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\begin{center}
\vskip 1cm{\LARGE\bf Touchard Polynomials, Partial Bell Polynomials and Polynomials of Binomial Type \\
}
\vskip 1cm
\large
Miloud Mihoubi and Mohammed Said Maamra\footnote{Research supported by LAID3
Laboratory of USTHB University.} \\
University of Science and Technology Houari Boumediene (USTHB) \\
Faculty of Mathematics \\
PB 32 \\
El Alia, 16111, Algiers \\
Algeria\\
\href{mailto:miloudmihoubi@gmail.com}{\tt miloudmihoubi@gmail.com}\\
\href{mmihoubi@usthb.dz}{\tt mmihoubi@usthb.dz} \\
\href{mailto:mmaamra@yahoo.fr}{\tt mmaamra@yahoo.fr} \\
\end{center}

\vskip .2 in
\begin{abstract}
Touchard generalized the Bell polynomials in order to give some
combinatorial interpretation on permutations. Chrysaphinou
introduced and studied a class of polynomials related to Touchard's
generalization. In the present paper, we establish some relations
between Touchard polynomials, Bell polynomials and the polynomials of
binomial type. Several identities and relations with Stirling numbers
are obtained.
\end{abstract}




\vskip .2 in



\section{Introduction}

Among the partition polynomials, the partial Bell polynomials,
introduced by Bell \cite{4}, play an important role in different
application frameworks. Several properties and identities are given,
see \cite{1,2,3,5}. Another partition polynomials, called Touchard
polynomials, introduced by Touchard \cite{6}, present an extension of
the partial Bell polynomials. Some algebraic, combinatorial and
probabilistic properties of these polynomials are studied by Touchard
\cite{6}, Chrysaphinou \cite{7}, Charalambides \cite{1}, Kuzmin and
Leonova \cite{8}. In this paper, we give some relations between
Touchard polynomials and partial Bell polynomials. We exploit these
relations and the polynomials of binomial type to derive some
identities for these polynomials.

In this context, let $( x_{i};i\geq1) $ \ and $( y_{i};i\geq1) $ be two sequences of real
numbers. 
The Touchard polynomials
\[T_{n,k}(x_j;y_j):=T_{n,k}(x_{1},\ldots,x_{n};y_{1},\ldots,y_{n}), \ n\geq k\geq0 ,\]
are defined by their bivariate generating function
\begin{equation*}
\underset{n\geq0}{\sum}\underset{k=0}{\overset{n}{\sum}}T_{n,k}(x_j;y_j) u^{k}\frac{t^{n}}{n!}=\exp\Biggl( u
\underset{i\geq1}{\sum}x_{i}\frac{t^{i}}{i!}+\underset{i\geq 1}{\sum}y_{i}
\frac{t^{i}}{i!}\Biggr) .
\end{equation*}
The vertical generating function of Touchard polynomials, for fixed $k,$
is given by
\begin{equation*}
\underset{n\geq k}{\sum}T_{n,k}(x_j;y_j)
\frac{t^{n}}{n!}=\frac{1}{k!}\Biggl( \underset{i\geq1}{\sum }x_{i}\frac{t^{i}
}{i!}\Biggr) ^{k}\exp\Biggl( \underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}
\Biggr) ,\text{ }k=0,1,\ldots.
\end{equation*}
The partial Bell polynomials are given by
\[B_{n,k}(x_j):=B_{n,k}(x_{1},\ldots,x_{n})=T_{n,k}(x_{1},\ldots,x_{n};0,\ldots,0), \ n\geq k\geq0 .\]
The polynomials of binomial type $( f_{n}( x) ) $ are defined by
\begin{equation*}
\underset{n\geq0}{\sum}f_{n}( x) \frac{u^{n}}{n!}=\Biggl(
\underset{n\geq0}{\sum}f_{n}( 1) \frac{u^{n}}{n!}\Biggr)
^{x},\text{ \ with }f_{0}( x) =1 \text{ \ and } \ f_1(x)\neq 0 \text{ \ for } x\neq0.
\end{equation*}
For a real number $a$ we consider in the following the sequence $\left( f_{n}(x;a)\right) $ defined by
\begin{equation*}
f_{n}(x;a):=\frac{x}{an+x}f_{n}\left( an+x\right) .
\end{equation*}
The sequence $\left( f_{n}(x;a)\right) $ is also of binomial type, see \cite{3}, and
for more details on sequences of binomial type see \cite{10}. \\
We use the following notation and hypothesis: 

