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\begin{document}

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\begin{center}
\vskip 1cm{\LARGE\bf Some Congruences for the Partial Bell \\
\vskip .1in 
Polynomials} 
\vskip 1cm
\large Miloud Mihoubi\footnote{Research
research supported by LAID3 Laboratory of USTHB University.}\\
University of Science and Technology Houari Boumediene\\
Faculty of Mathematics\\
P. O. Box 32 \\
16111 El-Alia, Bab-Ezzouar, Algiers \\
Algeria\\
\href{mailto:miloudmihoubi@hotmail.com}{\tt miloudmihoubi@hotmail.com}\\

\end{center}

\vskip .2in

\begin{abstract}
Let $B_{n,k}$ and $A_{n}=\sum_{j=1}^{n}B_{n,j}$ with $A_0=1$ be,
respectively, the $(n,k)^{\rm th}$ partial and the $n^{\rm th}$ complete
Bell polynomials with indeterminate arguments $x_1,x_2,\ldots$. 
Congruences for $A_{n}$ and $B_{n,k}$ with respect to a prime number have
been studied by several authors. In the present paper, we propose
some results involving congruences for $B_{n,k}$ when the arguments
are integers. We give a relation between Bell polynomials and we apply it
to several congruences. The obtained congruences are
connected to binomial coefficients.
\end{abstract}

\vskip .2in

\newtheorem{examp}[theorem]{Application}
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\section{Introduction}

Let $x_{1},x_{2},\ldots $ denote indeterminates. Recall that the partial
Bell polynomials $B_{n,k}\left( x_{1},x_{2},\ldots \right) $ are given
by
\begin{equation}
B_{n,k}\left( x_{1},x_{2},\ldots \right) =\sum \frac{n!}{k_{1}!k_{2}!\cdots}\left(
\frac{x_{1}}{1!}\right) ^{k_{1}}\left( \frac{x_{2}}{2!}\right)
^{k_{2}}\ldots \left( \frac{x_{n-k+1}}{\left( n-k+1\right) !}\right)
^{k_{n-k+1}},  \label{0}
\end{equation}%
where the summation takes place over all integers
$k_{1},k_{2},\ldots \geq 0$ such that
\begin{equation*}
k_{1}+2k_{2}+\cdots +\left( n-k+1\right) k_{n-k+1}=n\text{ \ and \ }%
k_{1}+k_{2}+\cdots +k_{n-k+1}=k.
\end{equation*}%
For references, see Bell \cite{8}, Comtet \cite{5} and Riordan \cite{9}.

Congruences for Bell polynomials have been studied by several
authors. Bell \cite{8} and Carlitz \cite{6} give some
congruences for complete Bell polynomials. In this paper, we propose some
congruences for partial Bell polynomials when the arguments are
integers. Indeed, we give a relation between Bell polynomials, given by
Theorem \ref{a} below, and we use it in the first part of the paper,
and with connection of the results of
Carlitz \cite{6} in the second part,
to deduce some congruences for partial Bell polynomials.
Some applications to Stirling numbers of the first and second kind and to the
binomial coefficients are given.

\section{Main results}

The next theorem gives an interesting relation between Bell polynomials. We use it to establish
some congruences for partial Bell polynomials.

\begin{theorem}
\label{a}Let $\left\{ x_{n}\right\} $ be a real sequence. Then for
$n,r,k$ integers with $n,r,k\geq 1,$ we have
\begin{equation}
x_{1}^{k}\sum_{j=1}^{n}B_{n,j}\left( y_{1},y_{2},\ldots\right) \left(
k-nr\right) ^{j-1}=x_{1}^{nr}\dfrac{B_{n+k,k}\left(
x_{1},x_{2},x_{3},\ldots\right) }{k\binom{n+k}{k}}  \label{s1}
\end{equation}%
with $y_{n}=\dfrac{B_{\left( r+1\right) n,nr}\left(
x_{1},x_{2},x_{3},\ldots\right) }{nr\binom{\left( r+1\right) n}{nr}}.$
\end{theorem}

For $k=nr+s,$ Identity (\ref{s1}) becomes

\begin{remark}
Let $\left\{ x_{n}\right\} $ be a real sequence. Then for $%
n,r,s$ integers with $n,r\geq 1,$ we get
\begin{equation}
x_{1}^{s}A_{n}\left( sy_{1},sy_{2},\ldots\right) =\frac{s}{nr+s}\frac{B_{\left(
r+1\right) n+s,nr+s}\left( x_{1},x_{2},x_{3},\ldots\right) }{\binom{\left(
r+1\right) n+s}{nr+s}},\ \ s\geq -nr+1.  \label{s2}
\end{equation}%
For $s\geq 0,$ we obtain Proposition 8 in \cite{1}, (see also
\cite{10}).
\end{remark}

\begin{theorem}
\label{d}Let $k,s$ be a nonnegative integers and $p$ be a prime
number. Then for any sequence $\left\{ x_{j}\right\} $ of integers
we have
\begin{equation*}
\left( k+s+1\right) B_{sp,k+s+1}\left( x_{1},x_{2},\ldots\right)\equiv
0\ (\mod p).
\end{equation*}
\end{theorem}

\begin{example}
If we denote by $s\left( n,k\right) $ and $S\left( n,k\right) $ for
Stirling numbers of first and second kind respectively, then
from the well-known identities
\begin{equation*}
B_{n,k}\left( 0!,-1!,2!,\ldots\right) =s\left( n,k\right) \text{ \ and \ }%
B_{n,k}\left( 1,1,1,\ldots\right) =S\left( n,k\right)
\end{equation*}%
when $x_{n}=1$ or $x_{n}=\left( -1\right) ^{n-1}\left( n-1\right) !$
in Theorem \ref{d} we obtain
\begin{equation*}
\left( k+s+1\right) S\left( sp,k+s+1\right) \equiv \left(
k+s+1\right) s\left( sp,k+s+1\right) \equiv 0\ (\mod p).
\end{equation*}
\end{example}

