Poly-Cauchy Numbers with Higher Level

Abstract

In this article, mainly from the analytical aspect, we introduce poly-Cauchy numbers with higher levels (level s) as a kind of extensions of poly-Cauchy numbers with level 2 and the original poly-Cauchy numbers and investigate their properties. Such poly-Cauchy numbers with higher levels are yielded from the inverse relationship with an s-step function of the exponential function. We show such a function with recurrence relations and give the expressions of poly-Cauchy numbers with higher levels. Poly-Cauchy numbers with higher levels can be also expressed in terms of iterated integrals and a combinatorial summation. Poly-Cauchy numbers with higher levels for negative indices have a double summation formula. In addition, Cauchy numbers with higher levels can be also expressed in terms of determinants.

1. Introduction

The Stirling numbers with higher level (level s) were first studied by Tweedie [1] in 1918. Namely, those of the first kind ${\left[\left[\genfrac{}{}{0pt}{}{n}{k}\right]\right]}_s$ and the second kind ${\left\{\left\{\genfrac{}{}{0pt}{}{n}{k}\right\}\right\}}_s$ appeared as

$$x(x+1^s)(x+2^s)\cdots (x+{(n-1)}^s)=\sum _{k=0}^n{\left[\left[\genfrac{}{}{0pt}{}{n}{k}\right]\right]}_s{x}^k$$

and

$$x^n=\sum _{k=0}^n{\left\{\left\{\genfrac{}{}{0pt}{}{n}{k}\right\}\right\}}_{s}x(x-1^s)(x-2^s)\cdots \left(x-{(k-1)}^s\right),$$

respectively. They satisfy the recurrence relations

$${\left[\left[\genfrac{}{}{0pt}{}{n}{k}\right]\right]}_s={\left[\left[\genfrac{}{}{0pt}{}{n-1}{k-1}\right]\right]}_s+{(n-1)}^s{\left[\left[\genfrac{}{}{0pt}{}{n-1}{k}\right]\right]}_s$$

and

$${\left\{\left\{\genfrac{}{}{0pt}{}{n}{k}\right\}\right\}}_s={\left\{\left\{\genfrac{}{}{0pt}{}{n-1}{k-1}\right\}\right\}}_s+k{\left\{\left\{\genfrac{}{}{0pt}{}{n-1}{k}\right\}\right\}}_s$$

with ${\left[\left[\genfrac{}{}{0pt}{}{0}{0}\right]\right]}_s={\left\{\left\{\genfrac{}{}{0pt}{}{0}{0}\right\}\right\}}_s=1$ and ${\left[\left[\genfrac{}{}{0pt}{}{n}{0}\right]\right]}_s={\left\{\left\{\genfrac{}{}{0pt}{}{n}{0}\right\}\right\}}_s=0$ ($n\ge 1$). When $s=1$, they are the original Stirling numbers of both kinds. When $s=2$, they have been often studied as central factorial numbers of both kinds (see, e.g., [2]). The concept introduced by Tweedie This concept was used by Bell [3] to show a generalization of Lagrange and Wilson theorems. However, such generalized Stirling numbers have been forgotten or ignored for a long time.

Recently in [4,5], the Stirling numbers with higher levels have been rediscovered and studied more deeply, in particular, from the aspects of combinatorics. On the other hand, in [6], by using the Stirling numbers of the first kind with level 2, poly-Cauchy numbers with level 2 are introduced as a kind of generalizations of the original poly-Cauchy numbers, which may be interpreted as a kind of generalizations of the classical Cauchy numbers. In [7], by using the Stirling numbers of the second kind with level 2, poly-Bernoulli numbers with level 2 are introduced as a kind of generalizations of the original poly-Bernoulli numbers [8]. In [9], other poly-generalized numbers, which are called polycosecant numbers, are introduced and studied. This result leads to a variant of multiple zeta values of level 2 [10], which forms a subspace of the space of alternating multiple zeta values. However, no generalized Stirling number is considered in [9].

Another of the most famous generalized Stirling numbers is the r-Stirling number [11], which has meaningful relations with harmonic numbers from the summation formulas [12,13,14]. By using r-Stirling numbers, so-called various r-numbers are introduced.

It is remarkable to see that the original poly-Cauchy numbers (with level 1, ref. [15]), which may be also yielded by the logarithm function (an 1-step function) with the inverse relation of the exponential function. This can be said to be an analytical definition. Then, poly-Cauchy numbers with level 2 may be yielded or defined from the inverse relation about the hyperbolic sine function, which is a 2-step function of the exponential function [6]. Then, it would be a natural question how the poly-Cauchy numbers with level 3, 4, and generally level s can be defined by any functions (3, 4 and generally s-step functions, respectively) in a natural way.

In combinatorial ways, just as poly-Cauchy number with level 2 arises from the relationship with the Stirling numbers with level 2, poly-Cauchy number with level 3, 4 and generally level s could be hoped to arise from the Stirling numbers with level 3, 4 and generally level s, respectively. However, in the case of 3 or higher level, it is not easy to define and describe most of the properties including both combinatorial and analytical meanings naturally as well as those with levels 1 and 2. For example,

$${\left\{\left\{\genfrac{}{}{0pt}{}{n}{k}\right\}\right\}}_s=\frac{s}{\left(sk\right)!}\sum _{j=1}^k{(-1)}^{k-j}\left(\genfrac{}{}{0pt}{}{sk}{k-j}\right)j^{sn}$$

holds for $s=1,2$ and does not for $s\ge 3$ ([5]).

The purpose of this paper is to define poly-Cauchy numbers with higher level (level s) from the analytical implications and investigate their properties. Such poly-Cauchy numbers with higher levels are yielded from the inverse relationship with an s-step function of the exponential function. We show such a function with recurrence relations and give the expressions of poly-Cauchy numbers with higher levels. Poly-Cauchy numbers with higher levels can be also expressed in terms of iterated integrals and a combinatorial summation. Poly-Cauchy numbers with higher levels for negative indices have a double summation formula. In addition, Cauchy numbers with higher levels can be also expressed in terms of determinants.

