% EJC papers *must* begin with the following two lines. \documentclass[12pt]{article} \usepackage{e-jc} % Please remove all other commands that change parameters such as % margins or pagesizes. % only use standard LaTeX packages % only include packages that you actually need % we recommend these ams packages \usepackage{amsthm,amsmath,amssymb} % we recommend the graphicx package for importing figures \usepackage{graphicx} % use this command to create hyperlinks (optional and recommended) \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} % use these commands for typesetting doi and arXiv references in the bibliography \newcommand{\doi}[1]{\href{http://dx.doi.org/#1}{\texttt{doi:#1}}} \newcommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{\texttt{arXiv:#1}}} % all overfull boxes must be fixed; % i.e. there must be no text protruding into the margins % declare theorem-like environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{fact}[theorem]{Fact} \newtheorem{observation}[theorem]{Observation} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{open}[theorem]{Open Problem} \newtheorem{problem}[theorem]{Problem} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{note}[theorem]{Note} \def\Inv{\text{\rm Inv}} \newcommand{\N}{{\mathbb N}} \newcommand{\B}{{\mathbb B}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title{\bf Eulerian Numbers Associated with Arithmetical Progressions} % input author, affilliation, address and support information as follows; % the address should include the country, and does not have to include % the street address \author{Jos\'e L. Ram\'{\i}rez\\ \small Departamento de Matem\'aticas \\[-0.8ex] \small Universidad Nacional de Colombia\\[-0.8ex] \small Bogot\'a, Colombia\\ \small\tt jlramirezr@unal.edu.co\\ \and Sergio N. Villamar\'in \qquad Diego Villamizar\\ \small Department of Mathematics\\[-0.8ex] \small Tulane University\\[-0.8ex] \small New Orleans, LA 70118\\ \small\tt \{svillamaringomez, dvillami\}@tulane.edu } % \date{\dateline{submission date}{acceptance date}\\ % \small Mathematics Subject Classifications: comma separated list of % MSC codes available from http://www.ams.org/mathscinet/freeTools.html} \date{\dateline{Jan 1, 2012}{Jan 2, 2012}\\ \small Mathematics Subject Classifications: 11B83, 05A15, 05A19} \begin{document} \maketitle \begin{abstract} In this paper, we give a combinatorial interpretation of the $r$-Whitney-Eulerian numbers by means of coloured signed permutations. This sequence is a generalization of the well-known Eulerian numbers and it is connected to $r$-Whitney numbers of the second kind. Using generating functions, we provide some combinatorial identities and the log-concavity property. Finally, we show some basic congruences involving the $r$-Whitney-Eulerian numbers. \bigskip\noindent \textbf{Keywords:} Eulerian number, $r$-Whitney number, $r$-Whitney-Eulerian number, combinatorial identities, unimodality \end{abstract} \section{Introduction} The Eulerian numbers were introduced by Euler in a noncombinatorial way. Euler was trying to obtain a formula for the alternating sum $\sum_{i=1}^mi^n(-1)^i$ (cf.