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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\LARGE\bf 
Enumeration of the Partitions of an Integer \\
\vskip .05in
into Parts of a Specified Number of Different \\
\vskip .16in
Sizes and Especially Two Sizes
}
\vskip 1cm
\large
Nesrine Benyahia Tani\\
Algiers 3 University\\
Faculty of Economics and Management Sciences\\
2 Ahmed Waked Street \\
Dely Brahim, Algiers \\
Algeria\\
\href{mailto:benyahiatani@yahoo.fr}{\tt benyahiatani@yahoo.fr}\\ 
\ \\
Sadek Bouroubi\footnote{Both authors were supported by the laboratory LAID3.} \\
University of Science and Technology Houari Boumediene \\
Faculty of Mathematics\\
P. O. Box 32 \\
16111 El-Alia, Bab-Ezzouar, Algiers\\
Algeria\\
\href{mailto:sbouroubi@usthb.dz}{\tt sbouroubi@usthb.dz} \\
\href{mailto:bouroubis@yahoo.fr}{\tt bouroubis@yahoo.fr} \\
\end{center}

\vskip .2 in

\begin{abstract}
A partition of a non-negative integer $n$ is a way of writing $n$ as a
sum of a nondecreasing sequence of parts. The present paper provides
the number of partitions of an integer $n$ into parts of a specified
number of different sizes. We establish new formulas for such
partitions with particular interest to the number of partitions of $n$
into parts of two sizes. A geometric application is given at the end of
this paper.
\end{abstract}

\section{Introduction and definitions}
Let $n$ and $k$ be integers. A partition of $n$ into $k$ parts is an
integral solution of the system
 \[  \begin{cases}
   n=n_{1}+\cdots+n_{k};\\
  1\leq n_{1}\leq\cdots\leq n_{k}.\\
    \end{cases}
 \]

Euler was the first to undertake the problem of counting an integer's
partitions. Since then, mathematicians have been more and more
interested in integer partitions and their fascinating properties. In
fact, in the theory of integer partitions, various restrictions on the
nature of partitions are often considered. One may require that the
$n_i$'s be distinct, odd or even, or that $n$ must be split into
exactly $k$ parts, etc. More on integer partitions can be found in
\cite{and, bou, ben, charal, co} and \cite{redem, Pak, wilf}.

Here, we are interested in partitions of the number $n$ into parts of
precisely $s$ different sizes. Extending prior results, we derive
several identities linking this kind of partitions to the number of
divisors $\tau(n)$. In addition, we obtain new recurrence formulas to
count the number of such partitions and a new identity to count the
number of partitions of an integer into two sizes of parts.

Let $t(n,k,s)$ be the number of partitions of $n$ into $k$ parts of precisely $s$ different sizes, $k = s,\ldots,n-\frac{s(s-1)}{2}$, it is an
integral solution of the system
\begin{eqnarray}\label{S1}
   \begin{cases}
 n=a_{1}\ n_{1}+\cdots+a_{s}\ n_{s};  \\
 1\leq n_{1}<\cdots<n_{s}; \\
   a_{1}+\cdots+a_{s}=k;  \\
a_{1},\ldots, a_{s}\geq 1.\\
    \end{cases}
 \end{eqnarray}
The total number of partitions of $n$ into $s$ different sizes of parts is denoted
$t(n,s)$ (see \seqnum{A002133} for $t(n,2)$). 