2. Definitions

For integers n and k with $n\ge 0$, poly-Cauchy numbers${\mathcal{C}}_{n,s}^{\left(k\right)}$with levels ($s\ge 1$) are defined by

$${\mathrm{Lif}}_{s,k}\left({\mathfrak{AF}}_s\left(t\right)\right)=\sum _{n=0}^{\infty }{\mathcal{C}}_{n,s}^{\left(k\right)}\frac{t^n}{n!},$$

(1)

where

$${\mathrm{Lif}}_{s,k}\left(z\right)=\sum _{m=0}^{\infty }\frac{z^{sm}}{\left(sm\right)!{(sm+1)}^k}.$$

The function ${\mathfrak{AF}}_s\left(t\right)$ is the inverse function of

$${\mathfrak{F}}_s\left(t\right)=\sum _{m=0}^{\infty }\frac{t^{sm+1}}{(sm+1)!}.$$

When $s=1$, ${\mathcal{C}}_{n,1}^{\left(k\right)}=c_n^{\left(k\right)}$ are the original poly-Cauchy numbers [15,16], defined by

$${\mathrm{Lif}}_k\left({\mathfrak{AF}}_1\left(t\right)\right)=\sum _{n=0}^{\infty }{c}_n^{\left(k\right)}\frac{t^n}{n!},$$

where ${\mathrm{Lif}}_{1,k}\left(z\right)={\mathrm{Lif}}_k\left(z\right)$ is the polylogarithm factorial function (or polyfactorial function) and ${\mathfrak{AF}}_1\left(t\right)=\log (t+1)$ is the inverse function of

$$\sum _{m=0}^{\infty }\frac{t^{m+1}}{(m+1)!}=e^t-1.$$

When $k=1$, $c_n=c_n^{\left(1\right)}$ are the original Cauchy numbers defined by

$${\mathrm{Lif}}_1\left(\log (t+1)\right)=\frac{t}{\log (t+1)}=\sum _{n=0}^{\infty }{c}_n\frac{t^n}{n!}.$$

When $s=2$, ${\mathcal{C}}_{n,2}^{\left(k\right)}={\mathbb{C}}_n^{\left(k\right)}$ are poly-Cauchy numbers with level 2 [6], defined by

$${\mathrm{Lif}}_{2,k}\left({\mathfrak{AF}}_2\left(t\right)\right)=\sum _{n=0}^{\infty }{\mathbb{C}}_n^{\left(k\right)}\frac{t^n}{n!},$$

where ${\mathfrak{AF}}_2\left(t\right)=\mathrm{arcsinh}t$ is the inverse function of

$$\sum _{m=0}^{\infty }\frac{t^{2m+1}}{(2m+1)!}=\sinh t$$

When $k=1$, ${\mathbb{C}}_n={\mathbb{C}}_n^{\left(1\right)}$ are Cauchy numbers with level 2, defined by

$${\mathrm{Lif}}_{2,1}\left(\mathrm{arcsinh}t\right)=\frac{t}{\mathrm{arcsinh}t}=\sum _{n=0}^{\infty }{\mathbb{C}}_n\frac{t^n}{n!}.$$

When $k=1$ and $s=3$,

$${\mathrm{Lif}}_{3,1}\left(z\right)=\frac{e^z+{\omega }^2{e}^{\omega z}+\omega e^{{\omega }^{2}z}}{3z}=\frac{{\mathfrak{F}}_3\left(z\right)}{z},$$

where

$${\mathfrak{F}}_3\left(z\right)=\sum _{m=0}^{\infty }\frac{z^{3m+1}}{(3m+1)!}.$$

and $\omega =(1+\sqrt{-3})/2$, satisfying ${\omega }^3=1$. Note that a similar function to $1/{\mathfrak{F}}_3\left(z\right)$ is studied in [17].

For an arbitrary $s\ge 1$ and $k=1$, we have

$${\mathrm{Lif}}_{s,1}\left(z\right)=\sum _{m=0}^{\infty }{\displaystyle \frac{z^{sm}}{(sm+1)!}}={\displaystyle \frac{1}{sz}}\prod _{j=0}^{s-1}{\zeta }^j{e}^{{\zeta }^{s-j}z}={\displaystyle \frac{{\mathfrak{F}}_s\left(z\right)}{z}},$$

where $\zeta =e^{2\pi i/s}$, is the s-th root of the identity. The function ${\mathrm{Lif}}_{s,1}\left(z\right)$ becomes the s-step exponential function.

3. Basic Results

When $s=3$,

$$\begin{array}{c}\hfill {\mathfrak{AF}}_3\left(z\right)=z-\frac{1}{24}{z}^4+\frac{17}{2520}{z}^7-\frac{389}{259200}{z}^{10}+\frac{85897}{222393600}{z}^{13}-\frac{887731}{8211456000}{z}^{16}\\ \hfill +\frac{762918737}{23870702592000}{z}^{19}-\frac{16283723339}{1658385653760000}{z}^{22}+\cdots .\end{array}$$

When $s=4$,

$$\begin{array}{c}\hfill {\mathfrak{AF}}_4\left(z\right)=z-\frac{1}{120}{z}^5+\frac{25}{72576}{z}^9-\frac{1655}{83026944}{z}^{13}+\frac{32633}{24320507904}{z}^{17}\\ \hfill -\frac{4046837}{41098797121536}{z}^{21}+\frac{95346434209}{12477594806098329600}{z}^{25}\\ \hfill -\frac{13496484991405}{21884703082311982252032}{z}^{29}+\frac{7594510992880985}{148224339331565182966038528}{z}^{33}\\ \hfill -\frac{4010591254856244071}{921362493285009177316895490048}{z}^{37}\\ \hfill +\frac{116831353234301926949}{310374651792009578002102307782656}{z}^{41}-\cdots .\end{array}$$

In general, for the inverse function of ${\mathfrak{F}}_s\left(z\right)$, we have the following.

Proposition 1.

$${\mathfrak{AF}}_s\left(z\right)=d_{0}z-d_1{z}^{s+1}+d_2{z}^{2s+1}-\cdots +{(-1)}^n{d}_n{z}^{sn+1}+\cdots ,$$

where the coefficients $d_i$ satisfy the recurrence relation

$$d_n=\sum _{m=0}^{n-1}{(-1)}^{n-m-1}{d}_m\sum _{\genfrac{}{}{0pt}{}{i_1+\cdots +i_{sm+1}=n-m}{i_1,\cdots ,i_{sm+1}\ge 0}}\frac{1}{(s{i}_1+1)!\cdots (s{i}_{sm+1}+1)!}(n\ge 1)$$

(2)

with $d_0=1$.

Proof. 

