0   81    0  243    0   729     0
          1:  1  0  2  0  5   0  14   0  41    0  122    0  365     0  1094
          0:  0  1  0  2  0   5   0  14   0   41    0  122    0   365     0
             1: 2: 3: 4: 5:  6:  7:  8:  9:  10:  11:  12:  13:   14:   15: [n]

Avoiding $132$ and $213\dots k$
-------------------------------

\[ex2\] Let us find $I_{213}(x)$; let $T'=\{213\}$ and $T=\{21\}$, so Theorem \[thg\] gives $$I_{213}(x)=\frac{1}{1-x^2S_{21}(x^2)}+\frac{x}{1-x^2S_{21}(x^2)}I_{21}(x),$$ where by definitions $S_{21}(x)=I_{21}(x)=\frac{1}{1-x}$, hence $$I_{213}(x)=\frac{1+x}{1-2x^2},$$ which means the number of involutions $I_{213}(n)$ is given by $2^{[n/2]}$ for all $n\geq 0$.

Similarly, $I_{2134}(x)=\frac{1}{1-x-x^2}$, so the number of involutions $I_{2134}(n)$ is given by $F_n$ the $n$th Fibonacci number.

We can easily prove by a combinatorial way that $I_{213}(n)$ is given by $2^{[n/2]}$.\
An involution $\pi$ avoiding both $132$ and $213$ of length $n$ can be either $(n+1-i)(n+2-i)\dots n \pi' 12\dots i$ with $i\geq1$ or $12\dots n$ such that $\pi'$ is also an involution avoiding both $132$ and $213$ (of length $n-2i$, and if we subtract $i$ to each element). We code this recursive decomposition of an involution $\pi$ avoiding both $132$ and $213$ by a word of nonnegative integers formed by the successive positive numbers $i$ and whose last nonnegative integer is (the smallest integer of) the number of the fixed points in $\pi$ divided by $2$. This coding is clearly bijective.\
For example, the involutions $\epsilon$, $1$, $12$, $21$, $123$, $321$, $1234$, $4231$, $3412$ and $4321$ are respectively coded by $0$, $0$, $1$, $10$, $1$, $10$, $2$, $11$, $20$ and $110$. Moreover, the involution $21 \linebreak[0] \ 19 \linebreak[0] \ 20
\linebreak[0] \ 16 \linebreak[0] \ 17 \linebreak[0] \ 18
\linebreak[0] \ 15 \linebreak[0] \ 14 \linebreak[0] \ 9
\linebreak[0] \ 10 \linebreak[0] \ 11 \linebreak[0] \ 12
\linebreak[0] \ 13 \linebreak[0] \ 8 \linebreak[0] \ 7
\linebreak[0] \ 4 \linebreak[0] \ 5 \linebreak[0] \ 6
\linebreak[0] \ 2 \linebreak[0] \ 3 \linebreak[0] \ 1$ in $S_{21}(132,213)$ is coded by $123112$.\
Thus, involutions avoiding both $132$ and $213$ of length $n$ are coded by words $w = w_1 w_2 \ldots w_{l-1} w_l$ with $l \geq 1$, $w_j \geq 1$ for all $1 \leq j < l$, $w_l \geq 0$ and $\sum_{j=1}^{l} w_j = \lfloor \frac{n}{2} \rfloor$. Trivially, words $w$ are in bijection with words $w_1 w_2 \ldots w_{l-1}
(w_l+1)$ which are compositions of $\lfloor \frac{n}{2} \rfloor +
1$ into $l$ positive parts enumerated by $2^{\lfloor \frac{n}{2}
\rfloor}$.

The case of varying $k$ is more interesting. As an extension of Example \[ex2\] let us consider the case $T=\{213\dots k\}$. Similarly as Theorem \[th12k\] we have

\[th21k\] For all $k\geq 1$, $$I_{213\dots k}(x)=\frac{1}{x\cdot U_k\left( \frac{1}{2x} \right)}
          \sum_{j=0}^{k-1} U_j\left( \frac{1}{2x} \right).$$

Therefore, Theorem \[th12k\] and Theorem \[th21k\] yields $I_{123\dots k}(n)=I_{213\dots k}(n)$. We establish a bijection for this result.

\[12k=2134k\] There is a bijection between involutions avoiding both $132$ and $12\ldots k$ of length $n$ and involutions avoiding both $132$ and $2134\ldots k$ of length $n$, for any $k \geq 3$.\
Moreover, two involutions in bijection have the same number of fixed points $p$ for all $0 \leq p \leq k-3$ whereas the involutions avoiding both $132$ and $12\ldots k$ having $k-2$ or $k-1$ fixed points correspond to the involutions avoiding both $132$ and $2134\ldots k$ having $k-2$ or more fixed points.