$D\equiv \frac{d}{dx},\ D^{k}\equiv\frac{d^{k}}{dx^{k}},\
D_{x=x_{0}}^{k}\equiv \frac{d^{k}}{dx^{k}}\vert _{x=x_{0}}.$ 

For $n&lt;0,$ we set $f_{n}( x) =0,$ $T_{n,k}( x_{j};\ y_{j}) =0$ and $B_{n,k}(
x_{j}) =0.$ 

For $x\in\mathbb{R}$, where $\mathbb{R}$ is the set of real numbers, we set
\begin{equation*}
\binom{x}{k}:=\frac{x( x-1) \cdots ( x-k+1)}{k!} \text{ for\ }
k=1,2,\ldots,\ \ \binom{x}{0}=1\text{ and \ }\binom{x}{k}=0\text{ \ otherwise.
}
\end{equation*}
Also, for all nnnegative integers $n, m$ we put
\begin{equation*}
\begin{array}{l}
1_{( m\mid n) }=
\begin{cases}
1, \text{ if }m\text{ divides }n; \\
0, \text{ otherwise;}
\end{cases}
\text{ and } \ \ \ 1_{( n\geq m) }=
\begin{cases}
1,\text{ if } n\geq m; \\
0,\text{ otherwise.}
\end{cases}
\end{array}
\end{equation*}

\section{The main results}

In this section, we establish some relations between the Touchard
polynomials, partial Bell polynomials and polynomials of binomial type. Furthermore,
we use these relations to develop several identities for Touchard polynomials. We start with the following theorem:

\begin{theorem}
\label{T3}Let $n,k,m$ be integers such that $n\geq k\geq1, m\geq1; \ a$ be a real number and $(x_n)$ be a sequence of
real numbers. Then
\begin{equation*}
T_{n,k}\Biggl( x_{j};\ -m( j-1) !a^{j/m}1_{( m\mid j) }\Biggr) =B_{n,k}(
x_{j}) -am!\binom{n}{m}B_{n-m,k}( x_{j}).
\end{equation*}
\end{theorem}

\begin{proof}
Let $y_{n}=-m( n-1) !a^{n/m}1_{( m\mid n) }.$ \\ For $a=0$ the theorem is trivial, otherwise, for $|t|&lt; |a|^{-1/m}$ we have
\[
\exp\Biggl(  \underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}\Biggr)  =\exp\Biggl(
-\underset{j\geq1}{\sum}a^{j}\frac{t^{mj}}{j}\Biggr)  =\exp(  \ln(
1-at^{m})  )  =1-at^{m}.
\]
Then
\begin{align*}
\underset{n\geq k}{\sum}T_{n,k}(  x_{j};y_{j})  \frac{t^{n}}{n!}  &amp;
=\frac{1}{k!}\Biggl(  \underset{i\geq1}{\sum}x_{i}\frac{t^{i}}{i!}\Biggr)
^{k}\exp\Biggl(  \underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}\Biggr) \\
&amp; =\frac{1}{k!}\Biggl(  \underset{i\geq1}{\sum}x_{i}\frac{t^{i}}{i!}\Biggr)
^{k}(  1-at^{m}) \\
&amp; =\underset{n\geq k}{\sum}\biggr(  B_{n,k}(  x_{j})  -a\frac
{n!}{(  n-m)  !}B_{n-m,k}(  x_{j})  \biggl)  \frac
{t^{n}}{n!}.
\end{align*}
Hence the theorem is proved.
\end{proof}

