\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf An Analytic Approach to Special Numbers \\ \vskip .10in and Polynomials } \vskip 1cm \large Grzegorz Rz\c{a}dkowski\\ Department of Finance and Risk Management \\ Warsaw University of Technology\\ Ludwika Narbutta 85\\ 00-999 Warsaw Poland\\ \href{mailto:g.rzadkowski@wz.pw.edu.pl}{\tt g.rzadkowski@wz.pw.edu.pl} \\ \end{center} \vskip .2 in \def \bangle{ \atopwithdelims \langle \rangle} \begin{abstract} The purpose of this article is to present, in a simple way, an analytical approach to special numbers and polynomials. The approach is based on derivative polynomials. The paper is, to some extent, a review article, although it contains some new elements. In particular, it seems that some integral representations for Bernoulli numbers and Bernoulli polynomials are new. \end{abstract} \section{Introduction} Let $u=u(z)$ be a holomorphic function defined in a domain $D_{u} \subset \mathbb{C}$ which fulfills the Riccati differential equation with constant coefficients \begin{equation}\label{R} u'=r(u-a)(u-b), \end{equation} where $r,a,b$ are real or complex numbers $r\neq 0, \; a \neq b$. Let $v=v(z)$ be a holomorphic function defined in a domain $D_{v} \subset \mathbb{C}$ which is related with $u(z)$ and fulfills the following differential equation \begin{equation}\label{R2} v'=rv\left(u-\frac{a+b}{2}\right), \end{equation} where $a,b,r,u(z)$ are as in (\ref{R}). Examples of such pairs of functions and equations are as follows: \begin{enumerate} \item $u(z)=\tan z, \quad u'(z)=u^{2}+1,\quad v(z)= \sec z, \quad v'=vu,$ \item $u(z)=\tanh z, \quad u'(z)=-u^{2}+1,\quad v(z)= 1/\cosh z, \quad v'=-vu,$ \item $u(z)=\cot z, \quad u'(z)=-u^{2}-1,\quad v(z)= \csc z, \quad v'=-vu,$ \item $u(z)=\coth z, \quad x'(z)=u^{2}-1,\quad v(z)= 1/\sinh z, \quad v'=vu,$ \item $u(z)=1/(1+e^{ z}), \quad u'(z)=u^{2}-u, \quad v(z)= e^{z/2}/(1+e^{ z}), \quad \\ v'=v(u-1/2),$\label{l} \item $u(z)=1/(1+e^{ -z}), \quad u'(z)=-u^{2}+u, \quad v(z)= e^{-z/2}/(1+e^{-z}), \\ v'=-v(u-1/2),$ \item more generally the logistic function: $u(z)=q/(1+pe^{ -sz}), \quad \\ u'(z)=\frac{s}{q}(q-u)u, \quad v(z)=qe^{-sz/2}/(1+pe^{ -sz}), \quad v'(z)=\frac{s}{q}v(q/2-u)$ (with $p>1,\; q>0,\; s>0$). \label{log} \end{enumerate} We will consider also the following generalization of equation (\ref{R2}) \begin{equation}\label{R2bis} v'=rv\left(u-\frac{a+b}{2}+d\right), \end{equation} where $d$ is a real or complex number. Such a system of differential equations has been investigated by Hoffman \cite{Ho} (instead of equations (\ref{R2}-\ref{R2bis}) he regarded $v'=vu$) and by Franssens \cite{F2} (who investigated the equation $v'=-vu$). Let $\{a_1,a_2,\ldots ,a_n\}$ be a permutation of the set $\{1,2,\ldots ,n\}$. Then $\{a_{j},a_{j+1}\}$ is an ascent of the permutation if $a_j< a_{j+1}$. The Eulerian number $\displaystyle {n \bangle k} $ is defined as the number of permutations of the set $\{1,2,\ldots ,n\}$ having $k$ permutation ascents, see \cite[p.