\begin{theorem}
\label{b}Let $n,k,s$ be integers with $n\geq k\geq 1,$ $s\geq 1$ and
$p$ be a prime number. Then for any sequence $\left\{ x_{j}\right\} $
of integers with $x_{1}$ not a multiple of $p$ we have
\begin{equation}
\left.
\begin{array}{l}
\dfrac{B_{n+sp,k+sp}\left( x_{1},x_{2},x_{3},\ldots\right) }{\left(
k+sp\right) \binom{n+sp}{k+sp}}\equiv x_{1}^{s}\dfrac{B_{n,k}\left(
x_{1},x_{2},x_{3},\ldots \right) }{k\binom{n}{k}}\ (\mod p)
\text{ \ if }p&gt;n-k+1\bigskip \\
x_{1}^{n}\dfrac{B_{n+sp,sp}\left( x_{1},x_{2},x_{3},\ldots \right) }{s\binom{n+sp%
}{sp}}\equiv x_{1}^{s}\dfrac{B_{\left( p+1\right) n,np}\left(
x_{1},x_{2},x_{3},\ldots \right) }{n\binom{\left( p+1\right) n}{np}}\ (\mod p^{2}) \text{ \ \ \ if \ }p&gt;n+1.%
\end{array}%
\right.  \label{a4}
\end{equation}
\end{theorem}

\begin{example}
If we consider the cases $k=1$ and $k=2$ in Theorem \ref{b} we
obtain
\begin{equation*}
\left.
\begin{array}{l}
B_{n+sp,1+sp}\left( x_{1},x_{2},x_{3},\ldots \right) \equiv x_{1}^{s}x_{n}\
(\mod p) \ \text{for }p&gt;n,\medskip \\
B_{n+sp,2+sp}\left( x_{1},x_{2},x_{3},\ldots \right) \equiv \dfrac{x_{1}^{s}}{2}%
\dsum_{j=1}^{n-1}\binom{n}{j}x_{j}x_{n-j}\ (\mod p) \text{
\ for }p&gt;n-1.%
\end{array}%
\right.
\end{equation*}%
Then, when $x_{n}=1$ or $x_{n}=\left( -1\right) ^{n-1}\left(
n-1\right) !$ we obtain
\begin{equation*}
\left.
\begin{array}{l}
S\left( n+sp,1+sp\right) \equiv 1\ (\mod p) \ \text{for }%
p&gt;n, \bigskip \\
s\left( n+sp,1+sp\right) \equiv \left( -1\right) ^{n-1}\left(
n-1\right) !\
(\mod p) \ \text{for }p&gt;n, \bigskip \\
S\left( n+sp,2+sp\right) \equiv 2^{n-1}-1\ (\mod p) \ \text{%
for }p&gt;n-1\text{ and} \bigskip \\
s\left( n+sp,2+sp\right) \equiv \left( -1\right) ^{n-1}\frac{n\left(
n+1\right) ^{2}}{2}\ (\mod p) \ \text{for }p&gt;n-1,%
\end{array}%
\right.
\end{equation*}
\end{example}

\begin{theorem}
\label{c}Let $n,k,s,p$ be integers with $n\geq k\geq 1,\ s\geq 1,\
p\geq 1$. Then for any sequence $\left\{ x_{j}\right\} $ of integers
with $x_{1}$ not a multiple of $p$ we have
\begin{equation}
\left.
\begin{array}{l}
\dfrac{B_{\left( s+1\right) n,sn}\left( x_{1},2x_{2},3x_{3},\ldots\right) }{%
\binom{\left( s+1\right) n}{sn}}\equiv sx_{1}^{n\left( s-1\right) }\dfrac{%
B_{2n,n}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{\binom{2n}{n}}\ (\mod n^{2}) ,\medskip \\
\dfrac{B_{n+sp,k+sp}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{\left(
k+sp\right) \binom{n+sp}{k+sp}}\equiv x_{1}^{s}\dfrac{B_{n,k}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{n}{k}}\ (\mod p) ,\medskip \\
x_{1}^{n}\dfrac{B_{n+sp,sp}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{s\binom{%
n+sp}{sp}}\equiv x_{1}^{s}\dfrac{B_{\left( p+1\right) n,np}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{n\binom{\left( p+1\right) n}{np}}\ (\mod n^{2}).%
\end{array}%
\right.  \label{5}
\end{equation}
\end{theorem}

\begin{example}
\label{g}Belbachir et al. \cite{o} have proved that
\begin{equation}
B_{n,k}\left( 1!,2!,\ldots ,\left( q+1\right) !,0,\ldots \right) =\dfrac{n!}{%
k!}\dbinom{k}{n-k}_{q},  \label{t1}
\end{equation}%
then, for $s\geq 1$ and $p\nmid j,$ the two last congruences of
(\ref{5}) and Identity (\ref{t1}) prove that
\begin{equation*}
\dbinom{k+sp}{j}_{q}\equiv \dbinom{k}{j}_{q}\ (\mod p) \text{
\ and \ }j\dbinom{sp}{j}_{q}\equiv s\dbinom{jp}{j}_{q}\ (\mod p^{2}).
\end{equation*}
\end{example}

\begin{corollary}
\label{e}Let $n,k,s$ be integers with $n\geq k\geq 1$ and $p$ be a prime
number. Then for any sequence $\left\{ x_{j}\right\} $ of integers with $%
x_{1}$ not a multiple of $p$ we have
\begin{equation*}
\left.
\begin{array}{l}
\dfrac{B_{\left( p+1\right) n,np}\left( x_{1},x_{2},\ldots \right) }{n\binom{%
\left( p+1\right) n}{np}}\equiv x_{1}^{n-1}\dfrac{B_{n+p,p}\left(
x_{1},x_{2},x_{3},\ldots \right) }{\binom{n+p}{p}}\ \bigskip (\mod p^{2}) \ \ \text{if }p&gt;n+1, \\
\dfrac{B_{\left( p+1\right) n,np}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{n%
\binom{\left( p+1\right) n}{np}}\equiv x_{1}^{n-1}\dfrac{B_{n+p,p}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{\binom{n+p}{p}}\ (\mod p^{2}).%
\end{array}%
\right.
\end{equation*}
\end{corollary}

