On q-generalized (r, s)-Stirling numbers

Abstract

In this paper by using q-numbers and r-shift, the (q, r)-Stirling numbers with level s are studied. One of the main aims is to give several identities in their transforms. We also give some applications to the values of a certain kind of q-multiple zeta functions.

1 Introduction

There are many types of q-generalizations of Stirling numbers of both kinds. The q-Stirling numbers satisfy certain recurrence relations and generating functions that are similar to those of the classical Stirling numbers. In addition, there are also many different definitions and notation. We refer the paper [4] and references therein. Combinatorial interpretations of the q-Stirling numbers have been given in [3]. The q-Stirling numbers have important applications in various areas of mathematics and physics, including combinatorics, representation theory, and statistical mechanics. They arise in the study of partition functions, quantum groups, and vertex operator algebras, among other topics. In addition, (q, t)-generalizations are introduced in [3] and developed in [9]. Apart from q-generalizations, r-Stirling numbers are introduced in [2] and are developed by many researchers. They are given combinatorically with a condition on r, and functionally shift by r.

Recently, in [6] and [7], Stirling numbers of the first kind with higher level and those of the second kind are studied, respectively. These Stirling numbers are originally introduced in Tweedie [14], but his works have been ignored or forgotten for a long time. In [5], the transform by Stirling numbers with higher level is studied by extending the results in [13]. The transformations by Stirling numbers allow us to convert between different combinatorial structures. This makes it possible to apply the properties and identities of one set of Stirling numbers to another, which can simplify calculations and provide insights into the structure of the objects being studied. In this paper, we consider a q-generalization of the r-Stirling numbers of both kinds with higher level, denoted by

$$\begin{aligned} \left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}\quad \hbox {and}\quad \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}, \end{aligned}$$

respectively. If $r=s=1$,

$$\begin{aligned} \left[ n\atop k\right] _q=\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(1,1)}\quad \hbox {and}\quad \left\{ n\atop k\right\} _q=\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(1,1)} \end{aligned}$$

are the q-Stirling numbers of both kinds, respectively. If $r=1$ and $q\rightarrow 1$,

$$\begin{aligned} \left[ \!\!\left[ n\atop k\right] \!\!\right] ^{(s)}=\left[ \!\!\left[ n\atop k\right] \!\!\right] _1^{(1,s)}\quad \hbox {and}\quad \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} ^{(s)}=\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _1^{(1,s)} \end{aligned}$$

are the Stirling numbers of both kinds with higher level (level s), respectively.¹ If $r=s=1$ and $q\rightarrow 1$,

$$\begin{aligned} \left[ n\atop k\right] =\left[ \!\!\left[ n\atop k\right] \!\!\right] _1^{(1,1)}\quad \hbox {and}\quad \left\{ n\atop k\right\} =\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _1^{(1,1)} \end{aligned}$$

are the classical Stirling numbers of both kinds, respectively. In addition, if $s=1$ and $q\rightarrow 1$, the r-Stirling numbers of both kinds ( [2]) are reduced. In such a generalized case, we consider the transformation by a q-generalization of the r-the Stirling numbers of the first kind with higher level:

$$\begin{aligned} b_n=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}a_k \end{aligned}$$

with its inverse

$$\begin{aligned} a_n=\sum _{k=r}^n(-1)^{n-k}\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}b_k, \end{aligned}$$

and vice versa, that comes from the orthogonal relation.

As an application, we give several properties of a certain kind of q-multiple zeta functions.

2 Preliminaries

Let $[n]_q$ denote the q-number, defined by

$$\begin{aligned}{}[n]_q=\frac{q^n-1}{q-1}\quad (q\ne 1). \end{aligned}$$

Its q-factorial is given by $[n]_q!=[n]_q[n-1]_q\cdots [1]_q$ with $[0]_q!=1$. Let r be a positive integer. The q-version of r-Stirling numbers of the first kind with higher level are denoted by $\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}$, and appear in the coefficients in the expansion

$$\begin{aligned} (x)_{n,q}^{(r,s)}=\sum _{k=0}^n(-1)^{n-k}\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}x^k, \end{aligned}$$