The expression can be obtained by the following process. First, put ${\mathfrak{F}}_s^{-1}\left(z\right):={\mathfrak{AF}}_s\left(z\right)$ as

$${\mathfrak{F}}_s^{-1}\left(z\right)=d_{0}z-d_1{z}^{s+1}+d_2{z}^{2s+1}-\cdots +{(-1)}^n{d}_n{z}^{sn+1}+\cdots .$$

(3)

Then we can find $d_0=1,d_1,d_2,\dots$ as follows. For convenience, put

$$H_{s,n}\left(j\right):=\sum _{\genfrac{}{}{0pt}{}{i_1+\cdots +i_{sn+1}=j}{i_1,\cdots ,i_{sn+1}\ge 0}}\frac{1}{(s{i}_1+1)!\cdots (s{i}_{sn+1}+1)!}.$$

Since ${\mathfrak{F}}_s^{-1}\left({\mathfrak{F}}_s\left(z\right)\right)={\mathfrak{F}}_s\left({\mathfrak{F}}_s^{-1}\left(z\right)\right)=z$, we see that

$$\begin{array}{cc}\hfill z& =\sum _{n=0}^{\infty }{(-1)}^n{d}_n{\left(\sum _{m=0}^{\infty }\frac{z^{sm+1}}{(sm+1)!}\right)}^{sn+1}\hfill \\ \hfill & =\sum _{n=0}^{\infty }{(-1)}^n{d}_n\sum _{m=0}^{\infty }{H}_{s,n}\left(m\right)z^{sn+sm+1}\hfill \\ \hfill & =\sum _{n=0}^{\infty }{(-1)}^n{d}_n\sum _{l=n}^{\infty }{H}_{s,n}(l-n)z^{sl+1}\hfill \\ \hfill & =\sum _{n=0}^{\infty }\left(\sum _{m=0}^n{(-1)}^m{d}_m{H}_{s,m}(n-m)\right)z^{sn+1}.\hfill \end{array}$$

Hence, for $n\ge 1$

$$\sum _{m=0}^n{(-1)}^m{d}_m{H}_{s,m}(n-m)=0$$

with $d_0=1$. The exact values of $d_0,d_1,d_2,\dots$ can be obtained by the recurrence relation (2). Some values of $H_{s,n}\left(j\right)$ for smaller j can be given as follows.

$$\begin{array}{cc}\hfill H_{s,n}\left(0\right)& =1,\hfill \\ \hfill H_{s,n}\left(1\right)& =\frac{sn+1}{(s+1)!},\hfill \\ \hfill H_{s,n}\left(2\right)& =\frac{sn+1}{(2s+1)!}+\frac{1}{{\left((s+1)!\right)}^2}\left(\genfrac{}{}{0pt}{}{sn+1}{2}\right),\hfill \\ \hfill H_{s,n}\left(3\right)& =\frac{sn+1}{(3s+1)!}+\frac{(sn+1)\left(sn\right)}{(2s+1)!(s+1)!}+\frac{1}{{\left((s+1)!\right)}^3}\left(\genfrac{}{}{0pt}{}{sn+1}{3}\right),\hfill \\ \hfill H_{s,n}\left(4\right)& =\frac{sn+1}{(4s+1)!}+\frac{(sn+1)\left(sn\right)}{(3s+1)!(s+1)!}+\frac{1}{{\left((2s+1)!\right)}^2}\left(\genfrac{}{}{0pt}{}{sn+1}{2}\right)\hfill \\ \hfill & +\frac{sn+1}{(2s+1)!{\left((s+1)!\right)}^2}\left(\genfrac{}{}{0pt}{}{sn}{2}\right)+\frac{1}{{\left((s+1)!\right)}^4}\left(\genfrac{}{}{0pt}{}{sn+1}{4}\right).\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill d_1& =H_{s,0}\left(1\right)=\frac{1}{(s+1)!},\hfill \\ \hfill d_2& =-H_{s,0}\left(2\right)+d_1{H}_{s,1}\left(1\right)=\frac{1}{(s+1)!r!}-\frac{1}{(2s+1)!},\hfill \\ \hfill d_3& =H_{s,0}\left(3\right)-d_1{H}_{s,1}\left(2\right)+d_2{H}_{s,2}\left(1\right)\hfill \\ \hfill & =\frac{3r+2}{2{\left((s+1)!\right)}^{2}s!}-\frac{3s+2}{(2s+1)!(s+1)!}+\frac{1}{(3s+1)!},\hfill \\ \hfill d_4& =-H_{s,0}\left(4\right)+d_1{H}_{s,1}\left(3\right)-d_2{H}_{s,2}\left(2\right)+d_3{H}_{s,3}\left(1\right)\hfill \\ \hfill & =\frac{(4s+3)(2s+1)}{3{\left((s+1)!\right)}^{3}s!}+\frac{1}{(2s+1)!\left(2s\right)!}-\frac{4s+3}{\left(2s\right)!{\left((s+1)!\right)}^2}\hfill \\ \hfill & +\frac{2(2s+1)}{(3s+1)!(s+1)!}-\frac{1}{(4s+1)!},\hfill \end{array}$$

(4)

□

Thus, by the definition (1), explicit expressions of ${\mathcal{C}}_{n,s}^{\left(k\right)}$ for each concrete s and small n can be achieved. For $s=3$, we have

$$\begin{array}{cc}\hfill {\mathcal{C}}_{0,3}^{\left(k\right)}& =1,\hfill \\ \hfill {\mathcal{C}}_{3,3}^{\left(k\right)}& =\frac{1}{4^k},\hfill \\ \hfill {\mathcal{C}}_{6,3}^{\left(k\right)}& =-\left(\genfrac{}{}{0pt}{}{6}{4}\right)\frac{1}{4^k}+\frac{1}{7^k},\hfill \\ \hfill {\mathcal{C}}_{9,3}^{\left(k\right)}& =\frac{3(35·1+79)}{8}\left(\genfrac{}{}{0pt}{}{9}{7}\right)\frac{1}{4^k}-\left(\genfrac{}{}{0pt}{}{9}{4}\right)\frac{1}{7^k}+\frac{1}{{10}^k},\hfill \\ \hfill {\mathcal{C}}_{12,3}^{\left(k\right)}& =-\frac{9·22·153}{4}\left(\genfrac{}{}{0pt}{}{12}{10}\right)\frac{1}{4^k}+\frac{3(35·2+79)}{8}\left(\genfrac{}{}{0pt}{}{12}{7}\right)\frac{1}{7^k}-\left(\genfrac{}{}{0pt}{}{12}{4}\right)\frac{1}{{10}^k}+\frac{1}{{13}^k}.\hfill \end{array}$$