In order to establish this result we consider a generating tree for the involutions avoiding both $132$ and $2134\ldots k$ which is characterized by the same succession system $(*)$ given in Subsection \[subsec21\] characterizing a generating tree for the involutions avoiding both $132$ and $12\ldots k$.\
So let $\pi$ be an involution avoiding both $132$ and $2134\ldots k$ of length $n$ and let $q=|\{\pi(x)=x\}|$ be the number of fixed points of $\pi$. The label $(p)$ of $\pi$ is defined by $p=q$ if $q \leq k-3$ or by $p=k-2$ if $q \geq k-2$ and $(n+k) \bmod 2 = 0$ or by $p=k-1$ if $q \geq k-2$ and $(n+k) \bmod 2 = 1$. Of course, the empty involution of length $0$ has label $(0)$. We obtain $\sigma$ an involution avoiding both $132$ and $2134\ldots k$ of length $n+1$ by applying the following rules:

-   If $p \in [0,k-3]$, we have $\pi=\pi'\pi''$ with $|\pi'|=\frac{n-p}{2}$, and then $\sigma$ obtained by inserting a fixed point between $\pi'$ and $\pi''$ has label $(p+1)$.

-   If $p=k-2$, we have $\pi=\pi'x\pi''$ with $\pi(x)=x=\frac{n+4-k}{2}$, and then $\sigma$ obtained by inserting a fixed point between $\pi'$ and $x$ has label $(k-1)$.

-   If $p \in [1,k-3]$, we have $\pi=\pi'\pi''x\pi'''$ with $|\pi'|=\frac{n-p}{2}$, $\pi(x)=x$ and $\pi(y) \neq y$ for all $1 \leq y < x$. Then $\sigma$ obtained by modifying the first fixed point $x$ by a cycle starting between $\pi'$ and $\pi''$ (and ending in $x$) has label $(p-1)$.

-   If $p=k-1$, we have $\pi=\pi'x(x+1)\pi''$ with $\pi(x)=x=\frac{n+3-k}{2}$, and then $\sigma$ obtained by inserting a fixed point between $\pi'$ and $x$ has label $(k-2)$.

-   If $p=k-2$, we have $\pi=\pi'(x-j)(x-j+1)\dots(x+j)\pi''$ with $j
    \geq 0$, $\pi(x)=x=\frac{n+4-k}{2}$, $|\pi'|=\frac{n-k}{2}+1-j$ and $e>x$ for all $e \in \pi'$. Then $\sigma$ obtained by modifying the $2j+1$ fixed points between $\pi'$ and $\pi''$ by $j+1$ consecutive cycles each of difference (between the index and the value) $j+1$ that is $(\pi_1'+1)(\pi_2'+1)\dots(\pi_{\frac{n-k}{2}+1-j}'+1)
    (\frac{n-k}{2}+3)(\frac{n-k}{2}+4)\dots(\frac{n-k}{2}+3+j)
    (\frac{n-k}{2}-j+2)(\frac{n-k}{2}-j+3)\dots(\frac{n-k}{2}+2)
    \pi''$ has label $(k-3)$.

There is a bijection between permutations avoiding both $132$ and $12\ldots k$ of length $n$ and permutations avoiding both $132$ and $2134\ldots k$ of length $n$, for any $k \geq 3$.

By Proposition \[prom\], we deduce that permutations $\pi$ avoiding both $132$ and $12\ldots k$ (respectively $2134\ldots k$) of length $n$ are in bijection with involutions without fixed points $(\pi^{-1}+n) \pi$ avoiding both $132$ and $12\ldots k$ (respectively $2134\ldots k$) of length $2n$. Moreover, a particular case of Theorem \[12k=2134k\] establishes a one-to-one correspondence between involutions avoiding both $132$ and $12\ldots k$ without fixed points and involutions avoiding both $132$ and $2134\ldots k$ without fixed points.

Avoiding $132$ and $(d+1(d+2)\dots k12\dots d$
----------------------------------------------

\[ex22\] By Proposition \[prom\] it is easy to obtain for $n\geq1$, $$I_{231}=n;\quad  I_{321}=[n/2]+1.$$

We consider a combinatorial approach to show Example \[ex22\]. Clearly, we have that involutions avoiding both $132$ and $231$ of length $n$ are $i (i-1) \ldots 1 (i+1) (i+2) \ldots n$ for all $1 \leq i \leq n$ and that involutions avoiding both $132$ and $321$ of length $n$ are $(i+1) (i+2) \ldots (2i) 1 2 \ldots i (2i+1) (2i+2) \ldots n$ for all $0 \leq i \leq \lfloor \frac{n}{2} \rfloor$.

As an extension of Example \[ex22\] let us consider the case $T=\{[k,d]\}$, where $[k,d]=(d+1,d+2,\dots,k,1,2,\dots,d)$.

\[thkd\] For any $k\geq 2$, $k/2\geq d\geq 1$, $$I_{[k,d]}=\frac{1}{x(U_d(t)-U_{d-1}(t))}\left[
                U_{d-1}(t)+\frac{U_{k-2d-1}(t)}{U_{k-d}(t)U_{k-d-1}(t)}\sum_{j=0}^{k-d-1} U_j(t)
                \right],\quad t=\frac{1}{2x}.$$

