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\begin{center}
\vskip 1cm{\LARGE\bf 
Linear Recurrences for $r$-Bell Polynomials
}
\vskip 1cm
\large
Miloud Mihoubi\footnote{This research is supported by the PNR project
8/u160/3172.} and Hac\`{e}ne Belbachir\footnote[2]{The research is partially
supported by the LITIS Laboratory of Rouen University and the PNR project
8/u160/722.}\\
USTHB, Faculty of Mathematics \\
RECITS Laboratory,  DG-RSDT \\
P. O. Box 32 \\ 
16111 El-Alia \\
Bab-Ezzouar \\
Algiers \\
Algeria\\
\href{mailto:mmihoubi@usthb.dz}{\tt mmihoubi@usthb.dz} \\
\href{mailto:miloudmihoubi@gmail.com}{\tt miloudmihoubi@gmail.com} \\
\href{mailto:hbelbachir@usthb.dz}{\tt hbelbachir@usthb.dz} \\
\href{mailto:hacenebelbachir@gmail.com}{\tt hacenebelbachir@gmail.com} \\
\end{center}

\vskip .2 in

\vskip .2 in

\begin{abstract}
Letting\textbf{\ }$B_{n,r}$ be the $n$-$th$ $r$-Bell polynomial, it is well known
that $B_{n}(x)$ admits\ specific integer coordinates in the 
two bases $ \{x^{i}\}  _{i}$\ and $ \{  xB_{i}(
x )    \}  _{i}$ according to, respectively, the Stirling numbers and the
binomial coefficients. Our aim is to prove that the sequences $B_{n+m,r}
 (  x )  $ and $B_{n,r+s} (  x )  $ admit a binomial recurrence coefficient
 in different bases of the
$\mathbb{Q}$-vector space formed by polynomials of $\mathbb{Q} [X].$
\end{abstract}

\section{Introduction}

In different ways, Belbachir and Mihoubi \cite{bel3} and Gould and
Quaintance \cite{gou} showed that the
Bell polynomial $B_{n+m}$ admits integer
coordinates in the bases $  \{  x^{i}B_{j} (  x )    \}
_{i,j}$. Xu and Cen \cite{aim} extended the latter in some particular cases of
complete Bell polynomials. Also, the second
author and Bencherif \cite{bel1,bel2} established that Chebyshev polynomials of first and
second kind, and more generally bivariate polynomials associated with
recurrence sequences of order two, including Jacobsthal polynomials, Vieta
polynomials, Morgan-Voyce polynomials and others, admit remarkable integer
coordinates in a specific bases. Some recurrence relations on Bell numbers and
polynomials are given by Spivey
\cite{spy} and some other relations by Sun and Wu
\cite{yid}. What about $r$-Bell polynomials?

The $r$-Bell polynomials
$  \{  B_{n,r}  \}  _{n\geq0}$ are defined by their generating function%

\begin{equation*}
\sum_{n\geq0}B_{n,r} (  x )  \frac{t^{n}}{n!}=\exp  (  x (
e^{t}-1 )+rt),
\end{equation*}
and satisfy the generalized Dobinsky formula%
\begin{equation}
B_{n,r} (  x )  =\exp  (  -x )  \sum_{i=0}^{\infty}%
\dfrac{ (  i+r )  ^{n}}{i!}x^{i}.\label{a}%
\end{equation}
It is well known that $B_{n,r} (  x )  $ admits integer coordinates
in the following two: bases $  \{  x^{i}  \}  _{i}$\ and $  \{
B_{i} (  x )    \}  _{i}$ as%
\begin{equation}
B_{n,r} (  x )  =\sum_{i=0}^{n} {n+r \brace i+r}%
_{\!\!r}x^{i}\  \text{and }B_{n,r} (  x )  =\sum_{i=0}^{n}\dbinom{n}%
{i}r^{n-i}B_{i} (  x )  ,\label{b}%
\end{equation}
according to, respectively, the $r$-Stirling numbers of the second kind and the
binomial coefficients, see for example \cite{mez1}. For a general overview of
the $r$-Stirling numbers, one can see \cite{bro,car1,car2,mih2}. An extension
of $r$-Stirling numbers of the second kind and the $r$-Bell polynomials is
given in \cite{mih1}. In the sequel, we refer to \cite{belbel,belbou} for some
properties and recurrence relations of $r$-Lah numbers.

