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\begin{center}
\vskip 1cm{\LARGE\bf 
Number of Dissections of the Regular  \\
\vskip .1in
$n$-gon by Diagonals
}
\vskip 1cm
\large
Joris N. Buloron, Roberto B. Corcino, and Jay M. Ontolan\\
Cebu Normal University\\
Cebu City, Philippines\\
\href{mailto:jorisbuloron@yahoo.com}{\tt jorisbuloron@yahoo.com} 
\end{center}

\vskip .2 in

\begin{abstract}
This paper presents a formula for the distinct dissections by diagonals
of a regular $n$-gon modulo the action of the dihedral group. This
counting includes dissection with intersecting or non-intersecting
diagonals. We utilize a corollary of the Cauchy-Frobenius theorem,
which involves counting of cycles. We also give an explicit formula for
the prime number case. We give as a remark the number of distinct
dissections, modulo the action of the cyclic group of finite order.
\end{abstract}

\section{Introduction}

The theory of polygon dissection has proven to be a rich area of
mathematical thoughts. Cayley derived the number of ways to dissect an
$n-$gon  using a specified number of diagonals. Other mathematicians
gave proofs of older formulas involving polygon dissections using new
techniques, such as generating functions, Legendre polynomials, and
Lagrange inversion \cite{ca}. Przytycki and Sikora showed relationships
between polygon dissections and special types of numbers, such as the
Catalan numbers \cite{ps}. Explicit formulas for  dissections of a
regular polygon using non-intersecting diagonals were derived in a
paper of Bowman and Regev \cite{br}. More recently, Siegel counted the
number of dissections of a regular $n$-gon using non-intersecting
diagonals in his thesis \cite{s}.

The main aim of this paper is to count the number of distinct
dissections of an unlabeled regular $n$-gon by diagonals modulo the
dihedral group. We consider both intersecting and non-intersecting
diagonals in our counting. To do this, we first label the vertices of
the polygon and determine which dissections of this labeled $n$-gon are
the same up to the canonical action of the dihedral group of degree
$n$. We present the following definition:

\begin{definition}
Let $n\geq 3$. A regular polygon with $n$ vertices is called an $n$-gon. A {\em diagonal} of an $n$-gon is a segment extending from a vertex to a non-adjacent vertex. A {\em dissection} of the $n$-gon is any set of crossing or non-crossing diagonals of the $n$-gon. A dissection without any diagonal is an {\em empty dissection}.
\end{definition}

The main result of this paper is anchored on a consequence of the Cauchy-Frobenius theorem \cite[Corollary\ 1.7A, p.\ 26]{dm}. We give it below as Lemma \ref{cf}.

\begin{lemma}
Let $G$ be a finite group acting on a finite set $\Delta$. Suppose $\Gamma$ is a non-empty finite set and $\Fun(\Delta,\Gamma)$ is the set of all functions from $\Delta$ to $\Gamma$, then $G$ acts on $\Fun(\Delta,\Gamma)$ by 
$$f^x(\delta)=f(\delta^{x^{-1}})\,\,\,(\forall f\in \Fun(\Delta,\Gamma),x\in G,\delta\in\Delta.)$$
In addition, the number of orbits of this action is equal to
$$\frac{1}{|G|}\left(\sum_{g\in G}|\Gamma|^{c(g)}\right)$$
where $c(g)$ counts the number of cycles of $g$ as it acts on $\Delta$, including the trivial cycles, if they exist. 
\label{cf}
\end{lemma} 

\section{Preliminaries}

Let $[n]=\left\{1,2,\ldots,n\right\}$ be the set of vertices of a regular $n$-gon. It is well-known that the {\em dihedral group} of degree $n$, with presentation $D_n=\langle r,s:r^n=1=s^2,srs=r^{-1}\rangle$, acts on $[n]$ in a natural way. This is obvious when we express the elements of $D_n$ as permutations of $[n]$ corresponding to the symmetries of an $n$-gon, i.e., $D_n\leq\Sym([n])$. Here, $r$ is the $\frac{2\pi}{n}$-rotation and $s$  is the reflection along the axis through center and vertex 1.

\begin{definition}
Let $i,j\in [n]$ be vertices of the $n$-gon. If $i<j$, then we define the {\em cycle length} of $i$ and $j$ as follows: 
$$d(\left\{i,j\right\})=\min\left\{j-i,n-(j-i)\bmod n\right\}.$$  
\end{definition}

