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\begin{document}
\begin{center}
\uppercase{\bf Inversion sequences avoiding a quadruple of length-3 patterns}
\vskip 20pt
{\bf David Callan}\\
{\smallit Department of Statistics, University of Wisconsin, Madison, WI 53706, USA}\\
{\tt callan@stat.wisc.edu}\\ 
\vskip 10pt
{\bf Toufik Mansour}\\
{\smallit Department of Mathematics, University of Haifa, 3103301 Haifa, Israel}\\
{\tt tmansour@univ.haifa.ac.il}\\ 
\end{center}
\vskip 20pt
\centerline{\smallit Received: , Revised: , Accepted: , Published: } % We will fill in the dates
\vskip 30pt


\centerline{\bf Abstract}
\noindent
An \emph{inversion sequence} is a sequence of integers $e=e_0\dotsb e_n$ which satisfies $0\le e_i \leq i$, for all $i=0,1,\ldots,n$. For a set of patterns $B$, let $\I_n(B)$ be the set of inversion sequences of length $n$ that avoid all the patterns from~$B$. We say that two sets of patterns $B$ and $C$ are \emph{I-Wilf-equivalent} if $|\I_n(B)|=|\I_n(C)|$, for all $n\geq0$.
In this paper, we show that the number of I-Wilf-equivalences among quadruples of length-3 patterns is at least $212$ and at most $215$, where three open cases remain.

\pagestyle{myheadings}
\markright{\smalltt INTEGERS: 23 (2023)\hfill}
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\baselineskip=12.875pt
\vskip 30pt

\section{Introduction}
Any word $e=e_0\cdots e_n$ such that $0\le e_i \leq i$ for all $i=0,1,\ldots,n$ is called an \emph{inversion sequence} \cite{CMSW,MS2015} of length $n$. We denote the set of inversion sequences of length $n$ by $\I_n$.

We say that a word $u=u_1\dotsb u_n$ is \emph{order-isomorphic} to a word $v=v_1\dotsb v_n$ if for
every pair of indices $i,j\in[n]$, we have $u_i<u_j$ ($u_i=u_j$) if and only if $v_i<v_j$ ($v_i=v_j$). We say that the word
$w$ \emph{contains} a word $p=p_1\dotsb p_k$ if $w$ contains a subsequence of length $k$ which is order-isomorphic to~$p$. Otherwise, we say that $w$
{\em avoids} $p$. We denote the set of all inversion sequences of length $n$ that avoid a pattern $p$ by $\I_n(p)$. For a set of patterns $P$, define $\I_n(P)=\cap_{p\in P}\I_n(p)$, for all $n\geq0$. We say that two sets of patterns $P$ and $Q$ are \emph{I-Wilf-equivalent}, and we write $P\isim Q$, if $|\I_n(P)|=|\I_n(Q)|$, for all $n\neq0$.

Pattern avoidance in inversion sequences was initiated in \cite{CMSW,MS2015}. Subsequently, several researchers studied the number of I-Wilf-equivalences for single, pairs of, and triples of length-$3$ patterns:

\begin{itemize}
\item Martinez and Savage \cite{MaSa} generalized and extended the notion of pattern-avoidance for inversion sequences to triples of binary relations that lead to new conjectures and open problems. In particular, some of these results are related to quadruples of length-3 patterns as mentioned in Table \ref{tabnon4} (also, see \cite{CL}). Most conjectures in this work have been solved in \cite{AC,CL,Lin,LinKim}, where both generating functions and bijections were employed.

\item The results of \cite{B,CJL,MaSa,YanLin} determined all the I-Wilf-equivalence classes of pairs of length-3 patterns. So, it showed that there are 48 Wilf classes among 78 pairs of length-3 patterns; for a complete list of the classes with open cases in terms of enumeration see Tables 1 and 2 in \cite{YanLin}. Kotsireas, Mansour, and Yıdırım \cite{KMY} enumerated of some of these open cases (see Remark \ref{remarkkmy}).

\item Callan, Jelínek, and Mansour \cite{CJM} showed that the number of I-Wilf-equivalence classes among triples of length-3
patterns is $137$, $138$ or~$139$. In particular, it remains
to prove $\{101,102,110\}\isim\{021,100,101\}$ and $\{100,110,201\}\isim\{100,120,210\}$.
\end{itemize}


We refer the reader to \cite{CJM} and references therein. The aim of this paper is to prove the following result.

\begin{theorem}\label{thm}
The number of I-Wilf-equivalence classes among quadruple length-3 patterns is at least $212$ and at most $215$
\end{theorem}

Based on numerical results, Table \ref{tabn}, and Theorem \ref{thm}, we conjecture the following.
\begin{conjecture}\label{conj}
We conjecture
\begin{itemize}
\item Class 152: $\{010,100,102,210\}\isim\{011,201,210\}$;
\item Class 166: $\{010,100,110,201\}\isim\{010,101,120,201\}$;
\item Class 207: $\{100,101,110,201\}\isim\{101,110,120,210\}$.
\end{itemize}
\end{conjecture}

Note that we checked the conjecture up to $n=13$ and we commented on theses three cases in Table \ref{tabnon4} as ``still {\bf Open}". So if we assume that the conjecture is true, then Theorem \ref{thm} shows that there are exactly $212$ I-Wilf-equivalence classes among quadruple length-3 patterns.

\section{The strategy to prove Theorem \ref{thm}}
Define
$P=\{000,001,010,011,012,021,100,101,102,110,120,201,210\}$
to be the set of all length-3 patterns. The goal of this paper (which we don't quite achieve) is to prove there are exactly $212$ I-Wilf-equivalences among $B\subset P$ with $|B|=4$. Since the number of subsets $B$ of $P$ with $|B|=4$ is $715$, it seems impossible to reach our goal by constructing explicit bijections between classes of inversion sequences. The way out is to combine several steps as follows.
\medskip

\noindent{\bf First step}: We find all the sequences $\{|\I_n(B)|\}_{n=0}^9$, for all $B\subset P$ with $|B|=4$. 
%After sorting these sequences according to alphabet order, we list all these sequences in Table \ref{tabn}. Note that 
Table \ref{tabn} in the Appendix below divides the 715 4-subsets of $P$ into 212 classes, where the first column of this table assigns the number of the class. Theorem \ref{thm} is equivalent to proving that the classes in Table \ref{tabn} are exactly the I-Wilf-equivalences among quadruples of length-3 patterns.
\medskip

\noindent{\bf Second step}: Let $C$ be any class in Table \ref{tabn}. We say that $C$ is {\em trivial} if $C$ contains exactly one subset. Otherwise, $C$ is {\em nontrivial}. Since each trivial class in Table \ref{tabn} is an I-Wilf-equivalence, we need to consider only the nontrivial classes in Table \ref{tabn}. There are exactly 120 trivial classes, denoted by $T$ in the first column in Table \ref{tabn}. Thus, it remains to consider $212-120=92$ nontrivial classes. Table \ref{tabnon4} below contains only the 92 nontrivial classes, where in its first column we retain the class number from Table \ref{tabn}. \medskip

{\bf Third step}: Let $B$ be any set of patterns in $P$. We say that $B$ is {\em reducible} if there exists $C\subsetneqq B$ such that $\I_n(B)=\I_n(C)$, for all $n\geq0$. In this context we write $C\rsim B$. Clearly, $C\rsim B$ implies $C\isim B$.

\begin{theorem}\label{thred2}
We have
\begin{align*}
\{001\}\rsim\{001,101\},&&\{001\}\rsim\{001,102\},&&\{001\}\rsim\{001,201\},&&
\{011\}\rsim\{011,101\},\\
\{011\}\rsim\{011,110\},&&\{012\}\rsim\{012,102\},&&\{012\}\rsim\{012,120\},&&
\{021\}\rsim\{021,201\},\\
\{021\}\rsim\{021,210\}.
\end{align*}
\end{theorem}
\begin{proof}
Since the proofs are similar, we show only the equivalence $\{001\}\rsim\{001,101\}$. Clearly, $\I_n(\{001,101\})\subseteq\I_n(\{001\})$. So it remains to show that
$\I_n(\{001\})\subseteq\I_n(\{001,101\})$. Let $\pi=\pi_0\pi_1\cdots\pi_n\in\I_n(001)$ with $n\geq3$ (clearly, the statement holds for $n\leq 2$) and assume that $\pi$ contains $101$. Thus, there exists $0\leq i<j<k\leq n$ such that $\pi_i=\pi_k>\pi_j$. By induction we prove that $\pi_j\geq m$ and $i\geq m+1$, for any $m=0,1,\ldots,n-3$.

Clearly, the claim holds for $m=0$. Since $\pi_0=0$, we have $i\geq1$. If $\pi_j=0$ then $\pi$ contains $001$ (as $\pi_0\pi_j\pi_k$), so $\pi_i>\pi_j\geq 1$ and then $i\geq2$. So the claim holds for $m=1$. Assume that the claim holds for $m$ and let us prove it for $m+1$.
Suppose, for a contradiction, that $\pi\in\I_n(\{001\})$ is such that there exists $0\leq i<j<k\leq n$ with $\pi_i=\pi_k>\pi_j\geq m$ and $i\geq m+1$.
Since $\pi$ avoids $001$, all the letters left of $\pi_i$ are different, so $\pi_s=s$ for all $s=0,1,\ldots,m$. If $\pi_j=m$, then $\pi$ contains $\pi_m\pi_j\pi_k=mm\pi_k$, that is, it contains $001$. Thus, $\pi_j\geq m+1$. Since $\pi$ is an inversion sequence and $\pi_i>\pi_j$, we have $i\geq m+2$.
Hence, by induction on $m$, we have that $\pi$ satisfies the claim for $m=n-3$, thus $\pi=01\cdots n-3\pi_{n-2}\pi_{n-1}\pi_n$ and $\pi_n>\pi_{n-1}\geq n-3$. Since $\pi$ avoids $001$, we have that $\pi_{n-1}\geq n-2$, so $\pi_{n-2}>n-2$, which contradicts the fact that $\pi_s\leq s$ for all $s=0,1,\ldots,n$ ($\pi\in\I_n$). Thus, $\pi$ avoids $101$, which completes the proof.
\end{proof}

In the next theorem, we describe all the subsets of 3 patterns in $P$ reducible to subsets of $2$ patterns in $P$ that are not obtained from Theorem \ref{thred2}. Here, we omit the proof.

\begin{theorem}\label{thred3}
We have
\begin{itemize}
\item $\{000,001\}\rsim\{000,001,100\}$;
\item $\{000,011\}\rsim\{000,011,\tau\}$, for any $\tau=100,201,210$;
\item $\{000,012\}\rsim\{000,012,\tau\}$, for any $\tau=201,210$;
\item $\{000,021\}\rsim\{000,021,100\}$;
\item $\{001,010\}\rsim\{001,010,\tau\}$, for any $\tau=021,100,110,120,210$;
\item $\{001,011\}\rsim\{001,011,021\}$;
\item $\{001,012\}\rsim\{001,012,021\}$;
\item $\{001,110\}\rsim\{001,110,210\}$;
\item $\{001,120\}\rsim\{001,120,210\}$;
\item $\{010,011\}\rsim\{010,011,100\}$;
\item $\{010,012\}\rsim\{010,012,\tau\}$, for any $\tau=101,201$;
\item $\{010,021\}\rsim\{010,021,\tau\}$, for any $\tau=100,101,102,110,120$.
\end{itemize}
\end{theorem}

In the next theorem, we describe all the reducible subsets of 4 patterns in $P$ to subsets of 3 patterns in $P$ that are not obtained from Theorems \ref{thred2} and \ref{thred3}. Again, we omit the proof.
\begin{theorem}\label{thred4}
We have
\begin{itemize}
\item $\{001,011,012\}\rsim\{001,011,012,210\}$;
\item $\{001,011,100\}\rsim\{001,011,100,210\}$;
\item $\{001,012,100\}\rsim\{001,012,100,210\}$;.
\end{itemize}
\end{theorem}

Note that when we have a reducible subset of four patterns in $P$, we can consider the references \cite{YanLin,MaSa,CJM}. These references considered the I-Wilf-equivalences and enumerations of $|\I_n(B)|$ whenever $B$ is any pair or triple of patterns from $P$. In Table \ref{tabnon4}, for given $B$ in a class $C$, if the computations for $\I_n(B)$ are not simple, we write $B=B'$ and cite Theorems \ref{thred2}-\ref{thred4}, where $B$ is reducible to $B'$.
\medskip

{\bf Fourth step}: First, following \cite{KMY}, we define the generating tree (see \cite{W}) $\T(B)$. Set $\I_B=\cup_{n=0}^{\infty}\I_n(B)$.  The tree $\T(B)$ is understood to be empty if there is no inversion sequence of arbitrary length avoiding the set $B$, that is, $0\in B$.  Otherwise, the root can always be taken as $0$. Starting with this root which stays at level $0$, we construct the remainder of the nodes of the tree $\T(B)$ as follows: the children of $e_0e_1\cdots e_{n-1}\in\I_{n-1}(B)$ are obtained from the set $\{e_0e_1\cdots e_{n-1}e_n\mid e_n=0,1,\ldots,n\}$ by obeying the pattern-avoiding restrictions of the patterns in $B$.

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

We are now ready to describe the details of the fourth step which is based on the algorithm of \cite{KMY}. Let $C$ be any nontrivial class in Table \ref{tabn}. For each subset $B\in C$, we run the main algorithm of \cite{KMY}, call it Algorithm KMY, for guessing and proving (if possible) the rules of the generating tree $\T(B)$. Then, we translate these rules into a system of equations and we solve for $F_B(x)=\sum_{n\geq0}|\I_n(B)|x^{n+1}$. For examples, we refer the reader to \cite{CJM,KMY,MY2}. See Table \ref{tabnon4} for all the generating trees $\T(B)$ that we obtained and all the corresponding generating functions $F_B(x)=\sum_{n\geq0}|\I_n(B)|x^{n+1}$.

\begin{remark}\label{remarkkmy}
In \cite{KMY}, Kotsireas, Mansour, and Yıdırım suggested an algorithm for guessing and proving (if possible) the rules of the generating tree $\T(B)$. In particular, they solved six open cases (see Tables 1 and 2 in \cite{YanLin}) for such pattern classes: $\I_n(000, 021)$, $\I_n(102, 021)$, $\I_n(100,012)$, $\I_n(120,210)$, Wilf-equivalent $\I_n(011,201)$ and $\I_n(011,210)$, and Wilf-equivalent $\I_n(100, 021)$ and $\I_n(110, 021)$. Moreover, they extended the algorithm to the case of restricted growth sequences (for pattern avoidance on restricted growth sequences, see \cite{Mb}) and presented an explicit formula for the generating function for the number of restricted growth sequences of length $n$ that avoid either \{12313,12323\}, $\{12313,12323,12333\}$, or $\{123\cdots\ell1\}$.
\end{remark}

We end this section with Table \ref{tabnon4} which presents the 92 nontrivial classes (see second column). In the third and fourth columns of the table, we present the rules of the generating tree $\T(B)$ (the root is always $0$) and the corresponding generating function $F_B(x)$, whenever $B$ is any set in the first column of the table.
\vspace{10pt}