~\cite{Foata}). Explicitly, \emph{Eulerian numbers} $A(n,k)$ can be defined by the recurrence relation \cite{Comtet} \begin{align}\label{ec2} A(n,k)&=(n-k+1)A(n-1,k-1) + kA(n-1,k), \quad n\geq 1, \, k\geq 2, \end{align} with the initial values $A(n,1)=1$ for $n\geq 0$ and $A(0,k)=0$ if $k\geq 2$. Eulerian numbers can also be computed by the following expression \begin{align}\label{ec3} A(n,k)&=\sum_{i=0}^{k}S(n,i)i!\binom{n-i}{k-i}(-1)^{k-i}, \end{align} where $S(n,m)$ are the Stirling numbers of the second kind. Another interesting identity involving Eulerian numbers is called \emph{Worpitzky's identity} \begin{align*} x^n=\sum_{k=1}^{n}\binom{x+k-1}{n}A(n,k),\quad n\geq 1. \end{align*} It is well-known that Eulerian numbers have a combinatorial interpretation in term of permutations. In particular, the Eulerian number $A(n,k)$ counts the number of permutations $\pi=\pi_1\pi_2\cdots \pi_n$ with $k-1$ descents, that is $k-1=|\{i\in[n-1]: \pi_i>\pi_{i+1} \}|$. The \emph{Eulerian polynomials} are defined by $$A_n(x):=\sum_{k=1}^nA(n,k)x^k,$$ with $A_0(x)=1$. These polynomials satisfy the following relation for any non-negative integer $n$~\cite[p. 245]{Comtet}. \[\frac{A_n(x)}{(1-x)^{n+1}}=\sum_{k=0}^\infty k^nx^k.\] The Eulerian numbers and their generalizations have been studied extensively (cf.~\cite{Pet}). In the present article, we are interested in a recently generalization called \emph{$r$-Whitney-Eulerian numbers} and denoted by $A_{m,r}(n,k)$ in~\cite{MEZRAM}. This new sequence is defined by the expression \begin{align} A_{m,r}(n,k)=\sum_{j=0}^{n}W_{m,r}(n,j)m^jj!\binom{n-j}{k-j}(-1)^{k-j}, \label{defEulWhit} \end{align} where $W_{m,r}(n,k)$ are the $r$-Whitney numbers of the second kind. The \emph{$r$-Whitney numbers of the second kind} $W_{m,r}(n,k)$ were defined by Mez\H{o}~\cite{Mezo} as the connecting coefficients between some special polynomials. Specifically, for non-negative integers $n, k$ and $r$ with $n\geq k \geq 0$ and for any integer $m>0$ \begin{align} (mx+r)^n&=\sum_{k=0}^{n}m^kW_{m,r}(n,k)x^{\underline{k}},\label{defW} \end{align} where $x^{\underline{n}}=x(x-1)\cdots (x-n+1)$ if $n\geq 1$ and $x^{\underline{0}}=1$. The $r$-Whitney numbers of the second kind satisfy the recurrence~\cite{Mezo} \begin{align} W_{m,r}(n,k)&=W_{m,r}(n-1,k-1) + (km+r)W_{m,r}(n-1,k). \label{rec2} \end{align} Note that if $(m,r)=(1,0)$ we obtain the Stirling numbers of the second kind, if $(m,r)=(1,r)$ we have the $r$-Stirling (or noncentral Stirling) numbers~\cite{Broder}, and if $(m,r)=(m,1)$ we have the Whitney numbers \cite{Ben}. For more details on $r$-Whitney numbers see for example~\cite{CJ, CCMR, MRS, Merca3, Ram, MT, RamSha, RamSha2}. From \eqref{defEulWhit} and recurrence \eqref{rec2} we obtain that the $r$-Whitney-Eulerian numbers satisfy the recurrence relation \begin{align}\label{ec8} A_{m,r}(n,k)=(km+r)A_{m,r}(n-1,k)+(m(n-(k-1))-r) A_{m,r}(n-1,k-1), \end{align} with the initial values $A_{m,r}(0,0)=1$, $A_{m,r}(n,k)=0 $ if $k\geq n+1$ or $k\leq-1$. If $(m,r)=(1,0)$ we recover the Eulerian numbers $A(n,k)$. If $(m,r)=(1,r)$ we obtain the cumulative numbers studied by Dwyer~\cite{Dwy1, Dwy2}, see also the Euler-Frobenius numbers studied by Gawronski and Neuschel~\cite{GN}. If $(m,r)=(q+1,1)$ we obtain the $q$-Eulerian numbers studied by Brenti~\cite{Brenti}. The \emph{$r$-Whitney-Eulerian polynomials} are defined by \[A_{n,m,r}(x):=\sum_{k=0}^n A_{m,r}(n,k)x^k.