Form $\Delta_n=\left\{\left\{i,j\right\}:d(\left\{i,j\right\})\geq 2\right\}$. This is simply the set of all diagonals of the $n$-gon and it can be shown that $|\Delta_n|=\frac{n^2-3n}{2}$. Moreover, the group $D_n$ acts on $\Delta_n$ in a natural way. Observe that $\left\{i,j\right\}\in\Delta_n$ if and only if $i$ and $j$ are non-adjacent. Since each element of $D_n$ only {\em rotates} or {\em reflects} the $n$-gon, then for $x\in D_n$

$$d(\left\{i^x,j^x\right\})=d(\left\{i,j\right\}).$$

It can then be proven that the map $\Delta_n\times D_n\rightarrow
\Delta_n$ defined by

$$\left\{i,j\right\}^g=\left\{i^g,j^g\right\}$$
is an action. Let us denote the corresponding permutation representation of this action by $\rho:D_n\rightarrow \Sym(\Delta_n)$. That is, $\rho(r)$ and $\rho(s)$ are permutations of the set $\Delta_n$ satisfying the following:
\begin{enumerate} 
\item[$i$.] $\rho(r)(\left\{i,j\right\})=\left\{i+1\bmod n,j+1\bmod n\right\}$; 
\item[$ii$.] $\rho(s)(\left\{i,j\right\})=\left\{2-i\bmod n,2-j\bmod n\right\}$.
\end{enumerate} 

Consider the family $\Fun(\Delta_n,\Gamma)$ where $\Gamma=\left\{0,1\right\}$. We can view each function $f\in \Fun(\Delta_n,\Gamma)$ as a way of dissecting the $n$-gon. Here, $f(\left\{i,j\right\})=1$ means that there exists a diagonal from vertex $i$ to $j$. Otherwise, $i$ and $j$ are not connected by any diagonal. The action of an element $x\in D_n$ on $\Fun(\Delta_n,\Gamma)$ can be viewed as either rotating or reflecting the dissection $f$ to $f^x$ preserving the form of the dissection. Consequently, every orbit of this action represents a certain way of dissecting an $n$-gon. This only means that counting the distinct orbits is equivalent to counting the number of distinct dissections of the $n$-gon modulo the dihedral group.  

\begin{proposition}
The number $\gamma(n)$ of distinct dissections of an $n$-gon modulo the dihedral action is 
$$\gamma(n)=\frac{1}{2n}\left(\sum_{g\in D_n}2^{c(g)}\right)$$
where $c(g)$ counts the number of cycles of $g$ as it acts on $\Delta_n$, including the trivial cycles whenever they exist.
\label{prop1} 
\end{proposition}

\section{Result}

The following observation will be used to prove the succeeding claims:

\begin{proposition}
Let $n> 4$ be a natural number. Then $\rho[D_n]\cong D_n$.
\label{mono}
\end{proposition}
\begin{proof}
Let $r,s$ be the generators of $D_n$. When we express $\rho(r)$ as a product of disjoint cycles, we see that $\left(\left\{1,3\right\}\,\,\left\{2,4\right\}\,\,\left\{3,5\right\}\,\,\ldots\,\,\left\{n-1,1\right\}\,\,\left\{n,2\right\}\right)$ is one of these cycles. Since this cycle is of length $n$ and $|\rho(r)|\leq n$, then the length of each cycle is at most $n$ and so $|\rho(r)|=n$.

We now show that $|\rho(s)|=2$. Since $|s|=2$, then $|\rho(s)|$ divides 2 and so the length of each cycle is at most two. If $n$ is odd then $\rho(s)$ sends $\left\{1,\frac{n+1}{2}\right\}$ to $\left\{1,\frac{n+3}{2}\right\}$ and this creates a cycle of length two. If $n$ is even, $\rho(s)$ sends $\left\{1,\frac{n}{2}\right\}$ to $\left\{1,\frac{n+4}{2}\right\}$ and again, this makes a cycle of length two. Hence, $|\rho(s)|=2$.