If $s$ is specified, then $t(n,k,s)=0$ if $k\leq s-1$ and either $k>n-\frac{s(s-1)}{2}$ or $n<\max\{k,\frac{s(s+1)}{2}\}$. Then we have
\begin{equation}\label{eq1}
t(n,s)=\sum_{k=s}^{\frac{2n-s(s-1)}{2}}
t(n,k,s)=\sum_{k\geq 1}t(n,k,s).
\end{equation}
For instance, if $s=1$, then $k\geq1$, $n\geq k$, and
\begin{equation}\label{eq2}
t(n,k,1)= \begin{cases}
1,&\text{if $n$ is a multiple of $k$;}  \\
0,&\text{otherwise}.\\
    \end{cases}
\end{equation}
Therefore
\begin{equation*}
   \underset{n\geq k} \sum t(n,k,1)\ q^n = \dfrac{q^k}{1-q^k}\cdot
\end{equation*}
Also, it is easy to see that $$t(n,2,2)=\left\lfloor\dfrac{n-1}{2}\right\rfloor,$$
where $\left\lfloor x\right\rfloor$ is the greatest integer $\leq x$. So, we have
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
 \underset{n\geq k} \sum t(n,2,2)\ q^n &=& q^3 + q^4 + 2 q^5 + 2 q^6 + 3 q^7 + 3 q^8+ \cdots \\
  &=& \dfrac{q^3}{1-q}+ \dfrac{q^5}{1-q}+ \dfrac{q^7}{1-q}+\cdots\\
  &=& \dfrac{q^3}{(1-q)(1-q^2)}\cdot\\
\end{eqnarray*}
P. A. MacMahon \cite{ma} was the first mathematician interested in this kind of partitions. Also, Emeric Deutsch studied the number of partitions of $n$ into exactly two odd sizes of parts (see \seqnum{A117955})  and the number of partitions of $n$ into exactly two sizes of parts, one odd and one even (see \seqnum{A117956}).
\section{Preliminary results}
Throughout the remainder of the paper, let $\tau(k)$ and let
$\tau_{d\downarrow}(k)$ be respectively the number of positive divisors of
$k$ and the number of positive divisors of $k$ less than or equal to
$d$.

In this section we state some recurrence formulas involving the number $t(n,k,s)$. The main identity of the present work is based on the following results:
\begin{theorem}\label{the1}
Let $n$, $k$ and $s$ be integers. For $k\geq s\geq 2$, $n\geq k+\frac{s(s-1)}{2}$ and $n\geq\max\{k,\frac{s(s+1)}{2}\}$, we have
\begin{eqnarray}\label{eq3}
t(n,k,s)=\underset{i=1} {\overset{\lfloor\frac{2n-s(s-1)}{2k}\rfloor}\sum}{
\underset{j=1}{\overset{k-s+1}\sum}{t(n-k i,k-j,s-1)}},
\end{eqnarray}
and
\begin{eqnarray}\label{eq4}
t(n,k,2)=\dsum_{i=1}^{\lfloor\frac{n-1}{k}\rfloor}
\tau_{k-1\downarrow}(n-k
i).
\end{eqnarray}
\end{theorem}
\begin{proof}
Note that every part $n_{i}$ in System (\ref{S1}), for
$i=2,\ldots,s$, can be written as $n_{i}=n_{1}+d_{i}$,
$d_{i}\geq 1$. Considering $n_{1}$ and $a_{1}$ as parameters, System (\ref{S1}) can be rewritten as follows:
\begin{eqnarray}\label{S2}
 \begin{cases}
n-kn_{1}=a_{2}\ d_{2} + \cdots +a_{s}\ d_{s};  \\
1\leq d_{2}<\cdots<d_{s};\\
a_{2}+\cdots+a_{s}=k-a_{1};\\
a_{1},\ldots, a_{s}\geq 1.
    \end{cases}
\end{eqnarray}

Hence, we get
\begin{equation*}
    t(n,k,s)=\sum_{n_{1}\in \mathcal{N}}\sum_{a_{1}\in \mathcal{A}}t(n-k
n_{1},k-a_{1},s-1),
\end{equation*}
where $\mathcal{N}$ and $\mathcal{A}$ are the sets containing the
values of $n_{1}$ and $a_{1}$ respectively.

The smallest values of $n_{1}$ and $a_{1}$ is 1. The
largest value of $n_{1}$ is found by setting $a_{i+1}=1$
\textrm{and}\ $d_{i+1}=i, \textrm{for}\ i=1,\ldots,s-1$, in the first
equation of System (\ref{S2}). Then, we get
$$1\leq n_{1}\leq \left\lfloor\frac{2n-s(s-1)}{2k}\right\rfloor.$$
Setting $a_{i}=1,
\textrm{for}\ i=2,\ldots,s$ in the third equation of System (\ref{S2}), one can see that
the largest value of $a_{1}$ is $k-s+1$.

