关于高阶富比尼多项式的一些新的恒等式

Some New Identities on Fubini Polynomials of Higher Order

摘要: 探讨高阶富比尼多项式的性质。利用生成函数、建立偏微分方程以及复分析等方法和技巧，给出关于高阶富比尼多项式的一些新的递推关系、封闭计算公式及恒等式。特别是包括：高阶富比尼多项式与贝努利数的关系式，它满足的微分–差分方程以及高阶富比尼多项式的一种无穷项和的计算公式。作为新的恒等式的应用，研究了高阶富比尼多项式的同余性，特别是将学者Diagana等深入使用p-adic变换和p-adic积分理论而得到的关于富比尼数的恒等式及同余性结果，推广到更一般的富比尼多项式情形，并提出一个微分算子的计算问题。

Abstract: The properties of Fubini polynomials of higher order are investigated. Using generating functions, partial-differential equations, complex analysis and related techniques, we establish several new recurrence relations, closed-form expressions and identities for these polynomials. In particular, we derive explicit connections between Fubini polynomials and Bernoulli numbers of higher order, a differential-difference equation satisfied by Fubini polynomials of higher order, and an infinite-series evaluation formula. As applications of the new identities we study congruences for Fubini polynomials of higher order, extending the identities and congruences for Fubini numbers obtained by Diagana et al. via p-adic Laplace transforms and p-adic integration to the more general setting of Fubini polynomials. Finally, we propose a computational problem involving a specific differential operator.

1. 引言

富比尼数和多项式分别是一类特殊的重要的数和多项式，它们在数学，特别是组合数学、数论及概率论中有很多重要的应用，如在组合数计算、无穷项和计算、幂和及zeta函数值的计算中。至今一直有文章讨论它们的性质、各种推广及其各种应用[1]-[8]。

高阶富比尼多项式$F_n^{\left(r\right)}\left(y\right)$ 由下列函数定义[3]：

$f=f\left(t,y,r\right)=:{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r={\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$ (1)

特别地，当$r=1$ 时，$F_n^{\left(1\right)}\left(y\right)=:F_n\left(y\right)$ 称为富比尼多项式；当$y=1$ 时，$F_n^{\left(r\right)}\left(1\right)=:f_n^{\left(r\right)}$ 称为高阶富比尼数；当$r=1$ ，$y=1$ 时，$F_n^{\left(1\right)}\left(1\right)=:f_n$ 称为富比尼数。富比尼多项式又称为几何多项式。高阶富比尼多项式具有显式表达式[3]

$F_n^{\left(r\right)}\left(y\right)={\displaystyle \sum _{k=0}^n\left\{\begin{array}{c}n\\ k\end{array}\right\}}\cdot {\left(r\right)}_{\overline{k}}\cdot y^k$ (2)

其中$\left\{\begin{array}{c}n\\ k\end{array}\right\}$ 为第二类Stirling数，${\left(r\right)}_{\overline{k}}=:r\cdot \left(r+1\right)\cdots \left(r+k-1\right)$ 。

Ahlbach等[1]给出了高阶富比尼数$f_n^{\left(r\right)}$ 的组合解释，即“带隔板的偏好排列总数”的组合数，并利用组合方法证明高阶富比尼数与富比尼数之间满足关系式

$f_n^{\left(r+1\right)}=\frac{1}{2^r\cdot r!}{\displaystyle \sum _{k=0}^r\left[\begin{array}{c}r+1\\ k+1\end{array}\right]\cdot }{f}_{n+k}$ (3)

Boyadzhiev等[7] 利用指数多项式给出高阶富比尼多项式的一种积分表示，由指数多项式的性质，研究了高阶富比尼多项式的性质，并将之应用于涉及黎曼zeta函数值的级数的研究。

Diagana等[8]深入使用p-adic拉普拉斯变换和p-adic积分，研究并得到高阶富比尼数的新的恒等式及高阶富比尼数的同余性结果。

Kargin [9]讨论了富比尼多项式两项乘积的和及积分的一些公式，给出富比尼多项式新的显式公式，并给出它们的一些应用。Kargin [10]利用富比尼多项式给出p-Bernoulli数的一种积分表示，并用于研究p-Bernoulli数性质，以及讨论了p-Bernoulli数与Hurwitz-Lerch函数的关系。

本文探讨高阶富比尼多项式的性质。利用生成函数、建立偏微分方程以及复分析中柯西积分定理与柯西高阶导数公式，残数定理与残数计算等方法和技巧，给出关于高阶富比尼多项式的一些新的递推关系、封闭计算公式及恒等式。特别是包括：高阶富比尼多项式$F_n^{\left(r\right)}\left(y\right)$ 与贝努利数$B_n^{\left(r\right)}$ 的关系式，它满足的微分–差分方程

$y\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)+r\cdot y\cdot F_n^{\left(r\right)}\left(y\right)=F_{n+1}^{\left(r\right)}\left(y\right)$ (4)

以及高阶富比尼多项式的一种无穷项和的计算公式

${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{F_{kn}^{\left(r\right)}\left(y\right)}{\left(kn\right)!}}=\frac{1}{k}{\displaystyle \sum _{i=0}^{k-1}f\left({\omega }_i\right)}$ (5)

作为新的恒等式的应用，研究了高阶富比尼多项式的同余恒等式，特别是将学者Diagana等深入使用p-adic拉普拉斯变换和p-adic积分理论而得到的关于富比尼数的恒等式及同余性结果，推广到更一般的富比尼多项式情形，即有恒等式

$\left[y^q-{\left(1+y\right)}^q\right]F_n\left(y\right)={\left(1+y\right)}^q\cdot {\displaystyle \sum _{k=0}^{n-1}{F}_k\left(y\right)\cdot {\left(-q\right)}^{n-k}\cdot \left(\begin{array}{l}n\\ k\end{array}\right)}+{\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^q{\left(1+y\right)}^{j-1}\cdot y^{q-j}\cdot j^n}$ (6)

以及当$p$ 不能整除$\left(1+y\right)$ ，且$r$ 满足$1\le r\le p-1$ 时，有同余恒等式

${\left(1+y\right)}^{r-1}\cdot F_{p-1}^{\left(r\right)}\left(y\right)\equiv {\left(1+y\right)}^{r-1}-1\left(\mathrm{mod}p\right)$ (7)