Finally, we obtain

$$\rho(s)\rho(r)\rho(s)=\rho(srs)=\rho(r^{-1})=\rho(r)^{-1}.$$ 
\end{proof}


For $x\in D_n$, we now count the number of cycles in the decomposition of $\rho(x)$. We make use of the well-known properties of permutations stated as Lemma \ref{cycles}.

\begin{lemma}
Let $\alpha\in \Sym([n])$ such that $\alpha=c_1c_2\cdots c_l$, where $c_i$'s are disjoint cycles, then 
$$|\alpha|=\lcm(\length(c_i):i\in\left\{1,2,\ldots,l\right\}).$$ 
If $\alpha=(a_1\,\,a_2\,\,\ldots\,\,a_k)$, then the number of disjoint cycles of $\alpha^t$, where $1\leq t\leq k$, is $\gcd(k,t)$. 
\label{cycles}
\end{lemma}

\begin{lemma}
Let $n\geq 4$. For $i\in\left\{1,2,\ldots,n\right\}$,

\begin{displaymath} 
c(r^i)=\begin{cases}
\left(\frac{n-4}{2}\right)\gcd(n,i)+\gcd(\frac{n}{2},i), & \text{if $n$ is even}; \\
\left(\frac{n-3}{2}\right)\gcd(n,i), & \text{if $n$ is odd.}
\end{cases}
\end{displaymath}

\label{lem3}
\end{lemma}

\begin{proof}
We start with $n=4$. Then $\Delta_4=\left\{\left\{1,3\right\},\left\{2,4\right\}\right\}$, $i\in\left\{1,2,3,4\right\}$ and we obtain the following computations:
\begin{enumerate}
\item[$i=1$.] $\rho(r)=\left(\left\{1,3\right\}\,\,\left\{2,4\right\}\right)$ and so $c(r)=1=\left(\frac{4-4}{2}\right)\gcd(4,1)+\gcd(\frac{4}{2},1)$;
\item[$i=2$.] $\rho(r^2)=\left(\left\{1,3\right\}\right)\left(\left\{2,4\right\}\right)=1_{\Delta_4}$ and so $c(r^2)=2=\left(\frac{4-4}{2}\right)\gcd(4,2)+\gcd(\frac{4}{2},2)$;
\item[$i=3$.] $\rho(r^3)=\left(\left\{1,3\right\}\,\,\left\{2,4\right\}\right)$ and so $c(r^3)=1=\left(\frac{4-4}{2}\right)\gcd(4,3)+\gcd(\frac{4}{2},3)$;
\item[$i=4$.] $\rho(r^4)=\rho(1_{[4]})=1_{\Delta_4}=\left(\left\{1,3\right\}\right)\left(\left\{2,4\right\}\right)$ and so $c(r^4)=2=\left(\frac{4-4}{2}\right)\gcd(4,4)+\gcd(\frac{4}{2},4)$.
\end{enumerate}

We let $n>4$ and consider two cases. Firstly, assume $n$ is even. The elements of $\Delta_n$ can be partitioned according to different cycle lengths and we get the following cycle decomposition:
 
$$\rho(r)=\underbrace{\left(\left\{1,3\right\}\,\,\left\{2,4\right\}\,\,\ldots\,\,\left\{n,2\right\}\right)}_{n-\text{cycle}}\underbrace{\left(\left\{1,4\right\}\,\,\left\{2,5\right\}\,\,\ldots\,\,\left\{n,3\right\}\right)}_{n-\text{cycle}}\ldots$$
$$\underbrace{\left(\left\{1,n/2\right\}\,\,\left\{2,(n+2)/2\right\}\,\,\ldots\,\,\left\{n,(n-2)/2\right\}\right)}_{n-\text{cycle}}\underbrace{\left(\left\{1,(n+2)/2\right\}\,\,\left\{2,(n+4)/2\right\}\,\,\ldots\,\,\left\{n/2,n\right\}\right)}_{n/2-\text{cycle}}$$
in which there are $\frac{n-4}{2}$ $n$-cycles and only one $\frac{n}{2}$-cycle. For $i\in\left\{1,2,\ldots,n\right\}$:

$$\rho(r^i)=\left(\left\{1,3\right\}\,\,\left\{2,4\right\}\,\,\ldots\,\,\left\{n,2\right\}\right)^i\left(\left\{1,4\right\}\,\,\left\{2,5\right\}\,\,\ldots\,\,\left\{n,3\right\}\right)^i\ldots$$
$$\left(\left\{1,n/2\right\}\,\,\left\{2,(n+2)/2\right\}\,\,\ldots\,\,\left\{n,(n-2)/2\right\}\right)^i\left(\left\{1,(n+2)/2\right\}\,\,\left\{2,(n+4)/2\right\}\,\,\ldots\,\,\left\{n/2,n\right\}\right)^i.$$