Proposition \[prom\] yields, in the second case the generating function for the number of involutions $[k,d]$-avoiding permutations is $xI_{[k,d]}(x)$. In the first case, we assume that $\gamma$ either $(1)$ avoiding $12\dots (k-d)$, or $(2)$ containing $12\dots (k-d)$. In $(1)$, $\beta$ and $\delta$ avoiding $12\dots (k-d-1)$, so the generating function for these number of involutions is $x^2R_{k-d-1}(x^2)I_{12\dots(k-d)}(x)$ (similarly Theorem \[th12k\]). In $(2)$, $\beta$ and $\delta$ avoiding $12\dots (d-1)$, so the generating function for these number of involutions is $x^2R_{d-1}(x^2)(I_{[k,d]}(x)-I_{12\dots(k-d)}(x))$ (the generating function for the number of involutions in $S_n(132,[k,d])$ such containing $12\dots(k-d)$ is given $I_{[k,d]}(x)-I_{12\dots(k-d)}(x)$). Therefore $$\begin{array}{ll}
I_{[k,d]}(x)&=1+xI_{[k,d]}(x)+x^2R_{k-d-1}(x^2)I_{12\dots(k-d)}(x)+\\
        &+x^2R_{d-1}(x^2)(I_{[k,d]}(x)-I_{12\dots(k-d)}(x)),
\end{array}$$ which means that $$I_{[k,d]}(x)=\frac{1}{1-x-x^2R_{k-d-1}(x^2)}\cdot(1+x^2I_{12\dots (k-d)}(x)(R_{k-d-1}(x^2)-R_{d-1}(x^2))).$$ Hence, by use the identities $R_k(x)=\frac{1}{1-xR_{k-1}(x)}$ and $R_a(x)-R_b(x)=\frac{U_{a-b-1}(t)}{\sqrt{x}U_a(t)U_b(t)}$, the theorem holds.

\[ex3412\] Theorem \[thkd\] yields for $k=4$ and $d=2$, the number of involutions $I_{3412}(n)$ is given by $F_n$ the $n$th Fibonacci number.

We consider a combinatorial approach to show Example \[ex3412\]. An involution $\pi$ avoiding both $132$ and $3412$ of length $n$ can be written $i\pi' 1(i+1)(i+2)\cdots n$ with $1\leq i\leq n$ such that $\pi'$ is also an involution avoiding both $132$ and $3412$ (of length $i-2$, and if we subtract $1$ to each element). We code $\pi$ by a word of $\{a, b^2\}^\ast$ of length $n$ in that way: $a$ if $\pi_i=i$, $b^2$ if $\pi_i<i$ and nothing if $\pi_i>i$ for all $1\leq i\leq n$. This coding is clearly bijective.\
Following [@MV2] we say that $\tau\in S_k$ is a [*wedge*]{} pattern if it can be represented as $\tau=(\tau^1,\rho^1,\dots,\tau^r,\rho^r)$ so that each of $\tau^i$ is nonempty, $(\rho^1,\rho^2,\dots,\rho^r)$ is a layered permutation of $1,\dots,s$ for some $s$, and $(\tau^1,\tau^2,\dots,\tau^r)=(s+1,s+2,\dots,k)$. For example, $645783912$ and $456378129$ are wedge patterns.

For a further generalization of Theorem \[th12k\], Theorem \[th21k\] and [@MV2 Th. 2.6], consider the following definition. We say that $\tau\in S_{2l}$ is a [*double-wedge*]{} pattern if there exist a wedge pattern $\sigma\in S_{l-1}$ such that $$\tau=(\sigma^{-1}+l,2l,\sigma,l)\ \mbox{or}\ \tau=(\sigma+l,2l,\sigma^{-1},l).$$ For example, the double-wedge patterns of length $10$ are $6789(10)1234\linebreak[0] 5$, $7689(10)\linebreak[0]
213\linebreak[0] 4\linebreak[0] 5\linebreak[0]$, $7869(10)\linebreak[0] 31245\linebreak[0]$, $7896(10)\linebreak[0]
41235\linebreak[0]$, $8679(10)\linebreak[0] 23145\linebreak[0]$, $8796(10)\linebreak[0] 42135\linebreak[0]$, $8967(10)\linebreak[0]
34125\linebreak[0]$, $9678(10)\linebreak[0] 23415\linebreak[0]$ and $9768(10)\linebreak[0] 32415\linebreak[0]$.

\[dwedge\] For any double-wedge pattern $\tau\in S_{2l}(132)$ $$I_\tau(x)=I_{12\dots(2l)}(x)=\frac{R_l(x^2)}{1-xR_l(x^2)}.$$

First of all, let us find the generating function $I_\rho(x)$ where $\rho=(\sigma^{-1},2l,\sigma,l)$. By use Proposition \[prom\] we obtain in the first case $xI_\rho(x)$, and in the second case $x^2S_\sigma(x^2)I_\rho(x)$ where $S_\sigma(x^2)$ is the generating function for the number of permutations in $S_n(132,\sigma)$, therefore ($1$ for the empty permutation) $$I_\rho(x)=1+xI_\rho(x)+x^2S_\sigma(x^2)I_\rho(x).$$ On the other hand, Mansour and Vainshtein proved $S_\sigma(x)=R_{l-1}(x)$ for any wedge pattern $\sigma$, so $$I_\rho(x)=\frac{1}{1-x-x^2R_{l-1}(x^2)}.$$ By use the identity $R_l(x)=\frac{1}{1-xR_{l-1}(x)}$ we have $$I_\rho(x)=\frac{R_l(x)}{1-xR_l(x^2)}.$$ Now, let us find $I_{12\dots 2l}(x)$ in terms of $R_j(x)$. So, by use the identity $$\sum_{j=0}^{2l}U_j(t)=\frac{U_{2l}(t)U_{l-1}(t)}{U_l(t)-U_{l-1}(t)}$$ and use the symmetric inverse operation, the first part of the theorem holds.

\[extwedge\] For any wedge pattern $\sigma\in S_{l-1}$ the generating function for the number of permutations in $S_n(132,(\sigma^{-1}+l,2l,\sigma,l,2l+1,\dots,k))$ (or $S_n(132,(\sigma+l,2l,\sigma^{-1},l,2l+1,\dots,k))$, or $S_n(132,(\sigma+l,2l,\sigma,l,2l+1,\dots,k))$) is given by $R_{k}(x)$, for all $k\geq 2l$.

