% 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} % use this command to create hyperlinks (optional and recommended) \usepackage{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} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{remark}[theorem]{Remark} \newcommand\mm{{\rm mod\ }} \newcommand{\cpk}{{\rm cpk\,}} \newcommand{\cval}{{\rm cval\,}} \newcommand{\des}{{\rm des\,}} \newcommand{\ades}{{\rm ades\,}} \newcommand{\mdn}{{\mathcal D}_n} \newcommand{\mbn}{{\mathcal B}_n} \newcommand{\ma}{\mathcal{A}} \newcommand{\msn}{\mathfrak{S}_n} \newcommand{\exc}{{\rm exc\,}} \newcommand{\fix}{{\rm fix\,}} \newcommand{\cyc}{{\rm cyc\,}} \newcommand{\bn}{{\rm\bf n}} \newcommand{\rz}{{\rm RZ}} \newcommand{\sn}{S_n} \newcommand{\lrf}[1]{\lfloor #1\rfloor} \newcommand{\lrc}[1]{\lceil #1\rceil} \newcommand{\as}{{\rm as\,}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % if needed include a line break (\\) at an appropriate place in the title \title{\bf A family of two-variable derivative polynomials for tangent and secant} % 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{Shi-Mei Ma\thanks{Supported by the National Natural Science Foundation of China (11126217).}\\ \small School of Mathematics and Statistics\\[-0.8ex] \small Northeastern University at Qinhuangdao\\[-0.8ex] \small Hebei 066004, P.R. China\\ \small\tt shimeima@yahoo.com.cn } % \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{May 10, 2012}{Jan 4, 2013}\\ \small Mathematics Subject Classifications: 05A05, 05A15} \begin{document} \maketitle % E-JC papers must include an abstract. The abstract should consist of a % succinct statement of background followed by a listing of the % principal new results that are to be found in the paper. The abstract % should be informative, clear, and as complete as possible. Phrases % like "we investigate..." or "we study..." should be kept to a minimum % in favor of "we prove that..." or "we show that...". Do not % include equation numbers, unexpanded citations (such as "[23]"), or % any other references to things in the paper that are not defined in % the abstract. The abstract will be distributed without the rest of the % paper so it must be entirely self-contained. \begin{abstract} In this paper we introduce a family of two-variable derivative polynomials for tangent and secant. Generating functions for the coefficients of this family of polynomials are studied. In particular, we establish a connection between these generating functions and Eulerian polynomials. % keywords are optional \bigskip\noindent \textbf{Keywords:} Derivative polynomials; Eulerian polynomials; Tangent function; Secant function \end{abstract} \section{Introduction} Throughout this paper, denote by $D$ the differential operator $\frac{d}{dx}$. Let $y=\tan(x)$, and let $z=\sec(x)$. Then $D(y)=z^2$ and $D(z)=yz$. An important tangent identity is given by $$1+y^2=z^2.