If we set $x_{n}=nf_{n-1}( x;b) $ in Theorem \ref{T3} and use Proposition 1
of \cite{3}, we obtain:

\begin{corollary}
\label{C4}Let $( f_{n}( x) ) $ be a sequence of binomial type of polynomials. We have
\begin{equation*}
T_{n,k}\Biggl( jf_{j-1}( x;b) ;\ -m( j-1) !a^{j/m}1_{( m\mid j) }\Biggr) =
\frac{n!}{k!}\Biggl( \frac{f_{n-k}( kx;b) }{( n-k) !}-a\frac {f_{n-m-k}(
kx;b) }{( n-m-k) !}\Biggr).
\end{equation*}
\end{corollary}

\begin{example}
For $f_{n}( x) =x^{n}$ in Corollary \ref{C4} we get
\begin{align*}
&amp; T_{n,k}\Biggl( jx( b( j-1) +x) ^{j-2};\ -m( j-1) !a^{j/m}1_{( m\mid j) }
\Biggr) \\
&amp; =x\frac{n!}{( k-1) !}\Biggl( \frac{( b( n-k) +kx) ^{n-k-1}}{( n-k) !}-a
\frac{( b( n-m-k) +kx) ^{n-m-k-1}}{( n-m-k) !}1_{( n\geq m+k) }\Biggr) ,
\end{align*}
and for $f_{n}( x) =n!\binom{x}{n}$ in Corollary \ref{C4} we get
\begin{align*}
&amp; T_{n,k}\Biggl( xj!\frac{\binom{b( j-1) +x}{j-1}}{b( j-1) +x};\ -m( j-1)
!a^{j/m}1_{( m\mid j) }\Biggr) \\
&amp; =x\frac{n!}{( k-1) !}\Biggl( \frac{\binom{b( n-k) +kx}{n-k}}{b( n-k) +kx}-a
\frac{\binom{b( n-m-k) +kx}{n-m-k}}{b( n-m-k) +kx}1_{( n\geq m+k) }\Biggr) .
\end{align*}
\end{example}

As above, for particular cases of Touchard polynomials, the following proposition gives another expression in term of
polynomials of binomial type.

\begin{proposition}
\label{P1}Let $b, \alpha$ be two real numbers and $( f_{n}( x) ) $ be a
sequence of binomial type of polynomials. We have
\begin{equation*}
T_{n,k}\Biggl( jf_{j-1}( x) ;\ \alpha Df_{j}( 0) \Biggr) =\binom{n}{k}
f_{n-k}( kx+\alpha) ,
\end{equation*}
or more generally
\begin{align*}
&amp; T_{n,k}\Biggl( jx\frac{f_{j-1}( b( j-1) +x) }{b( j-1) +x};\ \alpha\frac{
f_{j}( bj) }{bj}\Biggr)  \\
&amp; =\binom{n}{k}( kx+\alpha) \frac{f_{n-k}( b( n-k) +kx+\alpha) }{b( n-k)
+kx+\alpha},\text{ }b\neq0.
\end{align*}
\end{proposition}