\ 267]{GKP}. For example for $n=3$ the permutation $\{1,2,3\}$ has two ascents, namely $\{1,2\}$ and $\{2,3\}$, and $\{3,2,1\}$ has no ascents. Each of the other four permutations of the set has exactly one ascent. Thus $\displaystyle {3 \bangle 0} =1 $, $\displaystyle {3 \bangle 1} =4 $, and $\displaystyle {3 \bangle 2} =1 $. It is well known that Eulerian numbers satisfy the following relations: {\setlength\arraycolsep{2pt} \begin{eqnarray} {n \bangle k} &=& {n \bangle n-k-1} , \nonumber \\ {n+1 \bangle k} &=& (k+1) {n \bangle k} +(n-k+1) {n \bangle k-1} , \label{eq1}\\ {n \bangle k} &=& \sum\limits_{j=0}^{k}(-1)^{j}{n+1 \choose j}(k-j+1)^{n}. \label{eq2} \end{eqnarray}} The Eulerian polynomial $E_{n}(x),\; n=0,1,2,\ldots $ is defined, by Comtet \cite{C}, by the formula \begin{equation}\label{Ep} E_{n}(x)= \sum\limits_{k=0}^{n-1}{n \bangle k}x^{k+1}\;\;\textrm{for}\; n\ge 1,\quad E_0(x)=1. \end{equation} There is a slightly different definition, used e.g., by Foata \cite{Fo}, of the Eulerian polynomial $A_{n}(x)$ i.e., \begin{equation}\label{Ep2} A_{n}(x)= \sum\limits_{k=0}^{n-1}{n \bangle k}x^{k}, \quad A_0(x)=1. \end{equation} Thus $E_{n}(x)=xA_{n}(x)$ for $n\ge 1$, $A_{1}(x)= A_{0}(x)=E_{0}(x)\equiv 1$. The MacMahon numbers $(M_{n,k})$ are defined in papers \cite{M, F}, by the recurrence formula \begin{equation}\label{MM} M_{n,k} = (2k-1)M_{n-1,k} + (2n-2k+1)M_{n-1,k-1}, \end{equation} where $1\le k\le n,\;\; M(n,1)=M(n,n)=1,\; n=1,2,\ldots $. The MacMahon polynomial $M_{n}(x),\; n=0,1,2,\ldots $ is defined as follows \begin{equation}\label{MMp} M_{n}(x)=\sum\limits_{k=1}^{n+1}M_{n+1,k}\:x^{k-1}. \end{equation} \section{Derivative polynomials} The following theorem has been discussed during the Conference ICNAAM 2006 in Greece and it appeared in my paper \cite{Rz}. For convenience of the reader we give it with an inductive proof. Independently the theorem has been considered and proved, with a proof based on generating functions, by Franssens \cite{F2} (see also \cite{Rz1}). \begin{theorem}\label{th1} If a function $u(z)$ satisfies equation (\ref{R}), then the $n$th derivative of $u(z)$ can be expressed by the following formula: \begin{equation}\label{f} u^{(n)}(z) = r^{n}\sum\limits_{k=0}^{n-1} {n \bangle k} (u-a)^{k+1}(u-b)^{n-k} \end{equation} where $n=2,3,\ldots $. \end{theorem} \begin{proof} By (\ref{R}) we get \[ u''(t) = r[(u-a)+(u-b)]u'(z)=r^{2}[(u-a)(u-b)^{2}+(u-a)^{2}(u-b)], \] which establishes (\ref{f}) for $n=2$. Let us assume that for an integer $n\ge 2$ formula (\ref{f}) holds. Using recurrence formula (\ref{eq1}) in the last step of the following calculation we get {\setlength\arraycolsep{2pt} \begin{eqnarray} && u^{(n+1)}(z)= r^{n}\frac{d}{dz}\sum\limits_{k=0}^{n-1} {n \bangle k} (u-a)^{k+1}(u-b)^{n-k}\nonumber \\ &&= r^{n+1}\sum\limits_{k=0}^{n-1} {n \bangle k} \left[(k\!+\!1)(u\!-\!a)^{k+1}(u\!-\!b)^{n-k+1}\!+\!(n\!-\!k)(u\!-\!a)^{k+2}(u\!-\!b)^{n-k}\right]\nonumber \\ && =r^{n+1}\left[ {n \bangle 0} (u\!