To prove (\ref{eq4}) we apply (\ref{eq3}) with $s=2$,
\begin{eqnarray*}
t(n,k,2)=\sum_{i=1}^{\lfloor\frac{n-1}{k}\rfloor} \sum_{j=1}^{k-1}t(n-k
i,j,1).
\end{eqnarray*}
So by (\ref{eq2}) we get
\begin{eqnarray*}
\sum_{j=1}^{k-1}t(n-k i,j,1)=\tau_{k-1\downarrow}(n-k i).
\end{eqnarray*}
This implies (\ref{eq4}).
\end{proof}
Theorem \ref{the1} allows the easy recovery of known
identities such as,
\begin{eqnarray}\label{eq5}
t(n,2,2)=\left\lfloor\dfrac{n-1}{2}\right\rfloor.
\end{eqnarray}
Also, it allows to deduce some new values for $t(n,k,2)$.
For instance, for $k=3\ldots 6$, we have
\begin{corollary}\label{cor1}
For $n\geq3$, we have
\begin{eqnarray*}
t(n,3,2)= \begin{cases}
\dfrac{n-3}{3}+\left\lfloor\dfrac{n-3}{6}\right\rfloor, &\textrm{\ if\ }\  n \equiv0\ ({\rm mod }\ 3);\\
\ \\
\dfrac{n-1}{3}+\left\lfloor\dfrac{n-1}{6}\right\rfloor, &\textrm{\ if\ }\  n \equiv1\ ({\rm mod }\ 3);\\
\ \\
\dfrac{n-2}{3}+\left\lfloor\dfrac{n+1}{6}\right\rfloor, &\textrm{\ if\ }\  n \equiv2\ ({\rm mod }\ 3).
    \end{cases}
\end{eqnarray*}

\begin{eqnarray*}
t(n,4,2)= \begin{cases}
\dfrac{n-4}{2}+\left\lfloor\dfrac{n-4}{12}\right\rfloor, &\textrm{\ if\ }\  n \equiv0\ ({\rm mod }\ 4);\\
\ \\
\dfrac{n-1}{4}+\left\lfloor\dfrac{n-1}{12}\right\rfloor, &\textrm{\ if\ }\  n \equiv1\ ({\rm mod }\ 4);\\
\ \\
\dfrac{n-2}{2}+\left\lfloor\dfrac{n+2}{12}\right\rfloor, &\textrm{\ if\ }\  n \equiv2\ ({\rm mod }\ 4);\\
\ \\
\dfrac{n-3}{4}+\left\lfloor\dfrac{n+5}{12}\right\rfloor, &\textrm{\ if\ }\  n \equiv3\ ({\rm mod }\ 4).\\
\end{cases}
\end{eqnarray*}

\begin{eqnarray*}
t(n,5,2)=\begin{cases}
\dfrac{n-5}{5}+\left\lfloor\dfrac{n-5}{10}\right\rfloor+\left\lfloor\dfrac{n-5}{15}\right\rfloor+\left\lfloor\dfrac{n-5}{20}\right\rfloor, &\textrm{\ if\ }\  n \equiv0\ ({\rm mod }\ 5);\\
\ \\
\dfrac{n-1}{5}+\left\lfloor\dfrac{n-1}{10}\right\rfloor+\left\lfloor\dfrac{n+4}{15}\right\rfloor+\left\lfloor\dfrac{n-1}{20}\right\rfloor, &\textrm{\ if\ }\  n \equiv1\ ({\rm mod }\ 5);\\ 
\ \\
\dfrac{n-2}{5}+\left\lfloor\dfrac{n+3}{10}\right\rfloor+\left\lfloor\dfrac{n-2}{15}\right\rfloor+\left\lfloor\dfrac{n+3}{20}\right\rfloor, &\textrm{\ if\ }\  n \equiv2\ ({\rm mod }\ 5);\\
\ \\
\dfrac{n-3}{5}+\left\lfloor\dfrac{n-3}{10}\right\rfloor+\left\lfloor\dfrac{n+2}{15}\right\rfloor+\left\lfloor\dfrac{n+7}{20}\right\rfloor, &\textrm{\ if\ }\  n \equiv3\ ({\rm mod }\ 5);\\ 
\ \\
\dfrac{n-4}{5}+\left\lfloor\dfrac{n+1}{10}\right\rfloor+\left\lfloor\dfrac{n+1}{15}\right\rfloor+\left\lfloor\dfrac{n+11}{20}\right\rfloor, &\textrm{\ if\ }\  n \equiv4\ ({\rm mod }\ 5).
\end{cases}
\end{eqnarray*}