\begin{proof}
Let
\begin{equation*}
1+\underset{n\geq1}{\sum}f_{n}(  x)  \frac{t^{n}}{n!}
=\exp\biggl(  x\underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}\biggr)
\end{equation*}
be the exponential generating function of the sequence $(  f_{n}(  x)
)  $ and $x_{n}=nf_{n-1}(  x) .$ \\ Necessarily $y_{n}=Df_{n}(  0) .$ We have
\begin{align*}
\underset{n\geq k}{\sum}T_{n,k}\biggl( x_j ;\ \alpha
y_{j}\biggr)  \frac{t^{n}}{n!}  &amp; =\frac{1}{k!}\biggl(  \underset{i\geq1}{\sum
}x_{i}\frac{t^{i}}{i!}\biggr)  ^{k}\exp\biggl(  \alpha\underset{i\geq1}{\sum
}y_{i}\frac{t^{i}}{i!}\biggr) \\
&amp; =\frac{t^{k}}{k!}\exp\biggl(  (  kx+\alpha)  \underset{i\geq
1}{\sum}y_{i}\frac{t^{i}}{i!}\biggr) \\
&amp; =\frac{t^{k}}{k!}\underset{n\geq0}{\sum}f_{n}(  kx+\alpha)
\frac{t^{n}}{n!}\\
&amp; =\underset{n\geq k}{\sum}\binom{n}{k}f_{n-k}(  kx+\alpha)
\frac{t^{n+k}}{n!}.
\end{align*}
Then, we obtain
\[
T_{n,k}\biggl(  jf_{j-1}(  x)  ;\alpha Df_{j}(  0)
\biggr)  =\binom{n}{k}f_{n-k}(  kx+\alpha)  .
\]
To finish the proof, replace $f_{n}(  x)  $ by $f_{n}(
x;b)  $ in the last identity.
\end{proof}

\begin{example}
For $f_{n}( x) =x^{n}$ in Proposition \ref{P1} we get
\begin{equation*}
T_{n,k}\biggl(  xj( b( j-1) +x) ^{j-2};\ \alpha ( j-1) ( bj) ^{j-2}\biggr)  =
\binom{n}{k}( kx+\alpha) ( b( n-k) +kx+\alpha) ^{n-k-1},
\end{equation*}
and for $f_{n}( x) =n!\binom{x}{n}$ in Proposition \ref{P1} we get
\begin{align*}
&amp; T_{n,k}\biggl( \frac{j!x}{b( j-1) +x}\binom{b( j-1) +x}{j-1};\ \alpha( -1)
^{j-1}( j-1) !\biggr) \\
&amp; =\frac{n!}{k!}\frac{kx+\alpha}{n( n-k) +kx+\alpha}\binom{n( n-k) +kx+\alpha
}{n-k}.
\end{align*}
\end{example}

Hence we may state the following:

\begin{corollary}
\label{C1}Let $r, s, p$ be a nonnegative integers, $r\geq1,$ and $( x_{n}) $ be a
sequence of real numbers with $x_{1}=1.$ We have
\begin{align*}
&amp; T_{n,k}\biggl( \frac{js}{( r(j-1) +s)}
\frac{B_{( r+1) ( j-1) +s,\ r( j-1) +s}( x_{i}) }{\binom{( r+1) ( j-1) +s}{r( j-1) +s}};\ \frac{p}{rj}\frac{B_{( r+1) j,\ rj}(x_{i}) }{\binom{( r+1) j}{rj}}\biggr)  \\
&amp; =\binom{n}{k} \frac{ks+p}{ r( n-k) +ks+p}\frac{B_{( r+1) ( n-k) +ks+p,\ r( n-k) +ks+p}( x_{i})
}{\binom{( r+1) ( n-k) +ks+p}{r( n-k) +ks+p}}.
\end{align*}
\end{corollary}

\begin{proof}
For $x_2\neq 0,$ let $\{  f_{n}(  x)  \}  $ be a sequence of binomial type
such that $f_{n}(  1)  =\frac{x_{n+1}}{n+1}$.  \\  From the
known identity $B_{n,k}(  jf_{j-1}(  1)  )  =\dbinom{n}{k}f_{n-k}(k)$ we get
\begin{equation*}
f_{n}(  k)  =\binom{n+k}{k}^{-1}B_{n+k,k}(  x_{i})
,\;\;n\geq0,k\geq1.
\end{equation*}
Take $b=r,$ $x=s$ and $\alpha=p$ in Proposition \ref{P1} to get
\begin{equation*}
T_{n,k}\biggl(  js\frac{f_{j-1}(  r(  j-1)  +s)
}{r(  j-1)  +s};\ p\frac{f_{j}(  rj)  }{rj}\biggr)
=\binom{n}{k}(  ks+p)  \frac{f_{n-k}(  r(  n-k)
+ks+p)  }{r(  n-k)  +ks+p}.
\end{equation*}
Therefore, it suffices to use the identity $f_{n}(  k)  =\binom{n+k}{k}^{-1}B_{n+k,k}(  x_{i})$
to express in the last identity $f_{i-1}(  r(  i-1)  +s)  \text{ and } f_{n-k}(r(  n-k)  +ks+p)$
by the partial Bell polynomials. \\
For the case $x_2=0$ the corollary remains true by continuity.
\end{proof}

