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\begin{center}
\vskip 1cm{\LARGE\bf 
$n$-Color Odd Self-Inverse Compositions
}
\vskip 1cm
\large
Yu-hong Guo\footnote{This work is supported by the
National Natural Science Foundation of China (Grant No.\ 11461020)
and the Fund of the Education Department of
Gansu Province (No.\ 2010-04.)}\\
School of Mathematics and Statistics \\
Hexi University \\
Gansu, Zhangye, 734000 \\
P. R. China \\
\href{mailto:gyh7001@163.com}{\tt gyh7001@163.com}\\
\end{center}

\vskip .2 in

\begin{abstract}
An $n$-color odd self-inverse composition is an $n$-color
self-inverse composition with odd parts. In this paper, we obtain
generating functions, explicit formulas, and recurrence formulas for
$n$-color odd self-inverse compositions.
\end{abstract}

\section{Introduction}\label{sec1}

In the classical theory of partitions, compositions were first
defined
 by MacMahon $\cite{a}$ as ordered partitions. For example, there are
 $5$ partitions and $8$ compositions of $4$. The partitions are
 $4$, $31$, $22$, $211$, $1111$ and the compositions are $4$, $31$, $13$, $22$, $211$, $121$,
 $112$, $1111$.

Agarwal and Andrews $\cite{b}$ defined an $n$-color partition as a
partition in which a part of size $n$ can come in $n$ different
colors. They denoted different colors by subscripts: $n_{1}$,
$n_{2}$, $\ldots$, $n_{n}$.  In analogy with MacMahon's ordinary
compositions, Agarwal $\cite{c}$ defined an
 $n$-color composition as an $n$-color ordered partition. Thus,
 for example, there are $8$ $n$-color compositions of $3$, viz.,
 \[
3_{1}, 3_{2},3_{3}, 2_{1}1_{1}, 2_{2}1_{1}, 1_{1}2_{1},1_{1}2_{2},
1_{1}1_{1}1_{1}.
\]

More properties of $n$-color compositions were given in
$\cite{d,e}$.

\begin{definition}${(\cite{a})}$
A composition is said to be self-inverse when the parts of the
composition read from  left to right are identical with the parts when read
from right to left.
\end{definition}

In analogy with the definition above for classical self-inverse
compositions, Narang and Agarwal $\cite{f}$ defined an $n$-color
self-inverse composition and gave some properties of them.

\begin{definition}${(\cite{f})}$
An $n$-color odd composition is an $n$-color  composition with odd
parts.
\end{definition}

For example there are $8$ $n$-color self-inverse compositions of
$4$, viz.,
 \[
4_{1}, 4_{2},4_{3},4_{4}, 2_{1}2_{1},
2_{2}2_{2},1_{1}2_{1}1_{1},1_{1}2_{2}1_{1}.
\]

In 2010, the author $\cite{g}$ also defined an $n$-color even self-inverse
composition and gave some properties.

\begin{definition}${(\cite{g})}$
An $n$-color even composition is an $n$-color composition whose parts
are even.
\end{definition}

\begin{definition}${(\cite{g})}$
An $n$-color even composition whose parts read from left to right are
identical with when read from right to left is called an $n$-color
even self-inverse composition.
\end{definition}

Thus, for example, there are $8$ $n$-color even self-inverse
compositions of $4$, viz.,
\[
4_{1}, 4_{2}, 4_{3}, 4_{4}, 2_{1}2_{1}, 2_{1}2_{2}, 2_{2}2_{1},
2_{2}2_{2}.
\]
And there are $6$ $n$-color even self-inverse compositions of $4$,
viz.,
\[
4_{1}, 4_{2},4_{3},4_{4}, 2_{1}2_{1}, 2_{2}2_{2}.
\]

Recently, the author $\cite{h}$ studied $n$-color odd compositions.

\begin{definition}${(\cite{h})}$
An $n$-color odd composition is an $n$-color composition whose parts
are odd.
\end{definition}