By Lemma \ref{cycles}, we obtain

$$c(r^i)=\left(\frac{n-4}{2}\right)\gcd(n,i)+\gcd(n/2,i).$$

Secondly, take $n$ to be odd. Similar to the first case, the elements of $\Delta_n$ can be partitioned according to different cycle lengths. We obtain the following:
$$\rho(r)=\underbrace{\left(\left\{1,3\right\}\,\,\left\{2,4\right\}\,\,\ldots\,\,\left\{n,2\right\}\right)}_{n-\text{cycle}}\underbrace{\left(\left\{1,4\right\}\,\,\left\{2,5\right\}\,\,\ldots\,\,\left\{n,3\right\}\right)}_{n-\text{cycle}}\ldots$$
$$\underbrace{\left(\left\{1,(n+1)/2\right\}\,\,\left\{2,(n+3)/2\right\}\,\,\ldots\,\,\left\{n,(n-1)/2\right\}\right)}_{n-\text{cycle}}$$
in which there are $\frac{n-3}{2}$ $n$-cycles. As with the above, we can compute the following:
$$c(r^i)=\left(\frac{n-3}{2}\right)\gcd(n,i).$$
\end{proof}

\begin{lemma}
Let $n\geq 4$ and $s_v\in D_n\backslash \langle r\rangle$ be a reflection with axis passing through the center and a vertex. Then 

\begin{displaymath}
c(s_v)=\begin{cases}
\frac{n^2-2n}{4}, & \text{if $n$ is even}; \\
\frac{n^2-2n-3}{4}, & \text{if $n$ is odd.}
\end{cases}
\end{displaymath}

\label{lem4}
\end{lemma}

\begin{proof}
Note that the case $n=4$ is an easy computation. We consider two cases for $n>4$. Firstly, take $n$ to be even. The axis of $s_v$ is the  diagonal $\left\{i,i+\frac{n}{2}\bmod n\right\}$. Form 
$$\Delta_o=\left\{\left\{i-k\bmod n,i+k\bmod n\right\}:k\in\left\{1,2,\ldots,\frac{n-2}{2}\right\}\right\}.$$ 
Observe that $(i\pm k\bmod n)^{s_v}=i\mp k\bmod n$ and preserves both $i$ and $i+\frac{n}{2}\bmod n$. This implies that $s_v$ fixes setwise each element of $\Delta_o\cup\left\{\left\{i,i+\frac{n}{2}\bmod n\right\}\right\}$. Let $\left\{\alpha,\beta\right\}$ be an element of $\Delta_n\backslash\left(\Delta_o\cup\left\{\left\{i,i+\frac{n}{2}\bmod n\right\}\right\}\right)$, we consider three subcases. Let $\alpha=i$. It follows that $\beta\in\left\{i\pm k\bmod n: k\in\left\{2,\ldots,\frac{n-2}{2}\right\}\right\}$. If $\beta=i+k\bmod n$ then $\left\{i,i+k\bmod n\right\}^{s_v}=\left\{i,i-k\bmod n\right\}$. If $\beta=i-k\bmod n$ then $\left\{i,i-k\bmod n\right\}^{s_v}=\left\{i,i+k\bmod n\right\}$. Similar argument when $\alpha=i+\frac{n}{2}\bmod n$. Suppose $\left\{\alpha,\beta\right\}\cap\left\{i,i+\frac{n}{2}\bmod n\right\}$. It implies that $\alpha,\beta\in\left\{i\pm k\bmod n:k\in\left\{1,2,\ldots,\frac{n-2}{2}\right\}\right\}$. If $\alpha=i+k_1\bmod n$ and $\beta=i+k_2\bmod n$ where $k_1,k_2\in\left\{1,2,\ldots,\frac{n-2}{2}\right\}$, then $\left\{i+k_1\bmod n,i+k_2\bmod n\right\}^{s_v}=\left\{i-k_1\bmod n,i-k_2\bmod n\right\}$. Similar argument can be used for $\alpha=i-k_1\bmod n$ and $\beta=i-k_2\bmod n$. Without loss of generality, assume $\alpha=i-k_1\bmod n$ and $\beta=i+k_2\bmod n$. It means that $k_1\neq k_2$ and so $\left\{i-k_1\bmod n,i+k_2\bmod n\right\}^{s_v}=\left\{i+k_1\bmod n,i-k_2\bmod n\right\}$. In all these subcases, we obtain $\left\{\alpha,\beta\right\}^{s_v}\neq\left\{\alpha,\beta\right\}$.