Let $\tau=(\sigma^{-1}+l,2l,\sigma,l)$ and let $S_\tau(x)$ be the generating function for the number of permutations in $S_n(132,\tau)$. By [@MV2 Th. 1] we have $$S_\tau(x)=1+x(S_\tau(x)-S_\sigma^{-1}(x))S_\sigma(x)+xS_\sigma^{-1}(x)S_\tau(x).$$ On the other hand, by [@MV2 Th. 2.6] and $\sigma$ a wedge pattern in $S_{l-1}(132)$ we have $S_{\sigma^{-1}}(x)=S_\sigma(x)=R_{l-1}(x)$. Therefore, by use the identity $R_l(x)=\frac{1}{1-xR_{l-1}(x)}$ we get $$S_\tau(x)=\frac{R_l(x)(1-xR_{l-1}(x)R_l(x))}{1-xR_l^2(x)}.$$ By use the definitions of Chebyshev polynomials of the second kind it is easy to see $$\frac{R_l(x)(1-xR_{l-1}(x)R_l(x))}{1-xR_l^2(x)}=R_{2l}(x),$$ hence by use again Theorem [@MV2 Th. 1] we have $S_{(\tau,2l+1,\dots,k)}(x)=R_k(x)$. Similarly we obtain the other cases.

As a corollary of Theorem \[dwedge\] we have

\[idwedge\] For any double wedge pattern $\tau\in I_{2l}(132)$ $$I_{(\tau,2l+1,2l+2,\dots,k)}(x)=I_{12\dots k}(x).$$

Since, if $S_\beta(x)=S_\gamma(x)$ and $I_\beta(x)=I_\gamma(x)$, then Theorem \[thg\] yields $I_{\tau'}(x)=I_{\beta'}(x)$, and by use [@MV2 Th. 1] we have $S_{\tau'}(x)=S_{\rho'}(x)$, where $\tau'=(\tau_1,\dots,\tau_p,p+1)$ and $\rho'=(\rho_1,\dots,\rho_p,p+1)$ two patterns in $S_{p+1}$. Hence, the theorem holds by use Theorem \[dwedge\], Theorem \[extwedge\], and induction on $p$.

In view of Theorem \[dwedge\] and Theorem \[idwedge\] it is a challenge to find a bijective proof.

Avoiding $132$ and two other patterns
-------------------------------------

Now, let us restrict more than two patterns ($132$ and two other patterns).

\[ex3\] Let us find $I_{123,213}(x)$; let $T'=\{123,213\}$ and $T=\{12,21\}$, so Theorem \[thg\] gives $$I_{123,213}(x)=\frac{1}{1-x^2S_{12,21}(x^2)}+\frac{x}{1-x^2S_{12,21}(x^2)}I_{12,21}(x),$$ where by definitions $S_{12,21}(x)=I_{12,21}(x)=1+x$, hence $$I_{123,213}(x)=\frac{1+x+x^2}{1-x^2-x^4},$$ which means the number of involutions $I_{123,213}(2n)$ is given by $F_{n+1}$, and $I_{123,213}(2n+1)$ is given by $F_{n}$ for all $n\geq 0$, where $F_m$ is the $m$th Fibonacci number.

We consider a combinatorial approach to show Example \[ex3\] by distinguishing the cases of odd and even length.\
An involution $\pi$ avoiding $132$, $123$ and $213$ of length $2n+1$ can be written either $(2n+1) \pi' 1$ or $(2n) (2n+1) \pi''
2 1$ or $1$ (if $n=0$) such that $\pi'$ and $\pi''$ are also involutions avoiding $132$, $123$ and $213$ (of length $2n-1$ for $\pi'$ if we subtract $1$ to each element, of length $2n-3$ for $\pi''$ if we subtract $2$ to each element). We code $\pi$ by a word of $\{a,b^2\}^*$ of length $n$ in that way: $a$ if $\pi_i=2n+2-i$, $b^2$ if $\pi_i=2n+1-i$ and nothing if $\pi_i=2n+3-i$ for all $1\leq i\leq n$. This coding is clearly bijective.\
An involution $\pi$ avoiding $132$, $123$ and $213$ of length $2n$ can be written either $(2n) \pi' 1$ (that includes $2 1$ if $n=1$) or $(2n-1) (2n) \pi'' 2 1$ or $1 2$ (if $n=1$) or the empty involution (if $n=0$) such that $\pi'$ and $\pi''$ are also involutions avoiding $132$, $123$ and $213$ (of length $2n-2$ for $\pi'$ if we subtract $1$ to each element, of length $2n-4$ for $\pi''$ if we subtract $2$ to each element). We code $\pi$ by a word of $\{a,b^2\}^*$ of length $n+1$ in that way: $a$ if $\pi_i=2n+1-i$ for all $1\leq i\leq n-1$, $b^2$ if $\pi_n=n+1$, $b^2$ if $\pi_i=2n-i$ for all $1\leq i\leq n-2$, $b^2a$ if $\pi_{n-1}=n+1$ and $aa$ if $\pi_n=n$. Moreover, the empty involution is coded by $a$. This coding is clearly bijective.