从而包含Diagana等的结果作为特殊的$y=1$ 时的情形。并且由导出的递推公式以及著名的Mellin导数公式启发，提出一个微分算子的计算问题：

问题：给出微分算子

$T^m={\left[y\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}+r\cdot y\right]}^m$ (8)

的显式计算公式。

2. 高阶富比尼多项式的一些新的恒等式

首先，我们利用生成函数方法，给出高阶富比尼多项式与高阶贝努利数的关系式。高阶贝努利数由如下函数定义：

${\left(\frac{t}{{\text{e}}^t-1}\right)}^r={\displaystyle \sum _{n=0}^{\infty }{B}_n^{\left(r\right)}\frac{t^n}{n!}}$ (9)

定理1 高阶富比尼多项式与高阶贝努利数关系式

${\displaystyle \sum _{k=0}^r\left(\begin{array}{c}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^n{B}_l^{\left(r\right)}}\cdot F_{n-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{c}n\\ l\end{array}\right)=0$ ，当$0\le n<r$ 时 (10)

$F_n^{\left(r\right)}\left(y\right)=\frac{1}{y^r}\cdot \frac{n!}{\left(n+r\right)!}\cdot {\displaystyle \sum _{k=0}^r\left(\begin{array}{c}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^{n+r}{B}_l^{\left(r\right)}}\cdot F_{n+r-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{c}n+r\\ l\end{array}\right)$ , $n\ge 0$ (11)

证 高阶富比尼多项式的定义式(1)为：

$f=f\left(t,y,r\right)=:{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r={\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$

另一方面，由高阶贝努利数的定义式(9)

$\begin{array}{c}f={\left[\frac{y\left({\text{e}}^t-1\right)}{y\left({\text{e}}^t-1\right)}\right]}^r\cdot f={\left[\frac{1}{y\left({\text{e}}^t-1\right)}\right]}^r\cdot {\left[\frac{y\left({\text{e}}^t-1\right)-1+1}{1-y\left({\text{e}}^t-1\right)}\right]}^r\\ =\frac{1}{t^r{y}^r}{\left[\frac{t}{{\text{e}}^t-1}\right]}^r\cdot {\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}-1\right]}^r\\ =\frac{1}{t^r{y}^r}{\left[\frac{t}{{\text{e}}^t-1}\right]}^r\cdot {\displaystyle \sum _{k=0}^r\left(\begin{array}{l}r\\ k\end{array}\right)}\cdot {\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^k{\left(-1\right)}^{r-k}\\ =\frac{1}{t^r{y}^r}{\displaystyle \sum _{k=0}^r\left(\begin{array}{l}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}\cdot {\left[\frac{t}{{\text{e}}^t-1}\right]}^r\cdot {\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^k\\ =\frac{1}{t^r{y}^r}{\displaystyle \sum _{k=0}^r\left(\begin{array}{l}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}\cdot {\displaystyle \sum _{n=0}^{\infty }{B}_n^{\left(r\right)}\cdot \frac{t^n}{n!}}\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(k\right)}\left(y\right)\frac{t^n}{n!}}\\ =\frac{1}{t^r{y}^r}\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{k=0}^r\left(\begin{array}{l}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^n{B}_l^{\left(r\right)}}\cdot F_{n-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{l}n\\ l\end{array}\right)\right]\frac{t^n}{n!}}\end{array}$

当$0\le n<r$ 时，因$t=0$ 不是极点，故有恒等式

${\displaystyle \sum _{k=0}^r\left(\begin{array}{l}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^n{B}_l^{\left(r\right)}}\cdot F_{n-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{l}n\\ l\end{array}\right)=0$

故$f$ 可写为

$\begin{array}{c}f=\frac{1}{t^r{y}^r}\cdot {\displaystyle \sum _{n=r}^{\infty }\left[{\displaystyle \sum _{k=0}^r\left(\begin{array}{c}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^n{B}_l^{\left(r\right)}}\cdot F_{n-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{c}n\\ l\end{array}\right)\right]\frac{t^n}{n!}}\\ =\frac{1}{t^r{y}^r}\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{k=0}^r\left(\begin{array}{c}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^{n+r}{B}_l^{\left(r\right)}}\cdot F_{n+r-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{c}n+r\\ l\end{array}\right)\right]\frac{t^{n+r}}{\left(n+r\right)!}}\\ =\frac{1}{y^r}\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{k=0}^r\left(\begin{array}{c}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^{n+r}{B}_l^{\left(r\right)}}\cdot F_{n+r-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{c}n+r\\ l\end{array}\right)\right]\frac{t^n}{\left(n+r\right)!}}\\ =\frac{1}{y^r}\cdot {\displaystyle \sum _{n=0}^{\infty }\left[\frac{n!}{\left(n+r\right)!}\cdot {\displaystyle \sum _{k=0}^r\left(\begin{array}{l}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^{n+r}{B}_l^{\left(r\right)}}\cdot F_{n+r-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{c}n+r\\ l\end{array}\right)\right]\frac{t^n}{n!}}\end{array}$

比较两边$\frac{t^n}{n!}$ 的系数，得到

$F_n^{\left(r\right)}\left(y\right)=\frac{1}{y^r}\cdot \frac{n!}{\left(n+r\right)!}\cdot {\displaystyle \sum _{k=0}^r\left(\begin{array}{c}r\\ k\end{array}\right)}\cdot {\left(-1\right)}^{r-k}{\displaystyle \sum _{l=0}^{n+r}{B}_l^{\left(r\right)}}\cdot F_{n+r-l}^{\left(k\right)}\left(y\right)\cdot \left(\begin{array}{c}n+r\\ l\end{array}\right)$

证毕。

定理2 富比尼多项式满足下列递推关系式

$\left[y^q-{\left(1+y\right)}^q\right]F_n\left(y\right)={\left(1+y\right)}^q\cdot {\displaystyle \sum _{k=0}^{n-1}{F}_k\left(y\right)\cdot {\left(-q\right)}^{n-k}\cdot \left(\begin{array}{l}n\\ k\end{array}\right)}+{\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^q{\left(1+y\right)}^{j-1}\cdot y^{q-j}\cdot j^n}$ (6)