For $s=4$, since

$$\begin{array}{cc}\hfill & \frac{{\left({\mathfrak{AF}}_4\left(x\right)\right)}^{4m}}{\left(4m\right)!}\hfill \\ \hfill & =\frac{x^{4m}}{\left(4m\right)!}-\left(\genfrac{}{}{0pt}{}{4m+4}{5}\right)\frac{x^{4m+4}}{(4m+4)!}+\frac{2(126m+281)}{5}\left(\genfrac{}{}{0pt}{}{4m+8}{9}\right)\frac{x^{4m+8}}{(4m+8)!}\hfill \\ \hfill & -\frac{8(6006m^2+40183m+67157)}{5}\left(\genfrac{}{}{0pt}{}{4m+12}{13}\right)\frac{x^{4m+12}}{(4m+12)!}\hfill \\ \hfill & +\frac{16(12864852m^3+172143972m^2+767355367m+1139488217)}{45}\left(\genfrac{}{}{0pt}{}{4m+16}{17}\right)\frac{x^{4m+16}}{(4m+16)!}-\cdots ,\hfill \end{array}$$

we have

$$\begin{array}{cc}\hfill {\mathcal{C}}_{0,4}^{\left(k\right)}& =1,\hfill \\ \hfill {\mathcal{C}}_{4,4}^{\left(k\right)}& =\frac{1}{5^k},\hfill \\ \hfill {\mathcal{C}}_{8,4}^{\left(k\right)}& =-\left(\genfrac{}{}{0pt}{}{8}{5}\right)\frac{1}{5^k}+\frac{1}{9^k},\hfill \\ \hfill {\mathcal{C}}_{12,4}^{\left(k\right)}& =\frac{2(126·1+281)}{5}\left(\genfrac{}{}{0pt}{}{12}{9}\right)\frac{1}{5^k}-\left(\genfrac{}{}{0pt}{}{12}{5}\right)\frac{1}{9^k}+\frac{1}{{13}^k},\hfill \\ \hfill {\mathcal{C}}_{16,4}^{\left(k\right)}& =-\frac{8(6006·1^2+40183·1+67157)}{5}\left(\genfrac{}{}{0pt}{}{16}{13}\right)\frac{1}{5^k}\hfill \\ \hfill & +\frac{2(126·2+281)}{5}\left(\genfrac{}{}{0pt}{}{16}{9}\right)\frac{1}{9^k}-\left(\genfrac{}{}{0pt}{}{16}{5}\right)\frac{1}{{13}^k}+\frac{1}{{17}^k}.\hfill \end{array}$$

4. Iterated Integrals

Similarly to the cases of the poly-Cauchy numbers with levels 1 and 2 ([6,15]), Cauchy numbers with higher levels have an expression in terms of iterated integrals.

Since

$${\displaystyle \frac{d}{dz}}\left(z{\mathrm{Lif}}_{s,k}\left(z\right)\right)={\mathrm{Lif}}_{s,k-1}\left(z\right),$$

we have

$${\displaystyle \frac{d}{dz}}\left(z{\mathrm{Lif}}_{s,k}\left(z\right)\right)={\displaystyle \frac{d}{dz}}\left(\sum _{n=0}^{\infty }{\displaystyle \frac{z^{3n+1}}{\left(sn\right)!{(sn+1)}^k}}\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{z^{sn}}{\left(sn\right)!{(sn+1)}^{k-1}}}={\mathrm{Lif}}_{s,k-1}\left(z\right).$$

Therefore,

$${\mathrm{Lif}}_{s,k-1}\left(z\right)={\displaystyle \frac{1}{z}}{\int }_0^z{\mathrm{Lif}}_{s,k-1}\left(z\right)dz.$$

By iteration, we get

$${\mathrm{Lif}}_{s,k}\left(z\right)=\underset{k-1}{\underbrace{{\displaystyle \frac{1}{z}}{\int }_0^z{\displaystyle \frac{1}{z}}{\int }_0^z\cdots {\displaystyle \frac{1}{z}}{\int }_0^z}}{\mathrm{Lif}}_{s,1}\left(z\right)\underset{k-1}{\underbrace{dz\cdots dz}}.$$

Putting $z={\mathfrak{AF}}_s\left(t\right)$, we get

$${\mathrm{Lif}}_{s,k}\left({\mathfrak{AF}}_s\left(t\right)\right)={\displaystyle \frac{1}{{\mathfrak{AF}}_s\left(t\right)}}\underset{k-1}{\underbrace{{\int }_0^t{\displaystyle \frac{{\mathfrak{G}}_s\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}\cdots {\int }_0^t}}{\displaystyle \frac{t{\mathfrak{G}}_s\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}\underset{k-1}{\underbrace{dt\cdots dt}},$$

where

$$\begin{array}{cc}\hfill {\mathfrak{G}}_s\left(z\right)& =\frac{d}{dz}{\mathfrak{AF}}_s\left(z\right)=\sum _{n=0}^{\infty }{(-1)}^n(sn+1)d_n{z}^{sn}\hfill \\ \hfill & =1-\frac{1}{s!}{z}^s+\frac{1}{\left(2s\right)!}\left(\left(\genfrac{}{}{0pt}{}{2s+1}{s}\right)-1\right)z^{2s}\hfill \\ \hfill & -\frac{1}{\left(3s\right)!}\left(\frac{1}{2}\left(\genfrac{}{}{0pt}{}{3s+2}{s+1,s+1,s}\right)-\left(\genfrac{}{}{0pt}{}{3s+2}{s+1}\right)+1\right)z^{3s}\hfill \\ \hfill & +\frac{1}{\left(4s\right)!}\left(\frac{1}{6}\left(\genfrac{}{}{0pt}{}{4s+3}{s+1,s+1,s+1,s}\right)+\left(\genfrac{}{}{0pt}{}{4s+1}{2s}\right)\right.\hfill \\ \hfill & \left.-\frac{1}{2}\left(\genfrac{}{}{0pt}{}{4s+3}{2s+1,s+1,s+1}\right)+\left(\genfrac{}{}{0pt}{}{4s+2}{s+1}\right)-1\right)z^{4s}-\cdots ,\hfill \end{array}$$

where $\left(\genfrac{}{}{0pt}{}{n}{s_1,\cdots ,s_m}\right)=\frac{n!}{\left(s_1\right)!\cdots \left(s_m\right)!}$ denotes the multinomial coefficient with $n=s_1+\cdots +s_m$.