-\!a)(u\!-\!b)^{n+1} +\sum\limits_{k=1}^{n-1}\left((k\!+\!1) {n \bangle k} +(n\!-\!k\!+\!1) {n \bangle k\!-\!1} \right)\right.\nonumber \\ &&\hspace{11mm}\left. \times (u\!-\!a)^{k+1}(u\!-\!b)^{n-k+1} + {n \bangle n\!-\!1} (u\!-\!a)^{n+1}(u\!-\!b)\right] \nonumber \\ && =r^{n+1}\sum\limits_{k=0}^{n} {n\!+\!1 \bangle k} (u-a)^{k+1}(u-b)^{n-k+1},\nonumber \end{eqnarray}} which ends the proof. \end{proof} The following two theorems are connected with solutions of equations (\ref{R2}) and (\ref{R2bis}) respectively and are due to Franssens \cite{F2}. Theorem \ref{th2} is a particular case of Theorem \ref{th3}. We write them down here in a slightly different form than in \cite{F2}. Franssens proved the theorems by using generating functions but they can be proved also by induction, similarly as Theorem \ref{th1}. \begin{theorem}\label{th2} If functions $u=u(z)$ and $v=v(z)$ are any solutions of the equations (\ref{R}), (\ref{R2}) respectively, then the $n$th derivative of $v(z)$ is equal: \begin{equation}\label{g} v^{(n)}(z)=v\frac{r^{n}}{2^{n}}\sum\limits_{k=1}^{n+1}M_{n+1,k}(u-a)^{n+1-k}(u-b)^{k-1}. \end{equation} \end{theorem} Denote by $Q_{n}(u; a, b),\;n=0,1,2,\ldots $ the polynomial (of order $n$) standing on the right hand side of equation (\ref{g}) i.e., \[Q_{n}(u; a, b)=\sum\limits_{k=1}^{n+1}M_{n+1,k}(u-a)^{n+1-k}(u-b)^{k-1}. \] \begin{theorem}\label{th3} If functions $u=u(z)$ and $v=v(z)$ are any solutions of the equations (\ref{R}), (\ref{R2bis}) respectively, then the $n$th derivative of $v(z)$ is equal: \begin{equation}\label{gd} v^{(n)}(z)=v\frac{r^{n}}{2^{n}}\sum\limits_{k=0}^{n}{n \choose k} (2d)^{k}Q_{n-k}(u; a, b). \end{equation} \end{theorem} The polynomial $Q_{n}(u; a, b)$ is related to MacMahon polynomial (\ref{MMp}) by the formula \[M_{n}(x)=\frac{Q_{n}(u; a, b)}{(u-b)^{n}}\left|_{\frac{u-a}{u-b}=x}\right. . \] Similarly we denote by $P_{n+1}(u;a,b),\; n=1,2,\ldots$ the polynomial (of order $n+1$) standing on the right hand side of equation (\ref{f}). Thus \[P_{n+1}(u;a,b)=\sum\limits_{k=0}^{n-1} {n \bangle k} (u-a)^{k+1}(u-b)^{n-k},\; n=1,2,\ldots \quad P_{1}(u)=u-a. \] Obviously the polynomial $P_{n+1}(u;a,b)$ can be rearranged into the Eulerian polynomial $E_{n}(x),\;\;n=1,2,\ldots$ using the following formula: \begin{equation} E_{n}(x)=\frac{P_{n+1}(u;a,b)}{(u-b)^{n+1}}\left|_{\frac{u-a}{u-b}=x}\right. \end{equation} Polynomials $(P_{n}(u;a,b))$ and $(Q_{n}(u;a,b))$ are called the derivative polynomials. They have been introduced by Hoffman \cite{Ho} who used them to calculate some integrals with parameters and for summing some series, without giving any explicit formula for the coefficients. The polynomials were recently intensively studied; see e.g., \cite{Bo, DC, F, Rz1, Rz2}. \section{Generating functions for Eulerian polynomials} It is easy to find the closed form of the following exponential generating function; see \cite{Ho, F2}: \begin{equation}\label{gf} F(u,t)= u + rP_{2}(u;a,b)t+r^{2}P_{3}(u;a,b)\frac{t^2}{2!