\begin{eqnarray*}
t(n,6,2)=\begin{cases}
\dfrac{n-6}{2}+\left\lfloor\dfrac{n-6}{12}\right\rfloor+\left\lfloor\dfrac{n-6}{30}\right\rfloor, &\textrm{\ if\ }\  n \equiv0\ ({\rm mod }\ 6);\\
\ \\
\dfrac{n-1}{6}+\left\lfloor\dfrac{n-1}{30}\right\rfloor, &\textrm{\ if\ }\  n \equiv1\ ({\rm mod }\ 6);\\
\ \\
\dfrac{n-2}{3}+\left\lfloor\dfrac{n-2}{12}\right\rfloor+\left\lfloor\dfrac{n+4}{30}\right\rfloor, &\textrm{\ if\ }\  n \equiv2\ ({\rm mod }\ 6);\\
\ \\
\dfrac{n-3}{3}+\left\lfloor\dfrac{n+9}{30}\right\rfloor, &\textrm{\ if\ }\  n \equiv3\ ({\rm mod }\ 6);\\ 
\ \\
\dfrac{n-4}{3}+\left\lfloor\dfrac{n+2}{12}\right\rfloor+\left\lfloor\dfrac{n+14}{30}\right\rfloor, &\textrm{\ if\ }\  n \equiv4\ ({\rm mod }\ 6);\\a
\ \\
\dfrac{n-5}{6}+\left\lfloor\dfrac{n+19}{30}\right\rfloor, &\textrm{\ if\ }\ n \equiv5\ ({\rm mod }\ 6).
\\
\end{cases}
\end{eqnarray*}
\end{corollary}
\begin{proof}
The results follow immediately from Theorem \ref{the1}.
\end{proof}
\begin{corollary}\label{cor2}
For $n\geq3$ and $\lceil\frac{n+1}{2}\rceil\leq k\leq n-1$, we have
\begin{eqnarray*}
t(n,k,2)=\tau(n-k).
\end{eqnarray*}
\end{corollary}
\begin{proof}
On the one hand, the sum in (\ref{eq4}) is reduced to one element if
$1\leq\frac{n-1}{k}<2$, i.e.,
$$\frac{n-1}{2}<k\leq n-1.$$
On the other hand, $\tau_{k-1\downarrow}(n-k)=\tau(n-k)$, if
$k-1\geq n-k$, i.e.,
$$k\geq \frac{n+1}{2}\ \cdot$$
Hence the result follows.
\end{proof}

\begin{remark} From Corollary \ref{cor2}, for $n\geq2k+1$ and
$k\geq 1$, we have
\begin{equation*}
t(n,n-k,2)=\tau(k).
\end{equation*}
For instance, for $n\geq27$, we get
$$t(n,n-13,2)=\tau(13)=2.$$
\end{remark}

\section{Main identity}
The aim of this section is to derive an explicit formula for
$t(n,k,2)$. Before giving the next Theorem, we introduce some notation. Let
\begin{itemize}
\item $\varphi_{i}(j)=
\begin{cases}
1,& \text{   if }j\equiv0\ (\text{mod}\ i);\\
0,& \text{   otherwise}.%
\end{cases}$


\item $\chi _{k}\left( i,j\right) =\begin{cases}
0,& \text{   if }i\neq 0\text{ and }\gcd(k,j)\neq 1\text{ and }\gcd(i,j)=1;   \\
1,& \text{   otherwise.}%
\end{cases}$