Moreover we can express the Laurent series of ${\mathfrak{G}}_s\left(t\right)/{\mathfrak{AF}}_s\left(t\right)$, in fact,

$${\displaystyle \frac{{\mathfrak{G}}_s\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}={\displaystyle \frac{{\mathfrak{AF}}_s^{\prime }\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}$$

with ${\mathfrak{AF}}_s\left(t\right)=t{\mathfrak{D}}_s\left(t\right)$ and ${\mathfrak{D}}_s\left(0\right)\ne 0$. Hence

$${\displaystyle \frac{{\mathfrak{G}}_s\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}={\displaystyle \frac{{\mathfrak{D}}_s\left(t\right)+t{\mathfrak{D}}_s^{\prime }\left(t\right)}{t{\mathfrak{D}}_s\left(t\right)}}={\displaystyle \frac{1}{t}}+{\displaystyle \frac{{\mathfrak{D}}_s^{\prime }\left(t\right)}{{\mathfrak{D}}_s\left(t\right)}}={\displaystyle \frac{1}{t}}+\frac{d}{dt}\log {\mathfrak{D}}_s\left(t\right).$$

From (3), we have

$${\mathfrak{D}}_s\left(t\right)=d_0-d_1{t}^s+d_2{t}^{2s}-d_3{t}^{3s}+\cdots =d_0+u\left(t\right),$$

with $d_0=1$ and $u\left(t\right)=-d_1{t}^s+d_2{t}^{2s}-d_3{t}^{3s}+\cdots$. So,

$$\log {\mathfrak{D}}_s\left(t\right)=\log (1+u\left(t\right))=\sum _{k=1}^{\infty }{(-1)}^{k-1}{\displaystyle \frac{u{\left(t\right)}^k}{k}}$$

Therefore,

$$\begin{array}{c}\hfill \log {\mathfrak{D}}_s\left(t\right)=(-d_1{t}^s+d_2{t}^{2s}-d_3{t}^{3s}+\cdots )-{\displaystyle \frac{1}{2}}{(-d_1{t}^s+d_2{t}^{2s}-d_3{t}^{3s}+\cdots )}^2+\\ \hfill +{\displaystyle \frac{1}{3}}{(-d_1{t}^s+d_2{t}^{2s}-d_3{t}^{3s}+\cdots )}^3+\cdots .\end{array}$$

So, it follows that

$$\begin{array}{c}\hfill \log {\mathfrak{D}}_s\left(t\right)=-d_1{t}^s+\left(d_2-{\displaystyle \frac{d_1^2}{2}}\right)t^{2s}+\left(-d_3+d_1{d}_2-{\displaystyle \frac{d_1^3}{3}}\right)t^{3s}\\ \hfill +\left(d_4-d_1{d}_3-{\displaystyle \frac{d_2^2}{2}}+d_1^2{d}_2-{\displaystyle \frac{d_1^4}{4}}\right)t^{4s}+\cdots ,\end{array}$$

yielding the expression

$$\begin{array}{c}\hfill \frac{d}{dt}\log {\mathfrak{D}}_s\left(t\right)=-s{d}_1{t}^{s-1}+2s\left(d_2-{\displaystyle \frac{d_1^2}{2}}\right)t^{2s-1}+3s\left(-d_3+d_1{d}_2-{\displaystyle \frac{d_1^3}{3}}\right)t^{3s-1}\\ \hfill +4s\left(d_4-d_1{d}_3-{\displaystyle \frac{d_2^2}{2}}+d_1^2{d}_2-{\displaystyle \frac{d_1^4}{4}}\right)t^{4s-1}+\cdots .\end{array}$$

After substituting the vales of $d_n$, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\mathfrak{G}}_s\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}& ={\displaystyle \frac{1}{t}}-{\displaystyle \frac{s}{(s+1)!}}{t}^{s-1}+2s\left({\displaystyle \frac{2s+1}{2{\left((s+1)!\right)}^2}}-\frac{1}{(2s+1)!}\right)t^{2s-1}\hfill \\ \hfill & +3s\left(\frac{3s+1}{(2s+1)!(s+1)!}-\frac{(3s+2)(3s+1)}{6{\left((s+1)!\right)}^3}-\frac{1}{(3s+1)!}\right)t^{3s-1}\hfill \\ \hfill & +4s\left(\frac{4s+1}{(3s+1)!(s+1)!}-\frac{4s+1}{\left(2s\right)!{\left((s+1)!\right)}^2}+\frac{4s+1}{2{\left((2s+1)!\right)}^2}\right.\hfill \\ \hfill & \left.+\frac{(4s+3)(4s+1)(2s+1)}{12{\left((s+1)!\right)}^4}-\frac{1}{(4s+1)!}\right)t^{4s-1}+\cdots .\hfill \end{array}$$

(5)

Proposition 2.

We have

$$\sum _{n=0}^{\infty }{\mathcal{C}}_{n,s}^{\left(k\right)}\frac{t^n}{n!}={\displaystyle \frac{1}{{\mathfrak{AF}}_s\left(t\right)}}\underset{k-1}{\underbrace{{\int }_0^t{\displaystyle \frac{{\mathfrak{G}}_s\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}\cdots {\int }_0^t}}{\displaystyle \frac{t{\mathfrak{G}}_s\left(t\right)}{{\mathfrak{AF}}_s\left(t\right)}}\underset{k-1}{\underbrace{dt\cdots dt}},$$

where ${\mathfrak{G}}_s\left(z\right)=\frac{d}{dz}{\mathfrak{AF}}_s\left(z\right)$ and a more precise expression of ${\mathfrak{G}}_s\left(t\right)/{\mathfrak{AF}}_s\left(t\right)$ is given in (5).

5. An Explicit Expression

If we know the coefficients $d_n$ ($n\ge 0$) appeared in ${\mathfrak{AF}}_s\left(t\right)$ in Proposition 1, we can get an expression of ${\mathcal{C}}_{n,s}^{\left(k\right)}$.

Theorem 1.