\] For non-negative integers $r, n$ and positive $m$, it is known \cite{MEZRAM} that they satisfy the following identity \[\sum_{i=0}^{\infty}(mi+r)^nx^i=\frac{A_{n,m,r}(x)}{(1-x)^{n+1}},\] and their exponential generating function is \begin{align} \label{gfunpol} \sum_{n=0}^\infty A_{n,m,r}(x)\frac{y^n}{n!}=\frac{(1-x)\exp(ry(1-x))}{1-x\exp(my(1-x))}. \end{align} For a similar class of Eulerian numbers connected to the Whitney numbers see the papers of Rahmani~\cite{Rahmani} and Mez\H{o}~\cite{Mezo3}. In the present article, we give a combinatorial interpretation of the $r$-Whitney-Eulerian numbers by means of coloured signed permutations. Afterwards, we find several combinatorial identities in terms of this new sequence. Moreover, we prove that the $r$-Whitney-Eulerian numbers are log-concave and therefore unimodal. Finally, we establish some interesting congruences involving this sequence. \section{Combinatorial Interpretation} A \emph{signed permutation} on $[n]$ is a map $$\sigma:[n] \mapsto \{\pm1, \pm2, \dots, \pm n\}$$ which is bijective and $|\sigma|$ is a permutation ($|\sigma|$ is defined by $|\sigma|(i)=|\sigma(i)|$ for all $i\in[n]$). We denote by $B_n$ the set of all signed permutations. For any signed permutation $\sigma\in B_n$, we define a \emph{descent} to be a position $i$ such that $\sigma(i+1)<\sigma(i)$ with $i\in[n-1]\cup \{0\}$. We define $\sigma(0)=0$. For example, if $\sigma=3(-2)54(-1)$ then 1, 3 and 4 are the descents. If $\sigma=(-3)(-2)5(-4)1$ then $0$ and $3$ are the descents. The number of descents of a signed permutation $\sigma$ is denoted by $\text{des}_B(\sigma)$. An \emph{inversion} of a signed permutation $\sigma$ in $B_n$ is a pair $(i,j)$ such that $i|\sigma(j)|$. The set of all inversion of $\sigma$ is denoted by $\Inv_B(\sigma)$. For example, if $\sigma=(-3)(-2)5(-4)1$ then $$\Inv_B(\sigma)=\{(1,2), (1,5), (2,5), (3,4), (3,5), (4,5)\}.$$ A signed permutation $\sigma\in B_n$ is \emph{$(m,r)$-coloured} if it satisfies the following conditions: \begin{itemize} \item If $(i,\ell)\notin \Inv_B(\sigma)$ for all $\ell\geq i$, and $\sigma(i)>0$ then $\sigma(i)$ is coloured with one of $r$ colors. But, if $\sigma(i)<0$ then it is coloured with one of $m-r$ colours. \item If the above inversion property does not hold, then we colour $\sigma(i)$ with one of $m-1$ colours providing that $\sigma(i)<0$, but if $\sigma(i)$ is positive we coloured it with one colour. \end{itemize} Let $n, k, m, r \geq 0$ be integers with $m\geq r$. Let $\B_{n,k}^{(m,r)}$ denote the set of $(m,r)$-coloured signed permutations of $B_n$ with $k$ descents. \begin{theorem} For any integers $n, k, m, r \geq 0$, with $m\geq r$ we have $$|\B_{n,k}^{(m,r)}|=A_{m,r}(n,k).