\begin{example}
As in Application \ref{g}, we have
\begin{equation*}
\dbinom{jp}{j}_{q}\equiv j\dbinom{p}{j}_{q}\ (\mod p^{2}).
\end{equation*}
\end{example}

\begin{theorem}
\label{CC5}Let $k\geq 2,\ j\geq 1$ be integers and $p$ be an odd
prime number. Then for any sequence of integers $\left\{
x_{j}\right\} $ we have
\begin{equation}
\left.
\begin{array}{l}
\dfrac{B_{p^{j}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{p^{j}+k}{%
k}}\equiv x_{1}^{k-1}x_{p^{j}+1}\ (\mod p) \text{ \ \ if }%
p\nmid kx_{1},\medskip \\
\dfrac{B_{\left( r+1\right) p^{j},p^{j}r}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{p^{j}r\binom{\left( r+1\right) p^{j}}{p^{j}r%
}}\equiv x_{1}^{r-1}\left( x_{p^{j}+1}-x_{p^{j-1}+1}\right) \ (\mod p) \text{ \ \ if}\ p\nmid x_{1}.%
\end{array}%
\right.  \label{a5}
\end{equation}
\end{theorem}

\begin{example}
As in Application (\ref{g}), let $j=1$ in the second congruence of Theorem %
\ref{CC5}. Then
\begin{equation*}
\dfrac{\left( p-1\right) !}{r}\dbinom{pr}{p}_{q}\equiv -1\ (\mod p).
\end{equation*}
\end{example}

\begin{theorem}
\label{CC6}Let $k\geq 2,\ j\geq 1$ be integers and $p$ be an odd
prime number. Then for any sequence of integers $\left\{
x_{j}\right\} $ we have
\begin{equation*}
\left.
\begin{array}{l}
\dfrac{B_{2p^{j}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{2p^{j}+k%
}{k}}\equiv x_{1}^{k-2}\left( \left( k-1\right)
x_{p^{j}+1}^{2}+x_{1}x_{2p^{j}+1}\right) \ (\mod p) \text{
\ if }p\nmid kx_{1}\medskip \\
\dfrac{B_{2\left( r+1\right) p^{j},2p^{j}r}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{2p^{j}r\binom{2\left( r+1\right) p^{j}}{%
2p^{j}r}}\equiv x_{1}^{2r-2}\left( x_{1}x_{2p^{j}+1}-x_{p^{j}+1}^{2}\right)
\ (\mod p) \text{ \ if }p\nmid x_{1}.%
\end{array}%
\right.
\end{equation*}
\end{theorem}

\begin{remark}
Similarly to the last proofs, one can exploit the results of Carlitz
\cite{6} with connection to Theorem \ref{a} to obtain more
congruences for partial Bell polynomials.
\end{remark}

\section{Proof of the main results}

\begin{proof}[Proof of Theorem \protect\ref{a}]
Let $\left\{ x_{n}\right\} $ be a sequence of real numbers with
$x_{1}:=1$ and let $\left\{ f_{n}\left( x\right) \right\} $ be a
sequence of polynomials defined by
\begin{equation*}
f_{n}\left( x\right) =\dsum_{j=1}^{n}B_{n,j}\left( \dfrac{x_{2}}{2},\dfrac{%
x_{3}}{3},\dfrac{x_{4}}{4}\ldots \right) \left( x\right) _{j},
\end{equation*}%
with $f_{0}\left( x\right) =1,\left( x\right) _{j}:=x\left( x-1\right)
\cdots \left( x-j+1\right) $ for $j\geq 1$ and $\left( x\right) _{0}:=1.$\newline
We have $nf_{n-1}\left( 1\right) =n\dsum_{j=1}^{n-1}B_{n-1,j}\left( \dfrac{%
x_{2}}{2},\dfrac{x_{3}}{3},\ldots \right) \left( 1\right) _{j}=x_{n}$ and $%
D_{x=0}f_{1}\left( 0\right) =1\neq 0.$\newline
It is well known that $\left\{ f_{n}\left( x\right) \right\} $ presents a
sequence of binomial type, see \cite{5}. Then, from Proposition 1 in \cite{1}
we have%
\begin{equation}
y_{n}=\frac{1}{nr\binom{\left( r+1\right) n}{nr}}B_{\left( r+1\right)
n,nr}\left( 1,x_{2},x_{3},\ldots \right) =\frac{f_{n}\left( nr\right) }{nr}%
=D_{x=0}f_{n}\left( x;r\right) ,  \label{s4}
\end{equation}%
where $\left\{ f_{n}\left( x;a\right) \right\} $ is a sequence of binomial
type defined by
\begin{equation}
f_{n}\left( x;a\right) :=\dfrac{x}{an+x}f_{n}\left( an+x\right)  \label{s5}
\end{equation}%
with $a$ is a real number, see \cite{1}. From Proposition 1 in \cite{1} we
have also
\begin{equation}
\dfrac{B_{n+k,k}\left( 1,x_{2},x_{3},\ldots \right) }{k\binom{n+k}{k}}=\frac{%
f_{n}\left( k\right) }{k}=\frac{f_{n}\left( k-nr;r\right) }{k-nr},
\label{s6}
\end{equation}%
but from \cite{3} we can write $f_{n}\left( k-nr;r\right) $ as%
\begin{equation}
f_{n}\left( k-nr;r\right) =\sum_{j=1}^{n}B_{n,j}\left( D_{x=0}f_{1}\left(
x;r\right) ,D_{x=0}f_{2}\left( x;r\right) ,\ldots \right) \left( k-nr\right)
^{j}.  \label{s7}
\end{equation}%
Then, by substitution (\ref{s7}) in (\ref{s6}) and by using (\ref{s4}) we obtain
\begin{equation}
\dfrac{B_{n+k,k}\left( 1,x_{2},x_{3},\ldots \right) }{k\binom{n+k}{k}}%
=\sum_{j=1}^{n}B_{n,j}\left( y_{1},y_{2},\ldots \right) \left( k-nr\right)
^{j-1}.  \label{s8}
\end{equation}%
We can verify that Identity (\ref{s1}) is true for $x_{1}=0,$ and, for $%
x_{1}\neq 0$ it can be derived from (\ref{s8}) by replacing $x_{n}$ by $%
\dfrac{x_{n}}{x_{1}}$ and by using the well known identities
\begin{equation}
\left.
\begin{array}{l}
B_{n,k}\left( xa_{1},xa_{2},xa_{3},\ldots \right) \ =x^{k}B_{n,k}\left(
a_{1},a_{2},a_{3},\ldots \right) \text{ \ and} \bigskip \\
B_{n,k}\left( xa_{1},x^{2}a_{2},x^{3}a_{3},\ldots \right)
=x^{n}B_{n,k}\left(
a_{1},a_{2},a_{3},\ldots \right) ,%
\end{array}%
\right.  \label{s9}
\end{equation}%
where $\left\{ a_{n}\right\} $ is any real sequence.\medskip
\end{proof}

