关于第三类退化的Poly-Cauchy多项式的组合恒等式

Some Identities on Degenerate Poly-Cauchy Polynomials of the Third Kind

摘要: 本文利用发生函数和Riordan阵研究了第三类退化的Poly-Cauchy多项式相关的恒等式。首先，运用发生函数方法给出第三类退化的Poly-Cauchy多项式的性质，从而得到了关于第三类退化的Poly-Cauchy多项式的一些组合恒等式。其次，应用Riordan阵法，建立了第三类退化的Poly-Cauchy多项式与两类Stirling数、Lab数、Bell数之间的一些关系式。

Abstract: In this paper, using generating functions and Riordan arrays, we establish some identities involving the degenerate Poly-Cauchy polynomials of the third kind. Using the generating functions, we explore some properties of the degenerate Poly-Cauchy polynomials of the third kind, and obtain some combinatorial identities involving the degenerate Poly-Cauchy polynomials of the third kind. In addition, using Riordan arrays, we give some interesting relations involving degenerate Poly-Cauchy polynomials of the third kind with the Stirling numbers of both kinds, the Lab numbers and the Bell numbers.

1. 预备知识

在文献 [1] 中，Cauchy多项式 $C_n\left(x\right)$ 的发生函数定义为：

${\displaystyle \underset{n\ge 0}{\sum }{C}_n}\left(x\right)\frac{t^n}{n!}=\frac{t}{\ln \left(1+t\right)}{\left(1+t\right)}^x$. (1)

其中，当 $x=0$ 时，称 $C_n\left(0\right)=C_n$ 为Cauchy数。

首先，我们引入第三类退化的Poly-Cauchy多项式的定义，其发生函数的定义为(文献 [2] )：

${\displaystyle \underset{n\ge 0}{\sum }{C}_{n,\lambda ,3}}\left(x\right)\frac{t^n}{n!}=\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}$. (2)

其次，我们给出本文中用到的几种组合序列的发生函数的定义：即第一类Stirling数 $s\left(n,k\right)$ ；第一类无符号Stirling数 $\left|s\left(n,k\right)\right|$ ；第二类Stirling数 $S\left(n,k\right)$ ；Lab数 $L\left(n,k\right)$ ；第一类Bell数 $B\left(n,k\right)$ 以及第二类Bell数 $\beta \left(n,k\right)$ 的发生函数定义如下(文献 [3] )：

${\displaystyle \underset{n\ge k}{\sum }s\left(n,k\right)\frac{t^n}{n!}}=\frac{1}{k!}{\ln }^k\left(1+t\right)$, (3)

${\displaystyle \underset{n\ge k}{\sum }\left|s\left(n,k\right)\right|\frac{t^n}{n!}}=\frac{1}{k!}{\left(\ln \frac{1}{1-t}\right)}^k$, (4)

${\displaystyle \underset{n\ge k}{\sum }S\left(n,k\right)\frac{t^n}{n!}}=\frac{1}{k!}{\left({\text{e}}^t-1\right)}^k$, (5)

${\displaystyle \underset{n\ge k}{\sum }L\left(n,k\right)\frac{t^n}{n!}}=\frac{1}{k!}{\left(\frac{-t}{1+t}\right)}^k$, (6)

${\displaystyle \underset{n\ge k}{\sum }B\left(n,k\right)\frac{t^n}{n!}}=\frac{1}{k!}{\left(\exp \left({\text{e}}^t-1\right)-1\right)}^k$, (7)

${\displaystyle \underset{n\ge k}{\sum }\beta \left(n,k\right)\frac{t^n}{n!}}=\frac{1}{k!}{\ln }^k\left(1+\ln \left(1+t\right)\right)$. (8)

下面引进一个引理：

引理1 令 $D=\left(g\left(t\right),f\left(t\right)\right)={\left\{d_{n,k}\right\}}_{n,k\in N}$ 为一个Riordan阵，令 $h\left(t\right)={\displaystyle {\sum }_{k\ge 0}{h}_k{t}^k}$ 为序列 ${\left\{h_n\right\}}_{n\in N}$ 的发生函数，则有(文献 [3] )

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{d}_{n,k}{h}_k}=\left[t^n\right]g\left(t\right)h\left(f\left(t\right)\right)=\left[t^n\right]g\left(t\right)h\left(y\right)\left(y=f\left(t\right)\right)$. (9)