$$ \end{theorem} \begin{proof} Let $b_{n,k}^{(m,r)}=|\B_{n,k}^{(m,r)}|$. We are going to prove that the numbers $b_{n,k}^{(m,r)}$ satisfy the same recurrence that $A_{m,r}(n,k)$ with the same initial values. Indeed, note that any $(m,r)$-coloured signed permutation of $[n]$ with $k$ descents can be obtained from a $(m,r)$-coloured signed permutation $\pi'$ of $[n-1]$ with $k$ or $k-1$ descents by inserting the entries $n$ or $-n$ into $\pi'$. In the first case, we have to put the entry $n$ at the end of $\pi'$, or we have to put the entries $n$ or $-n$ between two entries that form one of the $k$ descents of $\pi'$. Then we have the following possibilities: $$(r+k+k(m-1))b_{n-1,k}^{(m,r)}=(km+r)b_{n-1,k}^{(m,r)}.$$ In the second case, we have to put the entries $n$ or $-n$ at the beginning of $\pi'$, or we have to put the entry $-n$ at the end of $\pi'$ or we have to insert $n$ or $-n$ between one of the $(n-2)-(k-1)=n-k-1$ ascents of $\pi'$. Hence we have the following possibilities $$(1+(m-1)+(m-r)+(n-k-1)+(n-k-1)(m-1))b_{n-1,k-1}^{(m,r)}=(m(n-(k-1))-r)b_{n-1,k-1}^{(m,r)}.$$ Therefore $$b_{n,k}^{(m,r)}=(km+r)b_{n-1,k}^{(m,r)}+(m(n-(k-1))-r)b_{n-1,k-1}^{(m,r)},$$ and the theorem is proved. \end{proof} \begin{example} Let $n=2, m=3$ and $r=2$. The $m-1=2$ different colours of the elements will be fixed as \textcolor{red}{red} and \textcolor{green}{green}; the $r=2$ different colours of the elements will be fixed as \textcolor{cyan}{cyan} and \textcolor{blue}{blue}; while the $m-r=1$ colours of the elements will be fixed as \textcolor{magenta}{magenta}. Therefore, $A_{3,2}(2,0)=4, A_{3,2}(2,1)=13$ and $A_{3,2}(2,2)=1$, where the coloured signed permutations are in Table \ref{tab}. \begin{table}[ht] \centering \begin{tabular}{|c|p{9cm}|} \hline Descents & Coloured signed permutations \\ \hline 0 & $\textcolor{cyan}{1}\textcolor{cyan}{2}, \quad \textcolor{blue}{1}\textcolor{cyan}{2}, \quad \textcolor{cyan}{1}\textcolor{blue}{2}, \quad \textcolor{blue}{12}.$ \\ \hline 1 & 2\textcolor{cyan}{1},\quad 2\textcolor{blue}{1}, \quad 2\textcolor{magenta}{(-1)}, \quad \textcolor{cyan}{1}\textcolor{magenta}{(-2)}, \quad \textcolor{blue}{1}\textcolor{magenta}{(-2)}, \quad \textcolor{magenta}{(-1)}\textcolor{cyan}{2}, \quad \textcolor{magenta}{(-1)}\textcolor{blue}{2}, \newline \textcolor{green}{(-2)}\textcolor{cyan}{1}, \textcolor{green}{(-2)}\textcolor{blue}{1}, \quad \textcolor{red}{(-2)}\textcolor{cyan}{1},\quad \textcolor{red}{(-2)}\textcolor{blue}{1}, \quad \textcolor{green}{(-2)}\textcolor{magenta}{(-1)}, \quad \textcolor{red}{(-2)}\textcolor{magenta}{(-1)}. \\ \hline 2 & \textcolor{magenta}{(-1)(-2)}. \\ \hline \end{tabular} \caption{(3,2)-Coloured signed permutations of size 2.} \label{tab} \end{table} \end{example} \begin{theorem} The following identity holds $$ m^nn!=\sum_{k=0}^{n}A_{m,r}(n,k).