Thus, for example, there are $7$ $n$-color odd compositions of $4$,
viz.,
\[
3_{1}1_{1}, 3_{2}1_{1},3_{3}1_{1}, 1_{1}3_{1}, 1_{1}3_{2},
1_{1}3_{3},1_{1}1_{1}1_{1}1_{1}.
\]

In this paper, we shall study $n$-color odd self-inverse
compositions.

\begin{definition}
An $n$-color odd composition whose parts read from left to right are
identical with when read from right to left is called an $n$-color odd
self-inverse composition.
\end{definition}

Thus, for example, there are $4$ $n$-color odd self-inverse
compositions of $6$, viz.,
\[
3_{1}3_{1}, 3_{2}3_{2}, 3_{3}3_{3}, 1_{1}1_{1}1_{1}1_{1}1_{1}1_{1}.
\]
\par
 In section \ref{sec2} we shall give explicit formulas, recurrence formulas, generating functions
 for $n$-color odd self-inverse compositions.

The author $\cite{h}$ proved the following theorems.

 \begin{theorem}${(\cite{h})}$
Let $C_{o}(m,q)$ and $C_{o}(q)$ denote the enumerative generating
functions for $C_{o}(m,\nu)$ and $C_{o}(\nu)$, respectively, where
$C_{o}(m,\nu)$ is the number of $n$-color odd compositions of $\nu$
into $m$ parts and $C_{o}(\nu)$ is the number of $n$-color odd
compositions of $\nu$. Then

\begin{eqnarray}\label{squaresum}
&amp;&amp;C_{o}(m,q)=\frac{q^{m}(1+q^{2})^{m}}{(1-q^{2})^{2m}},\\
&amp;&amp;C_{o}(q)=\frac{q+q^{3}}{1-q-2q^{2}-q^{3}+q^{4}},\\
&amp;&amp;C_{o}(m,\nu)=\sum_{i+j=\frac{\nu-m}{2}}{{2m+i-1}\choose{2m-1}}{{m}\choose{j}}, \label{a}\\
&amp;&amp;C_{o}(\nu)=\sum_{m \leq \nu}
~\sum_{i+j=\frac{\nu-m}{2}}{{2m+i-1}\choose{2m-1}}{{m}\choose{j}}
\label{b}.
\end{eqnarray}
where $(\nu-m)$ is even, and $(\nu-m) \geq 0$; $0\leq i,j$ are
integers.
\end{theorem}

\begin{theorem}${(\cite{h})}$ \label{th1}
Let $C_{o}(\nu)$ denote the number of $n$-color  odd compositions of
$\nu$. Then
\begin{eqnarray*}
&amp;&amp;C_{o}(1)=1, C_{o}(2)=1, C_{o}(3)=4, C_{o}(4)=7~~and\\
&amp;&amp;C_{o}(\nu) = C_{o}(\nu-1)+2C_{o}(\nu-2)+C_{o}(\nu-3)-C_{o}(\nu-4)~for~\nu&gt;4.
\end{eqnarray*}
\end{theorem}

\section{Main results}\label{sec2}

In this section, we first prove the following explicit formulas for
the number of $n$-color odd self-inverse compositions.
\begin{theorem}\label{th2}
Let $S(O, \nu)$ denote the number of $n$-color odd self-inverse
compositions of $\nu$. Then
\begin{eqnarray*}
(1)~~S(O,
2\nu+1)=(2\nu+1)+\sum_{t=1}^{2\nu-1}\sum_{m\leq\frac{2\nu+1-t}{2}}\sum_{i+j=\frac{2\nu+1-t-2m}{4}}t{{2m+i-1}\choose{2m-1}}{{m}\choose{j}},
\end{eqnarray*}
where $\nu=0,1,2,\ldots$;~~~$t=2k+1,k=0,1,2,\ldots,(\nu-1)$;~~$0
\leq\frac{2\nu+1-t-2m}{2}$ is even; $0\leq i,j$ are integers.
\begin{eqnarray*}
(2) ~~S(O, 2\nu)=\sum_{m \leq \nu}
~\sum_{i+j=\frac{\nu-m}{2}}{{2m+i-1}\choose{2m-1}}{{m}\choose{j}},
\end{eqnarray*}
where $0 \leq\nu-m$ is even, and $0\leq i,j$ are integers.
\end{theorem}