证 设

$\begin{array}{c}f=f\left(t,y\right)=:\frac{1}{1-y\left({\text{e}}^t-1\right)}={\displaystyle \sum _{n=0}^{\infty }{F}_n\left(y\right)\frac{t^n}{n!}}\\ =\frac{{\left(1+y\right)}^q-{\left(y{\text{e}}^t\right)}^q}{1-y\left({\text{e}}^t-1\right)}\cdot \left[-{\left(y{\text{e}}^t\right)}^{-q}\right]+\frac{{\left(1+y\right)}^q\cdot {\left(y{\text{e}}^t\right)}^{-q}}{1-y\left({\text{e}}^t-1\right)}\\ =P_1+P_2\end{array}$

其中$P_1=\frac{{\left(1+y\right)}^q-{\left(y{\text{e}}^t\right)}^q}{1-y\left({\text{e}}^t-1\right)}\cdot \left[-{\left(y{\text{e}}^t\right)}^{-q}\right]$ ，$P_2=\frac{{\left(1+y\right)}^q\cdot {\left(y{\text{e}}^t\right)}^{-q}}{1-y\left({\text{e}}^t-1\right)}$ 。

由n次方差公式，可得：

$\begin{array}{c}{P}_1=-{\displaystyle \sum _{j=1}^q{\left(1+y\right)}^{j-1}\cdot {\left(y{\text{e}}^t\right)}^{q-j}\cdot {\left(y{\text{e}}^t\right)}^{-q}}=-{\displaystyle \sum _{j=1}^q{\left(1+y\right)}^{j-1}\cdot {\left(y{\text{e}}^t\right)}^{-j}}\\ =-{\displaystyle \sum _{j=1}^q\frac{{\left(1+y\right)}^{j-1}}{y^j}\cdot {\text{e}}^{-jt}}=-{\displaystyle \sum _{j=1}^q\frac{{\left(1+y\right)}^{j-1}}{y^j}\cdot {\displaystyle \sum _{n=0}^{\infty }{\left(-j\right)}^n\frac{t^n}{n!}}}\\ =-{\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{j=1}^q\frac{{\left(1+y\right)}^{j-1}}{y^j}}\cdot {\left(-j\right)}^n\right]\frac{t^n}{n!}}\end{array}$

$\begin{array}{c}{P}_2=\frac{{\left(1+y\right)}^q\cdot {\left(y{\text{e}}^t\right)}^{-q}}{1-y\left({\text{e}}^t-1\right)}=\frac{{\left(1+y\right)}^q}{y^q}\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n\left(y\right)\frac{t^n}{n!}}\cdot {\displaystyle \sum _{n=0}^{\infty }{\left(-q\right)}^n\frac{t^n}{n!}}\\ =\frac{{\left(1+y\right)}^q}{y^q}\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{k=0}^n{F}_k\left(y\right){\left(-q\right)}^{n-k}\left(\begin{array}{l}n\\ k\end{array}\right)}\right]\frac{t^n}{n!}}\end{array}$

比较两边$\frac{t^n}{n!}$ 的系数，可得：

$\begin{array}{c}{F}_n\left(y\right)=-\left[{\displaystyle \sum _{j=1}^q\frac{{\left(1+y\right)}^{j-1}}{y^j}\cdot }{\left(-j\right)}^n\right]+\frac{{\left(1+y\right)}^q}{y^q}\cdot {\displaystyle \sum _{k=0}^n{F}_k\left(y\right){\left(-q\right)}^{n-k}\left(\begin{array}{l}n\\ k\end{array}\right)}\\ =\frac{{\left(1+y\right)}^q}{y^q}\cdot {\displaystyle \sum _{k=0}^n{F}_k\left(y\right){\left(-q\right)}^{n-k}\left(\begin{array}{l}n\\ k\end{array}\right)}+{\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^q\frac{{\left(1+y\right)}^{j-1}}{y^j}}\cdot j^n\end{array}$

$y^q\cdot F_n\left(y\right)={\left(1+y\right)}^q\cdot {\displaystyle \sum _{k=0}^n{F}_k\left(y\right)\cdot {\left(-q\right)}^{n-k}\cdot \left(\begin{array}{l}n\\ k\end{array}\right)}+{\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^q{\left(1+y\right)}^{j-1}\cdot y^{q-j}\cdot j^n}$

$\left[y^q-{\left(1+y\right)}^q\right]F_n\left(y\right)={\left(1+y\right)}^q\cdot {\displaystyle \sum _{k=0}^{n-1}{F}_k\left(y\right)\cdot {\left(-q\right)}^{n-k}\cdot \left(\begin{array}{l}n\\ k\end{array}\right)}+{\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^q{\left(1+y\right)}^{j-1}\cdot y^{q-j}\cdot j^n}$

证毕。

推论1 富比尼数满足下列递推关系式

$\left[1-2^q\right]f_n=2^q\cdot {\displaystyle \sum _{k=0}^{n-1}{f}_k\cdot {\left(-q\right)}^{n-k}\cdot \left(\begin{array}{l}n\\ k\end{array}\right)}+{\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^q{2}^{j-1}\cdot j^n}$ (12)

证 在定理2中，特别令$y=1$ ，并且由$F_n\left(1\right)=f_n$ ，则得到推论1。

注：这是文[8]中深入地利用p-adic拉普拉斯变换和p-adic积分理论而得到的结果。

定理3 高阶富比尼多项式满足下列递推关系式

$F_{n+1}^{\left(r\right)}\left(y\right)=y\cdot \left[{\displaystyle \sum _{l=1}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)\left(\frac{l}{n-l+1}+r\right)}\right]+r\cdot y\cdot F_0^{\left(r\right)}\left(y\right)$ (13)

证 高阶富比尼多项式的定义式为

$f=f\left(t,y,r\right)=:{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r={\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$ (1)

等式两边关于$t$ 求导，得到

${f^{\prime }}_t={\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}\cdot r\cdot y\cdot {\text{e}}^t$ (14)

变形得到

$\left(1+y\right)\cdot {f^{\prime }}_t-y\cdot {f^{\prime }}_t\cdot {\text{e}}^t=r\cdot y\cdot f\cdot {\text{e}}^t$