$$ In 1995, Hoffman~\cite{Hoffman95} considered two sequences of {\it derivative polynomials} defined respectively by \begin{equation*}\label{derivapoly-1} D^n(y)=P_n(y)\quad {\text and}\quad D^n(z)=z Q_n(y) \end{equation*} for $n\geq 0$. From the chain rule it follows that the polynomials $P_n(u)$ satisfy $P_0(u)=u$ and $P_{n+1}(u)=(1+u^2)P_n'(u)$, and similarly $Q_0(u)=1$ and $Q_{n+1}(u)=(1+u^2)Q_n'(u)+uQ_n(u)$. The first few of the polynomials $P_n(u)$ are $$P_1(u)=1+u^2,P_2(u)=2u+2u^3,P_3(u)=2+8u^2+6u^4.$$ There is a wealth of literature on derivative polynomials (see~\cite{Carlitz72,Cvijovic09,Franssens07,Hoffman99,Josuat10,Ma12,Ma122} for instance). Let $[n]=\{1,2,\ldots,n\}$, and let $\msn$ denote the the set of permutations of $[n]$. A permutation $\pi=\pi(1)\pi(2)\cdots\pi(n)\in\msn$ is {\it alternating} if $\pi(1)>\pi(2)<\cdots \pi(n)$. In other words, $\pi(i)<\pi({i+1})$ if $i$ is even and $\pi(i)>\pi({i+1})$ if $i$ is odd. It is well known~\cite{Andre79} that the {\it Euler numbers} $E_n$ defined by $$y+z=\sum_{n=0}^\infty E_n\frac{x^n}{n!}$$ count alternating permutations in $\msn$. The study of Euler numbers is a topic in combinatorics (see~\cite{Stanley}). Since the tangent is an odd function and the secant is an even function, we have $$y=\sum_{n=0}^\infty E_{2n+1}\frac{x^{2n+1}}{(2n+1)!} \quad{\text and} \quad z=\sum_{n=0}^\infty E_{2n}\frac{x^{2n}}{(2n)!}.$$ For this reason $E_{2n+1}$ is called a {\it tangent number} and $E_{2n}$ is called a {\it secant number}. Let $S(x)=y+z$. Clearly, $S(0)=1$. It is easy to verify that \begin{equation}\label{D-1} 2D(S(x))=1+S^2(x). \end{equation} Differentiation of~(\ref{D-1}) gives \begin{equation}\label{D-2} 2^2D^2(S(x))=2S(x)+2S^3(x). \end{equation} A second differentiation gives $2^3D^3(S(x))=2+8S^2(x)+6S^4(x)$. Now we present a connection between $S(x)$ and $P_n(u)$ \begin{proposition} For $n\geq 0$, we have $2^nD^n(S(x))=P_n(S(x))$. \end{proposition} \begin{proof} We proceed by induction on $n$. It suffices to consider the case $n\geq 3$. Assume that the statement is true for $n=k$. Then \begin{align*} 2^{k+1}D^{k+1}(S(x))&=2D(P_k(S(x)))\\ &=2P_k'(S(x))D(S(x))\\ &=(1+S^2(x))P_k'(S(x))\\ &=P_{k+1}(S(x)). \end{align*} Thus the statement is true for $k+1$, as desired. \end{proof} Writing the derivative polynomials in terms of $y$ and $z$ as follows: \begin{equation*} D^n(y)=\sum_{k=0}^{\lrf{\frac{n-1}{2}}}W_{n,k}y^{n-2k-1}z^{2k+2}, \end{equation*} \begin{equation*} D^n(z)=\sum_{k=0}^{\lrf{\frac{n}{2}}}W_{n,k}^{\textit{l}}y^{n-2k}z^{2k+1}, \end{equation*} we observed that the coefficients $W_{n,k}$ and $W_{n,k}^{\textit{l}}$ have simple combinatorial interpretations (see~\cite{Ma12}). The coefficient $W_{n,k}$ is the number of permutations in $\msn$ with $k$ interior peaks, where an interior peak of $\pi$ is an index $2\leq i\leq n-1$ such that $\pi(i-1)<\pi(i)>\pi(i+1)$. The coefficient $W_{n,k}^{\textit{l}}$ is the number of permutations in $\msn$ with $k$ left peaks, where a left peak of $\pi$ is either an interior peak or else the index $1$ in the case $\pi(1)>\pi(2)$ (see~\cite{Petersen09} for instance). This paper is organized as follows. In Section~\ref{Section-2}, we collect some notation, definitions and results that will be needed in the rest of the paper. In Section~\ref{Section-3}, we establish a connection between the Eulerian numbers and the expansion of $(Dy)^n(y)$. In Section~\ref{Section-4}, we establish a connection between the Eulerian numbers of type $B$ and the expansion of $(Dy)^n(z)$. In Section~\ref{Section-5}, some polynomials related to $(yD)^n(y)$ and $(yD)^n(z)$ are studied. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Preliminaries}\label{Section-2} %\hspace*{\parindent} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% A {\it descent} of a permutation $\pi\in\msn$ is a position $i$ such that $\pi(i)>\pi(i+1)$. Denote by $\des(\pi)$ the number of descents of $\pi$. Then the equations \begin{equation*} A_n(x)=\sum_{\pi\in\msn}x^{\des(\pi)+1}=\sum_{k=1}^nA(n,k)x^{k}, \end{equation*} define the {\it Eulerian polynomials} $A_n(x)$ and the {\it Eulerian numbers} $A(n,k)$. Set $A_0(x)=1$. The exponential generating function for $A_n(x)$ is \begin{equation}\label{Axz} A(x,t)=\sum_{n\geq 0}A_n(x)\frac{t^n}{n!}=\frac{1-x}{1-xe^{t(1-x)}}. \end{equation} The numbers $A(n,k)$ satisfy the recurrence relation \begin{equation}\label{recu-Euleriannum} A(n+1,k)=kA(n,k)+(n-k+2)A(n,k-1) \end{equation} with the initial conditions $A(0,0)=1$ and $A(0,k)=0$ for $k\geq 1$ (see~\cite[A008292]{Sloane}). The first few of the Eulerian polynomials $A_n(x)$ are $$A_0(x)=1,A_1(x)=x,A_2(x)=x+x^2,A_3(x)=x+4x^2+x^3.$$ %It is well known that %\begin{equation}\label{symmetric} %A_n(x)=x^{n+1}A_n\left(\frac{1}{x}\right). %\end{equation} An explicit formula for $A(n,k)$ is given as follows: \begin{equation*} A(n,k)=\sum_{i=0}^k(-1)^i\binom{n+1}{i}(k-i)^n. \end{equation*} The {\it hyperoctahedral group} $B_n$ is the group of signed permutations of the set $\pm[n]$ such that $\pi(-i)=-\pi(i)$ for all $i$, where $\pm[n]=\{\pm1,\pm2,\ldots,\pm n\}$. Let $${B}_n(x)=\sum_{k=0}^nB(n,k)x^{k}=\sum_{\pi\in B_n}x^{\des_B(\pi)},$$ where $$\des_B=\#\{i\in\{0,1,2,\ldots,n-1\}|\pi(i)>\pi({i+1})\},$$ with $\pi(0)=0$. The polynomial $B_n(x)$ is called an {\it Eulerian polynomial of type $B$}, while $B(n,k)$ is called an {\it Eulerian number of type $B$} (see~\cite[A060187]{Sloane}). The first few of the polynomials ${B}_n(x)$ are $$B_0(x)=1,B_1(x)=1+x,B_2(x)=1+6x+x^2,B_3(x)=1+23x+23x^2+x^3.