(1)

where for $r,s\ge 1$, $(x)_{n,q}^{(r,s)}$ is defined by

$$\begin{aligned} (x)_{n,q}^{(r,s)}:=x^r\prod _{i=r}^{n-1}\bigl (x-([i]_q)^s\bigr )\quad (n>r) \end{aligned}$$

with $(x)_{r,q}^{(r,s)}=x^r$. When $r=s=1$, $s_q(n,k)=(-1)^{n-k}\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(1,1)}$ are the signed q-Stirling numbers of the first kind (see, e.g., [4]), and $\left[ \!\!\left[ n\atop k\right] \!\!\right] _q=\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(1,1)}$ are the unsigned q-Stirling numbers of the first kind. When $r=1$ and $q\rightarrow 1$, $\left[ \!\!\left[ n\atop k\right] \!\!\right] ^{(s)}=\left[ \!\!\left[ n\atop k\right] \!\!\right] _1^{(1,s)}$ are the (unsigned) Stirling numbers of the first kind with higher level ( [6]). When $r=s=1$ and $q\rightarrow 1$, $\left[ n\atop k\right] =\left[ \!\!\left[ n\atop k\right] \!\!\right] _1^{(1,1)}$ are the unsigned Stirling numbers of the first kind.

The q-version of r-Stirling numbers of the second kind with higher level are denoted by $\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}$, and appear in the coefficients in the expansion

$$\begin{aligned} x^n=\sum _{k=0}^n\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}(x)_{k,q}^{(r,s)}. \end{aligned}$$

(2)

When $r=1$ and $q\rightarrow 1$, $\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} ^{(s)}=\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(1,s)}$ are the Stirling numbers of the second kind with higher level, studied in [7]. When $r=s=1$, $S_q(n,k)=\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q=(-1)^{n-k}\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(1,1)}$ are the signed q-Stirling numbers of the second kind (see, e.g., [4]). When $r=s=1$ and $q\rightarrow 1$, $\left\{ n\atop k\right\} =\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _1^{(1,1)}$ are the classical Stirling numbers of the second kind.

From the definitions in (1) and (2), the following orthogonal relations are yielded.

Theorem 1

$$\begin{aligned} {}&\sum _{k=0}^{\max \{n,m\}}(-1)^{n-k}\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}\left\{ \!\!\left\{ k\atop m\right\} \!\!\right\} _q^{(r,s)}=\delta _{n,m}\,, \end{aligned}$$

(3)

$$\begin{aligned} {}&\sum _{k=0}^{\max \{n,m\}}(-1)^{k-m}\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}\left[ \!\!\left[ k\atop m\right] \!\!\right] _q^{(r,s)}=\delta _{n,m}\,, \end{aligned}$$

(4)

where $\delta _{n,m}$ is the Kronecker delta.

Proof of Theorem 1

By (1) and (2), we have

$$\begin{aligned} x^n&=\sum _{k=r}^n\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}(x)_{k,q}^{(r,s)}\\&=\sum _{k=r}^n\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}\sum _{m=0}^k(-1)^{k-m}\left[ \!\!\left[ k\atop m\right] \!\!\right] _q^{(r,s)} x^m\\&=\sum _{m=0}^n\sum _{k=m}^n(-1)^{k-m}\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}\left[ \!\!\left[ k\atop m\right] \!\!\right] _q^{(r,s)} x^m\,. \end{aligned}$$

Comparing the coefficients on both sides, we obtain

$$\begin{aligned} {}&{} \sum _{k=m}^n(-1)^{k-m}\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}\left[ \!\!\left[ k\atop m\right] \!\!\right] _q^{(r,s)} =\delta _{n,m}, \end{aligned}$$

which is (4). Another identity (3) is yielded from (4). $\square$

3 First kind

The recurrence relation is given by

$$\begin{aligned} \left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}=\left[ \!\!\left[ n-1\atop k-1\right] \!\!\right] _q^{(r,s)}+\bigl ([n-1]_q\bigr )^s\left[ \!\!\left[ n-1\atop k\right] \!\!\right] _q^{(r,s)} \end{aligned}$$

(5)

with

$$\begin{aligned}&\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}=0\quad (0\le k\le r,\, n\ge k),\\ {}&\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}=0\quad (n<k),\quad \left[ \!\!\left[ 0\atop 0\right] \!\!\right] _q^{(r,s)}=1\,. \end{aligned}$$