\begin{proof}[Proof of Theorem \protect\ref{d}]
We prove that $kB_{sp,k}\equiv 0$ $(\mod p) ,$ $k\geq s+1.$
From the identities
\begin{equation}
\dbinom{sp}{j}\equiv 0\ (\mod p) ,\text{ for }p\nmid j%
\text{ and }\dbinom{sp}{pj}\equiv \dbinom{s}{j}\ (\mod p) ,
\label{s110}
\end{equation}%
and from the recurrence relation given by
\begin{equation*}
kB_{n,k}=\underset{j}{\dsum }\dbinom{n}{j}x_{j}B_{n-j,k-1}
\end{equation*}%
with $B_{n,k}:=B_{n,k}\left( x_{1},x_{2},\ldots \right) $ and $x_{j}=0$ for $%
j\leq 0,$ we obtain \newline \bigskip
$\left( k+1\right) B_{sp,k+1}=\underset{j}{\dsum }\dbinom{sp}{j}%
x_{j}B_{sp-j,k}\equiv \dsum_{j=1}^{s}\dbinom{s}{j}x_{jp}B_{\left(
s-j\right) p,k}\ (\mod p) .$\newline \bigskip
Then, for $s=0,$ we get $kB_{0,k}\equiv 0\ (\mod p) ,$ $%
k\geq 0.$\newline \bigskip For $s=1,$ we get $\left( k+1\right)
B_{p,k+1}\equiv x_{p}B_{0,k}\equiv 0\ (\mod p),$ $k\geq
1.$\newline \bigskip For $s=2,$ the last congruences imply that
\newline $\left( k+1\right) B_{2p,k+1}\equiv
2x_{p}B_{p,k}+x_{2p}B_{0,k}=0\ (\mod p),$ $k\geq 2$ and
$p\nmid k. \bigskip $\newline  The induction on $s$ proves that
$kB_{sp,k}\equiv 0\ (\mod p) $ when $k\geq s+1$.\medskip
\end{proof}

\begin{proof}[Proof of Theorem \protect\ref{b}]
From \cite{5} we have
\begin{equation}
B_{n,k}\left( x_{1},x_{2},\ldots \right) =\frac{n!}{\left( n-k\right) !}%
\sum_{j=0}^{k}B_{n-k,k-j}\left( \frac{x_{2}}{2},\frac{x_{3}}{3},\frac{x_{4}}{%
4},\ldots \right) \frac{x_{1}^{j}}{j!},\ \ n\geq k\geq 1.  \label{xx}
\end{equation}%
Then, for $i\in \left\{ 1,\ldots ,n\right\} ,$ the last identity and Identities (%
\ref{s9}) imply
\begin{equation*}
\left.
\begin{array}{c}
t_{i}=\left( \left( n+1\right) !\right) ^{i}y_{i}=\dfrac{\left( \left(
n+1\right) !\right) ^{i}}{ir\binom{\left( r+1\right) i}{ir}}B_{\left(
r+1\right) i,ir}\left( x_{1},x_{2},\ldots ,x_{i-j+1}\right) =\medskip \\
\dsum_{j=1}^{i}\dfrac{\left( ir-1\right) !}{\left( ir-j\right) !}%
x_{1}^{ir-j}B_{i,j}\left( \dfrac{\left( n+1\right) !}{2}x_{2},\dfrac{\left(
\left( n+1\right) !\right) ^{2}}{3}x_{3},\ldots,\dfrac{\left( \left( n+1\right)
!\right) ^{i-j}}{i-j+1}x_{i-j+1}\right) ,%
\end{array}%
\right.
\end{equation*}%
from which we deduce that $t_{1},\ldots ,t_{n}$ are integers, and then $%
B_{n,1}\left( t_{1},t_{2},\ldots \right) ,\ldots ,B_{n,n}\left(
t_{1},t_{2},\ldots \right) $ are also integers. Therefore, by using the
second identity of (\ref{s9}), Identity (\ref{s1}) becomes
\begin{equation*}
x_{1}^{k}\sum_{j=1}^{n}B_{n,j}\left( t_{1},t_{2},\ldots \right) \left(
k-nr\right) ^{j-1}=x_{1}^{nr}\frac{\left( \left( n+1\right) !\right)
^{n}B_{n+k,k}\left( x_{1},x_{2},x_{3},\ldots \right) }{k\binom{n+k}{k}}.
\end{equation*}%
Hence, when we replace $k$ by $\alpha +sp$ in the last identity we
obtain
\begin{equation*}
x_{1}^{\alpha +sp}\sum_{j=1}^{n}B_{n,j}\left( t_{1},t_{2},\ldots \right) \left(
\alpha +sp-nr\right) ^{j-1}=x_{1}^{nr}\frac{\left( \left( n+1\right)
!\right) ^{n}B_{n+\alpha +sp,\alpha +sp}\left( x_{1},x_{2},x_{3},\ldots \right)
}{\left( \alpha +sp\right) \binom{n+\alpha +sp}{\alpha +sp}},
\end{equation*}%
and when we reduce modulo $p$ in the last identity
we obtain
\begin{equation*}
x_{1}^{nr}\dfrac{\left( \left( n+1\right) !\right) ^{n}B_{n+\alpha
+sp,\alpha +sp}\left( x_{1},x_{2},x_{3},\ldots \right) }{\left( \alpha
+sp\right) \binom{n+\alpha +sp}{\alpha +sp}}\equiv x_{1}^{\alpha
+s}\sum_{j=1}^{n}B_{n,j}\left( t_{1},t_{2},\ldots \right) \left( \alpha
-nr\right) ^{j-1}\ (\mod p).
\end{equation*}%
But from (\ref{s1}) we have \medskip \newline $x_{1}^{\alpha
}\dsum_{j=1}^{n}B_{n,j}\left( t_{1},t_{2},\ldots \right) \left( \alpha
-nr\right) ^{j-1}=\left( \left( n+1\right) !\right)
^{n}x_{1}^{\alpha }\dsum_{j=1}^{n}B_{n,j}\left(
y_{1},y_{2},\ldots \right) \left( \alpha -nr\right) ^{j-1}$

