Generating Trees Unifying Several Classes of Four-Letter Pattern-Avoiding Descent Sequences

Abstract

A descent sequence is a word of nonnegative integers satisfying that the first term is zero, and each subsequent term is at most one plus the number of descents among the preceding terms, where a descent is a position at which a term is greater than the term immediately following it. In this work, we study descent sequences subject to the additional restriction of avoiding a given pattern of length four. We analyze seven distinct avoidance classes and provide enumerative results for each of them. Our approach is based on the construction of generating trees with one or two labels, from which we derive succession rules and corresponding systems of recurrence relations. These recurrences are then used to compute explicit generating functions for the number of descent sequences of length n avoiding either 0001, 0010, 0011, 0012, 0021, 0110, 0112, 0123, or 0132.

1. Introduction

Pattern avoidance has been widely investigated across a variety of combinatorial frameworks such as permutations, set partitions, ascent sequences, inversion sequences, and descent-like structures. This line of research is motivated by deep connections with enumerative combinatorics, generating functions, bijective constructions, and Wilf-equivalence phenomena. Among these objects, descent sequences form an interesting family closely linked to Catalan-type structures and other restricted integer sequences that naturally arise in enumeration problems. Investigating pattern avoidance in this setting extends classical avoidance theory to a broader class of combinatorial sequences and sheds light on their structural and enumerative behavior. Standard references include [1,2,3].

A descent in an integer sequence $w_1{w}_2\dots w_m$ is an index j such that $w_j>w_{j+1}$. We call a sequence of nonnegative integers ${\pi }_1{\pi }_2\dots {\pi }_n$ a descent sequence of length n if it satisfies:

• the initial condition ${\pi }_1=0$,
• for each i with $2\le i\le n$, the inequality ${\pi }_i\le \mathrm{des}({\pi }_1{\pi }_2\dots {\pi }_{i-1})+1$ holds,

where $\mathrm{des}(·)$ counts the number of descents in the given prefix. For instance, 010203 satisfies these conditions and is a descent sequence, whereas 01103 fails to satisfy them. These sequences were first introduced by Callan [4], and later studied jointly by the author and Callan in the context of avoidance of pairs of length-three patterns [5].

We now recall the notion of pattern containment for integer sequences. Let $w=w_1{w}_2\dots w_n$ be a word over the integers. A pattern $\tau ={\tau }_1\dots {\tau }_m$ is a word over $\{0,1,\dots ,\mathit{\ell }\}$ in which every symbol in this alphabet appears at least once. We say that w contains $\tau$ if there exist indices $1\le f_1<f_2<\dots <f_m\le n$ such that the subsequence $w_{f_1}{w}_{f_2}\dots w_{f_m}$ is order-isomorphic to $\tau$. Equivalently, for any $i,j$, the relations between $w_{f_i}$ and $w_{f_j}$ (less than, greater than, or equal) match exactly those between ${\tau }_i$ and ${\tau }_j$. If no such subsequence exists, then w is said to avoid $\tau$. As an illustration, the descent sequence 010201303 realizes every pattern of length three but avoids the pattern 0321. We write ${\mathcal{D}}_n({\tau }^{\left(1\right)},\dots ,{\tau }^{\left(m\right)})$ for the set of descent sequences of length n that avoid all patterns ${\tau }^{\left(1\right)},\dots ,{\tau }^{\left(m\right)}$. Two pattern sets T and R are said to be D-Wilf-equivalent, written $T\stackrel{d}{\sim }R$, if $|{\mathcal{D}}_n\left(T\right)| = |{\mathcal{D}}_n\left(R\right)|$ for all $n\ge 0$.

Callan and Mansour [5] determined the classification of D-Wilf-equivalence classes: there are 9 classes for single patterns of length three, 23 classes for pairs of length-three patterns, and 69 classes for patterns of length four. They also obtained explicit generating functions in several cases, including

$$F_{1002}\left(x\right)=F_{1012}\left(x\right)={\displaystyle \frac{x(1-2x)}{(1-x)(1-3x+x^2)}},F_{0100}\left(x\right)=F_{0101}\left(x\right)=F_{0102}\left(x\right)={\displaystyle \frac{x}{1-2x}}.$$

Here $F_{\tau }\left(x\right)$ denotes the generating function counting nonempty descent sequences in ${\mathcal{D}}_n\left(\tau \right)$.

In this work, we focus on computing $F_{\tau }\left(x\right)$ for several patterns of length four. For each equivalence class, we construct a generating tree with one or two labels, extract functional equations from the associated succession rules, and solve them to obtain closed-form generating functions. In particular, we enumerate descent sequences avoiding any of the patterns 0001, 0010, 0011, 0012, 0021, 0110, 0112, 0123, and 0132. (See also [1,2,3] for related results on pattern avoidance in permutations and set partitions.)

The structure of the paper is as follows. Section 2 introduces the framework of generating trees and outlines the general strategy. In Section 2.1, Section 2.2, Section 2.3, Section 2.4, Section 2.5, Section 2.6 and Section 2.7, we treat the individual patterns 0010, 0011, 0110, 0112, 0123, 0132, and 0001 separately. In each case, we derive appropriate succession rules and compute the corresponding generating functions.

2. Generating Trees and Proofs

We now describe the generating tree $\mathcal{T}\left(P\right)$ associated with pattern-avoiding descent sequences in $\mathcal{D}\left(P\right)={\bigcup }_{n\ge 1}{\mathcal{D}}_n\left(P\right)$, following the construction in [5] and related works such as [6,7,8,9,10,11,12].

The tree is rooted at the single vertex labeled 0 at level 1. Suppose ${\pi }_1\dots {\pi }_{n-1}\in {\mathcal{D}}_{n-1}\left(P\right)$. Its children are all sequences obtained by appending a value ${\pi }_n$ satisfying ${\pi }_n\in \{0,1,\dots ,\mathrm{des}({\pi }_1\dots {\pi }_{n-1})+1\}$, provided that the resulting sequence still avoids all patterns in P.

For a node $\pi$, let $\mathcal{T}(P;\pi )$ denote the subtree rooted at $\pi$ that contains all of its descendants. Two nodes $\pi$ and ${\pi }^{\prime }$ are considered equivalent, written as $\pi \sim {\pi }^{\prime }$, if their corresponding subtrees are isomorphic as ordered rooted trees.

We then form a reduced tree $\mathcal{T}\left[P\right]$ by identifying each node with the first representative (in a breadth-first, left-to-right traversal) of its isomorphism class. From this point onward, we treat $\mathcal{T}\left(P\right)$ and $\mathcal{T}\left[P\right]$ as the same object.

Example 1.

Let $P=\left\{0021\right\}$. The children of 0 in the generating tree $\mathcal{T}\left(P\right)$ are 00 and 01, which leads to the succession rule $0⇝00,01$.

Clearly, the children of $00\in \mathcal{T}\left(P\right)$ are 000 and 001. The descent sequence $000{\pi }^{\prime }$ avoids P if and only if the descent sequence $00{\pi }^{\prime }$ avoids P, so $000\sim 00$. Also, the descent sequence $001{\pi }^{\prime }$ avoids P if and only if the descent sequence $00{\pi }^{\prime }$ avoids P (by removing the third letter in $001{\pi }^{\prime }$), so $001\sim 00$. Thus, we have the succession rule $00⇝00,00$. Similarly, we have $01⇝010,01$ and $010⇝010,00,010$. Hence, the generating tree $\mathcal{T}\left(P\right)$ is determined by the following succession rules: $0⇝00,01$, $00⇝00,00$, $01⇝010,01$, and $010⇝010,00,010$.

For a given set P, we aim to obtain an explicit expression for the generating function

$$F_P\left(x\right)=\sum _{n\ge 1}\left(\mathrm{number}\mathrm{of}\mathrm{nodes}\mathrm{at}\mathrm{level}n\mathrm{of}\mathcal{T}\left(P\right)\right)x^n.$$

To this end, we introduce the more refined quantity

$$F_{P;\pi }\left(x\right)=\sum _{n\ge 1}\left(\mathrm{number}\mathrm{of}\mathrm{nodes}\mathrm{at}\mathrm{level}n\mathrm{of}\mathcal{T}(P;\pi )\right)x^n,$$

where $\pi \in D\left(P\right)$. It is immediate that $F_P\left(x\right)=F_{P;0}\left(x\right)$. Furthermore, any succession rule of the form

$$v⇝v^{\left(1\right)},\dots ,v^{\left(k\right)}\mathrm{with}v,v^{\left(1\right)},\dots ,v^{\left(k\right)}\in D\left(P\right),$$

corresponds to the functional equation

$$F_{P;v}\left(x\right)=x+x{\displaystyle \sum _{i=1}^k}{F}_{P;v^{\left(i\right)}}\left(x\right).$$

Let $P=\left\{0012\right\}$. By Example 1, we have

$$\begin{array}{cc}\hfill F_{P;0}\left(x\right)& =x+x{F}_{P;00}\left(x\right)+x{F}_{P;01}\left(x\right),\hfill \\ \hfill F_{P;00}\left(x\right)& =x+2x{F}_{P;00}\left(x\right),\hfill \\ \hfill F_{P;01}\left(x\right)& =x+x{F}_{P;010}\left(x\right)+x{F}_{P;01}\left(x\right),\hfill \\ \hfill F_{P;010}\left(x\right)& =x+2x{F}_{P;010}\left(x\right)+x{F}_{P;00}\left(x\right).\hfill \end{array}$$

From the solution of this linear system, we obtain the following result.