From the recurrence relation (5), we can see some initial values:

$$\begin{aligned}&\left[ \!\!\left[ n\atop r\right] \!\!\right] _q^{(r,s)}=\left( \frac{[n-1]_q!}{[r-1]_q!}\right) ^s,\quad \left[ \!\!\left[ n\atop r+1\right] \!\!\right] _q^{(r,s)}=\left( \frac{[n-1]_q!}{[r-1]_q!}\right) ^s\sum _{j=r}^{n-1}\frac{1}{\bigl ([j]_q\bigr )^{s}}\,,\\ {}&\left[ \!\!\left[ n\atop r+2\right] \!\!\right] _q^{(r,s)}=\frac{1}{2}\left( \frac{[n-1]_q!}{[r-1]_q!}\right) ^s\left( \left( \sum _{j=r}^{n-1}\frac{1}{\bigl ([j]_q\bigr )^{s}}\right) ^2-\sum _{j=r}^{n-1}\frac{1}{\bigl ([j]_q\bigr )^{2 s}}\right) \\ {}&\qquad \qquad \quad =\left( \frac{[n-1]_q!}{[r-1]_q!}\right) ^s\sum _{r\le i<j\le n-1}\frac{1}{\bigl ([i]_q[j]_q\bigr )^s}\,,\\ {}&\left[ \!\!\left[ n\atop n\right] \!\!\right] _q^{(r,s)}=1,\quad \left[ \!\!\left[ n\atop n-1\right] \!\!\right] _q^{(r,s)}=\sum _{j=r}^{n-1}\bigl ([j]_q\bigr )^s,\\ {}&\left[ \!\!\left[ n\atop n-2\right] \!\!\right] _q^{(r,s)}=\sum _{r\le i<j\le n-1}\bigl ([i]_q[j]_q\bigr )^s\,. \end{aligned}$$

In general, by using the recurrence relation (5), we have expressions with combinatorial summations.

Theorem 2

For $r\le m\le n-1$ and $r\ge 1$, we have

$$\begin{aligned} \left[ \!\!\left[ n\atop m\right] \!\!\right] _q^{(r,s)}=\left( \frac{[n-1]_q!}{[r-1]_q!}\right) ^s\sum _{r\le i_1<\dots <i_{m-1}\le n-1}\frac{1}{\bigl ([i_1]_q\dots [i_{m-1}]_q\bigr )^s}. \end{aligned}$$

For $n-m\ge r$ and $r\ge 1$, we have

$$\begin{aligned} \left[ \!\!\left[ n\atop n-m\right] \!\!\right] _q^{(r,s)}&=\sum _{r\le i_1<\dots <i_m\le n-1}\bigl ([i_1]_q\dots [i_m]_q\bigr )^s\\ {}&=\sum _{r\le i_1\le i_2\le \dots \le i_m\le n-m}([i_1]_q[i_2+1]_q\cdots [i_m+m-1]_q)^s\\ {}&=\sum _{i_m=r}^{n-m}([i_m+m-1]_q)^s\sum _{i_{m-1}=r}^{i_m}([i_{m-1}+m-2]_q)^s\\ {}&\qquad \cdots \sum _{i_{2}=r}^{i_3}([i_{2}+1]_q)^s\sum _{i_{1}=r}^{i_2}([i_{1}]_q)^s\,. \end{aligned}$$

4 Second kind

The recurrence relation is given by

$$\begin{aligned} \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}=\left\{ \!\!\left\{ n-1\atop k-1\right\} \!\!\right\} _q^{(r,s)}+\bigl ([k]_q\bigr )^s\left\{ \!\!\left\{ n-1\atop k\right\} \!\!\right\} _q^{(r,s)} \end{aligned}$$

(6)

with

$$\begin{aligned}&\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}=0\quad (0\le k\le r-1,~n\ge k),\quad \left\{ \!\!\left\{ 0\atop 0\right\} \!\!\right\} _q^{(r,s)}=1,\\ {}&\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}=0\quad (n\le k)\,. \end{aligned}$$