}+\cdots. \end{equation} For convenience of the reader we give the calculation for (\ref{gf}). Let $u=u(z)$ be a solution of the equation (\ref{R}). By the Taylor formula for the function $u=u(z)$ we have {\setlength\arraycolsep{2pt} \begin{eqnarray*} F(u(z),t)&=& u(z) + rP_{2}(u(z);a,b)t+r^{2}P_{3}(u(z);a,b)\frac{t^2}{2!}+\cdots \\ &=& u(z)+u'(z)t+u''(z)\frac{t^2}{2!}+\cdots =u(z+t), \end{eqnarray*}} and \begin{equation}\label{gr1} F(u,t)= u(z(u)+t). \end{equation} For example if $a=0,\; b=1,\; u=1/(1+\exp(z)),\; \exp(z)=(1-u)/u$ we get \begin{equation}\label{gr2} F(u,t)= \frac{1}{1+e^{ z(u)+t}}=\frac{1}{1+\frac{1-u}{u}\:e^{t}}=\frac{u}{u+(1-u)e^{t}}. \end{equation} The generating function (\ref{gr2}) can be used for calculation of the exponential generating function for the Eulerian polynomials (\ref{Ep}) \begin{equation}\label{gr3} E_{0}(x) +E_{1}(x)y + E_{2}(x)\frac{y^2}{2!}+E_{3}(x)\frac{y^3}{3!}+\cdots . \end{equation} In order to do it let us observe, that we obtain the generating function (\ref{gr3}) by substituting in the expression \[\frac{F(u,t)-u}{u-1}+1, \] where $F(u,t)$ is given by the formula (\ref{gr2}), $u/(u-1)=x$ and $(u-1)t=y$ (that is $1/(u-1)=x-1$, $x/(x-1)=u$, $t=(x-1)y$). We compute {\setlength\arraycolsep{2pt} \begin{eqnarray*} &&\frac{F(u,t)-u}{u-1}+1=\frac{F(u,t)-1}{u-1}=\frac{\frac{u}{u+(1-u)e^{t}}-1}{u-1}=\frac{e^{t}}{u+(1-u)e^{t}}\\ &&=\frac{e^{(x-1)y}}{\frac{x}{x-1}-\frac{1}{x-1}e^{(x-1)y}}=\frac{1-x}{1-xe^{(1-x)y}}. \end{eqnarray*}} Therefore the generating function for the Eulerian polynomials $(E_{n}(x))$ is \begin{equation}\label{gr4} E_{0}(x) +E_{1}(x)y + E_{2}(x)\frac{y^2}{2!}+E_{3}(x)\frac{y^3}{3!}+\cdots =\frac{1-x}{1-xe^{(1-x)y}}. \end{equation} Formula (\ref{gr4}) gives immediately the generating function for the Eulerian polynomials $(A_{n}(x))$ defined by (\ref{Ep2}). We have {\setlength\arraycolsep{2pt} \begin{eqnarray} &&A_{0}(x) +A_{1}(x)y + A_{2}(x)\frac{y^2}{2!}+A_{3}(x)\frac{y^3}{3!}+\cdots \nonumber\\ && =\left(\frac{1-x}{1-xe^{(1-x)y}}-1 \right)\frac{1}{x} +1=\frac{x-1}{x-e^{(x-1)y}} \label{gr5} \end{eqnarray}} Foata \cite{Fo} notices that formula (\ref{gr5}) was known to Euler. \section{Some other classical formulas concerning Eulerian polynomials} The approach to Eulerian numbers and polynomials presented here is useful in obtaining other known results. For example the following classical formula concerning Eulerian polynomials \begin{equation}\label{c1} E_{n}(x)=\sum\limits_{k=1}^{n-1}{n \choose k} E_{k}(x)(x-1)^{n-1-k}+E_{1}(x)(x-1)^{n-1} \end{equation} $n=1,2,\ldots $ or equivalently expressed in terms of the polynomials$(A_{n}(x))$: \begin{equation}\label{c2} A_{n}(x)=\sum\limits_{k=0}^{n-1}{n \choose k} A_{k}(x)(x-1)^{n-1-k} \end{equation} is an easy consequence of the following lemma. \begin{lemma} If a function $u=f(z)$ fulfills the equation $u'=u(u-1)$ then for any $n=1,2,\ldots$ \begin{equation}\label{fn} f^{(n)}(z)=(f(z)-1)\sum\limits_{k=0}^{n-1}{n \choose k}f^{(k)}(z). \end{equation} \end{lemma} \begin{proof} The proof is by induction with respect to $n$. For $n=1$ formula (\ref{fn}) is obviously true and let us suppose that it holds for a positive integer $n$. We have {\setlength\arraycolsep{2pt} \begin{eqnarray} &&\hspace{-7mm}f^{(n+1)}(z)=f'(z)\sum\limits_{k=0}^{n-1}{n \choose k}f^{(k)}(z)+(f(z)-1)\sum\limits_{k=0}^{n-1}{n \choose k}f^{(k+1)}(z)\nonumber\\ &&=(f(z)\!