\item $W_{k}=[W_{k}(i,j)]$, $0\leq i\leq k-1$, $1\leq j\leq k-1$  be a matrix, whose elements are given by

$$W_{k}\left(i,j\right)=\begin{cases}
d,&\text{  if } i\in I_{k,j}(d)\text{ and } \chi_{k}\left(i,j\right)=1; \\
j,&\text{  otherwise}.%
\end{cases}$$
where, $0\leq d\leq \dfrac{j}{\gcd(k,j)}-1$ and

$I_{k,j}(d)=\left\{i=\left(\left\lfloor\dfrac{d k-1}{j}\right\rfloor+a\right)j-d k \text{  /  } 1\leq a\leq \left\lfloor\dfrac{(d+1)k-1}{j}\right\rfloor-\left\lfloor\dfrac{d k-1}{j}\right\rfloor \right\}.$

\end{itemize}

\begin{remark}
The construction of matrix $W_k$ is special, it is filled  column by column as follows:
\begin{enumerate}
 \item Case $\chi_{k}\left(i,j\right)=1$\\
 Each value of the parameter $d$ generates some values of the parameter $a$, which in return produce the values of the lines $i$, this process allows to define the elements of the concerned lines.
 \item Case $\chi_{k}\left(i,j\right)=0$\\
 The empty elements are replaced by the number $j$ of the column.
\end{enumerate}
\end{remark}
For example, for $k=6$, we get
\begin{equation*}
    W_{6}=\left[\begin{array}{ccccc}
                       0 & 0 & 0 &0&0\\
                       0 & 2 & 3 &4&4 \\
                       0 & 0 & 3 &1&3\\
                       0 & 2 & 0 &4&2\\
                       0 & 0 & 3 &0&1\\
                       0 & 2 & 3 &4&0\\
                \end{array}
          \right].
\end{equation*}
The formulas of Corollary \ref{cor1} are special cases that motivate the following generalization.
\begin{theorem}\label{the2}
For $n\geq3$, $n\equiv i\ ({\rm mod}\ k)$, $2\leq k \leq n-1$, we have
\begin{center}
$t(n,k,2) =\begin{cases}
\dsum_{j=1}^{k-1}\left\lfloor\dfrac{\gcd(k,j)}{k j}(n-k)\right\rfloor, & \text{  if } i=0;\\
\dsum_{j=1}^{k-1}\chi_{k}(i,j)\left\lfloor1+ \gcd(k,j)\dfrac{n-i-k-k W_{k}(i,j)}{k j} \right\rfloor, & \text{  otherwise.}
\end{cases} $
\end{center}
\end{theorem}