Using definitions and Theorem \[thg\] it is easy to see the following.

\[th12k213\] For all $k\geq 1$, $$I_{123\dots k, 213}(x)=I_{(k-1)\dots 21k, 123}(x)=\frac{1+x+x^2+\dots+x^{k-1}}{1-x^2-x^4-\dots-x^{2(k-1)}}.$$

\[ex312decr\] Using Proposition \[prom\] it is easy to see for $n\geq 1$, $$I_{213,321}(n)=\frac{1}{2}((-1)^n+3),\quad
I_{213,4321}(n)=[n/2]+1.$$

We consider a combinatorial approach to show Example \[ex312decr\]. Clearly, we have that involutions avoiding $132$, $213$ and $321$ of length $n$ are $1 2 \ldots n$ and also $(m+1) (m+2) \ldots n 1 2 \ldots m$ if $n=2m$ with $m \geq 1$. We also have that involutions avoiding $132$, $213$ and $4321$ of length $n$ are $(n+1-i) (n+2-i) \ldots n (i+1) (i+2) \ldots (n-i) 1 2 \ldots i$ for all $0 \leq i \leq \lfloor \frac{n}{2} \rfloor$.

Avoiding $132$ and containing another pattern
=============================================

Let $I_\tau^r(n)$ denote the number of involutions in $S_n(132)$ containing $\tau$ exactly $r$ times, and let $I_\tau^r(x)=\sum_{n\geq 0}I_\tau^r(n)x^n$ be the corresponding generating function. Let us start by the following example.

\[exbb0\] By Proposition \[prom\] it is easy to see $$I_{12}^1(x)=xI_{1}^1(x)+x^2I_{12}^1(x),$$ which means $I_{12}^1(x)=\frac{x^2}{1-x^2}$.

As extension of Example \[exbb0\] let us consider the case $\tau=12\dots k$.

\[th1\_12\] For all $k\geq 1$; $$I_{12\dots k}^{1}=\frac{1}{U_k\left( \frac{1}{2x} \right)}.$$

By Proposition \[prom\] we have for $n\geq k$, $$I_{12\dots k}^{1}(n)=I_{12\dots (k-1)}^{1}(n-1)+\sum_{j=1}^{[n/2]} s_{12\dots (k-1)}(j-1)I_{12\dots k}^{1}(n-2j),$$ where $s_{12\dots k}(j-1)$ is the number of $12\dots k$-avoiding permutations in $S_{j-1}(132)$. Besides $I_{12\dots k}^{1}(n)=0$ for all $n\leq k-1$, and $I_{12\dots k}^{1}(k)=1$. Similarly as proof of Theorem \[th12k\] we have $$I_{12\dots k}^{1}(x)=xR_{k}(x^2)I_{12\dots (k-1)}^{1}(x).$$ Hence, by induction on $k$ with initial condition $I_1^{1}=x$, the theorem holds.

Similarly as Theorem \[th1\_12\] we have an explicit formula when $\tau=213\dots k$ or $\tau=23\dots k1$.

For all $k\geq 2$; $$I_{213\dots k}^{1}=\frac{1-x^2}{U_k\left( \frac{1}{2x} \right)},
    \quad
    I_{23\dots k1}^1(x)=\frac{x^3}{(1-x)U_{k-2}\left( \frac{1}{2x} \right)}.$$

More generally, by Proposition \[prom\] and the argument proof of Theorem \[th12k\] we get

\[th12kr\] For any $k,r\geq 1$ $$I_{12\dots k}^{r}(x)=xI_{12\dots (k-1)}^{r}(x)+x^2\sum_{2a+b=r} S_{12\dots (k-1)}^{a}(x^2)I_{12\dots k}^{b}(x),$$ where $S_{12\dots (k-1)}^{a}(x)$ is the generating function for the number of permutations in $S_n$ containing $12\dots (k-1)$ exactly $a$ times.

In [@Kr] found an explicit formula for $S_{12\dots k}^{r}(x)$, so Theorem \[th12kr\] yields a recurrence for $I_{12\dots
k}^{r}(x)$. For example, following [@Kr] ([@MV1 Th. 3.1]) we have a recurrence for $I_{12\dots k}^{r}(x)$ where $r=1,2,\dots,2k$.

Let $k\geq 1$; for $r=1,2,\dots,2k$ $$I_{12\dots k}^{r}(x)=xI_{12\dots (k-1)}^r(x)+x^2R_{k-1}(x^2)I_{12\dots k}^{r}(x)
    +x^2\sum_{2a+b=r,\ a>0} x^{a-1}I_{12\dots k}^{b}(x)\frac{U_{k-1}^{a-1}\left( \frac{1}{2x} \right)}{U_k^{a+1}\left( \frac{1}{2x} \right)}.$$

The above Theorem yields for $r=2$ an explicit formula for $I_{12\dots k}^{2}(x)$.