\begin{example}
\label{E1}By Corollary \ref{C1} and the identity $B_{n,k}( 1!,2!,\ldots,( q+1)
!,0,\ldots) =\frac{n!}{k!}\binom{k}{n-k}_{q},$ see \cite{9}, we get
\begin{align*}
&amp; T_{n,k}\biggl( \frac{j!s}{r( j-1) +s}\binom{r( j-1) +s}{j-1}_{q};\ \frac{(
j-1) !p}{r}\binom{rj}{j}_{q}\biggr) \\
&amp; =\frac{n!}{k!}\frac{ks+p}{r( n-k) +ks+p}\binom{r( n-k) +ks+p}{n-k}_{q},
\end{align*}
where $\dbinom{k}{n}_{q}$ is the coefficients defined by $
(1+x+x^{2}+\cdots+x^{q})^{k}=\sum\limits_{n\geq0}\binom{k}{n}_{q}x^{n}.$ \\
Also, for $x_n=1,$  $ x_n=(-1)^{n-1}(n-1)!,$ $ x_n=(n-1)!$ or $x_n=n!$ we get identities related Touchard polynomials to Stirling numbers of the first and second kind.
\end{example}

The following two propositions give relations between Touchard polynomials and
the successive derivatives of polynomials of binomial type.

\begin{proposition}
\label{P2}Let $b$ be a real number, $b\neq0,$ and $( f_{n}( x) ) $ be a sequence of
binomial type of polynomials. We have
\begin{equation*}
T_{n,k}\biggl(  Df_{j}( 0) ;\ xDf_{j}( 0) \biggr)
=\frac{1}{k!}D^{k}f_{n}( x) ,
\end{equation*}
or more generally
\begin{equation*}
T_{n,k}\biggl( \frac{f_{j}( bj) }{bj};\ x\frac{f_{j}( bj) }{bj}\biggr) =
\frac{1}{k!}D^{k}\biggl( \frac{x}{bn+x}f_{n}( bn+x) \biggr ) .
\end{equation*}
\end{proposition}

\begin{proof}
Let $(  f_{n}(  x))  $ be a sequence of binomial type defined as in the proof of Proposition \ref{P1}
and $x_{n}=nf_{n-1}(  x)  .$ We have $y_{n}=Df_{n}(  0) $ and
\[
\underset{n\geq k}{\sum}D^{k}f_{n}(  x)  \frac{t^{n}}{n!}=\biggl(
\underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}\biggr)  ^{k}\exp\biggl(
x\underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}\biggr)  =k!\underset{n\geq
k}{\sum}T_{n,k}(  y_{j};\ xy_{j})  \frac{t^{n}}{n!}.
\]
Then
\[T_{n,k}\biggl(  y_{j}  ;\ xy_{j} \biggr )=
T_{n,k}\biggl(  Df_{j}(  0)  ;\ xDf_{j}(  0) \biggr )
=\frac{1}{k!}D^{k}f_{n}(  x)  .
\]
After that, replace $f_{n}(  x)  $ by $f_{n}(x;b)  $ in the last identity.
\end{proof}