由于${\text{e}}^t={\displaystyle \sum _{n=0}^{\infty }\frac{t^n}{n!}}$ ，代入上面等式，可得：

$\left(1+y\right)\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_{n+1}^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}-y\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_{n+1}^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{t^n}{n!}}=r\cdot y\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{t^n}{n!}}$

$\left(1+y\right)\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_{n+1}^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}-y\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{l=0}^n{F}_{l+1}^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)}\right]\frac{t^n}{n!}}=r\cdot y\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{l=0}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)}\right]\frac{t^n}{n!}}$

比较两边$\frac{t^n}{n!}$ 的系数可得：

$\left(1+y\right)\cdot F_{n+1}^{\left(r\right)}\left(y\right)-y\cdot {\displaystyle \sum _{l=0}^n{F}_{l+1}^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)}=r\cdot y\cdot {\displaystyle \sum _{l=0}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)}$

可变形得到

$F_{n+1}^{\left(r\right)}\left(y\right)-y\cdot {\displaystyle \sum _{l=1}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l-1\end{array}\right)}=r\cdot y\cdot {\displaystyle \sum _{l=0}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l\end{array}\right)}$

合并同类项可得：

$F_{n+1}^{\left(r\right)}\left(y\right)=y\cdot \left\{{\displaystyle \sum _{l=1}^n{F}_l^{\left(r\right)}\left(y\right)\left[\left(\begin{array}{c}n\\ l-1\end{array}\right)+r\cdot \left(\begin{array}{c}n\\ l\end{array}\right)\right]}\right\}+r\cdot y\cdot F_0^{\left(r\right)}\left(y\right)$

由组合数计算可得：

$F_{n+1}^{\left(r\right)}\left(y\right)=y\cdot \left[{\displaystyle \sum _{l=1}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)\left(\frac{l}{n-l+1}+r\right)}\right]+r\cdot y\cdot F_0^{\left(r\right)}\left(y\right)$

证毕。

定理4 高阶富比尼多项式满足下列递推关系式

$F_{n+1}^{\left(r\right)}\left(y\right)=\left[n+\left(n+r\right)\cdot y\right]\cdot F_n^{\left(r\right)}\left(y\right)+\left(1+y\right)\cdot {\displaystyle \sum _{l=1}^{n-1}{F}_l^{\left(r\right)}\left(y\right){\left(-1\right)}^{n-l}\left(\begin{array}{c}n\\ l-1\end{array}\right)}$ (15)

证 由(14)式${f^{\prime }}_t={\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}\cdot r\cdot y\cdot {\text{e}}^t$ 变形为$\left(1+y\right){f^{\prime }}_t\cdot {\text{e}}^{-t}-y\cdot {f^{\prime }}_t=r\cdot y\cdot f$ ，由${\text{e}}^{-t}={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{t^n}{n!}}$ ，代入上面等式得到

$\left(1+y\right)\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_{n+1}^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}\cdot {\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{t^n}{n!}}-y\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_{n+1}^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}=r\cdot y\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$

从而得到

$\left(1+y\right)\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{l=0}^n{F}_{l+1}^{\left(r\right)}\left(y\right){\left(-1\right)}^{n-l}\left(\begin{array}{l}n\\ l\end{array}\right)}\right]\frac{t^n}{n!}}-y\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_{n+1}^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}=r\cdot y\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$

比较两边$\frac{t^n}{n!}$ 的系数可得：

$\left(1+y\right)\cdot {\displaystyle \sum _{l=0}^n{F}_{l+1}^{\left(r\right)}\left(y\right){\left(-1\right)}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)}-y\cdot F_{n+1}^{\left(r\right)}\left(y\right)=r\cdot y\cdot F_n^{\left(r\right)}\left(y\right)$

化简得到

$F_{n+1}^{\left(r\right)}\left(y\right)+\left(1+y\right)\cdot {\displaystyle \sum _{l=0}^{n-1}{F}_{l+1}^{\left(r\right)}\left(y\right){\left(-1\right)}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)}=r\cdot y\cdot F_n^{\left(r\right)}\left(y\right)$

变形得到

$F_{n+1}^{\left(r\right)}\left(y\right)=r\cdot y\cdot F_n^{\left(r\right)}\left(y\right)+\left(1+y\right)\cdot {\displaystyle \sum _{l=1}^n{F}_l^{\left(r\right)}\left(y\right){\left(-1\right)}^{n-l}\left(\begin{array}{c}n\\ l-1\end{array}\right)}$

合并同类项后可得：

$F_{n+1}^{\left(r\right)}\left(y\right)=\left[n+\left(n+r\right)\cdot y\right]\cdot F_n^{\left(r\right)}\left(y\right)+\left(1+y\right)\cdot {\displaystyle \sum _{l=1}^{n-1}{F}_l^{\left(r\right)}\left(y\right){\left(-1\right)}^{n-l}\left(\begin{array}{c}n\\ l-1\end{array}\right)}$

证毕。

定理5 高阶富比尼多项式满足下列递推关系式

$\frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)-y\cdot {\displaystyle \sum _{l=0}^{n-1}\frac{\text{d}}{\text{d}y}{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)}=r\cdot {\displaystyle \sum _{l=0}^{n-1}{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{l}n\\ l\end{array}\right)}$ (16)

证 由高阶富比尼多项式的定义式(1)，等式两边关于$y$ 求导，即得

${f^{\prime }}_y={\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}\cdot r\cdot \left({\text{e}}^t-1\right)$ (17)

$\left(1+y\right)\cdot {f^{\prime }}_y-y\cdot {f^{\prime }}_y\cdot {\text{e}}^t=-r\cdot f+r\cdot f\cdot {\text{e}}^t$

代入相应的幂级数展开式得到

$\left(1+y\right)\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}-y\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{t^n}{n!}}=-r\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}+r\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{t^n}{n!}}$

$\left(1+y\right)\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}-y\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{l=0}^n\frac{\text{d}}{\text{d}y}{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l\end{array}\right)}\right]\frac{t^n}{n!}}=-r\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}+r\cdot {\displaystyle \sum _{n=0}^{\infty }\left[{\displaystyle \sum _{l=0}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l\end{array}\right)}\right]\frac{t^n}{n!}}$