Our aim is to prove that the polynomials $B_{n+m,r}$ and $B_{n,r+s}$ admit a binomial recurrence
coefficient in the families
\begin{equation*}
  \{  x^{i}B_{n,j+r} (  x )    \}  _{i,j},\    \{
x^{i}B_{n,i+r} (  x )    \}  _{i},\    \{  x^{i}B_{j,r} (
x )    \}  _{j},\text{ }  \{  B_{j,s} (  x )    \}
_{j}\text{ and }  \{  x^{i}B_{j} (  x )    \}  ,
\end{equation*}
of the basis of the $\mathbb{Q} $-vector space formed by polynomials of $ \mathbb{Q} [X]$  .

\section{Main results}
Mez\H{o} \cite[Thm.\ 7.1]{mez1} showed that the $r$-Bell polynomials
satisfy the following recurrence relation%
\begin{equation*}
B_{n,r+1} (x)=\sum_{i=0}^{n}{n \choose i}B_{i,r} (x )  .
\end{equation*}
This can be generalized as follows.

\begin{theorem}
\label{P1}Decomposition of $B_{n,r+s} (  x )  $ into the family of
basis $\{B_{i,r}(x)\}_{i}.$ For all nonnegative
integers $n,\ r$ and $s$, we have%
\begin{equation*}
B_{n,r+s}(x)=\sum_{i=0}^{n}{n \choose i}r^{n-i}B_{i,s} (x)  .
\end{equation*}
\end{theorem}

\begin{proof}
Use (\ref{a}) to get
\begin{equation}
\frac{d^{s}}{dx^{s}}(\exp(x)B_{n,r}(x)
 )  =\exp  (  x )  B_{n,r+s} (  x )  .\label{2}
\end{equation}
Using the following identity \cite{mez1}
\begin{equation}
B_{n,r} (x)=\sum_{i=0}^{n}{n \choose i}r^{n-i}B_{i} (
x )  ,\label{6}
\end{equation}
we obtain
\begin{equation*}
\frac{d^{s}}{dx^{s}} (  \exp  (  x )  B_{n,r} (  x )
 )  =\sum_{i=0}^{n}{n \choose i}r^{n-i}\frac{d^{s}}{dx^{s}} (
\exp  (  x )  B_{i} (  x )   )  ,
\end{equation*}
and, applying property (\ref{2}), we obtain the desired identity.

We give now a combinatorial proof: let $x$ be a positive integer (a number of
colors). By the definition of the $r$-Bell numbers, $B_{n,r+s} (
x )$ gives the number of partitions of an $(n+r+s)$-element set, with the
restriction that the first $r+s$ elements are in distinct subsets (these are
called distinguished elements from now on). Moreover, the blocks not
containing distinguished elements are colored with one of the $x$ colors.

We can construct such partitions in the following way: from the $n$
non-distinguished elements we put $n-i$ into the blocks of $r$ distinguished
elements. To do this, we have ${n \choose n-i}={n \choose i}$\ possibilities
choosing those $n-i$ elements. Then, we put these elements into the above
mentioned blocks, which can happen on $r^{n-i}$ ways. Then the remaining
$n+s-(n-i)=s+i$ elements have to form a partition in which $s$ elements go to
different blocks and the other blocks are colored with one of the $x$ colors.
The number of these possibilities is exactly $B_{i,s} (x)$. The left and right hand sides coincide for any positive integer $x$, so they coincide for any $x\in \mathbb{R}$.
\end{proof}

\begin{corollary}
\label{C1}For all nonnegative integers $n,\ k,\ r$ and $s$, we have
\begin{eqnarray}
   {n+r+s \brace k+r+s}_{\!\!r+s}&amp;=&amp; \frac{1}{k!}\sum_{j=0}^{n-k}{{s \choose j}}{n+r \brace j+k+r}_{\!\!r} (j+k )!, \label{cca}\\
  {n+r+s \brace k+r+s}_{\!\!r+s} &amp;=&amp; \sum_{i=k}^{n}{n \choose i}{i+r \brace k+r}_{\!\!r}s^{n-i}. \label{cbc}
\end{eqnarray}
\end{corollary}