2. 第三类退化的Poly-Cauchy多项式与一些组合数间的关系

定理1 设n为非负整数，则有

$C_{n,\lambda ,3}\left(x+y\right)={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right)}}{\left(\frac{y}{\lambda }\right)}_j{\lambda }^{j}s\left(k,j\right)C_{n-k,\lambda ,3}\left(x\right)$. (10)

证明：由第三类退化的Poly-Cauchy多项式的发生函数(2)，可得

$\begin{array}{l}{\displaystyle \underset{n\ge 0}{\sum }{C}_{n,\lambda ,3}}\left(x+y\right)\frac{t^n}{n!}=\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x+y}{\lambda }}\\ =\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{y}{\lambda }}\\ =\left({\displaystyle \underset{n\ge 0}{\sum }{C}_{n,\lambda ,3}}\left(x\right)\frac{t^n}{n!}\right)\left({{\displaystyle \underset{j\ge 0}{\sum }\left(\frac{y}{\lambda }\right)}}_j\text{}\frac{{\lambda }^j{\ln }^j\left(1+t\right)}{j!}\right)\\ =\left({\displaystyle \underset{n\ge 0}{\sum }{C}_{n,\lambda ,3}}\left(x\right)\frac{t^n}{n!}\right)\left({{\displaystyle \underset{j\ge 0}{\sum }\left(\frac{y}{\lambda }\right)}}_j{\lambda }^j{\displaystyle \underset{n\ge j}{\sum }s\left(n,j\right)}\frac{t^n}{n!}\right)\\ =\left({\displaystyle \underset{n\ge 0}{\sum }{C}_{n,\lambda ,3}}\left(x\right)\frac{t^n}{n!}\right)\left({\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{n}{\sum }}{\left(\frac{y}{\lambda }\right)}_j{\lambda }^{j}s\left(n,j\right)}}\frac{t^n}{n!}\right)\\ ={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right)}}{\left(\frac{y}{\lambda }\right)}_j{\lambda }^{j}s\left(k,j\right)C_{n-k,\lambda ,3}\left(x\right)\frac{t^n}{n!}}\end{array}$

比较等式两边 $\frac{t^n}{n!}$ 的系数，可完成定理的证明。

定理2 设n为非负整数，则有

$C_{n,\lambda ,3}\left(x+1\right)+C_{n,\lambda ,3}\left(x\right)={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}}{\left(\frac{2}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)s\left(n,k\right)$. (11)

证明：由式(2)，可得

$\begin{array}{l}{\displaystyle \underset{n\ge 0}{\sum }{C}_{n,\lambda ,3}\left(x+1\right)+C_{n,\lambda ,3}\left(x\right)\frac{t^n}{n!}}\\ =\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x+1}{\lambda }}+\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\\ =\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}+1\right)\\ =\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{2}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\end{array}$

$=\lambda \left({{\displaystyle \underset{k\ge 1}{\sum }\left(\frac{2}{\lambda }\right)}}_k\frac{{\lambda }^k{\ln }^k\left(1+t\right)}{k!}\right)\frac{1}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}$

$\begin{array}{l}=\lambda \left({{\displaystyle \underset{k\ge 1}{\sum }\left(\frac{2}{\lambda }\right)}}_k\frac{{\lambda }^k{\ln }^{k-1}\left(1+t\right)}{k!}\right)\frac{\lambda \ln \left(1+t\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\\ =\lambda \left({{\displaystyle \underset{k\ge 0}{\sum }\left(\frac{2}{\lambda }\right)}}_{k+1}\frac{1}{k+1}\frac{{\lambda }^k{\ln }^k\left(1+t\right)}{k!}\right)\left({\displaystyle \underset{k\ge 0}{\sum }{C}_k\left(\frac{x}{\lambda }\right)}\frac{{\lambda }^k{\ln }^k\left(1+t\right)}{k!}\right)\\ =\lambda {{\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}\left(\frac{2}{\lambda }\right)}}_{j+1}\frac{{\lambda }^k}{j+1}{C}_{k-l}\left(\frac{x}{\lambda }\right)\frac{{\ln }^k\left(1+t\right)}{k!}\\ ={{\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}\left(\frac{2}{\lambda }\right)}}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-l}\left(\frac{x}{\lambda }\right){\displaystyle \underset{n\ge k}{\sum }s\left(n,k\right)}\frac{t^n}{n!}\\ ={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)s}\left(n,k\right)}}{\left(\frac{2}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-l}\left(\frac{x}{\lambda }\right)\frac{t^n}{n!}\end{array}$