$$ The numbers $B(n,k)$ satisfy the recurrence relation \begin{equation}\label{Bnk-Euleriannum} B(n+1,k)=(2k+1)B(n,k)+(2n-2k+3)B(n,k-1), \end{equation} with the initial conditions $B(0,0)=1$ and $B(0,k)=0$ for $k\geq 1$. An explicit formula for $B(n,k)$ is given as follows: \begin{equation*} B(n,k)=\sum_{i=0}^k(-1)^{i}\binom{n+1}{i}(2k-2i+1)^{n} \end{equation*} for $0\leq k\leq n$ (see~\cite{Eriksen00} for details). For $n\geq 0$, we always assume that $$(Dy)^{n+1}(y)=(Dy)(Dy)^n(y)=D(y(Dy)^n(y)),$$ $$(Dy)^{n+1}(z)=(Dy)(Dy)^n(z)=D(y(Dy)^n(z)),$$ $$(yD)^{n+1}(y)=(yD)(yD)^n(y)=yD((yD)^n(y)),$$ $$(yD)^{n+1}(z)=(yD)(yD)^n(z)=yD((yD)^n(z)).$$ Clearly, $(Dy)^n(y+z)=(Dy)^n(y)+(Dy)^n(z)$. For $n\geq 1$, we define \begin{equation*}\label{def-derivative-2} (Dy)^n(y+z)=\sum_{k=0}^{2n}J(2n,k)y^{2n-k}z^{k+1}. \end{equation*} In Section~\ref{Section-3} and Section~\ref{Section-4}, we respectively obtain that $$J(2n,2k-1)=2^nA(n,k), \quad 1\leq k\leq n,$$ and $$J(2n,2k)=B(n,k), \quad 0\leq k\leq n.$$ Let $J_n(x)=\sum_{k=0}^{2n}J(2n,k)x^k$ for $n\geq 1$. Then $xJ_n(x)=2^nA_n(x^2)+xB_n(x^2)$. Therefore, from~\cite[Theorem 3]{Ma12}, we have \begin{equation}\label{JnxAnx} xJ_n(x)=(1+x)^{n+1}A_n(x). \end{equation} Using~(\ref{JnxAnx}), we get the following proposition. \begin{proposition} For $n\geq 1$, we have $$(Dy)^n(y+z)=(y+z)^{n+1}\sum_{k=1}^nA(n,k)y^{n-k}z^k.$$ \end{proposition} %\hspace*{\parindent} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{On the expansion of $(Dy)^n(y)$}\label{Section-3} %\hspace*{\parindent} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% For $n\geq 1$, we define \begin{equation}\label{def-derivative-1} (Dy)^n(y)=\sum_{k=1}^{n}E(n,k)y^{2n-2k+1}z^{2k}. \end{equation} \begin{theorem}\label{thm11} For $1\leq k\leq n$, we have %\begin{equation*}\label{BnkAnk} $E(n,k)=2^nA(n,k)$. %\end{equation*} \end{theorem} \begin{proof} Note that $D(y^2)=2yz^2$. Then $E(1,1)=2A(1,1)$. Since $$D(y(Dy)^n(y))=2\sum_{k=1}^nkE(n,k)y^{2n-2k+3}z^{2k}+ 2\sum_{k=1}^n(n-k+1)E(n,k)y^{2n-2k+1}z^{2k+2},$$ there follows \begin{equation}\label{T-recurrence} E(n+1,k)=2(kE(n,k)+(n-k+2)E(n,k-1)). \end{equation} By comparing~(\ref{recu-Euleriannum}) with~(\ref{T-recurrence}), we obtain the desired result. \end{proof} Let \begin{equation*}\label{Def-22} F_n(y)=(Dy)^n(y)=\sum_{k=0}^nF(n,k)y^{2k+1}. \end{equation*} Then $F_{n+1}(y)=D(yF_n(y))$. Hence \begin{equation}\label{bny-recu} F_{n+1}(y)=(1+y^2)F_n(y)+y(1+y^2)F'_n(y) \end{equation} with initial value $F_0(y)=y$. Set $F_n(y)=2^na_n(y)$ and $a_n(y)=\sum_{k=0}^na(n,k)y^{2k+1}$. It follows from Theorem~\ref{thm11} that \begin{equation}\label{explicit-any} a_n(y)=\sum_{k=1}^nA(n,k)y^{2n-2k+1}(1+y^2)^k. \end{equation} Equating the coefficients of $y^{2n-2k+1}$ on both sides of~(\ref{explicit-any}), we obtain $$a(n,n-k)=\sum_{i=k}^n\binom{i}{k}A(n,i).