\begin{proof}
From the definition of $B_{n,r} (x)$ given by (\ref{b}), we have
\begin{equation*}
\frac{d^{s}}{dx^{s}} (\exp(x)B_{n,r} (  x )
 )  =\sum_{i=0}^{n}{n+r \brace i+r}%
_{\!\!r}\frac{d^{s}}{dx^{s}} (  x^{i}\exp  (  x )   ),
\end{equation*}
and upon using the Leibniz formula, one obtains
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
   B_{n,r+s} (  x )&amp;=&amp; \sum_{i=0}^{n}\sum_{k=0}^{i}{{s \choose k}}\frac{i!}{ (  i-k )  !} {n+r \brace i+r}_{\!\!r}x^{i-k} \\
   &amp;=&amp; \sum_{i=0}^{n}\sum_{l=0}^{i}{{s \choose i-l}}\frac{i!}{l!}{n+r \brace i+r}_{\!\!r}x^{l} \\
   &amp;=&amp; \sum_{l=0}^{n}x^{l}\sum_{i=l}^{n}{{s \choose i-l}}\frac{i!}{l!}{n+r \brace i+r}_{\!\!r}
\end{eqnarray*}
The identity \eqref{cca} follows by identification using the definition
of $B_{n,r+s} (x)$, and the fact that the elements $1,2,\ldots,r+s$ are
in different parts.

We have a combinatorial interpretation as follows: for $j=0,\ldots,s,$ there
are ${s \choose s-j}={s \choose j}$ ways to form $s-j$ singletons using the
elements in $  \{  1,\ldots,s  \}  $ and there are $
{n+r \brace k+r+j}_{\!\!r}$ ways to partition the set $  \{  s+1,\ldots,n+r+s  \}  $ into
$ (  k+r+s )  - (  s-j )  =k+r+j$ subsets such that the
elements of the set $  \{  s+1,\ldots,s+r  \}  $ are in different
subsets. The $j$ elements of the set $  \{  1,\ldots,s  \}  $ not
already used can be inserted in the $ (  k+r+j )  -r=k+j$ subsets in
\begin{equation*}
 (  k+j )  \cdots  (   (  k+j )  -j+1 )
=\frac{ (  k+j )  !}{k!}
\end{equation*}
ways. Then the number of partitions of the set $  \{  1,\ldots
,n+r+s  \}  $ into $k+r+s$ subsets such that the elements of the set
$  \{  1,\ldots,r+s  \}  $ are in different subsets is
\begin{equation*}
{n+r+s \brace k+r+s}
_{\!\!r+s}=\underset{j=0}{\overset{s}{\sum}}{s \choose j}{n+r \brace k+r+j}%
_{\!\!r}\frac{ (  k+j )  !}{k!}.
\end{equation*}
For the identity \eqref{cbc}, using the definition of $B_{n,r} (  x )
$ and Theorem \ref{P1} gives
\begin{eqnarray*}
   \sum_{k=0}^{n}{n+r+s \brace k+r+s}_{\!\!r+s}x^{k}&amp;=&amp; B_{n,r+s} (  x ) \\
     &amp;=&amp; \sum_{i=0}^{n}{n \choose i}s^{n-i}B_{i,r} (  x )  \\
     &amp;=&amp; \sum_{i=0}^{n}{n \choose i}s^{n-i}\sum_{k=0}^{i}{i+r \brace k+r}_{\!\!r}x^{k} \\
     &amp;=&amp; \sum_{k=0}^{n}x^{k}\sum_{i=k}^{n}{n \choose i}{i+r \brace k+r}_{\!\!r}s^{n-i}.
\end{eqnarray*}
Then, by identification, we obtain the identity \eqref{cbc} of the corollary.

We also give a combinatorial proof for this identity: from the $n$
non-distinguished elements $i$ go to the $k+r$ blocks which contain the first
$r$ distinguished elements: ${n \choose i}
{i+r \brace k+r}_{\!\!r}$ possibilities.
The remaining $n-i$ elements go to the $s$ additional
distinguished blocks, in $s^{n-i}$ ways. (So the $k+r+s$ blocks are guaranteed).
Finally we sum the $i$ disjoint cases.
\end{proof}
We note that the formula \eqref{cbc} is immediate from \cite[Lemma 13]{bro} with appropriate substitutions.