Proposition \ref{mono} and Lemma \ref{cycles} assure that the length of every cycle in $\rho(s_v)$ is at most two. The above results tell us that each element of $\Delta_o\cup\left\{i,i+\frac{n}{2}\bmod n\right\}$ creates an $1$-cycle in $\rho(s_v)$, while each element of $\Delta_n\backslash\left(\Delta_o\cup\left\{i,i+\frac{n}{2}\bmod n\right\}\right)$ creates a $2$-cycle. Hence,
$$c(s_v)=\frac{n^2-2n}{4}.$$

For the second case, assume $n$ is an odd integer. The axis of $s_v$ is the segment extending from vertex $i$ to the midpoint of the edge $\left\{i+(n-1)/2\bmod n,i-(n-1)/2\bmod n\right\}$. Form 
$$\Delta_o=\left\{\left\{i+k\bmod n,i-k\bmod n\right\}:k\in\left\{1,2,\ldots,\frac{n-1}{2}\right\}\right\}.$$ 
Observe that $i^{s_v}=i$ and $(i\pm k\bmod n)^{s_v}=i\mp k\bmod n$. Thus, each element of $$\Delta_o\backslash\left\{\left\{i+\frac{n-1}{2}\bmod n,i-\frac{n-1}{2}\bmod n\right\}\right\}$$ creates an $1$-cycle in $\rho(s_v)$. Let $\left\{\alpha,\beta\right\}\in\Delta_n\backslash\Delta_o$. We consider two subcases. Without loss of generality, assume $\alpha=i$. It follows that $\beta\in\left\{i\pm k\bmod n:k\in\left\{2,\ldots,(n-1)/2\right\}\right\}$ and either $\left\{i,i+k\bmod n\right\}^{s_v}=\left\{i,i-k\bmod n\right\}$ or $\left\{i,i-k\bmod n\right\}^{s_v}=\left\{i,i+k\bmod n\right\}$. Let $i\notin\left\{\alpha,\beta\right\}$. It means that $\alpha,\beta\in\left\{i\pm k\bmod n:k\in\left\{1,2,\ldots,(n-1)/2\right\}\right\}$. As with the above, we always obtain $\left\{\alpha,\beta\right\}^{s_v}\neq\left\{\alpha,\beta\right\}$ in different subcases.

Since the length of each cycle of $\rho(s_v)$ is at most two, then the two subcases above imply that every $\left\{\alpha,\beta\right\}\in\Delta_n\backslash\Delta_o$ creates a $2$-cycle in $\rho(s_v)$. Hence,
$$c(s_v)=\frac{n^2-2n-3}{4}.$$
\end{proof}

\begin{lemma}
Let $n\geq 6$ be even. Suppose $s_e\in D_n\backslash \langle r\rangle$ to be a reflection with axis passing through the origin and midpoints of opposing edges. Then 
$$c(s_e)=\frac{n^2-2n-4}{4}$$
\label{lem5}
\end{lemma}
\begin{proof}
The axis of $s_e$ is the segment extending from the midpoint of an edge $\left\{i,i+1\bmod n\right\}$ to the midpoint of $\left\{i-\left(\frac{n}{2}-1\right)\bmod n,i+\frac{n}{2}\bmod n\right\}$. We note that for $j\in [n]$, $j^{s_e}=(2i+1)-j\bmod n$. Let 
$$\Delta_o=\left\{\left\{i+k\bmod n,i-k+1\bmod n\right\}:k\in\left\{2,3,\ldots,(n-2)/2\right\}\right\}.$$ 
It should be noted that $s_e$ fixes setwise each element of $\Delta_o$ and creates an $1$-cycle in $\rho(s_e)$.

For $\left\{\alpha,\beta\right\}\in\Delta_n\backslash\Delta_o$, there exists $k\in\left\{1,2,\ldots,\frac{n}{2}\right\}$ such that if $\alpha=i+k\bmod n$, then $\beta\in[n]\backslash\left\{i+k\bmod n,i-k+1\bmod n\right\}$ and so 
$$\left\{i+k\bmod n,\beta\right\}^{s_e}=\left\{i-k+1\bmod n,\beta^{s_e}\right\}\neq \left\{\alpha,\beta\right\}.$$
Also, if $\alpha=i-k+1\bmod n$ then $\beta\in[n]\backslash\left\{i+k\bmod n,i-k+1\bmod n\right\}$ and so
$$\left\{i-k+1\bmod n,\beta\right\}^{s_e}=\left\{i+k\bmod n,\beta^{s_e}\right\}\neq \left\{\alpha,\beta\right\}.$$
Hence, each element of $\Delta_n\backslash \Delta_o$ creates a $2$-cycle of $\rho(s_e)$. That is, 
$$c(s_e)=\frac{n^2-2n-4}{4}.$$
\end{proof} 

