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\begin{document}
\title{Wilf classes for weak ascent sequences avoiding a pair or triple of length-3 patterns}
\author{David Callan}
\address{Department of Statistics, University of Wisconsin, Madison, USA}
\email{callan@stat.wisc.edu}
\author{Toufik Mansour}
\address{Department of Mathematics, University of Haifa, 3103301 Haifa, Israel}
\email{tmansour@univ.haifa.ac.il}

\begin{abstract} A \emph{weak ascent sequence} is a word $\pi=\pi_1\pi_2\cdots\pi_n$ over the
set of nonnegative integers such that $\pi_1=0$ and $\pi_i\leq1+\wasc(\pi_1\pi_2\cdots\pi_{i-1})$ for $i=2, \dots, n$, where $\wasc(\pi_1\pi_2\cdots\pi_m)$ is the numbers of \emph{weak ascents} in the
word  $\pi_1\pi_2\cdots\pi_m$, that is, the number of positions $j$ such that $\pi_j\leq \pi_{j+1}$.
Here we obtain some enumerative results for weak ascent sequences avoiding a set of two or three 3-letter patterns, leading to a conjecture for the number of Wilf equivalence classes for weak ascent sequences, called $A$-Wilf equivalence classes, avoiding a pair (respectively, triple) of 3-letter patterns. The main tool is the use of generating trees. Some cases are treated using bijective methods. 
\end{abstract}
\subjclass[2010]{05A05, 05A15}
\keywords{Weak ascent sequence; Generating tree; Kernel method; Pattern avoidance; Wilf equivalence class}
\maketitle

\section{Introduction}
An {\it ascent} in a sequence $s_1s_2\cdots s_m$ is an index $1\leq j\leq m-1$ such that $s_j<s_{j+1}$. An {\it ascent sequence} $a_1a_2\cdots a_n$ is one consisting of
non-negative integers satisfying $a_1=0$ and for all $i$ with $1<i\leq n$, $a_i\leq \as(a_1a_2\cdots a_{i-1})+1$,
where $\as(a_1a_2\cdots a_k)$ is the number of ascents in the sequence $a_1a_2\cdots a_k$. For example, the sequence $01013224$ is an ascent sequence, whereas $01003$ is not. Bousquet-M\`{e}lou, Claesson, Dukes, and Kitaev \cite{Bous} connected ascent sequences to $(2 + 2)$-free posets. Since then, ascent sequences have been considered in a series of papers where connections to many other combinatorial structures have been found.

The concept of ascent sequence has been extended to weak ascent sequences, where B\'enyi, Claesson, and Dukes \cite{BCD} found a relation between weak ascent sequences and permutations avoiding a length-4 bivincular pattern, restricted upper-triangular binary matrices, and factorial posets that are weakly $(3+1)$-free. A \emph{weak ascent sequence} is a word $\pi=\pi_1\pi_2\cdots\pi_n$ over the set of non negative integers such that $\pi_1=0$ and $\pi_i\leq1+\wasc(\pi_1\pi_2\cdots\pi_{i-1})$ for $i=2, \dots, n$, where $\wasc(\pi_1\pi_2\cdots\pi_m)$ is the numbers of \emph{weak ascents} in the word  $\pi_1\pi_2\cdots\pi_m$, that is, the number of positions $j$ such that $\pi_j\leq \pi_{j+1}$.  We denote the set of all weak ascent sequences of length $n$ by $\A_n$. For example, $\A_3=\{000, 001, 002, 010, 011, 012\}$
%\begin{align*}
%\A_3=\{000, 001, 002, 010, 011, 012\}\,,
%\end{align*}
while $\{0010,0011,0012,0013\} \subset\A_4$, $0014 \notin \A_4$.
A \emph{pattern} $\pi$ is a word containing each of the letters in $\{0,1, \dots , \ell-1\}$ at least once for some integer $\ell\geq 1$.  A weak ascent sequence $w=w_1w_2\cdots w_n\in \A_n$ \emph{contains} a pattern $\pi=\pi_1\cdots\pi_k $  if there is a subsequence $w_{i_1}w_{i_2}\cdots w_{i_k}$ of $w$, $i_1<i_2<\cdots<i_k$, which is order-isomorphic to $\pi$. We say that $w$ \emph{avoids} the pattern $\pi$ if $w$ does not contain any occurrence of $\pi$. For example, the weak ascent sequence $00203413\in \A_{8}$ contains the pattern $011$ and avoids the pattern $321$.  We denote the set of weak ascent sequences of length $n$ that avoid $\pi$ by $\A_n(\pi)$. Further, for a set of patterns $B$, we set $\A_n(B)=\cap_{\pi\in B}\A_n(\pi)$.

The goal of this work is to study weak ascent sequences avoiding a set of two or three length-3 patterns. This study is analogous to the one conducted by Duncan and Steingr\'imsson on ascent sequences \cite{DS,DM}. More precisely, we say that two sets of patterns $P$ and $Q$ are $A$-Wilf equivalent, denoted $P\asim Q$, if
$|\A_n(P)|=|\A_n(Q)|$ for every $n\geq0$.
Let $\textrm{WAS}_k$ denote the number of $A$-Wilf equivalence classes for weak ascent sequences avoiding $k$ distinct patterns of length three.
The main result of this paper can be formulated briefly as follows.
\begin{theorem}\label{thm}
We have $59\leq \textrm{WAS}_2\leq61$ and $150\leq \textrm{WAS}_3\leq155.$
\end{theorem}
\begin{conjecture}\label{conj}
We have $\textrm{WAS}_2=59$ and $ \textrm{WAS}_3=150.$
\end{conjecture}

\section{Generating trees}\label{gen}
We apply here the generating tree method introduced by West \cite{W}. Given a fixed set $B$ of patterns of length at least $2$, let $\A(B)$ denote the set of all non-empty weak ascent sequences that avoid $B$. We define the generating tree $\T(B)$ to be the plain tree with root $0$ as follows, see \cite{BMR}. Starting with this root which stays at level $1$, we construct the remainder of the nodes of the tree $\T(B)$ recursively as follows. The children of $x_1\cdots x_n\in\A_n(B)$ are those sequences from the set $$\{x_1\cdots x_{n+1}\mid x_{n+1}=0,1,\ldots,\#\wasc(x_1\cdots x_n)+1\}$$ that avoid the patterns in $B$.

We define an equivalence relation on the nodes of $\T(B)$. Let $\T(B;w)$ be the subtree consisting of the weak ascent sequence $w$ as the root and its descendants in $\T(B)$.  We say that $w$ is {\em equivalent} to $w'$ if and only if $\T(B;w)\cong\T(B;w')$ (in the sense of plain trees). Let $\T'(B)$ be the same tree as $\T(B)$, where we replace each node $w$ by the first node $w'\in\T(B)$ from top to bottom and from left to right in $\T(B)$ such that $\T(B;w)\cong\T(B;w')$. From now, we consider $\T'(B)$ and $\T(B)$ identical.

For each node $w$ in $\T(B)$, we define $W_w(x)$ be the generating function for the number nodes at level $n$ in the subtree $\T(B;w)$ of $\T(B)$, where the root stay at level $1$, namely,
$$W_w(x)=\sum_{n\geq1}|\{\mbox{nodes at level $n$ in the subtree $\T(B;w)$ of $\T(B)$}\}|\cdot x^n.$$

The basic outline of the generating tree method is the following.
\begin{itemize}
\item[(1)] Let $B$ be any set of patterns and let $D$ be any positive number (here we use $D=5$).
\item[(2)] We find the first $D$ levels of the generating tree $\T(B)$.
\item[(3)] By (2), we guess all the succession rules of $\T(B)$.
\item[(4)] Based on (3), we try to prove these succession rules (for instance, see Example \ref{ex1}). If we fail, then we increase $D$ by $1$ and go back to Step (2). Otherwise, the succession rules of the generating tree $\T(B)$ are found.
\item[(5)] Each rule $v\rightsquigarrow w^{(1)},\ldots,w^{(s)}$ of $\T(B)$, we translate to an equation as $W_v(x)=x+x\sum_{j=1}^sW_{w^{(i)}}(x)$. Then, we try to solve the obtained system of the equations for $W_0(x)=\sum_{n\geq1}|\A_n(B)|x^n$.
\end{itemize}

\begin{example}\label{ex1}
Let $B=\{001,101\}$. Based on the first $5$ levels of the generating tree $\T(B)$, we obtain the following rules
\begin{align*}
&0\rightsquigarrow 00,01,&&00\rightsquigarrow 000,\\
&01\rightsquigarrow 00,011,012,&&000\rightsquigarrow 0000,\\
&011\rightsquigarrow 000,0111,&&012\rightsquigarrow 000,011,0122,0123,\\
&0000\rightsquigarrow 00000,&&0111\rightsquigarrow 0000,01111,\\
&0122\rightsquigarrow 0000,0111,01222,&&0123\rightsquigarrow 0000,0111,0122,01233,01234,\\
&00000\rightsquigarrow 000000,&&01111\rightsquigarrow 00000,011111,\\
&01222\rightsquigarrow 00000,01111,012222,&&01233\rightsquigarrow 00000,01111,01222,012333,\\
&01234\rightsquigarrow 00000,01111,01222,01233,012344,012345.
\end{align*}
Thus, we guess that the generating tree $\T(B)$ has the following rules
\begin{align*}
&a_1\rightsquigarrow a_2,b_1,&& a_m\rightsquigarrow a_{m+1},\\ &b_m\rightsquigarrow a_{m+1},c_{1,m},\ldots,c_{m-1,2},c_{m,2},b_{m+1},&&c_{m,j}\rightsquigarrow a_{m+j},c_{1,m+j-1},\ldots,c_{m-1,j+1},c_{m,j+1},
\end{align*}
where $a_m=0^m$, $b_m=01\cdots m$, $c_{m,j}=01\cdots(m-1)m^j$. This can be explained as follows:
\begin{itemize}
\item The only children of $a_1$ are $a_2$ and $b_1$, so $a_1\rightsquigarrow a_2,b_1$ holds.
\item The only child of $a_m$ with $m\geq2$ is $a_{m+1}$ (remember, we avoid $B$), so $a_m\rightsquigarrow a_{m+1}$ holds.
\item The children of $b_m$ are $b_mj$ with $j=0,\ldots,m+1$. Note that $\T(B;b_m0)\cong\T(B;a_{m+1}$, $b_m(m+1)=b_{m+1}$, $b_mm=c_{m,2}$, $\T(B,b_m(m-1))\cong\T(B;c_{m-1,2})$, and $\T(B,b_mj)\cong\T(B;01\cdots (j-1)j^{m+1-j})=\T(B;c_{j,m+1-j})$ with $j=1,2,\ldots,m-2$. So, $b_m\rightsquigarrow a_{m+1},c_{1,m},\ldots,c_{m-1,2},c_{m,2},b_{m+1}$ holds.
\item Similarly, the last two rules are holds.
\end{itemize}
Let $A_m(x)=W_{a_m}(x)$, $B_m(x)=W_{b_m}(x)$, and $C_{m,j}(x)=W_{c_{m,j}}(x)$. Thus, the rules can be translated to
\begin{align*}
A_1(x)&=x+xA_2(x)+xB_1(x),\\
A_m(x)&=x+xA_{m+1}(x),\\
B_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^{m-1}C_{i,m+1-i}(x)+xC_{m,2}(x)+xB_{m+1},\\
C_{m,j}(x)&=x+xA_{m+j}(x)+x\sum_{i=1}^{m-1}C_{i,m+j-i}(x)+xC_{m,j+1}(x).
\end{align*}
By induction on $m$, we have $A_m(x)=\frac{x}{1-x}$ and $C_{m,j}(x)=\frac{x}{(1-x)^{m+1}}$. Thus,
$$B_m(x)=\frac{x}{1-x}+x\sum_{i=1}^{m-1}\frac{x}{(1-x)^{i+1}}+\frac{x^2}{(1-x)^{m+1}}+xB_{m+1}(x).$$
Hence, $B_1(x)=\frac{x}{(1-x)(1-2x)}$, which implies that $$W_0(x)=A_1(x)=x+\frac{x^2}{1-x}+\frac{x^2}{(1-x)(1-2x)}=\frac{x}{1-2x}.$$
\end{example}

\section{Avoiding a pair of patterns of length $3$}\label{secpairs}
A {\em candidate class} is a maximal subset $C$ of the set of all pairs (in next section, triples) of patterns of length $3$ such that for any $P, P'\in C$, $|\A_n(P)| =|\A_n(P')|$, for all $n=1,2,\ldots,11$. A candidate class is called {\em trivial} if it contains exactly one pair of patterns, otherwise, it is called {\em nontrivial}. Thus a trivial class certainly constitutes a (singleton) $A$-Wilf equivalence class. By Table \ref{tab2} there are exactly $48$ trivial and $11$ nontrivial candidate classes. The aim of this section is to show that the $11$ nontrivial candidate classes of pairs of length $3$ patterns are indeed $11$ $A$-Wilf equivalences. Table \ref{tab1} describes all these $11$ nontrivial candidate classes including the corresponding succession rules of the generating trees and explicit formulas (if they exist) for the generating function. In classes 15 and 21 bijective methods are used. The two open cases, classes 41 and 43, are in blue and indicated by the double question marks.

\begingroup
\allowdisplaybreaks
{\fontsize{7}{8}\selectfont
\begin{longtable}[c]{|l|l|R|L|}
\caption{Rules of generating trees for weak ascent sequences avoiding a pair of length-$3$ patterns.\label{tab1}}\\ \hline
\multicolumn{4}{|c|}{Beginning of Table \ref{tab1}}\\ \hline
Class&$B$ pair& Rules of $\T(B)$&$G_B(x)$/Reference \\ \hline
\endfirsthead  \hline
\multicolumn{4}{|c|}{Continuation of Table \ref{tab1}}\\ \hline
Class&$B$ pair& Rules of $\T(B)$&$G_B(x)$/Reference  \\ \hline
\endhead \hline
\endfoot \hline
\multicolumn{4}{|c|}{End of Table \ref{tab1}}\\ \hline\hline
\endlastfoot
1&\{001,010\}&&\\
 &\{001,011\}&$01\cdots m\rightsquigarrow 0^{m+2},01\cdots(m+1)$; $0^m\rightsquigarrow 0^{m+1}$&\\\cline{2-3}
 &\{001,012\}&$01^m\rightsquigarrow 0^{m+2},01^{m+1}$; $0^m\rightsquigarrow 0^{m+1}$&$\frac{x}{(1-x)^2}$\\\hline
%1,2,3,4,5,6,7,8,9,10,11

5&\{000,011\}&$0\rightsquigarrow a_0,01$; $a_m\rightsquigarrow a_{m+1},b_{m+2}$; $01\rightsquigarrow010,b_1$; $b_m\rightsquigarrow c_m,b_{m+1}$; $c_m\rightsquigarrow c_{m+1}$; $010\rightsquigarrow c_2$, where $a_m=0012\cdots m$, $b_m=0012\cdots(m-2)m$, $c_m=0012\cdots(m-2)m(m-1)$&\\\cline{2-3}
 &\{001,021\}&&\\
 &\{001,120\}&$a_1\rightsquigarrow a_2,c_1$; $a_m\rightsquigarrow a_{m+1}$; $c_m\rightsquigarrow a_{m+1},b_{m+1},c_{m+1}$; $b_m\rightsquigarrow a_{m+1},b_{m+1}$, where $a_m=0^m$, $b_m=01^m$, $c_m=01\cdots m$&\\\cline{2-3}
 &\{001,110\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow (a_m)^m,a_{m+1},b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$&$\frac{x}{1-x}+\frac{x^2}{(1-x)^3}$\\\hline
%1,2,4,7,11,16,22,29,37,46,56

8&\{001,101\}&&\\
 &\{001,102\}&&\\
 &\{001,201\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+1},c_{1,m},\ldots,c_{m-1,2},c_{m,2},b_{m+1}$; $c_{m,j}\rightsquigarrow a_{m+j},c_{1,m+j-1},\ldots,c_{m-1,j+1},c_{m,j+1}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_{m,j}=01\cdots(m-1)m^j$&\\\cline{2-3}
 &\{010,012\}&&\\
 &\{011,012\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},\ldots,b_{m,j-1},b_{m+1,j}$, where $a_m=0^m$, $b_{m,j}=a_mj$&Example \ref{ex1}, Section \ref{sec8}; $\frac{x}{1-2x}$\\\hline
%1,2,4,8,16,32,64,128,256,512,1024

11&\{012,021\}&$a_m\rightsquigarrow a_{m+1},(b_m)^m$; $b_m\rightsquigarrow b_m,b_{m+1}$, where $a_m=0^m$, $b_m=a_m1$&\\\cline{2-3}
  &\{012,110\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},\ldots,b_{m,j},c_m$; $c_m\rightsquigarrow c_{m+1}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_m=a_m11$&Section \ref{sec11}; $\frac{x(2x^2-2x+1)}{(1-x)^2(1-2x)}$\\\hline
%1,2,5,12,27,58,121,248,503,1014,2037

13&\{012,201\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},c_{m,2},\ldots,c_{m,j},b_{m+1,j}$; $c_{m,j}\rightsquigarrow d_m,c_{m,2},\ldots,c_{m,j-1},c_{m+1,j}$; $d_m\rightsquigarrow d_{m+1}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mm(m-1)$, $d_m=a_m210$&\\\cline{2-3}
  &\{012,210\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,j},(d_m)^{j-1},b_{m+1,j}$; $d_m\rightsquigarrow d_{m+1}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $d_m=a_m21$&Section \ref{sec13}; $\frac{x(2x^4-5x^3+7x^2-4x+1)}{(1-x)^4(1-2x)}$\\\hline
%1,2,5,13,33,80,185,411,885,1862,3853

15&\{011,102\}&&See Section \ref{sec15}\\\cline{2-3}
  &\{012,102\}&&\\
  &\{012,120\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},\ldots,b_{m,j}, b_{m+1,j}$, where $a_m=0^m$, $b_{m,j}=a_mj$; &See \cite{BMR}, $\frac{x(1-x)}{1-3x+x^2}$\\\hline
%1,2,5,13,34,89,233,610,1597,4181,10946

21&\{010,021\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m+1,j},\ldots,b_{m+1,m+1}$, where $a_m=0^m$, $b_{m,j}=a_mj$&\\\cline{2-3}
  &\{011,201\}&&Section \ref{sec21}, $\frac{1-\sqrt{1-4x}}{2x}-1$\\\hline
%1,2,5,14,42,132,429,1430,4862,16796,58786

24&\{011,101\}&&\\
  &\{011,110\}&$\asim\{011\}$&See \cite{BMR}\\\hline
%1,2,5,14,43,143,510,1936,7775,32869,145665

{\color{blue}41}&\{102,110\}&\raisebox{-1.1mm}{??}&\\
  &\{102,120\}&  & See A279561\\\hline
%1,2,6,21,77,287,1079,4082,15522,59280,227240

{\color{blue}43}&\{021,201\}&&\\
  &\{021,210\}&$\asim\{021\}$&See \cite{BMR}\\\cline{2-3}
  &\{101,102\}&??&\\\hline
%1,2,6,21,80,322,1347,5798,25512,114236,518848

57&\{100,201\}&&\\
  &\{101,201\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow c_{m,j,0},\ldots,c_{m,j,j-1},b_{m+1,j},\ldots,b_{m+1,m+1}$; $c_{m,j,i}\rightsquigarrow c_{m,j,0},\ldots, c_{m,j,i-1}$, $c_{m+1,j,i},c_{m+1,j+1,i+1},\ldots,c_{m+1,m+1,m+1+i-j}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j,i}=a_mji$&\\\hline
%1,2,6,22,92,422,2074,10754,58202,326240,1882960
\end{longtable}}

\subsection{Class 8}\label{sec8}
In Example \ref{ex1} we showed that the generating functions for the number weak ascent sequences of length $n$ that avoid  $\{001,101\}$ is given by $\frac{x}{1-2x}$. Thus, it remains to study the case $B=\{010,012\}$. Let $A_m(x)$ (respectively, $B_m(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(B)$ with a root $a_m$ (respectively, $b_m$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+x\sum_{i=1}^{j-1}B_{m,i}(x)+xB_{m+1,j}.
\end{align*}
By induction on $j$, we have $B_{m,j}(x)=\frac{x}{(1-x)^j}$. Thus,
$$A_m(x)=x+xA_{m+1}(x)+x\sum_{i=1}^m\frac{x}{(1-x)^i},$$
Hence, $A_1(x)=\frac{x}{1-2x}$.