Theorem 1.

The generating function $F_{0012}\left(x\right)$ for the number of descent sequences of length n that avoid 0012 is given by

$${\displaystyle \frac{x(1-3x+2x^2+x^3)}{(1-x){(1-2x)}^2}}.$$

Note that ${\mathcal{D}S}_n\left(021\right)={\mathcal{D}S}_n\left(0021\right)$ for all $n\ge 1$. Hence, by [5], we have the following formula.

Theorem 2.

The generating function $F_{0021}\left(x\right)$ for the number of descent sequences of length n that avoid 0021 is given by

$${\displaystyle \frac{{(1-x)}^2-\sqrt{(1-x)(1-3x-x^2-x^3)}}{2x(1-x)}}.$$

We now apply the previously described technique to derive explicit formulas for the remaining six patterns of length 4, as presented in the next six subsections. Note that the proof of a succession rule of the form $v⇝v^{\left(1\right)},\dots ,v^{\left(k\right)}$ typically involves appending a letter $s_j$ to v and verifying that if $v{s}_j$ avoids the given pattern, then $v{s}_j\sim v^{\left(j\right)}$ for all $j=1,2,\dots ,s$.

2.1. The Pattern 0010

Lemma 1.

The generating tree $\mathcal{T}\left(\right\{0010\left\}\right)$ is determined by the following succession rules:

$$\begin{array}{cccc}0⇝00,01,\hfill & 00⇝00,001,\hfill & 001⇝001,\hfill & 01⇝a_1,b_1,\hfill \\ b_1⇝f_1,b_1,\hfill & a_m⇝a_m,d_m,e_{m+1},\hfill & b_m⇝f_m,b_m,\hfill & c_m⇝c_m,\hfill \\ d_m⇝d_m,c_{m+1},\hfill & e_m⇝a_m,b_m,\hfill & f_m⇝f_m,d_m,c_{m+1},\hfill & \end{array}$$

where $a_m=01021\dots m(m-1)$, $b_m=a_{m-1}mm$, $c_m=a_{m-1}(m-1)m$, $d_m=a_{m}m$, $e_m=a_{m-1}m$, and $f_m=a_{m-1}mm(m-1)$.

Proof. 

Clearly, the children of $0\in \mathcal{T}\left(P\right)$ are 00 and 01. Thus, we obtain the succession rule $0⇝00,01$. The children of $00\in \mathcal{T}\left(P\right)$ are 000 and 001. Note that a descent sequence $000{\pi }^{\prime }$ avoids P if and only if $00{\pi }^{\prime }$ avoids P, which implies that $000\sim 00$. Thus, we obtain the succession rule $00⇝00,001$.

The only child of $001\in \mathcal{T}\left(P\right)$ is 0010. Moreover, a descent sequence $0010{\pi }^{\prime }$ avoids P if and only if $001{\pi }^{\prime }$ avoids P, which implies that $0010\sim 001$. Hence, we obtain the succession rule $001⇝001$.

Clearly, the children of $01\in \mathcal{T}\left(P\right)$ are $a_1=010$ and $b_1=011$, and the children of $b_1\in \mathcal{T}\left(P\right)$ are $f_1=0110$ and $0111\sim b_1$. Thus, we have the succession rules $01⇝a_1,b_1$ and $b_1⇝f_1,b_1$.

The children of $a_m=01021\dots m(m-1)\in \mathcal{T}\left(P\right)$ are $a_{m}0\sim a_m$, $d_m=01021\dots m$$(m-1)m$, and $e_{m+1}=01021\dots m(m-1)(m+1)$. Thus, we obtain the succession rule $a_m⇝a_m,d_m,e_{m+1}$. Moreover, the children of $b_m=01021\dots (m-1)(m-2)mm\in \mathcal{T}\left(P\right)$ are $f_m=b_m(m-1)$ and $b_{m}m\sim b_m$. Thus, we obtain the succession rule $b_m⇝f_m,b_m$.

Similarly, we have $c_m⇝c_m$, $d_m⇝d_m,c_{m+1}$, $e_m⇝a_m,b_m$, and $f_m⇝f_m,d_m,c_{m+1}$. □

Lemma 2.

We have

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{F}_{\{0010;0\}}\left(x\right)\hfill & =x+x{F}_{\{0010;00\}}\left(x\right)+x{F}_{\{0010;01\}}\left(x\right),\hfill \\ F_{\{0010;00\}}\left(x\right)\hfill & =x+x{F}_{\{0010;00\}}\left(x\right)+x{F}_{\{0010;001\}}\left(x\right),\hfill \\ F_{\{0010;001\}}\left(x\right)\hfill & =x+x{F}_{\{0010;001\}}\left(x\right),\hfill \\ F_{\{0010;01\}}\left(x\right)\hfill & =x+xA\left(0\right)+xB\left(0\right),\hfill \\ B\left(0\right)\hfill & =x+xF\left(0\right)+xB\left(0\right),\hfill \end{array}\right.\end{array}$$

(1)

where $A\left(v\right),B\left(v\right)$ and $F\left(v\right)$ satisfies the following system of equations:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}A\left(v\right)\hfill & ={\displaystyle \frac{x}{1-v}}+xA\left(v\right)+xD\left(v\right)+{\displaystyle \frac{x}{v}}(E\left(v\right)-E\left(0\right)),\hfill \\ B\left(v\right)\hfill & ={\displaystyle \frac{x}{1-v}}+xF\left(v\right)+xB\left(v\right),\hfill \\ C\left(v\right)\hfill & ={\displaystyle \frac{x}{1-v}}+xC\left(v\right),\hfill \\ D\left(v\right)\hfill & ={\displaystyle \frac{x}{1-v}}+xD\left(v\right)+{\displaystyle \frac{x}{v}}(C\left(v\right)-C\left(0\right)),\hfill \\ E\left(v\right)\hfill & ={\displaystyle \frac{x}{1-v}}+xA\left(v\right)+xB\left(v\right),\hfill \\ F\left(v\right)\hfill & ={\displaystyle \frac{x}{1-v}}+xF\left(v\right)+xD\left(v\right)+{\displaystyle \frac{x}{v}}(C\left(v\right)-C\left(0\right))\hfill \end{array}\right.\end{array}$$

(2)

with $K\left(v\right)={\sum }_{m\ge 1}{F}_{\left\{0010\right\};k_m}\left(x\right)v^{m-1}$, for all $K\in \{A,B,C,D,E,F\}$.

Proof. 

By translating the first five succession rules in Lemma 1 into equations, we obtain (1).