For integers n and k with $n\ge 0$,

$${\mathcal{C}}_{sn,s}^{\left(k\right)}=\sum _{m=0}^n\frac{{(-1)}^{n-m}\left(sn\right)!}{\left(sm\right)!{(sm+1)}^k}\sum _{\genfrac{}{}{0pt}{}{i_1+\cdots +i_{sm}=n-m}{i_1,\cdots ,i_{sm}\ge 0}}{d}_{i_1}\cdots d_{i_{sm}}{t}^{sn}.$$

Proof. 

By the definition in (1), we have

$$\begin{array}{cc}\hfill \sum _{n=0}^{\infty }{\mathcal{C}}_{n,s}^{\left(k\right)}\frac{t^n}{n!}& =\sum _{n=0}^{\infty }{\mathcal{C}}_{sn,s}^{\left(k\right)}\frac{t^{sn}}{\left(sn\right)!}\hfill \\ \hfill & =\sum _{m=0}^{\infty }\frac{1}{\left(sm\right)!{(sm+1)}^k}{\left(\sum _{l=0}^{\infty }{(-1)}^l{d}_l{t}^{sl+1}\right)}^{sm}\hfill \\ \hfill & =\sum _{m=0}^{\infty }\frac{1}{\left(sm\right)!{(sm+1)}^k}\hfill \\ \hfill & \times \sum _{n=m}^{\infty }\sum _{\genfrac{}{}{0pt}{}{i_1+\cdots +i_{sm}=n-m}{i_1,\cdots ,i_{sm}\ge 0}}{(-1)}^{i_1+\cdots +i_{sm}}{d}_{i_1}\cdots d_{i_{sm}}{t}^{(s{i}_1+1)+\cdots +(s{i}_{sm}+1)}\hfill \\ \hfill & =\sum _{n=0}^{\infty }\sum _{m=0}^n\frac{1}{\left(sm\right)!{(sm+1)}^k}\sum _{\genfrac{}{}{0pt}{}{i_1+\cdots +i_{sm}=n-m}{i_1,\cdots ,i_{sm}\ge 0}}{(-1)}^{n-m}{d}_{i_1}\cdots d_{i_{sm}}{t}^{sn}.\hfill \end{array}$$

Comparing the coefficients on both sides, we get the desired result. □

6. Some Expressions of Poly-Cauchy Numbers with Higher Levels for Negative Indices

The poly-Bernoulli numbers ${\mathbb{B}}_n^{\left(k\right)}$ [8], defined by

$$\frac{{\mathrm{Li}}_k(1-e^{-t})}{1-e^{-t}}=\sum _{n=0}^{\infty }{\mathbb{B}}_n^{(-k)}\frac{t^n}{n!},$$

where

$${\mathrm{Li}}_k\left(z\right)=\sum _{n=1}^{\infty }\frac{z^n}{n^k}$$

is the polylogarithm function, satisfy the duality formula ${\mathbb{B}}_n^{(-k)}={\mathbb{B}}_k^{(-n)}$ for $n,k>0$, because of the symmetric formula

$$\sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{\mathbb{B}}_n^{(-k)}\frac{x^n}{n!}\frac{y^k}{k!}=\frac{e^{x+y}}{e^x+e^y-e^{x+y}}.$$

Though the corresponding duality formula does not hold for the original poly-Cauchy numbers (ref. [16], Proposition 1) and poly-Cauchy numbers with level 2 (ref. [6], Theorem 4.1), we still have the double summation formula of poly-Cauchy numbers with higher level.

Theorem 2.

For nonnegative integers n and k,

$$\sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{\mathcal{C}}_{sn,s}^{(-sk)}\frac{x^{sn}}{\left(sn\right)!}\frac{y^{sk}}{\left(sk\right)!}=\frac{1}{s^2}\sum _{j=0}^{s-1}\sum _{h=0}^{s-1}{e}^{{\zeta }^{j}y}{\left({\mathfrak{BF}}_s\left(x\right)\right)}^{{\zeta }^h{e}^{{\zeta }^{j}y}},$$

where ${\mathfrak{BF}}_s\left(x\right)=e^{{\mathfrak{AF}}_s\left(x\right)}$ and ζ is the s-th root of unity as $\zeta =e^{2\pi i/s}=\cos (2\pi /s)+i\sin (2\pi /s)$.

Proof. 

From the definition in (1), we have

$$\begin{array}{cc}\hfill & \sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{\mathcal{C}}_{sn,s}^{(-sk)}\frac{x^{sn}}{\left(sn\right)!}\frac{y^{sk}}{\left(sk\right)!}=\sum _{k=0}^{\infty }{\mathrm{Lif}}_{s,k}\left({\mathfrak{AF}}_s\left(x\right)\right)\frac{y^{sk}}{\left(sk\right)!}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\sum _{m=0}^{\infty }\frac{{\left({\mathfrak{AF}}_s\left(x\right)\right)}^{sm}}{\left(sm\right)!}{(sm+1)}^{sk}\frac{y^{sk}}{\left(sk\right)!}\hfill \\ \hfill & =\sum _{m=0}^{\infty }\frac{{\left({\mathfrak{AF}}_s\left(x\right)\right)}^{sm}}{\left(sm\right)!}\frac{1}{s}\sum _{j=0}^{s-1}{e}^{{\zeta }^j(sm+1)y}\hfill \\ \hfill & =\frac{1}{s}\sum _{j=0}^{s-1}{e}^{{\zeta }^{j}y}\sum _{m=0}^{\infty }\frac{e^{{\zeta }^{j}y}{\mathfrak{AF}}_s\left(x\right)}{\left(sm\right)!}\hfill \\ \hfill & =\frac{1}{s^2}\sum _{j=0}^{s-1}\sum _{h=0}^{s-1}{e}^{{\zeta }^{j}y}{e}^{{\zeta }^h{e}^{{\zeta }^{j}y}{\mathfrak{AF}}_s\left(x\right)},\hfill \end{array}$$

yielding the desired result. □

7. Cauchy Numbers with Higher Level

When $k=1$ in (1), ${\mathcal{C}}_{n,s}={\mathcal{C}}_{n,s}^{\left(1\right)}$ are the Cauchy numbers with higher level, defined by

$$\frac{t}{{\mathfrak{AF}}_s\left(t\right)}=\sum _{n=0}^{\infty }{\mathcal{C}}_{n,s}\frac{t^n}{n!}.$$

(6)

In this section, we shall show some properties of ${\mathcal{C}}_{n,s}={\mathcal{C}}_{n,s}^{\left(1\right)}$. First, we give its determinant expression. A similar expression for the hypergeometric Cauchy numbers is given in [18].