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \right. =\left( \left( n+1\right) !\right) ^{n}x_{1}^{nr}%
\dfrac{B_{n+\alpha ,\alpha }\left( x_{1},x_{2},x_{3},\ldots \right)
}{\alpha \binom{n+\alpha }{\alpha }},$\newline from which the last
congruence becomes
\begin{equation*}
\dfrac{\left( \left( n+1\right) !\right) ^{n}B_{n+\alpha +sp,\alpha
+sp}\left( x_{1},x_{2},x_{3},\ldots \right) }{\left( \alpha +sp\right) \binom{%
n+\alpha +sp}{\alpha +sp}}\equiv x_{1}^{s}\left( \left( n+1\right)
!\right) ^{n}\dfrac{B_{n+\alpha ,\alpha }\left(
x_{1},x_{2},x_{3},\ldots \right) }{\alpha \binom{n+\alpha }{\alpha }}\
(\mod p).
\end{equation*}%
Now, when we replace $n$ by $n-\alpha ,$ the last congruence becomes
\begin{equation*}
\dfrac{\left( \left( n-\alpha +1\right) !\right) ^{n}B_{n+sp,\alpha
+sp}\left( x_{1},x_{2},x_{3},\ldots \right) }{\left( \alpha +sp\right) \binom{%
n+sp}{\alpha +sp}}\equiv x_{1}^{s}\left( \left( n-\alpha +1\right) !\right)
^{n}\dfrac{B_{n,\alpha }\left( x_{1},x_{2},x_{3},\ldots \right) }{\alpha \binom{n%
}{\alpha }}\ (\mod p).
\end{equation*}%
Then, if $p&gt;n-\alpha +1$ we obtain
\begin{equation*}
\dfrac{B_{n+sp,\alpha +sp}\left( x_{1},x_{2},\ldots \right) }{\left(
\alpha +sp\right) \binom{n+sp}{\alpha +sp}}\equiv
x_{1}^{s}\dfrac{B_{n,\alpha }\left( x_{1},x_{2},x_{3},\ldots \right)
}{\alpha \binom{n}{\alpha }}\ (\mod p).
\end{equation*}%
For the second part of theorem, when we replace $k$ by $sr$ in (\ref{s1}) we
get%
\begin{equation*}
x_{1}^{sr}\sum_{j=1}^{n}B_{n,j}\left( t_{1},t_{2},\ldots \right) r^{j-1}\left(
s-n\right) ^{j-1}=x_{1}^{nr}\dfrac{\left( \left( n+1\right) !\right)
^{n}B_{n+sr,sr}\left( x_{1},x_{2},x_{3},\ldots \right) }{sr\binom{n+sr}{sr}},
\end{equation*}%
and, because $B_{n,j}\left( t_{1},t_{2},\ldots \right) $ ($1\leq j\leq
n$) are integers, the last identity proves that \medskip \newline
$x_{1}^{nr}\dfrac{\left( \left( n+1\right) !\right)
^{nr}B_{n+sr,sr}\left( x_{1},x_{2},x_{3},\ldots \right)
}{sr\binom{n+sr}{sr}}\equiv x_{1}^{sr}z_{n}\medskip $

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \  \right. \equiv x_{1}^{sr}\dfrac{\left(
\left( n+1\right) !\right) ^{nr}B_{\left( r+1\right) n,nr}\left(
x_{1},x_{2},x_{3},\ldots\right) }{nr\binom{\left( r+1\right) n}{nr}}\
(\mod r). $\newline Let $r=p&gt;n+1$ be a prime number. Now,
because the expressions
\begin{equation*}
\dfrac{\left( \left( n+1\right) !\right) ^{np}B_{n+sp,sp}\left(
x_{1},x_{2},x_{3},\ldots\right) }{sp\binom{n+sp}{sp}}\text{ and }\frac{\left(
\left( n+1\right) !\right) ^{np}B_{\left( p+1\right) n,np}\left(
x_{1},x_{2},x_{3},\ldots\right) }{np\binom{\left( p+1\right) n}{np}}
\end{equation*}%
are integers, we obtain
\begin{equation*}
x_{1}^{n}\dfrac{B_{n+sp,sp}\left( x_{1},x_{2},x_{3},\ldots\right) }{s\binom{n+sp%
}{sp}}\equiv x_{1}^{s}\dfrac{B_{\left( p+1\right) n,np}\left(
x_{1},x_{2},x_{3},\ldots \right) }{n\binom{\left( p+1\right) n}{np}}\
(\mod p^{2}).
\end{equation*}
\end{proof}

