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

1. Introduction

A descent in a sequence of integers $w_1{w}_2\dots w_m$ is a pair of adjacent entries $w_j{w}_{j+1}$ such that $w_j>w_{j+1}$. A sequence ${\pi }_1{\pi }_2\dots {\pi }_n$ of non-negative integers is called a descent sequence of length n if it satisfies the following conditions:

• ${\pi }_1=0$,
• for all i with $2\le i\le n$, we have ${\pi }_i\le \mathrm{des}({\pi }_1{\pi }_2\dots {\pi }_{i-1})+1$,

where $\mathrm{des}({\pi }_1{\pi }_2\dots {\pi }_m)$ denotes the number of descents in the sequence ${\pi }_1{\pi }_2\dots {\pi }_m$. For example, the word 010203 is a descent sequence, but 01103 is not. Descent sequences were introduced by Callan [3]. Later, the author and Callan [5] studied descent sequences avoiding two patterns of length three.

We recall the notion of pattern avoidance for descent sequences. Let $w=w_1{w}_2\dots w_n$ be a sequence of integers. Let $\tau ={\tau }_1\dots {\tau }_m$ be a pattern, that is, a word of length m over the alphabet $\{0,1,\dots ,\ell \}$ that includes each symbol $0,1,\dots ,\ell$ at least once, for some $\ell \ge 0$ and $m\ge 1$. We say that the sequence wcontains the pattern $\tau$ if there exists a subsequence $w_{f_1}{w}_{f_2}\dots w_{f_m}$ of w, with $1\le f_1<f_2<\dots <f_m\le n$, such that $w_{f_i}X{w}_{f_j}$ if and only if ${\tau }_{i}X{\tau }_j$, for all $1\le i,j\le m$ and $X\in \{<,>,=\}$. If no such subsequence exists, we say that wavoids the pattern $\tau$. For example, the descent sequence 010201303 contains all patterns of length three, but avoids the pattern 0321. We denote by ${\mathcal{D}S}_n\left(\{{\tau }^{\left(1\right)},\dots ,{\tau }^{\left(m\right)}\}\right)$, or simply ${\mathcal{D}S}_n({\tau }^{\left(1\right)},\dots ,{\tau }^{\left(m\right)})$, the set of descent sequences of ${\mathcal{D}S}_n$ that avoid ${\tau }^{\left(j\right)}$ for all $j=1,2,\dots ,m$. Two collections of patterns T and R are called D-Wilf-equivalent, denoted $T\stackrel{d}{\sim }R$, if the number of descent sequences of ${\mathcal{D}S}_n\left(T\right)$ equals the number of descent sequences of ${\mathcal{D}S}_n\left(R\right)$ for all $n\ge 0$.

Callan and Mansour [5] established that the D-Wilf-equivalence classification consists of 9 classes for one pattern of length three, 23 classes for pairs of length-three patterns, and 69 classes for patterns of length four. In particular, they showed that $F_{1002}\left(x\right)=F_{1012}\left(x\right)={\displaystyle \frac{x(1-2x)}{(1-x)(1-3x+x^2)}}$ and $F_{0100}\left(x\right)=F_{0101}\left(x\right)=F_{0102}\left(x\right)={\displaystyle \frac{x}{1-2x}}$. Here, $F_{\tau }\left(x\right)$ denotes the generating function for the number of nonempty descent sequences of ${\mathcal{D}S}_n\left(\tau \right)$.

In this paper, we study the generating function $F_{\tau }\left(x\right)$ for seven different patterns of length four. Specifically, for each class, we construct a generating tree with 1–2 labels, derive a system of recurrence relations from the set of the succession rules, and find an explicit formula for the generating function counting nonempty descent sequences of ${\mathcal{D}S}_n$ that avoid either 0001, 0010, 0011, 0012, 0021, 0110, 0112, 0123, or 0132.

2. Generating Trees and the Strategy for the Proofs

We start by recalling the generating tree $\mathcal{T}\left(P\right)$ for pattern-avoiding descent sequences of $\mathcal{D}S\left(P\right)={\cup }_{n\ge 1}{\mathcal{D}S}_n\left(P\right)$ as described in [5] (extension of [1,2,4,6,7,8,9]), for any set of patterns P. The tree $\mathcal{T}\left(P\right)$ begins with the root labeled 0, which is positioned at level 1, and the children of a sequence ${\pi }_1\dots {\pi }_{n-1}\in {\mathcal{D}S}_{n-1}\left(P\right)$ are obtained from the set

$$\{\pi ={\pi }_1{\pi }_2\dots {\pi }_{n-1}{\pi }_n\mid {\pi }_n=0,1,\dots ,\mathrm{des}({\pi }_1\dots {\pi }_{n-1})+1,\pi avoidsP\},$$

Define $\mathcal{T}(P;\pi )$ as the subtree of $\mathcal{T}\left(P\right)$ rooted at $\pi$ which includes all of its descendants. For any two nodes $\pi ,{\pi }^{\prime }\in \mathcal{D}S\left(P\right)$, we say that the subtrees $\mathcal{T}(P;\pi )$ and $\mathcal{T}(P;{\pi }^{\prime })$ are isomorphic, and write $\pi \sim {\pi }^{\prime }$, if they are isomorphic as left-right ordered plane trees. Let $\mathcal{T}\left[P\right]$ be the tree obtained from $\mathcal{T}\left(P\right)$ by replacing each node $\pi$ with the first node ${\pi }^{\prime }\in \mathcal{D}S\left(P\right)$, in a top-to-bottom and left-to-right order, for which $\pi \sim {\pi }^{\prime }$. From this point forward, we treat $\mathcal{T}\left(P\right)$ and $\mathcal{T}\left[P\right]$ as the same.

 Example 1.