Translating the succession rule $a_m⇝a_m,d_m,e_{m+1}$ into an equation in terms of $F_{P;\pi }\left(x\right)$ yields

$$F_{P;a_m}\left(x\right)=x+x{F}_{P;a_m}\left(x\right)+x{F}_{P;d_m}\left(x\right)+x{F}_{P;e_{m+1}}\left(x\right).$$

Multiplying both sides by $v^{m-1}$ and summing over $m\ge 1$, we obtain

$$\begin{array}{cc}\hfill \sum _{m\ge 1}{F}_{P;a_m}\left(x\right)v^{m-1}& =\sum _{m\ge 1}x{v}^{m-1}+x\sum _{m\ge 1}{F}_{P;a_m}\left(x\right)v^{m-1}+x\sum _{m\ge 1}{F}_{P;d_m}\left(x\right)v^{m-1}\hfill \\ \hfill & +x\sum _{m\ge 1}{F}_{P;e_{m+1}}\left(x\right)v^{m-1}.\hfill \end{array}$$

Therefore,

$$A\left(v\right)={\displaystyle \frac{x}{1-v}}+xA\left(v\right)+xD\left(v\right)+{\displaystyle \frac{x}{v}}(E\left(v\right)-E\left(0\right)).$$

Similarly, the remaining equations can be derived from the corresponding succession rules. □

By solving (2) for $B\left(v\right),C\left(v\right),D\left(v\right),F\left(v\right)$, we obtain

$$\begin{array}{cccc}\hfill & B\left(v\right)={\displaystyle \frac{(1-2x+3x^2-x^3)x}{{(1-x)}^4(1-v)}},\hfill & \hfill & C\left(v\right)={\displaystyle \frac{x}{(1-x)(1-v)}},\hfill \\ \hfill & D\left(v\right)={\displaystyle \frac{x}{{(1-x)}^2(1-v)}},\hfill & \hfill & F\left(v\right)={\displaystyle \frac{x}{{(1-x)}^3(1-v)}}.\hfill \end{array}$$

By these expressions, (2) reduces to

$$\begin{array}{cc}\hfill (1-x-{\displaystyle \frac{x^2}{v}})A\left(v\right)& =-{\displaystyle \frac{x^2}{v}}A\left(0\right)+{\displaystyle \frac{({(1-x)}^2+x^3)x}{{(1-x)}^4(1-v)}}.\hfill \end{array}$$

This class of functional equations can be solved systematically via the kernel method; see Banderier et al. [13] for details. The equation is first rearranged so that the unknown generating function is multiplied by a polynomial kernel in x and v. The choice $v={\displaystyle \frac{x^2}{1-x}}$ is made so that this kernel vanishes, thereby eliminating the unknown term and allowing us to solve for $A\left(0\right)$ explicitly. In this case, we obtain

$$A\left(0\right)={\displaystyle \frac{({(1-x)}^2+x^3)x}{(1-x-x^2){(1-x)}^4}}.$$

Hence, by using the expression of $A\left(0\right)$, we have

$$\begin{array}{cccc}\hfill A\left(v\right)={\displaystyle \frac{({(1-x)}^2+x^3)x}{(1-x-x^2){(1-x)}^4(1-v)}},& \hfill & \hfill E\left(v\right)={\displaystyle \frac{(1-2x+2x^2+x^3-x^4)x}{(1-x-x^2){(1-x)}^3(1-v)}},& \hfill \end{array}$$

By using the expressions for $A\left(v\right)$, $B\left(v\right)$, and $F\left(v\right)$, we obtain explicit formulas for $A\left(0\right)$, $B\left(0\right)$, and $F\left(0\right)$. Hence, by solving (1) for $F_{0010;0}\left(x\right)$, we obtain the following result.

Theorem 3.

The generating function $F_{0010}\left(x\right)$ for the number of descent sequences of length n that avoid 0010 is given by

$${\displaystyle \frac{(x^3+x^2-2x+1)x}{(1-x-x^2){(1-x)}^3}}.$$

2.2. The Pattern 0011

As in the proof of Lemma 1, one can show that the generating tree $\mathcal{T}\left(0011\right)$ is determined by the following succession rules:

$$\begin{array}{ccc}0⇝00,01,\hfill & 00⇝00,001,\hfill & 001⇝0010,\hfill \\ 0010⇝0010,00102,\hfill & 00102⇝001020,\hfill & 001020⇝001020,j_3,\hfill \\ 01⇝010,01,\hfill & 010⇝010,0101,0102,\hfill & 0101⇝01010,00102,\hfill \\ 01010⇝01010,b_2,010103,\hfill & 010103⇝f_2,a_3,\hfill & 0102⇝01010,b_2,\hfill \\ a_m⇝e_m,i_{m+1}(m\ge 3),\hfill & b_m⇝f_m,j_{m+1}(m\ge 2),\hfill & c_m⇝f_{m-1},a_m(m\ge 3),\hfill \\ d_m⇝e_{m-1},b_{m+1}(m\ge 3),\hfill & e_m⇝e_m,b_{m+1},c_{m+1}(m\ge 2),\hfill & f_m⇝f_m,a_{m+1},d_{m+1},(m\ge 2),\hfill \\ g_m⇝g_m,j_{m+1}(m\ge 4),\hfill & h_m⇝h_m,i_{m+1}(m\ge 3),\hfill & i_m⇝g_m(m\ge 4),\hfill \\ j_m⇝h_m,(m\ge 3),\hfill & & \end{array}$$

where $a_m=01010320540\dots (m-2)(m-3)0m(m-1)$, $b_m=010210430\dots (m-2)$$(m-3)0m(m-1)$, $c_m=a_{m-2}0m$, $d_m=b_{m-2}0m$, $e_m=a_{m}0$, $f_m=b_{m}0$, $g_m=a_{m-1}m0$, $h_m=b_{m-1}m0$, $i_m=a_{m-1}m$, and $j_m=b_{m-1}m$.

Define $K\left(v\right)={\sum }_{m\ge 3}{F}_{\left\{0011\right\};k_m}\left(x\right)v^{m-2}$ for $K\in \{B,E,F\}$, $K\left(v\right)={\sum }_{m\ge 3}{F}_{\left\{0011\right\};k_m}\left(x\right)v^{m-3}$ for $K\in \{A,C,D,H,J\}$, and $K\left(v\right)={\sum }_{m\ge 3}{F}_{\left\{0011\right\};k_m}\left(x\right)v^{m-4}$ for $K\in \{G,I\}$. Similar to the arguments in the proof of Lemma 2, we can translate these succession rules into the following equations:

$$\begin{array}{c}\hfill \begin{array}{c}{F}_{\left\{0011\right\};0}\left(x\right)=x+x{F}_{\left\{0011\right\};00}\left(x\right)+x{F}_{\left\{0011\right\};01}\left(x\right),\hfill \\ F_{\left\{0011\right\};00}\left(x\right)=x+x{F}_{\left\{0011\right\};00}\left(x\right)+x{F}_{\left\{0011\right\};001}\left(x\right),\hfill \\ F_{\left\{0011\right\};001}\left(x\right)=x+x{F}_{\left\{0011\right\};0010}\left(x\right),\hfill \\ F_{\left\{0011\right\};0010}\left(x\right)=x+x{F}_{\left\{0011\right\};0010}\left(x\right)+x{F}_{\left\{0011\right\};00102}\left(x\right),\hfill \\ F_{\left\{0011\right\};00102}\left(x\right)=x+x{F}_{\left\{0011\right\};001020}\left(x\right),\hfill \\ F_{\left\{0011\right\};001020}\left(x\right)=x+x{F}_{\left\{0011\right\};001020}\left(x\right)+x{F}_{\left\{0011\right\};j_3}\left(x\right),\hfill \\ F_{\left\{0011\right\};01}\left(x\right)=x+x{F}_{\left\{0011\right\};010}\left(x\right)+x{F}_{\left\{0011\right\};01}\left(x\right),\hfill \\ F_{\left\{0011\right\};010}\left(x\right)=x+x{F}_{\left\{0011\right\};010}\left(x\right)+x{F}_{\left\{0011\right\};0101}\left(x\right)+x{F}_{\left\{0011\right\};0102}\left(x\right),\hfill \\ F_{\left\{0011\right\};0101}\left(x\right)=x+x{F}_{\left\{0011\right\};01010}\left(x\right)+x{F}_{\left\{0011\right\};00102}\left(x\right),\hfill \\ F_{\left\{0011\right\};01010}\left(x\right)=x+x{F}_{\left\{0011\right\};01010}\left(x\right)+x{F}_{\left\{0011\right\};b_2}\left(x\right)+x{F}_{\left\{0011\right\};010103}\left(x\right),\hfill \\ F_{\left\{0011\right\};010103}\left(x\right)=x+x{F}_{\left\{0011\right\};f_2}\left(x\right)+x{F}_{\left\{0011\right\};a_3}\left(x\right),\hfill \\ F_{\left\{0011\right\};0102}\left(x\right)=x+x{F}_{\left\{0011\right\};01010}\left(x\right)+x{F}_{\left\{0011\right\};b_2}\left(x\right),\hfill \end{array}\end{array}$$

