% EJC papers *must* begin with the following two lines. \documentclass[12pt]{article} \usepackage{e-jc} \specs{P1.48}{25(1)}{2018} % 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{amsmath,amsthm,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, U.S.A.\\ \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{Jul 28, 2017}{Feb 17, 2018}{Mar 2, 2018}\\ \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 recent 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.\qedhere \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.\qedhere \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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