$$ \end{theorem} \begin{proof} Let $\sigma\in \B_{n,k}^{(m,r)}$. Consider the permutation $|\sigma|$ defined by $|\sigma|(i)=|\sigma(i)|$ for all $i\in[n]$. Let $P_{\sigma}=\{i\in[n]: (i,j)\notin \Inv(|\sigma|) \ \text{for any} \ j>i\}$. We suppose that $\ell=|P_{\sigma}|$, and suppose there are $t$ negative positions of these $\ell$ $(0\leq t \leq \ell)$, then these negative positions can be coloured with one of $m-r$ colours, while the $\ell-t$ positive positions can be coloured with one of $r$-colours. Therefore by the product rule we have $$\sum_{t=0}^\ell\binom{\ell}{t}(m-r)^tr^{\ell-t}\sum_{t=0}^{n-\ell}\binom{n-\ell}{t}(m-1)^t1^{n-\ell-t}=m^\ell m^{n-\ell}=m^n$$ ways to colour each fixed permutation. So, summing over all possible non-signed permutations we get the desired identity. \end{proof} \section{Some Combinatorial Identities} The goal of the current section is to extend some well-known identities for the classical Eulerian numbers to the $r$-Whitney-Eulerian numbers. \begin{theorem} For $n, k\geq 0$, we have the following identity $$ W_{m,r}(n,k)=\frac{1}{m^kk!}\sum_{i=0}^{k}A_{m,r}(n,i)\binom{n-i}{k-i}. $$ \end{theorem} \begin{proof} The proof follows by showing that the right side of the identity have the same recurrence relations as the $r$-Whitney numbers of the second kind. \end{proof} The $r$-Whitney-Eulerian numbers are not symmetric as the classical Eulerian numbers \linebreak ($A(n,k)=A(n,n-k+1)$). However, we note that $A_{m,r}(n,n-k-1)=\widehat{A}_{m,r}(n,k)$, where $\widehat{A}_{m,r}(n,k)$ are the generalized Eulerian numbers defined by Xiong et al.~\cite{Xiong}. From above relation and Lemmas 7 and 8 of \cite{Xiong}, we obtain a generalization of the Worpitzky's identity. \begin{theorem} For $n \geq 0$, we have the identities $$(mx+r)^n=\sum_{k=0}^nA_{m,r}(n,k)\binom{x+n-k}{n}=\sum_{k=1}^{n+1}A_{m,r}(n,n-k+1)\binom{x+k-1}{n}.$$ \end{theorem} Theorem \ref{teo6} gives a generalization of the well-known identity for the Eulerian numbers (cf.~\cite[p. 243]{Comtet}) $$A(n,k)=\sum_{i=0}^{k}(-1)^i(k-i)^n\binom{n+1}{i}.$$ \begin{theorem}\label{teo6} For $n, k\geq 0$, we have the identity \begin{align*} A_{m,r}(n,k)&=\sum_{i=0}^{k}(-1)^i[(k-i)m+r]^n\binom{n+1}{i}. \end{align*} \end{theorem} \begin{proof} By using the generating function \eqref{gfunpol} we have \begin{multline*} \sum_{n=0}^\infty\sum_{k=0}^\infty A_{m,r}(n,k)x^k\frac{y^n}{n!}=\frac{(1-x)\exp(ry(1-x))}{1-x\exp(my(1-x))}=(1-x)\exp(ry(1-x))\sum_{i=0}^\infty x^i e^{imy(1-x)}\\ =(1-x)\sum_{i=0}^\infty x^i e^{y(1-x)(im+r)}=\sum_{i=0}^\infty \sum_{n=0}^\infty (1-x)^{n+1}(im+r)^nx^i\frac{y^n}{n!}\\ =\sum_{i=0}^\infty \sum_{n=0}^\infty \sum_{\ell=0}^{n+1}\binom{n+1}{\ell}(-1)^\ell(im+r)^nx^{i+\ell}\frac{y^n}{n!}. \end{multline*} Comparing the coefficients on both sides, we get the desired result. \end{proof} Above identity gives us special values when $k$ is small: \begin{align*} A_{m,r}(n,0)&=r^n, \quad A_{m,r}(n,1)=(m+r)^n-r^n(n+1), \\ A_{m,r}(n,2)&=(2m+r)^n-(m+r)^n(n+1)+r^n\binom{n+1}{2}. \end{align*} Finally, by using the generating function \eqref{gfunpol} we find a relation between the $r$-Whitney-Eulerian polynomials and the classical Eulerian polynomials. \begin{theorem} For $n\geq 0$, we have the following identity $$A_{n,m,r}(x)=\sum_{j=0}^n\binom{n}{j}m^jr^{n-j}A_j(x)(1-x)^{n-j}.