比较等式两边 $\frac{t^n}{n!}$ 的系数，可得定理的证明。

在定理2中，令 $x=0$，可得如下推论。

推论2 设n为非负整数，则有

$C_{n,\lambda ,3}\left(1\right)+C_{n,\lambda ,3}\left(0\right)={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}}{\left(\frac{2}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}s\left(n,k\right)$. (12)

定理3 设n为非负整数，则有

$C_{n,\lambda ,3}\left(x-1\right)={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)\left(-1\right){}^j}}{\langle \frac{1}{\lambda }\rangle }_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)s\left(n,k\right)$. (13)

证明：由式(2)，可得

$\begin{array}{l}{\displaystyle \underset{n\ge 0}{\sum }{C}_{n,\lambda ,3}\left(x-1\right)\frac{t^n}{n!}}=\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x-1}{\lambda }}\\ =\frac{\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}\frac{{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}}{{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}}\\ =\frac{\lambda \left(1-{\left(1+\lambda \ln \left(1+t\right)\right)}^{-\frac{1}{\lambda }}\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\\ =\frac{-\lambda \left({\left(1+\lambda \ln \left(1+t\right)\right)}^{-\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\\ =\frac{-\lambda {\displaystyle {\sum }_{k\ge 1}{\left(-\frac{1}{\lambda }\right)}_k\frac{{\lambda }^k{\ln }^k\left(1+t\right)}{k!}}}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\end{array}$

$\begin{array}{l}=-\lambda {\displaystyle \underset{k\ge 0}{\sum }{\left(-1\right)}^{k+1}{\langle \frac{1}{\lambda }\rangle }_{k+1}\frac{{\lambda }^k{\ln }^k\left(1+t\right)}{\left(k+1\right)!}}\frac{\lambda \ln \left(1+t\right)}{\ln \left(1+\lambda \ln \left(1+t\right)\right)}{\left(1+\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}\\ =\left(-\lambda {\displaystyle \underset{k\ge 0}{\sum }{\left(-1\right)}^{k+1}{\langle \frac{1}{\lambda }\rangle }_{k+1}\frac{{\lambda }^k}{k+1}\frac{{\ln }^k\left(1+t\right)}{k!}}\right)\left({\displaystyle \underset{k\ge 0}{\sum }{\lambda }^k{C}_k\left(\frac{x}{\lambda }\right)\frac{{\ln }^k\left(1+t\right)}{k!}}\right)\\ =-\lambda {\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}{\left(-1\right)}^{j+1}{\langle \frac{1}{\lambda }\rangle }_{j+1}\frac{{\lambda }^k}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\frac{{\ln }^k\left(1+t\right)}{k!}}\\ ={\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}{\left(-1\right)}^j{\langle \frac{1}{\lambda }\rangle }_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}{\displaystyle \underset{n\ge k}{\sum }s\left(n,k\right)\frac{t^n}{n!}}\\ ={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)\left(-1\right){}^j}}{\langle \frac{1}{\lambda }\rangle }_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)s\left(n,k\right)\frac{t^n}{n!}}\end{array}$

比较等式两边 $\frac{t^n}{n!}$ 的系数，可以得到结论(13)。

在定理3中，令 $x=0$，即可得如下推论。

推论3 设n为非负整数，则有

$C_{n,\lambda ,3}\left(-1\right)={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)\left(-1\right){}^j}}{\langle \frac{1}{\lambda }\rangle }_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}s\left(n,k\right)$. (14)

定理4 设n为非负整数，则有

${\left(-1\right)}^n{C}_{n,\lambda ,3}(x)={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}}{\left(-1\right)}^k{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\left|s\left(n,k\right)\right|$. (15)

证明：由式(2)，可得

$\begin{array}{l}{\displaystyle \underset{n\ge 0}{\sum }{\left(-1\right)}^n{C}_{n,\lambda ,3}\left(x\right)\frac{t^n}{n!}}=\frac{\lambda \left({\left(1+\lambda \ln \left(1-t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1-t\right)\right)}{\left(1+\lambda \ln \left(1-t\right)\right)}^{\frac{x}{\lambda }}\\ =\frac{\lambda \left({\left(1-\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1-\lambda \ln \left(\frac{1}{1-t}\right)\right)}{\left(1-\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{x}{\lambda }}\\ =\lambda \frac{{\displaystyle {\sum }_{k\ge 1}{\left(-\lambda \right)}^k{\left(\frac{1}{\lambda }\right)}_k\frac{1}{k!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}}{\ln \left(1-\lambda \ln \left(\frac{1}{1-t}\right)\right)}{\left(1-\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{x}{\lambda }}\\ =\lambda {\displaystyle \underset{k\ge 0}{\sum }{\left(-\lambda \right)}^k{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{1}{\left(k+1\right)!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\frac{-\lambda \ln \left(\frac{1}{1-t}\right)}{\ln \left(1-\lambda \ln \left(\frac{1}{1-t}\right)\right)}{\left(1-\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{x}{\lambda }}\end{array}$