\begingroup
\allowdisplaybreaks
{\fontsize{5}{3}\selectfont
\begin{longtable}[c]{|l|l|R|L|}
\caption{Succession rules for the generating trees $\T'(B)$ and generating functions $F_B(x)$ for pattern set $B\in C$, where $C$ is a class in Table \ref{tabn}. Note that we distinguished those pattern sets $B$ with same generating tree by dashed lines and we denote the pattern set $B$ at line $d$ in class C by $C(d)$.\label{tabnon4}}\\ \hline
\multicolumn{4}{|c|}{Beginning of Table \ref{tabnon4}}\\ \hline
Class&$B$&Rules of $\T'(B)$&$F_B(x)$\\ \hline
\endfirsthead  \hline
\multicolumn{4}{|c|}{Continuation of Table \ref{tabnon4}}\\ \hline Class&$B$&Rules of $\T'(B)$&$F_B(x)$\\ \hline
\endhead \hline
\endfoot \hline
\multicolumn{4}{|c|}{End of Table \ref{tabnon4}}\\ \hline\hline
\endlastfoot
1&\{000,001,010,012\}&&\\
 &\{000,001,011,012\}&$0\rightsquigarrow 00,01$; $01\rightsquigarrow00$&$x+2x^2+x^3$\\\hline
2&\{000,001,010,011\}&&\\
 &\{001,010,011,012\}&$0\rightsquigarrow00,01$; $01\rightsquigarrow01$&$x+2x^2+\frac{x^3}{1-x}$\\\hline
4&\{000,001,012,021\}&&\\
 &\{000,001,012,100\}&&\\
 &\{000,001,012,101\}&&\\
 &\{000,001,012,102\}&&\\
 &\{000,001,012,120\}&&\\
 &\{000,001,012,201\}&&\\
 &\{000,001,012,210\}&$0\rightsquigarrow00,01$; $01\rightsquigarrow00,011$; $011\rightsquigarrow00$&\\\cdashline{2-3}
 &\{000,010,011,012\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow01,002$; $002\rightsquigarrow01$&$x+2x^2+2x^3+x^4$\\\hline
5&\{000,001,011,120\}&$0\rightsquigarrow00,01$; $01\rightsquigarrow00,012$; $012\rightsquigarrow012$&\\\cdashline{2-3}
 &\{001,011,012,100\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow010$&$x+2x^2+2x^3+\frac{x^4}{1-x}$\\\hline
6&\{000,001,010,021\}&&\\
 &\{000,001,010,100\}&&\\
 &\{000,001,010,101\}&&\\
 &\{000,001,010,102\}&&\\
 &\{000,001,010,110\}&&\\
 &\{000,001,010,120\}&&\\
 &\{000,001,010,201\}&&\\
 &\{000,001,010,210\}&&\\
 &\{000,001,011,021\}&&\\
 &\{000,001,011,100\}&&\\
 &\{000,001,011,101\}&&\\
 &\{000,001,011,102\}&&\\
 &\{000,001,011,110\}&&\\
 &\{000,001,011,201\}&&\\
 &\{000,001,011,210\}&$0\rightsquigarrow00,0$&\\\cdashline{2-3}
 &\{001,010,011,021\}&&\\
 &\{001,010,011,100\}&&\\
 &\{001,010,011,101\}&&\\
 &\{001,010,011,102\}&&\\
 &\{001,010,011,110\}&&\\
 &\{001,010,011,120\}&&\\
 &\{001,010,011,201\}&&\\
 &\{001,010,011,210\}&&\\
 &\{001,010,012,021\}&&\\
 &\{001,010,012,100\}&&\\
 &\{001,010,012,101\}&&\\
 &\{001,010,012,102\}&&\\
 &\{001,010,012,110\}&&\\
 &\{001,010,012,120\}&&\\
 &\{001,010,012,201\}&&\\
 &\{001,010,012,210\}&&\\
 &\{001,011,012,021\}&&\\
 &\{001,011,012,101\}&&\\
 &\{001,011,012,102\}&&\\
 &\{001,011,012,110\}&&\\
 &\{001,011,012,120\}&&\\
 &\{001,011,012,201\}&&\\
 &\{001,011,012,210\}&$0\rightsquigarrow(00)^2$; $00\rightsquigarrow00$&$x+\frac{2x^2}{1-x}$\\\hline
8&\{000,011,012,100\}&&\\
 &\{000,011,012,101\}&&\\
 &\{000,011,012,102\}&&\\
 &\{000,011,012,110\}&&\\
 &\{000,011,012,120\}&&\\
 &\{000,011,012,201\}&&\\
 &\{000,011,012,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,01$; $01\rightsquigarrow001$&$x+2x^2+3x^3+x^4$\\\hline
10&\{001,011,100,120\}&&\\
  &\{001,012,100,110\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow010,00$&$x+2x^2+3x^3+\frac{2x^4}{1-x}$\\\hline
11&\{000,010,012,100\}&&\\
  &\{000,010,012,110\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow01,002$; $01\rightsquigarrow011$; $002\rightsquigarrow01,011$&$x+2x^2+3x^3+3x^4+x^5$\\\hline
12&\{000,010,012,101\}&&\\
  &\{000,010,012,102\}&&\\
  &\{000,010,012,120\}&&\\
  &\{000,010,012,201\}&&\\
  &\{000,010,012,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow01,002$; $01\rightsquigarrow011$; $002\rightsquigarrow01,0022$; $0022\rightsquigarrow01$&$x+2x^2+3x^3+3x^4+2x^5+x^6$\\\hline
14&\{000,001,021,110\}&&\\
  &\{000,001,110,120\}&$0\rightsquigarrow00,01$; $01\rightsquigarrow(00)^2,01$&\\\cdashline{2-3}
  &\{001,011,021,120\}&&\\
  &\{001,011,101,120\}&&\\
  &\{001,011,102,120\}&&\\
  &\{001,011,110,120\}&&\\
  &\{001,011,120,201\}&&\\
  &\{001,011,120,210\}&&\\
  &\{001,012,021,110\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow(00)^2$&\\\cdashline{2-3}
  &\{001,011,021,100\}&&\\
  &\{001,011,100,101\}&&\\
  &\{001,011,100,102\}&&\\
  &\{001,011,100,110\}&&\\
  &\{001,011,100,201\}&&\\
  &\{001,011,100,210\}&&\\
  &\{001,012,021,100\}&&\\
  &\{001,012,100,101\}&&\\
  &\{001,012,100,102\}&&\\
  &\{001,012,100,120\}&&\\
  &\{001,012,100,201\}&&\\
  &\{001,012,100,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow010,01$&\\\cdashline{2-3}
  &\{001,012,101,110\}&&\\
  &\{001,012,102,110\}&&\\
  &\{001,012,110,120\}&&\\
  &\{001,012,110,201\}&&\\
  &\{001,012,110,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow(00)^2$&$x+2x^2+\frac{3x^3}{1-x}$\\\hline
15&\{000,001,021,100\}&&\\
  &\{000,001,021,101\}&&\\
  &\{000,001,021,102\}&&\\
  &\{000,001,021,201\}&&\\
  &\{000,001,021,210\}&&\\
  &\{000,001,100,120\}&&\\
  &\{000,001,101,120\}&&\\
  &\{000,001,102,120\}&&\\
  &\{000,001,120,201\}&&\\
  &\{000,001,120,210\}&$0\rightsquigarrow00,01$; $01\rightsquigarrow00,011,01$; $011\rightsquigarrow00$&$x+2x^2+3x^3+\frac{4x^4}{1-x}$\\\hline
16&\{000,001,100,110\}&&\\
  &\{000,001,101,110\}&&\\
  &\{000,001,102,110\}&&\\
  &\{000,001,110,201\}&&\\
  &\{000,001,110,210\}&$a_m\rightsquigarrow(00)^{m+1},a_{m+1}$, where $a_m=01\cdots m$&\\\cdashline{2-3}
  &\{000,010,011,021\}&$0\rightsquigarrow0,01$; $01\rightsquigarrow01$&\\\cdashline{2-3}
  &\{001,010,021,100\}&&\\
  &\{001,010,021,101\}&&\\
  &\{001,010,021,102\}&&\\
  &\{001,010,021,110\}&&\\
  &\{001,010,021,120\}&&\\
  &\{001,010,021,201\}&&\\
  &\{001,010,021,210\}&&\\
  &\{001,010,100,101\}&&\\
  &\{001,010,100,102\}&&\\
  &\{001,010,100,110\}&&\\
  &\{001,010,100,120\}&&\\
  &\{001,010,100,201\}&&\\
  &\{001,010,100,210\}&&\\
  &\{001,010,101,102\}&&\\
  &\{001,010,101,110\}&&\\
  &\{001,010,101,120\}&&\\
  &\{001,010,101,201\}&&\\
  &\{001,010,101,210\}&&\\
  &\{001,010,102,110\}&&\\
  &\{001,010,102,120\}&&\\
  &\{001,010,102,201\}&&\\
  &\{001,010,102,210\}&&\\
  &\{001,010,110,120\}&&\\
  &\{001,010,110,201\}&&\\
  &\{001,010,110,210\}&&\\
  &\{001,010,120,201\}&&\\
  &\{001,010,120,210\}&&\\
  &\{001,010,201,210\}&&\\
  &\{001,011,021,101\}&&\\
  &\{001,011,021,102\}&&\\
  &\{001,011,021,110\}&&\\
  &\{001,011,021,201\}&&\\
  &\{001,011,021,210\}&&\\
  &\{001,011,101,102\}&&\\
  &\{001,011,101,110\}&&\\
  &\{001,011,101,201\}&&\\
  &\{001,011,101,210\}&&\\
  &\{001,011,102,110\}&&\\
  &\{001,011,102,201\}&&\\
  &\{001,011,102,210\}&&\\
  &\{001,011,110,201\}&&\\
  &\{001,011,110,210\}&&\\
  &\{001,011,201,210\}&&\\
  &\{001,012,021,101\}&&\\
  &\{001,012,021,102\}&&\\
  &\{001,012,021,120\}&&\\
  &\{001,012,021,201\}&&\\
  &\{001,012,021,210\}&&\\
  &\{001,012,101,102\}&&\\
  &\{001,012,101,120\}&&\\
  &\{001,012,101,201\}&&\\
  &\{001,012,101,210\}&&\\
  &\{001,012,102,120\}&&\\
  &\{001,012,102,201\}&&\\
  &\{001,012,102,210\}&&\\
  &\{001,012,120,201\}&&\\
  &\{001,012,120,210\}&&\\
  &\{001,012,201,210\}&$0\rightsquigarrow00,0$; $00\rightsquigarrow 00$&\\\cdashline{2-3}
  &\{010,011,012,021\}&$a_m\rightsquigarrow a_{m+1},(01)^m$, where $a_m=0^m$&$\frac{x}{(1-x)^2}$\\\hline
17&\{000,001,100,210\}&&\\
  &\{000,001,101,210\}&&\\
  &\{000,001,102,210\}&&\\
  &\{000,001,201,210\}&$a_0\rightsquigarrow b_0,a_1$; $a_m\rightsquigarrow b_0^m,b_m,a_{m+1}$; $b_m\rightsquigarrow b_0^m$, where $a_m=01\cdots m$, $b_m=a_mm$&\\\cdashline{2-3}
  &\{000,010,011,102\}&$0\rightsquigarrow00,01$, $00\rightsquigarrow00,002$, $01\rightsquigarrow01$, $002\rightsquigarrow0021,002$&$\frac{x(1+x^3)}{(1-x)^2}$\\\hline
19&\{000,001,100,101\}&&\\
  &\{000,001,100,102\}&&\\
  &\{000,001,100,201\}&&\\
  &\{000,001,101,102\}&&\\
  &\{000,001,101,201\}&&\\
  &\{000,001,102,201\}&$a_m\rightsquigarrow a_{m+1},b_0,\ldots,b_m$; $b_m\rightsquigarrow b_0,\ldots,b_{m-1}$, where $a_m=01\cdots m$, $b_m=a_mm$&\\\cdashline{2-3}
  &\{000,010,011,120\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00,0$; $01\rightsquigarrow01$&\\\cdashline{2-3}
  &\{010,011,012,100\}&&\\
  &\{010,011,012,101\}&&\\
  &\{010,011,012,102\}&&\\
  &\{010,011,012,110\}&&\\
  &\{010,011,012,120\}&&\\
  &\{010,011,012,201\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow b_1,\ldots,b_{m-1}$, where $a_m=0^m$, $b_m=a_mm$&$\frac{x(1+x)}{1-x-x^2}$\\\hline
20&\{000,010,011,100\}&&\\
  &\{000,010,011,101\}&&\\
  &\{000,010,011,110\}&&\\
  &\{000,010,011,201\}&&\\
  &\{000,010,011,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow(00)^2$; $01\rightsquigarrow01$&$\frac{x(1-x-x^2)}{(1-x)(1-2x)}$\\\hline
30&\{000,010,101,120\}&&\\
  &\{000,010,110,120\}&&See Subsection \ref{secc30}\\\hline%1,2,4,10,27,79,247,816,2822
36&\{000,010,100,201\}&&\\
  &\{000,010,100,210\}&&See Subsection \ref{secc36}\\\hline%1,2,4,10,28,87,295,1071,4121
37&\{000,010,101,201\}&&\\
  &\{000,010,101,210\}&&See Subsection \ref{secc37}\\\hline%1,2,4,10,28,88,302,1116,4386
39&\{000,012,021,101\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow(001)^2$; $01\rightsquigarrow010,001$; $001\rightsquigarrow010$&\\\cdashline{2-3}
  &\{000,012,021,110\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow(001)^2$; $01\rightsquigarrow001,011$; $001\rightsquigarrow011$&$x+2x^2+4x^3+3x^4$\\\hline
41&\{000,012,021,100\}&&\\
  &\{000,012,021,102\}&&\\
  &\{000,012,021,120\}&&\\
  &\{000,012,021,201\}&&\\
  &\{000,012,021,210\}&$0\rightsquigarrow(00)^2$; $00\rightsquigarrow(001)^2$; $001\rightsquigarrow0011$&\\\cdashline{2-3}
  &\{000,012,100,110\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,002$; $01\rightsquigarrow001,011$; $001\rightsquigarrow011$; $002\rightsquigarrow(011)^2$&$x+2x^2+4x^3+4x^4$\\\hline
42&\{000,012,100,101\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,01$; $01\rightsquigarrow001,010$; $001\rightsquigarrow010$&\\\cdashline{2-3}
  &\{000,012,102,110\}&&\\
  &\{000,012,110,120\}&&\\
  &\{000,012,110,201\}&&\\
  &\{000,012,110,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,01$, $01\rightsquigarrow001,011$; $001\rightsquigarrow011$&$x+2x^2+4x^3+4x^4+x^5$\\\hline
43&\{000,012,101,102\}&&\\
  &\{000,012,101,120\}&&\\
  &\{000,012,101,201\}&&\\
  &\{000,012,101,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,002$ $01\rightsquigarrow010,001$ $001\rightsquigarrow010$'; $002\rightsquigarrow001,0022$; $0022\rightsquigarrow001$,&$x+2x^2+4x^3+4x^4+2x^5+x^6$\\\hline
44&\{000,012,100,102\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,002$; $01\rightsquigarrow001,011$; $001\rightsquigarrow011$; $002\rightsquigarrow(011)^2$&\\\cdashline{2-3}
  &\{000,012,100,120\}&&\\
  &\{000,012,100,201\}&&\\
  &\{000,012,100,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,002$; $01\rightsquigarrow(001)^2$; $001\rightsquigarrow0011$; $002\rightsquigarrow0011,001$&$x+2x^2+4x^3+5x^4+x^5$\\\hline
45&\{000,012,102,120\}&&\\
  &\{000,012,102,201\}&&\\
  &\{000,012,102,210\}&&\\
  &\{000,012,120,201\}&&\\
  &\{000,012,120,210\}&&\\
  &\{000,012,201,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow001,002$; $01\rightsquigarrow(001)^2$; $001\rightsquigarrow0011$; $002\rightsquigarrow001,0022$; $0022\rightsquigarrow001$&$x+2x^2+4x^3+5x^4+2x^5+x^6$\\\hline
46&\{000,011,021,102\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00,002$; $01\rightsquigarrow010,01$; $002\rightsquigarrow002$&\\\cdashline{2-3}
  &\{000,011,102,120\}&$0\rightsquigarrow0,01$; $01\rightsquigarrow010,012$; $012\rightsquigarrow012$&\\\cdashline{2-3}
  &\{001,021,100,110\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$;
$01\rightsquigarrow010,00,01$&\\\cdashline{2-3}
  &\{001,021,100,120\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$;
$01\rightsquigarrow010,011,012$; $011\rightsquigarrow010,011$;
$012\rightsquigarrow00,012$&\\\cdashline{2-3}
  &\{001,021,110,120\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow(00)^2,012$; $012\rightsquigarrow00,012$&\\\cdashline{2-3}
  &\{001,100,110,120\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow010,00,01$&$\frac{x(1+x^2-x^3)}{(1-x)^2}$\\\hline
47&\{000,011,021,120\}&$0\rightsquigarrow(00)^2$; $00\rightsquigarrow00,002$; $002\rightsquigarrow002$&\\\cdashline{2-3}
  &\{000,011,100,102\}&&\\
  &\{000,011,101,102\}&&\\
  &\{000,011,102,110\}&&\\
  &\{000,011,102,201\}&&\\
  &\{000,011,102,210\}&$0\rightsquigarrow0,01$; $01\rightsquigarrow010,01$&\\\cdashline{2-3}
  &\{001,021,101,110\}&&\\
  &\{001,021,102,110\}&&\\
  &\{001,021,110,201\}&&\\
  &\{001,021,110,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$;
  $01\rightsquigarrow(00)^2,01$&\\\cdashline{2-3}
  &\{001,021,101,120\}&&\\
  &\{001,021,102,120\}&&\\
  &\{001,021,120,201\}&&\\
  &\{001,021,120,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$;
$01\rightsquigarrow00,(011)^2$; $011\rightsquigarrow00,011$&\\\cdashline{2-3}
  &\{001,021,100,201\}&&\\
  &\{001,021,100,210\}&&\\
  &\{001,100,101,110\}&&\\
  &\{001,100,102,110\}&&\\
  &\{001,100,110,201\}&&\\
  &\{001,100,110,210\}&$a_m\rightsquigarrow (010)^m,00,a_{m+1}$; $00\rightsquigarrow00$, where $a_m=01\cdots m$&\\\cdashline{2-3}
  &\{001,021,100,101\}&&\\
  &\{001,021,100,102\}&&\\
  &\{001,100,101,120\}&&\\
  &\{001,100,102,120\}&&\\
  &\{001,100,120,201\}&&\\
  &\{001,100,120,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow010,011,01$; $011\rightsquigarrow010,011$&\\\cdashline{2-3}
  &\{001,101,110,120\}&&\\
  &\{001,102,110,120\}&&\\
  &\{001,110,120,201\}&&\\
  &\{001,110,120,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow(00)^2,01$&\\\cdashline{2-3}
  &\{011,012,021,100\}&$a_m\rightsquigarrow(01)^m,a_{m+1}$; $01\rightsquigarrow010$, where $a_m=0^m$&$\frac{x(1+x^2)}{(1-x)^2}$\\\hline
48&\{000,011,021,100\}&&\\
&\{000,011,021,101\}&&\\
&\{000,011,021,110\}&&\\
&\{000,011,021,201\}&&\\
&\{000,011,021,210\}&$0\rightsquigarrow0,00$; $00\rightsquigarrow00,002$; $002\rightsquigarrow002$&\\\cdashline{2-3}
&\{001,100,101,210\}&&\\
&\{001,100,102,210\}&&\\
&\{001,100,201,210\}&$a_m\rightsquigarrow(010)^m,b_m,a_{m+1}$; $b_m\rightsquigarrow(010)^m,b_m$, where $a_m=01\cdots m$, $b_m=a_mm$&\\\cdashline{2-3}
&\{001,101,102,110\}&&\\
&\{001,101,110,201\}&&\\
&\{001,101,110,210\}&&\\
&\{001,102,110,201\}&&\\
&\{001,102,110,210\}&&\\
&\{001,110,201,210\}&$a_m\rightsquigarrow a_{m+1},(00)^m$; $00\rightsquigarrow00$, where $a_m=0^m$, &\\\cdashline{2-3}
&\{001,021,101,102\}&&\\
&\{001,021,101,201\}&&\\
&\{001,021,101,210\}&&\\
&\{001,021,102,201\}&&\\
&\{001,021,102,210\}&&\\
&\{001,021,201,210\}&&\\
&\{001,101,102,120\}&&\\
&\{001,101,120,201\}&&\\
&\{001,101,120,210\}&&\\
&\{001,102,120,201\}&&\\
&\{001,102,120,210\}&&\\
&\{001,120,201,210\}&$0\rightsquigarrow00,01$; $00\rightsquigarrow00$; $01\rightsquigarrow00,011,01$; $011\rightsquigarrow00,011$&\\\cdashline{2-3}
&\{010,012,021,100\}&&\\
&\{010,012,021,101\}&&\\
&\{010,012,021,102\}&&\\
&\{010,012,021,110\}&&\\
&\{010,012,021,120\}&&\\
&\{010,012,021,201\}&&\\
&\{010,012,021,210\}&&\\
&\{011,012,021,101\}&&\\
&\{011,012,021,102\}&&\\
&\{011,012,021,110\}&&\\
&\{011,012,021,120\}&&\\
&\{011,012,021,201\}&&\\
&\{011,012,021,210\}&$a_m\rightsquigarrow a_{m+1},(01)^m$; $01\rightsquigarrow01$, where $a_m=0^m$&$\frac{x(1-x+x^2)}{(1-x)^3}$\\\hline
50&\{000,011,100,120\}&&\\
&\{000,011,101,120\}&&\\
&\{000,011,110,120\}&&\\
&\{000,011,120,201\}&&\\
&\{000,011,120,210\}&$0\rightsquigarrow0,01$; $01\rightsquigarrow0,012$; $012\rightsquigarrow012$&\\\cdashline{2-3}
&\{001,100,101,102\}&&\\
&\{001,100,101,201\}&&\\
&\{001,100,102,201\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m,c_m$; $b_m\rightsquigarrow b_1,\ldots,b_{m-1}$; $c_m\rightsquigarrow b_1,\ldots,b_m,c_m$, where $a_m=01\cdots m$, $b_m=a_m(m-1)$, $c_m=a_mm$&\\\cdashline{2-3}
&\{011,012,100,201\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow010,b_1,\ldots,b_{m-1}$, where $a_m=0^m$, $b_m=a_mm$&$\frac{x}{(1-x)(1-x-x^2)}$\\\hline
51&\{011,012,100,101\}&&\\
  &\{011,012,100,102\}&&\\
  &\{011,012,100,110\}&&\\
  &\{011,012,100,120\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_1\rightsquigarrow010$; $b_2\rightsquigarrow b_1^2$; $b_m\rightsquigarrow c_m,b_1,\ldots,b_{m-1}$; $c_m\rightsquigarrow c_1,b_1,c_3,\ldots,c_{m-1}$, where $a_m=0^m$, $b_m=a_mm$, $c_m=b_m0$&$\frac{x(1-x+x^2+x^4)}{(1-x)^3}$\\\hline
54&\{001,101,102,210\}&&\\
  &\{001,101,201,210\}&&\\
  &\{001,102,201,210\}&$a_m\rightsquigarrow b_0^m,b_m,a_{m+1}$; $b_m\rightsquigarrow b_0^m,b_m$, where $a_m=01\cdots m$, $b_m=a_mm$&\\\cdashline{2-3}
  &\{010,012,110,210\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow b_1^m$, where $a_m=0^m$, $b_m=a_mm$&\\\cdashline{2-3}
  &\{010,012,100,210\}&&\\
  &\{011,012,101,210\}&&\\
  &\{011,012,102,210\}&&\\
  &\{011,012,110,210\}&&\\
  &\{011,012,120,210\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow(0021)^{m-1},b_m$, where $a_m=0^m$, $b_m=a_mm$&$\frac{x(1-2x+2x^2)}{(1-x)^4}$\\\hline
55&\{010,012,100,101\}&&\\
 &\{010,012,100,102\}&&\\
 &\{010,012,100,120\}&&\\
 &\{010,012,100,201\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow c_2,\ldots,c_m,b_m$; $c_m\rightsquigarrow c_2,\ldots,c_{m-1}$, where $a_m=0^m$, $b_m=a_mm$, $c_m=b_m(m-1)$&\\\cdashline{2-3}
 &\{010,012,101,110\}&&\\
 &\{010,012,102,110\}&&\\
 &\{010,012,110,120\}&&\\
 &\{010,012,110,201\}&&\\
 &\{011,012,101,201\}&&\\
 &\{011,012,102,201\}&&\\
 &\{011,012,110,201\}&&\\
 &\{011,012,120,201\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow b_1,b_1,\ldots,b_{m-1}$, where $a_m=0^m$, $b_m=a_mm$&$\frac{x(1-3x+4x^2-x^3-x^4+x^5)}{(1-x)^2(1-x-x^2)}$\\\hline
56&\{010,012,101,210\}&&\\
  &\{010,012,102,210\}&&\\
  &\{010,012,120,210\}&&\\
  &\{010,012,201,210\}&$a_m\rightsquigarrow a_{m+1,}b_1,\ldots,b_m$; $b_m\rightsquigarrow b_1^{m-1},b_m$, where $a_m=0^m$, $b_m=a_mm$&$\frac{x(1-3x+4x^2-2x^3+x^4)}{(1-x)^5}$\\\hline
57&\{000,011,100,101\}&&\\
&\{000,011,100,110\}&&\\
&\{000,011,100,201\}&&\\
&\{000,011,100,210\}&&\\
&\{000,011,101,110\}&&\\
&\{000,011,101,201\}&&\\
&\{000,011,101,210\}&&\\
&\{000,011,110,201\}&&\\
&\{000,011,110,210\}&&\\
&\{000,011,201,210\}&$0\rightsquigarrow0,0$&\\\cdashline{3-3}
&\{001,101,102,201\}&$a_m\rightsquigarrow a_{m+1},b_0,\ldots,b_m$; $b_m\rightsquigarrow b_0,\ldots,b_m$, where $a_m=01\cdots m$, $b_m=a_mm$&\\\cdashline{2-3}
&\{010,011,021,100\}&&\\
&\{010,011,021,101\}&&\\
&\{010,011,021,102\}&&\\
&\{010,011,021,110\}&&\\
&\{010,011,021,120\}&&\\
&\{010,011,021,201\}&&\\
&\{010,011,021,210\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow b_1,\ldots,b_m$, where $a_m=0^m$, $b_m=a_m1$&\\\cdashline{2-3}
&\{010,012,101,102\}&&\\
&\{010,012,101,120\}&&\\
&\{010,012,101,201\}&&\\
&\{010,012,102,120\}&&\\
&\{010,012,102,201\}&&\\
&\{010,012,120,201\}&&\\
&\{011,012,101,102\}&&\\
&\{011,012,101,110\}&&\\
&\{011,012,101,120\}&&\\
&\{011,012,102,110\}&&\\
&\{011,012,102,120\}&&\\
&\{011,012,110,120\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots b_m$; $b_m\rightsquigarrow b_1,\cdots,b_m$, where $a_m=0^m$, $b_m=a_mm$&$\frac{x}{1-2x}$\\\hline
61&\{010,011,100,102\}&&\\
  &\{010,011,101,102\}&&\\
  &\{010,011,102,110\}&$\rsim\{010,011,102\}$ -- Theorems \ref{thred2}-\ref{thred3}& See \cite{CJM}\\\hline
62&\{000,010,021,100\}&&\\
  &\{000,010,021,101\}&&\\
  &\{000,010,021,102\}&&\\
  &\{000,010,021,110\}&&\\
  &\{000,010,021,120\}&&\\
  &\{000,010,021,201\}&&\\
  &\{000,010,021,210\}&$\rsim\{000,010,021\}$ -- Theorems \ref{thred2}-\ref{thred3}& See \cite{CJM}\\\hline%Motzkin
63&\{010,011,120,201\}&&\\
  &\{010,011,120,210\}&Theorem \ref{thC63}& Theorem \ref{thC63}\\\hline%:1,2,4,9,22,57,153,421,1179,
64&\{010,011,100,120\}&&\\
  &\{010,011,101,120\}&&\\
  &\{010,011,110,120\}&$\rsim\{010,011,120\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{CJM}\\\hline%:1,2,4,9,22,58,161,467,1402,
66&\{010,011,100,201\}&&\\
  &\{010,011,100,210\}&&\\
  &\{010,011,101,201\}&&\\
  &\{010,011,101,210\}&&\\
  &\{010,011,110,201\}&&\\
  &\{010,011,110,210\}&$\rsim\{010,011,201\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{CJM}\\\hline%:1,2,4,9,23,65,198,639,2160,
67&\{010,011,100,101\}&&\\
  &\{010,011,100,110\}&&\\
  &\{010,011,101,110\}&$\rsim\{010,011\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{YanLin}\\\hline%:1,2,4,9,23,66,210,733,2781,
68&\{012,021,100,101\}&$a_m\rightsquigarrow a_{m+1},(01)^m$; $01\rightsquigarrow010,01$, where $a_m=0^m$&\\\cdashline{2-3}
  &\{012,021,100,110\}&&\\
  &\{012,021,101,110\}&$a_m\rightsquigarrow a_{m+1},(01)^m$; $01\rightsquigarrow(010)^2$; $010\rightsquigarrow010$, where $a_m=0^m$&$\frac{x(1-x+2x^2)}{(1-x)^3}$\\\hline%:1,2,5,10,17,26,37,50,65,
71&\{012,021,100,102\}&&\\
  &\{012,021,100,120\}&&\\
  &\{012,021,100,201\}&&\\
  &\{012,021,100,210\}&$\rsim\{012,021,100\}$ -- Theorems \ref{thred2}-\ref{thred3}&\\\cdashline{2-3}
  &\{012,021,101,102\}&&\\
  &\{012,021,101,120\}&&\\
  &\{012,021,101,201\}&&\\
  &\{012,021,101,210\}&$\rsim\{012,021,101\}$ -- Theorems \ref{thred2}-\ref{thred3}&\\\cdashline{2-3}
  &\{012,021,102,110\}&&\\
  &\{012,021,110,120\}&&\\
  &\{012,021,110,201\}&&\\
  &\{012,021,110,210\}&$\rsim\{012,021,110\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{CJM}\\\hline%:1,2,5,11,21,36,57,85,121,
74&\{012,100,102,110\}&&\\
  &\{012,100,110,120\}&$\rsim\{012,100,110\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{CJM}\\\hline%:1,2,5,11,22,39,66,108,175,
77&\{012,100,101,201\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow (010)^2,b_1,\ldots,b_{m-1}$, where $a_m=0^m$, $b_m=a_mm$&\\\cdashline{2-3}
  &\{012,101,110,201\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow c_1,\ldots,c_m,b_m$; $c_m\rightsquigarrow c_1\ldots c_{m-1}$; $010\rightsquigarrow010$, where $a_m=0^m$, $b_m=a_mm$, $c_m=a_mm(m-1)$&$\frac{x(1-x+x^2+x^3)}{(1-x)^2(1-x-x^2)}$\\\hline%:1,2,5,11,22,41,73,126,213,
80&\{012,100,101,102\}&&\\
  &\{012,100,101,120\}&$\rsim\{012,100,101\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{CJM}\\\hline%:1,2,5,11,23,45,85,156,281,
81&\{011,021,100,102\}&$a_m\rightsquigarrow a_{m+1},b_1,\ldots,b_m$; $b_m\rightsquigarrow010,b_1,\ldots,b_m$, where $a_m=0^m$, $b_m=a_m1$&\\\cdashline{2-3}
  &\{012,101,102,110\}&&\\
  &\{012,101,110,120\}&$\rsim=\{012,101,110\}$ -- Theorems \ref{thred2}-\ref{thred3}, $a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow c_mb_1\cdots b_{m-1}c_1$, $c_m\rightsquigarrow c_mb_1\cdots b_{m-1}$, where $a_m=0^m$, $b_m=a_mm$, $c_m=b_m0$& $\frac{x(1-x+x^2)}{(1-x)(1-2x)}$\\\hline%:1,2,5,11,23,47,95,191,383,
84&\{012,100,201,210\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow (010)(0021)^{m-1}b_m$, $010\rightsquigarrow010$, where $a_m=0^m$, $b_m=a_mm$&\\\cdashline{2-3}
  &\{012,110,201,210\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow b_1(011)^m$, $011\rightsquigarrow011$, where $a_m=0^m$, $b_m=a_mm$&$\frac{x(3x^2-2x+1)}{(x-1)^4}$\\\hline%:1,2,5,12,25,46,77,120,177,
85&\{000,021,100,102\}&$\rsim\{000,021,102\}$ -- Theorems \ref{thred2}-\ref{thred3}& See \cite{CJM}\\\cdashline{2-3}
  &\{000,021,102,201\}&&\\
  &\{000,021,102,210\}&$a_0\rightsquigarrow b_0c_1$, $a_m\rightsquigarrow b_ma_m\cdots a_1(002)$, $b_m\rightsquigarrow a_{m+1}\cdots a_1(002)$, $c_m\rightsquigarrow(0101)d_mc_m\cdots c_2(012)$, $d_m\rightsquigarrow(0101)c_{m+1}\cdots c_2(012)$, $010\rightsquigarrow0101$, $012\rightsquigarrow(0101)d_1(012)$, where $a_m=0^2\cdots(m-1)^2m$, $b_m=a_mm$, $c_m=01^2\cdots(m-1)^2m$, $d_m=c_mm$&$(1-x-x^2-x^3-(1+x^2)\sqrt{1-2x-3x^2})/(2x^2)-x^2+x^4$\\\hline%:1,2,5,12,25,60,148,374,962,
86&\{012,100,102,210\}&&\\
  &\{012,100,120,210\}&$\rsim\{012,100,210\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\cdashline{2-3}
  &\{012,101,201,210\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow(010)^mb_m$, $010\rightsquigarrow010$, where $a_m=0^m$, $b_m=a_mm$& $\frac{x(1-3x+5x^2-3x^3+x^4)}{(1-x)^5}$\\\cdashline{2-3}
  &\{012,102,110,210\}&&\\
  &\{012,110,120,210\}&$\rsim\{012,110,210\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%:1,2,5,12,26,51,92,155,247,
87&\{012,100,102,201\}&&\\
  &\{012,100,120,201\}&$\rsim\{012,100,201\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
  &\{012,102,110,201\}&&\\
  &\{012,110,120,201\}&$\rsim\{012,110,201\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%:1,2,5,12,26,51,93,161,269,
89&\{012,101,102,210\}&&\\
  &\{012,101,120,210\}&$\rsim\{012,101,210\}$ -- Theorem \ref{thred2}& See \cite{CJM}\\\hline%:1,2,5,12,27,57,113,211,373,
90&\{011,021,100,120\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow b_1\cdots b_m(012)$, $012\rightsquigarrow012$, where $a_m=0^m$, $b_m=0^m1$&\\\cdashline{2-3}
  &\{011,021,101,102\}&&\\
  &\{011,021,102,110\}&&\\
  &\{011,021,102,201\}&&\\
  &\{011,021,102,210\}&$\rsim\{011,021,102\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\cdashline{2-3}
  &\{011,102,120,201\}&Theorem \ref{thC90}&Theorem \ref{thC90}\\\cdashline{2-3}
  &\{011,102,120,210\}&Theorem \ref{thC90}&Theorem \ref{thC90}\\\cdashline{2-3}
  &\{012,021,102,120\}&&\\
  &\{012,021,102,201\}&&\\
  &\{012,021,102,210\}&&\\
  &\{012,021,120,201\}&&\\
  &\{012,021,120,210\}&&\\
  &\{012,021,201,210\}&$\rsim\{012,021\}$ -- Theorem \ref{thred2}&See \cite{YanLin}\\\cdashline{2-3}
  &\{012,101,102,201\}&&\\
  &\{012,101,120,201\}&$\rsim\{012,101,201\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\cdashline{2-3}
  &\{012,102,110,120\}&$\rsim\{012,110\}$ -- Theorem \ref{thred2}&See \cite{YanLin},  $\frac{x(1-2x+2x^2)}{(1-x)^2(1-2x)}$\\\hline%:1,2,5,12,27,58,121,248,503,
91&\{011,101,102,120\}&&\\
  &\{011,102,110,120\}&$\rsim\{011,102,120\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%:1,2,5,12,28,64,144,320,704,
96&\{000,021,101,120\}&&\\
  &\{000,021,110,120\}&$a_0\rightsquigarrow b_0(01)$, $01\rightsquigarrow a_1b_0(002)$, $002\rightsquigarrow b_0(002)$, $a_m\rightsquigarrow b_ma_m\cdots a_1(002)$, $b_m\rightsquigarrow a_{m+1}\cdots a_1(002)$, where $a_m=0^2\cdots(m-1)^2m$, $b_m=a_mm$&$((x-1)(x+2)\sqrt{1-2x-3x^2}+x^3-4x^2-3x+2)/(2x^2)$\\\hline%:1,2,5,12,29,72,182,468,1220,
97&\{011,100,101,102\}&&\\
  &\{011,100,102,110\}&$\rsim\{011,100,102\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%:1,2,5,12,30,75,190,483,1235,
102&\{012,102,201,210\}&&\\
   &\{012,120,201,210\}&$\rsim\{012,201,210\}$ -- Theorem \ref{thred2}& See \cite{CJM}\\\hline%1,2,5,13,32,73,156,318,629,
104&\{000,021,100,120\}&&\\
   &\{000,021,120,201\}&&\\
   &\{000,021,120,210\}&$\rsim\{000,021,120\}$ -- Theorems  \ref{thred2}-\ref{thred3}\\\hline%:1,2,5,13,32,81,207,537,1409,
107&\{012,102,120,201\}&$\rsim\{012,201\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
&\{012,102,120,210\}&$\rsim\{012,210\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,5,13,33,80,185,411,885,
108&\{011,021,100,101\}&&\\
   &\{011,021,100,110\}&&\\
   &\{011,021,100,201\}&&\\
   &\{011,021,100,210\}&$\rsim\{011,021,100\}$ -- Theorems \ref{thred2}&See \cite{CJM}\\\cdashline{2-3}
   &\{011,021,101,120\}&&\\
   &\{011,021,110,120\}&&\\
   &\{011,021,120,201\}&&\\
   &\{011,021,120,210\}&$\rsim\{011,021,120\}$ -- Theorems \ref{thred2}&See \cite{CJM} \\
   &\{011,101,102,210\}&&\\
   &\{011,102,110,210\}&$\rsim\{011,102,210\}$ -- Theorems \ref{thred2}&See \cite{CJM}\\\hline%:1,2,5,13,33,81,193,449,1025,
109&\{011,101,102,201\}&&\\
   &\{011,102,110,201\}&$\rsim\{011,102,201\}$ -- Theorem \ref{thred2}&See  \cite{CJM}\\\hline%:1,2,5,13,33,82,201,489,1185,
110&\{000,021,100,110\}&&\\
   &\{000,021,110,201\}&&\\
   &\{000,021,110,210\}&$\rsim\{000,021,110\}$ -- Theorems \ref{thred2} and \ref{thred3}&See \cite{CJM}\\\hline%:1,2,5,13,33,84,215,556,1453,
114&\{000,021,100,101\}&&\\
   &\{000,021,101,201\}&&\\
   &\{000,021,101,210\}&$\rsim\{000,021,101\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{CJM}\\\hline%:1,2,5,13,35,96,267,750,2123,
115&\{011,100,120,201\}&&\\
   &\{011,100,120,210\}&Theorem \ref{thC115}& Theorem \ref{thC115}\\\hline%:1,2,5,13,35,96,268,758,2167,
117&\{011,100,101,120\}&&\\
   &\{011,100,110,120\}&$\rsim\{011,100,120\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%:1,2,5,13,36,103,306,935,2933,
124&\{000,021,100,201\}&&\\
   &\{000,021,100,210\}&&\\
   &\{000,021,201,210\}&$\rsim\{000,021\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{YanLin}\\\hline%:1,2,5,14,39,111,317,911,2627,
127&\{011,101,120,201\}&&\\
   &\{011,110,120,201\}&$\rsim\{011,120,201\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
   &\{011,101,120,210\}&&\\
   &\{011,110,120,210\}&$\rsim\{011,120,210\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%:1,2,5,14,41,123,375,1156,3590,
128&\{010,021,100,101\}&&\\
   &\{010,021,100,102\}&&\\
   &\{010,021,100,110\}&&\\
   &\{010,021,100,120\}&&\\
   &\{010,021,100,201\}&&\\
   &\{010,021,100,210\}&&\\
   &\{010,021,101,102\}&&\\
   &\{010,021,101,110\}&&\\
   &\{010,021,101,120\}&&\\
   &\{010,021,101,201\}&&\\
   &\{010,021,101,210\}&&\\
   &\{010,021,102,110\}&&\\
   &\{010,021,102,120\}&&\\
   &\{010,021,102,201\}&&\\
   &\{010,021,102,210\}&&\\
   &\{010,021,110,120\}&&\\
   &\{010,021,110,201\}&&\\
   &\{010,021,110,210\}&&\\
   &\{010,021,120,201\}&&\\
   &\{010,021,120,210\}&&\\
   &\{010,021,201,210\}&$\rsim\{010,021\}$ -- Theorems \ref{thred2}-\ref{thred3}&\\\cdashline{2-3}
   &\{011,021,101,110\}&&\\
   &\{011,021,101,201\}&&\\
   &\{011,021,101,210\}&&\\
   &\{011,021,110,201\}&&\\
   &\{011,021,110,210\}&&\\
   &\{011,021,201,210\}&$\rsim\{011,021\}$ -- Theorems \ref{thred2}-\ref{thred3}&See \cite{YanLin}\\\hline%:1,2,5,14,42,132,429,1430,4862,
130&\{011,100,101,201\}&&\\
   &\{011,100,110,201\}&$\rsim\{011,100,201\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
   &\{011,100,101,210\}&&\\
   &\{011,100,110,210\}&$\rsim\{011,100,210\}$ -- Theorem \ref{thred2}& See \cite{CJM}\\\hline%:1,2,5,14,42,133,441,1521,5425,
133&\{000,101,120,201\}&&\\
   &\{000,101,120,210\}&&See Subsection \ref{secc133}\\\hline%:1,2,5,14,43,142,495,1800,6781
152&\{010,100,102,210\}&&  \\\cdashline{2-3}
   &\{011,101,201,210\}&&\\
   &\{011,110,201,210\}&$\rsim\{011,201,210\}$ -- Theorem \ref{thred2}&See \cite{CJM}, $152(1)\isim152(2)$ still {\bf Open}\\\hline%1,2,5,15,50,176,638,2354,8789,
156&\{010,100,101,120\}&&\\
   &\{010,100,110,120\}&&See Subsection \ref{secc156}\\\hline%1,2,5,15,50,180,685,2723,11207,
164&\{010,100,120,201\}&&\\
   &\{010,110,120,201\}&&See Subsection \ref{secc164}\\\hline%12,5,15,51,189,745,3077,13180,
166&\{010,100,110,201\}&&\\
   &\{010,100,110,210\}&&\\
   &\{011,101,110,201\}&$\rsim\{011,210\}$ -- Theorem \ref{thred2}&See \cite{YanLin}\\
   &\{011,101,110,210\}&$\rsim\{011,210\}$ -- Theorem \ref{thred2}&See \cite{YanLin}\\
   &\{010,101,110,201\}&$a_m\rightsquigarrow a_{m+1}a_mb_{m,2}\cdots b_{m,m}$, $b_{m,j}\rightsquigarrow a_{m+2-j}^2b_{m+3-j,2}\cdots b_{m,j-1}b_{m,j}\cdots b_{m,m}$, where $a_m=0^m$, $b_{m,j}=a_mj$&\\\cdashline{2-3}
   &\{010,100,120,210\}&&\\
   &\{010,101,120,201\}&&\\
   &\{010,110,120,210\}&$a_m\rightsquigarrow a_{m+1},a_m,b_{m,2},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow a_{m+2-j},a_{m+1-j},b_{m+3-j,2},\ldots b_{m+1,j},$ $b_{m+1-j,2},\ldots,b_{m+1-j,m+1-j}$, where $a_m=0^m$, $b_{m,j}=a_mj$&\\\cdashline{2-3}
   &\{010,101,120,210\}&$a_m\rightsquigarrow a_{m+1},a_m,b_{m,2},\ldots,b_{m,m}$, $b_{m,j}\rightsquigarrow a_{m+2-j},a_{m+1-j},c_{m+3-j,2},\ldots,c_{m,j},$ $b_{m+1,j},b_{m+1-j,2},\ldots,b_{m+1-j,m+1-j}$; $c_{m,j}\rightsquigarrow c_{m+4-j,3},\ldots,c_{m+1,j},a_{m+3-j},a_{m+2-j},$ $b_{m+2-j,2},\ldots,b_{m+2-j,m+2-j}$, where $a_m=0^m$, $b_{m,j}=a_mj$&$166(7)\isim166(9)$ holds by Subsection \ref{secc166}, $166(1)\isim166(6)$ still {\bf Open} \\\hline%1,2,5,15,51,189,746,3091,13311
168&\{000,100,101,201\}&&\\
   &\{000,100,110,210\}&&Subsection \ref{secc168} \cite{MaSa,YanLin}\\\hline%1,2,5,15,51,191,772,3320,15032
169&\{010,100,101,201\}&&\\
   &\{010,100,101,210\}&Theorem \ref{thC169}&Open\\\hline%1,2,5,15,51,191,773,3334,15161
172&\{010,100,201,210\}&&\\
   &\{010,101,201,210\}&$a_m\rightsquigarrow a_{m+1}a_mb_{m,2}\cdots b_{m,m}$, $b_{m,j}\rightsquigarrow(a_{m+2-j})^{j-1}b_{m+1,j}b_{m,j}\cdots b_{m,m}$, where $a_m=0^m$, $b_{m,j}=a_mj$&Theorem \ref{thC172}\\\hline%1,2,5,15,52,201,841,3726,17213,
174&\{021,100,102,120\}&&\\
   &\{021,101,102,120\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow(010)b_{m+1}c_1\cdots c_m$, $c_m\rightsquigarrow c_1\cdots c_{m+1}$, $010\rightsquigarrow010$, where $a_m=0^m$, $b_m=a_m1$, $c_m=a_m12$&\\\cdashline{2-3}
   &\{021,102,110,120\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow(010)c_1\cdots c_m(012)$, $c_m\rightsquigarrow c_1\cdots c_{m+1}(012)$, $010\rightsquigarrow(010)(0101)$, $0101\rightsquigarrow0101$, $012\rightsquigarrow c_1(012)$, where $a_m=0^m$, $b_m=a_m1$, $c_m=a_m12$&$-\sqrt{1-4x}/(2x)-(2x^5-11x^4+16x^3-14x^2+6x-1)/(2x(1-x)^4)$\\\hline%1,2,6,18,52,152,464,1486,4946,
178&\{021,102,110,201\}&&\\
   &\{021,102,110,210\}&$\rsim\{021,102,110\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,6,19,57,168,506,1585,5165,
179&\{021,102,120,201\}&&\\
   &\{021,102,120,210\}&$\rsim\{021,102,120\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,6,19,58,174,528,1649,5328,
180&\{021,100,102,201\}&&\\
   &\{021,100,102,210\}&$\rsim\{021,100,102\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,6,19,59,183,580,1893,6347,
181&\{021,101,102,201\}&&\\
   &\{021,101,102,210\}&$\rsim\{021,101,102\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,6,19,60,191,619,2048,6909,
183&\{021,100,101,120\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow c_mb_{m+1}c_1\cdots c_m$, $c_m\rightsquigarrow c_1\cdots c_{m+1}$, where $a_m=0^m$, $b_m=a_m1$, $c_m=a_m10$&\\\cdashline{2-3}
   &\{021,100,110,120\}&&\\
   &\{021,101,110,120\}&$a_m\rightsquigarrow a_{m+1}b_1\cdots b_m$, $b_m\rightsquigarrow c_mc_1\cdots c_m(012)$, $c_m\rightsquigarrow c_1\cdots c_{m+1}(012)$, $012\rightsquigarrow(012)c_1$, where $a_m=0^m$, $b_m=a_m1$, $c_m=a_m10$&$-(2-3x)\sqrt{1-4x}/(2x(1-x))+(4x^2-7x+2)/(2x(1-x))$\\\hline%1,2,6,19,61,199,661,2234,7668
189&\{100,102,120,201\}&Theorem \ref{thC189}&\\\cdashline{2-3}
   &\{102,110,120,201\}&Theorem \ref{thC189}&Theorem \ref{thC189}\\\hline%1,2,6,20,68,231,788,2711,9423,
190&\{021,100,110,201\}&&\\
   &\{021,100,110,210\}&$\rsim\{021,100,110\}$-- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,6,20,68,232,794,2732,9468
191&\{021,100,120,201\}&&\\
   &\{021,100,120,210\}&$\rsim\{021,100,120\}$ -- Theorem
   \ref{thred2}&\\\cdashline{2-3}
   &\{021,101,120,201\}&&\\
   &\{021,101,120,210\}&$\rsim\{021,101,120\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
   &\{021,110,120,201\}&&\\
   &\{021,110,120,210\}&$\rsim\{021,110,120\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
   &\{100,102,120,210\}&$\rsim\{100,102,120\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
   &\{101,102,120,201\}&&\\
   &\{101,102,120,210\}&$\rsim\{101,102,120\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
   &\{102,110,120,210\}&$\rsim\{102,110,120\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,6,20,68,233,805,2807,9879
196&\{021,100,101,201\}&&\\
   &\{021,100,101,210\}&$\rsim\{021,100,101\}$ -- Theorem \ref{thred2}&\\\cdashline{2-3}
   &\{021,101,110,201\}&&\\
   &\{021,101,110,210\}&$\rsim\{021,100,110\}$ -- Theorem \ref{thred2}&See \cite{CJM}\\\hline%1,2,6,20,70,252,924,3432,12870,
203&\{021,120,201,210\}&$\rsim\{021,120\}$ -- Theorem \ref{thred2}&See \cite{YanLin}\\\cdashline{2-3}
   &\{101,102,201,210\}&&$(1-4x+\sqrt{1-8x+20x^2-16x^3})/(2(1-x)(1-4x))$, see \cite{YanLin,MaSa}\\\hline%1,2,6,21,77,287,1079,4082,15522
204&\{021,100,201,210\}&$\rsim\{021,100\}$ -- Theorem \ref{thred2}&\\
   &\{021,110,201,210\}&$\rsim\{021,110\}$ -- Theorem \ref{thred2}&See \cite{YanLin}\\\hline%1,2,6,21,78,297,1144,4433,17238
206&\{100,101,120,201\}&&\\
   &\{100,101,120,210\}&Theorem \ref{thC206}&\\\cdashline{2-3}
   &\{101,110,120,201\}&Theorem \ref{thC206}&$206(3)\isim206(1)$, see Subsection \ref{secc206}\\\hline%1,2,6,21,80,323,1363,5950,26671
{\bf207}&\{100,101,110,201\}&&\\
   &\{100,101,110,210\}&&Subsection \ref{secc207}\\\cdashline{2-3}
   &\{101,110,120,210\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow c_{m,j},\ldots,c_{m+1-j,1},a_{m+2-j},$ $b_{m+1-j,1},\ldots,b_{m+1-j,m+1-j}$; $c_{m,j}\rightsquigarrow c_{m+1,j},\ldots,c_{m+2-j,1},a_{m+2-j},$ $b_{m+1-j,1},\ldots,b_{m+1-j,m+1-j}$, where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_mj0$&$207(1)\isim207(3)$ still {\bf Open}\\\hline%1,2,6,21,80,324,1375,6052,27425
208&\{100,110,120,201\}&&\\
   &\{100,110,120,210\}&$a_m\rightsquigarrow a_{m+1},b_{m,1},\ldots,b_{m,m}$; $b_{m,j}\rightsquigarrow a_{m+2-j}^2,b_{m+2-j,1},\ldots,b_{m,j-1},$ $b_{m+1-j,1},\ldots,b_{m+1-j,m+1-j}$, where $a_m=0^m$, $b_{m,j}=a_mj$&Theorem \ref{thC208}\\\hline%1,2,6,21,82,343,1509,6893,32419
210&\{100,101,201,210\}&&\\
   &\{100,110,201,210\}&&\\
   &\{101,110,201,210\}&&See \cite{CM2,MaSa}\\\hline%1,2,6,22,90,394,1806,8558,41586
211&\{100,120,201,210\}&&\\
   &\{110,120,201,210\}&&See \cite{MaSa}\\\hline%1,2,6,22,90,396,1833,8801,43441
\end{longtable}}
\endgroup