For all $k\geq 1$, $$I_{12\dots k}^{2}(x)=\frac{1}{U_k\left( \frac{1}{2x} \right)}\sum_{i=1}^k
    \frac{\sum_{j=0}^{k-i} U_j\left( \frac{1}{2x} \right)}{U_{k+1-i}\left( \frac{1}{2x} \right) U_{k-i}\left( \frac{1}{2x} \right)}.$$

Containing $132$ once and avoiding another pattern
==================================================

We first relate involutions containing $132$ once to $132$-avoiding involutions.

\[bij132once-132\] There is a bijection $\Psi$ between involutions containing $132$ exactly once of length $n$ having $p$ fixed points with $1\leq p\leq n$ and $132$-avoiding involutions of length $n-2$ having also $p$ fixed points.

Let $\pi=\pi'xz\pi''y\pi'''$ with $\pi(x)=x$, $\pi(y)=z$ and $1+x=y<z$ be an involution containing $132$ once (that is subsequence $xzy$) of length $n$ having $p$ fixed points. We replace the subsequence $xzy$ by a fixed point between $\pi''$ and $\pi'''$ in order to obtain an $132$-avoiding involution of length $n-2$ having $p$ fixed points. Note that the only possibility to have exactly once $132$ subsequence is a cycle with a fixed point just to its left. Moreover, $y=x+1$ in order to forbid another $132$ subsequence and cycles are only allowed from $\pi'$ to $\pi''$ and from $\pi'$ to $\pi'''$ (and not from $\pi''$, $\pi'$, $\pi''$, $\pi'''$ respectively to $\pi'''$, $\pi'$, $\pi''$, $\pi'''$) whereas fixed points can uniquely be into $\pi'''$. Clearly the involution we obtain avoids $132$ and in particular, the fixed point $z-2$ cannot be a part of an $132$-subsequence because it cannot be the $3$ or $2$ (all the elements on its left are greater than it) and it cannot be the $1$ (there is no cycle starting on its right).\
Let $\sigma = \sigma' \sigma'' \sigma''' t \sigma''''$ with $\sigma(t) = t$ and $\sigma(i) \neq i$ for all $1 \leq i < t$ (that is $t$ is the first fixed point), $\sigma'(i) > t$ for all $1 \leq i \leq |\sigma'|$ (all the elements of $\sigma'$ are cycles ending into $\sigma''''$), $\sigma''(i) \in
[|\sigma'\sigma''|+1,t-1]$ for all $1 \leq i \leq |\sigma''|$ and $\sigma'''(i) \in [|\sigma'|+1,|\sigma'\sigma''|]$ for all $1 \leq
i \leq |\sigma'''|$ ($\sigma''\sigma'''$ is entirely constituted by cycles from $\sigma''$ to $\sigma'''$) be an $132$-avoiding involution of length $n-2$ having $p$ fixed points. We modify the fixed point $t$ by a cycle starting between $\sigma''$ and $\sigma'''$ (and ending between $\sigma'''$ and $\sigma''''$) and by adding a fixed point just to the right of $\sigma''$ in order to obtain an involution containing $132$ once of length $n$ having $p$ fixed points. Proposition \[prom\] leads immediately to the decomposition of $\sigma$. The involution we obtain contains $132$ exactly once that is the subsequence we modify and insert. There is no other $132$-subsequence and in particular, the fixed point inserted and the start of the new cycle cannot be the $3$ or $2$ of another $132$-subsequence (all the elements on their left are greater than them), the fixed point inserted and the start of the new cycle and the end of the new cycle cannot be the $1$ of another $132$-subsequence (there is no cycle starting on their right), the end of the new cycle cannot be the $3$ of another $132$-subsequence (because in that case the $2$ must be connected to $\sigma'$ and the $1$ must be the fixed point inserted or an element of $\sigma'''$ that forms an $231$-subsequence), and the end of the new cycle cannot be the $2$ of another $132$-subsequence (because in that case the $1$ must be an element of $\sigma'\sigma''$ or the start of the new cycle and the $3$ must be the fixed point inserted or an element of $\sigma'''$ that forms an $312$-subsequence excepted for the fixed point inserted and the new cycle).\
So we have established a bijection between $\pi$ an involution containing $132$ once and $\sigma$ an $132$-avoiding involution where $t=z-2$, $\pi'$ corresponds to $\sigma'\sigma''$, $\pi''=\sigma'''$ and $\pi'''$ corresponds to $\sigma''''$.