(3)

and

$$\begin{array}{c}\hfill \begin{array}{c}{F}_{\left\{0011\right\};a_m}\left(x\right)=x+x{F}_{\left\{0011\right\};e_m}\left(x\right)+x{F}_{\left\{0011\right\};i_{m+1}}\left(x\right),(m\ge 3),\hfill \\ F_{\left\{0011\right\};b_m}\left(x\right)=x+xF\left(v\right)+{\displaystyle \frac{x}{v}}(J\left(v\right)-J\left(0\right)),\hfill \\ F_{\left\{0011\right\};c_m}\left(x\right)=x+x{F}_{\left\{0011\right\};f_{m-1}}\left(x\right)+x{F}_{\left\{0011\right\};a_m}\left(x\right),(m\ge 3),\hfill \\ F_{\left\{0011\right\};d_m}\left(x\right)=x+x{F}_{\left\{0011\right\};e_{m-1}}\left(x\right)+x{F}_{\left\{0011\right\};b_{m+1}}\left(x\right),(m\ge 3),\hfill \\ F_{\left\{0011\right\};e_m}\left(x\right)=x+x{F}_{\left\{0011\right\};e_m}\left(x\right)+x{F}_{\left\{0011\right\};b_{m+1}}\left(x\right)+x{F}_{\left\{0011\right\};c_{m+1}}\left(x\right),(m\ge 2),\hfill \\ F_{\left\{0011\right\};f_m}\left(x\right)=x+x{F}_{\left\{0011\right\};f_m}\left(x\right)+x{F}_{\left\{0011\right\};a_{m+1}}\left(x\right)+x{F}_{\left\{0011\right\};d_{m+1}}\left(x\right),(m\ge 2),\hfill \\ F_{\left\{0011\right\};g_m}\left(x\right)=x+x{F}_{\left\{0011\right\};g_m}\left(x\right)+x{F}_{\left\{0011\right\};j_{m+1}}\left(x\right),(m\ge 4),\hfill \\ F_{\left\{0011\right\};h_m}\left(x\right)=x+x{F}_{\left\{0011\right\};h_m}\left(x\right)+x{F}_{\left\{0011\right\};i_{m+1}}\left(x\right),(m\ge 3),\hfill \\ F_{\left\{0011\right\};i_m}\left(x\right)=x+x{F}_{\left\{0011\right\};g_m}\left(x\right),(m\ge 4),\hfill \\ F_{\left\{0011\right\};j_m}\left(x\right)=x+x{F}_{\left\{0011\right\};h_m}\left(x\right),(m\ge 3).\hfill \end{array}\end{array}$$

(4)

By multiplying each equation in (4) by $v^{m-s}$ (where $s\in 2,3,4$ corresponds to the definitions of the generating functions $A\left(v\right),B\left(v\right),\dots ,J\left(v\right)$), we obtain

$$\begin{array}{c}\hfill \begin{array}{c}A\left(v\right)={\displaystyle \frac{x}{1-v}}+xE\left(v\right)+{\displaystyle \frac{x}{v}}(I\left(v\right)-I\left(0\right)),\hfill \\ B\left(v\right)={\displaystyle \frac{x}{1-v}}+xF\left(v\right)+{\displaystyle \frac{x}{v}}(J\left(v\right)-J\left(0\right)),\hfill \\ C\left(v\right)={\displaystyle \frac{x}{1-v}}+xF\left(v\right)+xA\left(v\right),\hfill \\ D\left(v\right)={\displaystyle \frac{x}{1-v}}+xE\left(v\right)+{\displaystyle \frac{x}{v}}(B\left(v\right)-B\left(0\right)),\hfill \\ E\left(v\right)={\displaystyle \frac{x}{1-v}}+xE\left(v\right)+{\displaystyle \frac{x}{v}}(B\left(v\right)-B\left(0\right))+{\displaystyle \frac{x}{v}}(C\left(v\right)-C\left(0\right)),\hfill \\ F\left(v\right)={\displaystyle \frac{x}{1-v}}+xF\left(v\right)+{\displaystyle \frac{x}{v}}(A\left(v\right)-A\left(0\right))+{\displaystyle \frac{x}{v}}(D\left(v\right)-D\left(0\right)),\hfill \\ G\left(v\right)={\displaystyle \frac{x}{1-v}}+xG\left(v\right)+{\displaystyle \frac{x}{v}}(J\left(v\right)-J\left(0\right)),\hfill \\ H\left(v\right)={\displaystyle \frac{x}{1-v}}+xH\left(v\right)+{\displaystyle \frac{x}{v}}(I\left(v\right)-I\left(0\right)),\hfill \\ I\left(v\right)={\displaystyle \frac{x}{1-v}}+xG\left(v\right),\hfill \\ J\left(v\right)={\displaystyle \frac{x}{1-v}}+xH\left(v\right).\hfill \end{array}\end{array}$$

(5)

By solving (5) throughout the algorithm in [14], we obtain

$$\begin{array}{cc}\hfill A\left(v\right)& =B\left(v\right)={\displaystyle \frac{(1+x)x}{(1-x-2x^2-x^3)(1-v)}},\hfill \\ \hfill C\left(v\right)& =D\left(v\right)={\displaystyle \frac{(1-x^2-x^3)x}{(1-x-x^2)(1-x-2x^2-x^3)(1-v)}},\hfill \\ \hfill E\left(v\right)& =F\left(v\right)={\displaystyle \frac{(1+x)(1-x^2-x^3)x}{(1-x-x^2)(1-x-2x^2-x^3)(1-v)}},\hfill \\ \hfill G\left(v\right)& =H\left(v\right)={\displaystyle \frac{(1+x)x}{(1-x-x^2)(1-v)}},\hfill \\ \hfill I\left(v\right)& =J\left(v\right)={\displaystyle \frac{x}{(1-x-x^2)(1-v)}}.\hfill \end{array}$$

Thus, by (3), we have

$$\begin{array}{cc}\hfill F_{\left\{0011\right\};0}\left(x\right)& ={\displaystyle \frac{(1-x^2-x^3)x}{(1-x)(1-x-2x^2-x^3)}},\hfill \\ \hfill F_{\left\{0011\right\};00}\left(x\right)& =F_{\left\{0011\right\};001020}\left(x\right)=F_{\left\{0011\right\};0010}\left(x\right)={\displaystyle \frac{(1+x)x}{1-x-x^2}},\hfill \\ \hfill F_{\left\{0011\right\};01}\left(x\right)& ={\displaystyle \frac{(1-x-x^2+x^3+x^4)x}{(1-x)(1-x-x^2)(1-x-2x^2-x^3)}},\hfill \\ \hfill F_{\left\{0011\right\};001}\left(x\right)& =F_{\left\{0011\right\};00102}\left(x\right)={\displaystyle \frac{x}{1-x-x^2}},\hfill \\ \hfill F_{\left\{0011\right\};010}\left(x\right)& =F_{\left\{0011\right\};01010}\left(x\right)=-{\displaystyle \frac{(1+x)(1-x^2-x^3)x}{(1-x-x^2)(1-x-2x^2-x^3)}},\hfill \\ \hfill F_{\left\{0011\right\};0101}\left(x\right)& ={\displaystyle \frac{(1+x)x}{1-x-2x^2-x^3}},\hfill \\ \hfill F_{\left\{0011\right\};0102}\left(x\right)& =F_{\left\{0011\right\};010103}\left(x\right)={\displaystyle \frac{(1-x^2-x^3)x}{(1-x-x^2)(1-x-2x^2-x^3)}}.\hfill \end{array}$$

Therefore, we can present the following result.

Theorem 4.

The generating function $F_{0011}\left(x\right)$ for the number of descent sequences of length n that avoid 0011 is given by

$${\displaystyle \frac{(1-x^2-x^3)x}{(1-x)(1-x-2x^2-x^3)}}.$$

2.3. The Pattern 0110

As in the proof of Lemma 1, one can show that the generating tree $\mathcal{T}\left(\right\{0110\left\}\right)$ is determined by the following succession rules

$$\begin{array}{cc}0⇝0,01,\hfill & 01⇝d_{1,0},d_{1,1},\hfill \\ a_m⇝a_m,b_{m+1},(m\ge 1),\hfill & b_m⇝d_{m,m-1},d_{m,m},(m\ge 2),\hfill \\ c_{m,j}⇝d_{m,j},d_{m,j+1},\dots ,d_{m,m},(0\le j\le m-2),\hfill & d_{m,m}⇝d_{m,m},(m\ge 1),\hfill \\ d_{m,j}⇝d_{m,j},d_{m,j+1},\dots ,d_{m,m-1},a_m,c_{m+1,j},(0\le j\le m-1),\hfill & \end{array}$$

where $a_m=0102\dots 0m0m$, $b_m=a_{m-1}m$, $c_{m,j}=0102\dots 0(m-1)jm$, and $d_{m,j}=0102\dots 0mj$. Note that by translating these succession rules into equations, we obtain

$$\begin{array}{cc}\hfill F_{P;0}\left(x\right)& =x+x{F}_{P;0}\left(x\right)+x{F}_{P;01}\left(x\right),\hfill \\ \hfill F_{P;01}\left(x\right)& =x+x{F}_{P;d_{1,0}}\left(x\right)+x{F}_{P;d_{1,1}}\left(x\right),\hfill \\ \hfill F_{P;a_m}\left(x\right)& =x+x{F}_{P;a_m}\left(x\right)+x{F}_{P;b_{m+1}}\left(x\right),(m\ge 1),\hfill \\ \hfill F_{P;b_m}\left(x\right)& =x+x{F}_{P;d_{m,m-1}}\left(x\right)+x{F}_{P;d_{m,m}}\left(x\right),(m\ge 2),\hfill \\ \hfill F_{P;c_{m,j}}\left(x\right)& =x+x{\displaystyle \sum _{i=j}^m}{F}_{P;d_{m,i}}\left(x\right),(0\le j\le m-2),\hfill \\ \hfill F_{P;d_{m,m}}\left(x\right)& =x+x{F}_{P;d_{m,m}}\left(x\right),(m\ge 1),\hfill \\ \hfill F_{P;d_{m,j}}\left(x\right)& =x+x{\displaystyle \sum _{i=j}^{m-1}}{F}_{P;d_{m,j}}\left(x\right)+x{F}_{P;a_m}\left(x\right)+x{F}_{P;c_{m+1,j}}\left(x\right),(0\le j\le m-1).\hfill \end{array}$$