\subsection{Class 11}\label{sec11}
{\bf Case \{012,021\}} Let $A_m(x)$ (respectively, $B_m(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,021)$ with a root $a_m$ (respectively, $b_m$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+mxB_m(x),\\
B_m(x)&=x+xB_m(x)+xB_{m+1}(x).
\end{align*}
By induction on $m$, we have $B_m(x)=\frac{x^2}{1-2x}$. Thus,
$$A_m(x)=x+xA_{m+1}(x)+\frac{mx^2}{1-2x}.$$
Hence, $A_1(x)=\frac{x(2x^2-2x+1)}{(1-x)^2(1-2x)}$.

{\bf Case \{012,110\}} Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_m(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,110)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_m$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+x\sum_{i=1}^jB_{m,i}(x)+xC_m(x),\\
C_m(x)&=x+xC_{m+1}(x).
\end{align*}
By induction on $m$, $C_m(x)=\frac{x}{1-x}$. So, by induction on $j$, we have that $B_{m,j}(x)=\frac{x}{(1-x)^{j+1}}$. Hence,
$$A_m(x)=x+xA_{m+1}(x)+x\sum_{i=1}^m\frac{x}{(1-x)^{i+1}},$$
which implies that $A_1(x)=\frac{x(2x^2-2x+1)}{(1-x)^2(1-2x)}$.

\subsection{Class 13}\label{sec13}
{\bf Case \{012,201\}} Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$, $D_m(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,201)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m,j}$, $d_m$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+xB_{m,1}(x)+x\sum_{i=2}^jC_{m,i}(x)+xB_{m+1,j}(x),\\
C_{m,j}(x)&=x+xD_m(x)+x\sum_{i=2}^{j-1}C_{m,i}(x)+xC_{m+1,j}(x),\\
D_m(x)&=x+xD_{m+1}(x).
\end{align*}
By induction on $m,j$, we have that $D_m(x)=\frac{x}{1-x}$, $C_{m,j}(x)=\frac{x}{(1-x)^j}$, and $B_{m,j}(x)=\frac{x}{(1-x)^{j+1}}+\frac{x^3}{(1-x)^2(1-2x)}$.
Hence, $$A_m(x)=x+xA_{m+1}+\sum_{j=1}^m\frac{x^2}{(1-x)^{j+1}}+\frac{mx^4}{(1-x)^2(1-2x)},$$
which implies that
$$A_1(x)=\frac{x(2x^4-5x^3+7x^2-4x+1)}{(1-x)^4(1-2x)}.$$

{\bf Case \{012,201\}} Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $D_m(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,210)$ with a root $a_m$ (respectively, $b_{m,j}$, $d_m$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+xB_{m,j}(x)+(j-1)xD_m(x)+xB_{m+1,j}(x),\\
D_m(x)&=x+xD_{m+1}(x).
\end{align*}
By induction on $m,j$, we have that $D_m(x)=\frac{x}{1-x}$ and $B_{m,j}(x)=\frac{x(jx-2x+1)}{(1-x)(1-2x)}$.
Hence, $$A_m(x)=x+xA_{m+1}+\sum_{j=1}^m\frac{x^2(jx-2x+1)}{(1-x)(1-2x)},$$
which implies that
$$A_1(x)=\frac{x(2x^4-5x^3+7x^2-4x+1)}{(1-x)^4(1-2x)}.$$

\subsection{Class 15}\label{sec15}
Note that $\{012,102\}\asim\{012\}\asim\{012,120\}$. Thus, by \cite{BMR}, we have that the generating function for the number of weak ascent sequences of length $n$ that avoid $\{012,\tau\}$ is given by $\frac{x(1-x)}{1-3x+x^2}$, that is, the number of weak ascent sequences of length $n$ that avoid $\{012,\tau\}$ is given by $F_{2n-1}$, where $F_n$ is the $n$th Fibonacci number (defined by $F_n=F_{n-1}+F_{n-2}$ with $F_0=0$ and $F_1=1$). So, it remains to show that $|\A_n(\{011,102\})|=F_{2n-1}$, for all $n\geq1$.

The following lemma is immediate from the definition of a weak ascent sequence.
\begin{lemma}\label{maxentry}
The maximum possible entry in position $j$ in a weak ascent sequence is $j-1$ for $j\ge 1$. \qed
\end{lemma}

For a weak ascent sequence, let $\m$ denote the number of 0s it contains and let $\r$ denote the number of its descents.
Suppose $w \in\A_{n}(011,102)$. % and $w$ is not all 0s.
The nonzero entries of $w$ are all distinct to avoid 011; say they form the \emph{positive-support} set of $w$, denoted $S=S(w)$. Then $1 \le \m \le n$ and $\v S\v=n-\m$. Say $\a =\max(S)$ (when $S\ne \emptyset$).
The first descent top (if there is one) must be $\a$ to avoid a 102 pattern.  Hence, $\r\le \m-1$ for suppose $w$ begins with $p\ge 1$ 0s followed by $q\ge 0$ increasing positive entries followed by $\a$, the first descent top. Then $\a \le p+q$ by the weak ascent condition. This implies there are at most $p$ descent tops and if there are $p$ descent tops, then another 0 is a descent bottom. In any case, (\# descents) $\le $ (\# 0s)$-1$ and $\r\le \m-1$. Also, there cannot be more descents than elements of $S$, so $\r \le n-\m$.

Now, $\a$ occurs at or before position $n-\r$ to allow space for $\r$ descents following $\a$. So $\a \le n-\r-1$ by Lemma \ref{maxentry}, and $\a\ge n-\m$ since the elements of $S$ are distinct.
Hence, $\a=n-\m+j$ for some $j \in \{0,1,\dots,\m-1-\r\}$
and $\a$ occurs at or after position $n-\m+j+1$ by Lemma \ref{maxentry}.


Say $d_1<d_2<\dots <d_{\r}$ are the positions of the descent tops. Then $w_{d_1}=\a$ as noted above and $w_{d_2}>\dots > w_{d_{\r}}>w_n$ for else two of these entries would form the 1,\,2 of a 102 pattern.

Now consider an illustrative example with $n=23,\ \m=14,\ \r=4,\ \{d_1,d_2,d_3,d_4\}=\{13,17,18,21\},\ S=\{1,2,4,5,7,8,9,11,12\}$ and $\a=12$.

\begin{figure}[htp]
{\fontsize{7}{3}\selectfont
\[
\begin{array}{|c|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}\hline
\textrm{position }i & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \\ \hline
w_i & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 11 & 12 & 0 & 5 & 8 & 9 & 7 & 0 & 0 & 4 & 1 & 2 \\ \hline \hline
\textrm{position }i &  \multicolumn{11}{c}{ }  &  & 13 & \multicolumn{3}{c|}{ }   & 17 & 18 & \multicolumn{2}{c|}{ }  & 21 &  & 23 \\ \hline
\textrm{skeleton} & \multicolumn{11}{c}{S=\{1,2,4,5,7,8,9,11,12\}}        &   & 12 &  \multicolumn{3}{c|}{ }  & 9 & 7 &  \multicolumn{2}{c|}{ }    & 4 &  & 2 \\ \hline
\end{array}
\]
\caption{A (011,102)-avoiding weak ascent sequence $w$ and its skeleton}}
\end{figure}

\begin{lemma}\label{skel}
For $w \in\A_n(011,102)$, knowledge of the positive-support $S$, the descent locations $d_1,d_2,\ldots,d_{\r}$, and $w_{d_1},w_{d_2}, \ldots, w_{d_{\r}},w_n$ is sufficient to recover $w$.
\end{lemma}
\begin{proof}
By assumption, the entries $w_{d_1},w_{d_2}, \ldots, w_{d_{\r}},w_n$ are already assigned, where $w_{d_1}=\max(S)=\a$ and $w_n$ may be 0. From the pool of unassigned entries, $$\{0^{\m}\}\cup S\backslash \{w_{d_1},w_{d_2}, \ldots, w_{d_{\r}},w_n\},$$ fill in the blank slots in $w$ as follows. In the rightmost blank slot, place the largest pool entry that is $\le$ its neighbor on the right. Then working right to left, fill each blank spot with the largest unused pool entry that is $\le$ its neighbor on the right (there always will be such an entry). This procedure is necessary and sufficient to avoid a 102 pattern. The result obviously avoids 011. The entry $\a=w_{d_1}$ already meets the weak ascent condition. So do all later entries because they are smaller than $\a$, and so do all earlier entries because they will form a run of 0s followed by an increasing sequence of entries less than $\a$.
\end{proof}

To count $\A_{n,\m,\r}(011,102):= \{w\in \A_{n}(011,102)\mid\,w$ contains $\m$ zeros
and has $\r$ descents\}, it suffices, in view of Lemma \ref{skel},
to count ``skeletons'' as illustrated in Figure 1.
With $\a=n-\m+j$ for some $j \in \{0,1,\ldots,m-1-\r\}$ as shown above,
the descent tops form an $\r$-element subset
of $\{n-\m+j-1,\dots,n-1\}$, so $\binom{\m-j-1}{\r}$ choices. The positive-support set $S$ is an $(n-\m)$-element
subset of $\{1,\dots,n-\m+j\}$ that includes $\a=n-\m+j$, so $\binom{n-\m+j-1}{n-\m-1}$ choices. Lastly, $w_{d_2}, \dots, w_{d_{\r}},w_n$ is an $\r$-element subset of $\{0\}\cup(S\backslash\{\a\})$,
where $\v S\v=n-\m$, so $\binom{n-\m}{\r}$ choices. Consequently,
\begin{align*}
\v \A_{n,\m,\r}(011,102)\v & =\binom{n-\m}{\r} \sum_{j=0}^{m-1-\r}\binom{n-\m+j-1}{n-\m-1}\binom{\m-j-1}{r} \\
 & = \binom{n-\m}{\r} \binom{n-1}{\m-1-\r}\, ,
\end{align*}
which implies
\begin{align*}
\v A_{n}(011,102)\v  & = \sum_{\m=1}^{n}  \sum_{\r=0}^{\min(n-\m,\m-1)}\binom{n-\m}{\r} \binom{n-1}{\m-1-\r}\\
 & = \sum_{\m=1}^{n} \binom{2n-\m-1}{\m-1}  = F_{2n-1}\, .
\end{align*}

\subsection{Class 21}\label{sec21}
{\bf Case \{010,021\}} Let $A_m(x)$ (respectively, $B_{m,j}(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(010,021)$ with a root $a_m$ (respectively, $b_{m,j}$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+x\sum_{i=j}^{m+1}B_{m+1,i}(x).
\end{align*}
Define $A(x;v)=\sum_{m\geq1}A_m(x)v^{m-1}$ and $B(x;v,u)=\sum_{m\geq1}\sum_{j=1}^mB_{m,j}(x)v^{m-1}u^{m-j}$. Then the above recurrences can be written as
\begin{align*}
A(x;v)&=\frac{x}{1-v}+\frac{x}{v}(A(x;v)-A(x;0))+xB(x;v,1),\\
B(x;v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{vu(1-u)}(B(x;v,u)-uB(x;vu,1))-\frac{x}{vu}B(x;v,0).
\end{align*}
It is not hard to see that the solution of this system is given by
\begin{align*}
A(x;v)&=\frac{(1-v)(1-\sqrt{1-4x})-2x}{2(v^2-v+x)},\\
B(x;v,u)&=\frac{(1-vu)(1-\sqrt{1-4x})-2x}{2(u^2v^2-uv+x)(1-v)}.
\end{align*}
Hence, $A(x;0)=\frac{1-\sqrt{1-4x}}{2x}-1$.

{\bf Case \{011,201\}} The aim here is to prove
$$\v \A_n(011,201) \v = C_n\textrm{, the $n$th Catalan number}.$$
Weak ascent sequences avoiding $011$ and $201$ can be characterized as weak ascent sequences in which all the nonzero entries are distinct, to avoid 011, and after each descent $b>a$, no later entry lies in the interval $[a+1,b-1]$, to avoid $201$.

We will define a bijection $\phi$ from $\A_n(011,201)$ to Dyck $n$-paths, well known to be counted by $C_n$.
First we recall some nomenclature about Dyck paths. A Dyck $n$-path is a lattice path of $n$
upsteps $U=(1,1)$ and  $n$ downsteps $D=(1,-1)$ that stays weakly
above the horizontal line, called \emph{\gl,} that joins its initial
and terminal points (vertices). We have the usual notions of size (number of upsteps), ascent (maximal sequence of contiguous upsteps), descent, height (above \gl), valley (contiguous $DU$), valley vertex (the vertex between the $D$ and $U$), and return downstep (that returns the path to \gl). A Dyck path with exactly one return downstep is \emph{indecomposable} and every nonempty Dyck path decomposes uniquely (at its returns) into a concatenation of indecomposable Dyck subpaths, its \emph{components}. To \emph{elevate} a Dyck path means to prepend an upstep and append a downstep, thereby producing an indecomposable Dyck path, sometimes called an elevated Dyck path. The \emph{terminal descent} is the maximal string of contiguous downsteps that ends the path.

The \emph{matching} downstep of an upstep $U_0$ in a Dyck path is the terminal downstep of the shortest Dyck subpath that starts at $U_0$. Note that upsteps/downsteps come in matching pairs. Also, the \emph{long-Dyck matching} downstep is the terminal downstep of the longest Dyck subpath that starts at $U_0$. Of course, more than one upstep may have the same long-Dyck matching downstep.


A  $U\!DU$ incident with \gl is called a \emph{low} $U\!DU$,
otherwise a \emph{high} $U\!DU$. An upstep whose
long-Dyck matching downstep is on the terminal descent is said to \emph{reach} the terminal descent. Note that all upsteps in the last component extending from height 1 to height 2 reach the terminal descent. These notions are illustrated in Figure X below.

\begin{figure}[htp]
\begin{center}
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\psset{unit=.4cm}
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\rput(6,-.6){\textrm{{\scriptsize $\uparrow$}}}
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\rput(6,-2.2){\textrm{{\scriptsize valley vertex}}}

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\rput(4.3,1.6){\textrm{{\scriptsize downstep}}}
\rput(4.3,2.2){\textrm{{\scriptsize return}}}

\rput(14,-.6){\textrm{{\scriptsize $\uparrow$}}}
\rput(14,-1.4){\textrm{{\scriptsize \gl}}}

%\rput(19,1.5){\textrm{{\scriptsize matching}}}
\rput(19,1.3){\textrm{{\scriptsize matching steps}}}

\rput(16,1.3){\textrm{{\scriptsize $\leftarrow$}}}
\rput(22,1.3){\textrm{{\scriptsize $\rightarrow$}}}

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\rput(27.4,2.5){\textrm{{\scriptsize $D_0$}}}
\end{pspicture}
\caption{A Dyck path of height 4 with 3 components, one low $U\!DU$, 2 high $U\!DU$s, 3 $DDU\!D$s, one $DDUU$, and terminal descent of length 3; $D_0$ is the long-Dyck matching downstep of $U_0$ (and of 3 other upsteps) and $U_0$ reaches the terminal descent.}\label{figX}
\end{center}
\end{figure}

The bijection $\phi$ sends noninitial 0s to valleys, descents to 0 to $DDU\!D$s, and descents to a nonzero
integer to $DDUU$s. In particular, the total number of descents in the avoider equals the number of $DDU$s
in the Dyck path. Also, $\phi$ sends trailing 0s to trailing $U\!D$s;
in particular, an avoider ends with a positive integer if and only if the corresponding Dyck path ends $DD$.
For an avoider $w \in\A_n(011,201)$, the construction of the Dyck path $P=\phi(w)$ is iterative and
produces a label on the initial vertex and on the valley vertex of each $U\!DU$ in
$P$ in such a way that all labels are distinct, say the labels form a set $L$, and with $r$ denoting
the number of descents in $w$, the interval $[0,r+1]$ is the disjoint union of $L$ and the distinct entries in $w$.
Thus $\v L \v + \v w\backslash \{0\} \v =r+1$.
Furthermore, the initial vertex gets label $r+1$ and with $m=\max(w)\ge 0$, low $U\!DU$s (at \gl) get a label $>m$ and high $U\!DU$s get a label $<m$.
Then $\phi(P)$ is taken to be the Dyck path (without the labels).

The base case is $\phi(0)=U\!D$ with label 1 at the initial vertex.
So, with $W_k=w_1\dots w_k$, suppose $\phi(W_k)$ has been found as the $U\!DU$-labelled Dyck path $P_k,\ 1\le k<n$, and set $m=\max(W_k),\ r= $ number of descents in $W_k,\ a=w_{k+1}$. Then, to construct $P_{k+1}=\phi(w_1w_2\dots w_{k+1})$,

\begin{enumerate}
\item
If $a=0<w_k$, append $U\!D$ to $P_k$ and leave labels unchanged.

\item If $a=0=w_k$ so that $a$ introduces a new weak ascent, append $U\!D$ to $P_k$ thereby introducing
a new $U\!DU$ at \gl. Label the new $U\!DU$ with $m+1$ and increment by 1 all existing labels that exceed $m$,
leaving labels $<m$ frozen in place.

\item If $a=r+1$, delete the label $r+1$ from the initial vertex, elevate $P_k$ and add label $r+2$ to the new initial vertex, leaving other labels unchanged.