For example, the involution $22 \linebreak[0] \ 19 \linebreak[0] \
17 \linebreak[0] \ 18 \linebreak[0] \ 16 \linebreak[0] \ 12
\linebreak[0] \ 11 \linebreak[0] \ 13 \linebreak[0] \ {\bf 9}
\linebreak[0] \ {\bf 14} \linebreak[0] \ 7 \linebreak[0] \ 6
\linebreak[0] \ 8 \linebreak[0] \ {\bf 10} \linebreak[0] \ 15
\linebreak[0] \ 5 \linebreak[0] \ 3 \linebreak[0] \ 4
\linebreak[0] \ 2 \linebreak[0] \ 20 \linebreak[0] \ 21
\linebreak[0] \ 1 \linebreak[0] \ 23$ containing $132$ once (the subsequence $9$ $14$ $10$) corresponds to the $132$-avoiding involution $20 \linebreak[0] \ 17 \linebreak[0] \ 15 \linebreak[0]
\ 16 \linebreak[0] \ 14 \linebreak[0] \ 10 \linebreak[0] \ 9
\linebreak[0] \ 11 \linebreak[0] \ 7 \linebreak[0] \ 6
\linebreak[0] \ 8 \linebreak[0] \ {\bf 12} \linebreak[0] \ 13
\linebreak[0] \ 5 \linebreak[0] \ 3 \linebreak[0] \ 4
\linebreak[0] \ 2 \linebreak[0] \ 18 \linebreak[0] \ 19
\linebreak[0] \ 1 \linebreak[0] \ 21$.

The number of involutions containing $132$ exactly once of length $n$ having $p$ fixed points with $1 \leq p \leq n$ is the ballot number $\binom{n-2}{\frac{n+p}{2}-1}-\binom{n-2}{\frac{n+p}{2}}$. Moreover, the number of involutions containing $132$ exactly once of length $n$ is $\binom{n-2}{\lfloor \frac{n-3}{2} \rfloor}$.

We immediately deduce this result from bijection $\Psi$ of Theorem \[bij132once-132\] and Corollary \[nb132ptfix\]. In fact, the number of involutions containing $132$ once of length $n$ is either the number of $132$-avoiding involutions of length $n-2$ if $n$ is odd or the number of $132$-avoiding involutions of length $n-2$ having more than one fixed point if $n$ is even.

Of course, some of the following results can immediately be obtained from bijection $\Psi$ of Theorem \[bij132once-132\] and results of Section \[sec2\].

Let $J_\tau(n)$ denote the number of involutions in $S_n(\tau)$ such containing $132$ exactly once, and let $J_\tau(x)=\sum_{n\geq
0} J_\tau(n)x^n$ be the corresponding generating function. The following proposition is the base of all the other results in this section, which holds immediately from definitions.

\[prom2\] Let $\pi$ an involution in $S_n$ such that contains $132$ exactly once, and let $\pi_j=n$. Then holds either

1.  or $\pi_n=n$;

2.  or $\pi=(\pi',n,\pi'',\pi''',j)$ where $1\leq j\leq n/2$, $\pi'''={\pi'}^{-1}$ and $\pi'$ avoids $132$.

3.  or $\pi=(\pi',m,2m+1,\pi'',m+1)$ where $n=2m+1$, $\pi''={\pi'}^{-1}$ and $\pi'\in S_{m-1}(132)$.

Another approach to find the generating function of involutions in $S_n$ containing $132$ exactly once is by use Proposition \[prom2\].

\[thcc1\] Let $C(t)$ be the generating function for the Catalan numbers; then $$J_\varnothing(x)=\frac{x^3C(x^2)}{1-x-x^2C(x^2)}.$$

According to Proposition \[prom2\] with terms of generating functions we get the following: the first part of the proposition yields $xJ_\varnothing(x)$, the second part of the proposition yields $x^2C(x^2)J_\varnothing(x)$, and the third part of the proposition gives $x^3C(x^2)$. Hence $$J_\varnothing(x)=xJ_\varnothing(x)+x^2C(x^2)J_\varnothing(x)+x^3C(x^2).$$

\[excc1\] From Proposition \[prom2\] it is easy to see that, the number of the involutions in $S_n(123)$ and containing $132$ exactly once is $2^{(n-3)/2}$ for $n$ odd, otherwise is $0$. Also, $J_{1234}(n)=F_{n-3}$ the $(n-3)$th Fibonacci number, $J_{12345}(n)=3^{[(n-3)/2]}$.

Again, the case of varying $k$ is more interesting. As an extension of Example \[excc1\] let us consider the case $\tau=12\dots k$.

\[thcc2\] For all $k\geq 1$, $$J_{12\dots k}(x)=\frac{x}{U_k\left( \frac{1}{2x} \right)} \sum_{j=1}^{k-2} U_{j}\left( \frac{1}{2x} \right).$$

Proposition \[prom2\] with use the generating function of permutations in $S_n(132,12\dots k)$ given by $R_k(x)$, yields $$J_{12\dots k}(x)=xJ_{12\dots(k-1)}+x^2R_{k-1}(x^2)J_{12\dots k}(x)+x^3R_{k-1}(x^2).$$ By use the relation $R_k(y)=1/(1-yR_{k-1}(y))$ we get that $$J_{12\dots k}(x)=xR_k(x^2)J_{12\dots(k-1)}(x)+x^3R_{k-1}(x^2)R_k(x^2),$$ so induction on $k$ with Example \[excc1\] gives the theorem.