比较两边$\frac{t^n}{n!}$ 的系数可得：

$\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)-y\cdot {\displaystyle \sum _{l=0}^n\frac{\text{d}}{\text{d}y}{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l\end{array}\right)}=-r\cdot F_n^{\left(r\right)}\left(y\right)+r\cdot {\displaystyle \sum _{l=0}^n{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l\end{array}\right)}$

化简即为

$\frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)-y\cdot {\displaystyle \sum _{l=0}^{n-1}\frac{\text{d}}{\text{d}y}{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l\end{array}\right)}=r\cdot {\displaystyle \sum _{l=0}^{n-1}{F}_l^{\left(r\right)}\left(y\right)\left(\begin{array}{c}n\\ l\end{array}\right)}$

注：定理3给出$r$ 阶富比尼多项式的一种递推关系，定理4给出的是$r$ 阶富比尼多项式的一种含有交错和的递推关系，而定理5给出$r$ 阶富比尼多项式的含有关于$y$ 的导数的一种递推关系。

定理6 生成函数$f$ 满足下列一阶偏微分方程

$y\left(1+y\right)\cdot {f^{\prime }}_y-{f^{\prime }}_t=-r\cdot y\cdot f$ (18)

证 由前面的等式(14) (17)，可知

$\begin{array}{c}左手边={\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}\cdot \left[y\left(1+y\right)\cdot r\cdot \left({\text{e}}^t-1\right)-r\cdot y\cdot {\text{e}}^t\right]\\ ={\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}\cdot r\cdot y\left[\left(1+y\right)\cdot \left({\text{e}}^t-1\right)-{\text{e}}^t\right]\\ ={\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}\cdot r\cdot y\left[y\cdot \left({\text{e}}^t-1\right)-1\right]\\ =-r\cdot y\cdot f=右手边\end{array}$

证毕。

高阶富比尼多项式满足下列微分–差分方程：

定理7 高阶富比尼多项式满足下列递推关系式

$y\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)+r\cdot y\cdot F_n^{\left(r\right)}\left(y\right)=F_{n+1}^{\left(r\right)}\left(y\right)$ (4)

证 由定理6，定义函数$f$ 满足偏微分方程(18)，故有

$y\left(1+y\right)\cdot {\displaystyle \sum _{n=0}^{\infty }\frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}-{\displaystyle \sum _{n=0}^{\infty }{F}_{n+1}^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}=-r\cdot y\cdot {\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$

比较两边$\frac{t^n}{n!}$ 的系数可得：

$y\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)-F_{n+1}^{\left(r\right)}\left(y\right)=-r\cdot y\cdot F_n^{\left(r\right)}\left(y\right)$

移项得到

$y\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}{F}_n^{\left(r\right)}\left(y\right)+r\cdot y\cdot F_n^{\left(r\right)}\left(y\right)=F_{n+1}^{\left(r\right)}\left(y\right)$

证毕。

注：定义算子$T=:y\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}+r\cdot y$ ，显然有$F_{n+1}^{\left(r\right)}\left(y\right)=T{F}_n^{\left(r\right)}\left(y\right)$ 及$F_{n+m}^{\left(r\right)}\left(y\right)=T^m{F}_n^{\left(r\right)}\left(y\right)$ ，自然提出下面的问题。

问题：给出算子

$T^m={\left[y\left(1+y\right)\cdot \frac{\text{d}}{\text{d}y}+r\cdot y\right]}^m$ (18)

的显式计算公式。

下面利用复分析的方法和技巧，来建立高阶富比尼多项式满足的关系式。

定理8 高阶富比尼多项式满足关系式

$F_n^{\left(r\right)}\left(y\right)={\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle \sum _{m=0}^n\left\{\begin{array}{c}m\\ k\end{array}\right\}\cdot \left(\begin{array}{c}n\\ m\end{array}\right)}\cdot F_{n-m}^{\left(r+l\right)}\left(y\right)$ (19)

特别地，令$l=1$ ，得到封闭关系式

$F_n^{\left(r+1\right)}\left(y\right)-y\cdot {\displaystyle \sum _{m=1}^n\left(\begin{array}{l}n\\ m\end{array}\right)}\cdot F_{n-m}^{\left(r+1\right)}\left(y\right)=F_n^{\left(r\right)}\left(y\right)$ (20)

证 设$f=f\left(t\right)=:f\left(t,y,r\right):={\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r={\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$ ，由柯西高阶导数公式得：$F_n^{\left(r\right)}\left(y\right)=\frac{n!}{2\pi i}{\displaystyle {\oint }_C\frac{f\left(t\right)}{t^{n+1}}\text{d}t}$ ，其中C是选定的以原点为心的圆周，取逆时针方向。故

$\begin{array}{c}{F}_n^{\left(r\right)}\left(y\right)=\frac{n!}{2\pi i}{\displaystyle {\oint }_C{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle {\oint }_C{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+l}\cdot {\left[1-y\left({\text{e}}^t-1\right)\right]}^l\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle {\oint }_{C}f\left(t,y,r+l\right)\cdot {\left[1-y\left({\text{e}}^t-1\right)\right]}^l\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle {\oint }_{C}f\left(t,y,r+l\right)\cdot {\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot {\left({\text{e}}^t-1\right)}^k\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle {\oint }_{C}f\left(t,y,r+l\right)\cdot \frac{{\left({\text{e}}^t-1\right)}^k}{k!}\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle {\oint }_{C}f\left(t,y,r+l\right)\cdot {\displaystyle \sum _{m\ge k}^{}\left\{\begin{array}{c}m\\ k\end{array}\right\}\cdot \frac{t^m}{m!}}\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle {\oint }_{C}f\left(t,y,r+l\right)\cdot {\displaystyle \sum _{m=0}^n\left\{\begin{array}{c}m\\ k\end{array}\right\}\cdot \frac{t^m}{m!}}\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle \sum _{m=0}^n\left\{\begin{array}{c}m\\ k\end{array}\right\}\cdot \frac{1}{m!}}\cdot {\displaystyle {\oint }_{C}f\left(t,y,r+l\right)\cdot \frac{1}{t^{n-m+1}}\text{d}t}\\ =\frac{n!}{2\pi i}{\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle \sum _{m=0}^n\left\{\begin{array}{c}m\\ k\end{array}\right\}\cdot \frac{1}{m!}}\cdot \frac{2\pi i}{\left(n-m\right)!}\cdot F_{n-m}^{\left(r+l\right)}\left(y\right)\\ ={\displaystyle \sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle \sum _{m=0}^n\left\{\begin{array}{c}m\\ k\end{array}\right\}\cdot \left(\begin{array}{c}n\\ m\end{array}\right)}\cdot F_{n-m}^{\left(r+l\right)}\left(y\right)\end{array}$