Theorem 3.

For $n\ge 1$,

$${\mathcal{C}}_{sn,s}=\left(sn\right)!\left|\begin{array}{ccccc}{d}_1& 1& 0& & \\ d_2& d_1& 1& & \\ ⋮& & \ddots & \ddots & 0\\ ⋮& & & d_1& 1\\ d_n& \cdots & \cdots & d_2& d_1\end{array}\right|,$$

where $d_n$ is the coefficient of $t^{sn+1}$ appeared in ${\mathfrak{AF}}_s\left(t\right)$ in Proposition 1.

Remark 1.

By using the values of d’s in (4), Theorem 3 yields

$$\begin{array}{cc}\hfill {\mathcal{C}}_{0,s}& =1,{\mathcal{C}}_{s,s}=\frac{1}{s+1},{\mathcal{C}}_{2s,s}=\frac{1}{2s+1}-\frac{s\left(2s\right)!}{{\left((s+1)!\right)}^2},\hfill \\ \hfill {\mathcal{C}}_{3s,s}& =\frac{1}{3s+1}-\frac{3s\left(3s\right)!}{(2s+1)!(s+1)!}+\frac{s(3s+1)!}{2{\left((s+1)!\right)}^3},\hfill \\ \hfill {\mathcal{C}}_{4s,s}& =\frac{1}{4s+1}-\frac{4s\left(4s\right)!}{(3s+1)!(s+1)!}-\frac{(8s+3)\left(4s\right)!}{(2s+1)!{\left((s+1)!\right)}^2}\hfill \\ \hfill & -\frac{2s\left(4s\right)!}{(2s+1)!{\left((s+1)!\right)}^2}+\frac{(4s+3)\left(4s\right)!}{\left(2s\right)!{\left(s+1)!\right)}^2}-\frac{s(8s^2+6s+1)\left(4s\right)!}{{\left((s+1)!\right)}^4},\cdots .\hfill \end{array}$$

Proof of Theorem 3.

From (6), we have

$$\begin{array}{cc}\hfill 1& =\left(\sum _{m=0}^{\infty }{\mathcal{C}}_{sm,s}\frac{t^{sm}}{\left(sm\right)!}\right)\left(\sum _{l=0}^{\infty }{(-1)}^l{d}_l{t}^{sl}\right)\hfill \\ \hfill & =\sum _{n=0}^{\infty }\sum _{l=0}^n\frac{{\mathcal{C}}_{sn-sl,s}}{(sn-sl)!}{(-1)}^l{d}_l{t}^{sn}.\hfill \end{array}$$

where the coefficients $d_0,d_1,\dots$ are also given in (3) with (4). Comparing the coefficients on both sides,

$$\sum _{l=0}^n\frac{{\mathcal{C}}_{sn-sl,s}}{(sn-sl)!}{(-1)}^l{d}_l=0(n\ge 1).$$

By the inversion relation

$$\begin{gathered} \begin{array}{cc}& \sum _{k=0}^n{(-1)}^{n-k}{\alpha }_{k}R(n-k)=0(n\ge 1)with{\alpha }_0=R\left(0\right)=1\hfill \\ ⟺\hfill & \end{array} \\ {\alpha }_n=\left|\begin{array}{cccc}R\left(1\right)& 1& & 0\\ R\left(2\right)& \ddots & \ddots & \\ ⋮& \ddots & \ddots & 1\\ R\left(n\right)& \cdots & R\left(2\right)& R\left(1\right)\end{array}\right|⟺R\left(n\right)=\left|\begin{array}{cccc}{\alpha }_1& 1& & 0\\ {\alpha }_2& \ddots & \ddots & \\ ⋮& \ddots & \ddots & 1\\ {\alpha }_n& \cdots & {\alpha }_2& {\alpha }_1\end{array}\right| \end{gathered}$$

(e.g., see [19]), we get the result as

$${\alpha }_n=d_n\mathrm{and}R\left(n\right)=\frac{{\mathcal{C}}_{sn,s}}{\left(sn\right)!}.$$

□

By the inversion formula shown in the above proof, we also have the following Corollary. Similar determinant expressions of Bernoulli, Cauchy and related numbers were found in [20]).

Corollary 1.

For $n\ge 1$,

$$d_n=\left|\begin{array}{cccc}\frac{{\mathcal{C}}_{s,s}}{s!}& 1& & 0\\ \frac{{\mathcal{C}}_{2s,s}}{\left(2s\right)!}& \frac{{\mathcal{C}}_{3s,s}}{\left(3s\right)!}& & \\ ⋮& & \ddots & 1\\ \frac{{\mathcal{C}}_{sn,s}}{\left(sn\right)!}& \cdots & \frac{{\mathcal{C}}_{2s,s}}{\left(2s\right)!}& \frac{{\mathcal{C}}_{s,s}}{s!}\end{array}\right|.$$

By Trudi’s formula

$$\begin{array}{cc}\hfill \left|\begin{array}{ccccc}{a}_1& a_2& \cdots & \cdots & a_m\\ a_0& a_1& \ddots & & ⋮\\ & \ddots & \ddots & \ddots & ⋮\\ & & \ddots & a_1& a_2\\ 0& & & a_0& a_1\end{array}\right|& \\ \hfill =& \sum _{t_1+2t_2+\cdots +m{t}_m=m}\left(\genfrac{}{}{0pt}{}{t_1+\cdots +t_m}{t_1,\cdots ,t_m}\right){(-a_0)}^{m-t_1-\cdots -t_m}{a}_1^{t_1}{a}_2^{t_2}\cdots a_m^{t_m}\hfill \end{array}$$

(refs. [21,22]; ref. [23], Volume 3, pp. 208–209, p. 214), we have a different expression of ${\mathcal{C}}_{n,s}$.