Similarly, we obtain another case $\tau=213\dots k$.

\[thcc22\] For all $k\geq 3$, $$J_{213\dots k}(x)=\frac{x}{U_k\left( \frac{1}{2x} \right)}
    \left[
    xU_2\left(\frac{1}{2x}\right) +\sum_{j=2}^{k-2} U_{j}\left( \frac{1}{2x} \right)
    \right].$$

Similarly as proof of Theorem \[thcc2\] with use the generating function for the number of $213\dots k$-avoiding permutations in $S_n(132)$ is given by $R_k(x)$ (see [@MV2]), we obtain that $$J_{213\dots k}(x)=xR_k(x^2)J_{213\dots(k-1)}(x)+x^3R_{k-1}(x^2)R_k(x^2),$$ and by induction with $J_{213}(x)=x^4R_3(x^2)$ (it is easy to see) the theorem holds.

Theorem \[thcc22\] yields $J_{2134}(2n+3)=J_{2134}(2n+4)=F_{2n}$ the $(2n)$th Fibonacci number for all $n \geq 0$.

\[excc2\] Proposition \[prom2\] yields, $J_{231}(n)=1$ for all $n\geq 1$, and $J_{2341}(n)=2^{[(n-1)/2]}-1$ for all $n\geq 1$.

Once again, the case of varying $k$ is more interesting. As an extension of Example \[excc2\] let us consider the case $\tau=23\dots k1$.

\[thcc3\] For all $k\geq 3$, $$J_{23\dots k1}(x)=\frac{x^2U_{k-3}\left( \frac{1}{2x} \right)}{(1-x)U_{k-2}\left( \frac{1}{2x} \right)}
       \left[ 1+\frac{1}{U_{k-1}\left( \frac{1}{2x} \right)}\sum_{j=1}^{k-3} U_j\left( \frac{1}{2x} \right) \right].$$

Similarly as proof Theorem \[thcc2\] we have that $$J_{23\dots k1}(x)=xJ_{23\dots k1}(x)+x^2R_{k-2}(x^2)J_{12\dots(k-1)}(x)+x^3R_{k-2}(x^2),$$ so by using Theorem \[thcc2\] the theorem holds.

More generally, we present an explicit expression when $\tau=[k,d]$ as follows.

\[thcc4\] For $k\geq 4$, $2\leq d\leq k/2$, $$J_{[k,d]}(x)=\frac{R_d(x^2)}{1-xR_d(x^2)}\left[
    x^2R_{k-d-1}(x^2)+\frac{x^2(R_{k-d-1}(x^2)-R_{d-1}(x^2))}{U_{k-d}\left(\frac{1}{2x}\right)}\sum_{j=1}^{k-d-2} U_j\left(\frac{1}{2x}\right)
    \right].$$

According to Proposition \[prom2\] in terms of generating functions we get the following. In first case $xJ_{[k,d]}(x)$. In the third case, if $\pi'$ contains $12\dots(k-d-1)$ then $\pi$ contains $[k,d]$ which is a contradiction, we get that $\pi'$ avoids $12\dots(k-d-1)$, hence $x^3R_{k-d-1}(x^2)$. Finally, in the second case, let us observe two subcases $\pi''$ contains $12\dots (k-d)$ or avoids $12\dots (k-d)$; so by use the same argument of the third case we get $$x^2R_{k-d-1}(x)J_{12\dots (k-d)}(x)+x^2R_{d-1}(x^2)(J_{[k,d]}(x)-J_{12\dots(k-d)}(x)).$$ Therefore, if we add all these cases we get $J_{[k,d]}(x)$. Hence, by Theorem \[thcc2\] this theorem holds.

Containing $132$ once and containing another pattern
====================================================

Let $J_\tau^r(n)$ denote the number of involutions in $S_n$ such containing $132$ exactly once and containing $\tau$ exactly $r$ times. Let $J_\tau^r(x)=\sum_{n\geq 0} J_\tau^r(n)x^n$ be the corresponding generating function. Let us start be the following result.

\[thd1\] For all $k\geq 1$, $$J_{12\dots k}^1(x)=0.$$

By Proposition \[prom2\] it is easy to see $$J_{12\dots k}^1(x)=xJ_{12\dots (k-1)}^1(x)+x^2R_{k-1}(x^2)J_{12\dots k}(x).$$ with $J_{12}^1(x)=0$, hence induction on $k$ gives the theorem.

Similarly as Theorem \[thd1\] we have another case where $\tau=23\dots k1$.

\[thd2\] For all $k\geq 1$, $$J_{23\dots k1}^1(x)=0.$$

\[exdd1\] Proposition \[prom2\] yields the following. The number of involutions $J_{21}^1(n)=1$ for $n\geq 3$, and $J_{213}(n)=2^{(n-8)/2}(1+(-1)^n)$.

Once again, the case of varying $k$ is more interesting. As an extension of Example \[exdd1\] let us consider the case $\tau=213\dots k$.

\[thd3\] For all $k\geq 3$, $J_{213\dots
    k}^1(x)=\frac{x(1-x^2)}{U_k\left(\frac{1}{2x}\right)}$.

By Proposition \[prom2\] it is easy to obtain $$J_{213\dots k}^1(x)=xJ_{213\dots (k-1)}^1(x)+x^2R_{k-1}(x^2)J_{213\dots k}(x).$$ with $J_{21}^1(x)=x^3/(1-x)$ (which is yield directly from definitions), hence induction on $k$ gives the theorem.

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