\section{Proof of Theorem \ref{thm}}
Let $B$ be any set of patterns and let $\T(B)$ be the generating tree for the class $\I_B$. The {\em length} of a node $v\in T(B)$ is defined to be the number of letters in $v$. For any $k\geq1$, let $D_k(B)$ be the multiset of all nodes of length $k$ at level $k-1$ in $\T(B)$. For each node $v\in D_k(B)$, we denote the multiset of all children of $v$ at level $k$ in $T(B)$ by $N_k(B;v)$. A generating tree $\T(B)$ is said to be {\it $d$-regular} (see \cite{MY2}) if there exists $k\geq1$ such that
\begin{itemize}
\item the number of different nodes in $D_r(B)$ equals $d$, for all $r>k$;
\item for any $v\in D_r(B)$ and $w\in N_r(B;v)$, the number of occurrences of $w$ in $N_r(B;v)$ does not depend on $r$, whenever $r>k$.
\end{itemize}
Clearly, $\T(B)$ is $0$-regular if and only if the set of all nodes of the generating tree $\T(B)$ is finite. For instance, the generating tree
\begin{align*}
0^m&\rightsquigarrow (0^{m+1})(01)\cdots(0^m1),\\
0^m1&\rightsquigarrow (010)(0021)^{m-1}(0^m1),\\
010&\rightsquigarrow010
\end{align*}
is $2$-regular. But the generating tree
\begin{align*}
0^m&\rightsquigarrow (0^{m+1})(0^m1)\cdots(0^mm),\\
0^mj&\rightsquigarrow (0^{m+1})(0^mj)(0^m(j+1))\cdots(0^mm),\quad 1\leq j\leq m
\end{align*}
and the generating tree
\begin{align*}
0^m&\rightsquigarrow (0^{m+1})(01)\cdots(0^m1),\\
0^m1&\rightsquigarrow (0^m1)^m(01)^2
\end{align*}
are not $d$-regular, for any $d$.  Several examples of $d$-regular generating trees are  presented in \cite{CJM,KMY,MY2} and \cite{MY2} contains a five step algorithm to obtain an explicit formula for the generating function $F_B(x)$ from a $d$-regular generating tree $\T(B)$. Accordingly, in this paper we omit the proof details for the generating function $F_B(x)$ whenever the generating tree $\T(B)$ is $d$-regular for some $d$. So, we only consider the classes $C$ in Table \ref{tabnon4} such that there exists $B\in C$ such that $\T(B)$ is not $d$-regular, for any $0\leq d\leq 6$.