$=\lambda \left({\displaystyle \underset{k\ge 0}{\sum }{\left(-\lambda \right)}^k{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{1}{\left(k+1\right)!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\right)\left({\displaystyle \underset{k\ge 0}{\sum }{\left(-\lambda \right)}^k{C}_k\left(\frac{x}{\lambda }\right)\frac{1}{k!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\right)$

$\begin{array}{l}=\lambda {\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}{\left(-\lambda \right)}^k{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{1}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\frac{1}{k!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\\ =\lambda {\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}{\left(-\lambda \right)}^k{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{1}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}{\displaystyle \underset{n\ge k}{\sum }\left|s\left(n,k\right)\right|\frac{t^n}{n!}}\\ ={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}}{\left(-1\right)}^k{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\left|s\left(n,k\right)\right|\frac{t^n}{n!}}\end{array}$

比较等式两边 $\frac{t^n}{n!}$ 的系数，可完成定理的证明。

定理5 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}S\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}={\displaystyle \underset{k=0}{\overset{n}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right){\left(\frac{1}{\lambda }\right)}_{k+1}\frac{{\lambda }^{n+1}}{k+1}{C}_{n-k}\left(\frac{x}{\lambda }\right)}$. (16)

证明：因为 $\left\{\frac{k!}{n!}S\left(n,k\right)\right\}=\left(1,{\text{e}}^t-1\right)$，再由引理1和式(2)，则有

$\begin{array}{l}{\displaystyle \underset{k=0}{\overset{n}{\sum }}S\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}=n!{\displaystyle \underset{k=0}{\overset{n}{\sum }}\frac{k!}{n!}S\left(n,k\right)\frac{C_{k,\lambda ,3}\left(x\right)}{k!}}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda \ln \left(1+y\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+y\right)\right)}{\left(1+\lambda \ln \left(1+y\right)\right)}^{\frac{xz}{\lambda }}\left(y={\text{e}}^t-1\right)\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda t\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda t\right)}{\left(1+\lambda t\right)}^{\frac{xz}{\lambda }}\\ =n!\left[t^n\right]{\displaystyle \underset{n\ge 1}{\sum }{\left(\frac{1}{\lambda }\right)}_n}\frac{{\left(\lambda t\right)}^n}{n!}\frac{\lambda }{\ln \left(1+\lambda t\right)}{\left(1+\lambda t\right)}^{\frac{xz}{\lambda }}\end{array}$

$\begin{array}{l}=n!\left[t^n\right]{\displaystyle \underset{n\ge 0}{\sum }{\left(\frac{1}{\lambda }\right)}_{n+1}}\frac{{\lambda }^{n+1}{t}^n}{\left(n+1\right)!}\frac{\lambda t}{\ln \left(1+\lambda t\right)}{\left(1+\lambda t\right)}^{\frac{xz}{\lambda }}\\ =n!\left[t^n\right]\left({\displaystyle \underset{n\ge 0}{\sum }{\left(\frac{1}{\lambda }\right)}_{n+1}}\frac{{\lambda }^{n+1}}{n+1}\frac{t^n}{n!}\right)\left({\displaystyle \underset{n\ge 0}{\sum }{\lambda }^n{C}_n\left(\frac{x}{\lambda }\right)}\frac{t^n}{n!}\right)\\ =n!\left[t^n\right]{\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right)}}{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{{\lambda }^{n+1}}{k+1}{C}_{n-k}\left(\frac{x}{\lambda }\right)\frac{t^n}{n!}\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right){\left(\frac{1}{\lambda }\right)}_{k+1}\frac{{\lambda }^{n+1}}{k+1}{C}_{n-k}(\; x\; \lambda \; )}\end{array}$

下面引进序列 ${\left\{f_n\right\}}_{n\in N}$ 和 ${\left\{g_n\right\}}_{n\in N}$ 的反演公式

$f_n={\displaystyle \underset{k=0}{\overset{n}{\sum }}S\left(n,k\right)g_k}\iff g_n={\displaystyle \underset{k=0}{\overset{n}{\sum }}s\left(n,k\right)f_k}$. (17)