\begin{proof}[Proof of Theorem \protect\ref{c}]
From Identity (\ref{xx}) we get
\begin{equation}
\frac{B_{n+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{n+k}{k}}%
=\sum_{j=1}^{k}\frac{\left( k-1\right) !}{\left( k-j\right)
!}B_{n,j}\left( x_{2},x_{3},x_{4},\ldots \right) x_{1}^{k-j},\ \ n,k\geq
1  \label{mm5}
\end{equation}%
and this implies that the numbers
\begin{equation}
z_{n}=\frac{B_{\left( r+1\right) n,nr}\left( x_{1},2x_{2},3x_{3},\ldots \right)
}{nr\binom{\left( r+1\right) n}{nr}},\ n\geq 1  \label{m5}
\end{equation}%
are integers, and then, the numbers $B_{n,j}\left( z_{1},z_{2},\ldots \right) $ ($%
1\leq j\leq n$) are also integers. From Identity (\ref{s2}), when we replace $%
r$ by $1$ and $s$ by $n\left( s-1\right) $ we obtain
\begin{equation}
B_{\left( s+1\right) n,sn}\left( x_{1},2x_{2},3x_{3},\ldots \right)
=x_{1}^{n\left( s-1\right) }ns\binom{\left( s+1\right) n}{sn}%
\sum_{j=1}^{n}B_{n,j}\left( \overline{z}_{1},\overline{z}_{2},\ldots \right)
\left( \left( s-1\right) n\right) ^{j-1},  \label{6}
\end{equation}%
with $\overline{z}_{n}:=\frac{1}{n\binom{2n}{n}}B_{2n,n}\left(
x_{1},2x_{2},\ldots \right) .$ \medskip \newline Furthermore, from
(\ref{6}), we have \medskip \newline
$\dfrac{B_{\left( s+1\right) n,sn}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{%
\binom{\left( s+1\right) n}{sn}}=nx_{1}^{n\left( s-1\right)
}s\sum_{j=1}^{n}B_{n,j}\left( z_{1},z_{2},\ldots \right) \left( \left(
s-1\right) n\right) ^{j-1}$

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\right. \equiv n\left\{ x_{1}^{n\left( s-1\right) }sz_{n}\right\} $

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\right. \equiv n\left\{ x_{1}^{n\left( s-1\right) }s\frac{1}{n\binom{2n}{n}}%
B_{2n,n}\left( x_{1},2x_{2},\ldots \right) \right\} $

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\right. \equiv x_{1}^{n\left( s-1\right) }s\frac{1}{\binom{2n}{n}}%
B_{2n,n}\left( x_{1},2x_{2},\ldots \right) \ (\mod n^{2}) ,$
i.e.,
\begin{equation*}
\frac{B_{\left( s+1\right) n,sn}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{%
\binom{\left( s+1\right) n}{sn}}\equiv sx_{1}^{n\left( s-1\right) }\frac{%
B_{2n,n}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{\binom{2n}{n}}\
(\mod n^{2}).
\end{equation*}%
For the second part of (\ref{5}), when we replace $k$ by $\alpha +sp$ in (%
\ref{s1}), we obtain
\begin{equation*}
x_{1}^{\alpha +sp}\sum_{j=1}^{n}B_{n,j}\left( z_{1},z_{2},\ldots \right) \left(
\alpha +sp-nr\right) ^{j-1}=x_{1}^{nr}\dfrac{B_{n+\alpha +sp,\alpha
+sp}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{\left( \alpha +sp\right) \binom{%
n+\alpha +sp}{\alpha +sp}}
\end{equation*}%
with $z_{n}$ is given by (\ref{m5}). Because the numbers
$B_{n,j}\left( z_{1},z_{2},\ldots \right) ,$ $1\leq j\leq n,$ are
integers, then when we reduce modulo $p$ in the last
identity we get
\begin{equation*}
x_{1}^{nr}\dfrac{B_{n+\alpha +sp,\alpha +sp}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{\left( \alpha +sp\right) \binom{n+\alpha +sp%
}{\alpha +sp}}\equiv x_{1}^{\alpha +s}\sum_{j=1}^{n}B_{n,j}\left(
z_{1},z_{2},\ldots \right) \left( \alpha -nr\right) ^{j-1}\ (\mod p)
\end{equation*}%
and by (\ref{s1}) the last congruence becomes
\begin{equation*}
\dfrac{B_{n+\alpha +sp,\alpha +sp}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{%
\left( \alpha +sp\right) \binom{n+\alpha +sp}{\alpha +sp}}\equiv x_{1}^{s}%
\dfrac{B_{n+\alpha ,\alpha }\left( x_{1},2x_{2},3x_{3},\ldots \right)
}{\alpha \binom{n+\alpha }{\alpha }}\ (\mod p).
\end{equation*}%
To terminate, it suffices to replace $n$ by $n-\alpha $ in the last
congruence.\newline \ \ \newline For the third part of (\ref{5}),
when we replace $k$ by $kr$ in (\ref{s1}), we obtain
\begin{equation*}
x_{1}^{kr}\sum_{j=1}^{n}B_{n,j}\left( z_{1},z_{2},\ldots \right) \left(
kr-nr\right) ^{j-1}=x_{1}^{nr}\dfrac{B_{n+kr,kr}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{rk\binom{n+kr}{kr}}
\end{equation*}%
and because the numbers $B_{n,j}\left( z_{1},z_{2},\ldots \right) ,$
$1\leq j\leq n,$ are integers, then when we reduce
modulo $r$ in the last identity we get
\begin{equation*}
x_{1}^{nr}\frac{B_{n+kr,kr}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{rk\binom{%
n+kr}{kr}}\equiv x_{1}^{kr}z_{n}\equiv \frac{x_{1}^{kr}B_{\left( r+1\right)
n,nr}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{nr\binom{\left( r+1\right) n}{nr%
}}\ (\mod p).
\end{equation*}%
Now, because%
\begin{equation*}
\dfrac{B_{n+kr,kr}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{rk\binom{n+kr}{kr}}%
\text{ and }\dfrac{x_{1}^{kr}B_{\left( r+1\right) n,nr}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{nr\binom{\left( r+1\right) n}{nr}}
\end{equation*}%
are integers and $x_{1}^{p}\equiv x_{1}\ (\mod p)$
for any prime number $p,$ then when we put $r=p,$ the last
congruence becomes
\begin{equation*}
x_{1}^{n}\frac{B_{n+kp,kp}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{kp\binom{%
n+kp}{kp}}\equiv x_{1}^{k}\frac{B_{\left( p+1\right) n,np}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{np\binom{\left( p+1\right) n}{np}}\
(\mod p).
\end{equation*}%
To complete this proof, it suffices to multiply the two sides of the last congruence by $%
p. $\medskip
\end{proof}