即得到关系式

$F_n^{\left(r\right)}\left(y\right)={\displaystyle \sum _{k=0}^l\left(\begin{array}{l}l\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle \sum _{m=0}^n\left\{\begin{array}{l}m\\ k\end{array}\right\}\cdot \left(\begin{array}{l}n\\ m\end{array}\right)}\cdot F_{n-m}^{\left(r+l\right)}\left(y\right)$

特别地，令$l=1$ ，得到封闭关系式

$F_n^{\left(r\right)}\left(y\right)={\displaystyle \sum _{k=0}^1\left(\begin{array}{l}1\\ k\end{array}\right)}\cdot {\left(-y\right)}^k\cdot k!\cdot {\displaystyle \sum _{m=0}^n\left\{\begin{array}{l}m\\ k\end{array}\right\}\cdot \left(\begin{array}{l}n\\ m\end{array}\right)}\cdot F_{n-m}^{\left(r+1\right)}\left(y\right)$

利用第二类Stirling的性质[11] [12]：$\left\{\begin{array}{c}m\\ 0\end{array}\right\}=1$ ，$m=0$ ；$\left\{\begin{array}{c}m\\ 0\end{array}\right\}=0$ ，$m\ge 1$ ；$\left\{\begin{array}{c}m\\ 1\end{array}\right\}=1$ ，$m\ge 1$ 可得：

$F_n^{\left(r\right)}\left(y\right)=F_n^{\left(r+1\right)}\left(y\right)-y\cdot {\displaystyle \sum _{m=1}^n\left(\begin{array}{l}n\\ m\end{array}\right)}\cdot F_{n-m}^{\left(r+1\right)}\left(y\right)$

证毕。

定理9 高阶富比尼多项式满足关系式

$F_n^{\left(r\right)}\left(y\right)=-r\cdot F_{n-1}^{\left(r\right)}\left(y\right)+r\cdot \left(1+y\right)\cdot F_{n-1}^{\left(r+1\right)}\left(y\right)$ (21)

证 由柯西高阶导数公式及分部积分技巧可得：

$\begin{array}{c}{F}_n^{\left(r\right)}\left(y\right)=\frac{n!}{2\pi i}{\displaystyle {\oint }_C{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r\cdot \frac{1}{t^{n+1}}\text{d}t}\\ =\frac{n!}{2\pi i}\cdot \left(-\frac{1}{n}\right)\cdot {\displaystyle {\oint }_C{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r\text{d}\frac{1}{t^n}}\\ =\frac{n!}{2\pi i}\cdot \frac{1}{n}\cdot {\displaystyle {\oint }_C\frac{1}{t^n}\text{d}{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r}\\ =\frac{n!}{2\pi i}\cdot \frac{1}{n}\cdot {\displaystyle {\oint }_C\left\{{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}\cdot \left(r\cdot y\cdot {\text{e}}^t\right)\right\}\cdot \frac{1}{t^n}\text{d}t}\\ =\frac{n!}{2\pi i}\cdot \frac{1}{n}\cdot {\displaystyle {\oint }_C\left\{r\cdot \left(1+y\right)\cdot {\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}-r\cdot {\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r\right\}\cdot \frac{1}{t^n}\text{d}t}\\ =\frac{\left(n-1\right)!}{2\pi i}\cdot {\displaystyle {\oint }_C\left\{r\cdot \left(1+y\right)\cdot {\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^{r+1}-r\cdot {\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r\right\}\cdot \frac{1}{t^n}\text{d}t}\\ =r\cdot \left(1+y\right)\cdot F_{n-1}^{\left(r+1\right)}\left(y\right)-r\cdot F_{n-1}^{\left(r\right)}\left(y\right)\end{array}$

证毕。

定理10 高阶富比尼多项式满足恒等式(求和公式)

${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{F_{kn}^{\left(r\right)}\left(y\right)}{\left(kn\right)!}}=\frac{1}{k}{\displaystyle \sum _{i=0}^{k-1}f\left({\omega }_i\right)}$ (5)

当$y\in \left(-\frac{\text{e}}{\text{e}-1},\frac{1}{\text{e}-1}\right)$ ，$k\in N^{+}$ 时，其中${\omega }_j,j=0,1,\cdots ,k-1$ 是方程$t^k+1=0$ 的$k$ 个根。

证 设$f=f\left(t\right)=:f\left(t,y,r\right)=:{\left[\frac{1}{1-y\left({\text{e}}^t-1\right)}\right]}^r={\displaystyle \sum _{n=0}^{\infty }{F}_n^{\left(r\right)}\left(y\right)\frac{t^n}{n!}}$ ，$\left|t\right|<\left|\ln \frac{1+y}{y}\right|$ ，由生成函数$f$ 的表达式易知，当$y\in \left(-\frac{\text{e}}{\text{e}-1},\frac{1}{\text{e}-1}\right)$ 时存在$\rho >1$ ，使得$f$ 在$\left|t\right|\le \rho$ 上解析，由柯西高阶导数公式及残数的求解方法：$\frac{F_{kn}^{\left(r\right)}\left(y\right)}{\left(kn\right)!}=\frac{1}{2\pi i}{\displaystyle {\oint }_{C:\left|t\right|=\rho }\frac{f\left(t\right)}{t^{kn+1}}}\text{d}t$ ，故