\item If $ m < a \le r$, elevate the terminal section of $P_k$ beginning at the vertex labelled $a$ (which will be at \gl) leaving label $a$ at \gl, then  increment by 1 all existing labels that are $\ge a$.

\item If $ 0<a<m$, insert $DU$ at the vertex labelled $a$ (which will be above \gl in the last component of $P_k$), locate the matching downstep $D_0$, say, of the newly
introduced $U=U_0$ and transfer the subpath between $U_0$ and $D_0$ inclusive (itself a Dyck path) to the end to form a new component.
\end{enumerate}
The process is illustrated in the example below.
\begin{figure}[htp]
\begin{center}
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\psline(0,0)(1,1)(2,0)(3,1)(4,2)(5,3)(6,4)(7,3)(8,4)(9,3)(10,2)(11,3)(12,2)(13,1)(14,0)(15,1)(16,2)(17,1)(18,2)(19,1)(20,0)
\psdots(0,0)(1,1)(2,0)(3,1)(4,2)(5,3)(6,4)(7,3)(8,4)(9,3)(10,2)(11,3)(12,2)(13,1)(14,0)(15,1)(16,2)(17,1)(18,2)(19,1)(20,0)
\psline[linestyle=dashed](0,0)(20,0)

\rput(0,-.6){\textrm{{\scriptsize $8$}}}

\rput(7,2.5){\textrm{{\scriptsize $1$}}}
\rput(2,-.6){\textrm{{\scriptsize $7$}}}
\rput(17,.5){\textrm{{\scriptsize $3$}}}

\rput(10,-1.8){\textrm{{\normalsize $0^3 20^3 564$}}}

\rput(10,-3.5){\textrm{{\normalsize The avoider $0^3 20^3 564$ has maximum entry $m=6$ and two eligible-to-append integers}}}

\rput(10,-4.9){\textrm{{\normalsize  that exceed 6, namely 7,8, and one that falls below 6, namely 3. The former label the}}}

\rput(9.5,-6.3){\textrm{{\normalsize origin and low $U\!DU$s. The latter label high $U\!DU$s that reach the terminal descent.}}}

\end{pspicture}

\caption{Construction of bijection $\phi$}\label{figm1}
\end{center}
\end{figure}


Next we show why this procedure works and how to invert it.
For an avoider $w\in\A_n(011,201)$ and its corresponding $U\!DU$-labeled Dyck path $P$, let $m =\max(w)$ and let $r$ denote the number of weak ascents in $w$, and say an integer $a$ is \emph{eligible} if $(w_1 \dots w_n\, a)\in\A_{n+1}(011,201)$. Then the eligible integers consist of 0 and the positive integers $<m$ that are not already in $w$ and do not lie between a descent top and a descent bottom in $w$ (equivalently, do not lie in the interval $w_{k+1}$ to $w_k$ whenever $w_k>w_{k+1}$) and the integers $m+1,m+2,\dots,r+1$.

\begin{lemma} \label{justify}
The $($maximum$)$ eligible integer $r+1$ is the label on the initial vertex, and the eligible integers $r,r-1,\dots,m+1$, in that order, are the labels on the valley vertices of the $U\!DU$s at \gl.
The eligible positive integers less than $m$
are the labels on the valley vertices of the $U\!DU$s above \gl for which the long-Dyck matching step is on the terminal descent of $P$, and these eligible positive integers occur in decreasing order left to right.
\end{lemma}
\begin{proof} The proof is by induction. We consider the various cases for appending an eligible integer to an avoider.

The result holds in case $a=0<w_n$ because there are no changes in label, the maximum entry ``$m$''  remains unchanged, all currently eligible positive integers below $m$ lose their eligiblity and all high $U\!DU$s fail to reach the terminal descent due to the newly introduced terminal $U\!D$. In case $a=0=w_n$, the only change in eligiblity is the addition of $r+2$ (due to the additional weak ascent) and the construction produces labels $m+1,\dots,r+1,r+2$ at \gl, while eligibility of integers below $m$ and labels above \gl are unchanged.

In case $a=r+1$, the integers $r,r-1,\dots,m+1$ label the low $U\!DU$s by the induction hypothesis. They remain eligible but are now less than the new maximum. Their $U\!DU$s get raised one unit, becoming high $U\!DU$s that reach the terminal descent because they are now in the last (and only) component between heights 1 and 2. Also, eligibility of integers below $m$ remains unchanged as do the high $U\!DU$ upsteps whose long-Dyck path reaches the terminal descent.

In case $m<a\le r$, the low $U\!DU$s labelled $a-1,a-2,\dots,m+1$ get raised one unit, thereby becoming high $U\!DU$s and, as in the previous case, they all reach the terminal descent. At the same time, these integers lose their eligibility. The new maximum letter in the avoider is $a$ and $a+1,a+2,\dots,r+1,r+2$ are the new labels on the low $U\!DU$s and the initial vertex. Other labels and eligibility are unchanged and so all is well.

Lastly, for the case $0<a<m$, where $a$ labels a high $U\!DU$, currently eligible positive integers between $a$ and $m$ lose their eligiblity while those below $a$ remain eligible. But the latter integers occur \emph{after} vertex $a$ and get transferred to a new last component where they still reach the terminal descent, and the former integers label $U\!DU$s that are no longer in the last component and so certainly don't reach the terminal descent.
\end{proof}
Due to Lemma \ref{justify}, the procedure works to map $\A_n(011,201)$ to Dyck $n$-paths.
To see that the map can be reversed, note that the 5 cases considered above to append an eligible integer to an existing avoider can be distinguished as follows. Case 3) occurs when the Dyck path has only
one component. Otherwise, let $C_{-2}$ denote the second last component and $C_{-1}$ the last component.
In case 1) $C_{-2}$ ends $DD$ and $C_{-1}=U\!D$, and in case 2) $C_{-2}=C_{-1}=U\!D$. This covers all
cases in which $C_{-1}=U\!D$. Case 4) elevates a segment following a $U\!D$ at \gl, so $C_{-2}=U\!D$
and $C_{-1}$ starts $UU$; in case 5) $C_{-2}$ ends $DD$ and $C_{-1}$ starts $UU$. Thus
from $\phi(w_1\dots w_{n-1}w_n)$, we can retrieve $\phi(w_1\dots w_{n-1})$ and, by induction, $w_1\dots w_{n-1}$. The eligible integers to extend $w_1\dots w_{n-1}$ can then be ascertained and, according to Lemma \ref{justify}, used to label all of the low $U\!DU$s and those of the high $U\!DU$s that reach the terminal descent in $\phi(w_1\dots w_{n-1})$. Then the actual
eligible integer $w_n$ can readily be found from the action that took $\phi(w_1\dots w_{n-1})$ to $\phi(w_1 \dots w_{n-1}w_n)$.


\section{Avoiding a triple of patterns of length $3$}\label{sectriple}
By Table \ref{tab4} there are exactly $110$ trivial and $40$ nontrivial candidate classes. The aim of this section is to show that the $40$ nontrivial candidate classes of triples of length $3$ patterns are indeed $40$ $A$-Wilf equivalences. Table \ref{tab2} describes all these $40$ nontrivial candidate classes including the corresponding succession rules of the generating trees and explicit formulas (if they exist) for the generating function. The classes 47, 63, 95, 105, 114, 140 are treated using bijective methods. The four open cases, classes 120, 123, 132, and 139, are in blue and indicated by the double question marks.


\begingroup
\allowdisplaybreaks
{\fontsize{7}{6}\selectfont
\begin{longtable}[c]{|l|l|R|L|}
\caption{Rules of generating trees for weak ascent sequences avoiding a triple of length-$3$ patterns.\label{tab3}}\\ \hline
\multicolumn{4}{|c|}{Beginning of Table \ref{tab3}}\\ \hline
Class&$B$ triple& Rules of $\T(B)$&$G_B(x)$/Reference \\ \hline
\endfirsthead  \hline
\multicolumn{4}{|c|}{Continuation of Table \ref{tab3}}\\ \hline
Class&$B$ triple& Rules of $\T(B)$&$G_B(x)$/Reference  \\ \hline
\endhead \hline
\endfoot \hline
\multicolumn{4}{|c|}{End of Table \ref{tab3}}\\ \hline\hline
\endlastfoot
2&\{000,001,010\}&$a_m\rightsquigarrow b_m,a_{m+1}$, where $a_m=01\cdots m$, $b_m=a_mm$&\\\cline{2-3}
 &\{000,001,011\}&$a_m\rightsquigarrow b_m,a_{m+1}$, where $a_m=01\cdots m$, $b_m=a_m0$&\\\cline{2-3}
 &\{001,010,011\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$&\\\cline{2-3}
 &\{001,010,012\}&&\\
 &\{001,011,012\}&$a_1\rightsquigarrow a_2,a_2$; $a_m\rightsquigarrow a_{m+1}$, where $a_m=0^m$&$\frac{x(1+x)}{1-x}$\\\hline

5&\{001,011,100\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow c_m,b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_m=b_m0$&\\\cline{2-3}
 &\{001,011,120\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_1\rightsquigarrow a_2,b_2$; $b_m\rightsquigarrow b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$&\\\cline{2-3}
 &\{001,012,100\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow c_m,b_{m+1}$, where $a_m=0^m$, $b_m=01^m$, $c_m=b_m0$&\\\cline{2-3}
 &\{001,012,110\}&$a_1\rightsquigarrow a_2,01$; $a_m\rightsquigarrow a_{m+1}$; $01\rightsquigarrow a_2,a_3$, where $a_m=0^m$&$\frac{x(1+x+x^2)}{1-x}$\\\hline

6&\{000,001,021\}&$0\rightsquigarrow 00,a_1$; $a_m\rightsquigarrow b_{m-1},c_m,a_{m+1}$; $c_m\rightsquigarrow b_m$, where $a_m=01\cdots m$, $c_m=01\cdots mm$, $b_m=c_m0$&\\\cline{2-3}
 &\{000,001,120\}&$0\rightsquigarrow 00,a_1$; $a_m\rightsquigarrow b_{m-1},c_m,a_{m+1}$; $c_m\rightsquigarrow b_m$, where $a_m=01\cdots m$, $c_m=01\cdots mm$, $b_m=c_m(m-1)$&$\frac{x(1+x+x^2+x^3)}{1-x}$\\\hline

7&\{000,001,110\}&$0\rightsquigarrow 00,a_1$; $a_m\rightsquigarrow(b_{m-1})^m,b_m,a_{m+1}$, where $a_m=01\cdots m$, $b_m=a_mm$&\\\cline{2-3}
&\{001,010,021\}&&\\
&\{001,010,100\}&&\\
&\{001,010,101\}&&\\
&\{001,010,102\}&&\\
&\{001,010,110\}&&\\
&\{001,010,120\}&&\\
&\{001,010,201\}&&\\
&\{001,010,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+2},b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$&\\\cline{2-3}
&\{001,011,021\}&&\\
&\{001,011,101\}&&\\
&\{001,011,102\}&&\\
&\{001,011,110\}&&\\
&\{001,011,201\}&&\\
&\{001,011,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+1},b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$&\\\cline{2-3}
&\{001,012,021\}&&\\
&\{001,012,101\}&&\\
&\{001,012,102\}&&\\
&\{001,012,120\}&&\\
&\{001,012,201\}&&\\
&\{001,012,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+1},b_{m+1}$, where $a_m=0^m$, $b_m=01^m$&$\frac{x}{(1-x)^2}$\\\hline

10&\{000,001,100\}&&\\
&\{000,001,101\}&&\\
&\{000,001,102\}&&\\
&\{000,001,201\}&$0\rightsquigarrow 00,a_1$; $a_m\rightsquigarrow b_{m-1,0},\ldots,b_{m-1,m-2},c_{m-1},c_m,a_{m+1}$; $b_{m,j}\rightsquigarrow b_{m,0},\ldots,b_{m,j-1}$; $c_m\rightsquigarrow b_{m,0},\ldots,b_{m,m-1}$, where $a_m=01\cdots m$, $b_{m,j}=a_mmj$, $c_m=a_mm$&\\\cline{2-3}
&\{010,011,012\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\cdots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},\ldots,b_{m,j-1}$&$\frac{x(1+x)}{1-x-x^2}$\\\hline
%1,2,3,5,8,13,21,34,55,89\\\hline

21&\{000,012,100\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,002$; $01\rightsquigarrow010,001$; $001\rightsquigarrow0011$; $002\rightsquigarrow0021,0022$; $010\rightsquigarrow0021$; $0022\rightsquigarrow0011$&\\\cline{2-3}
&\{000,012,102\}&&\\
&\{000,012,120\}&&\\
&\{000,012,201\}&&\\
&\{000,012,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,002$; $01\rightsquigarrow010,001$; $001\rightsquigarrow0011$; $002\rightsquigarrow001,0022$; $010\rightsquigarrow0101$; $0022\rightsquigarrow00221$; $00221\rightsquigarrow 002211$&$x+2x^2+4x^3+5x^4+2x^6+x^7$\\\hline

22&\{000,011,021\}&$a_m\rightsquigarrow b_m,a_{m+1}$; $b_0\rightsquigarrow d_1,c_2$; $b_m\rightsquigarrow c_{m+1}$; $c_m\rightsquigarrow c_{m+1}$; $d_m\rightsquigarrow d_{m+1},c_{m+2}$, where $a_m=01\cdots m$, $b_m=a_m0$, $c_m=0a_{m-2}m$, $d_m=0a_m$&\\\cline{2-3}
&\{000,011,102\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow a_1,b_2$; $01\rightsquigarrow010,b_2$; $a_m\rightsquigarrow a_{m+1},b_{m+2}$; $b_m\rightsquigarrow c_m,b_{m+1}$, where $a_m=0012\cdots m$, $b_m=a_{m-2}m$, $c_m=b_m(m-1)$&\\\cline{2-3}
&\{000,011,120\}&$0\rightsquigarrow a_0,01$; $01\rightsquigarrow010,012$; $010\rightsquigarrow012$; $012\rightsquigarrow d_3$; $a_m\rightsquigarrow a_{m+1},b_{m+2}$; $b_m\rightsquigarrow c_m,d_{m+1}$; $c_m\rightsquigarrow d_{m+1}$; $d_m\rightsquigarrow d_{m+1}$, where $a_m=001\cdots m$, $b_m=a_{m-2}m$, $c_m=b_m(m-1)$, $d_m=a_{m-3}(m-1)m$&\\\cline{2-3}
&\{001,021,100\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_1\rightsquigarrow c_1,b_2,d_2$; $b_m\rightsquigarrow c_m,b_{m+1}$; $d_m\rightsquigarrow c_m,b_{m+1},d_{m+1}$, where $a_m=0^m$, $b_m=01^m$, $c_m=b_m0$, $d_m=01\cdots m$&\\\cline{2-3}
&\{001,021,110\}&&\\
&\{001,110,120\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+1},a_{m+2},b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$&\\\cline{2-3}
&\{001,021,120\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_1\rightsquigarrow a_2,b_2,c_2$; $b_m\rightsquigarrow a_{m+1},b_{m+1}$; $c_m\rightsquigarrow a_{m+2},c_{m+1}$, where $a_m=0^m$, $b_m=01^m$, $c_m=01\cdots m$&\\\cline{2-3}
&\{001,100,110\}&$a_1\rightsquigarrow a_2,b_1$; $b_m\rightsquigarrow(c_m)^m,a_{m+2},b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_m=b_m0$&\\\cline{2-3}
&\{001,100,120\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow c_m,d_{m+1},b_{m+1}$; $d_m\rightsquigarrow c_m,d_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_m=01^m0$, $d_m=01^m$&$\frac{x(1+x^2)}{(1-x)^2}$\\\hline
%1,2,4,6,8,10,12,14,16,18\\\hline

23&\{000,011,100\}&&\\
&\{000,011,101\}&&\\
&\{000,011,110\}&&\\
&\{000,011,201\}&&\\
&\{000,011,210\}&$0\rightsquigarrow a_0,01$; $01\rightsquigarrow010$; $010\rightsquigarrow c_2$; $a_m\rightsquigarrow a_{m+1},b_{m+2}$; $b_m\rightsquigarrow c_m,b_{m+1}$; $c_m\rightsquigarrow c_{m+1}$, where $a_m=001\cdots m$, $b_m=a_{m-2}m$, $c_m=b_m(m-1)$&\\\cline{2-3}
&\{001,021,101\}&&\\
&\{001,021,102\}&&\\
&\{001,021,201\}&&\\
&\{001,021,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+1},c_{m+1},b_{m+1}$; $c_m\rightsquigarrow a_{m+1},c_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_m=01^m$&\\\cline{2-3}
&\{001,100,210\}&$a_1\rightsquigarrow a_2,d_1$; $a_m\rightsquigarrow a_{m+1}$; $d_m\rightsquigarrow(c_m)^m,b_{2,m},d_{m+1}$; $b_{i,j}\rightsquigarrow(c_{i+j-1})^j,b_{i+1,j}$, where $a_m=0^m$, $b_{i,j}=01\cdots(j-1)j^i$, $c_m=01^m0$, $d_m=01\cdots m$&\\\cline{2-3}
&\{001,101,110\}&&\\
&\{001,102,110\}&&\\
&\{001,110,201\}&&\\
&\{001,110,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow(a_{m+1})^m,a_{m+2},b_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$&\\\cline{2-3}
&\{001,101,120\}&&\\
&\{001,102,120\}&&\\
&\{001,120,201\}&&\\
&\{001,120,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+1},c_{m+1},b_{m+1}$; $c_m\rightsquigarrow a_{m+1},c_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_m=01^m$&\\\cline{2-3}
&\{010,012,021\}&&\\
&\{011,012,021\}&$a_m\rightsquigarrow a_{m+1},(b_m)^m$; $b_m\rightsquigarrow b_{m+1}$, where $a_m=0^m$, $b_m=a_m1$&$\frac{x(1-x+x^2)}{(1-x)^3}$\\\hline
%&1,2,4,7,11,16,22,29,37,46\\\hline

24&\{001,100,101\}&&\\
&\{001,100,102\}&&\\
&\{001,100,201\}&$a_1\rightsquigarrow a_2,d_1$; $a_m\rightsquigarrow a_{m+1}$; $d_m\rightsquigarrow c_{m,1},\ldots,c_{1,m},b_{2,m},d_{m+1}$; $b_{i,j}\rightsquigarrow c_{i+j-1,1},\ldots,c_{i,j},b_{i+1,j}$; $c_{i,j}\rightsquigarrow c_{i+j-1,1},\ldots,c_{i+1,j-1}$, where $a_m=0^m$, $d_m=01\cdots m$, $b_{i,j}=01\cdots(j-1)j^i$, $c_{i,j}=01\cdots(j-1)j^i(j-1)$&$\frac{x}{(1-x)(1-x-x^2)}$\\\hline
%1,2,4,7,12,20,33,54,88,143\\\hline