To solve this system of recurrence relations, we define

$$\begin{array}{cc}\hfill A\left(v\right)& =\sum _{m\ge 1}{F}_{\left\{0110\right\};a_m}\left(x\right)v^{m-1},\hfill \\ \hfill B\left(v\right)& =\sum _{m\ge 2}{F}_{\left\{0110\right\};b_m}\left(x\right)v^{m-2},\hfill \\ \hfill C(u,v)& =\sum _{m\ge 2}{\displaystyle \sum _{j=0}^{m-2}}{F}_{\left\{0110\right\};c_{m,j}}\left(x\right)v^{m-2}{u}^{m-2-j},\hfill \\ \hfill D(u,v)& =\sum _{m\ge 2}{\displaystyle \sum _{j=0}^{m-2}}{F}_{\left\{0110\right\};d_{m,j}}\left(x\right)v^{m-1}{u}^{m-1-j}.\hfill \end{array}$$

Then, the above succession rules can be written as

$$\begin{array}{cc}\hfill A\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xA\left(v\right)+xB\left(v\right),\hfill \\ \hfill B\left(v\right)& ={\displaystyle \frac{x}{(1-x)(1-v)}}+xD(0,v),\hfill \\ \hfill C(u,v)& ={\displaystyle \frac{x}{(1-v)(1-uv)}}+{\displaystyle \frac{x}{uv(1-u)}}(D(u,v)-uD(1,uv)-D(0,v)+uD(0,uv))\hfill \\ \hfill & +{\displaystyle \frac{x}{v(1-u)}}(D(0,v)-D(0,uv))+{\displaystyle \frac{x^2}{(1-x)(1-u)}}(1/(1-v)-u/(1-uv)),\hfill \\ \hfill D(u,v)& ={\displaystyle \frac{x}{(1-v)(1-uv)}}+{\displaystyle \frac{x}{1-u}}(D(u,v)-uD(1,uv))\hfill \\ \hfill & +{\displaystyle \frac{x}{1-u}}(A\left(v\right)-uA\left(uv\right))+xC(u,v).\hfill \end{array}$$

To solve this system, we guessed that

$$D(0,v)={\displaystyle \frac{(1-x)M\left(x\right)-1}{x(1-x)(1-v)}},$$

based on the initial coefficients of $x^n$ in the generating function $D(0,v)$, where

$$M\left(x\right)={\displaystyle \frac{1-x-\sqrt{1-2x-3x^2}}{2x^2}}$$

is the generating function for the Motzkin numbers (see Sequence A001006 in [15]). We now proceed to solve the system under this assumption. From the system, we have that

$$\begin{array}{cc}\hfill A\left(v\right)& ={\displaystyle \frac{1-x-\sqrt{1-2x-3x^2}}{2x(1-x)(1-v)}},\hfill \\ \hfill B\left(v\right)& ={\displaystyle \frac{1-x-2x^2-\sqrt{1-2x-3x^2}}{2x^2(1-v)}}.\hfill \end{array}$$

Thus, by finding $C(u,v)$ from the equation of $C(u,v)$, and substituting it into the equation of $D(u,v)$, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{uv+vx+x^2-v}{v(1-u)}}D(u,v/u)& ={\displaystyle \frac{x(v+x)}{uv(1-u)}}D(1,v)\hfill \\ \hfill & +{\displaystyle \frac{(x+v-1)\sqrt{1-2x-3x^2}+vx-v-x^2-2x+1}{2x(1-x)v(1-v/u)(1-v)}}.\hfill \end{array}$$

(6)

This class of functional equations can be solved systematically via the kernel method; see Banderier et al. [13] for details. The equation is first rearranged so that the unknown generating function is multiplied by a polynomial kernel in x, v, and u. The choice $u=(vx+x^2-v)/v$ is made so that this kernel vanishes, thereby eliminating the unknown term and allowing us to solve for $D(1,v)$ explicitly. In this case, we obtain

$$D(1,v)={\displaystyle \frac{(x+v-1)\sqrt{1-2x-3x^2}+vx-v-x^2-2x+1}{2x(1-x)(v^2+vx+x^2-v)(1-v)}}.$$

Using this expression with (6), we obtain

$$\begin{array}{cc}\hfill D(u,v)& ={\displaystyle \frac{(x+uv-1)\sqrt{1-2x-3x^2}+uvx-uv-x^2-2x+1}{2x(1-x)(u^2{v}^2+uvx+x^2-uv)(1-v)}},\hfill \\ \hfill C(u,v)& ={\displaystyle \frac{((x-1)u^2{v}^2+(x-2)(x-1)uv+x^3-x^2+2x-1)\sqrt{1-2x-3x^2}}{2x^2(1-x)(u^2{v}^2+uvx+x^2-uv)(1-v)(1-uv)}}\hfill \\ \hfill & +{\displaystyle \frac{(1-x^2)(1-2x)u^2{v}^2+(2x^4-3x^3+5x-2)uv+2x^5-x^4+x^2-3x+1}{2x^2(1-x)(u^2{v}^2+uvx+x^2-uv)(1-v)(1-uv)}}.\hfill \end{array}$$

Note that the expressions $A\left(v\right)$, $B\left(v\right)$, $C(u,v)$, and $D(u,v)$ (without any assumptions) satisfy the system, and therefore constitute its solution. Hence, from the first two succession rules, we observe

$$F_P\left(x\right)={\displaystyle \frac{x}{1-x}}+{\displaystyle \frac{x}{1-x}}(x+xD(0,0)+{\displaystyle \frac{x^2}{1-x}}),$$

which leads to the following result.

Theorem 5.

The generating function $F_{0110}\left(x\right)$ for the number of descent sequences of length n that avoid 0110 is given by

$${\displaystyle \frac{1-x-\sqrt{1-2x-3x^2}}{2x(1-x)}}.$$

2.4. The Pattern 0112

As in the proof of Lemma 1, one can show that the generating tree $\mathcal{T}\left(\right\{0112\left\}\right)$ is determined by the following succession rules

$$\begin{array}{cccc}\hfill & 0⇝0,a_1,\hfill & \hfill & a_m⇝c_m,b_1,b_2,\dots ,b_m,(m\ge 1),\hfill \\ \hfill & b_m⇝b_m,b_1,b_2,\dots ,b_m,(m\ge 1),\hfill & \hfill & c_m⇝c_m,b_1,b_2,\dots ,b_m,a_{m+1},(m\ge 1),\hfill \end{array}$$

where $a_m=0102\dots 0m$, $b_m=a_{m}m$, and $c_m=a_{m}0$. As before, we can translate these rules to equations:

$$\begin{array}{cc}\hfill F_{\left\{0112\right\};0}\left(x\right)& =x+x{F}_{\left\{0112\right\};0}\left(x\right)+xA\left(0\right),\hfill \\ \hfill A\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xC\left(v\right)+{\displaystyle \frac{x}{1-v}}B\left(v\right),\hfill \\ \hfill B\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xB\left(v\right)+{\displaystyle \frac{x}{1-v}}B\left(v\right),\hfill \\ \hfill C\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xC\left(v\right)+{\displaystyle \frac{x}{1-v}}B\left(v\right)+{\displaystyle \frac{x}{v}}(A\left(v\right)-A\left(0\right)).\hfill \end{array}$$

where $K\left(v\right)={\sum }_{m\ge 1}{F}_{P;k_m}\left(x\right)v^{m-1}$ for $K\in \{A,B,C\}$. Thus,

$$\begin{array}{cc}\hfill F_{\left\{0112\right\};0}\left(x\right)& ={\displaystyle \frac{x}{1-x}}(1+A\left(0\right)),\hfill \\ \hfill (1-{\displaystyle \frac{x^2}{v(1-x)}})A\left(v\right)& =-{\displaystyle \frac{x^2}{v(1-x)}}A\left(0\right)+{\displaystyle \frac{x}{1-2x-v+vx}}.\hfill \end{array}$$

This class of functional equations can be solved systematically via the kernel method; see Banderier et al. [13] for details. The equation is first rearranged so that the unknown generating function is multiplied by a polynomial kernel in x and v. The choice $v=x^2/(1-x)$ is made so that this kernel vanishes, thereby eliminating the unknown term and allowing us to solve for $A\left(0\right)$ explicitly. In this case, we obtain $A\left(0\right)={\displaystyle \frac{x}{1-2x-x^2}}$. Hence, by the first equation, we obtain the following result.