\begin{proof}[Proof of\ Corollary \protect\ref{e}]
From the first congruence of (\ref{a4}) when we replace $s$ by $s-1,$ $n$ by
$n+p$ and $k$ by $p$ we get%
\begin{equation*}
\dfrac{B_{n+sp,sp}\left( x_{1},x_{2},x_{3},\ldots \right) }{s\binom{n+sp}{sp}}%
\equiv x_{1}^{s-1}\dfrac{B_{n+p,p}\left( x_{1},x_{2},x_{3},\ldots \right) }{%
\binom{n+p}{p}}\ (\mod p^{2}) ,\ p&gt;n+1,\ s\geq 1,
\end{equation*}%
and by combining the last congruence and the second congruence of
(\ref{a4}) we obtain
\begin{equation*}
\dfrac{B_{\left( p+1\right) n,np}\left( x_{1},x_{2},x_{3},\ldots \right) }{n%
\binom{\left( p+1\right) n}{np}}\equiv
x_{1}^{n-1}\dfrac{B_{n+p,p}\left( x_{1},x_{2},x_{3},\ldots \right)
}{\binom{n+p}{p}}\ (\mod p^{2}) ,\ p&gt;n+1.
\end{equation*}%
Similarly, we use the second and the third congruences of (\ref{5}) to get
the second part of the corollary.
\end{proof}

\begin{proof}[Proof of Theorem \protect\ref{CC5}]
Identity (\ref{s1}) can be written as
\begin{equation}
\left.
\begin{array}{c}
x_{1}^{pr}\left( k-p\right) \dfrac{B_{p+k,k}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{p+k}{k}}=x_{1}^{k}A_{p}\left(
\left( k-p\right) z_{1},\left( k-p\right) z_{2},\ldots \right) , \\
\text{with }z_{n}=\dfrac{B_{\left( r+1\right) n,nr}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{nr\binom{\left( r+1\right) n}{nr}},\ k\geq
1.%
\end{array}%
\right.  \label{C22}
\end{equation}%
Bell \cite{8} showed, for any indeterminates $x_{1},x_{2},\ldots,$ that
\begin{equation}
A_{p}\left( x_{1},x_{2},x_{3},\ldots \right) \equiv x_{1}^{p}+x_{p}\text{ }%
(\mod p).  \label{C3}
\end{equation}%
Therefore, from (\ref{C3}) and (\ref{C22}), we obtain
\begin{equation*}
x_{1}^{pr}\left( k-p\right) \dfrac{B_{p+k,k}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{p+k}{k}}\equiv
x_{1}^{k}\left\{ \left( k-p\right) ^{p}z_{1}^{p}+\left( k-p\right)
z_{p}\right\} \ (\mod p) ,
\end{equation*}%
and Identity (\ref{mm5}) shows that $\dfrac{B_{p+k,k}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{p+k}{k}}$ and the terms of the
sequence $\left\{ z_{n};n\geq 1\right\} $ are integers. Now, because $%
z_{1}=x_{1}^{r-1}x_{2},$ then, when $k$ is not a multiple of $p,$ the
last congruence and Fermat little Theorem prove that
\begin{equation*}
x_{1}^{r}\dfrac{B_{p+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{p+k%
}{k}}\equiv x_{1}^{k-1}\left\{ x_{1}^{r}x_{2}+x_{1}y_{p}\right\} \
(\mod p).
\end{equation*}%
For $k=1$ in the last congruence we have
\begin{equation*}
y_{p}\equiv x_{1}^{r-1}x_{p+1}-x_{1}^{r-1}x_{2}\ (\mod p).
\end{equation*}%
The proof for $j=1$ results from the two last congruences.\newline
Assume now that the congruences given by (\ref{a5}) are true for the index $%
j$.\newline Carlitz \cite{8} showed, for any indeterminates
$x_{1},x_{2},\ldots ,$ that
\begin{equation*}
A_{p^{j}}\equiv
x_{1}^{p^{j}}+x_{p}^{p^{j-1}}+x_{p^{2}}^{p^{j-2}}+\cdots +x_{p^{j}}\
(\mod p).
\end{equation*}%
For $x_{1},x_{2},\ldots$ integers we obtain
\begin{equation*}
A_{p^{j}}\equiv x_{1}+x_{p}+x_{p^{2}}+\cdots +x_{p^{j}}\ (\mod p).
\end{equation*}%
Then, when we use Identity (\ref{C22}) and the fact that the sequence $%
\left\{ z_{n};n\geq 1\right\} $ is a sequence of integers, we obtain when $%
p\nmid kx_{1}$ \medskip \newline
$x_{1}^{r}\dfrac{B_{p^{j+1}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k%
\binom{p^{j+1}+k}{k}}\equiv x_{1}^{k}\left(
z_{1}+z_{p}+z_{p^{2}}+\cdots +z_{p^{j+1}}\right) $

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \right. \equiv x_{1}^{k}\left(
z_{1}+z_{p}+z_{p^{2}}+\cdots +z_{p^{j}}\right) +x_{1}^{k}z_{p^{j+1}}$

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \right. \equiv x_{1}^{r}\dfrac{B_{p^{j}+k,k}\left(
x_{1},2x_{2},3x_{3},\ldots \right)
}{k\binom{p^{j}+k}{k}}+x_{1}^{k}z_{p^{j+1}}$