26&\{001,101,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+2},c_{m+1},b_{m+1}$; $c_m\rightsquigarrow a_{m+1},c_{m+1}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_m=01^m$&\\\cline{2-3}
&\{001,102,210\}&&\\
&\{001,201,210\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow(a_{m+1})^m,c_{2,m},b_{m+1}$; $c_{i,j}\rightsquigarrow(a_{i+j})^j,c_{i+1,j}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_{ij}=01\cdots(j-1)j^i$&\\\cline{2-3}
&\{011,012,210\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,j},(c_m)^{j-1}$, where $a_m=0^m$, $b_m=a_mj$, $c_m=a_m21$&$\frac{x(1-2x+2x^2)}{(1-x)^4}$\\\hline
%1,2,4,8,15,26,42,64,93,130\\\hline

27&\{010,012,100\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow c_{m,2},\ldots,c_{m,j-1},b_{m+1,j}$; $c_{m,j}\rightsquigarrow c_{m,2},\ldots,c_{m,j-1}$, where $a_m=0^m$, $b_m=a_mj$, $c_{m,j}=a_mj(j-1)$&\\\cline{2-3}
&\{010,012,110\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},\ldots,b_{m,j-1},b_{m+1,j}$, where $a_m=0^m$, $b_m=a_mj$&\\\cline{2-3}
&\{011,012,201\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},b_{m,1},\ldots,b_{m,j-1}$, where $a_m=0^m$, $b_m=a_mj$&$\frac{x(1-x+x^3)}{(1-x)^2(1-x-x^2)}$\\\hline
%1,2,4,8,15,27,47,80,134,222\\\hline

29&\{001,101,102\}&&\\
&\{001,101,201\}&&\\
&\{001,102,201\}&$a_1\rightsquigarrow a_2,b_1$; $a_m\rightsquigarrow a_{m+1}$; $b_m\rightsquigarrow a_{m+1},c_{m,1},\ldots,c_{2,m-1},c_{2,m},b_{m+1}$; $c_{i,j}\rightsquigarrow a_{i+j},c_{i+j-1,1},\ldots,c_{i+1,j-1},c_{i+1,j}$, where $a_m=0^m$, $b_m=01\cdots m$, $c_{ij}=01\cdots(j-1)j^i$&\\\cline{2-3}
&\{010,011,021\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m+1,j+1},\ldots,b_{m+1,m+1}$, where $a_m=0^m$, $b_{m,j}=a_mj$&\\\cline{2-3}
&\{010,012,101\}&&\\
&\{010,012,102\}&&\\
&\{010,012,120\}&&\\
&\{010,012,201\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},\ldots,b_{m,j-1,b_{m+1,j}}$, where $a_m=0^m$, $b_{m,j}=a_mj$&\\\cline{2-3}
&\{011,012,101\}&&\\
&\{011,012,102\}&&\\
&\{011,012,110\}&&\\
&\{011,012,120\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow b_{m,1},\ldots,b_{m,j}$, where $a_m=0^m$, $b_{m,j}=a_mj$&$\frac{x}{1-2x}$\\\hline
%1,2,4,8,16,32,64,128,256,512\\\hline

35&\{010,011,100\}&&\\
&\{010,011,101\}&&\\
&\{010,011,110\}&$\asim\{010,011\}$&\\\hline
%1,2,4,9,22,58,164,494,1577,5311\\\hline

36&\{012,021,100\}&&\\
&\{012,021,101\}&$a_m\rightsquigarrow a_{m+1},(b_m)^m$; $b_m\rightsquigarrow b_{m+1},c_m$, $c_m\rightsquigarrow c_{m+1}$, where $a_m=0^m$, $b_m=a_m1$, $c_m=b_m0$&\\\cline{2-3}
&\{012,021,110\}&$a_m\rightsquigarrow a_{m+1},(b_m)^m$; $b_m\rightsquigarrow b_{m+1},c_m$, $c_m\rightsquigarrow c_{m+1}$, where $a_m=0^m$, $b_m=a_m1$, $c_m=b_m1$&$\frac{x(1-2x+3x^2-x^3)}{(1-x)^4}$\\\hline
%1,2,5,11,21,36,57,85,121,166\\\hline

41&\{012,100,210\}&&\\ &\{012,110,210\}&Section \ref{sec3x41}&Section \ref{sec3x41}\\\hline
%1,2,5,12,26,51,92,155,247,376\\\hline

42&\{012,100,201\}&&\\
&\{012,110,201\}&Section \ref{sec3x42}&Section \ref{sec3x42}\\\hline
%1,2,5,12,26,51,93,161,269,439\\\hline

43&\{012,100,102\}&&\\
&\{012,100,120\}&$\asim\{012,100\}$&\\\hline
%1,2,5,12,27,56,110,207,378,675\\\hline

45&\{011,021,102\}&&\\
&\{012,101,201\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow c_m,b_{m,1},\ldots,b_{m,j-1},b_{m+1,j}$; $c_m\rightsquigarrow c_{m+1}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_m=a_m10$&\\\cline{2-3}
&\{012,021,102\}&&\\
&\{012,021,120\}&&\\
&\{012,021,201\}&&\\
&\{012,021,210\}&$\asim\{012,021\}$&\\\cline{2-3}
&\{012,102,110\}&&\\
&\{012,110,120\}&$\asim\{012,110\}$&$\frac{x(2x^2-2x+1)}{(1-x)^2(1-2x)}$\\\hline
%1,2,5,12,27,58,121,248,503,1014\\\hline

47&\{011,021,100\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow c_{m,j},b_{m+1,j+1},\ldots,b_{m+1,m+1}$; $c_{m,j}\rightsquigarrow c_{m+1,j+1},\ldots,c_{m+1,m+1}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=b_{m,j}0$&\\\cline{2-3}
&\{011,021,120\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow c_{m,j},d_{m,j+1},\ldots,d_{m,m+1}$; $c_{m,j}\rightsquigarrow c_{m+1,j},d_{m,j+1},\ldots,d_{m,m+1}$; $d_{m,j}\rightsquigarrow d_{m+1,j+1},\ldots,d_{m+1,m+2}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=b_{m,j}0$, $d_{m,j}=a_m1j$&\\
&\{011,102,120\}&&Section \ref{sec3x47}\\\hline
%1,2,5,12,28,64,144,320,704,1536\\\hline

48&\{012,101,102\}&&\\
&\{012,101,120\}&$\asim\{012,101\}$&$\frac{x(1-x+x^2)}{1-3x+2x^2-x^3}$\\\hline
%%1,2,5,12,28,65,151,351,816,1897\\\hline

55&\{012,102,201\}&&\\
&\{012,120,201\}&$\asim\{012,201\}$&\\\cline{2-3}
&\{012,102,210\}&&\\
&\{012,120,210\}&$\asim\{012,210\}$&Section \ref{sec13}\\\hline
%1,2,5,13,33,80,185,411,885,1862\\\hline

58&\{000,021,100\}&&\\
&\{000,021,201\}&&\\
&\{000,021,210\}&$\asim\{000,021\}$&\\\hline
%1,2,5,13,33,87,232,629,1724,4772\\\hline

60&\{011,101,102\}&&\\
&\{011,102,110\}&$\asim\{011,102\}$&Class 15 in Table \ref{tab1}\\\cline{2-3}
&\{012,102,120\}&$\asim\{012\}$&see \cite{BMR}, $\frac{x(1-x)}{1-3x+x^2}$\\\hline
%1,2,5,13,34,89,233,610,1597,4181\\\hline

63&\{011,021,101\}&&\\
&\{011,021,110\}&&\\
&\{011,021,201\}&&\\
&\{011,021,210\}&$\asim\{011,021\}$&\\\cline{2-3}
&\{011,100,201\}&&See Section \ref{sec3x63}\\\hline
%%1,2,5,13,35,97,275,794,2327,6905\\\hline

64&\{011,101,120\}&&\\
&\{011,110,120\}&$\asim\{011,120\}$&\\\hline
%1,2,5,13,35,97,275,795,2335,6953\\\hline

68&\{011,100,101\}&&\\
&\{011,100,110\}&$\asim\{011,100\}$&\\\hline
%1,2,5,13,36,106,330,1083,3734,13483\\\hline

78&\{010,021,100\}&&\\
&\{010,021,101\}&&\\
&\{010,021,102\}&&\\
&\{010,021,110\}&&\\
&\{010,021,120\}&&\\
&\{010,021,201\}&&\\
&\{010,021,210\}&$\asim\{010,021\}$&\\\cline{2-3}
&\{011,101,201\}&&\\
&\{011,110,201\}&$\asim\{011,201\}$&$\frac{1-\sqrt{1-4x}}{2x}-1$; Class 21 in Table \ref{tab1}\\\hline
%%1,2,5,14,42,132,429,1430,4862,16796\\\hline

80&\{011,101,210\}&&\\
&\{011,110,210\}&$\asim\{011,210\}$&\\\hline
%1,2,5,14,42,133,440,1508,5320,19225\\\hline

95&\{010,101,120\}&&\\
&\{010,110,120\}&&Section \ref{sec3x95}\\\hline
%1,2,5,15,50,179,675,2650,10749,44805\\\hline

105&\{010,100,201\}&&\\
&\{010,101,201\}&&Section \ref{sec3x105}\\\hline
%1,2,5,15,51,190,759,3203,14129,64645\\\hline

114&\{021,100,120\}&&\\
&\{021,101,120\}&&\\
&\{021,110,120\}&&Section \ref{sec3x114}\\\hline
%1,2,6,19,61,199,661,2234,7668,26674\\\hline

117&\{021,102,201\}&&\\
&\{021,102,210\}&$\asim\{021,102\}$&\\\hline
%1,2,6,20,66,213,683,2211,7291,24552\\\hline

119&\{021,110,201\}&&\\
&\{021,110,210\}&$\asim\{021,110\}$&\\\hline
%1,2,6,20,69,242,857,3056,10959,39493\\\hline

{\color{blue}120}&\{021,120,201\}&&\\
&\{021,120,210\}&$\asim\{021,120\}$&\\\cline{2-3}
&\{102,110,120\}& ?? &  \\\hline
%1,2,6,20,69,242,859,3080,11140,40596\\\hline

{\color{blue}123}&\{021,100,201\}&&\\
&\{021,100,210\}&$\asim\{021,100\}$&\\\cline{2-3}
&\{101,102,110\} &?? & \\\hline
%1,2,6,20,70,252,924,3432,12870,48620\\\hline

124&\{021,101,201\}&&\\
&\{021,101,210\}&$\asim\{021,101\}$&\\\hline
%%1,2,6,20,71,264,1015,4002,16094,65758\\\hline

{\color{blue}132}&\{102,110,210\}&\raisebox{-1.1mm}{??}&\\
&\{102,120,210\}& &\\\hline
%1,2,6,21,76,276,1002,3641,13261,48451\\\hline

{\color{blue}139}&\{021,201,210\}&$\asim\{021\}$&\\ \cline{2-3}
&\{100,110,201\}&??&\\
&\{101,110,201\}& ?? &\\\hline
%1,2,6,21,80,322,1347,5798,25512,114236\\\hline

140&\{100,120,201\}&&\\
&\{110,120,201\}&&Section \ref{sec3x140}\\\hline
%1,2,6,21,80,322,1350,5837,25843,116581\\\hline

150&\{100,201,210\}&&\\
&\{101,201,210\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow (c_{m,j})^j,b_{m+1,j},\ldots,b_{m+1,m+1}$; $c_{m,j}\rightsquigarrow c_{m+1,j},d_{m,j,j+1},\ldots,d_{m,j,m+1}$; $d_{m,j,i}\rightsquigarrow(c_{m+1,i})^{j-i},d_{m+1,j,i},\ldots,d_{m+1,j,m+2}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mj0$, $d_{m,j,i}=a_mj0i$&\\\hline
%%1,2,6,22,90,394,1806,8558,41586,206098\\\hline
\end{longtable}}
\subsection{Class 41}\label{sec3x41}
{\bf Case \{012,100,210\}} By our algorithm, we obtain that the tree $\T(012,100,210)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow c_{m,j},(d_m)^{j-1},b_{m+1,j},\\
c_{m,j}&\rightsquigarrow (d_{m+1})^{j-1},c_{m+1,j},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mj0$, and $d_m=a_m21$.

Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$, $D_m(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,100,210)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m,j}$, $d_m$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+xC_{m,j}(x)+(j-1)xD_m(x)+xB_{m+1,j}(x),\\
C_{m,j}(x)&=x+(j-1)xD_{m+1}(x)+xC_{m+1,j}(x),\\
D_m(x)&=x.
\end{align*}
By induction on $m,j$, we have that $D_m(x)=x$, $C_{m,j}(x)=\frac{x(1+(j-1)x)}{1-x}$, and $B_{m,j}(x)=\frac{x(1+(j-1)x)}{(1-x)^2}$.
Hence, $$A_m(x)=x+xA_{m+1}+\sum_{j=1}^m\frac{x(1+(j-1)x)}{(1-x)^2},$$
which implies that
$$A_1(x)=\frac{x(2-2x+x^2)(1-3x+4x^2-x^3)}{(1-x)^5}.$$

{\bf Case \{012,110,210\}} By our algorithm, we obtain that the tree $\T(012,110,210)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow b_{m,j},(c_m)^{j-1},c_{m+1},\\
c_{m}&\rightsquigarrow c_{m+1},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, and $c_{m}=a_m11$.

Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m}(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,110,210)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m}$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+xB_{m,j}(x)+(j-1)xC_{m}(x)+xC_{m+1}(x),\\
C_{m}(x)&=x+xC_{m+1}(x).
\end{align*}
By induction on $m,j$, we have that $C_m(x)=\frac{x}{1-x}$ and $B_{m,j}(x)=\frac{x(1+(j-1)x)}{(1-x)^2}$.
Hence, $$A_m(x)=x+xA_{m+1}+\sum_{j=1}^m\frac{x(1+(j-1)x)}{(1-x)^2},$$
which implies that
$$A_1(x)=\frac{x(2-2x+x^2)(1-3x+4x^2-x^3)}{(1-x)^5}.$$

\subsection{Class 42}\label{sec3x42}
{\bf Case \{012,100,201\}} By our algorithm, we obtain that the tree $\T(012,100,201)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow c_m,d_{m,2},\ldots,d_{m,j},b_{m+1,j},\\
c_m&\rightsquigarrow c_{m+1},\\
d_{m,j}&\rightsquigarrow e_m,d_{m,2},\ldots,d_{m,j-1},\\
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_m=a_m10$, $d_{m,j}=a_mj(j-1)$, and $e_m=a_m210$.

Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_m(x)$, $D_{m,j}(x)$, $E_m(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,100,201)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_m$, $d_{m,j}$, $e_m$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+xC_m(x)+x\sum_{i=2}^jD_{m,i}(x)+xB_{m+1,j}(x),\\
C_m(x)&=x+xC_{m+1}(x),\\
D_{m,j}(x)&=x+xE_m(x)+x\sum_{i=2}^{j-1}D_{m,i}(x),\\
E_m(x)&=x.
\end{align*}
By induction on $m,j$, we have that $E_m(x)=x$, $D_{m,j}(x)=x(1+x)^(j-1)$, $C_m(x)=\frac{x}{1-x}$, and $B_{m,j}(x)=\frac{x^3}{(1-x)^2}+\frac{x(1+x)^j}{1-x}$.
Hence, $$A_m(x)=x+xA_{m+1}+x\sum_{j=1}^m(\frac{x^3}{(1-x)^2}+\frac{x(1+x)^j}{1-x}),$$
which implies that
$$A_1(x)=\frac{x(1-3x+4x^2-x^3-2x^4)}{(1-x)^4(1-x-x^2)}.$$

{\bf Case \{012,110,201\}} By our algorithm, we obtain that the tree $\T(012,110,201)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow b_{m,1},d_{m,2},\ldots,d_{m,j},c_m,\\
c_m&\rightsquigarrow c_{m+1},\\
d_{m,j}&\rightsquigarrow c_{m-1},d_{m,2},\ldots,c_m,
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_m=a_m11$, and $d_{m,j}=a_mj(j-1)$.

Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_m(x)$, $D_{m,j}(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(012,110,201)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_m$, $d_{m,j}$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+xB_{m,1}(x)+x\sum_{i=2}^jD_{m,i}(x)+xC_m(x),\\
C_m(x)&=x+xC_{m+1}(x),\\
D_{m,j}(x)&=x+xC_{m-1}(x)+x\sum_{i=2}^{j-1}D_{m,i}(x)+xC_m(x).
\end{align*}
By induction on $m,j$, we have that $D_{m,j}(x)=\frac{x(1+x)^(j-1)}{1-x}$, $C_m(x)=\frac{x}{1-x}$, and $B_{m,j}(x)=\frac{x^3}{(1-x)^2}+\frac{x(1+x)^j}{1-x}$.
Hence, $$A_m(x)=x+xA_{m+1}+x\sum_{j=1}^m(\frac{x^3}{(1-x)^2}+\frac{x(1+x)^j}{1-x}),$$
which implies that
$$A_1(x)=\frac{x(1-3x+4x^2-x^3-2x^4)}{(1-x)^4(1-x-x^2)}.$$

\subsection{Class 47}\label{sec3x47}
{\bf  Case \{011,102,120\}}:\label{(011,102,120)}

\begin{lemma} \label{char11}
Weak ascent sequences avoiding $011, 102$ and $120$ can be characterized as weak ascent sequences in which $(i)$ all the nonzero entries are distinct, to avoid $011$, $(ii)$ the entries greater than the last entry are decreasing, to avoid $120$, and $(iii)$ in case the last entry is nonzero, the entries less than the last entry are increasing and occur after all the $0$s, in both cases to avoid $102$. \qed
\end{lemma}
For example, $w=0^9\,9\,0\,8\,0\,0\,1\,2\,7\,4\,5$ (with the superscript denoting repetition) meets all three conditions in Lemma \ref{char11} because 9\,8\,7\,5 is decreasing and 1\,2\,4\,5 is increasing, and no 0 occurs after an entry smaller than 5.
\begin{lemma}\label{3cond}
Suppose $(w_i)_{i=1}^n$ is a sequence of nonnegative integers that meets all three conditions in Lemma \ref{char11}.
Then $w$ is a weak ascent sequence provided only that its largest entry $w_j$ satisfies $w_j\le j-1$.
\end{lemma}
\begin{proof}
Suppose $w_j\le j-1$. Then since $w_1\dots w_j$ must be weakly increasing by condition (iii),
$w_1\dots w_{j-1}$ contains $j-2$ ascents and so $w_j$ satisfies the weak ascent condition.
Entries after $w_j$, being smaller than $w_j$, satisfy the weak ascent condition automatically (a fortiore, if you prefer the Latin phrase in its British spelling), and for $\ell<j$, if $w_{\ell}$ is positive, then since the entries $w_i,\ \ell\le i\le j$, are increasing by conditions (i) and (iii), we have $w_{\ell}\le \ell-1$ and it too satisfies the weak ascent condition.
\end{proof}

Lemmas \ref{char11} and \ref{3cond} permit a direct count of $A_n(011,102,120)$ by last entry.
Fix $k \in [\hspace*{.1em}0,n-2\hspace*{.1em}]$ and let us count the sequences  $w\in A_n(011,102,120)$ with $w_n=k$.
Let $X$ denote the decreasing list of entries in $w$ that exceed $k$ and $Y$ denote the increasing list of nonzero entries in $w$ that fall below $k$.
Set $j=\v X \v$. If $j\ge 1$, then $\max(X)=k+(j-1)+i$ for some $i\ge 1$, because the smallest possible set comprising $X$ is $\{k+1,\dots,k+j\}$. For $w$ to satisfy the weak ascent condition, by Lemma \ref{3cond} this maximum, and hence all $j$ elements of $X$, must occur in position $k+j+i$ or later.
So, since the last position in $w$ is occupied by $k$, there are $\binom{n-k-j-i}{j}$ choices to place the elements of $X$. Also, $X$ consists of its maximum element $k+j-1+i$ and $j-1$ elements chosen from $k+1,\dots,k+j-2+i$, giving $\binom{j+i-2}{j-1}$ choices for the elements of $X$.