In different ways, Belbachir and Mihoubi \cite{bel3} and Gould and
Quaintance \cite{gou} showed that $B_{n+m} (  x )  $ admits a
recurrence relation according to the family $  \{  x^{i}B_{j} (
x )    \}  $ as follows:
\begin{equation}
B_{n+m} (  x )  =\sum_{k=0}^{n}\sum_{j=0}^{m}
{m \brace j}{n \choose k}j^{n-k}x^{j}B_{k} (  x )  ,\label{5}%
\end{equation}
In \cite{mez1}, Mez\H{o} cited the Carlitz identities \cite[eq.
(3.22--3.23)]{car1} given by

\begin{equation*}
B_{n+m,r}=\sum_{k=0}^{m} {m+r \brace k+r}
_{\!\!r}B_{n,k+r}\text{ and }B_{n,r+s}=\sum_{k=0}^{s}
{s+r \brack k+r}_{\!\!r} (  -1 )  ^{s-k}B_{n+k,r},
\end{equation*}
and established \cite{mez3}, by a combinatorial proof, the following identity%
\begin{equation*}
B_{n+m,r}=\sum_{k=0}^{n}\sum_{j=0}^{m} {m+r \brace j+r}%
_{\!\!r}{n \choose k} (  j+r )  ^{n-k}B_{k},
\end{equation*}
where $B_{n}=B_{n} (  1 )  $ is the number of ways to partition a set
of $n$ elements into non-empty subsets, $B_{n,r}=B_{n,r} (  1 )  $ is
the number of ways to partition a set of $n+r$ elements into non-empty subsets
such that the first $r$ elements are in different subsets and $%
{n \brace k}_{\!\!r}$ is an $r$-Stirling number of the second kind;
see \cite{bro,car1,car2}.
The following theorem generalizes these results.

\begin{theorem}
\label{T1} Decomposing  $B_{n+m,r} (  x )  $ into the family of
the basis $  \{  x^{k}B_{n,k+r} (  x )    \}  _{k},$ $  \{
x^{j}B_{k,r} (  x )    \}  _{j,k}$ and $  \{  x^{j}
B_{k} (  x )    \}  _{j,k}:$ for all nonnegative integers
$n,\ m,\ r$ and $s$, we have
\begin{eqnarray}
  B_{n+m,r} (x) &amp;=&amp; \sum_{k=0}^{m}{m+r \brace k+r}_{\!\!r}x^{k}B_{n,k+r} (x) \label{cbc1}\\
  B_{n+m,r} (x) &amp;=&amp; \sum_{k=0}^{n}\sum_{j=0}^{m}{m+r \brace j+r}_{\!\!r}{n \choose k}j^{n-k}x^{j}B_{k,r}(x) \label{cbc2}\\
  B_{n+m,r} (x) &amp;=&amp;  \sum_{k=0}^{n}\sum_{j=0}^{m}{m+r \brace j+r}_{\!\!r}{n \choose k}(j+r)^{n-k}x^{j}B_{k} (x) \label{cbc3}
\end{eqnarray}
Also, we have
\begin{equation}
x^{s}B_{n,r+s}(x)=\sum_{k=0}^{s}{s+r \brack k+r}
_{\!\!r}(-1)^{s-k}B_{n+k,r}(x)  \label{cbcc4}.
\end{equation}
\end{theorem}

\begin{proof}
For the identity \eqref{cbc1} we proceed as follows: the identity given in
\cite{bel3} and \cite{spy} can be written as follows
\begin{equation*}
B_{n+m} (  x )  =\sum_{i=0}^{n}\sum_{j=0}^{m}
{m \brace j}{n \choose i}j^{n-i}x^{j}B_{i} (  x )  =\sum_{j=0}^{m}
{m \brace j}x^{j}\sum_{i=0}^{n}{n \choose i}j^{n-i}B_{i,0} (  x ).
\end{equation*}
From Theorem \ref{P1}, we have $\sum_{i=0}^{n}{n \choose i}j^{n-i}
B_{i,s} (x)=B_{n,j+s} (x)$, $s\geq0$, then
\begin{equation*}
B_{n+m} (  x )  =\sum_{j=0}^{m}{m \brace j}x^{j}B_{n,j} (  x ),
\end{equation*}
and therefore
\begin{equation}
\frac{d^{r}}{dx^{r}} (  \exp  (  x )  B_{n+m} (  x )
 )  =\sum_{j=0}^{m}{m \brace j}\frac{d^{r}}{dx^{r}} (  x^{j}\exp  (  x )  B_{n,j} (
x )   )  .\label{4}%
\end{equation}
Now, using (\ref{a}), we get
\begin{equation}
\frac{d^{r}}{dx^{r}} (  \exp  (  x )  B_{n} (  x )
 )  =\exp  (  x )  B_{n,r} (  x ) \label{3}%
\end{equation}
and using (\ref{3}) and the Leibniz formula in (\ref{4}), we state that
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
   B_{n+m,r} (  x ) &amp;=&amp; \sum_{j=0}^{m} {m \brace j}\sum_{i=0}^{j}{r \choose i}\frac{j!}{ (j-i) !}x^{j-i}B_{n,j-i+r} (  x )\\
                    &amp;=&amp; \sum_{k=0}^{m}x^{k}B_{n,k+r} (  x )  \sum_{j=k}^{m}{m \brace j} {r \choose j-k}\frac{j!}{k!}.
\end{eqnarray*}
Let 