We now collect the properties from Lemmas \ref{lem3}, \ref{lem4} and \ref{lem5} and plug them in to the equation in Proposition \ref{prop1} to obtain our main result.
 
\begin{theorem}
Let $n\geq 3$. The number $\gamma(n)$ of distinct ways of dissecting an $n$-gon modulo the action of the dihedral group $D_n$ is:

\begin{displaymath}
\gamma(n)=\begin{cases}
\frac{1}{2n}\left(\left(\displaystyle\overset{n}{\underset{i=1}{\sum}}2^{\left(\frac{n-4}{2}\right)\gcd(n,i)+\gcd(\frac{n}{2},i)}\right) + \frac{n}{2}\left(2^{\frac{n^2-2n}{4}}+2^{\frac{n^2-2n-4}{4}}\right)\right), & \text{if $n$ is even}; \\
\frac{1}{2n}\left(\left(\displaystyle\overset{n}{\underset{i=1}{\sum}}2^{\left(\frac{n-3}{2}\right)\gcd(n,i)}\right) + n\left(2^{\frac{n^2-2n-3}{4}}\right)\right), & \text{if $n$ is odd.}
\end{cases}
\end{displaymath} 

\label{thm1}
\end{theorem}

\begin{corollary}
The number of dissections of a regular $p$-gon modulo the dihedral action, where $p$ is prime with $p\geq 3$, is 
$$\frac{(p-1)\cdot 2^{\frac{p-3}{2}}+2^{\frac{p^2-3p}{2}}+p\cdot2^{\frac{p^2-2p-3}{4}}}{2p}.$$
\label{thm2} 
\end{corollary}


\section{Remark}

The number $\gamma_c(n)$ of distinct ways of dissecting an $n$-gon modulo the action of the cyclic group $\langle (1\,\,2\,\,\ldots\,\,n)\rangle$ is

\begin{displaymath}
\gamma_c(n)=\begin{cases}
\frac{1}{n}\left(\displaystyle\overset{n}{\underset{i=1}{\sum}}2^{\left(\frac{n-4}{2}\right)\gcd(n,i)+\gcd(\frac{n}{2},i)}\right), & \text{if $n$ is even}; \\
\frac{1}{n}\left(\displaystyle\overset{n}{\underset{i=1}{\sum}}2^{\left(\frac{n-3}{2}\right)\gcd(n,i)}\right), & \text{if $n$ is odd.}
\end{cases}
\end{displaymath}

Moreover, when $n=p\geq 3$, then 

$$\gamma_c(p)=\frac{(p-1)\cdot 2^{\frac{p-3}{2}}+2^{\frac{p^2-3p}{2}}}{p}.$$

\section{Acknowledgment}

The authors would like to thank Cebu Normal University for its support while the research was being undertaken.

\begin{thebibliography}{5}
\bibitem{br} D. Bowman and A. Regev, Counting symmetry classes of dissections of a convex regular polygon, preprint, 2012, \url{http://arxiv.org/abs/1209.6270}.

\bibitem{ca} D. Callan, Polygon dissections and marked Dyck paths, preprint, 2005, \url{https://pdfs.semanticscholar.org/d405}.  

\bibitem{dm} J. Dixon and B. Mortimer, {\it Permutation Groups}, 
Springer-Verlag, 1996. 

\bibitem{ps} J. Przytycki and A. Sikora, Polygon dissections and Euler,
Fuss, Kirkman and Cayley numbers, preprint, 1998,
\url{https://arxiv.org/abs/math/9811086}.

\bibitem{s} A. Siegel,  Counting the number of distinct dissections of a regular $n$-gon, thesis, University of Akron, 2014.
\end{thebibliography}

\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05C25; Secondary 05C69.

\noindent \emph{Keywords: } 
$n$-gon, dissection, dihedral group.

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received September 13 2016; 
revised version received  May 9 2016; June 30 2017; July 18 2017.
Published in {\it Journal of Integer Sequences}, July 29 2017.

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\noindent
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