All told, the number of ways to place entries greater than $k$ is
\begin{equation}\label{dblsum}
1\textrm{ (for case $j=0$) }+\sum_{j\ge 1}\sum_{i\ge 1}\binom{j+i-2}{j-1} \binom{n-k-j-i}{j}\, .
\end{equation}
The inner sum can be evaluated using identity (5.26) of Table 169 in \cite{gkp}:
\[
\hspace*{25mm} \sum_{n-q\,\le \,i\, \le\, \ell-m}\binom{\ell-i}{m}\binom{q+i}{n} =\binom{\ell+q+1}{m+n+1}
\hspace*{18mm} \textrm{integers }\ell,m,n\ge 0.
\]
This useful identity has a fairly evident combinatorial interpretation: it counts $(m+n+1)$-element subsets of $[\hspace*{.1em}q+\ell+1\hspace*{.1em}]$ by the value $q+i+1$ of the $(n+1)$-th smallest element of the subset.
Putting $\ell=n-k-j,\,q=j-2,\,m=j,\,n=j-1$ in Knuth's identity, (\ref{dblsum}) becomes
\[
\sum_{j\ge 0} \binom{n-k-1}{2j} = 2^{n-k-2}\, .
\]
Now let $u(n,k)=\v \{w\in A_n(011,102,120):\ w_n=k\}\v,\ 0\le k \le n-1$. If $k=0$ and so no entries are smaller
than $k$, the preceding analysis shows $u(n,0)=2^{n-2}$. If $k\ge1$, the entries smaller than $k$ form an
arbitrary subset of $[\hspace*{.1em}k-1\hspace*{.1em}]$, right-justified in $w$, giving $2^{k-1}$ choices. So, for $k=n-1$ (with no entries
greater than $k$) we find $u(n,n-1)=2^{(n-1)-1} = 2^{n-2}$, and for $1\le k \le n-2$,
we have $u(n,k)=2^{k-1} \times 2^{n-k-2} = 2^{n-3}$. So $\v A_n(011,102,120) \v=2^{n-2}\textrm{ [\hspace*{.1em}for $k=0$\hspace*{.1em}]} +2^{n-2}\textrm{ [\hspace*{.1em}for $k=n-1$\hspace*{.1em}]} +\sum_{k=1}^{n-2}2^{n-3}$, and we obtain the following result.

\begin{theorem}\label{11} For all $n\ge2$, $\v A_n(011,102,120) \v = (n+2)2^{n-3}$.
\end{theorem}

{\bf Case \{011,021,120\}}: \label{(011,021,120)}
The following lemma evaluating the sum of the partial sums of row $m$ in Pascal's triangle is easy to establish by reversing the order of summation.
\begin{lemma}\label{partial}
For all $m\ge 0$, $\sum_{j=0}^m\sum_{i=0}^j \binom{m}{i}= (m+2)2^{m-1}$.
\end{lemma}

\begin{lemma}\label{char12}
Weak ascent sequences avoiding $011, 021$ and $120$ can be characterized as weak ascent sequences in which $(i)$ the nonzero entries are distinct, to avoid $011$, and increasing, to avoid $021$, and $(ii)$ there are at most two runs of $($contiguous$\hspace*{.1em})$ $0$s and, if two runs are present, then the initial run of \,$0$s is separated from the second run by a single nonzero entry, for else a $120$ is present.  \qed
\end{lemma}
For example, $0\,0\,0\,2\,0\,0\,0\,0\,0\,0\,4\,5\,10\,12\,13$ of length 15 is a weak ascent sequence,
and is in  $A_{15}(011,021,120)$ because $2\,4\,5\,10\,12\,13$ is increasing and only one entry separates the two runs of 0s.

It is easy to count the members of $A_n(011,021,120) $ with only one run of 0s:
if $w \in A_n(011,021,120) $ has only one run of 0s, the segment after this run is an arbitrary subset of $[1,n-1]$ in increasing order---$2^{n-1}$ choices and they all satisfy the weak ascent condition.

In the two-run case, we can count by the separating entry and its position: let $k\in[1,n-2]$ denote the first nonzero entry in $w$ and say it is in position $j \in[k+1,n-1]$ ($j\ge k+1$ to satisfy the weak ascent condition), so that $w_j=k$, $w_{j+1}=0$, and $w_{j+2}\dots w_n$ consists of a run of zero or more 0s followed by a subset of $\{k+1,\dots,n-2\}$ in increasing order (note that the only loss of a weak ascent in $w$ occurs at
$(w_j,w_{j+1})=(k,0)$ and so only $n-1$ is ruled out as a possible entry in $w$). Thus, the $n-j-1$ entries $w_{j+2}\dots w_n$ are determined by a subset of size up to $n-j-1$ chosen from $\{k+1,\dots,n-2\}$ which is a set of size $n-2-k$, yielding
\[
\binom{n-2-k}{0}+\binom{n-2-k}{1}+ \dots + \binom{n-2-k}{n-j-1}
\]
choices for given $j,k$ and they all satisfy the weak ascent condition.

Fix $k\in[1,n-2]$, set $m=n-2-k$, and sum over $j$ to get
\begin{align*}
\sum_{j=k+1}^{n-1}\sum_{i=0}^{n-j-1}\binom{n-2-k}{i} & = \sum_{j=0}^{n-2-k}\sum_{i=0}^{n-2-k-j}\binom{n-2-k}{i} \\
 & = \sum_{j=0}^m \sum_{i=0}^{m-j}\binom{m}{i} = (m+2)2^{m-1} = (n-k)2^{n-3-k}\, ,
\end{align*}
the next to last equality by Lemma \ref{partial}.
Consequently,
\begin{align*}
\v A_n(011,021,120) \v  & = 2^{n-1}\textrm{ [\hspace*{.1em}no descents\hspace*{.1em}] } +\ \sum_{k=1}^{n-2}(n-k)2^{n-3-k}  \textrm{ [\hspace*{.1em}one descent\hspace*{.1em}]}  \\
 &  = (n + 2) 2^{n - 3} \textrm{ for $n\ge 2$,}
\end{align*}
which leads to the following theorem.
\begin{theorem}\label{12} For all $n\ge2$, $\v A_n(011,021,120) \v = (n+2)2^{n-3}$.
\end{theorem}


{\bf Case \{011,021,100\}} \label{(011,021,100)}
Weak ascent sequences avoiding 011, 021 and 100 can be characterized as weak ascent sequences in which (i) the nonzero entries are distinct, to avoid 011, and increasing, to avoid 021, and (ii) there are at most two runs of  0s for else a 100 is present and further, if two runs are present, then  the second run consists of a single 0 for the same reason. For example, $0\,0\,0\,0\,0\,0\,0 \,2\, 4\, 5\, 9\, 0 \,10\,12\:13$ of length 15 is a weak ascent sequence, and is in  $A_{15}(011,021,100)$ because $2\,4\,5\,9\,10\,12\,13$ is increasing and the second run of 0s consists of a single 0.


Here is a bijection $\phi$ from $A_n(011,021,100)$ to $ A_n(011,021,120)$. The bijection $\phi$ is the identity on the intersection $A_n(011,021,100) \cap A_n(011,021,120)$. So suppose $$w \in A_n(011,021,100)\, \backslash \,A_n(011,021,120).$$ Then $w$ has the form $0^r \,y\,0\,z$ where $r\ge 1$, $y$ is a factor of length at least 2 (else $w \in A_n(011,021,120)$), $z$ is a possibly empty factor, and $yz$ is an increasing sequence of positive integers. Define the \emph{excess} of an increasing sequence of positive integers $u_1 u_2 \dots u_m$ to be $E(u_1 u_2 \dots u_m) := \sum_{i=1}^{m}(u_{i+1}-u_{i}-1)$.
Thus $E(u_1 u_2 \dots u_m)$  is a nonnegative integer that measures the deviation of $u$ from a sequence of consecutive integers. Now let $j=r+1$ so that $a:=w_j$ is the first nonzero entry in $w$, and let $k+1$
denote the position of the last 0 in $w$ so that $b:=w_k$ is the unique descent top in $w$.
Set $e = E(y)= E(w_j w_{j+1}\dots w_k)$.  Write $y$ as $axb$ where $x$ may be empty so
that $w=0^r\,a\,x\,b\,0\,z$. Then
\[
\phi(w)= 0^{r-e}\,a\,0^{e+2}\,x\,z\,,
\]
of the form $0^{s_1}\,a\,0^{s_2}\,z'$ and is a weak ascent sequence in $A_n(011,021,120)$ that contains a 100 pattern, as we leave the reader to check.

For example, with $w=0^7 \,2\, 4\, 5\, 9\, 0 \,10\,12\:13$, we have $r=7,\,a=2,\,b=9,\,j=8,\,k=11$
and $e=(b-a)-(k-j)=7-3=4.$ So $\phi(w)=0^3\,2\,0^6\,4\,5\,10\,12\,13$. Note that $\phi$ preserves the first
nonzero entry although not necessarily its position. Also, $\phi$ discards the one descent top in
$w$ and increases the number of 0s by 1.

To invert $\phi$, the excess $e$ is evident from the length of the second run of 0s, and then $r$
can be recovered from the length of the first run of 0s. Also, $a$ is the entry bracketed by 0s.
Next, if $z'=\epsilon$, then certainly $x=\epsilon$.  Otherwise, compute $E(az_1'),E(az_1'z_2'),\dots$. Then $x=\epsilon$ if
$E(az_1')>e$ and if $E(az_1')\le e$, choose the largest $i$ for which $E(az_1'\dots z_i')\le e$. Then $x=z_1'\dots z_i'$. This determination of $x$ follows from the inequality $e=E(w_1\dots w_{k-1}w_k)<E(w_1\dots w_{k-1}w_{k+2})$ which holds because $w_k<w_{k+2}$. Continuing, $j=r+1$ and $k=i+\v x\v+1$. Lastly, $w_k=b$ can be retrieved from the identity $e=b-a-(k-j)$. So $\phi$ is invertible. \qed

Hence, we can state the following result.
\begin{theorem} For all $n\ge2$, $\v A_n(011,021,100) \v = (n+2)2^{n-3}$.
\end{theorem}

\begin{remark}
By translating the rules of the $\T(011,021,100)$ and the rules $\T(011,021,120)\ ($see Table \ref{tab3}\hspace*{.25mm}$)$ to recurrence relations, we obtain that the generating function $$G_{\{011,021,100\}}(x)=G_{\{011,021,120\}}(x)=\frac{x(1-x)^2}{(1-2x)^2}.$$
\end{remark}


\subsection{Class 63}\label{sec3x63}
The first 4 triples in this class coincide with $(011,021)$-avoiders, treated in Section \ref{011and021}. The fifth triple is  treated in Section \ref{011triple} below. First, we review $UU\!DD$-free Dyck paths. See Section \ref{sec21} for a review of Dyck path terminology.
\subsubsection{$UU\!DD$-free Dyck paths} \label{uuddfree}
The set of Dyck paths of semilength $n$ that avoid $UU\!DD$ is denoted $\B_n$, counted by sequence A082582 in \cite{oeis}.
For example, $\B_1=\{U\!D\}$, $\B_2=\{U\!DU\!D\}$, and $\B_3=\{UU\!DU\!DD,\,U\!DU\!DU\!D\}$. In the rest of this section, we introduce notions regarding $UU\!DD$-free Dyck paths that will be used in Section \ref{011triple}.
For a $UU\!DD$-free Dyck path of height 1, the \emph{key} $D$ is the first $D$. In particular, the key downstep of $U\!D$ is its only $D$.
Now suppose $P$ is a $UU\!DD$-free Dyck path of height at least 2. Then $P$ contains a $UU$ and the last $UU$ in $P$ must be immediately followed by $D$ (or it's not the last $UU$), then $U$ (or a $UU\!DD$ is present), then $D$ (or it's not the last $UU$). So we always have the segment $UU\!DU\!D$ starting with the last $UU$.
The $D$ immediately following the last $UU$ is the \emph{key} $D$ in $P$, see Figure \ref{fig B}.
A downstep is \emph{special} if it is the matching downstep of the middle $U$ in a $DUU$ pattern, otherwise it is a \emph{normal} downstep. Note that every peak downstep is normal. There is a Dyck subpath, denoted $P(S_0)$, \emph{associated} with a special downstep $S_0$: the longest subpath $Q$ such that $QU_0RS_0$ is a Dyck subpath of $P$ where $U_0$ is the matching upstep for $S_0$. Note that $Q$ and $R$ are both nonempty because $S_0$ is special.  See Figure \ref{fig B}, where the blue lines indicate the matching upstep for each special downstep.
\begin{figure}[htp]
\begin{center}
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\psline[linecolor=blue,arrowsize=2.5pt 3.5]{<-}(2.8,.5)(17.2,.5)
\psline[linecolor=blue,arrowsize=2.5pt 3.5]{<-}(7.8,1.5)(14.2,1.5)

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\rput(9.9,4){\textrm{{\normalsize key $D$}}}
\rput(9.6,3.1){$\downarrow$}
\rput(6.8,2.1){$\searrow$}
\rput(7.8,2.8){$\searrow$}
\rput(10,-1.2){$P(S_1)=U\!DU\!D,\ P(S_2)=U\!D$}

\end{pspicture}
\caption{A $UU\!DD$-free Dyck path with 2 special downsteps}\label{fig B}
\end{center}
\end{figure}


Given a $UU\!DD$-free Dyck path, label all the normal downsteps in $P$ with $0,1,2,\dots$ in the order they are visited by the following recursively-defined traversal method. For a path of semilength 1,
traversal consists of visiting its only downstep. Otherwise, visit successive $D$s starting at
the first $D$ after the key $D$ until a special $D$, say $D_0$ is encountered. If and when this happens,
traverse $P(D_0)$ ($D_0$ will remain unlabelled), then visit the key $D$ of $P(D_0)$, then continue to the next $D$ after $D_0$ and proceed until the last $D$ of $P$ is dealt with. Finally, visit the key $D$ of $P$. See Figure \ref{fig C} below for a simple example and Figure \ref{fig D} for further examples.


The \emph{leg} of a downstep in a Dyck path is the longest subpath starting at that downstep and consisting entirely of downsteps. Similarly, the leg of an upstep is the longest subpath ending at that upstep and consisting entirely of upsteps.
We say a downstep labelled $m\ge 1$ in a $UU\!DD$-free Dyck path is \emph{available} if (i) it occurs at or after the key downstep and
(ii) the leg $L_1$ of its matching upstep is longer than its own leg
or both legs have the same length and the $L_1$ leg starts the path. The same criterion holds for the downstep labelled 0 to be available except that the ``matching'' upstep for downstep 0 is taken to be the first
(rather than second) $U$ in the $U\!DU\!D$ segment ending at downstep 0. Note the key downstep is always available because it is either in a $UU\!DU$ segment or is the second step in the path.
For example, in the labelled $UU\!DD$-free Dyck path below,
the downsteps labelled 0,1,2,7,9 are available but, for example, 8 is not available because both leg lengths are 1 and the upstep leg does not start the path, and 5 is not available because it occurs before the key downstep.

\begin{figure}[htp]
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\rput(21.78,0.78){\textrm{{\scriptsize $8$}}}

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\end{pspicture}
\caption{A labelled $UU\!DD$-free Dyck path with available steps 0,1,2,7,9;
the one special downstep is unlabelled}\label{fig C}
\end{figure}




\subsubsection{011,021}\label{011and021}
Weak ascent sequences avoiding 011 and 021 are easily characterized: they are the weak ascent sequences in which the nonzero entries are (strictly) increasing. Here is a recursively defined bijection $\psi: \A_n(011,021) \to \B_{n+1}$. The base case is $\psi(0)=U\!DU\!D$. By induction, suppose given $w=(w_i)_{i=1}^n
\in \A_{n}(011,021)$ and $P=\psi\big((w_i)_{i=1}^{n-1} \big) \in \B_n$.
If $w_n=0$, append $U\!D$ to $P$ to get $\psi(w)$.
Otherwise, set $k=\#\{\textrm{weak ascents in $w$}\} -w_n\ge 0$. Leave the first $k$ components of
$P$ intact and elevate the remainder of the path to get $\psi(w)$. The procedure is illustrated below.
The heavy dot in each of the first 4 paths indicates the vertex that starts the segment of the path
that gets elevated to produce the next path.

\begin{figure}[htp]
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\rput(29.5,.5){\textrm{{\footnotesize $\rightarrow$}}}


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\rput(24,-1){\textrm{{\normalsize $000$}}}
\end{pspicture}


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\end{pspicture}
\caption{Construction of bijection $\psi$}\label{figA}
\end{figure}

The construction of $\psi$ is validated by the following lemma, established by induction.
\begin{lemma}
Suppose $w=(w_i)_{i=1}^{n}\in  \A_{n}(011,021)$. Then
\[
\#\{\textrm{components  in }\psi(w)\} =
\begin{cases}
    2+\#\{\textrm{weak ascents in $w$}\} - \max{w}, & \text{if $w_n=0$;} \\
   1+\#\{\textrm{weak ascents in $w$}\} - w_n , & \text{if $w_n\ge 1$.}
\end{cases}
\]
\end{lemma}
To (recursively) invert $\psi$, given a $UU\!DD$-free Dyck path $P$, if $P$ ends $U\!D$, then delete this $U\!D$ to get a Dyck path $Q$ and append 0 to $\psi^{-1}(Q)$. If $P$ ends $DD$, remove the first and last steps of the last component of $P$ to get $Q$ and append $m$ to $\psi^{-1}(Q)$ where $m= 2+\#\{\textrm{weak ascents in $\psi^{-1}(Q)$}\} - \#\{\textrm{components  in }P\} $.