由式(17)和定理5立即可得如下等式。

定理6 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)s\left(n,k\right)}}=C_{n,\lambda ,3}\left(x\right)$. (18)

定理7 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}{\lambda }^{-k}S\left(n,k\right)S\left(k,j\right)C_{j,\lambda ,3}\left(x\right)}}={\lambda }^{-\left(n-1\right)}\frac{{\left(x+1\right)}^{n+1}-x^{n+1}}{n+1}$. (19)

证明：由定理5的证明可知

${\displaystyle \underset{k=0}{\overset{n}{\sum }}S\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}=\frac{\lambda \left({\left(1+\lambda t\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda t\right)}{\left(1+\lambda t\right)}^{\frac{x}{\lambda }}$. (20)

因为 $\left\{\frac{k!}{n!}S\left(n,k\right)\right\}=\left(1,{\text{e}}^t-1\right)$，再由式(9)和式(20)，可得

$\begin{array}{l}{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}{\lambda }^{-k}S\left(n,k\right)S\left(k,j\right)C_{j,\lambda ,3}}}\left(x\right)\\ =n!{\displaystyle \underset{k=0}{\overset{n}{\sum }}\frac{k!}{n!}{\lambda }^{-k}S\left(n,k\right)\frac{{\displaystyle {\sum }_{j=0}^{k}S\left(k,j\right)C_{j,\lambda ,3}\left(x\right)}}{k!}}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda y\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda y\right)}{\left(1+\lambda y\right)}^{\frac{x}{\lambda }}\left(y=\frac{{\text{e}}^t-1}{\lambda }\right)\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda \frac{{\text{e}}^t-1}{\lambda }\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \frac{{\text{e}}^t-1}{\lambda }\right)}{\left(1+\lambda \frac{{\text{e}}^t-1}{\lambda }\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\frac{\lambda \left({\text{e}}^{\frac{t}{\lambda }}-1\right)}{t}{\text{e}}^{\frac{xt}{\lambda }}=n!\left[t^{n+1}\right]\lambda \left({\text{e}}^{\frac{t}{\lambda }}-1\right){\text{e}}^{\frac{xt}{\lambda }}\\ ={\lambda }^{-\left(n-1\right)}\frac{{\left(x+1\right)}^{n+1}-x^{n+1}}{n+1}\end{array}$

由式(17)和定理7立即可得如下结论。

定理8 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}{\lambda }^{-n+k+1}\frac{{\left(x+1\right)}^{n+1}-x^{n+1}}{n+1}s\left(n,k\right)s\left(k,j\right)}}=C_{n,\lambda ,3}\left(x\right)$.(21)

定理9 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}S\left(n,k\right)L\left(k,j\right)C_{j,\lambda ,3}\left(x\right)}}={\displaystyle \underset{k=0}{\overset{n}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right){\left(\frac{1}{\lambda }\right)}_{k+1}{\left(-1\right)}^n\frac{{\lambda }^{n+1}}{k+1}{C}_{n-k}\left(\frac{x}{\lambda }\right)}$.(22)

证明：因为 $\left\{\frac{k!}{n!}L\left(n,k\right)\right\}=\left(1,\frac{-t}{1+t}\right)$，再由式(9)和式(2)，可得

${\displaystyle \underset{k=0}{\overset{n}{\sum }}L\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}=n!\left[t^n\right]\frac{\lambda \left({\left(1-\lambda \ln \left(1+t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1-\lambda \ln \left(1+t\right)\right)}{\left(1-\lambda \ln \left(1+t\right)\right)}^{\frac{x}{\lambda }}$, (23)

又由式(5)，可得

$\begin{array}{l}{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}S\left(n,k\right)L\left(k,j\right)C_{j,\lambda ,3}}\left(x\right)}=n!{\displaystyle \underset{k=0}{\overset{n}{\sum }}\frac{\frac{k!}{n!}{\displaystyle {\sum }_{j=0}^{k}L\left(k,j\right)C_{j,\lambda ,3}\left(x\right)}}{k!}S\left(n,k\right)}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1-\lambda \ln \left(1+y\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1-\lambda \ln \left(1+y\right)\right)}{\left(1-\lambda \ln \left(1+y\right)\right)}^{\frac{xz}{\lambda }}\left(y={\text{e}}^t-1\right)\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1-\lambda \ln \left(1+{\text{e}}^t-1\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1-\lambda \ln \left(1+{\text{e}}^t-1\right)\right)}{\left(1-\lambda \ln \left(1+{\text{e}}^t-1\right)\right)}^{\frac{xz}{\lambda }}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1-\lambda t\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1-\lambda t\right)}{\left(1-\lambda t\right)}^{\frac{xz}{\lambda }}\end{array}$

