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\title{\bf Eulerian Numbers Associated \\with Arithmetical Progressions}

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\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
}

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% \small Mathematics Subject Classifications: comma separated list of
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\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 $i<j$, but $|\sigma(i)|>|\sigma(j)|$. 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 $0<i<n$. It is well know that a sequence which is log-concave is also unimodal.   We first prove the following equality.
\begin{theorem}
For $n\geq 1$,  the $r$-Whitney-Eulerian polynomials satisfy the recurrence
\begin{align}\label{ec9}
A_{n,m,r}(x)&=(mx-mx^2)A'_{n-1,m,r}(x)+(r+(mn-r)x)A_{n-1,m,r}(x).
\end{align}
\end{theorem}
\begin{proof}
From recurrence (\ref{ec8}) we get
\begin{multline*}
A_{n,m,r}(x)=\sum_{k=0}^nA_{m,r}(n,k)x^k\\=\sum_{k=0}^n\left[ (km+r)A_{m,r}(n-1,k)x^k+(m(n-(k-1))-r)A_{m,r}(n-1,k-1)x^k\right]\\
=mx\sum_{k=0}^{n-1} kA_{m,r}(n-1,k)x^{k-1}+r\sum_{k=0}^{n-1}A_{m,r}(n-1,k)x^k+mnx\sum_{k=0}^{n-1}A_{m,r}(n-1,k)x^{k}\\
-mx^2\sum_{k=0}^{n-1}kA_{m,r}(n-1,k)x^{k-1}-rx\sum_{k=0}^{n-1}A_{m,r}(n-1,k)x^{k}\\
=(mx-mx^2)A'_{n-1,m,r}(x)+(r+(mn-r)x)A_{n-1,m,r}(x).\qedhere
\end{multline*}
\end{proof}

The log-concavity property  of the Eulerian numbers can be proved  by means of the real zero property of the Eulerian polynomials $A_n(x)$ (cf. \cite{Bona}). A sequence $\{a_0, a_1, \dots, a_n\}$ of the coefficients of a polynomial $f(x)=\sum_{k=0}^{n}a_kx^k$ of degree $n$ with only real zeros is called the \emph{P\'olya frequency sequence (PF)}.  It is well know that if a sequence is PF then it is  log-concave (cf. \cite{Bona}).  We are going to prove that the sequence $A_{m,r}(n,k)$ is a PF-sequence. To reach this aim, we first prove the following general lemma.

 \begin{lemma}
Let $(T_n(x))_n$ be a sequence of functions for $n\geq 0$ defined by
\begin{align*}
T_{n+1}(x)&=p_n(x)T_n(x)+q_n(x)T'_n(x)\\
T_0(x)&=T(x),
\end{align*}
for some sequence of functions $(p_n(x))_n, (q_n(x))_n$, then
\begin{align*}
T_{n+1}(x)&=r_n(x)\frac{d}{dx}(u_n(x)T_n(x)),
\end{align*}
where we define for some suitable real number  $\alpha$
$$r_n(x)=\frac{q_n(x)}{u_n(x)} \quad \text{ and }  \quad u_n(x)=e^{\int_\alpha^x\frac{p_n(t)}{q_n(t)}dt}.$$
\end{lemma}
\begin{proof}
Observe that
$$\frac{d}{dx}u_n(x)=\frac{p_n(x)}{q_n(x)}u_n(x).$$
Then
\begin{align*}
r_n(x)\frac{d}{dx}(u_n(x)T_n(x))&=\frac{q_n(x)}{u_n(x)}\frac{d}{dx}(u_n(x)T_n(x))\\
&=\frac{q_n(x)}{u_n(x)}T_n(x) \frac{d}{dx}u_n(x)+\frac{q_n(x)}{u_n(x)}T'_n(x)u_n(x)\\
&=\frac{q_n(x)}{u_n(x)}T_n(x)\left( \frac{p_n(x)}{q_n(x)}u_n(x)\right)+q_n(x)T'_n(x)\\
&=p_n(x)T_n(x)+q_n(x)T'_n(x)\\
&=T_{n+1}(x).\qedhere
\end{align*}
\end{proof}

\begin{theorem}\label{rzeros}
For $n\geq 1$, the $r$-Whitney-Eulerian polynomials $A_{n,m,r}(x)$ have only non-positive real roots if $m\geq r\geq 0$. Therefore $(A_{m,r}(n,k))_{k}$ is a PF-sequence.
\end{theorem}
\begin{proof}
The case in which $m=r$ is clear because $A_{m,m}(n,k)=m^nA(n,n-k-1)$. Let us assume that $m>r$, 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.


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\end{document}