$$ It follows from~(\ref{bny-recu}) that \begin{equation*}\label{ank} a(n+1,k)=(k+1)a(n,k)+ka(n,k-1). \end{equation*} Let $W_n(x)=\sum_{k=0}^na(n,k)x^{k+1}$. It is easy to verify that the polynomials $W_n(x)$ satisfy \begin{equation}\label{any-recurrence} W_{n+1}(x)=(x+x^2)W'_n(x), \end{equation} with initial value $W_{0}(x)=x$. %The first few terms of the polynomials $W_n(x)$ are given as follows: %\begin{align*} %W_1(x)&=x+x^2,\\ %%W_3(x)&=x+7x^2+12x^3+6x^4,\\ %W_4(x)&=x+15x^2+50x^3+60x^4+24x^5. %\end{align*} The triangular array $\{a(n,k)\}_{n\geq 0, 0\leq k\leq n}$ is called a {\it Worpitzky triangle} (see~\cite[A028246]{Sloane}). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\section{On the operator $((x+x^2)D)^n$}\label{Section-3} %\hspace*{\parindent} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In view of~(\ref{any-recurrence}), it is natural to consider the expansion of the operator $((x+x^2)D)^n$. We define \begin{equation}\label{def-Tnkx} ((x+x^2)D)^n=\sum_{k=1}^nG_{n,k}(x)(x+x^2)^kD^k \end{equation} for $n\ge 1$. Applying the operator $(x+x^2)D$ on the left of~(\ref{def-Tnkx}), we get \begin{equation}\label{Tnkx-1} G_{n+1,k}(x)=k(1+2x)G_{n,k}(x)+(x+x^2)D(G_{n,k}(x))+G_{n,k-1}(x). \end{equation} On the other hand, since $$D^k((x+x^2)D)=(x+x^2)D^{k+1}+k(1+2x)D^{k}+k(k-1)D^{k-1},$$ applying the operator $(x+x^2)D$ on the right of~(\ref{def-Tnkx}), we get \begin{equation}\label{Tnkx-2} G_{n+1,k}(x)=k(1+2x)G_{n,k}(x)+k(k+1)(x+x^2)G_{n,k+1}(x)+G_{n,k-1}(x). \end{equation} By comparing~(\ref{Tnkx-1}) with~(\ref{Tnkx-2}), we obtain $D(G_{n,k}(x))=k(k+1)G_{n,k+1}(x)$. Thus \begin{equation*}\label{TnkxD} G_{n,k}(x)=\frac{1}{k!(k-1)!}D^{k-1}(G_{n,1}(x)). \end{equation*} Thus $\deg G_{n,k}(x)=n-k$. Set $G_{n}(x)=G_{n,1}(x)$. Then~(\ref{Tnkx-1}) reduces to \begin{equation*}\label{Tnx-recu} G_{n+1}(x)=(1+2x)G_{n}(x)+(x+x^2)D(G_{n}(x)) \end{equation*} with initial value $G_1(x)=1$. Let $G_{n}(x)=\sum_{k=1}^nG(n,k)x^{k-1}$. It is easy to verify that \begin{equation}\label{Tnk-recurrence} G(n+1,k)=kG(n,k)+kG(n,k-1) \end{equation} with initial value $G(1,1)=1$. Recall that the {\it Stirling numbers of the second kind} $S(n,k)$ satisfy the recurrence relation \begin{equation}\label{recu-Stirling} S(n+1,k)=kS(n,k)+S(n,k-1) \end{equation} with initial conditions $S(0,0)=1$ and $S(n,0)=0$ for $n\geq 1$ (see~\cite[A008277]{Sloane}). By comparing~(\ref{Tnk-recurrence}) with~(\ref{recu-Stirling}), we