\begin{equation*}
a (  m,k,r )  =\sum_{j=k}^{m} {m \brace j}{r \choose j-k}\frac{j!}{k!}.
\end{equation*}
Then
\begin{eqnarray*}
\underset{m\geq0}{\sum}a (  m,k,r )  \frac{t^{m}}{m!}&amp;=&amp; \underset{j\geq k}{\sum}{r \choose j-k}\frac{j!}{k!}\underset{m\geq j}{\sum}
{m \brace j}\frac{t^{m}}{m!} \\
    &amp;=&amp; \frac{1}{k!}\underset{j\geq k}{\sum}{r \choose j-k} (\exp(t )-1 )^{j}\\
    &amp;=&amp; \frac{ (  \exp  (  t )  -1 )  ^{k}}{k!}\underset{j\geq 0}{\sum}{r \choose j} (  \exp  (  t )  -1 )  ^{j} \\
    &amp;=&amp; \frac{ (  \exp  (  t )  -1 )  ^{k}}{k!}\exp(rt ),
\end{eqnarray*}
which means that $a (  m,k,r )  ={m+r \brace k+r}%
_{\!\!r}\ $and $B_{n+m,r} (  x )  =\sum_{k=0}^{m}%
{m+r \brace k+r}_{\!\!r}x^{k}B_{n,k+r} (  x )  .$

For a combinatorial proof, we consider that there are $n+m$ non-distinguished
elements. From these we put $m$ and the $r$ distinguished elements into $k+r$
blocks, such that the $r$ distinguished elements are separated: there are $
{m+r \brace k+r}_{\!\!r}$ cases. We have to color the $k$ blocks not containing distinguished
elements, and this can happen $x^{k}$ ways. Then $n$ items remain. We can put
these elements into the already constructed blocks or into new blocks. We can
handle the already constructed blocks as distinguished elements. So we have
$n+k+r$ elements, of which $k+r$ are distinguished.
In addition, we have to color the
non-distinguished blocks. To do this, we have $B_{n,k+r}(x)$ possibilities.
Altogether, if $k$ is fixed, we have $
{m+r \brace k+r}_{\!\!r}x^{k}B_{n,k+r} (x)$ cases. We can sum over $k$.

For the identity \eqref{cbc2}, use Theorem \ref{P1} to replace $B_{n,k+r} (
x )$ by $\sum_{j=0}^{n}{n \choose j}k^{n-j}B_{j,r} (x)$.

For the identity \eqref{cbc3}, use relation (\ref{6}) to replace $B_{n,k+r} (
x )$ by $\sum_{j=0}^{n}{n \choose j} (k+r)^{n-j}B_{j}(x)$.

As a combinatorial proof, we can argue as follows:
from the $n$ elements we choose $k$
elements in ${n \choose k}$ ways and separate them. The remaining $m+r$
elements go to $j+r$ blocks, but $r$ elements stay in disjoint sets. This can
happen in ${m+r \brace j+r}_{\!\!r}$ ways.
We have to color the $j$ blocks; this is why the factor $x^{j}$
appears. The non-separated $n-k$ elements go to these blocks. This means
$(j+r)^{n-k}$ cases. Finally, the above $k$ separated items go to separated
and colored blocks; this is what $B_{k}(x)$ represents. We sum over the possible
values of $j$ and $k$. Again, the left- and right-hand sides coincide for any
positive integer $x$, so they coincide for any $x\in \mathbb{R}$.