\begin{proof}
\textbf{Case 1.} $n=kl$, i.e., $i=0$. Using Theorem
\ref{the1}, we get
\begin{eqnarray*}
 \begin{array}{lll}
   t(n,k,2)&=&\dsum_{h=1}^{l-1}\tau_{k-1\downarrow}(k h)\\
       &=&\dsum_{j=1}^{k-1}\dsum_{h=1}^{l-1}\varphi_{j}(k h).\\
   \end{array}
 \end{eqnarray*}
The divisors of $k h$ which are multiples of $j$ are of the form $\dfrac{k j d_h}{\gcd(k,j)}$. Then
\begin{eqnarray*}
 \begin{array}{lll}
   t(n,k,2)&=&\dsum_{j=1}^{k-1}\left\lfloor\dfrac{\gcd(k,j)}{j}(l-1)\right\rfloor\\
    &=&\dsum_{j=1}^{k-1}\left\lfloor\dfrac{\gcd(k,j)}{k j}(n-k)\right\rfloor.\\
 \end{array}
\end{eqnarray*}
\textbf{Case 2.} $n=kl+i$, $1\leq i\leq k-1$. Using Theorem \ref{the1}, we get
\begin{eqnarray*}
 \begin{array}{lll}
   t(n,k,2)&=&\dsum_{h=0}^{l-1}\tau_{k-1\downarrow}(kh+i)\\
    &=&\dsum_{i=1}^{k-1}\dsum_{h=0}^{l-1}\varphi_{j}(k h+i).\\
 \end{array}
\end{eqnarray*}
It is straightforward to verify that the divisors of $kh+i$ that are multiples of $j$ are of the following form
\begin{center}
$\chi_k (i,j)\left(\dfrac{k j d_h}{\gcd(k,j)} + i+ k W_{k}(i,j)\right)$.
\end{center}Hence,
\begin{eqnarray*}
 \begin{array}{lll}
   t(n,k,2)&=&\dsum_{j=1}^{k-1}\chi_k (i,j)\left\lfloor\dfrac{j+\gcd(k,j)\bigl(l-1-W_{k}(i,j)\bigr)}{j}\right\rfloor\\
             &=&\dsum_{j=1}^{k-1}\chi_k (i,j)\left\lfloor\dfrac{j+\gcd(k,j)\bigl(\frac{n-i}{k}-1-W_{k}(i,j)\bigr)}{j}\right\rfloor\\
             &=&\dsum_{j=1}^{k-1}\chi_k (i,j)\left\lfloor1+\dfrac{\gcd(k,j)}{k j}\bigl(n-i-k-k W_{k}(i,j)\bigr)\right\rfloor.\\
 \end{array}
\end{eqnarray*}
\end{proof}
\begin{example} $k=6$
\begin{enumerate}
\item For $i=0$, $n=6l$, we get
\begin{eqnarray*}
\begin{array}{lll}
t(6l,6,2)&=&\dsum_{j=1}^{5}\left\lfloor\dfrac{n-6}{6j}\gcd(6,j)\right\rfloor\\
&=&\dfrac{n-6}{2}+\left\lfloor\dfrac{n-6}{12}\right\rfloor+\left\lfloor\dfrac{n-6}{30}\right\rfloor.\\
\end{array}
\end{eqnarray*}
\item For $i=1$, $n=6l+1$ we get
\begin{eqnarray*}
\begin{array}{lll}
t(6l+1,6,2)&=&\dsum_{j=1}^{5}\chi_{6}(1,j)\left\lfloor1+\dfrac{\gcd(6,j)}{6j}\bigl(n-7-6W_{6}(1,j)\bigr)\right\rfloor\\
&=&\left\lfloor1+\dfrac{n-7}{6}\right\rfloor+\left\lfloor1 \  \ + \ \ \dfrac{n-7-24}{30}\right\rfloor\\
&=&\dfrac{n-1}{6}+\left\lfloor\dfrac{n-1}{30}\right\rfloor.\\
\end{array}
\end{eqnarray*}
\item For $i=2$, $n=6l+2$ we get
\begin{eqnarray*}
\begin{array}{rl}
t(6l+2,6,2)=&\dsum_{j=1}^{5}\chi_{6}(2,j)\left\lfloor1+\dfrac{\gcd(6,j)}{6j}\bigl(n-8-6W_{6}(2,j)\bigr)\right\rfloor\\
\ \\
=&\left\lfloor1+\dfrac{n-8}{6}\right\rfloor+\left\lfloor1+\dfrac{2(n-8)}{12}\right\rfloor+\left\lfloor1+\dfrac{2(n-8-6)}{24}\right\rfloor+\left\lfloor1+\dfrac{n-8-18}{30}\right\rfloor\\
\ \\
=&\dfrac{n-2}{3}+\left\lfloor\dfrac{n-2}{12}\right\rfloor+\left\lfloor\dfrac{n+4}{30}\right\rfloor.\\
\end{array}
\end{eqnarray*}