Theorem 6.

The generating function $F_{0112}\left(x\right)$ for the number of descent sequences of length n that avoid 0112 is given by

$${\displaystyle \frac{(1-x-x^2)x}{(1-x)(1-2x-x^2)}}.$$

2.5. The Pattern 0123

We start with the following result. For the proof, see Appendix A.

Lemma 3.

The generating tree $\mathcal{T}\left(\right\{0123\left\}\right)$ is determined by the following succession rules

$$\begin{array}{cccc}\hfill 0& ⇝& 0,a_1,\hfill & \\ \hfill a_m& ⇝& b_m,a_m,c_{m,2},c_{m,3},\dots ,c_{m,m},\hfill & m\ge 1,\hfill \\ \hfill b_m& ⇝& b_m,a_{m+1},f_{m,2},f_{m,3},\dots ,f_{m,m+1},\hfill & m\ge 1,\hfill \\ \hfill c_{m,j}& ⇝& d_{m,j},e_{m,j},c_{m+1,2},c_{m+1,3},\dots ,c_{m+1,j-1},c_{m,j},\hfill & 2\le j\le m,\hfill \\ \hfill d_{m,j}& ⇝& d_{m,j},e_{m,j},c_{m,2},c_{m,3},\dots ,c_{m,j-1},c_{m+1,j},\hfill & 2\le j\le m,\hfill \\ \hfill e_{m,j}& ⇝& d_{m+1,j},e_{m,j},c_{m,2},c_{m,3},\dots ,c_{m,j-1},c_{m+1,j},\hfill & 2\le j\le m,\hfill \\ \hfill f_{m,j}& ⇝& g_{m,j},h_{m,j},f_{m+1,2},f_{m+1,3},\dots ,f_{m+1,j-1},f_{m,j},\hfill & 2\le j\le m+1,\hfill \\ \hfill g_{m,j}& ⇝& g_{m,j},h_{m,j},f_{m,2},f_{m,3},\dots ,f_{m,j-1},f_{m+1,j},\hfill & 2\le j\le m+1,\hfill \\ \hfill h_{m,j}& ⇝& g_{m+1,j},h_{m,j},f_{m,2},f_{m,3},\dots ,f_{m,j-1},f_{m+1,j},\hfill & 2\le j\le m+1.\hfill \end{array}$$

where $a_m={\left(01\right)}^m$, $b_m=a_{m}0$, $c_{m,j}=a_{m}j$, $d_{m,j}=c_{m,j}0$, $e_{m,j}=c_{m,j}1$, $f_{m,j}=a_{m}0j$, $g_{m,j}=f_{m,j}0$, and $h_{m,j}=f_{m,j}1$.

Now, we define the following eight generating functions:

$$\begin{array}{cc}\hfill K\left(v\right)& =\sum _{m\ge 1}F(P;k_m)\left(x\right)v^{m-1},K\in \{A,B\},\hfill \\ \hfill K(v,u)& =\sum _{m\ge 2}{\displaystyle \sum _{j=2}^m}F(P;k_{m,j})\left(x\right)v^{m-2}{u}^{m-j},K\in \{C,D,E\},\hfill \\ \hfill K(v,u)& =\sum _{m\ge 1}{\displaystyle \sum _{j=1}^{m+1}}F(P;k_{m,j})\left(x\right)v^{m-1}{u}^{m+1-j},K\in \{F,G,H\}.\hfill \end{array}$$

Then, the above succession rules can be written as

$$\begin{array}{cc}\hfill F_P\left(x\right)& =x+x{F}_P\left(x\right)+xA\left(0\right),\hfill \\ \hfill A\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xB\left(v\right)+xA\left(v\right)+vxC(v,1),\hfill \\ \hfill B\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xB\left(v\right)+{\displaystyle \frac{x}{v}}(A\left(v\right)-A\left(0\right))+xF(v,1),\hfill \\ \hfill C(v,u)& ={\displaystyle \frac{x}{(1-v)(1-vu)}}+xD(v,u)+xE(v,u)\hfill \\ \hfill & +{\displaystyle \frac{x}{uv(1-u)}}(uC(v,1)-C(v,u)+(1-u)C(v,0))+xC(v,u),\hfill \\ \hfill D(v,u)& ={\displaystyle \frac{x}{(1-v)(1-vu)}}+xD(v,u)+xE(v,u)\hfill \\ \hfill & +{\displaystyle \frac{x}{1-u}}(C(v,1)-C(v,u))+{\displaystyle \frac{x}{uv}}(C(v,u)-C(v,0)),\hfill \\ \hfill E(v,u)& ={\displaystyle \frac{x}{(1-v)(1-vu)}}+{\displaystyle \frac{x}{uv}}(D(v,u)-D(v,0))+xE(v,u)\hfill \\ \hfill & +{\displaystyle \frac{x}{1-u}}(C(v,1)-C(v,u))+{\displaystyle \frac{x}{uv}}(C(v,u)-C(v,0)),\hfill \\ \hfill F(v,u)& ={\displaystyle \frac{x}{(1-v)(1-vu)}}+xG(v,u)+xH(v,u)\hfill \\ \hfill & +{\displaystyle \frac{x}{uv(1-u)}}(uF(v,1)-F(v,u)+(1-u)F(v,0))+xF(v,u),\hfill \\ \hfill G(v,u)& ={\displaystyle \frac{x}{(1-v)(1-vu)}}+xG(v,u)+xH(v,u)\hfill \\ \hfill & +{\displaystyle \frac{x}{1-u}}(F(v,1)-F(v,u))+{\displaystyle \frac{x}{uv}}(F(v,u)-F(v,0)),\hfill \\ \hfill H(v,u)& ={\displaystyle \frac{x}{(1-v)(1-vu)}}+{\displaystyle \frac{x}{uv}}(G(v,u)-G(v,0))+xH(v,u)\hfill \\ \hfill & +{\displaystyle \frac{x}{1-u}}(F(v,1)-F(v,u))+{\displaystyle \frac{x}{uv}}(F(v,u)-F(v,0)).\hfill \end{array}$$

Theorem 7.

We have

$$\begin{array}{cc}\hfill A\left(v\right)& ={\displaystyle \frac{x(x^2-3x+1)(3x-1)}{(x^3-6x^2+5x-1)(2vx-v-3x+1)}},\hfill \\ \hfill B\left(v\right)& =-{\displaystyle \frac{x(3x-1)(2x-1)}{(x^3-6x^2+5x-1)(2vx-v-3x+1)}},\hfill \\ \hfill C(v,u)& =-{\displaystyle \frac{vx}{(uv-1)(2vx-v-3x+1)}},\hfill \end{array}$$

and $C(v,u)=D(v,u)=E(v,u)=F(v,u)=G(v,u)=H(v,u)$.

Proof. 

Since the expressions of the generating functions $A\left(v\right)$, $B\left(v\right)$, $C(v,u)$, $D(v,u)$, $E(v,u)$, $F(v,u)$, $G(v,u)$, and $H(v,u)$ satisfy the system of equations, the proof is complete. □

Hence, the above theorem gives $A\left(0\right)={\displaystyle \frac{x(x^2-3x+1)}{1-5x+6x^2-x^3}}$. Thus, by the fact that $F_P\left(x\right)=x+x{F}_P\left(x\right)+xA\left(0\right)$, we obtain the following result.

Theorem 8.