\subsubsection{011,100,201} \label{011triple}
\textbf{Introduction.}
A weak ascent sequence avoids 011, 100, and 201 iff (i) the nonzero entries are distinct, (ii) 0 occurs
at most once as a descent bottom, and (iii) after a descent $ba$, no later entry lies between $a$ and $b$.
For $w\in A_{n-1}(011, 100, 201)$, say a number $m$ is \emph{available} (to extend $w$) if $wm \in A_n(011, 100, 201)$.
For example, with a superscript denoting repetition, if $w=0^6\hspace*{.1em}3\hspace*{.1em}6\hspace*{.1em}4\hspace*{.1em}7\hspace*{.1em}9$, the set of available numbers $m$ is $\M=\{0, 1, 2, 8, 10\}$.
Note that 0 is available if and only if all 0s in $w$  occur at the start, and by definition of weak ascent sequence, the maximum available number is $1+\#\textrm{ weak ascents}(w)$, which is always available because $\max(w) \le \#\textrm{ weak ascents}(w)$. In the next section, we recursively define a bijection $\phi: \A_n(011,100,201) \to \B_{n+1}$.
This bijection extends $\phi(w)$ to $\phi(wm)$ in a reversible way for each available number $m$ and relies on the correspondence between available numbers for an avoider and available labels in its corresponding $UU\!DD$-free Dyck path. As will be clear from the construction, $\phi$ also gives the following correspondences between avoiders $w$ and $UU\!DD$-free Dyck paths $\phi(w)$:
\begin{itemize}
\item \# \{labelled $D$s in $\phi(w)\} = 2+\textrm{ \# weak ascents}(w)$,
\item \# \{unlabelled $D$s in $\phi(w)\} = \# DUU$s in $\phi(w)$ = \# descents in $w$,
\item the labels on $\phi(w)$ comprise the interval
$[\hspace*{.1em}0,1+\textrm{ \# weak ascents}(w)\hspace*{.1em}]$,
\item The peak downsteps, being normal, are labelled and the labels consist of 0 on the downstep following the key downstep, $1+\textrm{ \# weak ascents}(w)$ on the key downstep, and the positive entries of $w$ on the remaining peak downsteps; in particular, \# peaks in $\phi(w)=2+\#$ positive entries in $w$,
\item the last entry in $w$ is the label on the last peak $D$ in $\phi(w)$.
\end{itemize}



\textbf{The bijection $\mbf{\phi: \A_n(011,100,201) \to \B_{n+1}}$.} \label{phi}
%Here we recursively define a bijection $\phi: \A_n(011,100,201) \to \B_{n+1}$.
\begin{theorem}\label{bijphi}
$($i$\,)$ The map $\phi$, presented recursively below, is a bijection from  $\A_n(011,100,201) $ to $\B_{n+1}$, and
$($ii$\,)$ for $n\ge 1$ and $w \in \A_n(011,100,201)$ the numbers available to extend $w$ coincide with the available labels for $\phi(w)$.
\end{theorem}
\begin{proof} The base case is $\phi(0)=U\!DU\!D$ with available numbers 0,1 and available labels also 0,1.

Now suppose $n\ge 2$, $w=(w_i)_{i=1}^n \in \A_{n}(011,100,201)$ and, inductively,
 $P=\phi\big((w_i)_{i=1}^{n-1} \big) \in \B_n$, and the available set
$\M$ to append an entry to $(w_i)_{i=1}^{n-1}$ coincides with the labels on the
available set of downsteps in $P$. Set $m=w_n\in \M$. First, suppose $m$ has its maximum possible value,
namely, $1+\#\textrm{ weak ascents}(w_i)_{i=1}^{n-1}$. In this case, append $U\!D$ to $P$ to get $\phi(w)$.
Next, suppose $0<m<1+\#\textrm{ weak ascents}(w_i)_{i=1}^{n-1}$.
Let $j$ denote the position of
the last downstep in the leg of the downstep labelled $m$. If $j=2n$, that is, if the $j$-th step is the last
step in $P$, insert a $U\!D$ just before the downstep labelled $m$ to get $\phi(w)$.   Otherwise, let $k>j$
denote the position of the last $U$ in $P$ and split $P$ into 5 segments $P_1,P_2,P_3,P_4,P_5$ ending at
positions $i-1,j,k-1,k+1,2n$ respectively. Form $P_2'$ by inserting a $U\!D$ just before the downstep labelled
$m$. Then $\phi(w)=P_1P_4P_3P_2'P_5$ and the inserted $U\!D$ becomes the last peak in $\phi(w)$.
The process in case $m=0$ is similar except instead of inserting a $U\!D$ just before the downstep
labelled $0$ (which would create a $UU\!DD$), the key $U\!DU\!D$ is elevated, that is,
replaced by $UU\!DU\!DD$.
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\caption{Construction of bijection $\phi$: avoiders and the corresponding path}\label{fig D}
\end{figure}

The construction of $\phi$ is illustrated for 4 avoiders in Figure \ref{fig D} below.  In the first two paths the heavy dot marks the vertex where $U\!D$ is inserted in the construction of the next path. Similarly, in the third path where $m=0$, the two heavy dots enclose the segment of the path that gets elevated in the next path.

In particular, for $n\ge 1$,
$\phi(0^n) = U^{n}DU\!D^{n}$,
for $012\dots \overline{n-1}$ the only available numbers are 0 and $n$,
and $\phi(012\dots \overline{n-1})= (U\!D)^{n+1}$ and
for $n\ge 3$, $\phi(012\dots \overline{n-2}\,0) =(U\!D)^{n-2}UU\!DU\!DD$. % $UU\!DU\!DD$.

If $m>\max(w_i)_{i=1}^{n-1}$, the only change in availablity between $\phi\left((w_i)_{i=1}^{n-1}\right)$ and $\phi\left((w_i)_{i=1}^{n-1}m\right)$ is the loss of $m$ and gain of $1+\#\textrm{ weak ascents}(w_i)_{i=1}^{n-1}$. On the other hand, if $m<\max(w_i)_{i=1}^{n-1}$, then $m<w_{n-1}$ and the only change in availablity is the loss of the numbers in $[\hspace*{.1em}m,w_{n-1}\hspace*{.1em}]$ that were previously available.

The cases $m=0$ and $m\ge 1$ can be distinguished because only in case $m=0$ is the last descent in $\phi(w)$ immediately preceded by $UU\!DU$. The case $m=1+\#\textrm{ weak ascents}(w_i)_{i=1}^{n-1}$ is easily inverted. The schematic diagrams in Figures \ref{phiforward} and \ref{phibackward} below depict what happens in case $1\le m\le \#\textrm{ weak ascents}(w_i)_{i=1}^{n-1}$ and  how to reverse the procedure (the case $m=0$ is similar). Here, $Q$ denotes the vertex where a $U\!D$ is inserted; the straight lines depict ascents and descents, and all ascents after $Q$ consist of just one step since the last $UU$ occurs before $Q$.

\begin{figure}[htp]
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\caption{A path $\phi(w)$ with insertion vertex $Q$ to construct $\phi(wm)$}\label{phiforward}
\end{figure}
Since $m$ is available, $Q$ joins 2 downsteps. The last downstep on $Q$'s descent and its matching upstep determine $P_2$. The last peak determines $P_4$. The availability criterion ensures that $P_1$ ends with an upstep or is empty. The segments $P_3$ and $P_5$ may be empty.
The resulting path is shown in Figure \ref{phibackward}
and  $P_1,P_2,P_3,P_4,P_5$ can all be recovered: the last peak determines the insertion vertex $Q$, the segment $P_2'P_5$ starts at the beginning of the ascent that contains the matching upstep for the downstep preceding the last peak, and the last $UU$ before $P_2'P_5$ determines $P_4$. Also, $P_5$ is determined since $P_2'$ is a Dyck path. Thus we can reverse $\phi$ (recursively) and $\phi$ is a bijection.
\begin{figure}[htp]
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\caption{The path $\phi(wm)$and how to recover $\phi(w)$}\label{phibackward}
\end{figure}
\end{proof}



\textbf{A direct description of $\mbf{\phi^{-1} }$.}
Given a $UU\!DD$-free Dyck path in $\B_{n+1}$, labelled as in Section \ref{uuddfree}, $\phi^{-1}(P)$ can be extracted (in reverse order) by the following procedure, the ``extraction procedure''.
\begin{enumerate}
\item Treat the descents after the key $D$ from bottom to top. (Note that the one-step descent consisting of the key $D$ is not treated.)

\item To treat a descent, first record the label on its topmost $D$ (necessarily a normal downstep). Then, if the descent has special downsteps, say $D_1,D_2,\dots,D_r$ in order top to bottom, record the label on the key $D$ of $P(D_1),P(D_2),\dots,P(D_r)$, then perform the entire extraction procedure on $P(D_r),P(D_{r-1}),\dots,P(D_1)$ in that order.


\end{enumerate}
After treating all descents, every peak downstep except the key downstep will have its label recorded. Reverse this list of labels and prepend $m$ 0s where $m= \#\,UU$s in $P$ to get $\phi^{-1}(P)$. We give an elaborate example in Figure \ref{direct} below. \newpage


\begin{figure}[htp]
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\caption{A path $P\in \B_{30}$, $\phi^{-1}(P)=0^{13}\ 2\ 5\ 3\ 0\ 7\ 10\ 12\ 14\ 13\ 16\ 17\ 15\ 8\ 21\ 19\ 6$}\label{direct}
\end{figure}



\subsection{Class 95}\label{sec3x95}
A \emph{left to right maximum},  \emph{LRmax} for short, in a weak ascent sequence $a=a_1a_2\cdots a_n$ is an entry $a_i$ such that $a_i>a_j$ for all $j<i$. If $a_i=m$ we say $m$ is the value
of the LRmax. Thus for $a=011022131$, the LRmax are $a_1=0,a_2=1,a_5=2,a_8=3$. So any weak ascent sequence $a\in\A_n$ can be decomposed uniquely as $m_1\pi^{(1)}\cdots m_k\pi^{(k)}$, where $m_1,\ldots,m_k$ are all of the LRmax entries in $a$; thus
$0=m_1<m_2<\cdots<m_k\leq n-1$ and $m_i>\pi^{(i)}$ (entrywise). We call this the LRmax decomposition of $a$.

Let $B=\{010,101,120\}$ and $C=\{010,110,120\}$. Denote the set of all words over the alphabet $\{0,1,2,\ldots\}$ by $W^*$. Let $s=s_1s_2\cdots s_n$ be any word in $W^*$ of length $n$. Let $s=m_1\theta^{(1)}\cdots m_k\theta^{(k)}$ be the LRmax decomposition of $s$. Then $s$ avoids $B$ if and only if for all $j=1,2,\ldots,k$,
\begin{itemize}
\item $m_j\geq\theta^{(j)}>m_{j-1}$, where $m_0=0$.
\item there exists $d_j\geq0$ such that $\theta^{(j)}=m_j^{d_j}\theta^{[j]}$ and $\theta^{[j]}$ avoids $B$.
\end{itemize}
Moreover, $s$ avoids $C$ if and only if for all $j=1,2,\ldots,k$,
\begin{itemize}
\item $m_j\geq\theta^{(j)}>m_{j-1}$, where $m_0=0$.
\item there exists $d_j\geq0$ such that $\theta^{(j)}=\theta^{[j]}m_j^{d_j}$ and $\theta^{[j]}$ avoids $C$.
\end{itemize}
We define recursively $\alpha(s)$ for any word $s\in W^*$ avoids $B$  by $m_1\alpha(\theta^{[1]})m_1^{d_1}\cdots m_k\alpha(\theta^{[k]})m_k^{d_k}$. Clearly, $s$ avoids $B$ if and only if $\alpha(s)\in W^*$ avoids $C$. By restricting to weak ascent sequences, we obtain that the bijection $\alpha$ leads to $B\asim C$.

\subsection{Class 105}\label{sec3x105}
Let $B=\{010,100,201\}$.  Any weak ascent sequence $b=m_1\theta^{(1)}\cdots m_k\theta^{(k)}\in\A_n(B)$ with LRmax $m_1,\ldots,m_k$ satisfies
\begin{itemize}
\item for all $j\geq1$, all the letters in $b$ that are smaller than $m_j$ are different and decreasing.
\item for all $j\geq1$, each letter of $b$ that smaller than $m_j$ is not belong to $\{m_1,\ldots,m_{j-1}\}$.
\end{itemize}
Let $C=\{010,100,201\}$.  Any weak ascent sequence $c=m_1\pi^{(1)}\cdots m_k\pi^{(k)}\in\A_n(C)$ with LRmax $m_1,\ldots,m_k$ satisfies
\begin{itemize}
\item for all $j\geq1$, all the letters in $c$ that are smaller than $m_j$ are forms a nonincreasing sequence.
\item for all $j\geq1$, each letter of $c$ that smaller than $m_j$ is not belong to $\{m_1,\ldots,m_{j-1}\}$
\item for all $j\geq1$, $\pi^{(j)}$ can be written as $m_j^{d_j}\pi^{[j]}$ with $d_j\geq0$ and $\pi^{[j]}<m_j$.
\end{itemize}

Let $c=m_1\pi^{(1)}\cdots m_k\pi^{(k)}\in\A_n(C)$ with LRmax $m_1,\ldots,m_k$. We define $b=\alpha(c)$ to be the same $c$, where we rewrite any maximal sequence of type $ss\cdots s=s^\ell$ with $\ell\geq2$ in $\pi^{[j]}$
as $sm_p^{\ell-1}$ such that $m_{p-1}<s<m_p$ (here $m_0=0$). Clearly, $b\in\A_n(B)$.

Let $b=m_1\theta^{(1)}\cdots m_k\theta^{(k)}\in\A_n(B)$ with LRmax $m_1,\ldots,m_k$. We define $c=\beta(b)$ to be the same $b$, where we rewrite any maximal sequence of type $sm_p^{\ell-1}$ with $\ell\geq2$ in $\theta^{(j)}$ as $s^{\ell}$ such that $m_{p-1}<s<m_p$ (here $m_0=0$). Clearly, $c\in\A_n(C)$.

Hence, there is a bijection between $\A_n(B)$ and $\A_n(C)$, which implies that $B\asim C$.


\subsection{Class 114}\label{sec3x114}
{\bf Case \{021,100,120\}} By our algorithm, we obtain that the tree $\T(021,100,120)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow c_{m,j},b_{m+1,j},d_{m,j+1},\ldots,d_{m,m+1},\\
c_{m,1}&\rightsquigarrow c_{m+1,1},d_{m,2},\ldots,d_{m,m+1},\\
c_{m,j}&\rightsquigarrow d_{m,j},\ldots,d_{m,m+1},\\
d_{m,j}&\rightsquigarrow d_{m+1,j},\ldots,d_{m+1,m+2},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mj0$, and $d_{m,j}=a_m1j$.

Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$, $D_{m,j}(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(021,100,120)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m,j}$, $d_{m,j}$). Thus,
\begin{align*}
A_m(x)&=x+ xA_{m+1}(x)+x\sum_{j=1}^mB_{m,j}(x),\\
B_{m,j}(x)&=x+xC_{m,j}(x)+xB_{m+1,j}(x)+x\sum_{i=j+1}^{m+1}D_{m,i}(x),\\
C_{m,1}(x)&=x+xC_{m+1,1}+x\sum_{i=2}^{m+1}D_{m,i}(x),\\
C_{m,j}(x)&=x+x\sum_{i=j}^{m+1}D_{m,i}(x),\\
D_{m,j}(x)&=x+x\sum_{i=j}^{m+2}D_{m+1,i}(x).
\end{align*}
Define $C(x)=\frac{1-\sqrt{1-4x}}{2x}$. By induction on $m,j$, we have that
$$D_{m,j}(x)=C_{m+1,j}(x)=xC^{m+3-j}(x).$$
Thus,
$$B_{m,j}(x)=x+xB_{m+1,j}(x)+x^2C^{m+2-j}(x)+x^2\sum_{i=j+1}^{m+1}C^{m+3-i}(x),$$
which equivalent to
$$B_{m,j}(x)=xB_{m+1,j}(x)+xC^{m+1-j}(x)(1+xC(x)).$$
By iterating infinity number of times (we assume $|x|<1/4$), we obtain that $B_{m,j}(x)=x(1+xC(x))C^{m+2-j}(x)$. Hence, by the recurrence of $A_m(x)$, we have
$$A_1(x)=\frac{x}{1-x}+x\sum_{m\geq1}\sum_{j=1}^m(1+xC(x))x^mC^{m+2-j}(x),$$
which implies
$$A_1(x)=x(1+xC(x))C^2(x)-\frac{x^2C(x)}{1-x}.$$
Hence, the generating function for the number of weak ascent sequences of length $n\geq0$ that avoid $\{021,100,120\}$ is given by $1+A_1(x)=1+x(1+xC(x))C^2(x)-\frac{x^2C(x)}{1-x}$, as required.

{\bf Case \{021,101,120\}} By our algorithm, we obtain that the tree $\T(021,100,120)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow c_{m,j},b_{m+1,j},d_{m,j+1},\ldots,d_{m,m+1},\\
c_{m,j}&\rightsquigarrow c_{m+1,j},d_{m,j+1},\ldots,d_{m,m+1},\\
d_{m,j}&\rightsquigarrow d_{m+1,j},\ldots,d_{m+1,m+2},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mj0$, and $d_{m,j}=a_m1j$. Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$, $D_{m,j}(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(021,100,120)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m,j}$, $d_{m,j}$). Thus,
\begin{align*}
A_m(x)&=x+ xA_{m+1}(x)+x\sum_{j=1}^mB_{m,j}(x),\\
B_{m,j}(x)&=x+xC_{m,j}(x)+xB_{m+1,j}(x)+x\sum_{i=j+1}^{m+1}D_{m,i}(x),\\
C_{m,j}(x)&=x+xC_{m+1,j}(x)+x\sum_{i=j+1}^{m+1}D_{m,i}(x),\\
D_{m,j}(x)&=x+x\sum_{i=j}^{m+2}D_{m+1,i}(x).
\end{align*}
As in the previous case, by induction on $m,j$, we have that
$$D_{m,j}(x)=C_{m+1,j}(x)=xC^{m+3-j}(x).$$
Then, we obtain that $B_{m,j}(x)=x(1+xC(x))C^{m+2-j}(x)$. Hence, by the recurrence of $A_m(x)$, we have
$$A_1(x)=x(1+xC(x))C^2(x)-\frac{x^2C(x)}{1-x}.$$
Hence, the generating function for the number of weak ascent sequences of length $n\geq0$ that avoid $\{021,101,120\}$ is given by $1+A_1(x)=1+x(1+xC(x))C^2(x)-\frac{x^2C(x)}{1-x}$, as required.