\begin{theorem}[\bf Class 63]\label{thC63}
We have $\{010,011,120,201\}\isim\{010,011,120,210\}$. Moreover, we have
\begin{align*}
&F_{\{010,011,120,201\}}(x)\\
&=\frac{x}{1-x}-\sum_{j\geq1}\frac{(1-x)(1-2x)x^{2j}\prod_{i=1}^{j} (x^i-2x+1)^2}{(x^{j+2}+1-2x)
\prod_{i=2}^{j+1}(x^i+1-2x)\prod_{i=2}^{j+1}(x^i-(1-2x)^2)}.
\end{align*}
\end{theorem}
\begin{proof}
Let $B=\{010,011,120,201\}$ or $B=\{010,011,120,210\}$. Then, the rules of the generating tree of $\T(B)$ are given by (here we used Algorithm KMY to guess and prove)
\begin{align*}
&a_m\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\
&b_{m,j}\rightsquigarrow b_{m,j-1}\cdots b_{m+2-j,1}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},
\end{align*}
where $a_m=0^m$ and $b_{m,j}=0^mj$ with $j=1,2,\ldots,m$.
Define $A_m(x)$ (respectively, $B_{m,j}(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T(B;a_m)$ (respectively, $\T(B;b_{m,j})$), where its root stays at level $0$. Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}(x)+x\sum_{j=1}^mB_{m,j}(x),\\
B_{m,j}(x)&=x+x\sum_{i=1}^{j-1}B_{m+i-j+1,i}(x)+x\sum_{i=1}^{m+1-j}B_{m+1-j,i}(x).
\end{align*}
Define $B_m(u)=\sum_{j=1}^mB_{m,j}(x)u^{m-j}$, $A(v)=\sum_{m\geq1}A_m(x)v^{m-1}$, and $B(v,u)=\sum_{m\geq1}B_m(u)v^{m-1}$. Then, the above recurrence can be written as
\begin{align}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xB(v,1),\label{eqC63a}\\
B(v,u)&=\frac{x}{(1-v)(1-uv)}+\frac{x}{u(1-v))}(B(v,u)-B(v,0))+\frac{x}{1-v}B(vu,1).\label{eqC63b}
\end{align}
By substituting $u=x/(1-v)$ into \eqref{eqC63b}, we have
$$B(v,0)=\frac{x}{1-v-xv}+\frac{x}{1-v}B(xv/(1-v),1).$$
Thus, by substituting $u=1$ into \eqref{eqC63b} and then solving for $B(v,1)$, we have
\begin{align*}
&B(v,1)\\
&=\frac{x(1-x-v)}{(1-v)(1-v-vx)(1-2x-v)}-\frac{x^2}{(1-v)(1-2x-v)}B(xv/(1-v),1).
\end{align*}
By iterating this equation indefinitely, we obtain for $|x|<1$,
\begin{align*}
&B(v,1)\\
&=-\sum_{j\geq0}\frac{(1-x)(1-x-v)x^{2j+1}\prod_{i=0}^{j} (vx^i-v-x+1)^2}{(vx^{j+2}+1-v-x)
\prod_{i=1}^{j+1}(vx^i+1-v-x)\prod_{i=1}^{j+1}(vx^i-(1-2x)(1-x-v))}.
\end{align*}
Hence, by taking $v=x$, \eqref{eqC63a} gives
$A(0)=\frac{x}{1-x}+xB(x,1)$, which completes the proof.
\end{proof}

\begin{theorem}[\bf Class 82]\label{thC82}
We have
$$F_{\{011,100,102,120\}}(x)=\frac{x(1-x-x^2)}{(1+x)(1-2x)}.$$
\end{theorem}
\begin{proof}
Let $B=\{011,100,102,120\}$. Then, the rules of the generating tree of $\T(B)$ is given by
\begin{align*}
&a_m\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\
&b_{m,1}\rightsquigarrow c_{1,0}d_mb_{m-1,1}\cdots b_{m-1,m-1},\\
&b_{m,j}\rightsquigarrow c_{j,0}\cdots c_{j,j-2}c_{j,0}d_{m+1-j}b_{m-j,1}\cdots b_{m-j,m-j},\quad j=2,3,\ldots,m-1,\\
&b_{m,m}\rightsquigarrow c_{m,0}\cdots c_{m,m-2}c_{m,0}d_1,\\
&c_{m,1}\rightsquigarrow c_{m-2,0}\cdots c_{m-2,m-4}c_{m-2,0},\\
&c_{m,j}\rightsquigarrow c_{j,0}\cdots c_{j,j-2}c_{j,0}c_{m-1-j,0}\cdots c_{m-1-j,m-3-j}c_{m-1-j,0},\quad j=1,2,\ldots,m-3,\\
&c_{m,m-2}\rightsquigarrow c_{m-2,0}\cdots c_{m-2,m-4}c_{m-2,0}c_{1,0},\\
&d_m\rightsquigarrow d_mb_{m-1,1}\cdots b_{m-1,m-1},\\
&c_{1,0}\rightsquigarrow c_{1,0},
\end{align*}
where $a_m=0^m$, $b_{m,j}=0^mj$ with $j=1,2,\ldots,m$, $c_{m,j}=0^mj0$, and $d_m=0^m12$. Define $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_{m,j}(x)$, $D_m(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T'(B;a_m)$ (respectively, $\T'(B;b_{m,j})$, $\T'(B;c_{m,j})$, $\T'(B;d_m)$), where its root stays at level $0$. 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\left(C_{j,0}+\sum_{i=0}^{j-2}C_{j,i}(x)\right)
+xD_{m+1-j}(x)+x\sum_{i=1}^{m-j}B_{m-j,i}(x),\\
C_{m,j}(x)&=x+xC_{1,0}(x)(\delta_{j=1}+\delta_{j=m-2})+x\left(c_{m-1-j,0}+\sum_{i=0}^{m-3-j}C_{m-1-j,i}(x)\right)\\
&+\left(C_{j,0}(x)+\sum_{i=0}^{j-2}C_{j,i}(x)\right),\\
D_m(x)&=x+xD_m(x)+x\sum_{i=1}^{m-1}B_{m-1,i}(x).
\end{align*}
Define $A(v)=\sum_{m\geq2}A_m(x)v^{m-1}$, $B(v,u)=\sum_{m\geq1}\sum_{j=1}^mB_{m,j}(x)u^{j-1}v^{m-1}$,\break $C(v,u)=\sum_{m\geq2}\sum_{j=0}^{m-2}C_{m,j}(x)u^jv^{m-1}$, and $D(v)=\sum_{m\geq1}D_m(x)v^{m-1}$. Then, the above recurrence can be written as
\begin{align}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xB(v,1),\label{eqC82a}\\
D(v)&=\frac{x}{1-v}+xD(v)+xvB(v,1),\label{eqC82d}\\
B(v,u)&=\frac{x}{(1-v)(1-uv)}+\frac{x}{1-v}(C(uv,1)+C(uv,0))\notag\\
&+\frac{x}{1-uv}D(v)+\frac{xv}{1-uv}B(v,1),\label{eqC82b}\\
C(v,u)&=\frac{xv}{(1-v)(1-uv)}+\frac{x^2v}{1-uv}\notag\\
&+\frac{x^2v^2u}{1-v}+\frac{xv}{1-uv}(C(v,1)+C(v,0))+\frac{xv^2u}{1-v}(C(uv,1)+C(uv,0)).\label{eqC82c}
\end{align}
By taking either $u=1$ or $u=0$ into \eqref{eqC82c}, we have a system of equations with variables $C(v,1)$ and $C(v,0)$. By solving this system, we have
$$C(v,1)=\frac{(x+1)xv}{(1-v)(1-x-2xv)},\quad
C(v,0)=\frac{(1+x-v-2vx)xv}{2v^2x+v^2-2vx-2v+1}.$$
By solving \eqref{eqC82d} for $D(v)$, we have
$$D(v)=\frac{x}{(1-v)(1-x)}+\frac{xv}{1-x}B(v,1).$$
By substituting expressions of $C(v,0)$, $C(v,1)$, and $D(v)$ into \eqref{eqC82b} with $u=1$, and then solving for $B(v,1)$, we obtain
$$B(v,1)=\frac{x(2vx+x^2+v-x-1)}{(2v^2x+v^2-2vx-2v+1)(v+x-1)}.$$
Then, by taking $u=x$ into \eqref{eqC82a} with using expression of $B(v,1)$, we complete the proof.
\end{proof}

By similar arguments as in the proof of Theorem \ref{thC82}, we obtain the following result.

\begin{theorem}[\bf Class 90]\label{thC90}
We have
$$F_{\{011,102,120,201\}}(x)=F_{\{011,102,120,210\}}(x)=\frac{x(1-2x+2x^2)}{(1-x)^2(1-2x)}.$$
Moreover, the rules of the generating tree $\{011,102,120,201\}$ are given by
\begin{align*}
&a_m\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\
&b_{m,j}\rightsquigarrow d_1^jc_{m+1-j,2}\cdots c_{m+1-j,c+2-j},\quad j=1,2,\\
&b_{m,j}\rightsquigarrow d_1^2b_{1,1}d_4\cdots d_jc_{m+1-j,2}\cdots c_{m+1-j,m+2-j},\quad j=3,4,\ldots,m-1\\
&b_{m,m}\rightsquigarrow d_1^3b_{1,1}d_4\cdots d_m,\\
&c_{m,j}\rightsquigarrow e_3\cdots e_jc_{m+2-j,2}\cdots c_{m+2-j,m+3-j},\quad j=2,3,\ldots,m,\\
&c_{m,m+1}\rightsquigarrow e_3\cdots e_{m+1}d_1,\\
&d_m\rightsquigarrow d_1^2b_{1,1}d_4\cdots b_{m-1},\\
&e_m\rightsquigarrow e_3\cdots e_{m-1},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_m1j$, $d_m=a_mm(m-1)$, and $e_m=a_m1m(m-1)$. The rules of the generating tree $\{011,102,120,210\}$ are given by
\begin{align*}
&a_m\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\
&b_{m,j}\rightsquigarrow d_je_2\cdots e_jc_{m+1-j,2}\cdots c_{m+1-j,m+2-j},\quad j=1,2,\ldots,m-1,\\
&b_{m,m}\rightsquigarrow d_me_2\cdots e_md_1,\\
&c_{m,j}\rightsquigarrow e_2\cdots e_{j-1}c_{m+2-j,2}\cdots c_{m+2-j,m+3-j},\quad j=2,3,\ldots,m,\\
&c_{m,m+1}\rightsquigarrow e_2\cdots e_md_1,\\
&d_m\rightsquigarrow d_me_2\cdots e_m,\\
&e_m\rightsquigarrow e_2\cdots e_{m-1},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, $c_{m,j}=a_m1j$, $d_m=a_mm0$, and $e_m=a_mm1$.
\end{theorem}

\begin{theorem}[\bf Class 115]\label{thC115}
We have $\{011,100,120,201\}\isim\{011,100,120,210\}$. Moreover,
\begin{align*}
&F_{\{011,100,120,201\}}(x)\\
&=\frac{x}{1-x}+\sum_{j\geq0}\frac{x^{j+4}(1+x)^j}{(v_j+2x-1)(1-v_j)^2} \prod_{i=0}^{j-1}\frac{1-x-v_i}{(1-v_i)(v_i+2x-1)(v_ix+v_i-1)}
\end{align*}
with $v_j=\frac{(1-x)x^{j+1}}{1-2x+x^{j+1}}$.
\end{theorem}
\begin{proof}
The rules of the generating trees $\T(\{011,100,120,201\})$ and  $\T(\{011,100,120,210\})$  are given by
\begin{align*}
&a_m\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\ &b_{m,j}\rightsquigarrow c_{m+1-j}c_{m-j}b_{m+2-j,1}\cdots b_{m,j-1}b_{m-j,1}\cdots b_{m-j,m-j},\\
&b_{m,m}\rightsquigarrow c_1b_{2,1}\cdots b_{m,m-1}(012),\\ &c_m\rightsquigarrow c_mb_{m,1}\cdots b_{m,m},\\
&012\rightsquigarrow012,
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$ with $j=1,2,\ldots,m$, and $c_m=a_m10$. Define $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_m(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T(B;a_m)$ (respectively, $\T(B;b_{m,j})$, $\T(B;c_m)$), where its root stays at level $0$. Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}+x\sum_{j=1}^mB_{m,j}(x),\\
C_m(x)&=x+xC_m(x)+x\sum_{j=1}^mB_{m,j}(x),\\
B_{m,j}(x)&=x+xC_{m+1-j}(x)+xC_{m-j}(x)+x\sum_{i=1}^{m-j}B_{m-j,i}(x)+x\sum_{i=1}^{j-1}B_{m+1-j+i,i}(x),\\
B_{m,m}(x)&=x+xC_1(x)+x\sum_{i=1}^{m-1}B_{i+1,i}(x)+\frac{x^2}{1-x},
\end{align*}
where $1\leq j\leq m-1$.
Define $F(v)=\sum_{m\geq1}F_m(x)v^{m-1}$ where $F\in\{A,C\}$ and $B(v,u)=\sum_{m\geq1}\sum_{j=1}^mB_{m,j}(x)u^{m-j}v^{m-1}$. Then, the above recurrence can be written as
\begin{align*}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xB(v,1),\\
C(v)&=\frac{x}{1-v}+xC(v)+xB(v,1),\\
B(v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x^2}{(1-x)(1-v)}+\frac{x}{1-v}(C(vu)+vuC(vu)+vuB(vu,1))\\
&+\frac{x}{u(1-v)}(B(v,u)-B(v,0)).
\end{align*}
By third equation with $u=1$, we obtain
$$B(v,0)=-\frac{(v^2+2xv-2v+2x+1)B(v,1)-x^2-x}{x(1-x)(1-v)}.$$
From second equation, we express the function $C(v)$ in terms of $B(v,1)$. Hence, the third equation with $u=vx/(1-x)$ gives
\begin{align}\label{eqC114a}
B(v,1)=a(v)B(vx/(1-v),1)+b(v),
\end{align}
where $a(v)=\frac{x^3}{(v+2x-1)(1-v)^2}$ and $b(v)=-\frac{x(x+1)(x-1+v)}{(1-v)(v+2x-1)(vx+v-1)}$. By iterating \eqref{eqC114a} infinite number of times (here we assumed $|x|<1$), we have
$$B(v,1)=\sum_{j\geq0} a(vx^j/(1-v(1-x^j)/(1-x)))\prod_{i=0}^{j-1}b(vx^i/(1-v(1-x^i)/(1-x))),$$
which, by equation of $A(v)$ with $v=x$, implies
$A(0)=\frac{x}{1-x}+xB(x,1)$,
which completes the proof.
\end{proof}