$\begin{array}{l}=n!\left[t^n\right]{\displaystyle \underset{n\ge 1}{\sum }{\left(\frac{1}{\lambda }\right)}_n}\frac{{\left(-\lambda t\right)}^n}{n!}\frac{\lambda }{\ln \left(1-\lambda t\right)}{\left(1-\lambda t\right)}^{\frac{xz}{\lambda }}\\ =n!\left[t^n\right]{\displaystyle \underset{n\ge 0}{\sum }{\left(\frac{1}{\lambda }\right)}_{n+1}}\frac{{\lambda }^{n+1}{\left(-t\right)}^n}{\left(n+1\right)!}\frac{-\lambda t}{\ln \left(1-\lambda t\right)}{\left(1-\lambda t\right)}^{\frac{xz}{\lambda }}\\ =n!\left[t^n\right]\left({\displaystyle \underset{n\ge 0}{\sum }{\left(-1\right)}^n{\left(\frac{1}{\lambda }\right)}_{n+1}}\frac{{\lambda }^{n+1}}{n+1}\frac{t^n}{n!}\right)\left({\displaystyle \underset{n\ge 0}{\sum }{\left(-\lambda \right)}^n{C}_n\left(\frac{x}{\lambda }\right)}\frac{t^n}{n!}\right)\\ =n!\left[t^n\right]{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{{\left(-1\right)}^n{\lambda }^{n+1}}{k+1}{C}_{n-k}\left(\frac{x}{\lambda }\right)\frac{t^n}{n!}\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}\left(\begin{array}{c}n\\ k\end{array}\right){\left(\frac{1}{\lambda }\right)}_{k+1}{\left(-1\right)}^n\frac{{\lambda }^{n+1}}{k+1}{C}_{n-k}(\; x\; \lambda \; )}\end{array}$

由式(17)和定理9，可得如下结论。

定理10 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\left(-\lambda \right)}^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)s\left(n,k\right)}}={\displaystyle \underset{k=0}{\overset{n}{\sum }}L\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}$. (24)

定理11 设n为非负整数，则有

$\begin{array}{l}{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}}\left|s\left(n,k\right)\right|\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left|s\left(n,k\right)\right|S\left(k,j\right)C_{j,\lambda ,3}\left(x\right)}}={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\left(-1\right)}^{n}L\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}\end{array}$ (25)

证明：由式(6)有

${\displaystyle \underset{n\ge k}{\sum }L\left(n,k\right)\frac{{\left(-t\right)}^n}{n!}}=\frac{1}{k!}{\left(\frac{t}{1-t}\right)}^k$, (26)

再由式(9)有

$\begin{array}{l}{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\left(-1\right)}^{n}L\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}=n!\left[t^n\right]\frac{\lambda \left({\left(1-\lambda \ln \left(1-t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1-\lambda \ln \left(1-t\right)\right)}{\left(1-\lambda \ln \left(1-t\right)\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}{\left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\lambda \frac{{\displaystyle {\sum }_{k\ge 1}{\lambda }^k{\left(\frac{1}{\lambda }\right)}_k\frac{1}{k!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}}{\ln \left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}{\left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\lambda {\displaystyle \underset{k\ge 0}{\sum }{\lambda }^k{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{1}{\left(k+1\right)!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\frac{\lambda \ln \left(\frac{1}{1-t}\right)}{\ln \left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}{\left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{x}{\lambda }}\end{array}$

$\begin{array}{l}=n!\left[t^n\right]\lambda \left({\displaystyle \underset{k\ge 0}{\sum }{\lambda }^k{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{1}{\left(k+1\right)!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\right)\left({\displaystyle \underset{k\ge 0}{\sum }{\lambda }^k{C}_k\left(\frac{x}{\lambda }\right)\frac{1}{k!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\right)\\ =n!\left[t^n\right]\lambda {\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\lambda }^k}{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{1}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\frac{1}{k!}{\left(\ln \left(\frac{1}{1-t}\right)\right)}^k}\\ =n!\left[t^n\right]\lambda {\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\lambda }^k}{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{1}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}{\displaystyle \underset{n\ge k}{\sum }\left|s\left(n,k\right)\right|\frac{t^n}{n!}}\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}}\left|s\left(n,k\right)\right|\end{array}$