-\!1)f(z)\sum\limits_{k=0}^{n-1}{n \choose k}f^{(k)}(z)+(f(z)\!-\!1)\sum\limits_{k=1}^{n}{n \choose k\!-\!1}f^{(k)}(z).\label{p1} \end{eqnarray}} By rearranging (\ref{fn}) to the form \[f(z)\sum\limits_{k=0}^{n-1}{n \choose k}f^{(k)}(z)=\sum\limits_{k=0}^{n}{n \choose k}f^{(k)}(z) \] and using it to the first sum of (\ref{p1}) we get {\setlength\arraycolsep{2pt} \begin{eqnarray*} f^{(n+1)}(z)&=&(f(z)-1)\sum\limits_{k=0}^{n}{n \choose k}f^{(k)}(z)+(f(z)-1)\sum\limits_{k=1}^{n}{n \choose k\!-\!1}f^{(k)}(z)\\ &=&(f(z)-1)\left(\sum\limits_{k=1}^{n}\left({n \choose k} + {n \choose k\!-\!1}\right)f^{(k)}(z)+f(z)\right)\\ &=&(f(z)-1)\sum\limits_{k=0}^{n}{n\!+\!1 \choose k}f^{(k)}(z), \end{eqnarray*}} and formula (\ref{fn}) is proved. \end{proof} Using Theorem \ref{th1} we see that formula (\ref{fn}) is equivalent to \[\sum\limits_{j=0}^{n-1} {n \bangle j}u^{j+1}(u-1)^{n-j}= (u-1)\left(\sum\limits_{k=1}^{n-1} {n\choose k}\sum\limits_{j=0}^{k-1} {k \bangle j}u^{j+1}(u-1)^{k-j}+u\right). \] By substituting here $u/(u-1)=x$, $u=x/(x-1)$, $u-1=1/(x-1)$ we get \[\sum\limits_{j=0}^{n-1} {n \bangle j}\frac{x^{j+1}}{(x-1)^{n+1}}= \frac{1}{x-1}\left(\sum\limits_{k=1}^{n-1} {n\choose k}\sum\limits_{j=0}^{k-1} {k \bangle j}\frac{x^{j+1}}{(x-1)^{k+1}}+\frac{x}{x-1}\right), \] hence we obtain the formula \[\frac{1}{ (x-1)^{n+1}}E_{n}(x)=\frac{1}{x-1}\left(\sum\limits_{k=1}^{n-1} {n\choose k}\frac{1}{(x-1)^{k+1}}E_{k}(x)+\frac{1}{x-1}E_{1}(x)\right), \] and the formulas (\ref{c1}) and (\ref{c2}) are proved. \section{Generating functions for the MacMahon polynomials}\label{s4} It is useful to get the generating function for the polynomials $(Q_{n}(u))$ as \[G(u,t)=Q_0(u;a,b)+\frac{r}{2}Q_{1}(u;a,b)t+\frac{r^{2}}{2^{2}}Q_{2}(u;a,b)\frac{t^{2}}{2!}+\frac{r^{3}}{2^{3}}Q_{3}(u;a,b)\frac{t^{3}}{3!}+\cdots \] Let the functions $u=u(z)$, $v=v(z)$ fulfill respectively equations (\ref{R}) and (\ref{R2}). Then using (\ref{g}) we have {\setlength\arraycolsep{2pt} \begin{eqnarray*} v(z)G(u(z),t) &=& v(z)+v(z)\frac{r}{2}Q_{1}(u(z);a,b)t+v(z)\frac{r^{2}}{2^{2}}Q_{2}(v(z);a,b)\frac{t^{2}}{2!}+\cdots\\ & =& v(z)+v'(z)t+v''(z)\frac{t^{2}}{2!}+\cdots = g(z+t). \end{eqnarray*}} For example if $u(z)=1/(1+e^{ z})$ and $v(z)= e^{z/2}/(1+e^{ z})$ we get \[\frac{e^{z/2}}{1+e^{z}}G(u(z), t)=\frac{e^{z/2}e^{t/2}}{1+e^{z}e^{t}}, \] hence \[G(u(z), t)=\frac{(1+e^{z})e^{t/2}}{1+e^{z}e^{t}}. \] Therefore in this case \begin{equation}\label{G} G(u, t)=\frac{(1+\frac{1-u}{u})e^{t/2}}{1+\frac{1-u}{u}e^{t}}=\frac{e^{t/2}}{u+(1-u)e^{t}}. \end{equation} The generating function (\ref{G}) can be used for calculation of the exponential generating function for the MacMahon polynomials (\ref{MMp}) \begin{equation}\label{G2} M_{0}(x) +\frac{1}{2}M_{1}(x)y + \frac{1}{2^{2}}M_{2}(x)\frac{y^2}{2!