\begin{theorem}[\bf Class 169]\label{thC169}
We have $\{010,100,101,201\}\isim\{010,100,101,210\}$.
\end{theorem}
\begin{proof}
Let $\T$ be the tree with a root $a_1$ and satisfies the following results
\begin{align*}
&a_m\rightsquigarrow a_{m+1}a_mb_{m,2}\cdots b_{m,m},\\
&b_{m,j}\rightsquigarrow c_{m+2-j,2}\cdots c_{m,j}b_{m+1,j}b_{m,j}\cdots b_{m,m},\quad 2\leq j\leq m,\\
&c_{m,2}\rightsquigarrow a_mb_{m,2}\cdots b_{m,m},\\
&c_{m,j}\rightsquigarrow c_{m+3-j,2}\cdots c_{m,j-1}b_{m,j-1}\cdots b_{m,m},\quad 3\leq j\leq m.
\end{align*}
Then $\T(\{010,100,101,201\})$ is given by $\T$ with $a_m=0^m$, $b_{m,j}=a_mj$, and $c_{m,j}=a_mj(j-1)$ and the tree  $\T(\{010,100,101,210\})$ is given by $\T$ with $a_m=0^m$, $b_{m,j}=a_mj$, and $c_{m,j}=a_mj1$. Hence, $\{010,100,101,201\}\isim\{010,100,101,210\}$.
\end{proof}

\begin{remark}
Let $B=\{010,100,101,201\}$. Define $A_m(x)$ (respectively, $B_{m,j}(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T(B;a_m)$ (respectively, $\T(B;b_{m,j})$), where its root stays at level $0$. Moreover, let $A(v)=\sum_{m\geq1}A_m(x)v^{m-1}$, $B(v,u)=\sum_{m\geq2}\sum_{j=2}^mB_{m,j}(x)u^{m-j}v^{m-2}$, and $C(v,u)=\sum_{m\geq2}\sum_{j=2}^mC_{m,j}(x)u^{m-j}v^{m-2}$. Then, the rules of the generating tree $\T(B)$ in the proof of Theorem \ref{thC169} imply
\begin{align*}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xA(v)+xvB(v,1),\\ B(v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{1-v}C(v,u)+\frac{x}{uv}(B(v,u)-B(v,0))\\
&+\frac{x}{1-u}(B(v,u)-uB(uv,1)),\\ C(v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{uv}(A(uv)-A(0))+xB(uv,1)\\
&+\frac{x}{u(1-v)}(C(v,u)-C(v,0))+\frac{x}{u(1-u)}(B(v,u)-uB(uv,1))\\
&-\frac{x}{u}B(v,0).
\end{align*}
Here, we failed to derive from these equations an explicit formula for $A(0)$.
\end{remark}

\begin{theorem}[\bf Class 172]\label{thC172}
We have $\{010,100,201,210\}\isim\{010,101,201,210\}$. Moreover,
$$F_{\{010,100,201,210\}}(x)=\frac{K(x)}{1-K(x)},$$
where
\begin{align*}
K(x)&=\frac{1}{4}(1+2x-x^2+(1-x)\sqrt{x^2-6x+1})\\
&-\frac{1}{2\sqrt{2}}\sqrt{(x+1)(x^2-4x+1)\sqrt{x^2-6x+1}+x^4-6x^3+4x^2-6x+1}.
\end{align*}
\end{theorem}
\begin{proof}
By Table \ref{tabnon4} (Class 172), we see that the generating tree $\T(\{010,100,201,210\})$ is the same as the generating tree  $\T(\{010,101,201,210\})$. Define $A_m(x)$ (respectively, $B_{m,j}(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T(B;a_m)$ (respectively, $\T(B;b_{m,j})$), where its root stays at level $0$. Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}+xA_m(x)+x\sum_{j=1}^mB_{m,j}(x),\\
B_{m,j}(x)&=x+(j-1)xA_{m+2-j}(x)+xB_{m+1;j}(x)+x\sum_{i=j}^mB_{m,i}(x),
\end{align*}
where $2\leq j\leq m$. 

Define $A(v)=\sum_{m\geq1}A_m(x)v^{m-1}$ and $B(v,u)=\sum_{m\geq2}\sum_{j=2}^mB_{m,j}(x)u^{m-j}v^{m-2}$. Then, the above recurrence can be written as
\begin{align}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xA(v)+xvB(v,1),\label{eq172a}\\
B(v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{vu(1-v)^2}(A(vu)-A(0))\notag\\
&+\frac{x}{uv}(B(v,u)-B(v,0))+\frac{x}{1-u}(B(v,u)-uB(vu,1)).\label{eq172b}
\end{align}
In order to solve \eqref{eq172a}-\eqref{eq172b}, we {\bf assume} that the generating functions $A(v)$ and $B(v,u)$ satisfy one extra equation
\begin{align}
B(v,1)&=\frac{A(v)}{1-v-\frac{1-x-\sqrt{1-6x+x^2}}{2}}.\label{eq172c}
\end{align}
Note that we guessed \eqref{eq172c} by looking at the first terms of the generating function $A(v)/B(v,1)$. Hence, by \eqref{eq172a} and \eqref{eq172c}, we have
$$\left(1-x-\frac{x}{v}-\frac{2xv}{1-2v+x+\sqrt{x^2-6x+1}}\right)A(v)=\frac{x}{1-v}-\frac{x}{v}A(0).$$
By taking $v=v_0=K(x)$, we obtain
$A(0)=\frac{v_0}{1-v_0}$, which, by using same equation, implies
$$A(v,1)=\frac{x(v_0-v)(1-2v+x+\sqrt{x^2-6x+1})}{(1-v)(1-v_0)((xv-v+x)\sqrt{x^2-6x+1}+(v+1)x^2+(1-2v)x+2v^2-v)}.$$
Thus, by \eqref{eq172c}, we have
$$B(v,1)=\frac{2x(v_0-v)}{(1-v)(1-v_0)((xv-v+x)\sqrt{x^2-6x+1}+(v+1)x^2+(1-2v)x+2v^2-v)}.$$
Now, by \eqref{eq172b}, we have
\begin{align*}
\left(1-\frac{x}{uv}-\frac{x}{1-u}\right)B(v,u)&=\frac{x}{(1-v)(1-vu)}+\frac{x}{vu(1-v)^2}(A(vu)-A(0))\notag\\
&-\frac{x}{uv}B(v,0)-\frac{xu}{1-u}B(vu,1).
\end{align*}
By taking $u=u_0=\frac{x+v-xv-\sqrt{(v-1)^2x^2-2vx+(1-2x)v^2}}{2v}$, we obtain
\begin{align*}
B(v,0)&=\frac{vu_0}{(1-v)(1-vu_0)}+\frac{1}{(1-v)^2}(A(vu_0)-A(0))
-\frac{vu_0^2}{1-u_0}B(vu_0,1),
\end{align*}
which leads to
\begin{align*}
B(v,u)&=\frac{\frac{x}{(1-v)(1-vu)}+\frac{x}{vu(1-v)^2}(A(vu)-A(0))
-\frac{x}{uv}B(v,0)-\frac{xu}{1-u}B(vu,1)}{1-\frac{x}{uv}-\frac{x}{1-u}},
\end{align*}
where we do not present the explicit expressions for $B(v,0)$ and $B(v,u)$ because they are too long.

Since $A(v)$ and $B(v,u)$ satisfy \eqref{eq172a}-\eqref{eq172c}, this completes the proof.
\end{proof}

\begin{theorem}[\bf Class 189]\label{thC189}
We have $\{100,102,120,201\}\isim\{102,110,120,201\}$. Moreover,
$$F_{\{100,102,120,201\}}(x)=\frac{1-3x-x^4-(1-x-2x^3-x^4)\sqrt{1-4x}+1)}{2x^2(2+x)(1-x)^2}.$$
\end{theorem}
\begin{proof}
We proceed by finding the generating function for each class.

(1) The rules of the generating trees $\T(\{100,102,120,201\})$ are given by
\begin{align*}
&a_m\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\ &b_{m,j}\rightsquigarrow c_1\cdots c_jb_{m+1,j}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 1\leq j\leq m,\\
&c_m\rightsquigarrow d_2\cdots d_{m-1}c_1,\\
&d_m\rightsquigarrow d_2\cdots d_{m-1},
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$ with $j=1,2,\ldots,m$, $c_m=a_mm(m-1)$, and $d_m=a_mm(m-1)(m-2)$. Define $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_m(x)$, $D_m(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T(B;a_m)$ (respectively, $\T(B;b_{m,j})$, $\T(B;c_m)$, $\T(B;d_m)$), where its root stays at level $0$. Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}+x\sum_{j=1}^mB_{m,j}(x),\\
B_{m,j}(x)&=x+x\sum_{i=1}^jC_i(x)+xB_{m+1,j}(x)+x\sum_{i=1}^{m+1-j}B_{m+1-j,i}(x),\\
C_m(x)&=x+xC_1(x)+x\sum_{i=2}^mD_i(x),\\
D_m(x)&=x+x\sum_{i=2}^{m-1}D_i(x),
\end{align*}
where $1\leq j\leq m$. Define $F(v)=\sum_{m\geq1}F_m(x)v^{m-1}$ where $F\in\{A,C\}$, $D(v)=\sum_{m\geq3}D_m(x)v^{m-2}$, $B(v,u)=\sum_{m\geq1}\sum_{j=1}^mB_{m,j}u^{m-j}v^{m-1}$. Then, by the fact that $C_1(x)=\frac{x}{1-x}$, the above recurrence can be written as
\begin{align*}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xB(v,1),\\
B(v,u)&=\frac{x(1+C(v))}{(1-v)(1-vu)}+\frac{x}{vu}(B(v,u)-B(v,0))+\frac{x}{1-v}B(vu,1),\\
C(v)&=\frac{x}{(1-x)(1-v)}+\frac{xv}{1-v}D(v),\\
D(v)&=\frac{x}{1-v}+\frac{xv}{1-v}D(v).
\end{align*}
Thus, $D(v)=\frac{x}{1-v-xv}$ and $C(v)=\frac{x(vx^2+v-1)}{(1-x)(1-v)(vx+v-1)}$. The equation of $B(v,u)$ with $u=x/v$ gives
$$B(v,0)=\frac{x(v^2x^2+vx^3-vx^2-v^2+2v-1)}{(1-v)^2(1-x)^2(vx+v-1)}+\frac{x}{1-v}B(x,1).$$
Therefore, by substituting expression of $B(v,0)$ into equation of $B(v,u)$ with $u=1$ and $v=\frac{1-\sqrt{1-4x}}{2}$, we have
$$B(x,1)=\frac{x(x^2\sqrt{1-4x}-x^2-2)}{(x^4-x^3-2x^2+3x-1)\sqrt{1-4x}+x^4+x^3-6x^2+5x-1)}.$$
Hence, by taking $v=\frac{1-\sqrt{1-4x}}{2}$ and solving for $A(0)$, we complete the proof.

(2) The rules of the generating trees $\T(\{102,110,120,201\})$ are given by
\begin{align*}
&a_m\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\ &b_{m,j}\rightsquigarrow c_1\cdots c_ja_{m+2-j}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 1\leq j\leq m,\\
&c_m\rightsquigarrow c_1(0101)^2d_3\cdots d_m,\\
&d_m\rightsquigarrow(0101)^2d_3\cdots d_m,\\
&0101\rightsquigarrow0101,
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$ with $j=1,2,\ldots,m$, $c_m=a_mm(m-1)$, and $d_m=a_mm(m-1)(m-2)$. Define $A_m(x)$ (respectively, $B_{m,j}(x)$, $C_m(x)$, $D_m(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T(B;a_m)$ (respectively, $\T(B;b_{m,j})$, $\T(B;c_m)$, $\T(B;d_m)$), where its root stays at level $0$. Define $B_m(x)=\sum_{j=1}^mB_{m,j}(x)$. Thus,
\begin{align*}
A_m(x)&=x+xA_{m+1}+xB_m(x),\\
B_{m,j}(x)&=x+x\sum_{i=1}^jC_i(x)+xA_{m+2-j}(x)+xB_{m+1-j}(x),\\
C_m(x)&=x+\frac{2x^2}{1-x}+xC_1(x)+x\sum_{i=3}^mD_i(x),\\
D_m(x)&=x+\frac{2x^2}{1-x}+x\sum_{i=3}^{m-1}D_i(x),
\end{align*}
where $1\leq j\leq m$. Define $F(v)=\sum_{m\geq1}F_m(x)v^{m-1}$ where $F\in\{A,B,C\}$ and $D(v)=\sum_{m\geq3}D_m(x)v^{m-3}$. Then, by the fact that $C_1(x)=\frac{x}{(1-x)^2}$, the above recurrence can be written as
\begin{align*}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xB(v),\\
B(v)&=\frac{x}{(1-v)^2}+\frac{x}{(1-v)^2}C(v)+\frac{x}{v(1-v)}(A(v)-A(0))+\frac{x}{1-v}B(v),\\
C(v)&=\frac{x}{(1-x)^2}+\frac{xv}{1-v}\left(1+\frac{2x}{1-x}+\frac{x}{(1-x)^2}\right)+\frac{xv^2}{1-v}D(v),\\
D(v)&=\frac{x}{1-v}+\frac{2x^2}{(1-x)(1-v)}+\frac{xv^2}{1-v}D(v).
\end{align*}
It is not hard to get explicit formulas for $D(v),C(v)$. Then by solving the second equation for $B(v)$ and substituting it into first equation, we obtain
\begin{align*}
&(v^2-v+x)A(v)+(v-1)xA(0)\\
&+\frac{xv(-2x^2+2x-1+(x^4+x^3+2x^2-5x+3)v)}{(vx+v-1)(1-x)^2(1-v)^2}\\
&+\frac{x(1-x)^2v^3(-(2x+3)+(1+x)v)}{(vx+v-1)(1-x)^2(1-v)^2}=0.
\end{align*}
Hence, by taking $v=\frac{1-\sqrt{1-4x}}{2}$ and solving for $A(0)$, we complete the proof.
\end{proof}


\begin{theorem}[\bf Class 208]\label{thC208}
We have  $\{100,110,120,201\}\isim\{100,110,120,210\}$. Moreover, the generating function $F_{\{100,110,120,201\}}(x)$ is given by
$$\frac{\frac{x}{1-x}+x\sum_{j\geq0}\frac{x^{2j+1}(1-w_j-x)(2(1+x)w_j^2-w_j-x)\prod_{i=0}^{j-1}(x-w_i)}{(1-w_j(1+x))\prod_{i=0}^{j}(w_i^2+(x-1)w_i+x)\prod_{i=0}^{j}(1-2w_i)\prod_{i=0}^{j}(1-w_i)}}
{1-2x\sum_{j\geq0}\frac{x^{2j+1}(1-w_j-x)\prod_{i=0}^{j-1}(x-w_i)}{\prod_{i=0}^{j}(w_i^2+(x-1)w_i+x)\prod_{i=0}^{j}(1-2w_i)\prod_{i=0}^{j-1}(1-w_i)}},$$
where $w_i=\frac{x^{i+1}}{1-x(1-x^i)/(1-x)}$ for all $i\geq0$.
\end{theorem}
\begin{proof}
By Table \ref{tabnon4}, we see that the generating tree $\T(\{100,110,120,201\})$ is the same as the generating tree
$\T(\{100,110,120,210\})$. Thus,  $\{100,110,120,201\}\isim\{100,110,120,210\}$.
Now fix $B=\{100,110,120,201\}$ (for example). Define $A_m(x)$ (respectively, $B_{m,j}(x)$) to be the generating function for the number of nodes at level $n\geq1$ for the subtree of $\T'(B;a_m)$ (respectively, $\T'(B;b_{m,j})$), where its root stays at level $0$.

Define $A(v)=\sum_{m\geq1}A_m(x)v^{m-1}$ and $B(v,u)=\sum_{m\geq1}\sum_{j=1}^mB_m(x)v^{m-1}u^{m-j}$. Then, as before, the rules of the generating tree $\T(B)$ can be written as
\begin{align}
A(v)&=\frac{x}{1-v}+\frac{x}{v}(A(v)-A(0))+xB(v,1),\label{eq208a}\\
B(v,u)&=\frac{x}{(1-v)(1-uv)}+\frac{2x}{uv(1-v)}(A(uv)-A(0))\notag\\
&+\frac{x}{u(1-v)}(B(v,u)-B(v,0))+\frac{x}{1-v}B(uv,1).\label{eq208b}
\end{align}
By finding $A(v)$ from \eqref{eq208a} and substituting it into \eqref{eq208b} after replacing $v$ by $v/u$, we obtain
\begin{align}
B(v/u,u)&=\frac{x}{(1-v)^2}+\frac{2x}{(1-v)(v-x)}\left(\frac{x}{1-v}-A(0)+xB(v,1)\right)\notag\\
&+\frac{x}{u-v}(B(v/u,u)-B(v/u,0))+\frac{xu}{u-v}B(v,1).\label{eq208c}
\end{align}
By taking $u=v+x$, we have
\begin{align*}
B(v/(v+x),0)&=\frac{x}{(1-v)^2}+\frac{2x}{(1-v)(v-x)}\left(\frac{x}{1-v}-A(0)+xB(v,1)\right)\\
&+(v+x)B(v,1),
\end{align*}
which, by replacing $v$ by $vx/(1-v)$, implies
\begin{align*}
B(v,0)&=\frac{(2(1+x)v^2-2(2+x)v+2)A(0)+vx-x}{(1-v)(1-2v)(1-v(1+x))}\\
&-\frac{xB(vx/(1-v),1)}{(1-v)(1-2v)}.
\end{align*}
Thus, by setting $u=1$ into \eqref{eq208b} and using expression of $B(v,0)$, we obtain
\begin{align*}
B(v,1)&=-\frac{(v-x)x^2}{(v^2+vx-v+x)(1-2v)(1-v)}B(vx/(1-v),1)\\
&+\frac{2x(1-v-x)}{(v^2+vx-v+x)(1-2v)}A(0)\\
&+\frac{x(1-v-x)(2v^2x+2v^2-v-x)}{(1-v)(1-2v)(1-v(1+x))(v^2+vx-v+x)}.
\end{align*}
By iterating this equation (here we assume that $|x|<1$), we have
\begin{align*}
B(v,1)&=2A(0)\sum_{j\geq0}
\frac{x^{2j+1}(1-v_j-x)\prod_{i=0}^{j-1}(x-v_i)}{\prod_{i=0}^{j}(v_i^2+(x-1)v_i+x)\prod_{i=0}^{j}(1-2v_i)\prod_{i=0}^{j-1}(1-v_i)}\\
&+\sum_{j\geq0}\frac{x^{2j+1}(1-v_j-x)(2(1+x)v_j^2-v_j-x)\prod_{i=0}^{j-1}(x-v_i)}{(1-v_j(1+x))\prod_{i=0}^{j}(v_i^2+(x-1)v_i+x)\prod_{i=0}^{j}(1-2v_i)\prod_{i=0}^{j}(1-v_i)}
\end{align*}
where $v_i=\frac{vx^i}{1-v(1-x^i)/(1-x)}$ for all $i\geq0$.

By \eqref{eq208a} with $v=x$, we have that $A(0)=\frac{x}{1-x}+xB(x,1)$,
which completes the proof.
\end{proof}

\section{Left to right maximum and combinatorial arguments}
{\ccc In this section, we use bijections based on decompositions of restricted inversion sequences according to left to right maximum to finish the proof for the rest of Theorem \ref{thm}. In order to do that, we need the following definitions and notation. }

A \emph{left to right maximum (LRmax)} in an inversion sequence $e=e_0e_1\dots e_n$ is an entry $e_i$ such that $e_i\ge e_j$ for all $j<i$. If $e_i=m$ we say $m$ is the \emph{value} of the LRmax. Thus for $e=01102211$, the LRmax are $e_0,e_1,e_2,e_4,e_5$ with values $0,1,1,2,2$ respectively. So any inversion sequence $e\in\I_n$ can be decomposed uniquely as $m_1\pi_1\cdots m_k\pi_k$
where $m_1,\dots,m_k$ are all of the LRmax entries in $e$; thus
$0\leq m_1\leq \cdots\leq m_k\leq n$ and $m_i > \pi_i$ (entrywise).
We call this the WLRmax decomposition of $e$ (W for weakly).

Say $\m_1<\m_2< \dots  <\m_k$ are the first occurrences of the (distinct) LRmax values of an inversion sequence $e$. Then $e$ can be decomposed  as $\m_1\pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ for some $k\ge1 $ with $\m_i \ge \pi_i$ (entrywise) for $1\le i \le k$. This is the \emph{LRmax decomposition} of $e$. For example, $0\,\,0\,0\,\circled{3}\,3\,1\,3\,\circled{4}\,2\,\circled{9}\,3$ is an LRmax decomposition with the first occurrence of each left to right maximum value circled (except for the initial 0).

An inversion sequence $e$ also has a unique decomposition as $e=0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$ with $r_i\ge 1$ for all $i$ where $m_2,\dots,m_k$ are the first occurrences of the nonzero LRmax values and
the first letter of $\pi_i$ (if present) is less than $m_i$ for $i=2,\dots,k$. We call this the \emph{strict LRmax decomposition} of $e$. For example, $00033429939=0^3 3^2 \pi_2 4^1 \pi_3 9^2 \pi_4$ with $\pi_2=\epsilon,\pi_3=2,\pi_4=39$.