\begin{proof}  $(1)$~~~Obviously, an odd number which is
$2\nu+1~(\nu=0,1,2,\ldots)$ can have odd self-inverse $n$-color
compositions only when the number of parts is odd. There are
$2\nu+1$ $n$-color odd self-inverse compositions when the number of
parts is only one. An odd self-inverse compositions of $2\nu+1$ into
$2m+1~(m\geq1)$ parts can be read as a central part, say, $t$ (where
$t$ is odd) and two identical odd $n$-color compositions of
$\frac{2\nu+1-t}{2}$ into $m$ parts on each side of the central
part. The number of odd $n$-color compositions of
$\frac{2\nu+1-t}{2}$ into $m$ parts is $C_{o}(m,\frac{2\nu+1-t}{2})$
by equation (\ref{a}). Now the central part can appear in $t$ ways.
Therefore, the number of $n$-color odd self-inverse compositions of
$2\nu+1$ is
\begin{eqnarray*}
S(O,
2\nu+1)&amp;=&amp;(2\nu+1)+\sum_{t=1}^{2\nu-1}\sum_{m\leq\frac{2\nu+1-t}{2}}t
C_{o}\left(m,\frac{2\nu+1-t}{2}\right)\\
&amp;=&amp;(2\nu+1)+\sum_{t=1}^{2\nu-1}\sum_{m\leq\frac{2\nu+1-t}{2}}\sum_{i+j=\frac{2\nu+1-t-2m}{4}}t{{2m+i-1}\choose{2m-1}}{{m}\choose{j}}.
\end{eqnarray*}
\par
~$(2)$ ~~For even numbers $2\nu~(\nu=1, 2, \ldots)$, we can have odd
self-inverse $n$-color compositions only when the number of parts is
even, and the two identical odd $n$-color compositions are exactly
odd $n$-color compositions of $\nu$, from equation (\ref{b}) we see that
the number of these is
\[
\sum_{m\leq
\nu}\sum_{i+j=\frac{\nu-m}{2}}{{2m+i-1}\choose{2m-1}}{{m}\choose{j}}.
\]
\par
Hence, the number of $n$-color odd self-inverse compositions of
$2\nu$  is
\begin{eqnarray*}
S(O, 2\nu)=\sum_{m\leq
\nu}\sum_{i+j=\frac{\nu-m}{2}}{{2m+i-1}\choose{2m-1}}{{m}\choose{j}}.
\end{eqnarray*}

We complete the proof of this theorem.
\end{proof}

From the proof of this theorem we can see that odd $n$ have
$n$-color odd self-inverse compositions where the number of parts is
odd. And even $n$ have $n$-color odd self-inverse
compositions where the number of parts is even. Let $S_{o}(\nu,m)$
denote the number of $n$-color odd self-inverse compositions of
$\nu$ into $m$ parts. Then we can get the following formula easily.
\[
 S_{o}(2k+1, 2l+1)=\sum_{t=1}^{2k-1}\sum_{i+j=\frac{2k+1-t-2l}{4}}{{2l+i-1}\choose{2l-1}}{{l}\choose{j}}.
 \]
where $t$ is odd, $k,l$ are integers and $k,l\geq 0$.
\[
  S_{o}(2k, 2l)=
\sum_{i+j=\frac{k-l}{2}}{{2l+i-1}\choose{2l-1}}{{l}\choose{j}}.
\]
where $k,l$ are integers and $k,l\geq 0$.