The generating function $F_{0123}\left(x\right)$ for the number of descent sequences of length n that avoid 0123 is given by

$${\displaystyle \frac{x(1-3x)}{1-5x+6x^2-x^3}}.$$

2.6. The Pattern 0132

As in the proof of Lemma 1, one can show that the generating tree $\mathcal{T}\left(\right\{0132\left\}\right)$ is determined by the following succession rules:

$$\begin{array}{cccc}\hfill & 0⇝0,a_1,\hfill & \hfill & a_m⇝b_m,a_m,c_m,c_{m-1},\dots ,c_2,\hfill \\ \hfill & b_m⇝b_m,a_{m+1},c_{m+1},c_m,\dots ,c_2,\hfill & \hfill & c_m⇝b_m,a_{m+1},c_m,c_{m-1},\dots ,c_2,\hfill \end{array}$$

where $a_m={\left(01\right)}^m$, $b_m=a_{m}0$, and $c_m=a_{m}2$.

Define $K\left(v\right)={\sum }_{m\ge 1}{F}_{P;k_m}\left(x\right)v^{m-1}$ with $K\in \{A,B\}$ and $C\left(v\right)={\sum }_{m\ge 2}{F}_{P;c_m}\left(x\right)v^{m-2}$. Then, the above succession rules lead to the following equations

$$\begin{array}{cc}\hfill F_P\left(x\right)& =x+x{F}_P\left(x\right)+xA\left(0\right),\hfill \\ \hfill A\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xB\left(v\right)+xA\left(v\right)+{\displaystyle \frac{vx}{1-v}}C\left(v\right),\hfill \\ \hfill B\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xB\left(v\right)+{\displaystyle \frac{x}{v}}(A\left(v\right)-A\left(0\right))+{\displaystyle \frac{x}{1-v}}C\left(v\right),\hfill \\ \hfill C\left(v\right)& ={\displaystyle \frac{x}{1-v}}+{\displaystyle \frac{x}{v}}(B\left(v\right)-B\left(0\right))+{\displaystyle \frac{x}{v^2}}(A\left(v\right)-A\left(0\right)-v{\displaystyle \frac{d}{dv}}A\left(v\right){\mid }_{v=0})+{\displaystyle \frac{x}{1-v}}C\left(v\right).\hfill \end{array}$$

Solving the last two equations for $B\left(v\right)$ and $C\left(v\right)$, and then substituting their expressions into the equation of $A\left(v\right)$, we obtain

$$\begin{array}{cc}\hfill & -\left({(1-x)}^2\right)v^3+(x^3-4x^2+3x-1)v^2+x^2(3-2x)v+x^3)A\left(v\right)\hfill \\ \hfill & =x^2(v^2+vx-2v-x)A\left(0\right)+v{x}^2(vx-v-x)\left(B\left(0\right)+{\displaystyle \frac{d}{dv}}A\left(v\right){\mid }_{v=0}\right)\hfill \\ \hfill & +vx(v{x}^2-vx-x^2+v).\hfill \end{array}$$

Let $v_1,v_2,v_3$ be the roots of the kernel equation ${(1-x)}^2)v^3+(x^3-4x^2+3x-1)v^2+x^2$$(3-2x)v+x^3=0$. Note that

$$\begin{array}{cc}\hfill v_1& =1-x-2x^2-6x^3-19x^4-66x^5+\dots ,\hfill \\ \hfill v_2& =-x\sqrt{x}+{\displaystyle \frac{3}{2}}{x}^2-{\displaystyle \frac{21}{8}}{x}^2\sqrt{x}+4x^3-{\displaystyle \frac{839}{128}}{x}^3\sqrt{x}+11x^4-{\displaystyle \frac{19757}{1024}}{x}^4\sqrt{x}+35x^5+\dots ,\hfill \\ \hfill v_3& =x\sqrt{x}+{\displaystyle \frac{3}{2}}{x}^2+{\displaystyle \frac{21}{8}}{x}^2\sqrt{x}+4x^3+{\displaystyle \frac{839}{128}}{x}^3\sqrt{x}+11x^4-{\displaystyle \frac{19757}{1024}}{x}^4\sqrt{x}+35x^5+\dots .\hfill \end{array}$$

Thus, by substituting $v=v_1$ and $v=v_2$ within the equation, which are chosen so as to annihilate the kernel, and solving for $A\left(0\right)$ and $B\left(0\right)+{\displaystyle \frac{d}{dv}}A\left(v\right){\mid }_{v=0}$, we obtain

$$A\left(0\right)=-{\displaystyle \frac{v_2{v}_3}{(x^2-2x+2)v_2{v}_3+x(1-x)(v_2+v_3)+x^2}}.$$

By the definitions we have that $v_1{v}_2{v}_3=-x^2/{(1-x)}^2$ and $v_1+v_2+v_3=1-x+x^2/{(1-x)}^2$. Thus,

$$A\left(0\right)={\displaystyle \frac{-x^2}{x^2(x^2-2x+2)+(1-x)(x^3-3x^2+2x-1)v_1+{(1-x)}^3{v}_1^2}}.$$

Hence, by $F_P\left(x\right)=x+x{F}_P\left(x\right)+xA\left(0\right)$, we obtain,

$$F_P\left(x\right)={\displaystyle \frac{x}{1-x}}\left(1+{\displaystyle \frac{-x^2}{x^2(x^2-2x+2)+(1-x)(x^3-3x^2+2x-1)v_1+{(1-x)}^3{v}_1^2}}\right).$$

By expressing $v_1$ in terms of $F_P\left(x\right)$, then substituting this expression into the kernel equation ${(1-x)}^2)v^3+(x^3-4x^2+3x-1)v^2+x^2(3-2x)v+x^3=0$, we have the following result.

Theorem 9.

The generating function $f=F_{0132}\left(x\right)$ for the number of descent sequences of length n that avoid 0132 satisfies

$$-x^3+x^2(x^2-3x+3)f+x(2x^3-5x^2+7x-3)f^2+{(1-x)}^4{f}^3=0.$$

Moreover,

$$F_{\left\{0132\right\}}\left(x\right)={\displaystyle \frac{x\left(3-7x+5x^2-2x^3-2\sqrt{xA}·\cos \left(\frac{1}{3}\arccos \left(\frac{\sqrt{x}B}{2A\sqrt{A}}\right)+\frac{\pi }{3}\right)\right)}{3{(1-x)}^4}},$$

where $A=x^5+x^4-10x^3+20x^2-14x+3$ and $B=2x^7-24x^6+72x^5-107x^4+90x^3-60x^2+34x-9$.

2.7. The Pattern 0001

We start with the following result. For the proof, see Appendix B.

Lemma 4.

The generating tree $\mathcal{T}\left(\right\{0001\left\}\right)$ is determined by the following succession rules