$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{F_{kn}^{\left(r\right)}\left(y\right)}{\left(kn\right)!}}={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{1}{2\pi i}{\displaystyle {\oint }_C\frac{f\left(t\right)}{t^{kn+1}}\text{d}t}\\ =\frac{1}{2\pi i}{\displaystyle {\oint }_{C}f\left(t\right)\cdot {\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{1}{t^{kn+1}}}\text{d}t\\ =\frac{1}{2\pi i}{\displaystyle {\oint }_C\frac{f\left(t\right)\cdot t^{k-1}}{t^k+1}\text{d}t}\\ ={\displaystyle \sum _{i=0}^{k-1}{\frac{f\left(t\right)\cdot t^{k-1}}{{\left(t^k+1\right)}^{\prime }}|}_{t={\omega }_i}}={\displaystyle \sum _{i=0}^{k-1}{\frac{f\left(t\right)}{k}|}_{t={\omega }_i}}=\frac{1}{k}{\displaystyle \sum _{i=0}^{k-1}f\left({\omega }_i\right)}\end{array}$

其中：${\omega }_j={\text{e}}^{i\cdot \frac{\left(2j+1\right)\pi }{k}}$ ，$j=0,1,\cdots ,k-1$ 是方程$t^k+1=0$ 的$k$ 个根。

3. 高阶富比尼多项式的同余性研究

为了深入理解一个著名数或多项式的性质和内在规律，常常从多种角度去考虑，如从组合、分析及数论等不同的角度去研究。同余性在组合计数简化和密码学中有重要应用。

本节将应用上节得到的关于高阶富比尼多项式的一些恒等式，研究高阶富比尼多项式同余性。

定理11 设$y\in \mathbb{Z}$ ，$p$ 为奇素数，则高阶富比尼多项式取值满足同余恒等式

$F_p^{\left(r\right)}\left(y\right)\equiv r\cdot y\left(\mathrm{mod}p\right)$ (22)

证 由公式(2)：令$n=p$ ，得到

$\begin{array}{c}{F}_p^{\left(r\right)}\left(y\right)={\displaystyle \sum _{k=0}^p\left\{\begin{array}{c}p\\ k\end{array}\right\}}\cdot {\left(r\right)}_{\overline{k}}\cdot y^k\\ ={\displaystyle \sum _{k=2}^{p-1}\left\{\begin{array}{c}p\\ k\end{array}\right\}}\cdot {\left(r\right)}_{\overline{k}}\cdot y^k+\left\{\begin{array}{c}p\\ 0\end{array}\right\}+\left\{\begin{array}{c}p\\ 1\end{array}\right\}\cdot r\cdot y+\left\{\begin{array}{c}p\\ p\end{array}\right\}\cdot {\left(r\right)}_{\overline{p}}\cdot y^p\end{array}$

由第二类Sterling数的性质：当$p$ 为奇素数时，$\left\{\begin{array}{c}p\\ k\end{array}\right\}\equiv 0\left(\mathrm{mod}p\right)$ ，$k=0,2,\cdots ,p-1$ . 故得到

$F_p^{\left(r\right)}\left(y\right)=\left\{\begin{array}{c}p\\ 0\end{array}\right\}+\left\{\begin{array}{c}p\\ 1\end{array}\right\}\cdot r\cdot y+\left\{\begin{array}{c}p\\ p\end{array}\right\}\cdot {\left(r\right)}_{\overline{p}}\cdot y^p\left(\mathrm{mod}p\right)$

利用性质$\left\{\begin{array}{c}n\\ 0\end{array}\right\}=0$ ，$\left\{\begin{array}{c}n\\ 1\end{array}\right\}=1$ ，$\left\{\begin{array}{c}n\\ n\end{array}\right\}=1$ ；$n>0$ ；由${\left(r\right)}_{\overline{p}}\equiv 0\left(\mathrm{mod}p\right)$，可得$F_p^{\left(r\right)}\left(y\right)\equiv r\cdot y\left(\mathrm{mod}p\right)$ 。

定理12 设$y\in \mathbb{Z}$ ，$p$ 是奇素数，则富比尼多项式取值满足同余恒等式

$F_{p-1}\left(y\right)\equiv \{\begin{array}{l}1,p|\left(1+y\right)\\ 0,其它\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\end{array}\left(\mathrm{mod}p\right)$ (23)

证 利用公式(6)，并令$q=p$ ，可得：

$\left[y^p-{\left(1+y\right)}^p\right]F_n\left(y\right)={\left(1+y\right)}^p\cdot {\displaystyle \sum _{k=0}^{n-1}{F}_k\left(y\right)\cdot {\left(-p\right)}^{n-k}\cdot \left(\begin{array}{l}n\\ k\end{array}\right)}+{\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^p{\left(1+y\right)}^{j-1}\cdot y^{p-j}\cdot j^n}$

$\left[y^p-{\left(1+y\right)}^p\right]F_n\left(y\right)\equiv {\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^{p-1}{\left(1+y\right)}^{j-1}\cdot y^{p-j}\cdot j^n}$ $\left(\mathrm{mod}p\right)$

当$n>0$ 时。由Fermat小定理可得：$a^p\equiv a\left(\mathrm{mod}p\right)$ ，故

$\left[y-\left(1+y\right)\right]F_n\left(y\right)\equiv {\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^{p-1}{\left(1+y\right)}^{j-1}\cdot y^{p-j}\cdot j^n}$ $\left(\mathrm{mod}p\right)$

$-F_n\left(y\right)\equiv {\left(-1\right)}^{n+1}\cdot {\displaystyle \sum _{j=1}^{p-1}{\left(1+y\right)}^{j-1}\cdot y^{p-j}\cdot j^n}$ $\left(\mathrm{mod}p\right)$

令$n=p-1$ 得到：

$F_{p-1}\left(y\right)\equiv {\displaystyle \sum _{j=1}^{p-1}{\left(1+y\right)}^{j-1}\cdot y^{p-j}\cdot j^{p-1}}$ $\left(\mathrm{mod}p\right)$

由Fermat小定理知：$j^{p-1}\equiv 1$ $\left(\mathrm{mod}p\right)$ $j=1,2,\cdots ,p-1$ ，可得：