而由式(14)和式(9)，还可以得到

$\begin{array}{l}{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left|s\left(n,k\right)\right|S\left(k,j\right)C_{j,\lambda ,3}\left(x\right)}}=n!{\displaystyle \underset{k=0}{\overset{n}{\sum }}\frac{k!}{n!}\left|s\left(n,k\right)\right|}\frac{{\displaystyle {\sum }_{j=0}^{k}S\left(k,j\right)C_{j,\lambda ,3}}\left(x\right)}{k!}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda y\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda y\right)}{\left(1+\lambda y\right)}^{\frac{x}{\lambda }}\left(y=\ln \left(\frac{1}{1-t}\right)\right)\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}{\left(1+\lambda \ln \left(\frac{1}{1-t}\right)\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1-\lambda \ln \left(1-t\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1-\lambda \ln \left(1-t\right)\right)}{\left(1-\lambda \ln \left(1-t\right)\right)}^{\frac{x}{\lambda }}\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}}\left|s\left(n,k\right)\right|\end{array}$

定理12 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)S\left(n,k\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}}={\displaystyle \underset{k=0}{\overset{n}{\sum }}B\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}$. (27)

证明：因为 $\left\{\frac{k!}{n!}B\left(n,k\right)\right\}=\left(1,\exp \left({\text{e}}^t-1\right)-1\right)$，再由式(9)和式(2)，可得

$\begin{array}{l}{\displaystyle \underset{k=0}{\overset{n}{\sum }}B\left(n,k\right)C_{k,\lambda ,3}\left(x\right)}=n!{\displaystyle \underset{k=0}{\overset{n}{\sum }}\frac{k!}{n!}B\left(n,k\right)\frac{C_{k,\lambda ,3}\left(x\right)}{k!}}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda \ln \left(1+y\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+y\right)\right)}{\left(1+\lambda \ln \left(1+y\right)\right)}^{\frac{x}{\lambda }}\left(y=\exp \left({\text{e}}^t-1\right)-1\right)\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda \ln \left(1+\exp \left({\text{e}}^t-1\right)-1\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+\exp \left({\text{e}}^t-1\right)-1\right)\right)}{\left(1+\lambda \ln \left(1+\exp \left({\text{e}}^t-1\right)-1\right)\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\frac{\lambda \left({\left(1+\lambda \left({\text{e}}^t-1\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \left({\text{e}}^t-1\right)\right)}{\left(1+\lambda \left({\text{e}}^t-1\right)\right)}^{\frac{x}{\lambda }}\end{array}$

$\begin{array}{l}=n!\left[t^n\right]\lambda {{\displaystyle \underset{k\ge 1}{\sum }\left(\frac{1}{\lambda }\right)}}_k\frac{{\lambda }^k{\left({\text{e}}^t-1\right)}^k}{k!}\frac{1}{\ln \left(1+\lambda \left({\text{e}}^t-1\right)\right)}{\left(1+\lambda \left({\text{e}}^t-1\right)\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\lambda {{\displaystyle \underset{k\ge 1}{\sum }\left(\frac{1}{\lambda }\right)}}_k\frac{{\lambda }^k{\left({\text{e}}^t-1\right)}^{k-1}}{k!}\frac{\lambda \left({\text{e}}^t-1\right)}{\ln \left(1+\lambda \left({\text{e}}^t-1\right)\right)}{\left(1+\lambda \left({\text{e}}^t-1\right)\right)}^{\frac{x}{\lambda }}\\ =n!\left[t^n\right]\lambda \left({{\displaystyle \underset{k\ge 0}{\sum }\left(\frac{1}{\lambda }\right)}}_{k+1}\frac{{\lambda }^k{\left({\text{e}}^t-1\right)}^{k-1}}{\left(k+1\right)!}\right)\left({\displaystyle \underset{k\ge 0}{\sum }{C}_k}\left(\frac{x}{\lambda }\right)\frac{{\lambda }^k{\left({\text{e}}^t-1\right)}^k}{k!}\right)\\ =n!\left[t^n\right]{\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}}{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\frac{{\left({\text{e}}^t-1\right)}^k}{k!}\end{array}$

$\begin{array}{l}=n!\left[t^n\right]{\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)}}{\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right){\displaystyle \underset{n\ge k}{\sum }S\left(n,k\right)\frac{t^n}{n!}}\\ =n!\left[t^n\right]{\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)S\left(n,k\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)}}\frac{t^n}{n!}}\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right)S\left(n,k\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}(\; x\; \lambda \; )}}\end{array}$