\subsection{Class 30: $\{000,010,101,120\}\isim\{000,010,110,120\}$}\label{secc30}
By the LRmax decomposition of any inversion sequence in either $\I_n(\{000,010,101,120\})$ or $\I_n(\{000$, $010,110,120\})$, we have the following lemma.
\begin{lemma}\label{lem30a}
Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then

(1) $e \in \I_n(\{000,010,101,120\})$ if and only
\begin{itemize}
\item[(i)] If a letter $m_i$ appears in $\pi_i$, then it appears as the leftmost letter of $\pi_i$, for all $i=1,2,\ldots,k$;
\item[(ii)] $\m_1\pi_1<\m_2\pi_2<\cdots<\m_k\pi_k$;
\item[(iii)] Each $\pi_i$ avoids $\{000,010,101,120\}$.
\end{itemize}
(2) $e \in \I_n(\{000,010,110,120\})$ if and only
\begin{itemize}
\item[(i)] If a letter $m_i$ appears in $\pi_i$, then it appears as the rightmost letter of $\pi_i$, for all $i=1,2,\ldots,k$;
\item[(ii)] $\m_1\pi_1<\m_2\pi_2<\cdots<\m_k\pi_k$;
\item[(iii)] Each $\pi_i$ avoids $\{000,010,110,120\}$.
\end{itemize}
\end{lemma}

Now, we are ready to define a recursive bijection $$f:\I_n(\{000,010,101,120\})\mapsto\I_n(\{000,010,110,120\}).$$ We define, $f(a)=a$, for any letter $0\leq a\leq n$. For any inversion sequence $e\in \I_n(\{000,010,101$, $120\})$ with LRmax decomposition $\m_1\pi_1\cdots \m_k\pi_k$, we define
$f(e)=\m_1\beta_1\cdots\m_k\beta_k$, where
\begin{itemize}
\item if $\pi_i=m_i\pi'_i$, then $\beta_i$ is defined to be $f(\pi'_i)m_i$,
\item otherwise, we define $\beta_i$ as $f(\pi_i)$.
\end{itemize}
For example, if $e=0022116655487$, then $$f(e)=0022f(11)66f(554)8f(7)=
0022116654587.$$ By Lemma \ref{lem30a}, we have that $e\in \I_n(\{000,010,101,120\})$ if and only if $f(e)\in \I_n(\{000,010,110,120\})$.

\subsection{Class 36: $\{000,010,100,201\}\isim\{000,010,100,210\}$}\label{secc36}
By the WLRmax decomposition of an inversion sequence in either $\I_n(\{000,010,100,201\})$ or $\I_n(\{000,010,100,210\})$, we have the following lemma.
\begin{lemma}\label{lem36a}
Suppose $e\in \I_n$ and $m_1 \pi_1 m_2 \pi_2 \dots m_k \pi_k$ is the WLRmax decomposition of $e$. Then

(1) $e\in\I_n(\{000,010,100,201\})$ if any only if
\begin{itemize}
\item[(i)] Each letter $m_i$ appears at most twice as a LRmax;
\item[(ii)] All the letters in $\pi_1\pi_2\cdots\pi_k$ are distinct;
\item[(iii)] All the letters that are smaller than $m_i$ in $\pi_i\cdots\pi_k$ forms a decreasing sequence.
\end{itemize}
(2) $e\in\I_n(\{000,010,100,210\})$ if any only if
\begin{itemize}
\item[(i)] Each letter $m_i$ appears at most twice as a LRmax;
\item[(ii)] All the letters in $\pi_1\pi_2\cdots\pi_k$ are distinct;
\item[(iii)] The letters that are smaller than $m_i$ in $\pi_i\cdots\pi_k$ form an increasing sequence.
\end{itemize}
\end{lemma}
Now, we are ready to define a bijection $$f:\I_n(\{000,010,100,201\})\mapsto\I_n(\{000,010,100,210\}).$$ Let $e\in\I_n(\{000,010,100,201\})$ with WLRmax decomposition $m_1\pi_1\cdots m_k\pi_k$. We reorder the letters of $e$ such that $m_1,\ldots,m_k$ stay in their positions, and reorder the letters of $\pi_1\cdots\pi_k$ such that $\pi_i\cdots\pi_k$ with $i=1,2,\ldots,k$ forms an increasing sequence. The result is $f(e)$. For example, if $e=00212548893$ then $f(e)=00212538894$.
Clearly, $e$ avoids $\{000,010,100,201\}$ if and only if $f(e)$ avoids $\{000,010,100,210\}$.

\subsection{Class 37: $\{000,010,101,201\}\isim\{000,010,101,210\}$}\label{secc37}
Using the strict LRmax decomposition, we have the following characterization of inversion sequences avoiding $\{000,010,101,201\}$ and $\{000,010,101,210\}$ respectively.
\begin{lemma}\label{37char} Suppose $e\in \I_n$ and $0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$ is the strict LRmax decomposition of $e$. So, $e \in \I_n(\{000,010,101,201\})$ $($respectively, $e \in \I_n(\{000,010,101,210\}))$ if and only if

\noindent $($i$\,)$ each letter in $e$ appears at most twice;\\
$($ii$\,)$ if a letter $x$ appears twice in $e$, then the two occurrences of $x$ are adjacent; \\
$($iii$\,)$ the entries in $\pi_i\dots \pi_k$ that are less than $m_i$ are weakly decreasing $($respectively, weakly increasing\,$)$ for $i=1,\dots,k$.
\end{lemma}
Here is the bijection.  In the strict LRmax decomposition $0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$ of $e \in \I_n(\{000,010$, $101, 201\})$, write $\pi_2\dots \pi_k$ as a list %$(a_{\ell}^{s_{\ell}})_{\ell=1}
$a_1^{s_1}a_2^{s_2}\dots a_{\ell}^{s_{\ell}}$
with superscripts $s_i=2$ to indicate repeated entries (otherwise $s_i=1$ and is omitted).
For example, for
\[
e=0\,0\,\circled{2}\,2\,\circled{4}\,\underline{3}\,\circled{7}\,
\underline{6\,6}\,\circled{10}\,\underline{9\,9\,8\,8\,5\,1} \in \I_{16}(\{000,010,101,201\})
\]
with the first occurrence of each nonzero LRmax
circled and the $\pi$'s underlined, we have $m_1=0,\,r_1=2,\,m_2=2,\,r_2=2,\,m_3=4,\,m_4=7,\,m_5=10$
and $\pi_2 = \epsilon,\,\pi_3=3,\,\pi_4=6\,6,\, \pi_5=9\,9\,8\,8\,5\,1$
and $a_1^{s_1}a_2^{s_2}\dots a_{\ell}^{s_{\ell}}=
3\,6^2\,9^2\,8^2\,5\,1$. In the expression $a_1^{s_1}a_2^{s_2}\dots a_{\ell}^{s_{\ell}}$, arrange
$a_1,\dots,a_{\ell}$ in increasing order while keeping the superscripts frozen in place. The example
yields $1\,3^2\,5^2\,6^2\,8\,9$. Replace each ``2'' superscript  with a duplicate entry to
get $1\,3\,3\,5\,5\,6\,6\,8\,9$ and split this list into a new set of $\pi$'s,
say $(\pi_{i}')_{i=2}^k$ with $\pi_{i}'$ of the same length as $\pi_i$ for each $i$.
Then the desired inversion sequence in $ \I_n(\{000,010,101,210\})$  has
strict LRmax decomposition $0^{r_1}m_2^{r_2} \pi'_2 \dots m_k^{r_k}\pi'_k$. In the example, we get
\begin{equation}\label{ex210}
0\,0\,\circled{2}\,2\,\circled{4}\,\underline{1}\,\circled{7}\,\underline{3\,3}\,\circled{10}\,\underline{5\,5\,6\,6\,8\,9}  \in \I_{16}(\{000,010,101,210\})
\end{equation}
with the first nonzero LRmax entries still circled and each $\pi_i'$ underlined for clarity.

To reverse the map, list the nonempty $\pi$'s as boxed entries with $m_i$ marked above the sequence
of boxes for $\pi_i$, so that each box is associated with an $m_i$. The preceding example \eqref{ex210} gives
\[
\begin{array}{lllll}
\,4 & & \,7 & & 10 \\[-1mm]
\boxed{1} & & \boxed{3}\boxed{3}  & &  \boxed{5}\boxed{5}\boxed{6}\boxed{6}\boxed{8}\framebox[4mm]{9}
\end{array}
\]
Now form a set $S$ of the distinct letters in $\pi_2\dots \pi_k$, here $S=\{1,3,5,6,8,9\}$, and erase the contents of the boxes except an ``x'' is inserted in each box that contained the second occurrence of a repeated letter:
\[
\begin{array}{lllll}
\,4 & & \,7 & & 10 \\[-1mm]
\boxed{\phantom{1}} & & \boxed{\phantom{1}}\framebox[4mm]{\red{x}\vphantom{1}}   & &  \boxed{\phantom{1}} \framebox[4mm]{\red{x}\vphantom{1}}\boxed{\phantom{1}}\framebox[4mm]{\red{x}\vphantom{1}} \boxed{\phantom{1}}\boxed{\phantom{1}}
\end{array}
\]
Next fill in the blank boxes left to right in turn with the letters of $S$ using the
largest available letter that is less than the $m_i$ associated with the box. The example yields
\[
\begin{array}{lllll}
\,4 & & \,7 & & 10 \\[-1mm]
\boxed{3} & & \boxed{6}\framebox[4mm]{\red{x}\vphantom{1}}   & &  \boxed{9} \framebox[4mm]{\red{x}\vphantom{1}}\boxed{8}\framebox[4mm]{\red{x}\vphantom{1}} \boxed{5}\boxed{1}
\end{array}
\]
Lastly, replace each ``x'' with the letter immediately to its left. This yields the $\pi_i$ sequence of the original inversion sequence.

\subsection{Class 168: $\{010,100,101,201\}\isim\{010,100,101,210\}$}\label{secc168}

This case is very similar to Case 37, see Subsection \ref{secc37}. Using the strict LRmax decomposition, we have the following characterization of inversion sequences avoiding $\{010,100,101,201\}$ and $\{010,100,101,210\}$ respectively.
\begin{lemma}\label{168char} Suppose $e\in \I_n$ and $0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$ is the strict LRmax decomposition of $e$. Then $e \in \I_n(\{010,100,101,201\})\,($respectively, $e \in \I_n(\{010,100,101,210\}))$ if and only if

\noindent $($i$\,)$ the letters in $\pi_2\dots \pi_k$ are all distinct ;\\
$($ii$\,)$ no $m_i$ appears as a letter in $\pi_2\dots \pi_k$ ; \\
$($iii$\,)$ the entries in the concatenation $\pi_i\dots \pi_k$ that are less than $m_i$ are  decreasing $($respectively, increasing\,$)$ for $i=1,\dots,k$.
\end{lemma}

Here is the bijection.  In the strict LRmax decomposition $0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$ of $e \in \I_n(\{010,100$, $101,201\})$, rearrange (if necessary) the letters of $\pi_2\dots\pi_k$ so that they are increasing. For example, with the $\pi_i$'s underlined and the first occurrence of each nonzero LRmax
circled,
\[\begin{array}{l}
0\,0\,0\,\circled{3}\,3\,3\,3\,\underline{2}\,\circled{7}\,7\,\underline{6}\,\circled{8}\,\underline{5\,4\,1} \in  \I_{15}(\{010,100,101,201\}) \\
\qquad\mapsto
0\,0\,0\,\circled{3}\,3\,3\,3\,\underline{1}\,\circled{7}\,7\,\underline{2}\,\circled{8}\,\underline{4\,5\,6} \in  \I_{15}(\{010,100,101,210\})\, .
\end{array}\]
The inverse mapping is obtained much as in Case 37 (but without the complication of repeated entries).


\subsection{Class 156: $\{010,100,101,120\}\isim\{ 010,100,110,120\}$}\label{secc156}

\begin{lemma}\label{lrmax} Suppose $e\in \I_n$ avoids $100$ and $b$ starts a $110$ or $101$ pattern in $e$. Then $b$ is a left to right maximum in $e$.
\end{lemma}
\begin{proof}
If $b$ starts a $110$ in $e$ but is not a left to right maximum, then there is a $c>b$ and an $a<b$ such that $cbba$ is a subsequence of $e$. But then $cbb$ is a forbidden $100$. Similarly for $101$.
\end{proof}

\begin{lemma}\label{facts} Suppose $e\in \I_n(\{010,100,120\})$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$.
Then \\
$($i$\,)$ $\m_1 \pi_1<\m_2 \pi_2< \dots <\m_k \pi_k$ $($entrywise$)$.\\
$($ii$\,)$ If $e$ avoids $101$, then all occurrences of $\m_i$ in $e$ occur as a single run at the
start of $\m_i \pi_i,\ 1\le i \le k$. \\
$($iii$\,)$  If $e$ avoids $110$, then all but the first occurrence of $\m_i$ occur as a single $($possibly vacuous$)$ run
at the end of $\pi_i,\ 1\le i \le k$.
\end{lemma}
\begin{proof}
(i) Certainly, $\m_i \pi_i<\m_{i+1}$. If $a\le \m_i$ for some entry $a$ in $\pi_{i+1}$, then $\m_i\m_{i+1}a$ is a $120$
if $a<\m_i$ and a $010$ if $a=\m_i$.
Parts (ii) and (iii) are clear.
\end{proof}
In view of Lemmas \ref{lrmax} and \ref{facts}, the following is a bijection
from $\I_n(\{010,100,101,120\})$ to $\I_n(\{010,100,110,120\})$:
in the LRmax decomposition $0\pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ of $e\in \I_n(\{010,100,101$, $120\})$,
for  each $i=2,\dots,k$ move all occurrences of $\m_i$ in $\pi_i$ from the start of $\pi_i$ to the end. For example,
\[
\begin{array}{lcl}
0\,0\,0\,\ 3\,\ 3\,3\,2\,1\,\ 6\ \,4\,5\,\ 8\ \,8\,7\,\ 9\ \,9\ 10 & = & \\
0\,0\,0\,\circled{3}\,3\,3\,2\,1\,\circled{6}\,4\,5\,\circled{8}\,8\,7\,\circled{9}\,9\,\circled{10} & \to & 0\,0\,0\,\circled{3}\,2\,1\,3\,3\,\circled{6}\,4\,5\,\circled{8}\,7\,8\,\circled{9}\,9\,\circled{10} \,.
\end{array}
\]
To reverse the map, move all occurrences of $\m_i$ in $\pi_i$ from the end of $\pi_i$ to the start.

\subsection{Class 133: $\{000,101,120,201\}\isim\{ 000,101,120,210\}$}\label{secc133}
We have the following characterizations whose straightforward proofs are left to the reader.
\begin{lemma}\label{133Achar} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{000,101,120,201\})$ if and only if

\noindent $($i$\,)$ $\m_i \le  \pi_{i+1}$ for $1\le i \le k-1$,\\
$($ii$\,)$ each $\pi_i$ is weakly decreasing, \\
$($iii$\,)$ no entry occurs three or more times, \\
$($iv$\,)$ if $e_i=m$ is both a LRmax and a descent top, then $e_i$ is the last occurrence of $m$ in $e$.
\end{lemma}

\begin{lemma}\label{133Bchar} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{000,101,120,210\})$ if and only if

\noindent $($i$\,)$ $\m_i \le  \pi_{i+1}$ for $1\le i \le k-1$,\\
$($ii$\,)$ for each $i$, if $\pi_i$ has a descent, then it has only one, it occurs right at the start and the descent top is the second occurrence of $\m_i$, \\
$($iii$\,)$ no entry occurs three or more times, \\
$($iv$\,)$ if $e_i=m$ is both a LRmax and a descent top, then $e_i$ is the last occurrence of $m$ in $e$.
\end{lemma}

Here is the bijection, with an obvious inverse. Given  $e\in \I_n(\{000,101,120,201\})$ with LRmax decomposition $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$, for each $i\ge 2$, reverse $\pi_i$ unless the first entry of $\pi$ is the second occurrence of $\m_i$ in which case leave this entry intact and reverse the rest of $\pi_i$. For example,
\[
\begin{array}{lcl}
0\,0\,\ 1\,\ 1\,\ 2\ \ 5\ \,4\,4\, 3\ \,9\,\ 9\,8 \,8\,7\, 6 \,6 & = & \\
0\,0\,\circled{1}\, 1\, \circled{2}\,\circled{5}\,4\,4\, 3 \,\circled{9}\,9\, 8 \,8\,7\, 6 \,6
 & \to & 0\,0\,\circled{1}\, 1\, \circled{2}\,\circled{5}\,3\,4\, 4 \,\circled{9}\,9\, 6 \,6\,7\, 8 \,8  \,.
\end{array}
\]



\subsection{Class 164: $\{010,100,120,201\}\isim\{ 010,110,120,201\}$}\label{secc164}
We have the following characterizations.
\begin{lemma}\label{164Achar} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{010,100,120,201\})$ if and only if

\noindent $($i$\,)$ $\m_i <  \pi_{i+1}$ for $1\le i \le k-1$, and\\
$($ii$\,)$ each $\m_i\pi_i$ starts with one or more occurrences of $\m_i$ and thereafter is decreasing except that it may end with zero or more occurrences of $m_i$.
\end{lemma}
\begin{proof}
If condition (i) is not met, then either $a=m_i$ occurs in $\pi_{i+1}$ and $m_i m_{i+1}a $ is a 010 or
$a<m_i$ occurs in $\pi_{i+1}$ and $m_i m_{i+1}a $ is a 120.
If $e \in \I_n(\{010,100,120,201\})$  and  $m_i=a$ say, then the entries other than $a$ in $\pi_i$ are decreasing, for else $m_i$ starts a 100 or a 201, and all occurrences of $a$ in $\pi_i$ are  at the start or the end for else $a$ is the ``2'' of a 120. Thus condition (ii) holds. We leave the reader to show the converse: that if the two conditions hold, then $e \in \I_n(\{010,100,120,201\})$.
\end{proof}
The next lemma has an analogous proof.
\begin{lemma}\label{164Bchar} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{010,110,120,201\})$ if and only if

\noindent $($i$\,)$ $\m_i <  \pi_{i+1}$ for $1\le i \le k-1$, and\\
$($ii$\,)$ each $\pi_i$ has the form $m_i^r w\, m_i^t$ with $r,t\ge 0,$ where $m_i>w$ and
$w$ is decreasing except that the last letter of $w$ may be repeated indefinitely.
\end{lemma}

Here is the bijection, with an obvious inverse. Suppose given  $e\in \I_n(\{010,100$, $120,201\})$
with LRmax decomposition $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$. By Lemma \ref{164Achar},
for each $i\ge 2$, $\pi_i$ has the form $m_i^r u_1 \dots u_{s}m_i^t$ with $m_i>u_1> \dots >u_{s}$ for some $r,s,t \ge 0$. Replace $\pi_i$ with $u_1 \dots u_{s-1}u_{s}^{r+1}m_i^t$ if $s \ge 1$ and leave $\pi$ unchanged if $s=0$. For example,
\[
\begin{array}{l}
0\,0\,0\,\circled{3}\,3\,3\,2\,1\,3\,3\,\circled{7}\,7\,6\,5\,4\,7\,\circled{8}\,\circled{10}\,10\,9\,10 \to
0\,0\,0\,\circled{3}\,2\,1\,1\,1\,3\,3\,\circled{7}\,6\,5\,4\,4\,7\,\circled{8}\,\circled{10}\ \,9\ 9\,10 \,.
\end{array}
\]
\subsection{Class 166: $\{010,101,120,201\}\isim\{010,101,120,210\}$}\label{secc166}
We have the following characterizations whose straightforward proofs are left to the reader.
\begin{lemma}\label{166R3char} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{010,101,120,201\})$ if and only if

\noindent $($i$\,)$ for $1\le i \le k-1$, we have $\m_i <  \pi_{i+1}$, and \\
$($ii$\,)$ each $\m_i \pi_i$  is weakly decreasing.
\end{lemma}
\begin{lemma}\label{166R4char} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{010,101,120,201\})$ if and only if

\noindent $($i$\,)$ for $1\le i \le k-1$, we have  $\m_i < \pi_{i+1}$, and \\
$($ii$\,)$ for each $i\ge 2$, all occurences of $b:=\m_i$ in $\pi_i$ occur at the start and the rest of $\pi_i$  is weakly increasing.
\end{lemma}
Here is the bijection, with an obvious inverse. For each $\pi_i$, set $b=\m_i$,
leave all occurences of $b$ at the start of $\pi_i$ intact, and reverse the remainder of $\pi_i$ to change  it from weakly decreasing to weakly increasing.
For example, with the first occurrence of each noninitial left-to-right maximum circled
\[
0\,\,0\,0\,\circled{3}\,3\,2\,2\,2\,1\,\circled{6}\,5\,4\,4 \to
0\,\,0\,0\,\circled{3}\,3\,1\,2\,2\,2\,\circled{6}\,4\,4\,5 \,.
\]

\subsection{Class 206: $\{101,100,120,201\}\isim\{101,110,120,201\}$}\label{secc206}
By the finding the generating trees for all pairs in Class 206, see Table \ref{tabnon4}, we can state the following result.
\begin{theorem}[\bf Class 206]\label{thC206}
We have

$(1)$ The rules of the generating tree $\T(\{100,101,120,201\})$ are
\begin{align*}
a_m&\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\
b_{m,j}&\rightsquigarrow a_{m+1-j}c_{m+2-j,2}\cdots c_{m,j}b_{m+1,j}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 1\leq j\leq m,\\
c_{m,j}&\rightsquigarrow a_{m+2-j}^2c_{m+3-j,2}\cdots c_{m,j-1}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 2\leq j\leq m,
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, and $c_{m,j}=a_mj(j-1)$.