Let $P=\left\{0021\right\}$. Clearly, the children of $0\in \mathcal{T}\left(P\right)$ 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 seek to derive an explicit expression for the generating function

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

To do so, we refine this definition by defining

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

where $\pi \in D\left(P\right)$. Clearly, we have $F_P\left(x\right)=F_{P;0}\left(x\right)$. Moreover, any succession rule of the form

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

is equivalent to the 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 system, we obtain the following result.

 Theorem 1.

The generating function $F_{0012}\left(x\right)$ 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 $F_{0021}\left(x\right)$ 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 $j=1,2,\dots ,s$. Since this verification process is routine, we omit the detailed proofs of the succession rules for the generating trees below and leave them to the interested reader.

2.1. The Pattern 0010

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

$$\begin{array}{cccccccc}\hfill & 0⇝00,01,\hfill & \hfill & 00⇝00,001,\hfill & \hfill & 001⇝001,\hfill & \hfill & 01⇝a_1,011,\hfill \\ \hfill & 011⇝f_1,b_1,\hfill & \hfill & a_m⇝a_m,d_m,e_{m+1},\hfill & \hfill & b_m⇝f_m,b_m,\hfill & \hfill & c_m⇝c_m,\hfill \\ \hfill & d_m⇝d_m,c_{m+1},\hfill & \hfill & e_m⇝a_m,b_m,\hfill & \hfill & f_m⇝f_m,d_m,c_{m+1},\hfill \end{array}$$

where $a_m=01021\dots m(m-1)$, $b_m=a_{m-1}mm$, $c_m=a_{m-1}(m-1)m$, $d_m=a_{m}a$, $e_m=a_{m-1}m$, and $f_m=a_{m-1}mm(m-1)$. By translating these succession rules to equations, we obtain

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{F}_{\{0010;0\}}\left(x\right)\hfill & =x+x{F}_{\{0010;00\}}\left(x\right)+x{F}_{\{0010;01\}}\left(x\right),\hfill \\ F_{\{0010;00\}}\left(x\right)\hfill & =x+x{F}_{\{0010;00\}}\left(x\right)+x{F}_{\{0010;001\}}\left(x\right),\hfill \\ F_{\{0010;001\}}\left(x\right)\hfill & =x+x{F}_{\{0010;001\}}\left(x\right),\hfill \\ F_{\{0010;01\}}\left(x\right)\hfill & =x+xA\left(0\right)+x{F}_{\{0010;011\}}\left(x\right),\hfill \\ F_{\{0010;011\}}\left(x\right)\hfill & x+xF\left(0\right)+xB\left(0\right),\hfill \end{array}\right.\end{array}$$

(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 $K\in \{A,B,C,D,E,F\}$. 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

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

Thus, by taking $v=x^2/(1-x)$, we obtain

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

By using expression of $A\left(0\right)$, we have

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

Substituting the expressions for $A\left(v\right)$, $B\left(v\right)$, and $F\left(v\right)$ into (1), we obtain a linear system. By solving this system yields the following result.

 Theorem 3.

The generating function $F_{0010}\left(x\right)$ is given by

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

2.2. The Pattern 0011

The generating tree $\mathcal{T}\left(\right\{0011\left\}\right)$ is determined by the following succession rules

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

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 $K\in \{G,I\}$. By translating these succession rules to equations (as before), and then solving the system, we obtain

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

and then

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

Therefore, we can present the following result.

 Theorem 4.

The generating function $F_{0011}\left(x\right)$ is given by

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

2.3. The Pattern 0110

The generating tree $\mathcal{T}\left(\right\{0110\left\}\right)$ is determined by the following succession rules

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

where $a_m=0102\dots 0m0m$, $b_m=a_{m-1}m$, $c_{m,j}=0102\dots 0(m-1)jm$, and $d_{m,j}=0102\dots 0mj$. Define $A\left(v\right)={\sum }_{m\ge 1}{F}_{\left\{0110\right\};a_m}\left(x\right)v^{m-1}$, $B\left(v\right)={\sum }_{m\ge 2}{F}_{\left\{0110\right\};b_m}\left(x\right)v^{m-2}$, and

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

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 [10]). 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

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

(3)

By taking $u=(vx+x^2-v)/v$, we have

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

Using this expression with (3), 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 $F_{0110}\left(x\right)$ is given by

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

2.4. The Pattern 0112

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,

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

By taking $v=x^2/(1-x)$, we obtain that $A\left(0\right)={\displaystyle \frac{x}{1-2x-x^2}}$. Hence, by the first equation, we obtain the following result.

 Theorem 6.

The generating function $F_{0112}\left(x\right)$ is given by

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

2.5. The Pattern 0123

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

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

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

$$\begin{array}{cc}\hfill F_P\left(x\right)& =x+x{F}_P\left(x\right)+xA\left(0\right),\hfill \\ \hfill A\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xB\left(v\right)+xA\left(v\right)+vxC(v,1),\hfill \\ \hfill B\left(v\right)& ={\displaystyle \frac{x}{1-v}}+xB\left(v\right)+{\displaystyle \frac{x}{v}}(A\left(v\right)-A\left(0\right))+xF(v,1)\hfill \end{array}$$

and

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

 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. 

It is very easy to check that 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 the equations. □

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 $F_{0123}\left(x\right)$ is given by

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

2.6. The Pattern 0132

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

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

So, by substituting $v=v_1$ and $v=v_2$ within the equation, 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 $f=F_{0132}\left(x\right)$ 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

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

$$\begin{array}{cc}\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},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},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},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},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},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},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},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},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},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},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},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},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},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},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+1,2j-1},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},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},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},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 ,f_{2m,2j-2},f_{2m,2j-1},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},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},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},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 this 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 $F_{0001}\left(x\right)$ is given by

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