immediately get the following result. \begin{proposition} For $1\leq k\leq n$, we have $G(n,k)=k!S(n,k)$. \end{proposition} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{On the expansion of $(Dy)^n(z)$}\label{Section-4} %\hspace*{\parindent} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %In this section, we will establish a connection between Eulerian numbers of type $B$ and %the expansion of $(Dy)^n(z)$. For $n\geq 0$, we define \begin{equation*}\label{def-derivative-2} (Dy)^n(z)=\sum_{k=0}^{n}H(n,k)y^{2n-2k}z^{2k+1}. \end{equation*} \begin{theorem} For $0\leq k\leq n$, we have $H(n,k)=B(n,k)$. \end{theorem} \begin{proof} Clearly, $H(0,0)=1$. Note that $D(yz)=y^2z+z^3$. Then $H(1,0)=B(1,0)$ and $H(1,1)=B(1,1)$. Note that $$(Dy)(Dy)^n(z)=\sum_{k=0}^n(1+2k)H(n,k)y^{2n-2k+2}z^{2k+1}+ \sum_{k=0}^n(2n-2k+1)H(n,k)y^{2n-2k}z^{2k+3}.$$ Then $$H(n+1,k)=(1+2k)H(n,k)+(2n-2k+3)H(n,k-1).$$ Hence the numbers $H(n,k)$ satisfy the same recurrence relation and initial conditions as $B(n,k)$, so they agree. \end{proof} Let $(Dy)^n(z)=zf_n(y)$. Using $(Dy)^{n+1}(z)=D(yzf_n(y))$, we get \begin{equation}\label{Rny-recu} f_{n+1}(y)=(1+2y^2)f_n(y)+y(1+y^2)f'_n(y) \end{equation} with initial value $f_0(y)=1$. %The first few terms of $f_n(y)$ are given as follows: %\begin{align*} %f_1(y)&=1+2y^2,\\ %f_2(y)&=1+8y^2+8y^4,\\ %f_3(y)&=1+26y^2+72y^4+48y^6,\\ %f_4(y)&=1+80y^2+464y^4+768y^6+384y^8. %\end{align*} Set $f_n(y)=\sum_{k=0}^nf({n,k})y^{2k}$. By~(\ref{Rny-recu}), we obtain \begin{equation*}\label{fnk-recu} f(n+1,k)=(1+2k)f(n,k)+2kf(n,k-1) \end{equation*} for $0\leq k\leq n$, with initial conditions $f(0,0)=1,f(0,k)=0$ for $k\geq 1$. It should be noted that $$(f(n,0),f(n,1),\ldots,f(n,n))$$ is the $f$-vector of the simplicial complex dual to the permutohedra of type $B$ of rank $n$ (see~\cite[A145901]{Sloane}). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Polynomials related to $(yD)^n(y)$ and $(yD)^n(z)$}\label{Section-5} %\hspace*{\parindent} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %Let us introduce a family of two-variable derivative polynomials. %\begin{definition} For $n\geq 1$, we define \begin{equation*} (yD)^n(y)=\sum_{k=1}^{n}M({n,k})y^{2k-1}z^{2n-2k+2}, \end{equation*} \begin{equation*} (yD)^n(z)=\sum_{k=1}^{n}N({n,k})y^{2k}z^{2n-2k+1}. \end{equation*} %\end{definition} \begin{theorem} For $1\leq k\leq n$, we have \begin{equation}\label{recurrence-1} M({n+1,k})=(2k-1)M({n,k})+(2n-2k+4)M({n,k-1}), \end{equation} \begin{equation}\label{recurrence-11} N({n+1,k})=2kN({n,k})+(2n-2k+3)N({n,k-1}). \end{equation} \end{theorem} \begin{proof} Note that $$(yD)(yD)^n(y)=\sum_{k=1}^n(2k-1)M({n,k})y^{2k-1}z^{2n-2k+4}+ \sum_{k=1}^n(2n-2k+2)M({n,k})y^{2k+1}z^{2n-2k+2}.