$(2)$ The rules of the generating tree $\T(\{100,101,120,210\})$ are
\begin{align*}
a_m&\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\
b_{m,j}&\rightsquigarrow a_{m+1-j}c_{m+2-j,2}\cdots c_{m,j}b_{m+1,j}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 1\leq j\leq m,\\
c_{m,j}&\rightsquigarrow a_{m+2-j}^2c_{m+3-j,2}\cdots c_{m,j-1}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 2\leq j\leq m,
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, and $c_{m,j}=a_mj0$.

$(3)$ The rules of the generating tree $\T(\{101,110,120,201\})$ are
\begin{align*}
a_m&\rightsquigarrow a_{m+1}b_{m,1}\cdots b_{m,m},\\
b_{m,j}&\rightsquigarrow a_{m+2-j}c_{m+1-j,1}\cdots c_{m,j}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 1\leq j\leq m,\\
c_{m,j}&\rightsquigarrow c_{m+2-j,1}\cdots c_{m-1,j-1}c_{m+2-j,1}a_{m+2-j}b_{m+1-j,1}\cdots b_{m+1-j,m+1-j},\quad 1\leq j\leq m,
\end{align*}
where $a_m=0^m$, $b_{m,j}=a_mj$, and $c_{m,j}=a_mj(j-1)$.
\end{theorem}

We have the following characterizations whose straightforward proofs are left to the reader.
\begin{lemma}\label{206Achar} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{101,100,120,201\})$ if and only if

\noindent $($i$\,)$ for $1\le i \le k-1$, we have $\m_i \le  \pi_{i+1}$  and, further, $\m_i < \pi_{i+1}$ unless
every entry of $\pi_i$ is equal to $m_i$, and \\
$($ii$\,)$ each $\m_i \pi_i$ starts with one or more occurrences of $\m_i$ and thereafter is strictly decreasing.
\end{lemma}
\begin{lemma}\label{206Bchar} Suppose $e\in \I_n$ and $\m_1 \pi_1 \m_2 \pi_2 \dots \m_k \pi_k$ is the LRmax decomposition of $e$. Then $e \in \I_n(\{101,110,120,201\})$ if and only if

\noindent $($i$\,)$ for $1\le i \le k-1$, we have $\m_i \le  \pi_{i+1}$  and, further, $\m_i < \pi_{i+1}$
unless every entry of $\pi_i$ is equal to $m_i$, and \\
$($ii$\,)$ each $\m_i\pi_i$  consists of a sequence of one or more letters that is strictly decreasing
except that the last letter may be repeated indefinitely.
\end{lemma}

Here is the bijection, with an obvious inverse. For each $\pi_i$, if all its letters are $\m_i$ leave $\pi_i$ intact, otherwise, let $a<\m_i$ denote its last letter, set $b=\m_i$, and transfer
all (if any) occurrences of $b$ at the start of $\pi_i$ to the end, changing them from $b$ to $a$. For example, with the first occurrence of each noninitial left-to-right maximum circled
\[0\,0\,\circled{2}\,2\,2\,1\,0\,\circled{5}\,4\,3\,\circled{8}\,8\,8\,\circled{9}\,9\,8 \to 0\,0\,\circled{2}\,1\,0\,0\,0\,\circled{5}\,4\,3\,\circled{8}\,8\,8\,\circled{9}\,8\,8\, .\]

\subsection{Class 207: $\{100,101,110,210\}\isim\{100,101,110,201\}$}\label{secc207}
We have the following characterization of inversion sequences avoiding $\{100,101,110\}$.
\begin{lemma}\label{prelimchar} Suppose $e\in \I_n$ and $0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$ is the strict LRmax decomposition of $e$. Then $e \in \I_n(\{100,101,110\})$ if and only if

\noindent $($i$\,)$ for $2\le i \le k$, either $m_i$ occurs only once in $e$ or each occurrence of $m_i$ is a weak right-to-left min in $e$ $($that is, $\le$ all following letters$)$, and \\
$($ii$\,)$ the concatenation $\pi_2 \dots \pi_k$ consists of distinct letters.
\end{lemma}

\begin{lemma}\label{207pt1and3char} Suppose $e\in \I_n(\{100,101,110\})$ and $0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$  is the strict LRmax decomposition of $e$. Then

\noindent $($i$\,)$ $e \in \I_n(\{100,101,110,210\})$
if and only if the concatenation $\pi_2 \dots \pi_k$ is $($strictly\,$)$ increasing, and \\
$($ii$\,)$ $e \in \I_n(\{100,101,110,201\})$
if and only if, for $i=2,\dots,k$, the entries in the concatenation $\pi_2 \dots \pi_k$ that are less than $m_i$ form a decreasing list.
\end{lemma}

Here is the bijection. In the strict LRmax decomposition $0^{r_1}m_2^{r_2} \pi_2 \dots m_k^{r_k}\pi_k$ of $e \in \I_n(\{100,101$, $110,210\})$, consider the $\pi$'s as a set of filled boxes, one entry in each box. Remove the entries from all boxes and refill the boxes left to right, using the largest available entry that is less than $m_i$ when the box is in $\pi_i$. For example, with the first occurrence of each left-to-right max circled and the $\pi$'s underlined.
\[
\begin{array}{l}
0^5\,\circled{6}\,\underline{0\,1}\,\circled{7}\,\circled{8}\,\underline{2\,3}\,\circled{12}\,\underline{4\ 5\ 10}\ \circled{15}\,\underline{11\,13\,14}\,\circled{16}\,16\,\circled{17} \\[2mm]
\qquad\mapsto0^5\,\circled{6}\,\underline{5\,4}\,\circled{7}\,\circled{8}\,\underline{3\,2}\,\circled{12}\,\underline{11\,10\,1}\,\circled{15}\,\underline{14\ 13\ 0}\ \circled{16}\,16\,\circled{17} \,.
\end{array}
\]
The inverse is obvious: identify the $\pi$'s and then rearrange their entries in increasing order.