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \right. \equiv x_{1}^{k-1}x_{p^{j}+1}+x_{1}^{k}z_{p^{j+1}}\ (\mod p)$. \medskip \newline For $k=1$ in the last congruence we
have
\begin{equation*}
x_{1}^{r}x_{p^{j+1}+1}\equiv x_{p^{j}+1}+x_{1}z_{p^{j+1}}\ (\mod p).
\end{equation*}%
From the two last congruences we deduce that
\begin{equation*}
\begin{array}{l}
\dfrac{B_{p^{j+1}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{p^{j}+k%
}{k}}\equiv x_{1}^{k-1}x_{p^{j+1}+1}\ (\mod p) \text{ \ \
if }p\nmid kx_{1},\medskip \\
\dfrac{B_{\left( r+1\right) p^{j+1},p^{j+1}r}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{p^{j+1}r\binom{\left( r+1\right) p^{j+1}}{%
p^{j+1}r}}\equiv x_{1}^{r-1}\left( x_{p^{j+1}+1}-x_{p^{j}+1}\right) \ (\mod p) \text{ \ if }p\nmid x_{1},
\end{array}
\end{equation*}
which completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \protect\ref{CC6}]
Carlitz \cite{8} showed, for any indeterminates $x_{1},x_{2},\ldots,$
that
\begin{equation*}
A_{2p^{j}}\equiv A_{p^{j}}^{2}+x_{2p^{j}}\ (\mod p).
\end{equation*}%
Then, for $x_{1},x_{2},\ldots $ integers we get
\begin{equation*}
A_{2p^{j}}\equiv \left(
x_{1}^{p^{j}}+x_{p}^{p^{j-1}}+\cdots+x_{p^{j}}\right)
^{2}+x_{2p^{j}}\equiv \left( x_{1}+x_{p}+\cdots +x_{p^{j}}\right)
^{2}+x_{2p^{j}}\ (\mod p) ,
\end{equation*}%
and, when we use Identity (\ref{C22}), we obtain
\begin{equation*}
x_{1}^{2p^{j}r}\left( k-2p^{j}r\right) \dfrac{B_{2p^{j}+k,k}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{2p^{j}+k}{k}}=x_{1}^{k}A_{2p^{j}}%
\left( \left( k-2p^{j}r\right) z_{1},\left( k-2p^{j}r\right)
z_{2},\ldots \right) ,
\end{equation*}%
and because $\left\{ z_{n};n\geq 1\right\} $ is a sequence of
integers, the last identity gives
\begin{equation*}
\left.
\begin{array}{c}
x_{1}^{2p^{j}r}\left( k-2p^{j}r\right) \dfrac{B_{2p^{j}+k,k}\left(
x_{1},2x_{2},3x_{3},\ldots \right) }{k\binom{2p^{j}+k}{k}}\equiv \\
x_{1}^{k}\left( \left( k-2p^{j}r\right) ^{2}\left(
z_{1}+z_{p}+z_{p^{2}}+\cdots+z_{p^{j}}\right) ^{2}+\left(
k-2p^{j}r\right) z_{2p^{j}}\right) \text{ }(\mod p).
\end{array}%
\right.
\end{equation*}%
\newline
From the proof of Theorem \ref{CC5}, the last congruence gives when $%
p\nmid kx_{1} \medskip $\newline $x_{1}^{k}\left(
x_{1}^{2r}\dfrac{B_{2p^{j}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right)
}{k\binom{2p^{j}+k}{k}}\right) \equiv kx_{1}^{2k}\left(
z_{1}+z_{p}+z_{p^{2}}+\cdots +z_{p^{j}}\right)
^{2}+x_{1}^{2k}z_{2p^{j}}$

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \right. \equiv k\left(
x_{1}^{r}\dfrac{B_{p^{j}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right)
}{k\binom{p^{j}+k}{k}}\right) ^{2}+x_{1}^{2k}z_{2p^{j}}$

$\left. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \right. \equiv k\left(
x_{1}^{r+k-1}x_{p^{j}+1}\right)
^{2}+x_{1}^{2k}z_{2p^{j}}$ $(\mod p) ,$ i.e.,
\begin{equation*}
x_{1}^{2r}\dfrac{B_{2p^{j}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k%
\binom{2p^{j}+k}{k}}\equiv x_{1}^{k}\left(
kx_{1}^{2r-2}x_{p^{j}+1}^{2}+z_{2p^{j}}\right) \ (\mod p).
\end{equation*}%
For $k=1$ in the last congruence we get
$x_{1}^{2r-1}x_{2p^{j}+1}\equiv
x_{1}^{2r-2}x_{p^{j}+1}^{2}+z_{2p^{j}},$ i.e.,
\begin{equation*}
z_{2p^{j}}\equiv x_{1}^{2r-2}\left(
x_{1}x_{2p^{j}+1}-x_{p^{j}+1}^{2}\right) \ (\mod p).
\end{equation*}%
Then%
\begin{equation*}
x_{1}^{2}\dfrac{B_{2p^{j}+k,k}\left( x_{1},2x_{2},3x_{3},\ldots \right) }{k%
\binom{2p^{j}+k}{k}}\equiv x_{1}^{k}\left( \left( k-1\right)
x_{p^{j}+1}^{2}+x_{1}x_{2p^{j}+1}\right) \ (\mod p).
\end{equation*}
\end{proof}

\section{Acknowledgements}The author thanks the anonymous referee for his/her careful reading and valuable comments and suggestions.

\begin{thebibliography}{9}
\bibitem{8} E. T. Bell, Exponential polynomials. {\it Ann. Math.}
{\bf 35} (1934), 258--277.

\bibitem{o} H. Belbachir, S. Bouroubi, and A. Khelladi, Connection between
ordinary multinomials, generalized Fibonacci numbers, partial Bell partition
polynomials and convolution powers of discrete uniform distribution. {\it Annales
Mathematicae et Informaticae} {\bf 35} (2008),  21--30.

\bibitem{6} L. Carlitz, Some congruences for the Bell polynomials. 
{\it Pacific J. Math.} {\bf  11} (1961), 1215--1222.

\bibitem{5} L. Comtet, \textit{Advanced Combinatorics.} D. Reidel Publishing
Company,\ Dordrecht-Holland, 1974.

\bibitem{1} M. Mihoubi, Bell polynomials and binomial type sequences.
{\it Discrete Math.} {\bf 308} (2008), 2450--2459.

\bibitem{10} M. Mihoubi, The role of binomial type sequences in determination
identities for Bell polynomials. To appear,  {\it Ars Combinatoria}.
Preprint available at  \href{http://arxiv.org/abs/0806.3468v1}{\tt http://arxiv.org/abs/0806.3468v1}.

\bibitem{9} J. Riordan, \textit{Combinatorial Identities}.  R. E. Krieger,
1979.

\bibitem{3} S. Roman, \textit{The Umbral Calculus}.  Academic Press, 1984.
\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}:  Primary 05A10;
Secondary 11B73, 11B75, 11P83.

\noindent \emph{Keywords: } Bell polynomials, congruences, Stirling
numbers, binomial coefficients.

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\noindent
Received January 10 2009;
revised version received April 29 2009.
Published in {\it Journal of Integer Sequences}, May 12 2009.

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