$$ Thus we obtain~(\ref{recurrence-1}). Similarly, we get~(\ref{recurrence-11}). \end{proof} From~(\ref{recurrence-1}) and~(\ref{recurrence-11}), we immediately get a connection between $M({n,k})$ and $N({n,k})$. \begin{corollary} For $1\leq k\leq n$, we have $M({n,k})=N({n,n-k+1})$. \end{corollary} Let $M_n(x)=\sum_{k=1}^nM({n,k})x^k$, and let $N_n(x)=\sum_{k=1}^nN({n,k})x^k$. Then we have \begin{equation}\label{MnxNnxSymm} M_n(x)=x^{n+1}N_n\left(\frac{1}{x}\right). \end{equation} Set $$R_n(y)=(yD)^n(y)=\sum_{k=0}^nR({n,k})y^{2k+1},~zT_n(y)=(yD)^n(z)=z\sum_{k=1}^nT({n,k})y^{2k}.$$ It is easy to verify that \begin{equation}\label{Rny-recurr} R_{n+1}(y)=y(1+y^2)R'_n(y), \end{equation} \begin{equation}\label{Galton} T_{n+1}(y)=y^2T_n(y)+y(1+y^2)T'_n(y). \end{equation} %The first few terms of $R_n(y)$ and $T_n(y)$ are given as follows: %\begin{align*} %R_1(y)&=y+y^3,\\ %R_2(y)&=y+4y^3+3y^5,\\ %R_3(y)&=y+13y^3+27y^5+15y^7,\\ %R_4(y)&=y+40y^3+174y^5+240y^7+105y^9,\\ %R_5(y)&=y+121y^3+990y^5+2550y^7+2625y^9+945y^{11};\\ %T_1(y)&=y^2,\\ %T_2(y)&=2y^2+3y^4,\\ %T_3(y)&=4y^2+18y^4+15y^6,\\ %T_4(y)&=8y^2+84y^4+180y^6+105y^8\\ %T_5(y)&=16y^2+360y^4+1500y^6+2100y^8+945y^{10}. %\end{align*} Equating the coefficient of $y^{2k+1}$ on both sides of~(\ref{Rny-recurr}), we get \begin{equation*}\label{Rny-recu-2} R({n+1,k})=(2k+1)R({n,k})+(2k-1)R({n,k-1}). \end{equation*} Equating the coefficient of $y^{2k}$ on both sides of~(\ref{Galton}), we get \begin{equation*}\label{GaltonTri} T({n+1,k})=2kT({n,k})+(2k-1)T({n,k-1}). \end{equation*} Clearly, $R({n,n})=T({n,n})=(2n-1)!!$, where $(2n-1)!!$ is the {\it double factorial number}. It should be noted that the triangular arrays $\{R({n,k})\}_{n\geq 1,0\leq k\leq n}$ and $\{T({n,k})\}_{n\geq 1,1\leq k\leq n}$ are both {\it Galton triangles} (see~\cite[A187075]{Sloane}), which has been studied by Neuwirth~\cite{Neuwirth01}. Now we present the following result. \begin{theorem}\label{RnyMny} For $n\geq 1$, we have \begin{equation*} R_n(y)=y^{2n+1}N_n\left(\frac{1+y^2}{y^2}\right)\quad {\text and}\quad T_n(y)= (1+y^2)^{n}N_n\left(\frac{y^2}{1+y^2}\right). \end{equation*} \end{theorem} \begin{proof} Recall that $z^2=y^2+1$. Then $$R_n(y)=\sum_{k=1}^{n}M({n,k})y^{2k-1}(y^2+1)^{n-k+1}= y^{-1}(1+y^2)^{n+1}M_n\left(\frac{y^2}{1+y^2}\right),$$ and $$T_n(y)=\sum_{k=1}^{n}N({n,k})y^{2k}(y^2+1)^{n-k}= (1+y^2)^{n}N_n\left(\frac{y^2}{1+y^2}\right).$$ It follows from~(\ref{MnxNnxSymm}) that $$(1+y^2)^{n+1}M_n\left(\frac{y^2}{1+y^2}\right)=y^{2n+2}N_n\left(\frac{1+y^2}{y^2}\right),$$ as desired. \end{proof} From Theorem~\ref{RnyMny}, we get $R_n(1)=N_n(2)$ and $T_n(1)=2^nN_n(\frac{1}{2})$. It follows from~(\ref{recurrence-11}) that \begin{equation}\label{recu-Nnx} N_{n+1}(x)=(2n+1)xN_n(x)+2x(1-x)N'_n(x) \end{equation} with initial value $N_0(x)=1$. The first few of the polynomials $N_n(x)$ are $$N_1(x)=x,N_2(x)=2x+x^2,N_3(x)=4x+10x^2+x^3.