\item For $i=3$, $n=6l+3$ we get
\begin{eqnarray*}
\begin{array}{lll}
t(6l+3,6,2)&=&\dsum_{j=1}^{5}\chi_{6}(3,j)\left\lfloor1+\dfrac{\gcd(6,j)}{6j}\bigl(n-9-6W_{6}(3,j)\bigr)\right\rfloor\\
\ \\
&=&\left\lfloor1+\dfrac{n-9}{6}\right\rfloor+\left\lfloor1+\dfrac{3(n-9)}{18}\right\rfloor+\left\lfloor1+\dfrac{n-9-12}{30}\right\rfloor\\ 
\ \\
&=&\dfrac{n-3}{3}+\left\lfloor\dfrac{n+9}{30}\right\rfloor.\\
\end{array}
\end{eqnarray*}
\item For $i=4$, $n=6l+4$ we get
\begin{eqnarray*}
\begin{array}{rll}
t(6l+4,6,2)=&\dsum_{j=1}^{5}\chi_{6}(4,j)\left\lfloor1+\dfrac{\gcd(6,j)}{6j}\bigl(n-10-6W_{6}(4,j)\bigr)\right\rfloor\\
\ \\
=&\left\lfloor1+\dfrac{n-10}{6}\right\rfloor+\left\lfloor1+\dfrac{2(n-10)}{12}\right\rfloor+\left\lfloor1+\dfrac{2(n-10)}{24}\right\rfloor+\left\lfloor1+\dfrac{n-10-6}{30}\right\rfloor\\
\ \\
=&\dfrac{n-4}{3}+\left\lfloor\dfrac{n+2}{12}\right\rfloor+\left\lfloor\dfrac{n+14}{30}\right\rfloor.\\
\end{array}
\end{eqnarray*}
\item For $i=5$, $n=6l+5$ we get
\begin{eqnarray*}
\begin{array}{lll}
t(6l+5,6,2)&=&\dsum_{j=1}^{5}\chi_{6}(5,j)\left\lfloor1+\dfrac{\gcd(6,j)}{6j}\bigl(n-11-6W_{6}(5,j)\bigr)\right\rfloor\\ 
\ \\
&=&\left\lfloor1+\dfrac{n-11}{6}\right\rfloor+\left\lfloor1+\dfrac{n-11}{30}\right\rfloor\\
\ \\
&=&\dfrac{n-5}{6}+\left\lfloor\dfrac{n+19}{30}\right\rfloor.\\
\end{array}
\end{eqnarray*}
\end{enumerate}
\end{example}
Using Theorem \ref{the2}, we obtain the following table for $n\leq 20$.
\begin{table}[H]
\begin{center}
\begin{tabular}{c|cccccccccccccccccc|c}
  $n\backslash k$&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&$t(n,2)$\\
  \hline
 $3$&1&&&&&&&&&&&&&&&&&&1\\
 $4$&1&1&&&&&&&&&&&&&&&&&2\\
 $5$&2&2&1&&&&&&&&&&&&&&&&5\\
 $6$&2&1&2&1&&&&&&&&&&&&&&&6\\
 $7$&3&3&2&2&1&&&&&&&&&&&&&&11\\
 $8$&3&3&2&2&2&1&&&&&&&&&&&&&13\\
 $9$&4&3&2&3&2&2&1&&&&&&&&&&&&17\\
 $10$&4&4&5&1&3&2&2&1&&&&&&&&&&&22\\
 $11$&5&5&3&4&2&3&2&2&1&&&&&&&&&&27\\
 $12$&5&4&4&3&3&2&3&2&2&1&&&&&&&&&29\\
 $13$&6&6&4&5&2&4&2&3&2&2&1&&&&&&&&37\\
 $14$&6&6&7&5&5&1&4&2&3&2&2&1&&&&&&&44\\
 $15$&7&6&4&3&4&4&2&4&2&3&2&2&1&&&&&&44\\
 $16$&7&7&7&5&6&4&3&2&4&2&3&2&2&1&&&&&55\\
 $17$&8&8&5&7&3&5&3&4&2&4&2&3&2&2&1&&&&59\\
 $18$&8&7&9&6&7&4&5&2&4&2&4&2&3&2&2&1&&&68\\
 $19$&9&9&6&7&3&7&3&4&3&4&2&4&2&3&2&2&1&&71\\
 $20$&9&9&9&5&7&5&8&3&3&3&4&2&4&2&3&2&2&1&81\\
\end{tabular}
\caption{$t(n,k,2)$,
 $2\leq k \leq19$, $3\leq n\leq 20$.}
\end{center}
\end{table}
Also, Theorem 6 allows us to obtain $t(n,2)$ for large values of $n$, the following table is introduced to illustrate a few.
\begin{table}[H]
\begin{center}
\begin{tabular}{c|ccccccccccccc}
  $n$&100&500&1000&1500&2000&2500&3000&3500&4000\\
  \hline%\hline
 $t(n,2)$&1135&11103&28340&54652&70128&91440&136790 &144687&169953 \\
    %\hline
\end{tabular}
\caption{Some values of $t(n,2)$.}
\end{center}
\end{table}