Now $S_{o}(\nu,m)$ with $\nu, m=1,2,\ldots,20$ is given in Tables
\ref{tab:time} and \ref{tab:time1}.

\begin{table}[htp]
\centering
\caption{$S_{o}(\nu,m)$ when both $\nu$ and $m$ are
odd}\label{tab:time}
\begin{tabular}{c|cccccccccc|c}
\hline
\backslashbox{$\nu$}{$m$} &amp;1  &amp;3  &amp;5  &amp;7  &amp;9  &amp;11 &amp;13 &amp;15 &amp;17 &amp;19 &amp;
$s_{\nu}$
\\
\hline
 1       &amp;1       &amp;  0       &amp;0       &amp;0        &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;1\\
 3       &amp;3       &amp;  1       &amp;0       &amp;0        &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;4\\
 5       &amp;5       &amp;  3       &amp;1       &amp;0        &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;9\\
 7       &amp;7       &amp;  8       &amp;3       &amp;1        &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;19\\
 9       &amp;9       &amp; 16       &amp;11      &amp;3        &amp;1       &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;40\\
11       &amp;11      &amp; 29       &amp;25      &amp;16       &amp;3       &amp;1       &amp;0       &amp;0       &amp;0       &amp;0       &amp;83\\
13       &amp;13      &amp; 49       &amp;56      &amp;34       &amp;17      &amp;3       &amp;1       &amp;0       &amp;0       &amp;0       &amp;173\\
15       &amp;15      &amp; 72       &amp;110     &amp;96       &amp;43      &amp;20      &amp;3       &amp;1       &amp;0       &amp;0       &amp;360\\
17       &amp;17      &amp;104       &amp;206     &amp;200      &amp;143     &amp;52      &amp;23      &amp;3       &amp;1       &amp;0       &amp;749\\
19       &amp;19      &amp;145       &amp;346     &amp;442      &amp;317     &amp;199     &amp;61      &amp;26      &amp;3       &amp;1       &amp;1559\\
\hline
\end{tabular}
\end{table}

From Tables \ref{tab:time} and \ref{tab:time1} we can see the
recurrence formulas for the number of the $n$-color odd self-inverse
compositions of $\nu$. So we prove the following recurrence
relations.

\begin{table}[h]
\centering
\caption{$S_{o}(\nu,m)$ when both $\nu$ and $m$ are even}\label{tab:time1}
\begin{tabular}{c|cccccccccc|c}
\hline
\backslashbox{$\nu$}{$m$} &amp;2  &amp;4  &amp;6  &amp;8  &amp;10  &amp;12 &amp;14 &amp;16 &amp;18 &amp;20 &amp;
$t_{\nu}$
\\
\hline
 2        &amp;1       &amp; 0       &amp;0        &amp;0        &amp;0       &amp;0       &amp;0       &amp;0       &amp;0       &amp;0      &amp;1\\
 4   &amp;0   &amp; 1   &amp;0   &amp;0    &amp;0   &amp;0   &amp;0   &amp;0   &amp;0   &amp;0   &amp;1\\
 6   &amp;3   &amp; 0   &amp;1   &amp;0    &amp;0   &amp;0   &amp;0   &amp;0   &amp;0   &amp;0   &amp;4\\
 8   &amp;0   &amp; 6   &amp;0   &amp;1    &amp;0   &amp;0   &amp;0   &amp;0   &amp;0   &amp;0   &amp;7\\
10   &amp;5   &amp; 0   &amp;9   &amp;0    &amp;1   &amp;0   &amp;0   &amp;0   &amp;0   &amp;0   &amp;15\\
12   &amp;0   &amp; 19  &amp;0   &amp;12   &amp;0   &amp;1   &amp;0   &amp;0   &amp;0   &amp;0   &amp;32\\
14   &amp;7   &amp; 0   &amp;42  &amp;0    &amp;15  &amp;0   &amp;1   &amp;0   &amp;0   &amp;0   &amp;65\\
16   &amp;0   &amp; 44  &amp;0   &amp;74   &amp;0   &amp;18  &amp;0   &amp;1   &amp;0   &amp;0   &amp;137\\
18   &amp;9   &amp;0    &amp;138 &amp;0    &amp;115 &amp;0   &amp;21  &amp;0   &amp;1   &amp;0   &amp;284\\
20   &amp;0   &amp;85   &amp;0   &amp;316  &amp;0   &amp;165 &amp;0   &amp;24  &amp;0   &amp;1   &amp;591\\
\hline
\end{tabular}
\end{table}