}+ \frac{1}{2^{3}}M_{3}(x)\frac{y^3}{3!}+\cdots . \end{equation} In order to do it let us observe, that we obtain the generating function (\ref{G2}) by substituting into $G(u,t)$ given by (\ref{G}), $u/(u-1)=x$ and $(u-1)t=y$ (that is $1/(u-1)=x-1$, $x/(x-1)=u$, $t=(x-1)y$). We compute \[\frac{e^{t/2}}{u+(1\!-\!u)e^{t}}=\frac{e^{(x-1)y/2}}{\frac{x}{x-1}+(1\!-\!\frac{x}{x-1})e^{(x-1)y}}= \frac{(x\!-\!1)e^{(x-1)y/2}}{x\!-\!e^{(x-1)y/2}}=\frac{(1\!-\!x)e^{(1-x)y/2}}{1\!-\!xe^{(1-x)y/2}}. \] Thus \[M_{0}(x) +\frac{1}{2}M_{1}(x)y + \frac{1}{2^{2}}M_{2}(x)\frac{y^2}{2!}+ \frac{1}{2^{3}}M_{3}(x)\frac{y^3}{3!}+\cdots =\frac{(1\!-\!x)e^{(1-x)y/2}}{1\!-\!xe^{(1-x)y/2}}, \] and \[M_{0}(x) +M_{1}(x)y + M_{2}(x)\frac{y^2}{2!}+ M_{3}(x)\frac{y^3}{3!}+\cdots =\frac{(1\!-\!x)e^{(1-x)y}}{1\!-\!xe^{(1-x)y}}. \] \section{Integral representations} In this section we use the Bernoulli numbers and Bernoulli polynomials. For their definition and a good introduction we recommend the book by Duren \cite[Ch.\ 11]{D}. We have proved \cite{Rz2} that for $n=1,2,3,\ldots$ \begin{equation}\label{i1} \int_{a}^{b}P_{n}(u;a,b)du=-(b-a)^{n+1}B_{n}, \end{equation} where $B_{n}$ is the $n$th Bernoulli number. Since $ P_{n}(u;a,b)$ is a polynomial i.e., an entire function, the (\ref{i1}) can be seen as integral over any curve (piecewise smooth), joining points $a$ and $b$. Formula (\ref{i1}) is important because it gives immediately the following Grosset--Veselov formula; see Grosset--Veselov \cite{GV} \begin{equation}\label{i2} B_{2m}=\frac{(-1)^{m-1}}{2^{2m+1}}\int_{-\infty}^{+\infty} \left(\frac{d^{m-1}}{dx^{m-1}} \frac{1}{\cosh^{2}x}\right)^{2}dx, \end{equation} which connects one--soliton solution of the KdV equation with Bernoulli numbers. Fairlie and Veselov \cite{FV} proved, by using the conservation laws, that KdV equation is directly related to the Faulhaber polynomials and the Bernoulli polynomials. The Faulhaber polynomials are well described by Knuth \cite{Kn}. Grosset and Veselov \cite{GV} demonstrated the formula (\ref{i2}) in two ways, using the cited results and then adapting an idea due to Logan described in the book \cite[Ch.\ 6.5]{GKP}. Boyadzhiev \cite{Bo3} gave an alternative proof of (\ref{i2}), based on the Fourier transform. He noted that this proof was independently suggested by Professor A. Staruszkiewicz (see also 'Note added in Proofs' at the end of \cite{GV}). In order to prove (\ref{i2}) let us observe that one of the solutions of the equation (\ref{R}), for $a=-1, b=1, r=-1$, is $u=\tanh z$. Since the image of the real line, under this function, is the interval $(-1,1)$ we have by (\ref{i1}) \begin{equation}\label{i3} \int_{-\infty}^{\infty}\frac{(\tanh z)^{(n-1)}}{\cosh^{2}z}\;dz=(-1)^{n-1}\int_{-1}^{1}P_{n}(u;-1,1)du=(-1)^{n}2^{n+1}B_{n}. \end{equation} Taking in (\ref{i3}) $n=2m$ and using, $(m-1)$-times, the formula of integration by parts for the leftmost integral, we get the Grosset--Veselov formula (\ref{i2}). There arises a natural question about similar calculation for other polynomials e.g., for $Q_{n}(u,a,b)$. Using formula (\ref{G}) we see that the generating function for polynomials $Q_{n}(u,a,b)$ in the case of $a=0, b=1, r=1$ is as follows: \[G(u,t)=Q_0(u;0,1))+\frac{1}{2}Q_{1}(u;0,1)t+\frac{1}{2^{2}}Q_{2}(u;0,1)\frac{t^{2}}{2!}+\cdots = \frac{e^{t/2}}{u+(1-u)e^{t}}, \] and therefore \begin{equation}\label{i4} \int_{0}^{1}G(u,t)du =e^{t/2}\int_{0}^{1}\frac{1}{u+(1-u)e^{t}}du=\frac{te^{t/2}}{e^{t}-1}. \end{equation} However since the generating function for the Bernoulli polynomials is \begin{equation}\label{gB} B_{0}(w)+B_{1}(w)t+B_{2}(w)\frac{t^{2}}{2!}+B_{3}(w)\frac{t^{3}}{3!}+\cdots =\frac{te^{wt}}{e^{t}-1}, \end{equation} then from (\ref{i4}) we get the following theorem. \begin{theorem} For $n=0,1,2,\ldots$ \begin{equation}\label{i5} \int_{0}^{1}Q_{n}(u;0,1)du=2^{n}B_{n}\left(\frac{1}{2}\right). \end{equation} \end{theorem} Since the polynomial $Q_{n}(u;a,b)$ is homogenous then by a suitable linear change of the variable in the integral (\ref{i5}) we get immediately \begin{equation}\label{i6} \int_{a}^{b}Q_{n}(u;a,b)du=2^{n}B_{n}\left(\frac{1}{2}\right)(b-a)^{n+1}. \end{equation} Let us recall that $B_{n}=B_{n}(0)$. Then in view of (\ref{i1}) and (\ref{i6}) the next natural question arises, which concerns the existence of a family of polynomials 'connecting' polynomials $P_{n}(u;a,b)$ and $Q_{n}(u;a,b)$ in the sense that the corresponding integrals would give the values of the Bernoulli polynomial at intermediate points between $0$ and $\frac{1}{2}$. Denote by $S_{n}(u; a,b,d),\;n=0,1,2,\ldots $ the polynomial (of order $n$) standing on the right hand side of the equation (\ref{gd}) i.e., \begin{equation}\label{gd2} S_{n}(u; a,b,d)=\sum\limits_{k=0}^{n}{n \choose k} (2d)^{k}Q_{n-k}(u; a, b). \end{equation} We will prove that $\{S_{n}(u; a,b,d)\}$ form the requested family of polynomials. A closed form formula for the following exponential generating function: \begin{equation}\label{gH} H(u,t)= S_{0}(u; a,b,d) + \frac{r}{2}S_{1}(u;a,b,d)t+\frac{r^{2}}{2^{2}}S_{2}(u;a,b,d)\frac{t^{2}}{2!}+\cdots, \end{equation} can be found similarly as in the previous cases. We assume that functions $u=u(z)$ and $v=v(z)$ are solutions of the equations (\ref{R}) and (\ref{R2bis}) respectively. Using the Taylor formula for the function $v=v(z)$ we have {\setlength\arraycolsep{2pt} \begin{eqnarray} v(z)H(u(z),t)&=& v(z)S_{0}(u(z); a,b,d) + v(z)\frac{r}{2}S_{1}(u(z);a,b,d)t \nonumber \\ &&+ v(z)\frac{r^{2}}{2^{2}}S_{2}(u(z);a,b,d)\frac{t^{2}}{2!}\cdots \nonumber \\ &=& v(z)+v'(z)t+v''(z)\frac{t^{2}}{2!}+\cdots =v(z+t).