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\section{Appendix}
\begingroup
\allowdisplaybreaks
{\fontsize{6}{6}\selectfont
\begin{longtable}[c]{|l|l|l|}
\caption{Sequences $\{|\I_n(B)|\}_{n=0}^9$, where $B$ any set of four patterns in $P$.\label{tabn}}\\ \hline
\multicolumn{3}{|c|}{Beginning of Table \ref{tabn}}\\ \hline
Class&$B$&$\{|\I_n(B)|\}_{n=0}^9$\\ \hline
\endfirsthead  \hline
\multicolumn{3}{|c|}{Continuation of Table \ref{tabn}}\\ \hline Class&$B$&$\{|\I_n(B)|\}_{n=0}^9$\\ \hline
\endhead \hline
\endfoot \hline
\multicolumn{3}{|c|}{End of Table \ref{tabn}}\\ \hline\hline
\endlastfoot
1&\{000,001,010,012\}, \{000,001,011,012\}&	1,2,1,0,0,0,0,0,0\\\hline
2&\{000,001,010,011\}, \{001,010,011,012\}&	1,2,1,1,1,1,1,1,1\\\hline
T3&\{000,001,012,110\}&	1,2,2,0,0,0,0,0,0\\\hline
4&\{000,001,012,021\},\{000,001,012,100\},\{000,001,012,101\}&\\
&\{000,001,012,102\},\{000,001,012,120\},\{000,001,012,201\}&\\
&\{000,001,012,210\},\{000,010,011,012\}&	1,2,2,1,0,0,0,0,0\\\hline
5&\{000,001,011,120\},\{001,011,012,100\}&	1,2,2,1,1,1,1,1,1\\\hline
6&\{000,001,010,021\},\{000,001,010,100\},\{000,001,010,101\}&\\	&\{000,001,010,102\},\{000,001,010,110\},\{000,001,010,120\}&\\
&\{000,001,010,201\},\{000,001,010,210\},\{000,001,011,021\}&	\\
&\{000,001,011,100\},\{000,001,011,101\},\{000,001,011,102\}&	\\
&\{000,001,011,110\},\{000,001,011,201\},\{000,001,011,210\}&	\\
&\{001,010,011,021\},\{001,010,011,100\},\{001,010,011,101\}&	\\
&\{001,010,011,102\},\{001,010,011,110\},\{001,010,011,120\}&	\\
&\{001,010,011,201\},\{001,010,011,210\},\{001,010,012,021\}&	\\
&\{001,010,012,100\},\{001,010,012,101\},\{001,010,012,102\}&	\\
&\{001,010,012,110\},\{001,010,012,120\},\{001,010,012,201\}&	\\
&\{001,010,012,210\},\{001,011,012,021\},\{001,011,012,101\}&	\\
&\{001,011,012,102\},\{001,011,012,110\},\{001,011,012,120\}&	\\
&\{001,011,012,201\},\{001,011,012,210\}&	1,2,2,2,2,2,2,2,2\\\hline
T7&\{000,011,012,021\}&	1,2,3,0,0,0,0,0,0\\\hline
8&\{000,011,012,100\},\{000,011,012,101\},\{000,011,012,102\}&	\\
&\{000,011,012,110\},\{000,011,012,120\},\{000,011,012,201\}&	\\
&\{000,011,012,210\}&	1,2,3,1,0,0,0,0,0\\\hline
T9&\{000,010,012,021\}&	1,2,3,2,0,0,0,0,0\\\hline
10&\{001,011,100,120\},\{001,012,100,110\}&	1,2,3,2,2,2,2,2,2\\\hline
11&\{000,010,012,100\},\{000,010,012,110\}&	1,2,3,3,1,0,0,0,0\\\hline
12&\{000,010,012,101\},\{000,010,012,102\},\{000,010,012,120\}&	\\
&\{000,010,012,201\},\{000,010,012,210\}&	1,2,3,3,2,1,0,0,0\\\hline
T13&\{000,001,021,120\}&	1,2,3,3,2,2,2,2,2\\\hline
14&\{000,001,021,110\},\{000,001,110,120\},\{001,011,021,100\}&	\\
&\{001,011,021,120\},\{001,011,100,101\},\{001,011,100,102\}&	\\
&\{001,011,100,110\},\{001,011,100,201\},\{001,011,100,210\}&	\\
&\{001,011,101,120\},\{001,011,102,120\},\{001,011,110,120\}&	\\
&\{001,011,120,201\},\{001,011,120,210\},\{001,012,021,100\}&	\\
&\{001,012,021,110\},\{001,012,100,101\},\{001,012,100,102\}&	\\
&\{001,012,100,120\},\{001,012,100,201\},\{001,012,100,210\}&	\\
&\{001,012,101,110\},\{001,012,102,110\},\{001,012,110,120\}&	\\
&\{001,012,110,201\},\{001,012,110,210\}&	1,2,3,3,3,3,3,3,3\\\hline
15&\{000,001,021,100\},\{000,001,021,101\},\{000,001,021,102\}&	\\
&\{000,001,021,201\},\{000,001,021,210\},\{000,001,100,120\}&	\\
&\{000,001,101,120\},\{000,001,102,120\},\{000,001,120,201\}&	\\
&\{000,001,120,210\}&	1,2,3,4,4,4,4,4,4\\\hline
16&\{000,001,100,110\},\{000,001,101,110\},\{000,001,102,110\}&	\\
&\{000,001,110,201\},\{000,001,110,210\},\{000,010,011,021\}&	\\
&\{001,010,021,100\},\{001,010,021,101\},\{001,010,021,102\}&	\\
&\{001,010,021,110\},\{001,010,021,120\},\{001,010,021,201\}&	\\
&\{001,010,021,210\},\{001,010,100,101\},\{001,010,100,102\}&	\\
&\{001,010,100,110\},\{001,010,100,120\},\{001,010,100,201\}&	\\
&\{001,010,100,210\},\{001,010,101,102\},\{001,010,101,110\}&	\\
&\{001,010,101,120\},\{001,010,101,201\},\{001,010,101,210\}&	\\
&\{001,010,102,110\},\{001,010,102,120\},\{001,010,102,201\}&	\\
&\{001,010,102,210\},\{001,010,110,120\},\{001,010,110,201\}&	\\
&\{001,010,110,210\},\{001,010,120,201\},\{001,010,120,210\}&	\\
&\{001,010,201,210\},\{001,011,021,101\},\{001,011,021,102\}&	\\
&\{001,011,021,110\},\{001,011,021,201\},\{001,011,021,210\}&	\\
&\{001,011,101,102\},\{001,011,101,110\},\{001,011,101,201\}&	\\
&\{001,011,101,210\},\{001,011,102,110\},\{001,011,102,201\}&	\\
&\{001,011,102,210\},\{001,011,110,201\},\{001,011,110,210\}&	\\
&\{001,011,201,210\},\{001,012,021,101\},\{001,012,021,102\}&	\\
&\{001,012,021,120\},\{001,012,021,201\},\{001,012,021,210\}&	\\
&\{001,012,101,102\},\{001,012,101,120\},\{001,012,101,201\}&	\\
&\{001,012,101,210\},\{001,012,102,120\},\{001,012,102,201\}&	\\
&\{001,012,102,210\},\{001,012,120,201\},\{001,012,120,210\}&	\\
&\{001,012,201,210\},\{010,011,012,021\}&	1,2,3,4,5,6,7,8,9\\\hline
17&\{000,001,100,210\},\{000,001,101,210\},\{000,001,102,210\}&	\\
&\{000,001,201,210\},\{000,010,011,102\}&	1,2,3,5,7,9,11,13,15\\\hline
T18&\{010,011,012,210\}&	1,2,3,5,8,12,17,23,30\\\hline
19&\{000,001,100,101\},\{000,001,100,102\},\{000,001,100,201\}&	\\
&\{000,001,101,102\},\{000,001,101,201\},\{000,001,102,201\}&	\\
&\{000,010,011,120\},\{010,011,012,100\},\{010,011,012,101\}&	\\
&\{010,011,012,102\},\{010,011,012,110\},\{010,011,012,120\}&	\\
&\{010,011,012,201\}&	1,2,3,5,8,13,21,34,55\\\hline
20&\{000,010,011,100\},\{000,010,011,101\},\{000,010,011,110\}&	\\
&\{000,010,011,201\},\{000,010,011,210\}&	1,2,3,5,9,17,33,65,129\\\hline
T21&\{000,010,102,120\}&	1,2,4,10,26,66,172,457,1225\\\hline
T22&\{000,010,102,110\}&	1,2,4,10,26,67,177,475,1287\\\hline
T23&\{000,010,100,102\}&	1,2,4,10,26,68,187,523,1486\\\hline
T24&\{000,010,101,102\}&	1,2,4,10,26,70,195,557,1619\\\hline
T25&\{000,010,102,210\}&	1,2,4,10,27,73,202,568,1612\\\hline
T26&\{000,010,102,201\}&	1,2,4,10,27,73,203,577,1667\\\hline
T27&\{000,010,100,120\}&	1,2,4,10,27,78,241,779,2617\\\hline
T28&\{000,010,100,110\}&	1,2,4,10,27,79,245,801,2743\\\hline
T29&\{000,010,101,110\}&	1,2,4,10,27,79,245,802,2757\\\hline
30&\{000,010,101,120\},\{000,010,110,120\}&	1,2,4,10,27,79,247,816,2822\\\hline
T31&\{000,010,100,101\}&	1,2,4,10,27,81,263,920,3441\\\hline
T32&\{000,010,120,201\}&	1,2,4,10,28,85,278,964,3493\\\hline
T33&\{000,010,120,210\}&	1,2,4,10,28,85,278,965,3505\\\hline
T34&\{000,010,110,201\}&	1,2,4,10,28,86,283,987,3609\\\hline
T35&\{000,010,110,210\}&	1,2,4,10,28,86,283,988,3625\\\hline
36&\{000,010,100,201\},\{000,010,100,210\}&	1,2,4,10,28,87,295,1071,4121\\\hline
37&\{000,010,101,201\},\{000,010,101,210\}&	1,2,4,10,28,88,302,1116,4386\\\hline
T38&\{000,010,201,210\}&	1,2,4,10,29,95,341,1308,5263\\\hline
39&\{000,012,021,101\},\{000,012,021,110\}&	1,2,4,3,0,0,0,0,0\\\hline
T40&\{000,012,101,110\}&	1,2,4,3,1,0,0,0,0\\\hline
41&\{000,012,021,100\},\{000,012,021,102\},\{000,012,021,120\}&	\\
&\{000,012,021,201\},\{000,012,021,210\},\{000,012,100,110\}&	1,2,4,4,0,0,0,0,0\\\hline
42&\{000,012,100,101\},\{000,012,102,110\},\{000,012,110,120\}&	\\
&\{000,012,110,201\},\{000,012,110,210\}&	1,2,4,4,1,0,0,0,0\\\hline
43&\{000,012,101,102\},\{000,012,101,120\},\{000,012,101,201\}&	\\
&\{000,012,101,210\}&	1,2,4,4,2,1,0,0,0\\\hline
44&\{000,012,100,102\},\{000,012,100,120\},\{000,012,100,201\}&	\\
&\{000,012,100,210\}&	1,2,4,5,1,0,0,0,0\\\hline
45&\{000,012,102,120\},\{000,012,102,201\},\{000,012,102,210\}&	\\
&\{000,012,120,201\},\{000,012,120,210\},\{000,012,201,210\}&	1,2,4,5,2,1,0,0,0\\\hline
46&\{000,011,021,102\},\{000,011,102,120\},\{001,021,100,110\}&	\\
&\{001,021,100,120\},\{001,021,110,120\},\{001,100,110,120\}&	1,2,4,5,6,7,8,9,10\\\hline
47&\{000,011,021,120\},\{000,011,100,102\},\{000,011,101,102\}&	\\
&\{000,011,102,110\},\{000,011,102,201\},\{000,011,102,210\}&	\\
&\{001,021,100,101\},\{001,021,100,102\},\{001,021,100,201\}&	\\
&\{001,021,100,210\},\{001,021,101,110\},\{001,021,101,120\}&	\\
&\{001,021,102,110\},\{001,021,102,120\},\{001,021,110,201\}&	\\
&\{001,021,110,210\},\{001,021,120,201\},\{001,021,120,210\}&	\\
&\{001,100,101,110\},\{001,100,101,120\},\{001,100,102,110\}&	\\
&\{001,100,102,120\},\{001,100,110,201\},\{001,100,110,210\}&	\\
&\{001,100,120,201\},\{001,100,120,210\},\{001,101,110,120\}&	\\
&\{001,102,110,120\},\{001,110,120,201\},\{001,110,120,210\}&	\\
&\{011,012,021,100\}&	1,2,4,6,8,10,12,14,16\\\hline
48&\{000,011,021,100\},\{000,011,021,101\},\{000,011,021,110\}&	\\
&\{000,011,021,201\},\{000,011,021,210\},\{001,021,101,102\}&	\\
&\{001,021,101,201\},\{001,021,101,210\},\{001,021,102,201\}&	\\
&\{001,021,102,210\},\{001,021,201,210\},\{001,100,101,210\}&	\\
&\{001,100,102,210\},\{001,100,201,210\},\{001,101,102,110\}&	\\
&\{001,101,102,120\},\{001,101,110,201\},\{001,101,110,210\}&	\\
&\{001,101,120,201\},\{001,101,120,210\},\{001,102,110,201\}&	\\
&\{001,102,110,210\},\{001,102,120,201\},\{001,102,120,210\}&	\\
&\{001,110,201,210\},\{001,120,201,210\},\{010,012,021,100\}&	\\
&\{010,012,021,101\},\{010,012,021,102\},\{010,012,021,110\}&	\\
&\{010,012,021,120\},\{010,012,021,201\},\{010,012,021,210\}&	\\
&\{011,012,021,101\},\{011,012,021,102\},\{011,012,021,110\}&	\\
&\{011,012,021,120\},\{011,012,021,201\},\{011,012,021,210\}&	1,2,4,7,11,16,22,29,37\\\hline
T49&\{011,012,100,210\}&	1,2,4,7,12,19,28,39,52\\\hline
50&\{000,011,100,120\},\{000,011,101,120\},\{000,011,110,120\}&	\\
&\{000,011,120,201\},\{000,011,120,210\},\{001,100,101,102\}&	\\
&\{001,100,101,201\},\{001,100,102,201\},\{011,012,100,201\}&	1,2,4,7,12,20,33,54,88\\\hline
51&\{011,012,100,101\},\{011,012,100,102\},\{011,012,100,110\}&	\\
&\{011,012,100,120\}&	1,2,4,7,13,23,41,72,126\\\hline
T52&\{011,012,201,210\}&	1,2,4,8,14,22,32,44,58\\\hline
T53&\{010,012,100,110\}&	1,2,4,8,14,23,36,55,83\\\hline
54&\{001,101,102,210\},\{001,101,201,210\},\{001,102,201,210\}&	\\
&\{010,012,100,210\},\{010,012,110,210\},\{011,012,101,210\}&	\\
&\{011,012,102,210\},\{011,012,110,210\},\{011,012,120,210\}&	1,2,4,8,15,26,42,64,93\\\hline
55&\{010,012,100,101\},\{010,012,100,102\},\{010,012,100,120\}&	\\
&\{010,012,100,201\},\{010,012,101,110\},\{010,012,102,110\}&	\\
&\{010,012,110,120\},\{010,012,110,201\},\{011,012,101,201\}&	\\
&\{011,012,102,201\},\{011,012,110,201\},\{011,012,120,201\}&	1,2,4,8,15,27,47,80,134\\\hline
56&\{010,012,101,210\},\{010,012,102,210\},\{010,012,120,210\}&	\\
&\{010,012,201,210\}&	1,2,4,8,16,31,57,99,163\\\hline
57&\{000,011,100,101\},\{000,011,100,110\},\{000,011,100,201\}&	\\
&\{000,011,100,210\},\{000,011,101,110\},\{000,011,101,201\}&	\\
&\{000,011,101,210\},\{000,011,110,201\},\{000,011,110,210\}&	\\
&\{000,011,201,210\},\{001,101,102,201\},\{010,011,021,100\}&	\\
&\{010,011,021,101\},\{010,011,021,102\},\{010,011,021,110\}&	\\
&\{010,011,021,120\},\{010,011,021,201\},\{010,011,021,210\}&	\\
&\{010,012,101,102\},\{010,012,101,120\},\{010,012,101,201\}&	\\
&\{010,012,102,120\},\{010,012,102,201\},\{010,012,120,201\}&	\\
&\{011,012,101,102\},\{011,012,101,110\},\{011,012,101,120\}&	\\
&\{011,012,102,110\},\{011,012,102,120\},\{011,012,110,120\}&	1,2,4,8,16,32,64,128,256\\\hline
T58&\{010,011,102,120\}&	1,2,4,9,20,45,100,221,484\\\hline
T59&\{010,011,102,210\}&	1,2,4,9,21,50,119,281,656\\\hline
T60&\{010,011,102,201\}&	1,2,4,9,21,50,120,289,697\\\hline
61&\{010,011,100,102\},\{010,011,101,102\},\{010,011,102,110\}&	1,2,4,9,21,51,126,316,799\\\hline
62&\{000,010,021,100\},\{000,010,021,101\},\{000,010,021,102\}&	\\
&\{000,010,021,110\},\{000,010,021,120\},\{000,010,021,201\}&	\\
&\{000,010,021,210\}&	1,2,4,9,21,51,127,323,835\\\hline
63&\{010,011,120,201\},\{010,011,120,210\}&	1,2,4,9,22,57,153,421,1179\\\hline
64&\{010,011,100,120\},\{010,011,101,120\},\{010,011,110,120\}&	1,2,4,9,22,58,161,467,1402\\\hline
T65&\{010,011,201,210\}&	1,2,4,9,23,64,186,551,1645\\\hline
66&\{010,011,100,201\},\{010,011,100,210\},\{010,011,101,201\}&	\\
&\{010,011,101,210\},\{010,011,110,201\},\{010,011,110,210\}&	1,2,4,9,23,65,198,639,2160\\\hline
67&\{010,011,100,101\},\{010,011,100,110\},\{010,011,101,110\}&	1,2,4,9,23,66,210,733,2781\\\hline
68&\{012,021,100,101\},\{012,021,100,110\},\{012,021,101,110\}&	1,2,5,10,17,26,37,50,65\\\hline
T69&\{012,100,101,110\}&	1,2,5,10,19,33,56,93,154\\\hline
T70&\{012,100,110,210\}&	1,2,5,11,21,35,53,75,101\\\hline
71&\{012,021,100,102\},\{012,021,100,120\},\{012,021,100,201\}&	\\
&\{012,021,100,210\},\{012,021,101,102\},\{012,021,101,120\}&	\\
&\{012,021,101,201\},\{012,021,101,210\},\{012,021,102,110\}&	\\
&\{012,021,110,120\},\{012,021,110,201\},\{012,021,110,210\}&	1,2,5,11,21,36,57,85,121\\\hline
T72&\{012,100,110,201\}&	1,2,5,11,21,36,58,90,137\\\hline
T73&\{000,021,102,120\}&	1,2,5,11,21,51,127,323,835\\\hline
74&\{012,100,102,110\},\{012,100,110,120\}&	1,2,5,11,22,39,66,108,175 \\\hline
T75&\{012,100,101,210\}&	1,2,5,11,22,40,67,105,156 \\\hline
T76&\{012,101,110,210\}&	1,2,5,11,22,41,72,120,191 \\\hline
77&\{012,100,101,201\},\{012,101,110,201\}&	1,2,5,11,22,41,73,126,213 \\\hline
T78&\{011,021,102,120\}&	1,2,5,11,22,42,79,149,284 \\\hline
T79&\{000,021,102,110\}&	1,2,5,11,22,52,128,324,836\\\hline
80&\{012,100,101,102\},\{012,100,101,120\}&	1,2,5,11,23,45,85,156,281\\\hline
81&\{011,021,100,102\},\{012,101,102,110\},\{012,101,110,120\}&	1,2,5,11,23,47,95,191,383\\\hline
T82&\{011,100,102,120\}&	1,2,5,11,25,55,121,263,569\\\hline T83&\{000,021,101,102\}&	1,2,5,11,25,60,148,374,962\\\hline
84&\{012,100,201,210\},\{012,110,201,210\}&	1,2,5,12,25,46,77,120,177\\\hline
T85&\{000,021,100,102\},\{000,021,102,201\},\{000,021,102,210\}&	1,2,5,12,25,60,148,374,962\\\hline
86&\{012,100,102,210\},\{012,100,120,210\},\{012,101,201,210\}&	\\
&\{012,102,110,210\},\{012,110,120,210\}&	1,2,5,12,26,51,92,155,247\\\hline
87&\{012,100,102,201\},\{012,100,120,201\},\{012,102,110,201\}&	\\
&\{012,110,120,201\}&	1,2,5,12,26,51,93,161,269 \\\hline
T88&\{012,100,102,120\}&	1,2,5,12,27,56,110,207,378\\\hline
89&\{012,101,102,210\},\{012,101,120,210\}&	1,2,5,12,27,57,113,211,373\\\hline
90&\{011,021,100,120\},\{011,021,101,102\},\{011,021,102,110\}&	\\
&\{011,021,102,201\},\{011,021,102,210\},\{011,102,120,201\}&	\\
&\{011,102,120,210\},\{012,021,102,120\},\{012,021,102,201\}&	\\
&\{012,021,102,210\},\{012,021,120,201\},\{012,021,120,210\}&	\\
&\{012,021,201,210\},\{012,101,102,201\},\{012,101,120,201\}&	\\
&\{012,102,110,120\}&	1,2,5,12,27,58,121,248,503 \\\hline
91&\{011,101,102,120\},\{011,102,110,120\}&	1,2,5,12,28,64,144,320,704 \\\hline
T92&\{012,101,102,120\}&	1,2,5,12,28,65,151,351,816 \\\hline
T93&\{011,100,102,210\}&	1,2,5,12,29,69,162,375,857 \\\hline
T94&\{011,100,102,201\}&	1,2,5,12,29,70,169,408,985 \\\hline
T95&\{000,021,101,110\}&	1,2,5,12,29,71,177,449,1157\\\hline
96&\{000,021,101,120\},\{000,021,110,120\}&	1,2,5,12,29,72,182,468,1220\\\hline
97&\{011,100,101,102\},\{011,100,102,110\}&	1,2,5,12,30,75,190,483,1235\\\hline
T98&\{000,102,110,120\}&	1,2,5,12,30,77,200,528,1408\\\hline
T99&\{000,101,102,120\}&	1,2,5,12,30,77,201,532,1424\\\hline
T100&\{000,101,102,110\}&	1,2,5,12,31,81,216,583,1590\\\hline
T101&\{000,100,102,120\}&	1,2,5,13,31,80,207,542,1439\\\hline
102&\{012,102,201,210\},\{012,120,201,210\}&	1,2,5,13,32,73,156,318,629\\\hline
T103&\{011,102,201,210\}&	1,2,5,13,32,75,170,377,824\\\hline
104&\{000,021,100,120\},\{000,021,120,201\},\{000,021,120,210\}&	1,2,5,13,32,81,207,537,1409\\\hline
T105&\{000,102,120,201\}&	1,2,5,13,32,84,215,566,1494\\\hline
T106&\{000,102,120,210\}&	1,2,5,13,32,84,217,575,1528\\\hline
107&\{012,102,120,201\},\{012,102,120,210\}&	1,2,5,13,33,80,185,411,885\\\hline
108&\{011,021,100,101\},\{011,021,100,110\},\{011,021,100,201\}&	\\
&\{011,021,100,210\},\{011,021,101,120\},\{011,021,110,120\}&	\\
&\{011,021,120,201\},\{011,021,120,210\},\{011,101,102,210\}&	\\
&\{011,102,110,210\}&	1,2,5,13,33,81,193,449,1025\\\hline
109&\{011,101,102,201\},\{011,102,110,201\}&	1,2,5,13,33,82,201,489,1185\\\hline
110&\{000,021,100,110\},\{000,021,110,201\},\{000,021,110,210\}&	1,2,5,13,33,84,215,556,1453\\\hline
111&\{000,102,110,201\}&    1,2,5,13,33,87,228,609,1636\\\hline
T112&\{011,101,102,110\}&	1,2,5,13,34,89,233,610,1597\\\hline
T113&\{000,102,110,210\}&	1,2,5,13,34,90,240,645,1745\\\hline
114&\{000,021,100,101\},\{000,021,101,201\},\{000,021,101,210\}&	1,2,5,13,35,96,267,750,2123\\\hline
115&\{011,100,120,201\},\{011,100,120,210\}&	1,2,5,13,35,96,268,758,2167\\\hline
T116&\{000,101,102,210\}&	1,2,5,13,36,101,288,827,2389\\\hline
117&\{011,100,101,120\},\{011,100,110,120\}&	1,2,5,13,36,103,306,935,2933 \\\hline
T118&\{000,100,101,102\}&	1,2,5,13,36,104,308,934,2881 \\\hline
T119&\{000,101,102,201\}&	1,2,5,13,37,107,321,979,3042 \\\hline
T120&\{000,101,110,120\}&	1,2,5,13,38,117,378,1275,4451\\\hline
T121&\{000,100,102,210\}&	1,2,5,14,38,107,304,868,2494\\\hline
T122&\{000,102,201,210\}&	1,2,5,14,38,108,301,854,2425\\\hline
T123&\{000,100,102,201\}&	1,2,5,14,38,110,323,972,2969\\\hline
124&\{000,021,100,201\},\{000,021,100,210\},\{000,021,201,210\}&	1,2,5,14,39,111,317,911,2627\\\hline
T125&\{011,120,201,210\}&	1,2,5,14,40,115,331,950,2713\\\hline
T126&\{011,100,201,210\}&	1,2,5,14,41,122,365,1094,3281\\\hline
127&\{011,101,120,201\},\{011,101,120,210\},\{011,110,120,201\}&	\\
&\{011,110,120,210\}&1,2,5,14,41,123,375,1156,3590\\\hline
128&\{010,021,100,101\},\{010,021,100,102\},\{010,021,100,110\}&\\
&\{010,021,100,120\},\{010,021,100,201\},\{010,021,100,210\}&\\
&\{010,021,101,102\},\{010,021,101,110\},\{010,021,101,120\}&\\
&\{010,021,101,201\},\{010,021,101,210\},\{010,021,102,110\}&\\
&\{010,021,102,120\},\{010,021,102,201\},\{010,021,102,210\}&\\
&\{010,021,110,120\},\{010,021,110,201\},\{010,021,110,210\}&\\
&\{010,021,120,201\},\{010,021,120,210\},\{010,021,201,210\}&\\
&\{011,021,101,110\},\{011,021,101,201\},\{011,021,101,210\}&\\
&\{011,021,110,201\},\{011,021,110,210\},\{011,021,201,210\}&	1,2,5,14,42,132,429,1430,4862\\\hline
T129&\{011,101,110,120\}&	1,2,5,14,42,132,431,1452,5026\\\hline
130&\{011,100,101,201\},\{011,100,101,210\},\{011,100,110,201\}&	\\
&\{011,100,110,210\}&	1,2,5,14,42,133,441,1521,5425\\\hline
T131&\{000,100,101,110\}&	1,2,5,14,42,134,451,1590,5834\\\hline
T132&\{000,100,101,120\}&	1,2,5,14,42,136,462,1632,5960\\\hline
133&\{000,101,120,201\},\{000,101,120,210\}&	1,2,5,14,43,142,495,1800,6781\\\hline
T134&\{000,101,110,201\}&	1,2,5,14,43,142,496,1811,6854\\\hline
T135&\{000,101,110,210\}&	1,2,5,14,43,142,497,1827,7008\\\hline
T136&\{011,100,101,110\}&	1,2,5,14,43,144,523,2048,8597\\\hline
T137&\{000,100,110,120\}&	1,2,5,14,44,149,533,2002,7810\\\hline
T138&\{000,110,120,201\}&	1,2,5,14,45,154,562,2144,8480 \\\hline
T139&\{000,110,120,210\}&	1,2,5,14,45,154,564,2161,8616 \\\hline
T140&\{010,100,102,120\}&	1,2,5,15,49,167,581,2049,7301 \\\hline
T141&\{010,102,110,120\}&	1,2,5,15,49,167,582,2058,7357 \\\hline
T142&\{010,101,102,120\}&	1,2,5,15,49,167,583,2068,7423 \\\hline
T143&\{010,100,102,110\}&	1,2,5,15,49,167,583,2071,7455 \\\hline
T144&\{010,101,102,110\}&	1,2,5,15,49,168,593,2135,7797 \\\hline
T145&\{010,100,101,102\}&	1,2,5,15,49,170,614,2285,8700 \\\hline
T146&\{000,100,120,201\}&	1,2,5,15,49,174,650,2533,10195\\\hline
T147&\{000,100,120,210\}&	1,2,5,15,49,174,652,2549,10311\\\hline
T148&\{010,102,120,201\}&	1,2,5,15,50,174,614,2178,7758\\\hline
T149&\{010,102,120,210\}&	1,2,5,15,50,174,616,2201,7919\\\hline
T150&\{010,102,110,201\}&	1,2,5,15,50,174,616,2202,7933\\\hline
T151&\{010,102,110,210\}&	1,2,5,15,50,174,617,2211,7983\\\hline
152&\{010,100,102,210\},\{011,101,201,210\},\{011,110,201,210\}&	1,2,5,15,50,176,638,2354,8789 \\\hline
T153&\{010,100,102,201\}&	1,2,5,15,50,176,639,2371,8953 \\\hline
T154&\{010,101,102,210\}&	1,2,5,15,50,177,649,2431,9230 \\\hline
T155&\{010,101,102,201\}&	1,2,5,15,50,177,651,2460,9489 \\\hline
156&\{010,100,101,120\},\{010,100,110,120\}&	1,2,5,15,50,180,685,2723,11207\\\hline
T157&\{010,101,110,120\}&	1,2,5,15,50,180,686,2736,11325\\\hline
T158&\{010,100,101,110\}&	1,2,5,15,50,180,689,2781,11773\\\hline
T159&\{000,120,201,210\}&	1,2,5,15,50,181,693,2767,11408\\\hline
T160&\{000,100,110,201\}&	1,2,5,15,50,181,697,2818,11845\\\hline
T161&\{000,100,101,210\}&	1,2,5,15,50,181,697,2822,11905\\\hline
T162&\{010,102,201,210\}&	1,2,5,15,51,184,679,2529,9474\\\hline
T163&\{000,110,201,210\}&	1,2,5,15,51,189,744,3059,12993\\\hline
164&\{010,100,120,201\},\{010,110,120,201\}&	1,2,5,15,51,189,745,3077,13180\\\hline
T165&\{000,101,201,210\}&	1,2,5,15,51,189,746,3084,13204\\\hline
166&\{010,100,110,201\},\{010,100,110,210\},\{010,100,120,210\}&	\\
&\{010,101,110,201\},\{010,101,120,201\},\{010,101,120,210\}&	\\
&\{010,110,120,210\},\{011,101,110,201\},\{011,101,110,210\}&	1,2,5,15,51,189,746,3091,13311\\\hline
T167&\{010,101,110,210\}&	1,2,5,15,51,189,747,3109,13511\\\hline
168&\{000,100,101,201\},\{000,100,110,210\}&	1,2,5,15,51,191,772,3320,15032\\\hline
169&\{010,100,101,201\},\{010,100,101,210\}&	1,2,5,15,51,191,773,3334,15161\\\hline
T170&\{010,110,201,210\}&	1,2,5,15,52,199,813,3477,15387\\\hline
T171&\{010,120,201,210\}&	1,2,5,15,52,199,815,3510,15711\\\hline
172&\{010,100,201,210\},\{010,101,201,210\}&	1,2,5,15,52,201,841,3726,17213\\\hline
T173&\{000,100,201,210\}&	1,2,5,16,58,230,965,4216,18970\\\hline
174&\{021,100,102,120\},\{021,101,102,120\},\{021,102,110,120\}&	1,2,6,18,52,152,464,1486,4946\\\hline
T175&\{021,100,102,110\}&	1,2,6,18,52,153,470,1508,5010\\\hline
T176&\{021,101,102,110\}&	1,2,6,18,53,158,486,1550,5109\\\hline
T177&\{021,100,101,102\}&	1,2,6,18,55,173,560,1858,6291\\\hline
178&\{021,102,110,201\},\{021,102,110,210\}&	1,2,6,19,57,168,506,1585,5165\\\hline
179&\{021,102,120,201\},\{021,102,120,210\}&	1,2,6,19,58,174,528,1649,5328\\\hline
180&\{021,100,102,201\},\{021,100,102,210\}&	1,2,6,19,59,183,580,1893,6347\\\hline
181&\{021,101,102,201\},\{021,101,102,210\}&	1,2,6,19,60,191,619,2048,6909\\\hline
T182&\{021,100,101,110\}&	1,2,6,19,61,198,651,2171,7345\\\hline
183&\{021,100,101,120\},\{021,100,110,120\},\{021,101,110,120\}&	1,2,6,19,61,199,661,2234,7668\\\hline
T184&\{100,102,110,120\}&	1,2,6,19,63,212,726,2521,8863\\\hline
T185&\{100,101,102,120\}&	1,2,6,19,63,213,733,2558,9034\\\hline
T186&\{101,102,110,120\}&	1,2,6,19,63,215,749,2650,9490\\\hline
T187&\{100,101,102,110\}&	1,2,6,19,64,222,788,2842,10378\\\hline
T188&\{021,102,201,210\}&	1,2,6,20,66,213,683,2211,7291\\\hline
189&\{100,102,120,201\},\{102,110,120,201\}&	1,2,6,20,68,231,788,2711,9423\\\hline
190&\{021,100,110,201\},\{021,100,110,210\}&	1,2,6,20,68,232,794,2732,9468\\\hline
191&\{021,100,120,201\},\{021,100,120,210\},\{021,101,120,201\}&	\\
&\{021,101,120,210\},\{021,110,120,201\},\{021,110,120,210\}&	\\
&\{100,102,120,210\},\{101,102,120,201\},\{101,102,120,210\}&	\\
&\{102,110,120,210\}&	1,2,6,20,68,233,805,2807,9879\\\hline
T192&\{100,102,110,201\}&	1,2,6,20,68,234,816,2882,10294 \\\hline
T193&\{100,102,110,210\}&	1,2,6,20,69,240,842,2979,10625 \\\hline
T194&\{101,102,110,210\}&	1,2,6,20,69,242,858,3068,11050 \\\hline
T195&\{101,102,110,201\}&	1,2,6,20,69,242,859,3080,11140 \\\hline
196&\{021,100,101,201\},\{021,100,101,210\},\{021,101,110,201\}&	\\
&\{021,101,110,210\}&	1,2,6,20,70,252,924,3432,12870 \\\hline
T197&\{100,101,102,210\}&	1,2,6,20,71,260,970,3662,13938 \\\hline
T198&\{100,101,102,201\}&	1,2,6,20,72,272,1064,4272,17504\\\hline
T199&\{100,101,110,120\}&	1,2,6,20,73,282,1140,4770,20526\\\hline
T200&\{102,110,201,210\}&	1,2,6,21,74,258,897,3131,11007\\\hline
T201&\{102,120,201,210\}&	1,2,6,21,75,265,927,3230,11268\\\hline
T202&\{100,102,201,210\}&	1,2,6,21,76,277,1016,3756,13998\\\hline
203&\{021,120,201,210\},\{101,102,201,210\}&	1,2,6,21,77,287,1079,4082,15522\\\hline
204&\{021,100,201,210\},\{021,110,201,210\}&	1,2,6,21,78,297,1144,4433,17238\\\hline
T205&\{021,101,201,210\}&	1,2,6,21,80,322,1347,5798,25512\\\hline
206&\{100,101,120,201\},\{100,101,120,210\},\{101,110,120,201\}&	1,2,6,21,80,323,1363,5950,26671\\\hline
207&\{100,101,110,201\},\{100,101,110,210\},\{101,110,120,210\}&	1,2,6,21,80,324,1375,6052,27425\\\hline
208&\{100,110,120,201\},\{100,110,120,210\}&	1,2,6,21,82,343,1509,6893,32419\\\hline
T209&\{101,120,201,210\}&	1,2,6,22,88,372,1644,7518,35266\\\hline
210&\{100,101,201,210\},\{100,110,201,210\},\{101,110,201,210\}&	1,2,6,22,90,394,1806,8558,41586\\\hline
211&\{100,120,201,210\},\{110,120,201,210\}&	1,2,6,22,90,396,1833,8801,43441\\\hline
212&\{000,100,102,110\}&	1,2,5,13,33,87,231,621,1686\\\hline
\end{longtable}}
\endgroup
\end{document}