\begin{theorem}\label{th3}
Let $s_{\nu}$ and $t_{\nu}$ denote the number of $n$-color  odd
self-inverse compositions for $2\nu+1$ and $2\nu$, respectively.
Then
\begin{eqnarray*}
(1)~~&amp;s_{0}&amp;=1,~ s_{1}=4, ~s_{2}=9,~ s_{3}=19 ~~and\\
&amp;s_{\nu}&amp;=s_{\nu-1}+2s_{\nu-2}+s_{\nu-3}-s_{\nu-4}~~~for~~~\nu&gt;3\\
(2)~~ &amp;t_{1}&amp;=1,~t_{2}=1, ~ t_{3}=4, ~t_{4}=7~~ and \\
&amp;t_{\nu}&amp;=t_{\nu-1}+2t_{\nu-2}+t_{\nu-3}-t_{\nu-4}~~~for~~~\nu&gt;4.
\end{eqnarray*}
\end{theorem}

\begin{proof} (Combinatorial)~~$(1)$~ To prove that
$s_{\nu}=s_{\nu-1}+2s_{\nu-2}+s_{\nu-3}-s_{\nu-4}$,  we split the
$n$-color odd self-inverse
compositions enumerated by $s_{\nu}+s_{\nu-4}$ into four classes:

\medskip

\noindent (A)~$s_{\nu}$ with $1_{1}$ on both ends.\\
(B)~$s_{\nu}$ with $3_{3}$ on both ends.\\
(C)~$s_{\nu}$ with $h_{t}$ on both ends, $h&gt;1$, $1\leq t \leq h-2$
and $n$-color odd self-inverse
compositions of $2\nu+1$ of form $(2\nu+1)_{u}$, $1\leq u \leq 2\nu-3$.\\
(D)~$s_{\nu}$ with $h_{t}$ on both ends except $3_{3}$, $h&gt;1$,
$h-1\leq t \leq h$, $(2\nu+1)_{u}$, $2\nu-2 \leq u \leq2\nu+1$ and
those enumerated by $s_{\nu-4}$. \vskip 10pt
\par
We transform the $n$-color odd self-inverse compositions in class
(A) by deleting $1_{1}$ on  both ends. This produces $n$-color odd
self-inverse compositions enumerated by $s_{\nu-1}$. Conversely, for
any $n$-color odd composition enumerated by $s_{\nu-1}$ we add
$1_{1}$  on both ends to produce the elements of the class (A). In
this way we establish that there are exactly $s_{\nu-1}$ elements in
the  class (A).
\par
Similarly, we can produce $s_{\nu-3}$ $n$-color odd self-inverse
compositions in class (B) by deleting $3_{3}$ on  both ends.
\par
 Next, we transform the $n$-color odd self-inverse compositions
in class (C) by subtracting $2$ from $h$, that is, replacing $h_{t}$
by $(h-2)_{t}$ and subtracting $4$ from $2\nu+1$ of $(2\nu+1)_{u}$,
$1\leq u \leq 2\nu-3 $. This transformation also establishes the
fact that there are exactly $s_{\nu-2}$ elements in class (C).
\par
Finally, we transform the elements in class (D) as follows: Subtract
$2_{2}$ from $h_{t}$ on both ends, that is, replace $h_{t}$ by
$(h-2)_{(t-2)}$, $h&gt;3$, $h-1\leq t\leq h$, while replace $h_{t}$ by
$(h-2)_{(t-1)}$ when $h=3$, $ t= 2$.  We will get those $n$-color
odd self-inverse compositions of $2\nu-3$ with $h_{t}$ on both
ends,~$h-1\leq t\leq h$ except self-inverse odd compositions in one
part.  We also replace $(2\nu+1)_{u}$ by $(2\nu-3)_{u-4}$, $2\nu-2
\leq u \leq 2\nu+1$. To get the remaining $n$-color odd compositions
from $s_{\nu-4}$ we add $2$ to both ends, that is, replace $h_{t}$
by $(h+2)_{t}$. For $n$-color odd self-inverse compositions into one
part we add $4$ ,that is, replace $(2\nu-7)_{t}$ by $(2\nu-3)_{t},
1\leq t \leq 2\nu-7 $. We see that the number of $n$-color odd
self-inverse compositions in class (D) is also equal to $s_{\nu-2}$.
Hence, we have $s_{\nu}+s_{\nu-4}=s_{\nu-1}+2s_{\nu-2}+s_{\nu-3}$.
viz., $s_{\nu}=s_{\nu-1}+2s_{\nu-2}+s_{\nu-3}-s_{\nu-4}$.