{\bf Case \{021,110,120\}} By our algorithm, we obtain that the tree $\T(021,100,120)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow c_{m,j},d_{m,j},\ldots,d_{m,m+1},\\
c_{m,j}&\rightsquigarrow c_{m+1,j},d_{m,j},\ldots,d_{m,m+1},\\
d_{m,j}&\rightsquigarrow d_{m+1,j},\ldots,d_{m+1,m+2},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mj0$, and $d_{m,j}=a_m1j$. Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$, $D_{m,j}(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(021,100,120)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m,j}$, $d_{m,j}$). Thus,
\begin{align*}
A_m(x)&=x+ xA_{m+1}(x)+x\sum_{j=1}^mB_{m,j}(x),\\
B_{m,j}(x)&=x+xC_{m,j}(x)+x\sum_{i=j}^{m+1}D_{m,i}(x),\\
C_{m,j}(x)&=x+xC_{m+1,j}(x)+x\sum_{i=j}^{m+1}D_{m,i}(x),\\
D_{m,j}(x)&=x+x\sum_{i=j}^{m+2}D_{m+1,i}(x).
\end{align*}
As in the previous case, by induction on $m,j$, we have that
$$D_{m,j}(x)=C_{m,j}(x)=xC^{m+3-j}(x).$$
Then, we obtain that $B_{m,j}(x)=x(1+xC(x))C^{m+2-j}(x)$. Hence, by the recurrence of $A_m(x)$, we have
$$A_1(x)=x(1+xC(x))C^2(x)-\frac{x^2C(x)}{1-x}.$$
Hence, the generating function for the number of weak ascent sequences of length $n\geq0$ that avoid $\{021,101,120\}$ is given by $1+A_1(x)=1+x(1+xC(x))C^2(x)-\frac{x^2C(x)}{1-x}$, as required.

\subsection{Class 140}\label{sec3x140}
Let $B=\{110,120,201\}$. Any weak ascent sequence $b=m_1\theta^{(1)}\cdots m_k\theta^{(k)}\in\A_n(B)$ with LRmax $m_1,\ldots,m_k$ satisfies
\begin{itemize}
\item for all $j\geq1$, $m_j\geq \theta^{(j)}\geq m_{j-1}$, where $m_0=0$.
\item removing all occurrences of $m_j$ from $\theta^{(j)}$, we obtain a nonincreasing sequence.
\item $\theta^{(j)}$ can be written as $\theta^{[j]}m_j\cdots m_j$ such that $\theta^{[j]}<m_j$.
\end{itemize}

Let $C=\{100,120,201\}$. Any weak ascent sequence $a=m_1\pi^{(1)}\cdots m_k\pi^{(k)}\in\A_n(C)$ with LRmax $m_1,\ldots,m_k$ satisfies
\begin{itemize}
\item for all $j\geq1$, $m_j\geq \pi^{(j)}\geq m_{j-1}$, where $m_0=0$.
\item removing all occurrences of $m_j$ from $\pi^{(j)}$, we obtain a decreasing sequence.
\end{itemize}

To show that $B\asim C$, we present a bijection $\alpha$ between the set of weak ascent sequences in $\A_n(B)$ with LRmax $m_1,\ldots,m_k$  and the set of weak ascent sequences in $\A_n(C)$ with LRmax $m_1,\ldots,m_k$. By the above two LRmax decompositions, to define $\alpha$, we have to map bijectively $\theta^{(j)}$ to $\pi^{(j)}$. Let $\theta^{(j)}=r_1^{s_1}r_2^{s_2}\cdots r_t^{s_t}m_j^k$, where $s_1,\ldots,s_t\geq1$, $t\geq0$, $m_j>r_1>\cdots>r_t\geq m_{j-1}$, and $k\geq0$. We define $\alpha(\theta^{(j)})=m_j^{s_1-1}s_1m_j^{s_2-1}s_2\cdots m_j^{s_t-1}s_tm_j^k$. Note that $\theta^{(j)}$ avoids $B$ and $\pi^{(j)}$ avoids $C$. Hence, by extending $\alpha$ to all $\theta^{(j)}$, we obtain that $\alpha$ is a bijection as required, which implies $B\asim C$.

\section{Open cases summarized}\label{open}
Recall $\textrm{WAS}_k$ is the number of $A$-Wilf equivalence classes for weak ascent sequences avoiding $k$ distinct patterns of length three.

In Section \ref{secpairs}, we showed that $59\leq \textrm{WAS}_2\leq61.$
To establish Conjecture \ref{conj} for pairs, one must show that (see Classes 41 and 43 in Table \ref{tab1})
$$\{102,110\}\asim\{102,120\}\mbox{ and }\{101,102\}\asim\{021\}.$$
Note that $|\A_n(102,110)|$ is given by sequence A279561.

In Section \ref{sectriple}, we showed that $150\leq \textrm{WAS}_3\leq155.$
To establish Conjecture \ref{conj} for triples, one must show that (see Classes 120, 123, 132, and 139 in Table \ref{tab2})
\begin{align*}
&\{102,110,120\}\asim\{021,120\},\\
&\{101,102,110\}\asim\{021,100\},\mbox{ so both counted by $\binom{2n}{n},$}\\
&\{102,110,210\}\asim\{102,120,210\},\\
&\{100,110,201\}\asim\{101,110,201\}\asim\{021\}.
\end{align*}


\section{Further results}\label{fur}
At first glance, one might wonder whether the algorithm for generating trees can aid in finding the generating function for weak ascent sequences in $\A_n(P)$, where $P$ is a trivial class. Below, we provide (only) two examples of such trivial classes of pattern pairs. Naturally, questions about the other trivial classes may arise, but these will be addressed in a future work.
\begin{theorem}
$($1\hspace*{.15em}$)$ The generating function for the number of weak ascent sequences in $\A_n(021,100)$ is given by $$\frac{x}{\sqrt{1-4x}}.$$

$($2\hspace*{.15em}$)$ The generating function for the number of weak ascent sequences in $\A_n(021,120)$ is given by $$\frac{-2+9x-7x^2-2x^3+(2-3x-x^2)\sqrt{1-4x})}{2(1-x)(1-4x-x^2)}.$$
\end{theorem}
\begin{proof}
(1) By our algorithm, we obtain that the tree $\T(021,100)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow c_{m,j},b_{m+1,j},\ldots,b_{m+1,m+1},\\
c_{m,j}&\rightsquigarrow c_{m+1,j},\ldots,c_{m+1,m+1},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, and $c_{m,j}=a_mj0$. Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(021,100)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m,j}$). Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{i=1}^mB_{m,i}(x),\\
B_{m,j}(x)&=x+xC_{m,j}(x)+x\sum_{i=j}^{m+1}B_{m+1,i}(x),\\
C_{m,j}(x)&=x+x\sum_{i=j}^{m+1}C_{m+1,i}(x).
\end{align*}
Define $A(x;v)=\sum_{m\geq1}A_m(x)v^{m-1}$,  $B(x;v,u)=\sum_{m\geq1}\sum_{j=1}^mB_{m,j}(x)v^{m-1}u^{m-j}$, and $C(x;v,u)=\sum_{m\geq1}\sum_{j=1}^mC_{m,j}(x)v^{m-1}u^{m-j}$. Thus, the above recurrences can be written as
\begin{align*}
A(x;v)&=\frac{x}{1-v}+\frac{x}{v}(A(x;v)-A(x;0))+xB(x;v,1),\\
B(x;v,u)&=\frac{x}{(1-v)(1-vu)}+xC(x;v,u)+\frac{x}{vu(1-u)}(B(x;v,u)-uB(x;vu,1)-(1-u)B(x;v,0)),\\
C(x;v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{vu(1-u)}(C(x;v,u)-uC(x;vu,1)-(1-u)C(x;v,0)).
\end{align*}
It is not hard to check that
\begin{align*}
C(x;v,u)&=\frac{(uv-1)\sqrt{1-4x}-uv-2x+1}{2(1-v)(u^2v^2-uv+x)},\\
B(x;v,u)&=-\frac{(u^3v^3+2(x^2+x-1)u^2v^2-(3x^2+x-1)uv+3x^2-x)}{2(u^2v^2-uv+x)^2(1-v)}\\
&-\frac{(uv-1)((2x^2+4x-1)uv(uv-1)-2x^3+5x^2-x)}{2(u^2v^2-uv+x)^2\sqrt{1-4x}(1-v)}.
\end{align*}
Hence, by taking $v=x$, we obtain that
$$A(x;0)=\frac{x}{1-x}+\frac{x(1-x-\sqrt{1-4x})}{(1-x)\sqrt{1-4x}}=\frac{x}{\sqrt{1-4x}},$$
as claimed.

(1) By our algorithm, we obtain that the tree $\T(021,120)$ with a root $a_1$ has the following rules
\begin{align*}
a_m&\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m},\\
b_{m,j}&\rightsquigarrow c_{m,j},b_{m+1,j},d_{m,j+1},\ldots,d_{m,m+1},\\
c_{m,j}&\rightsquigarrow c_{m+1,j},b_{m+1,j},d_{m,j+1},\ldots,d_{m,m+1},\\
d_{m,j}&\rightsquigarrow d_{m+1,j},\ldots,d_{m+1,m+2},\\
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mj0$, and $d_{m,j}=a_m1j$. Let $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$, $D_{m,j}$) be the generating function for the number of nodes at level $n\geq1$ in the subtree of the tree $\T(021,100)$ with a root $a_m$ (respectively, $b_{m,j}$, $c_{m,j}$, $d_{m,j}$).
Define $A(x;v)=\sum_{m\geq1}A_m(x)v^{m-1}$,  $B(x;v,u)=\sum_{m\geq1}\sum_{j=1}^mB_{m,j}(x)v^{m-1}u^{m-j}$, $C(x;v,u)=\sum_{m\geq1}\sum_{j=1}^mC_{m,j}(x)v^{m-1}u^{m-j}$, and $D(x;v,u)=\sum_{m\geq1}\sum_{j=2}^{m+1}C_{m,j}(x)v^{m-1}u^{m+1-j}$. Thus, the rules of $\T(021,120)$ can be written as
\begin{align*}
A(x;v)&=\frac{x}{1-v}+\frac{x}{v}(A(x;v)-A(x;0))+xB(x;v,1),\\
B(x;v,u)&=\frac{x}{(1-v)(1-vu)}+xC(x;v,u)+\frac{x}{uv}(B(x;v,u)-B(x;v,0))\\
&+\frac{x}{1-u}(D(x;v,u)-uD(x;vu,1)),\\
C(x;v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{uv}(B(x;v,u)+C(x;v,u)-B(x;v,0)-C(x;v,0))\\
&+\frac{x}{1-u}(D(x;v,u)-uD(x;vu,1)),\\
D(x;v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{vu(1-u)}(D(x;v,u)-uD(x;vu,1))-\frac{x}{vu}D(x;v,0).
\end{align*}
It is not hard to check that
\begin{align*}
D(x;v,u)&=\frac{1-2x-uv+(uv-1)\sqrt{1-4x}}{2(1-v)(u^2v^2-uv+x)},\\
C(x;v,u)&=\frac{(1+x)\sqrt{1-4x}+3x-1}{(1-v)(1-4x-x^2)(1-2vu+\sqrt{1-4x})},\\
B(x;v,u)&=\frac{((x+3)\sqrt{1-4x}+x-1)x}{(1-v)(1-4x-x^2)(1-2vu+\sqrt{1-4x})}.
\end{align*}
Hence, by taking $v=x$, we obtain that
$$A(x;0)=\frac{x}{1-x}+xB(x;x,1)=\frac{-2+9x-7x^2-2x^3+(2-3x-x^2)\sqrt{1-4x})}{2(1-x)(1-4x-x^2)},$$
as claimed.
\end{proof}


%---------------------------------------------------------------
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\newpage
\section*{Appendix}
\begingroup
\allowdisplaybreaks
{\fontsize{6}{8}\selectfont
\begin{longtable}[c]{|l|l|l|}
\caption{Weak ascent sequences avoiding a pair of length-3 patterns.\label{tab2}}\\ \hline
\multicolumn{3}{|c|}{Beginning of Table \ref{tab2}}\\ \hline
Class&$B$ triple& $\{|\A_n(B)|\}_{n=1}^{11}$ \\ \hline
\endfirsthead  \hline
\multicolumn{3}{|c|}{Continuation of Table \ref{tab2}}\\ \hline
Class&$B$ triple& $\{|\A_n(B)|\}_{n=1}^{11}$ \\ \hline
\endhead \hline
\endfoot \hline
\multicolumn{3}{|c|}{End of Table \ref{tab2}}\\ \hline\hline
\endlastfoot
1&\{001,010\}, \{001,011\}, \{001,012\}&1,2,3,4,5,6,7,8,9,10,11\\\hline

2&\{000,001\}&1,2,3,5,8,13,21,34,55,89,144\\\hline

3&\{000,010\}&1,2,4,10,28,85,277,964,3549,13737,55684\\\hline

4&\{000,012\}&1,2,4,5,2,1,0,0,0,0,0\\\hline

5&\{000,011\}, \{001,021\}, \{001,110\}&\\
&\{001,120\}&1,2,4,7,11,16,22,29,37,46,56\\\hline

6&\{001,100\}&1,2,4,7,12,20,33,54,88,143,232\\\hline

7&\{001,210\}&1,2,4,8,15,26,42,64,93,130,176\\\hline

8&\{001,101\}, \{001,102\}, \{001,201\}&\\
&\{010,012\}, \{011,012\}&1,2,4,8,16,32,64,128,256,512,1024\\\hline

9&\{010,011\}&1,2,4,9,22,58,164,494,1577,5311,18794\\\hline

10&\{012,100\}&1,2,5,12,27,56,110,207,378,675,1186\\\hline

11&\{012,021\}, \{012,110\}&1,2,5,12,27,58,121,248,503,1014,2037\\\hline

12&\{012,101\}&1,2,5,12,28,65,151,351,816,1897,4410\\\hline

13&\{012,201\}, \{012,210\}&1,2,5,13,33,80,185,411,885,1862,3853\\\hline

14&\{000,021\}&1,2,5,13,33,87,232,629,1724,4772,13313\\\hline

15&\{011,102\}, \{012,102\}, \{012,120\}&1,2,5,13,34,89,233,610,1597,4181,10946\\\hline

16&\{011,021\}&1,2,5,13,35,97,275,794,2327,6905,20705\\\hline

17&\{011,120\}&1,2,5,13,35,97,275,795,2335,6953,20948\\\hline

18&\{011,100\}&1,2,5,13,36,106,330,1083,3734,13483,50847\\\hline

19&\{000,102\}&1,2,5,14,40,121,373,1181,3796,12391,40902\\\hline

20&\{000,120\}&1,2,5,14,41,128,415,1397,4839,17207,62561\\\hline

21&\{010,021\}, \{011,201\}&1,2,5,14,42,132,429,1430,4862,16796,58786\\\hline

22&\{011,210\}&1,2,5,14,42,133,440,1508,5320,19225,70896\\\hline

23&\{000,110\}&1,2,5,14,43,141,489,1780,6763,26715,109372\\\hline

24&\{011,101\}, \{011,110\}&1,2,5,14,43,143,510,1936,7775,32869,145665\\\hline

25&\{000,101\}&1,2,5,14,44,150,548,2127,8703,37340,167266\\\hline

26&\{000,201\}&1,2,5,15,50,182,704,2866,12147,53271,240400\\\hline

27&\{000,210\}&1,2,5,15,50,183,717,2982,13028,59417,281255\\\hline

28&\{000,100\}&1,2,5,15,50,185,742,3205,14768,72176,372183\\\hline

29&\{010,102\}&1,2,5,15,51,186,707,2763,11024,44714,183830\\\hline

30&\{010,120\}&1,2,5,15,51,189,746,3093,13350,59607,274013\\\hline

31&\{010,110\}&1,2,5,15,51,189,747,3107,13484,60705,282302\\\hline

32&\{010,100\}&1,2,5,15,51,191,775,3367,15530,75559,385802\\\hline

33&\{010,101\}&1,2,5,15,51,191,775,3368,15549,75788,388083\\\hline

34&\{010,201\}&1,2,5,15,52,201,842,3750,17538,85402,430284\\\hline

35&\{010,210\}&1,2,5,15,52,201,843,3768,17741,87271,445739\\\hline

36&\{021,102\}&1,2,6,20,66,213,683,2211,7291,24552,84305\\\hline

37&\{021,110\}&1,2,6,20,69,242,857,3056,10959,39493,142955\\\hline

38&\{021,120\}&1,2,6,20,69,242,859,3080,11140,40596,148904\\\hline

39&\{021,100\}&1,2,6,20,70,252,924,3432,12870,48620,184756\\\hline

40&\{021,101\}&1,2,6,20,71,264,1015,4002,16094,65758,272208\\\hline

41&\{102,110\}, \{102,120\}&1,2,6,21,77,287,1079,4082,15522,59280,227240\\\hline

42&\{100,102\}&1,2,6,21,80,318,1305,5487,23535,102603,453400\\\hline

43&\{021,201\}, \{021,210\}, \{101,102\}&1,2,6,21,80,322,1347,5798,25512,114236,518848\\\hline

44&\{101,120\}&1,2,6,21,80,323,1365,5990,27128,126183,600506\\\hline

45&\{100,120\}&1,2,6,21,81,333,1437,6437,29705,140428,677247\\\hline

46&\{110,120\}&1,2,6,21,81,335,1461,6646,31295,151700,753824\\\hline

47&\{101,110\}&1,2,6,21,81,336,1478,6832,32976,165428,859527\\\hline

48&\{100,110\}&1,2,6,21,82,345,1537,7169,34739,173924,895993\\\hline

49&\{100,101\}&1,2,6,21,83,359,1668,8227,42697,231671,1307753\\\hline

50&\{102,210\}&1,2,6,22,87,351,1416,5681,22660,89961,355924\\\hline

51&\{102,201\}&1,2,6,22,87,354,1465,6154,26223,113236,494870\\\hline

52&\{110,201\}&1,2,6,22,90,394,1807,8575,41768,207678,1049909\\\hline

53&\{120,201\}&1,2,6,22,90,394,1811,8645,42517,214090,1098516\\\hline

54&\{120,210\}&1,2,6,22,90,396,1841,8940,44965,232747,1233821\\\hline

55&\{101,210\}&1,2,6,22,91,409,1956,9824,51359,277692,1545338\\\hline

56&\{110,210\}&1,2,6,22,91,410,1973,10012,53094,292273,1661525\\\hline

57&\{100,201\}, \{101,201\}&1,2,6,22,92,422,2074,10754,58202,326240,1882960\\\hline

58&\{100,210\}&1,2,6,22,92,424,2106,11102,61436,353980,2110366\\\hline

59&\{201,210\}&1,2,6,23,102,495,2549,13682,75714,428882,2474573\\\hline
\end{longtable}}
\endgroup