\begin{example}
For $f_{n}( x) =x^{n}$ in Proposition \ref{P2} we get
\begin{equation*}
T_{n,k}\biggl(  ( bj) ^{j-1};\ x( bj) ^{j-1}\biggr)
=\frac{n!}{( k!) ^{2}}( bn+x) ^{n-k-1}( b( n-1) +x) ,
\end{equation*}
and for $f_{n}( x) =n!\binom{x}{n}$ in Proposition \ref{P2} we get
\begin{equation*}
T_{n,k}\biggl(  \frac{j!}{bj}\binom{bj}{j};\ x\frac{j!}{bj}\binom{bj}{j}
\biggr)  =\frac{n!}{k!}D^{k}\biggl(  \frac{x}{bn+x}\binom{bn+x}{n}\biggr)  .
\end{equation*}
\end{example}

\begin{proposition}
\label{P3}Let $b, \alpha, \beta$ be real numbers, $r$ be a positive integer and $(
f_{n}( x) ) $ be a sequence of binomial type of polynomials. We have
\begin{align*}
&amp; T_{n,k}\biggl( jD_{z=0}^{r}( e^{\beta z}f_{j-1}( x+z) ) ;\ \alpha Df_{j}(
0) \biggr) \\
&amp; =\frac{( kr) !}{k!}T_{n,kr}\biggl( \beta1_{( j=1) }+jDf_{j-1}( 0) 1_{(
j\geq2) };\ ( kx+\alpha) Df_{j}( 0) \biggr) ,
\end{align*}
or more generally,
\begin{align*}
&amp; T_{n,k}\biggl( jD_{z=0}^{r}( \frac{( x+z) f_{j-1}( b( j-1) +x+z) }{b( j-1)
+x+z}e^{\beta z}) ;\ \alpha\frac{f_{j}( bj) }{bj}\biggr)  \\
&amp; =\frac{( kr) !}{k!}T_{n,kr}\biggl( \beta1_{( j=1) }+j\frac{f_{j-1}( b(
j-1) ) }{b( j-1) }1_{( j\geq2) };\ ( kx+\alpha) \frac{f_{j}( bj) }{bj}
\biggr) .
\end{align*}
\end{proposition}

\begin{proof}
Let $(  f_{n}(  x))  $ be a sequence of binomial type defined as in the proof of Proposition \ref{P1}
and $x_{n}=nD_{z=0}^{r}(  e^{\beta z}f_{n-1}(
x+z)  )  .$  We have $y_{n}=Df_{n}(  x)  (0)$ and
\begin{align*}
\frac{1}{k!}\biggl(  \underset{i\geq1}{\sum}x_{i}\frac{t^{i}}{i!}\biggr)  ^{k}
&amp; =\biggl(  \underset{i\geq1}{\sum}iD_{z=0}^{r}\biggl(  e^{\beta z}
f_{i-1}(  x+z)  \biggr)  \frac{t^{i}}{i!}\biggr)  ^{k}\\
&amp; =\frac{t^{k}}{k!}F(  t)  ^{kx}(  D_{z=0}^{r}(  e^{\beta
z}F(  t)  ^{z})  )  ^{k}\\
&amp; =\frac{t^{k}}{k!}F(  t)  ^{kx}(  \beta+\ln F(
t)  )  ^{kr}\\
&amp; =\frac{t^{k}}{k!}D_{z=0}^{kr}(  e^{\beta z}F(  t)
^{kx+z}) \\
&amp; =\frac{1}{k!}\biggl(  \beta t+\underset{i\geq2}{\sum}iy_{i-1}\frac{t^{i}}
{i!}\biggr)  ^{kr}\exp\biggl(  kx\underset{i\geq1}{\sum}y_{i}\frac{t^{i}}
{i!}\biggr)  .
\end{align*}
Then
\begin{align*}
\underset{n\geq k}{\sum}T_{n,k}\biggl(  x_j  ;\ \alpha y_{j}\biggr)  \frac{t^{n}}{n!}
&amp;=\frac{1}{k!}\biggl(  \underset{i\geq1}{\sum}x_{i}\frac{t^{i}}{i!}\biggr)
^{k}\exp\biggl(  \alpha\underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}\biggr) \\
&amp; =\frac{1}{k!}\biggl(  \beta t+\underset{i\geq2}{\sum}iy_{i-1}\frac{t^{i}}
{i!}\biggr)  ^{kr}\exp\biggl(  (  kx+\alpha)  \underset{i\geq1}
{\sum}y_{i}\frac{t^{i}}{i!}\biggr) \\
&amp; =\frac{(  kr)  !}{k!}\underset{n\geq k}{\sum}T_{n,kr}\biggl(
\beta1_{(  j=1)  }+jy_{j-1}1_{(  j\geq2)  };\ (
kx+\alpha)  y_{j}\biggr)  \frac{t^{n}}{n!}.
\end{align*}
Then, we obtain
\begin{align*}
&amp; T_{n,k}\biggl(  jD_{z=0}^{r}(  e^{\beta z}f_{j-1}(  x+z)
)  ;\alpha Df_{j}(  0)  \biggr) \\
&amp; =\frac{(  kr)  !}{k!}T_{n,kr}\biggl(  \beta1_{(  j=1)
}+jy_{j-1}1_{(  j\geq2)  };\ (  kx+\alpha)
Df_{j}(  0)\biggr)  .
\end{align*}
To finish the proof, replace $f_{n}(  x)  $ by $f_{n}(
x;b)  $ in the last identity.
\end{proof}