$(2)$ From Theorem \ref{th1} and Theorem \ref{th2}, we obtain the
recurrence formula of $t_{\nu}$ easily. Thus, we complete the proof.
\end{proof}

We easily get the following generating functions by the
recurrence relations.

\begin{corollary}\label{th7}
\begin{eqnarray*}
&amp;(1)&amp; ~~~~~~\sum_{\nu=0}^{\infty}s_{\nu}q^{\nu}=\frac{(1+q)^{3}}{1-q-2q^{2}-q^{3}+q^{4}}.\\
 &amp;(2)&amp; ~~~~~~\sum_{\nu=1}^{\infty}t_{\nu}q^{\nu}=\frac{q+q^{3}}{1-q-2q^{2}-q^{3}+q^{4}}.
\end{eqnarray*}
\end{corollary}

\section{Acknowledgments}\label{sec3}

 The author would like to thank the referee
for his/her suggestions and comments which have improved the
quality of this paper. 

\begin{thebibliography}{99}
\bibitem{b}{A. K. Agarwal and G. E. Andrews, Rogers-Ramanujan identities for partition with `$n$ copies of $n$', \emph{J. Combin. Theory Ser. A} {\bf 45} (1987), 40--49.}

\bibitem{c} {A. K. Agarwal, $N$-colour compositions, \emph{Indian J. Pure Appl. Math.} {\bf 31} (2000), 1421--1427.}

\bibitem{d} {A. K. Agarwal, An analogue of Euler's identity and new combinatorial  properties of $n$-colour compositions, \emph{J. Comput. Appl. Math.} {\bf 160} (2003), 9--15.}

\bibitem{j} G. E. Andrews, \emph{The Theory of Partitions}, 
Encyclopedia of Mathematics and Its Applications,
Cambridge University Press, 1998.

\bibitem{e} {Yu-Hong Guo, Some identities between partitions and compositions, \emph{Acta Math. Sinica (Chin. Ser.)} {\bf 50} (2007), 707--710.}

\bibitem{g} {Yu-Hong Guo, $N$-colour even self-inverse compositions, \emph{Proc. Indian Acad. Sci. Math. Sci.} {\bf 120} (2010), 27--33.}

\bibitem{h} Yu-Hong Guo, Some $n$-color compositions, \emph{J. Integer
Sequences} {\bf 15} (2012), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Guo/guo4.html}{Article 12.1.2}.

\bibitem{i} B. Hopkins, Spotted tilling and $n$-color compositions,
\emph{Integers} {\bf 12B} (2012/13), Article A6.

\bibitem{a}  P. A. MacMahon, \emph{Combinatory Analysis}, AMS Chelsea
Publishing, 2001.

\bibitem{f} G. Narang and A. K. Agarwal, $N$-colour self-inverse
compositions, \emph{Proc. Indian Acad. Sci. Math. Sci.} {\bf 116}
(2006), 257--266.


\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: 05A17.

\noindent \emph{Keywords: } $n$-color odd self-inverse composition,
generating function,  explicit formula, recurrence formula.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 18 2013;
revised versions received  May 5 2014; September 16 2014.
Published in {\it Journal of Integer Sequences}, November 4 2014.

\bigskip
\hrule
\bigskip

\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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