\label{g3} \end{eqnarray}} For example taking here $a=0, b=1, r=1$ and $\displaystyle u(z)=\frac{1}{1+e^{z}}$ the second equation (\ref{R2bis}) has the form \[v'(z)=v(z)\left(\frac{1}{1+e^{z}}-\frac{1}{2}+d\right), \] with a solution \begin{equation}\label{s2} v(z)=\frac{e^{(1/2+d)z}}{1+e^{z}}. \end{equation} Using (\ref{g3}) and (\ref{s2}) we get \[H(u(z),t)=\frac{v(z+t)}{v(z)}=\frac{(1+e^{z})e^{(1/2+d)t}}{1+e^{z}e^{t}} \] and putting here $\displaystyle u(z)=\frac{1}{1+e^{z}}$, $\displaystyle e^{z}=\frac{1-u}{u}$ we arrive at \begin{equation}\label{g4} H(u,t)=\frac{(1+\frac{1-u}{u})e^{(1/2+d)t}}{1+\frac{1-u}{u}e^{t}}=\frac{e^{(1/2+d)t}}{u+(1-u)e^{t}}. \end{equation} Then (\ref{g4}) yields \[\int_{0}^{1}H(u,t)du=\int_{0}^{1}\frac{e^{(1/2+d)t}}{u+(1-u)e^{t}}du=\frac{te^{(1/2+d)t}}{e^t-1}, \] and therefore using (\ref{gB}) we arrive at the formula ($n=0,1,2,\ldots$) \begin{equation}\label{i7} \int_{0}^{1}S_{n}(u;0,1,d)du=2^{n}B_{n}\left(\frac{1}{2}+d\right). \end{equation} In order to generalize (\ref{i7}) to the polynomial $S_{n}(u;a,b,d)$ we use formula (\ref{gd2}). Therefore by (\ref{i6}) we obtain {\setlength\arraycolsep{2pt} \begin{eqnarray*} &&\int_{a}^{b}S_{n}(u;a,b,d)du=\sum\limits_{k=0}^{n}{n \choose k} (2d)^{k}\int_{a}^{b}Q_{n-k}(u; a, b)du \\ &&= \sum\limits_{k=0}^{n}{n \choose k} (2d)^{k} 2^{n-k}B_{n-k}\left(\frac{1}{2}\right)(b-a)^{n-k+1} \\ &&=2^{n}(b-a)^{n+1} \sum\limits_{k=0}^{n}{n \choose k} \left(\frac{d}{b-a}\right)^{k}B_{n-k}\left(\frac{1}{2}\right)\\ &&=2^{n}(b-a)^{n+1} B_{n}\left(\frac{1}{2}+\frac{d}{b-a}\right). \end{eqnarray*}} At the very end of the above calculation we have used the following addition formula for the Bernoulli polynomials; see e.g., Temme \cite[p.\ 4]{T} \[B_{n}(x+y)=\sum\limits_{k=0}^{n}{n \choose k}B_{k}(x)y^{n-k}. \] Thus we have proved \begin{theorem} For $n=1,2,\ldots $ \[\int_{a}^{b}S_{n}(u;a,b,d)du=2^{n}(b-a)^{n+1} B_{n}\left(\frac{1}{2}+\frac{d}{b-a}\right). \] \end{theorem} Comparing the generating functions (with parameters $a=0,\: b=1,\: r=1,\: d=-1/2$): $F(u,t)$ (given by formula (\ref{gr2})) with $H(u,t)$ (formula (\ref{g4})) of the polynomials $\{P_{n}(u;0,1)\}$ and $\{S_{n}(u;0,1,-1/2)\}$ respectively we get also \begin{equation}\label{i8} \frac{P_{n+1}(u;0,1)}{u}=\frac{1}{2^{n}}S_{n}(u;0,1,-1/2). \end{equation} In particular, it follows from (\ref{i8}) that the coefficients of the polynomial $\displaystyle \frac{1}{2^{n}}S_{n}(u;0,1,-1/2)$ are all integer. \begin{thebibliography}{10} \bibitem{Bo} K. 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Temme, \emph{Special Functions}, John Wiley, 1996. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B73; Secondary 11B68. \noindent \emph{Keywords: } Eulerian number, Eulerian polynomial, MacMahon number, MacMahon polynomial, derivative polynomial, Bernoulli number, Bernoulli polynomial, integral representation. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A008292} and \seqnum{A060187}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 16 2015; revised versions received July 16 2015; July 19 2015. Published in {\it Journal of Integer Sequences}, July 29 2015. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}