Theorem 4.

$$\begin{array}{c}\hfill {\mathcal{C}}_{sn,s}=\left(sn\right)!\sum _{t_1+2t_2+\cdots +n{t}_n=n}\left(\genfrac{}{}{0pt}{}{t_1+\cdots +t_n}{t_1,\cdots ,t_n}\right){(-1)}^{n-t_1-\cdots -t_n}{\left(d_1\right)}^{t_1}{\left(d_2\right)}^{t_2}\cdots {\left(d_n\right)}^{t_n}\end{array}$$

and

$$\begin{array}{cc}\hfill d_n=\sum _{t_1+2t_2+\cdots +n{t}_n=n}\left(\genfrac{}{}{0pt}{}{t_1+\cdots +t_n}{t_1,\cdots ,t_n}\right){(-1)}^{n-t_1-\cdots -t_n}& \\ \hfill \times & {\left(\frac{{\mathcal{C}}_{s,s}}{s!}\right)}^{t_1}{\left(\frac{{\mathcal{C}}_{2s,s}}{\left(2s\right)!}\right)}^{t_2}\cdots {\left(\frac{{\mathcal{C}}_{sn,s}}{\left(sn\right)!}\right)}^{t_n}.\hfill \end{array}$$

8. A Recurrence Relation for ${\mathcal{C}}_{n,s}^{\left(k\right)}$ in Terms of ${\mathcal{C}}_{n,s}$

We can show a recurrence formula for ${\mathcal{C}}_{n,s}^{\left(k\right)}$ in terms of ${\mathcal{C}}_{n,s}^{(k-1)}$ and ${\mathcal{C}}_{n,s}$.

Theorem 5.

For integers n and k with $n\ge 0$ and $k\ge 1$,

$${\mathcal{C}}_{sn,s}^{\left(k\right)}=\left(sn\right)!\sum _{\nu =0}^n\sum _{m=0}^{\nu }\frac{{(-1)}^{\nu -m}(s\nu -sm+1)d_{\nu -m}{\mathcal{C}}_{sn-s\nu ,s}{\mathcal{C}}_{sm,s}^{(k-1)}}{(sn-s\nu )!\left(sm\right)!(s\nu +1)},$$

where $d_n$ is the coefficient of $t^{sn+1}$ appeared in ${\mathfrak{AF}}_s\left(t\right)$ in Proposition 1.

Remark 2.

Poly-Cauchy numbers $c_n^{\left(k\right)}$ have a recurrence formula (ref. [16], Theorem 7)

$$c_n^{\left(k\right)}=n!\sum _{\nu =0}^n\sum _{m=0}^{\nu }\frac{{(-1)}^{\nu -m}{c}_{n-\nu }{c}_m^{(k-1)}}{(n-\nu )!m!(\nu +1)}.$$

Poly-Cauchy numbers ${\mathbb{C}}_n^{\left(k\right)}$ with level 2 have a recurrence formula (ref. [6], Theorem 3.4)

$${\mathbb{C}}_{2n}^{\left(k\right)}=\left(2n\right)!\sum _{\nu =0}^n\sum _{m=0}^{\nu }{\left(-\frac{1}{4}\right)}^{\nu -m}\left(\genfrac{}{}{0pt}{}{2\nu -2m}{\nu -m}\right)\frac{{\mathbb{C}}_{2n-2\nu }{\mathbb{C}}_{2m}^{(k-1)}}{(2n-2\nu )!\left(2m\right)!(2\nu +1)}.$$

Proof of Theorem 5.

Similarly to the description in Section 4, we obtain

$$\begin{array}{cc}\hfill & \sum _{n=0}^{\infty }{\mathcal{C}}_{sn,s}^{\left(k\right)}\frac{x^{sn}}{\left(sn\right)!}={\mathrm{Lif}}_{s,k}\left({\mathfrak{AF}}_s\left(x\right)\right)\hfill \\ \hfill & =\frac{1}{{\mathfrak{AF}}_s\left(x\right)}{\int }_0^x{\mathrm{Lif}}_{s,k-1}\left({\mathfrak{AF}}_s\left(\sigma \right)\right){\mathfrak{G}}_s\left(\sigma \right)d\sigma \hfill \\ \hfill & =\left(\sum _{n=0}^{\infty }{\mathcal{C}}_{sn,s}\frac{x^{sn-1}}{\left(sn\right)!}\right){\int }_0^x\left(\sum _{m=0}^{\infty }{\mathcal{C}}_{sm,s}^{(k-1)}\frac{{\sigma }^{sm}}{\left(sm\right)!}\right)\left(\sum _{j=0}^{\infty }{(-1)}^j(sj+1)d_j{\sigma }^{rj}\right)d\sigma \hfill \\ \hfill & =\left(\sum _{n=0}^{\infty }{\mathcal{C}}_{sn,s}\frac{x^{sn-1}}{\left(sn\right)!}\right){\int }_0^x\left(\sum _{\nu =0}^{\infty }\sum _{m=0}^{\nu }{(-1)}^{\nu -m}(s\nu -sm+1)d_{\nu -m}\frac{{\mathcal{C}}_{sm,s}^{(k-1)}}{\left(sm\right)!}{\sigma }^{s\nu }\right)d\sigma \hfill \\ \hfill & =\left(\sum _{n=0}^{\infty }{\mathcal{C}}_{sn,s}\frac{x^{sn-1}}{\left(sn\right)!}\right)\left(\sum _{\nu =0}^{\infty }\sum _{m=0}^{\nu }{(-1)}^{\nu -m}(s\nu -sm+1)d_{\nu -m}\frac{{\mathcal{C}}_{sm,s}^{(k-1)}}{\left(sm\right)!}\frac{x^{s\nu +1}}{s\nu +1}\right)\hfill \\ \hfill & =\sum _{n=0}^{\infty }\sum _{\nu =0}^n\sum _{m=0}^{\nu }\frac{{(-1)}^{\nu -m}(s\nu -sm+1)d_{\nu -m}{\mathcal{C}}_{sn-s\nu ,s}{\mathcal{C}}_{sm,s}^{(k-1)}}{(sn-s\nu )!\left(sm\right)!(s\nu +1)}{x}^{sn}.\hfill \end{array}$$

Comparing the coefficients on both sides, we get the result. □

9. Conclusions

In this paper, we define poly-Cauchy numbers with higher level (level s) from the analytical implications, and study their properties. Such poly-Cauchy numbers with higher levels are yielded from the inverse relationship with an s-step function of the exponential function. When $s\ge 3$, the inverse function is not given using a known function, but it can be used to obtain the expressions and relations.

Poly-Bernoulli numbers with level 2 are defined and studied in [7]. Is it possible to introduce poly-Bernoulli numbers with higher levels? If so, is there any relation between them and poly-Cauchy numbers with higher levels?