\begin{example}
For $f_{n}( x) =x^{n}$ in Proposition \ref{P3} we get
\begin{align*}
&amp; T_{n,k}\biggl( jD_{z=0}^{r}( ( x+z) ( b( j-1) +x+z) ^{j-2}e^{\beta z})
;\alpha( bj) ^{j-1}\biggr) \\
&amp; =\frac{( kr) !}{k!}T_{n,kr}\biggl( \beta1_{( j=1) }+j( b( j-1) )
^{j-2}1_{( j\geq2) };\ ( kx+\alpha) ( bj) ^{j-1}\biggr) .
\end{align*}
\end{example}

\begin{proposition}
\label{P4}Let $(x_n)$ and $ (y_n)$ be two sequences of real numbers and $r$ be a positive integer. We have
\begin{align}
&amp; B_{n,k}\biggl(  \frac{r}{\binom{j+ r-1}{ r-1}}T_{j+r-1,r}(
x_i; y_i) \biggr ) =\frac{\binom{kr}{k}}{\binom{n+( r-1) k}{( r-1) k}} T_{n+(  r-1)  k,kr}(x_i; ky_i) \notag
\end{align}
\end{proposition}

\begin{proof}
Starting with the vertical generating function of Touchard polynomials to get
\begin{align*}
\biggl(  \underset{n\geq r}{\sum}T_{n,r}(  x_{i}; y_{i})
\frac{t^{n}}{n!}\biggr)  ^{k}  &amp; =\frac{1}{(  r!)  ^{k}}\biggl(
\underset{i\geq1}{\sum}x_{i}\frac{t^{i}}{i!}\biggr)  ^{kr}\exp\biggl(
k\underset{i\geq1}{\sum}y_{i}\frac{t^{i}}{i!}\biggr) \\
&amp; =\frac{(  kr)  !}{(  r!)  ^{k}}\underset{n\geq
kr}{\sum}T_{n,kr}(  x_{i}; ky_{i})  \frac{t^{n}}{n!}.
\end{align*}
Then
\[
T_{n,kr}(  x_{i}; ky_{i})  =\frac{(  r!)  ^{k}
k!}{(  kr)  !}B_{n,k}\biggl(  \underset{r-1}{\underbrace{0,\ldots
,0}},T_{r,r}(  x_{i}; y_{i})  ,T_{r+1,r}(  x_{i}
; y_{i})  ,\ldots\biggr)  ,
\]
and from \cite[p.\ 450]{1} we have
\[
B_{n,k}\biggl(  0,\ldots,0,a_{r},a_{r+1},\ldots\biggr)  =\frac{n!}{(  n-(
r-1)  k)  !}B_{n-(  r-1)  k,k}\biggl(  \frac
{i!a_{i+r-1}}{(  i+r-1)  !}\biggr)  .
\]
Then
\[
T_{n,kr}(  x_{j}; ky_{j})  =\frac{(  r!)  ^{k}
n!k!}{(  kr)  !(  n-(  r-1)  k)
!}B_{n-(  r-1)  k,k}\biggl(  \frac{j!}{(  r-1+j)
!}T_{r-1+j,r}(  x_{i}; y_{i})  \biggr)  .
\]
Change $n$ by $n+(r-1)k$ to finish the proof.
\end{proof}