$$ \end{theorem} \section{Unimodality and Log-Concavity Properties} In this section we prove the log-concavity and therefore the unimodality of the $r$-Whitney-Eulerian numbers. Recall that a finite sequence of non negative real numbers $\{a_k\}_{0\leq k\leq n}$ is said to be \emph{unimodal} if there is an index $i$ such that $a_0\leq a_1\leq \cdots \leq a_{i-1} \leq a_i \geq a_{i+1} \geq \cdots \geq a_{n-1} \geq a_n$. A sequence of real numbers is log-concave if $a_i^2\geq a_{i-1}a_{i+1}$ for $0r$, this implies $0<1-\frac{r}{m}$. Using our previous lemma and identity (\ref{ec9}) we have that \begin{align}\label{DifferentialRecursion} A_{n,m,r}(x)=mx^{1-\frac{r}{m}}(1-x)^{n+1}\frac{d}{dx}(x^{\frac{r}{m}}(1-x)^{-n}A_{n-1,m,r}(x)). \end{align} We now proceed by using induction over $n$. For $n=1$ we get $$A_{1,m,r}(x)=r+(m-r)x$$ which have only one real root being $$x=-\frac{r}{m-r}<0.$$ By the inductive hypothesis for $n-1$ the term $$x^{\frac{r}{m}}(1-x)^{-n}A_{n-1,m,r}(x)$$ has $n-1$ non-positive real roots plus the root in $x=0$. So by Rolle's Theorem the derivative of this term must have exactly $n-1$ non-positive real roots and by Equation (\ref{DifferentialRecursion}) the polynomial $A_{n,m,r}(x)$ must have $n-1$ non-positive real roots. Since complex roots appear in conjugate pairs the only choice for the last root of $A_{n,m,r}(x)$ is to be real and non positive since the polynomial $A_{n,m,r}(x)$ has positive coefficients. \end{proof} Therefore we have the following theorem. \begin{theorem} If $0\leq r \leq m$, the $r$-Whitney Eulerian sequence $(A_{m,r}(n,k))_k$ is log-concave and therefore unimodal. \end{theorem} Note that the proof of the Theorem \ref{rzeros} actually provide more information than what is stated. It also shows that the polynomials $A_{n,m,r}(x)$ and $A_{n-1,m,r}(x)$ are interlacing if $m\geq r$. Let $(r_i)_{i\in\N}$ and $(s_j)_{j\in\N}$ be the sequences of the real zeros of polynomials $f$ of degree $n$ and $g$ of degree $n-1$ in nonincreasing order, respectively. We say that $g$ \emph{interlaces} $f$ \cite{Liu}, denoted by $g \preccurlyeq f$ , if $$r_n \leqslant s_{n-1} \leqslant \cdots \leqslant s_2 \leqslant r_2 \leqslant s_1 \leqslant r_1.$$ So, by using the argument of the proof, we can state that \[A_{n-1,m,r}(x) \preccurlyeq A_{n,m,r}(x).\] \section{Some Congruences} In this section, we will show some properties regarding prime congruences over generalized Eulerian numbers. These results generalize those of Knopfmacher and Robbins \cite{KR}. We make use of the following lemmas \cite{KR}. \begin{lemma} If $p$ is a prime number and $\ell\geq 1$, $1\leq k \leq p^{\ell}-1$, then $$\binom{p^{\ell}}{k}\equiv 0 \pmod p.$$ \end{lemma} \begin{lemma} If $p$ is a prime number and $\ell\geq 1$, $1\leq k \leq p^\ell-1$, then $$\binom{p^\ell+1}{k}\equiv \begin{cases} 1 \pmod p, & \text{if \ } k=0, \, 1, \, p^l, \, p^l+1; \\ 0 \pmod p, & \text{if \ } 2\leq k\leq p^\ell-1. \end{cases}$$ \end{lemma} Remember that $A_{m,r}(n,n-k-1)=\widehat{A}_{m,r}(n,k)$. Now we can prove the main results of this section. \begin{theorem}\label{teo14} If $p$ is a prime number and $\ell \geq 1$, $1\leq k+1\leq p^\ell-1$, then $$\widehat{A}_{m,r}(p^\ell-1,k) \equiv \begin{cases} 1\pmod p, & \text{if \ } p \not | \ m(k+2)-r; \\ 0\pmod p, & \text{otherwise}. \end{cases}$$ \end{theorem} \begin{proof} From Theorem \ref{teo6} we can establish the identity \begin{align}\label{idd} \widehat{A}_{m,r}(n,k)&=\sum_{i=0}^{k+1}(-1)^i[(k+2-i)m-r]^n\binom{n+1}{i}. \end{align} Therefore \begin{align*} \widehat{A}_{m,r}(p^\ell-1,k)&=\sum_{i=0}^{k+1}(-1)^i[(k+2-i)m-r]^{p^\ell-1}\binom{p^\ell}{i} \\ &\equiv [m(k+2)-r]^{p^\ell-1}=([m(k+2)-r]^{p-1})^{\frac{p^\ell-1}{p-1}} \pmod p. \end{align*} From Fermat little's theorem we get the desired result. \end{proof} In particular, if $m=r=1$ we have the following congruence for the Eulerian numbers. \begin{corollary}[\cite{KR}, Theorem 1] If $p$ is a prime number and $\ell \geq 1$, $1\leq k \leq p^\ell-1$, then $$A(p^\ell-1,k) \equiv \begin{cases} 1\pmod p, & \text{if \ } p \not | k; \\ 0\pmod p, & \text{otherwise}. \end{cases}$$ \end{corollary} \begin{theorem} Let $n$ be an integer such that $n$ does not divide the integers $(k+2)m-r,\ (k+1)m-r$ and $m$. Then $n$ is prime if and only if $\widehat{A}_{m,r}(n-1,k)\equiv 1 \pmod n$, for $1\leq k+1\leq n-1$ and $m^{n-1}\equiv [(k+2)m-r]^{n-1}\equiv[(k+1)m-r]^{n-1}\equiv 1 \pmod n$. \end{theorem} \begin{proof} If we assume that $n$ is prime, then the implication follows from Theorem \ref{teo14}. For the converse observe that \begin{align*} (n-1)!&\equiv m^{n-1}(n-1)!=\sum_{k=0}^{n-1}\widehat{A}_{m,r}(n-1,k)\equiv \widehat{A}_{m,r}(n-1,0) + \sum_{k=1}^{n-1}1\\ &=[(k+2)m-r]^{n-1}-[(k+1)m-r]^{n-1}+(n-1)\equiv -1 \pmod n. \end{align*} From Wilson's Theorem we deduce that $n$ is a prime number. \end{proof} \begin{theorem} Suppose that $p$ does not divide $(k+2)m-r$ and $(k+1)m-r$. If $p$ is a prime number, $\ell\geq 1$, and $1\leq k+1 \leq p^\ell$, then $$\widehat{A}_{m,r}(p^\ell,k)\equiv m \pmod p.$$ \end{theorem} \begin{proof} From identity \eqref{idd} we have \begin{align*} \widehat{A}_{m,r}(p^\ell,k)&=\sum_{i=0}^{k+1}(-1)^i[(k+2-i)m-r]^{p^\ell}\binom{p^\ell+1}{i}\\ &\equiv [(k+2)m-r]^{p^\ell}-[(k+1)m-r]^{p^\ell}\\ &\equiv [(k+2)m-r]-[(k+1)m-r]\\ &\equiv m \pmod p. \end{align*} \end{proof} \begin{theorem} Suppose that $p$ does not divide $(k+2)m-r$ and $(k+1)m-r$. If $p$ is a prime number, $\ell\geq 1$ and $2\leq k+1 \leq p^\ell$, then $$\widehat{A}_{m,r}(p^\ell+1,k)\equiv 2m^2 \pmod p.$$ \end{theorem} \begin{proof} By recurrence (\ref{ec8}) we have $$ \widehat{A}_{m,r}(p^\ell+1,k)=((k+2)m-r)\widehat{A}_{m,r}(p^\ell,k)+(r+(p^m-k)m)\widehat{A}_{m,r}(p^\ell+1,k).$$ From the previous theorem we have \begin{align*} \widehat{A}_{m,r}(p^\ell+1,k)&\equiv ((k+2)m-r)m+(r+(p^m-k)m)m\equiv 2m^2 \pmod p. \end{align*} \end{proof} \subsection*{Acknowledgements} The authors would like to thank the anonymous referee for some useful comments. The research of Jos\'e L. Ram\'irez was partially supported by Universidad Nacional de Colombia, Project No. 37805. The second and the third authors are graduate students at Tulane University. \begin{thebibliography}{10} \bibitem{Ben} M.~Benoumhani. On Whitney numbers of Dowling lattices. \emph{Discrete Math.}, 159: 13--33, 1996. \bibitem{Bona} M.~B\'ona. \emph{Combinatorics of Permutations}. 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