$$ In particular, $N({n,1})=2^{n-1}$, $N({n,n})=1$ and $N_n(1)=(2n-1)!!$ for $n\geq 1$. There is a nice description of the polynomials $N_n(x)$ (see~\cite[A156919]{Sloane}): if $\vartheta=2xD$ and $r(x)=(1-x)^{-\frac{1}{2}}$, then $$\vartheta^n(r(x))=N_n(x)r(x)^{2n+1}.$$ In the following discussion, we consider some properties of the polynomials $N_n(x)$. The numbers $N({n,k})$ arise often in combinatorics and other branches of mathematics (see~\cite{Lehmer85} for instance). A {\it perfect matching} of $[2n]$ is a partition of $[2n]$ into $n$ blocks of size $2$. Analyzing the placement of $2n-1$ and $2n$, it is easy to verify that the number $N({n,k})$ counts perfect matchings of $[2n]$ with the restriction that only $k$ matching pairs have odd smaller entries (see~\cite[A185411]{Sloane}). For $n\geq 1$, an explicit formula for $N_n(x)$ is given as follows (see~\cite[A156919]{Sloane}): \begin{equation}\label{Explicit} N_n(x)=\sum_{k=1}^n2^{n-2k}\binom{2k}{k}k!S(n,k)x^k(1-x)^{n-k} \end{equation} where $S(n,k)$ is {\it the Stirling number of the second kind}. It follows from~(\ref{Explicit}) that $$N({n,k})=\sum_{i=1}^k(-1)^{k-i}2^{n-2i}\binom{2i}{i}\binom{n-i}{k-i}i!S(n,i).$$ Let $$N(x,t)=\sum_{n\geq 0}N_n(x)\frac{t^n}{n!}.$$ Using~(\ref{recu-Nnx}), the formal power series $N(x,t)$ satisfies the following partial differential equation: \begin{equation*}\label{diff-eq} (1-2xt)\frac{\partial N(x,t)}{\partial t}-2x(1-x)\frac{\partial N(x,t)}{\partial x}=xN(x,t). \end{equation*} By {\it the method of characteristics}~\cite{Wilf}, it is easy to derive an explicit formula: \begin{equation*} N(x,t)=e^{xt}\sqrt{\frac{1-x}{e^{2xt}-xe^{2t}}}. \end{equation*} Hence \begin{equation}\label{N2xt} N^2(x,t)=\frac{1-x}{1-xe^{2t(1-x)}}. \end{equation} Combining~(\ref{Axz}) and~(\ref{N2xt}), we get the following result. \begin{theorem} For $n\geq 0$, we have \begin{equation*} \sum_{k=0}^n\binom{n}{k}N_k(x)N_{n-k}(x)=2^nA_n(x). \end{equation*} \end{theorem} In the final part of this section, we present both central and local limit theorems for the coefficients of $N_n(x)$. As an application of a result~\cite[Theorem 2]{Ma08} on polynomials with only real zeros, the recurrence relation~(\ref{recu-Nnx}) enables us to show that the polynomials $\{N_n(x)\}_{n\geq 1}$ form a {\it Sturm sequence}. \begin{proposition}\label{RealZeros} For $n\geq 2$, the polynomial $N_n(x)$ has $n$ distinct real zeros, separated by the zeros of $N_{n-1}(x)$. \end{proposition} Let $\{a(n,k)\}_{0\leq k\leq n}$ be a sequence of positive real numbers. It has no {\it internal zeros} if there are no three indices $i