定理13 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\beta \left(n,k\right)}}={\displaystyle \underset{k=0}{\overset{n}{\sum }}{C}_{k,\lambda ,3}\left(x\right)s\left(n,k\right)}$. (28)

证明：由第三类退化的Poly-Cauchy多项式的发生函数(2)有

$\begin{array}{l}{\displaystyle \underset{k\ge 0}{\sum }{C}_{k,\lambda ,3}\left(x\right)\frac{{\ln }^k\left(1+t\right)}{k!}}={\displaystyle \underset{k\ge 0}{\sum }{C}_{k,\lambda ,3}\left(x\right)}{\displaystyle \underset{n\ge k}{\sum }s\left(n,k\right)\frac{t^n}{n!}}={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{C}_{k,\lambda ,3}\left(x\right)s\left(n,k\right)}\frac{t^n}{n!}}\\ =\frac{\lambda \left({\left(1+\lambda \ln \left(1+\ln \left(1+t\right)\right)\right)}^{\frac{1}{\lambda }}-1\right)}{\ln \left(1+\lambda \ln \left(1+\ln \left(1+t\right)\right)\right)}{\left(1+\lambda \ln \left(1+\ln \left(1+t\right)\right)\right)}^{\frac{x}{\lambda }}\\ =\lambda {\displaystyle \underset{k\ge 1}{\sum }{\left(\frac{1}{\lambda }\right)}_k\frac{{\lambda }^k{\ln }^k\left(1+\ln \left(1+t\right)\right)}{k!}}\frac{1}{\ln \left(1+\lambda \ln \left(1+\ln \left(1+t\right)\right)\right)}{\left(1+\lambda \ln \left(1+\ln \left(1+t\right)\right)\right)}^{\frac{x}{\lambda }}\\ =\lambda {\displaystyle \underset{k\ge 0}{\sum }{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{1}{k+1}\frac{{\lambda }^k{\ln }^k\left(1+\ln \left(1+t\right)\right)}{k!}}\frac{\lambda \ln \left(1+\ln \left(1+t\right)\right)}{\ln \left(1+\lambda \ln \left(1+\ln \left(1+t\right)\right)\right)}{\left(1+\lambda \ln \left(1+\ln \left(1+t\right)\right)\right)}^{\frac{x}{\lambda }}\end{array}$

$\begin{array}{l}=\lambda \left({\displaystyle \underset{k\ge 0}{\sum }{\left(\frac{1}{\lambda }\right)}_{k+1}\frac{{\lambda }^k}{k+1}}\frac{{\ln }^k\left(1+\ln \left(1+t\right)\right)}{k!}\right)\left({\displaystyle \underset{k\ge 0}{\sum }{C}_k}\left(\frac{x}{\lambda }\right)\frac{{\lambda }^k{\ln }^k\left(1+\ln \left(1+t\right)\right)}{k!}\right)\\ ={\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}}}\left(\frac{x}{\lambda }\right)\frac{{\ln }^k\left(1+\ln \left(1+t\right)\right)}{k!}\\ ={\displaystyle \underset{k\ge 0}{\sum }{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}}}\left(\frac{x}{\lambda }\right){\displaystyle \underset{}{\sum }\beta \left(n,k\right)\frac{t^n}{n!}}\\ ={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\beta }}\left(n,k\right)\frac{t^n}{n!}}\\ ={\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}k\\ j\end{array}\right){\left(\frac{1}{\lambda }\right)}_{j+1}\frac{{\lambda }^{k+1}}{j+1}{C}_{k-j}\left(\frac{x}{\lambda }\right)\beta \left(n,k\right)}}\end{array}$

比较等式两边 $\frac{t^n}{n!}$ 的系数，可完成定理的证明。

由式(17)和定理13，立即可得如下结论。

定理14 设n为非负整数，则有

${\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}{\displaystyle \underset{i=0}{\overset{j}{\sum }}\left(\begin{array}{c}j\\ i\end{array}\right)}{\left(\frac{1}{\lambda }\right)}_{i+1}\frac{{\lambda }^{j+1}}{i+1}{C}_{j-i}\left(\frac{x}{\lambda }\right)\beta \left(k,j\right)S\left(n,k\right)}}=C_{n,\lambda ,3}\left(x\right)$. (29)

基金项目

国家自然科学基金项目(11461050)，内蒙古自然科学基金项目(2020MS01020)。