\begingroup
\allowdisplaybreaks
{\fontsize{6}{8}\selectfont
\begin{longtable}[c]{|l|l|l|}
\caption{Weak ascent sequences avoiding a triple of length-3 patterns.\label{tab4}}\\ \hline
\multicolumn{3}{|c|}{Beginning of Table \ref{tab4}}\\ \hline
Class&$B$ triple& $\{|\A_n(B)|\}_{n=1}^{10}$ \\ \hline
\endfirsthead  \hline
\multicolumn{3}{|c|}{Continuation of Table \ref{tab4}}\\ \hline
Class&$B$ triple& $\{|\A_n(B)|\}_{n=1}^{11}$ \\ \hline
\endhead \hline
\endfoot \hline
\multicolumn{3}{|c|}{End of Table \ref{tab4}}\\ \hline\hline
\endlastfoot
1&\{000,001,012\}&1,2,2,1,0,0,0,0,0,0\\\hline

2&\{000,001,010\}, \{000,001,011\}&\\
&\{001,010,011\}, \{001,010,012\}&\\
&\{001,011,012\}&1,2,2,2,2,2,2,2,2,2\\\hline

3&\{000,011,012\}&1,2,3,1,0,0,0,0,0,0\\\hline

4&\{000,010,012\}&1,2,3,3,2,1,0,0,0,0\\\hline

5&\{001,011,100\}, \{001,011,120\}&\\
 &\{001,012,100\}, \{001,012,110\}&1,2,3,3,3,3,3,3,3,3\\\hline

6&\{000,001,021\}, \{000,001,120\}&1,2,3,4,4,4,4,4,4,4\\\hline

7&\{000,001,110\}, \{001,010,021\}&\\
&\{001,010,100\}, \{001,010,101\}&\\
&\{001,010,102\}, \{001,010,110\}&\\
&\{001,010,120\}, \{001,010,201\}&\\
&\{001,010,210\}, \{001,011,021\}&\\
&\{001,011,101\}, \{001,011,102\}&\\
&\{001,011,110\}, \{001,011,201\}&\\
&\{001,011,210\}, \{001,012,021\}&\\
&\{001,012,101\}, \{001,012,102\}&\\
&\{001,012,120\}, \{001,012,201\}&\\
&\{001,012,210\}&1,2,3,4,5,6,7,8,9,10\\\hline

8&\{000,001,210\}&1,2,3,5,7,9,11,13,15,17\\\hline

9&\{000,010,011\}&1,2,3,5,8,12,17,23,30,38\\\hline

10&\{000,001,100\}, \{000,001,101\}&\\
&\{000,001,102\}, \{000,001,201\}&\\
&\{010,011,012\}&1,2,3,5,8,13,21,34,55,89\\\hline

11&\{000,010,102\}&1,2,4,10,27,73,204,587,1716,5085\\\hline

12&\{000,010,120\}&1,2,4,10,27,76,224,686,2161,6977\\\hline

13&\{000,010,110\}&1,2,4,10,27,77,229,708,2260,7414\\\hline

14&\{000,010,100\}&1,2,4,10,27,78,241,786,2686,9587\\\hline

15&\{000,010,101\}&1,2,4,10,27,79,247,819,2859,10461\\\hline

16&\{000,010,201\}&1,2,4,10,28,85,275,936,3321,12203\\\hline

17&\{000,010,210\}&1,2,4,10,28,85,275,938,3347,12421\\\hline

18&\{000,012,021\}&1,2,4,4,0,0,0,0,0,0\\\hline

19&\{000,012,110\}&1,2,4,4,1,0,0,0,0,0\\\hline

20&\{000,012,101\}&1,2,4,4,2,1,0,0,0,0\\\hline

21&\{000,012,100\}, \{000,012,102\}&\\
&\{000,012,120\}, \{000,012,201\}&\\
&\{000,012,210\}&1,2,4,5,2,1,0,0,0,0\\\hline

22&\{000,011,021\}, \{000,011,102\}&\\
&\{000,011,120\}, \{001,021,100\}&\\
&\{001,021,110\}, \{001,021,120\}&\\
&\{001,100,110\}, \{001,100,120\}&\\
&\{001,110,120\}&1,2,4,6,8,10,12,14,16,18\\\hline

23&\{000,011,100\}, \{000,011,101\}&\\
&\{000,011,110\}, \{000,011,201\}&\\
&\{000,011,210\}, \{001,021,101\}&\\
&\{001,021,102\}, \{001,021,201\}&\\
&\{001,021,210\}, \{001,100,210\}&\\
&\{001,101,110\}, \{001,101,120\}&\\
&\{001,102,110\}, \{001,102,120\}&\\
&\{001,110,201\}, \{001,110,210\}&\\
&\{001,120,201\}, \{001,120,210\}&\\
&\{010,012,021\}, \{011,012,021\}&1,2,4,7,11,16,22,29,37,46\\\hline

24&\{001,100,101\}, \{001,100,102\}&\\
&\{001,100,201\}&1,2,4,7,12,20,33,54,88,143\\\hline

25&\{011,012,100\}&1,2,4,7,13,23,41,72,126,219\\\hline

26&\{001,101,210\}, \{001,102,210\}&\\
&\{001,201,210\}, \{011,012,210\}&1,2,4,8,15,26,42,64,93,130\\\hline

27&\{010,012,100\}, \{010,012,110\}&\\
&\{011,012,201\}&1,2,4,8,15,27,47,80,134,222\\\hline

28&\{010,012,210\}&1,2,4,8,16,31,57,99,163,256\\\hline

29&\{001,101,102\}, \{001,101,201\}&\\
&\{001,102,201\}, \{010,011,021\}&\\
&\{010,012,101\}, \{010,012,102\}&\\
&\{010,012,120\}, \{010,012,201\}&\\
&\{011,012,101\}, \{011,012,102\}&\\
&\{011,012,110\}, \{011,012,120\}&1,2,4,8,16,32,64,128,256,512\\\hline

30&\{010,011,102\}&1,2,4,9,21,51,126,316,799,2034\\\hline

31&\{000,010,021\}&1,2,4,9,21,51,127,323,835,2188\\\hline

32&\{010,011,120\}&1,2,4,9,21,51,127,324,842,2225\\\hline

33&\{010,011,201\}&1,2,4,9,22,57,154,429,1223,3550\\\hline

34&\{010,011,210\}&1,2,4,9,22,57,155,439,1286,3875\\\hline

35&\{010,011,100\}, \{010,011,101\}&\\
&\{010,011,110\}&1,2,4,9,22,58,164,494,1577,5311\\\hline

36&\{012,021,100\}, \{012,021,101\}&\\
&\{012,021,110\}&1,2,5,11,21,36,57,85,121,166\\\hline

37&\{012,100,110\}&1,2,5,11,22,39,66,108,175,283\\\hline

38&\{012,100,101\}&1,2,5,11,23,45,85,156,281,499\\\hline

39&\{012,101,110\}&1,2,5,11,23,47,95,191,383,767\\\hline

40&\{000,021,102\}&1,2,5,12,25,60,148,374,962,2511\\\hline

41&\{012,100,210\}, \{012,110,210\}&1,2,5,12,26,51,92,155,247,376\\\hline

42&\{012,100,201\}, \{012,110,201\}&1,2,5,12,26,51,93,161,269,439\\\hline

43&\{012,100,102\}, \{012,100,120\}&1,2,5,12,27,56,110,207,378,675\\\hline

44&\{012,101,210\}&1,2,5,12,27,57,113,211,373,628\\\hline

45&\{011,021,102\}, \{012,021,102\}&\\
&\{012,021,120\}, \{012,021,201\}&\\
&\{012,021,210\}, \{012,101,201\}&\\
&\{012,102,110\}, \{012,110,120\}&1,2,5,12,27,58,121,248,503,1014\\\hline

46&\{000,021,120\}&1,2,5,12,27,65,160,404,1038,2707\\\hline

47&\{011,021,100\}, \{011,021,120\}&\\
&\{011,102,120\}&1,2,5,12,28,64,144,320,704,1536\\\hline

48&\{012,101,102\}, \{012,101,120\}&1,2,5,12,28,65,151,351,816,1897\\\hline

49&\{000,021,110\}&1,2,5,12,28,67,164,411,1050,2726\\\hline

50&\{011,100,102\}&1,2,5,12,30,75,190,483,1235,3167\\\hline

51&\{011,100,120\}&1,2,5,12,30,76,197,518,1383,3737\\\hline

52&\{000,021,101\}&1,2,5,12,30,77,202,539,1458,3988\\\hline

53&\{012,201,210\}&1,2,5,13,32,73,156,318,629,1224\\\hline

54&\{000,102,120\}&1,2,5,13,32,85,223,599,1617,4410\\\hline

55&\{012,102,201\}, \{012,102,210\}&\\
&\{012,120,201\}, \{012,120,210\}&1,2,5,13,33,80,185,411,885,1862\\\hline

56&\{011,102,210\}&1,2,5,13,33,81,193,449,1025,2305\\\hline

57&\{011,102,201\}&1,2,5,13,33,82,201,489,1185,2866\\\hline

58&\{000,021,100\}, \{000,021,201\}&\\
&\{000,021,210\}&1,2,5,13,33,87,232,629,1724,4772\\\hline

59&\{011,120,201\}&1,2,5,13,34,89,233,610,1595,4161\\\hline

60&\{011,101,102\}, \{011,102,110\}&\\
&\{012,102,120\}&1,2,5,13,34,89,233,610,1597,4181\\\hline

61&\{011,120,210\}&1,2,5,13,34,90,241,651,1769,4826\\\hline

62&\{000,102,110\}&1,2,5,13,34,91,246,672,1850,5127\\\hline

63&\{011,021,101\}, \{011,021,110\}&\\
&\{011,021,201\}, \{011,021,210\}&\\
&\{011,100,201\}&1,2,5,13,35,97,275,794,2327,6905\\\hline

64&\{011,101,120\}, \{011,110,120\}&1,2,5,13,35,97,275,795,2335,6953\\\hline

65&\{011,100,210\}&1,2,5,13,35,98,284,847,2589,8085\\\hline

66&\{000,101,120\}&1,2,5,13,36,105,318,993,3183,10435\\\hline

67&\{000,101,110\}&1,2,5,13,36,105,318,995,3205,10591\\\hline

68&\{011,100,101\}, \{011,100,110\}&1,2,5,13,36,106,330,1083,3734,13483\\\hline

69&\{000,101,102\}&1,2,5,13,37,108,327,1010,3180,10155\\\hline

70&\{000,110,120\}&1,2,5,13,37,109,336,1067,3484,11640\\\hline

71&\{000,102,210\}&1,2,5,14,39,113,325,945,2747,8012\\\hline

72&\{000,100,102\}&1,2,5,14,39,115,347,1069,3351,10648\\\hline

73&\{000,102,201\}&1,2,5,14,39,116,345,1060,3289,10371\\\hline

74&\{000,100,120\}&1,2,5,14,40,121,377,1206,3943,13120\\\hline

75&\{011,201,210\}&1,2,5,14,41,123,375,1157,3603,11304\\\hline

76&\{000,120,201\}&1,2,5,14,41,127,405,1331,4465,15250\\\hline

77&\{000,120,210\}&1,2,5,14,41,127,407,1349,4587,15956\\\hline

78&\{010,021,100\}, \{010,021,101\}&\\
&\{010,021,102\}, \{010,021,110\}&\\
&\{010,021,120\}, \{010,021,201\}&\\
&\{010,021,210\}, \{011,101,201\}&\\
&\{011,110,201\}&1,2,5,14,42,132,429,1430,4862,16796\\\hline

79&\{000,110,201\}&1,2,5,14,42,132,429,1431,4872,16863\\\hline

80&\{011,101,210\}, \{011,110,210\}&1,2,5,14,42,133,440,1508,5320,19225\\\hline

81&\{000,100,110\}&1,2,5,14,42,134,448,1560,5627,20942\\\hline

82&\{000,110,210\}&1,2,5,14,43,140,480,1717,6374,24450\\\hline

83&\{000,101,210\}&1,2,5,14,43,141,488,1767,6650,25882\\\hline

84&\{000,100,101\}&1,2,5,14,43,143,504,1870,7253,29241\\\hline

85&\{011,101,110\}&1,2,5,14,43,143,510,1936,7775,32869\\\hline

86&\{000,101,201\}&1,2,5,14,44,148,528,1972,7647,30605\\\hline

87&\{000,201,210\}&1,2,5,15,49,171,621,2328,8935,34956\\\hline

88&\{000,100,201\}&1,2,5,15,49,173,644,2506,10101,41933\\\hline

89&\{000,100,210\}&1,2,5,15,49,174,657,2616,10884,47028\\\hline

90&\{010,102,120\}&1,2,5,15,50,175,627,2277,8347,30822\\\hline

91&\{010,102,110\}&1,2,5,15,50,175,628,2289,8436,31358\\\hline

92&\{010,100,102\}&1,2,5,15,50,177,650,2449,9410,36733\\\hline

93&\{010,101,102\}&1,2,5,15,50,178,662,2540,9977,39916\\\hline

94&\{010,100,120\}&1,2,5,15,50,179,674,2637,10635,43965\\\hline

95&\{010,101,120\}, \{010,110,120\}&1,2,5,15,50,179,675,2650,10749,44805\\\hline

96&\{010,100,110\}&1,2,5,15,50,179,676,2662,10849,45528\\\hline

97&\{010,101,110\}&1,2,5,15,50,179,677,2678,11009,46827\\\hline

98&\{010,100,101\}&1,2,5,15,50,181,702,2889,12533,57027\\\hline

99&\{010,102,210\}&1,2,5,15,51,185,692,2629,10076,38830\\\hline

100&\{010,102,201\}&1,2,5,15,51,185,693,2648,10277,40427\\\hline

101&\{010,110,201\}&1,2,5,15,51,188,731,2951,12254,52030\\\hline

102&\{010,120,201\}&1,2,5,15,51,188,731,2955,12317,52634\\\hline

103&\{010,110,210\}&1,2,5,15,51,188,733,2983,12570,54544\\\hline

104&\{010,120,210\}&1,2,5,15,51,188,733,2984,12583,54641\\\hline

105&\{010,100,201\}, \{010,101,201\}&1,2,5,15,51,190,759,3203,14129,64645\\\hline

106&\{010,100,210\}&1,2,5,15,51,190,760,3219,14292,66014\\\hline

107&\{010,101,210\}&1,2,5,15,51,190,760,3220,14311,66236\\\hline

108&\{010,201,210\}&1,2,5,15,52,200,825,3570,15995,73606\\\hline

109&\{021,102,110\}&1,2,6,19,57,168,506,1585,5165,17382\\\hline

110&\{021,102,120\}&1,2,6,19,58,174,528,1649,5328,17764\\\hline

111&\{021,100,102\}&1,2,6,19,59,183,580,1893,6347,21741\\\hline

112&\{021,101,102\}&1,2,6,19,60,191,619,2048,6909,23704\\\hline

113&\{021,100,110\}&1,2,6,19,61,198,651,2171,7345,25194\\\hline

114&\{021,100,120\}, \{021,101,120\}&\\
&\{021,110,120\}&1,2,6,19,61,199,661,2234,7668,26674\\\hline

115&\{021,101,110\}&1,2,6,19,62,207,703,2420,8424,29602\\\hline

116&\{021,100,101\}&1,2,6,19,63,216,759,2717,9867,36244\\\hline

117&\{021,102,201\}, \{021,102,210\}&1,2,6,20,66,213,683,2211,7291,24552\\\hline

118&\{100,102,120\}&1,2,6,20,69,240,842,2979,10628,38211\\\hline

119&\{021,110,201\}, \{021,110,210\}&1,2,6,20,69,242,857,3056,10959,39493\\\hline

120&\{021,120,201\}, \{021,120,210\}&\\
&\{102,110,120\}&1,2,6,20,69,242,859,3080,11140,40596\\\hline

121&\{101,102,120\}&1,2,6,20,69,243,869,3145,11491,42312\\\hline

122&\{100,102,110\}&1,2,6,20,70,248,891,3236,11866,43862\\\hline

123&\{021,100,201\}, \{021,100,210\}&\\
&\{101,102,110\}&1,2,6,20,70,252,924,3432,12870,48620\\\hline

124&\{021,101,201\}, \{021,101,210\}&1,2,6,20,71,264,1015,4002,16094,65758\\\hline

125&\{100,101,120\}&1,2,6,20,72,272,1067,4312,17848,75330\\\hline

126&\{101,110,120\}&1,2,6,20,72,274,1088,4470,18886,81693\\\hline

127&\{100,101,110\}&1,2,6,20,72,275,1098,4545,19395,84977\\\hline

128&\{100,101,102\}&1,2,6,20,73,280,1116,4572,19140,81510\\\hline

129&\{100,110,120\}&1,2,6,20,73,281,1128,4671,19829,85881\\\hline

130&\{102,110,201\}&1,2,6,21,75,267,951,3404,12268,44530\\\hline

131&\{102,120,201\}&1,2,6,21,76,274,979,3479,12351,43951\\\hline

132&\{102,110,210\}, \{102,120,210\}&1,2,6,21,76,276,1002,3641,13261,48451\\\hline

133&\{100,102,210\}&1,2,6,21,78,296,1133,4356,16797,64923\\\hline

134&\{101,102,210\}&1,2,6,21,78,297,1144,4433,17238,67184\\\hline

135&\{100,102,201\}&1,2,6,21,78,299,1176,4729,19378,80667\\\hline

136&\{101,102,201\}&1,2,6,21,79,311,1265,5275,22431,96900\\\hline

137&\{101,120,201\}&1,2,6,21,79,311,1270,5347,23074,101544\\\hline

138&\{101,120,210\}&1,2,6,21,79,312,1283,5459,23886,106912\\\hline

139&\{021,201,210\}, \{100,110,201\}&\\
&\{101,110,201\}&1,2,6,21,80,322,1347,5798,25512,114236\\\hline

140&\{100,120,201\}, \{110,120,201\}&1,2,6,21,80,322,1350,5837,25843,116581\\\hline

141&\{100,120,210\}&1,2,6,21,80,324,1374,6036,27263,125931\\\hline

142&\{110,120,210\}&1,2,6,21,80,324,1375,6052,27425,127273\\\hline

143&\{101,110,210\}&1,2,6,21,80,324,1377,6087,27807,130643\\\hline

144&\{100,110,210\}&1,2,6,21,81,334,1447,6514,30241,143993\\\hline

145&\{100,101,210\}&1,2,6,21,81,335,1459,6615,30980,149026\\\hline

146&\{100,101,201\}&1,2,6,21,82,346,1547,7236,35090,175268\\\hline

147&\{102,201,210\}&1,2,6,22,85,328,1253,4754,17994,68158\\\hline

148&\{110,201,210\}&1,2,6,22,89,382,1708,7871,37125,178363\\\hline

149&\{120,201,210\}&1,2,6,22,89,383,1727,8072,38777,190292\\\hline

150&\{100,201,210\}, \{101,201,210\}&1,2,6,22,90,394,1806,8558,41586,206098\\\hline
\end{longtable}}
\endgroup
\end{document}