\begin{example} Let $r,s,p,q$ be nonnegative integers with $q\geq 1.$ \\
Use Application \ref{E1} in Proposition \ref{P4} to obtain
\begin{align*}
&amp; B_{n,k}\biggl( \frac{j!( rs+p) }{r( j-1) +rs+p}\binom{r( j-1) +rs+p}{j-1}
_{q}\biggr) \\
&amp; =\frac{n!}{k!}\frac{k(rs+p)}{r( n-k) +k(rs+p)}\binom{r( n-k) +k(rs+p)}{n-k}_{q}.
\end{align*}
From the identity $T_{n,r}(  i!; (  i-1)  !s)  =\frac{n!}{r!}\binom
{n+s-1}{r+s-1}$  given in \cite[p.\ 453]{1} we obtain
\begin{align*}
B_{n,k}\biggl( j!\binom{j+r+s-2}{r+s-1}\biggr) &amp; =\frac{n!}{k!}\binom{n+k(r+s-1)-1}{k( r+s) -1}.
\end{align*}
From Theorems 15 and 16 given in \cite{11} we have
\begin{equation*}
T_{n,r}(  (  i-1)  !; (  i-1)  !s)  = \genfrac{[}{]}{0pt}{}{n+s}{r+s}_{s} \ \
\text{ and } \ \
T_{n,r}(  1; s1_{(  i=1)  })  =\genfrac{\{}{\}}{0pt}{}{n+s}{r+s}_{s}.
\end{equation*}
These identities and Proposition \ref{P4} give
\begin{align*}
B_{n,k}\biggl( \frac{r}{\binom{j+r-1}{r-1}}
\left[ \begin{array}{c}
j+r+s-1 \\
r+s
\end{array} \right]_{s}
\biggr) &amp; =\frac{\binom{kr}{k}}{\binom{n+( r-1) k}{( r-1) k}}
\left[
\begin{array}{c}
n+( r+s-1) k \\
( r+s) k
\end{array} \right]_{ks},  \\
B_{n,k}\biggl( \frac{r}{\binom{j+r-1}{r-1}}
\left\lbrace 
\begin{array}{c}
j+r+s-1 \\
r+s
\end{array} \right\rbrace_{s} \biggr) 
&amp; =\frac{\binom{kr}{k}}{\binom{n+( r-1) k}{( r-1) k}}
\left\lbrace 
\begin{array}{c}
n+( r+s-1) k \\
( r+s) k
\end{array} \right\rbrace_{ks}.
\end{align*}
\end{example}

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\bibitem{1} C. A. Charalambides, \textit{Enumerative Combinatorics.}
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\bibitem{5} M. Mihoubi, The role of binomial type sequences in
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A10; Secondary 05A99, 11B73, 11B75.

\noindent \emph{Keywords: } 
Touchard polynomials; partial Bell polynomials;
polynomials of binomial type.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 21 2010;
revised version received February 13 2011.
Published in {\it Journal of Integer Sequences}, March 25 2011.

\bigskip
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\noindent
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