\section{Application}
Let $\mathcal{P}_{n}$ be an $n$-side regular polygon. We say that an
inscribed quadrilateral in $\mathcal{P}_{n}$ is proper if none of
its sides belongs to $\mathcal{P}_{n}$.
\begin{theorem}\label{the3}
Let $n\geq9$ be an odd integer and let $\lozenge(n)$ be the number of
inscribed, non-isometric and proper quadrilaterals in
$\mathcal{P}_{n}$, using three equal chords. Then we have
$$\lozenge(n)=\begin{cases}
\dfrac{n-5}{4}+\left\lfloor\dfrac{n-5}{12}\right\rfloor,&\ \ \textrm{\ if\ }\  n \equiv1\ ({\rm mod }\ 4);\\
\ \\
\dfrac{n-7}{4}+\left\lfloor\dfrac{n+1}{12}\right\rfloor,&\ \ \textrm{\ if\ }\  n \equiv3\ ({\rm mod }\ 4).\\
\end{cases}
$$
\end{theorem}

\begin{proof}
The chords belonging to an inscribed quadrilateral in
$\mathcal{P}_{n}$, separate the number of vertices of
$\mathcal{P}_{n}$ into four parts, which do not include the quadrilateral vertices. In other words, each such quadrilateral generates a
partition of $n-4$ into four parts, using only two types of parts and vice versa. Then
$$\lozenge(n)=t(n-4,4,2),$$
and the result yields from Corollary \ref{cor1}.
\end{proof}

Figure 1, illustrates this idea in $\mathcal{P}_{19}$.
The first quadrilateral is generated by the partition $15=1+1+1+12$,
the second by $15=2+2+2+9$, the third by $15=3+3+3+6$ and the fourth
by $15=3+4+4+4$.\\
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\end{pspicture}
\begin{center}
     {Figure 1: The non-isometric proper quadrilaterals inscribed in $\mathcal{P}_{19}$, \\using three equal chords.}
\end{center}

\section{Acknowledgement}
The authors are indebted to the unknown referee for his valuable corrections and comments which
have improved the quality of the paper. The authors are also grateful to Professors L. Benaissa and N. Hannoun for their help.

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\bibitem{ben} S. Bouroubi and N. Benyahia Tani, A new identity for complete Bell polynomials based on a formula of Ramanujan, \textit{J. Integer Seq.} \textbf{12} (2009), \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL12/Bouroubi/bouroubi25.pdf} {Article 09.3.5}.

\bibitem{charal} A. Charalambos Charalambides, \textit{Enumerative Combinatorics}, Chapman \& Hall/CRC, 2002.

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\bibitem{ma} P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, \textit{Proc. London Math. Soc.}, \textbf{2} (1919), 75--113; Coll. Papers II, 303--341.%ok

\bibitem{redem} H. Rademacher, On the partition function $p(n)$, \textit{Proc. London, Math. Soc.}, \textbf{43} (1937), 241--254.

\bibitem{Pak} I. Pak, Partition bijections, a survey, \textit{Ramanujan J.}, \textbf{12} (2006), 5--75.

\bibitem{wilf} H. S. Wilf, Lectures on integer partitions, available at \hfill\newline
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: 
Primary 05A17; Secondary 11P83.

\noindent
\emph{Keywords:}  integer partitions,
partitions into parts of different sizes,
partitions into parts of two sizes, number of divisors.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A002133},
\seqnum{A117955}, and
\seqnum{A117956}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 4 2010;
revised version received  October 3 2010; January 31 2011; February 28 2011.
Published in {\it Journal of Integer Sequences}, March 25 2011.

\bigskip
\hrule
\bigskip

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