$\begin{array}{c}{F}_{p-1}\left(y\right)\equiv {\displaystyle \sum _{j=1}^{p-1}{\left(1+y\right)}^{j-1}\cdot y^{p-j}}\left(\mathrm{mod}p\right)\\ \equiv y^{p-1}{\displaystyle \sum _{j=1}^{p-1}{\left(\frac{1+y}{y}\right)}^{j-1}}\left(\mathrm{mod}p\right)\\ \equiv \left[{\left(1+y\right)}^{p-1}-y^{p-1}\right]\cdot y\left(\mathrm{mod}p\right)\\ \equiv {\left(1+y\right)}^{p-1}\cdot y-y^p\left(\mathrm{mod}p\right)\\ \equiv {\left(1+y\right)}^{p-1}\cdot y-y\left(\mathrm{mod}p\right)\\ \equiv {\left(1+y\right)}^p-{\left(1+y\right)}^{p-1}-y\left(\mathrm{mod}p\right)\\ \equiv \left(1+y\right)-{\left(1+y\right)}^{p-1}-y\left(\mathrm{mod}p\right)\\ \equiv 1-{\left(1+y\right)}^{p-1}\left(\mathrm{mod}p\right)\\ \equiv \{\begin{array}{l}1,p|\left(1+y\right)\\ 0,其它\end{array}\left(\mathrm{mod}p\right)\end{array}$

推论2 设$p$ 是奇素数，则富比尼数有同余恒等式

$f_{p-1}\equiv 0\left(\mathrm{mod}p\right)$ (24)

证 在定理12中，特别，令$y=1$ ，则有

$F_{p-1}\left(1\right)=f_{p-1}\equiv 0\left(\mathrm{mod}p\right)$

证毕。

注：这是文[8]中深入地利用p-adic拉普拉斯变换和p-adic积分理论而得到的结果。

定理13 设$y\in \mathbb{Z}$ ，$p$ 是奇素数，且$p$ 不能整除$\left(1+y\right)$ ，$r$ 是整数满足$1\le r\le p-1$ ，则高阶富比尼多项式取值满足下列同余恒等式

${\left(1+y\right)}^{r-1}\cdot F_{p-1}^{\left(r\right)}\left(y\right)\equiv {\left(1+y\right)}^{r-1}-1\left(\mathrm{mod}p\right)$ (7)

证 用数学归纳法：

1．当$r=1$ 时，$F_{p-1}^{\left(1\right)}\left(y\right)=F_{p-1}\left(y\right)\equiv 0\left(\mathrm{mod}p\right)$ ，命题成立。

2．归纳假设${\left(1+y\right)}^{r-1}\cdot F_{p-1}^{\left(r\right)}\left(y\right)\equiv {\left(1+y\right)}^{r-1}-1\left(\mathrm{mod}p\right)$ ，要证明${\left(1+y\right)}^r\cdot F_{p-1}^{\left(r+1\right)}\left(y\right)\equiv {\left(1+y\right)}^r-1\left(\mathrm{mod}p\right)$ 。

利用定理9递推公式(21)可知$r\left(1+y\right)\cdot F_{p-1}^{\left(r+1\right)}\left(y\right)=r\cdot F_{p-1}^{\left(r\right)}\left(y\right)+F_p^{\left(r\right)}\left(y\right)$ 成立，将该式两边同乘${\left(1+y\right)}^{r-1}$ ，并利用定理11可得：

$r{\left(1+y\right)}^r\cdot F_{p-1}^{\left(r+1\right)}\left(y\right)=r\cdot {\left(1+y\right)}^{r-1}\cdot F_{p-1}^{\left(r\right)}\left(y\right)+{\left(1+y\right)}^{r-1}\cdot F_p^{\left(r\right)}\left(y\right)$

$\begin{array}{c}r{\left(1+y\right)}^r\cdot F_{p-1}^{\left(r+1\right)}\left(y\right)\equiv r\cdot {\left(1+y\right)}^{r-1}\cdot F_{p-1}^{\left(r\right)}\left(y\right)+{\left(1+y\right)}^{r-1}\cdot r\cdot y\left(\mathrm{mod}p\right)\\ \equiv r\cdot \left[{\left(1+y\right)}^{r-1}-1\right]+{\left(1+y\right)}^{r-1}\cdot r\cdot y\left(\mathrm{mod}p\right)\\ \equiv r\cdot \left[{\left(1+y\right)}^r-1\right]\left(\mathrm{mod}p\right)\end{array}$

从而有${\left(1+y\right)}^r\cdot F_{p-1}^{\left(r+1\right)}\left(y\right)\equiv {\left(1+y\right)}^r-1\left(\mathrm{mod}p\right)$ ，故命题成立。

推论3 高阶富比尼数满足同余恒等式

$2^{r-1}\cdot f_{p-1}^{\left(r\right)}\equiv 2^{r-1}-1\left(\mathrm{mod}p\right)1\le r\le p-1$ (25)

证 在定理13中，特别，令$y=1$ ，则高阶富比尼数满足同余恒等式

$2^{r-1}\cdot f_{p-1}^{\left(r\right)}\equiv 2^{r-1}-1\left(\mathrm{mod}p\right)1\le r\le p-1$

证毕。

注：这是文[8]中Proposition 4.5给出的结果。

4. 总结与展望

本文探讨高阶富比尼多项式的性质。利用生成函数、建立偏微分方程以及复分析等方法和技巧，给出关于高阶富比尼多项式的一些新的递推关系、封闭计算公式及恒等式。特别是包括：高阶富比尼多项式与贝努利数的关系式，它满足的微分–差分方程以及高阶富比尼多项式的一种无穷项和的计算公式。作为新的恒等式的应用，研究了高阶富比尼多项式的同余性，特别是将学者Diagana等深入使用p-adic变换和p-adic积分理论而得到的关于富比尼数的恒等式及同余性结果，推广到更一般的富比尼多项式情形，并提出一个微分算子的计算问题。

未来的改进和值得研究的方向：利用已有高阶富比尼多项式的恒等式，进一步深入研究其同余性；研究高阶富比尼多项式与其它著名多项式和数的关系，例如与指数多项式[10] [13] [14]、弗罗贝尼乌斯–欧拉多项式的关系；研究与黎曼Zeta函数及其推广函数的联系；如何用组合方法将高阶富比尼数与富比尼数之间满足的关系式(3)推广到富比尼多项式情形；给出微分算子(8)的显式计算公式等。

基金项目

国家自然科学基金项目(12026420)；吉林省科技发展计划项目(YDZJ202201ZYTS627)。

NOTES

^*通讯作者。