For the identity \eqref{cbcc4} using (\ref{a}) and the following identity (see \cite{bro})
\begin{equation*}
\sum_{k=0}^{m} {m+r \brack k+r}_{\!\!r}x^{k}= (x+r)(x+r+1)  \cdots  (x+r+m-1),
\end{equation*}
we can write
\begin{eqnarray*}
  \sum_{k=0}^{s}{s+r \brack k+r}_{\!\!r} (-1)^{s-k}B_{n+k,r} (x) &amp;=&amp; (
-1 )^{s}\exp(-x)\sum_{i=0}^{\infty}(i+r)^{n}\dfrac{x^{i}}{i!}\sum_{k=0}^{s}{s+r \brack k+r}_{\!\!r}(-i-r)^{k}\\
   &amp;=&amp; (-1)^{s}\exp(-x)\sum_{i=0}^{\infty}(-i )(-i+1)\cdots(-i+s-1)(i+r)^{n}\dfrac{x^{i}}{i!}
\end{eqnarray*}
and this can be written as
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
  \exp(-x)\sum_{i=0}^{\infty}i (i-1)\cdots(i-s+1)(i+r)^{n}\dfrac{x^{i}}{i!} &amp;=&amp; x^{s}\exp(-x)\sum_{i=s}^{\infty}(i+r)
^{n}\dfrac{x^{i-s}}{ (i-s)!}  \\
   &amp;=&amp; x^{s}\exp(-x)\sum_{i=0}^{\infty}(i+r+s)^{n}\dfrac{x^{i}}{i!} \\
   &amp;=&amp; x^{s}B_{n,r+s}(x).
\end{eqnarray*}
\end{proof}

\begin{corollary}
\label{C2}For all nonnegative integers $n,\ m,\ k,\ r$ and $s$, we have
\begin{eqnarray}
   {n+m+r \brace k+r}_{\!\!r}&amp;=&amp; \sum_{j=0}^{\min \left(m,k\right)}{m+r \brace j+r}_{\!\!r}{n+j+r \brace k+r}_{\!\!j+r}, \label{abcd1} \\
  {n+r+s \brace k+r+s}_{\!\!r+s} &amp;=&amp; \sum_{j=0}^{s}(-1)^{s-j}{s+r \brack j+r}_{\!\!r}{n+j+r \brace k+s+r}_{\!\!r}. \label{abcd2}
\end{eqnarray}
\end{corollary}

\begin{proof}
For the identity \eqref{abcd1}, we have from Theorem \ref{T1}

\begin{equation*}
B_{n+m,r} (x)=\sum_{j=0}^{m}{m+r \brace j+r}_{\!\!r}x^{j}B_{n,j+r}(x).
\end{equation*}
Upon using (\ref{b}) to replace $B_{n,j+r} (x)$ by $\sum_{i=0}^{n}%
{n+j+r \brace i+j+r}_{\!\!j+r}x^{i},$ we can write
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
  B_{n+m,r} (x) &amp;=&amp; \sum_{j=0}^{m}{m+r \brace j+r}_{\!\!r}\sum_{i=0}^{n}{n+j+r \brace i+j+r}_{\!\!j+r}x^{i+j} \\
   &amp;=&amp; \sum_{k=0}^{n+m}x^{k}\sum_{j=0}^{\min \left(m,k\right)} {m+r \brace j+r}_{\!\!r}{n+j+r \brace k+r}_{\!\!j+r},
\end{eqnarray*}
and using the definition $B_{n+m,r} (x)=\sum_{k=0}^{n+m}{n+m+r \brace k+r}_{\!\!r}x^{k},$ the first identity
follows by identification.
The identity \eqref{abcd2} follows by the same way upon
using the fourth identity of Theorem \ref{T1}.
\end{proof}

\begin{remark}
One can proceed similarly, as in the proof of the Spivey's identity
\cite{spy} to obtain a combinatorial proof for the identity \eqref{cbc2} when
$x$ is a positive integer.
\end{remark}

\section{Acknowledgments}
The authors wish to thank the referee for a number of comments and suggestions that have
led to improvements in this paper.

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\hrule
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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B73; 
Secondary 05A10, 11B83.  

\noindent\emph{Keywords: } Bell and $r$-Bell polynomial, Stirling number,
$r$-Stirling number, recurrence relation.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000110}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 3 2013;
revised version received January 19 2014; January 29 2014; July 20 2014; September 6 2014; September 10 2014. 
Published in {\it Journal of Integer Sequences}, November 5 2014.

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\noindent
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