$$\begin{array}{ccc}\hfill 0& ⇝& 00,01,\hfill \\ \hfill 00& ⇝& 000,00,\hfill \\ \hfill 000& ⇝& 000,\hfill \\ \hfill 01& ⇝& a_1,d_1,\hfill \\ \hfill a_{2m+1}& ⇝& f_{m+1,0},g_{m+2,1},\dots ,f_{2m,2m-2},g_{2m+1,2m-1},\hfill \\ & & b_{2m+1,2m},f_{2m+1,2m+1},b_{2m+1,2m+2},(m\ge 0),\hfill \\ \hfill a_{2m}& ⇝& g_{m+1,0},f_{m+1,1},\dots ,g_{2m-1,2m-4},f_{2m-1,2m-3},\hfill \\ & & g_{2m,2m-2},b_{2m,2m-1},f_{2m,2m},b_{2m,2m+1},(m\ge 1),\hfill \\ \hfill b_{2m+1,2m}& ⇝& f_{m+1,0},g_{m+2,1},\dots ,f_{2m,2m-2},g_{2m+1,2m-1},\hfill \\ & & b_{2m+1,2m},(m\ge 0),\hfill \\ \hfill b_{2m+1,2m+2}& ⇝& f_{m+1,0},g_{m+2,1},\dots ,f_{2m,2m-2},g_{2m+1,2m-1},\hfill \\ & & f_{2m+1,2m},a_{2m+2},d_{2m+2},(m\ge 0),\hfill \\ \hfill b_{2m,2m-1}& ⇝& g_{m+1,0},f_{m+1,1},\dots ,g_{2m-1,2m-4},f_{2m-1,2m-3},\hfill \\ & & g_{2m,2m-2},b_{2m,2m-1},(m\ge 1),\hfill \\ \hfill b_{2m,2m+1}& ⇝& g_{m+1,0},f_{m+1,1},\dots ,g_{2m,2m-2},f_{2m,2m-1},\hfill \\ & & a_{2m+1},d_{2m+1},,(m\ge 1),\hfill \\ \hfill c_{2m+1}& ⇝& g_{m+1,0},f_{m+1,1},\dots ,g_{2m,2m-2},f_{2m,2m-1},\hfill \\ & & e_{2m+1},c_{2m+1},(m\ge 0),\hfill \\ \hfill c_{2m}& ⇝& f_{m,0},g_{m+1,1},\dots ,f_{2m-2,2m-4},g_{2m-1,2m-3},\hfill \\ & & f_{2m-1,2m-2},e_{2m},c_{2m},(m\ge 1),\hfill \\ \hfill d_{2m+1}& ⇝& g_{m+1,0},f_{m+1,1},\dots ,g_{2m,2m-2},f_{2m,2m-1},\hfill \\ & & f_{2m+1,2m+1},c_{2m+1},(m\ge 0),\hfill \\ \hfill d_{2m}& ⇝& f_{m,0},g_{m+1,1},\dots ,f_{2m-2,2m-4},g_{2m-1,2m-3},\hfill \\ & & f_{2m,2m},c_{2m},(m\ge 1),\hfill \\ \hfill e_{2m+1}& ⇝& f_{m+1,0},g_{m+2,1},\dots ,f_{2m,2m-2},g_{2m+1,2m-1},\hfill \\ & & f_{2m+1,2m},e_{2m+1},(m\ge 0),\hfill \\ \hfill e_{2m}& ⇝& g_{m+1,0},f_{m+1,1},\dots ,g_{2m,2m-2},f_{2m,2m-1},\hfill \\ & & e_{2m},(m\ge 1),\hfill \\ \hfill f_{2m+1,2m+1}& ⇝& f_{m+1,0},g_{m+2,1},\dots ,f_{2m,2m-2},g_{2m+1,2m-1},\hfill \\ & & f_{2m+1,2m},e_{2m+1},f_{2m+1,2m+1},(m\ge 0),\hfill \\ \hfill f_{2m,2m}& ⇝& g_{m+1,0},f_{m+1,1},\dots ,g_{2m,2m-2},f_{2m,2m-1},\hfill \\ & & e_{2m},f_{2m,2m},(m\ge 1),\hfill \\ \hfill f_{2m+1,2j+1}& ⇝& f_{2m+1-j,0},g_{2m+1-j,1},\dots ,f_{2m,2j-2},g_{2m,2j-1},\hfill \\ & & f_{2m+1,2j},f_{2m+1,2j+1},(0\le j\le m-1),\hfill \\ \hfill f_{2m+1,2j}& ⇝& g_{2m+2-j,0},f_{2m+2-j,1},\dots ,g_{2m+1,2j-2},f_{2m+1,2j-1},\hfill \\ & & f_{2m+1,2j},(0\le j\le m),\hfill \\ \hfill f_{2m,2j+1}& ⇝& f_{2m-j,0},g_{2m+1-j,1},\dots ,f_{2m-1,2j-1},g_{2m,2j-1},\hfill \\ & & f_{2m,2j},f_{2m,2j+1},(0\le j\le m-1),\hfill \\ \hfill f_{2m,2j}& ⇝& g_{2m+1-j,0},f_{2m+1-j,1},\dots ,g_{2m,2j-2},f_{2m,2j-1},\hfill \\ & & f_{2m,2j},(0\le j\le m-1),\hfill \\ \hfill g_{2m+1,2j+1}& ⇝& g_{2m+1-j,0},f_{2m+1-j,1},\dots ,g_{2m,2j-2},f_{2m,2j-1},\hfill \\ & & g_{2m+1,2j},g_{2m+1,2j+1},(0\le j\le m-1),\hfill \\ \hfill g_{2m+1,2j}& ⇝& f_{2m+1-j,0},g_{2m+2-j,1},\dots ,f_{2m,2j-2},g_{2m+1,2j-1},\hfill \\ & & g_{2m+1,2j},(0\le j\le m),\hfill \\ \hfill g_{2m,2j+1}& ⇝& g_{2m-j,0},f_{2m-j,1},\dots ,g_{2m-1,2j-2},f_{2m-1,2j-1},\hfill \\ & & g_{2m,2j},g_{2m,2j+1},(0\le j\le m-1),\hfill \\ \hfill g_{2m,2j}& ⇝& f_{2m-j,0},g_{2m+1-j,1},\dots ,f_{2m-1,2j-2},g_{2m,2j-1},\hfill \\ & & g_{2m,2j},(0\le j\le m-1),\hfill \end{array}$$

where $a_m=01021\dots m(m-1)$, $b_{m,j}=a_{m}j$, $c_m=a_{m-1}mmm$, $d_m=a_{m-1}mm$, $e_m=a_{m}mm$, $f_{m,j}=a_{m}m(m-1)\dots j$, and $g_{m,j}=a_m(m-2)(m-3)\dots j$.

Then, we translate these succession rules to equations, and solving the system (leaving to the interested reader), we obtain

$$\begin{array}{cccc}\hfill & F_{\left\{0001\right\};0}\left(x\right)={\displaystyle \frac{x(1-x+x^3+x^4)}{(1-x-x^2){(1-x)}^2}},\hfill & \hfill & F_{\left\{0001\right\};00}\left(x\right)={\displaystyle \frac{x}{{(1-x)}^2}},\hfill \\ \hfill & F_{\left\{0001\right\};000}\left(x\right)={\displaystyle \frac{x}{1-x}},\hfill & \hfill & F_{\left\{0001\right\};01}\left(x\right)={\displaystyle \frac{x(1-x+x^2+2x^3)}{(1-x-x^2){(1-x)}^2}},\hfill \\ \hfill & F_{\left\{0001\right\};a_m}\left(x\right)={\displaystyle \frac{x(1-x+x^3+x^4)}{(1-x-x^2){(1-x)}^{m+2}}},\hfill & \hfill & F_{\left\{0001\right\};c_m}\left(x\right)={\displaystyle \frac{x(1-x+x^2)}{{(1-x)}^{m+2}}},\hfill \\ \hfill & F_{\left\{0001\right\};d_m}\left(x\right)={\displaystyle \frac{x(1-x+2x^2)}{{(1-x)}^{m+2}}},\hfill & \hfill & F_{\left\{0001\right\};e_m}\left(x\right)={\displaystyle \frac{x}{{(1-x)}^{m+1}}},\hfill \\ \hfill & F_{\left\{0001\right\};b_{m,m-1}}\left(x\right)={\displaystyle \frac{x}{{(1-x)}^m}},\hfill & \hfill & F_{\left\{0001\right\};b_{m,m+1}}\left(x\right)={\displaystyle \frac{x(1-x+x^2+2x^3)}{(1-x-x^2){(1-x)}^{m+2}}},\hfill \\ \hfill & F_{\left\{0001\right\};f_{m,j}}\left(x\right)={\displaystyle \frac{x}{{(1-x)}^{j+1+\delta (m=j)}}},\hfill & \hfill & F_{\left\{0001\right\};g_{m,j}}\left(x\right)={\displaystyle \frac{x}{{(1-x)}^{j+1}}}.\hfill \end{array}$$

Let us give an example explaining how to confirm these formulas. For example, the rule $01⇝a_1,d_1$ is equivalent to

$$F_{\left\{0001\right\};01}\left(x\right)=x+x{F}_{\left\{0001\right\};a_1}\left(x\right)+x{F}_{\left\{0001\right\};d_1}\left(x\right).$$

Thus, we need to show

$${\displaystyle \frac{x(1-x+x^2+2x^3)}{(1-x-x^2){(1-x)}^2}}=x+{\displaystyle \frac{x^2(1-x+x^3+x^4)}{(1-x-x^2){(1-x)}^3}}+{\displaystyle \frac{x^2(1-x+2x^2)}{{(1-x)}^3}},$$

which it is true.

Therefore, we can present the following formula.

Theorem 10.

The generating function $F_{0001}\left(x\right)$ for the number of descent sequences of length n that avoid 0001 is given by

$${\displaystyle \frac{x(1-x+x^3+x^4)}{(1-x-x^2){(1-x)}^2}}.$$

3. Conclusions

In this paper, we studied pattern avoidance in descent sequences of ${\mathcal{D}}_n$ for several patterns of length four. Using the framework of generating trees, we developed a systematic approach based on labelled trees and succession rules. For each pattern, we derived a system of recurrence relations and solved it to obtain explicit formulas for the corresponding generating functions $F_{\tau }\left(x\right)$. Our results show that the generating tree method provides a unified and effective strategy for analyzing different pattern-avoidance classes. In particular, despite the diversity of the patterns considered, each case can be reduced to a manageable system of functional equations that leads to closed-form expressions.

A natural direction for future work is to extend this approach to other classes of patterns or longer patterns and to investigate possible Wilf-equivalences between additional families of pattern-avoiding descent sequences.