From the recurrence relation (6), we can see some initial values:

$$\begin{aligned}&\left\{ \!\!\left\{ n\atop r\right\} \!\!\right\} _q^{(r,s)}=([r]_q)^{(n-r)s},\quad \left\{ \!\!\left\{ n\atop r+1\right\} \!\!\right\} _q^{(r,s)}=\sum _{i=0}^{n-r-1}\bigl ([r+1]_q\bigr )^{(n-r-i-1)s}([r]_q)^{i s}\,,\\ {}&\left\{ \!\!\left\{ n\atop r+2\right\} \!\!\right\} _q^{(r,s)}=\sum _{j=0}^{n-r-2}\bigl ([r+2]_q\bigr )^{(n-r-j-2)s}\sum _{i=0}^j\bigl ([r+1]_q\bigr )^{(j-i)s}([r]_q)^{i s}\,,\\ {}&\left\{ \!\!\left\{ n\atop n\right\} \!\!\right\} _q^{(r,s)}=1,\quad \left\{ \!\!\left\{ n\atop n-1\right\} \!\!\right\} _q^{(r,s)}=\sum _{j=r}^{n-1}\bigl ([j]_q\bigr )^s,\\ {}&\left\{ \!\!\left\{ n\atop n-2\right\} \!\!\right\} _q^{(r,s)} =\sum _{r\le i\le j\le n-2}\bigl ([i]_q[j]_q\bigr )^s\,. \end{aligned}$$

In general, the q-Stirling numbers of the second kind with higher level can be expressed in terms of iterated summations.

Theorem 3

For $r+1\le k\le n$ and $r\ge 1$,

$$\begin{aligned} {}&{} \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}=\sum _{i_{k-r}=0}^{n-k}\bigl ([k]_q\bigr )^{(n-k-i_{k-r})s}\sum _{i_{k-r-1}=0}^{i_{k-r}}\bigl ([k-1]_q\bigr )^{(i_{k-r}-i_{k-r-1})s}\\{}&\qquad \qquad \qquad \cdots \sum _{i_2=0}^{i_3}\bigl ([r+2]_q\bigr )^{(i_{3}-i_{2})s}\sum _{i_1=0}^{i_2}\bigl ([r+1]_q\bigr )^{(i_{2}-i_{1})s}\bigl ([r]_q\bigr )^{i_{1} s}. \end{aligned}$$

For $n-k\ge r\ge 1$,

$$\begin{aligned} \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}=\sum _{r\le i_1\le i_2\le \cdots \le i_k\le n-k}([i_1]_q[i_2]_q\cdots [i_k]_q)^s. \end{aligned}$$

Remark

When $r=1$ and $q\rightarrow 1$, Theorem 3 is reduced to [7, Theorem 3.4].

Proof

Fix an integer $n\ge 2$. Assume that the identity is valid up to some k. For convenience, put

$$\begin{aligned} {}&{} t_n:=\sum _{i_{k-r}=0}^{n}\bigl ([k]_q\bigr )^{(n-i_{k-r})s}\sum _{i_{k-r-1}=0}^{i_{k-r}}\bigl ([k-1]_q\bigr )^{(i_{k-r}-i_{k-r-1})s}\\{}&{} \qquad \qquad \cdots \sum _{i_2=0}^{i_3}\bigl ([r+2]_q\bigr )^{(i_{3}-i_{2})s}\sum _{i_1=0}^{i_2}\bigl ([r+1]_q\bigr )^{(i_{2}-i_{1})s}\bigl ([r]_q\bigr )^{i_{1} s}. \end{aligned}$$

By the recurrence relation (6), we have

$$\begin{aligned}&\left\{ \!\!\left\{ n\atop k+1\right\} \!\!\right\} _q^{(r,s)}=t_{n-k-1}+\bigl ([k+1]_q\bigr )^s\left\{ \!\!\left\{ n-1\atop k+1\right\} \!\!\right\} _q^{(r,s)}\\ {}&=t_{n-k}+\bigl ([k+1]_q\bigr )^s\left( t_{n-k-1}+\bigl ([k+1]_q\bigr )^s\left\{ \!\!\left\{ n-2\atop k+1\right\} \!\!\right\} _q^{(r,s)}\right) \\ {}&=t_{n-k}+\bigl ([k+1]_q\bigr )^s t_{n-k-1}+\bigl ([k+1]_q\bigr )^{2 s}t_{n-k-2}\\ {}&\qquad +\cdots +\bigl ([k+1]_q\bigr )^{(n-k-1)s}\left\{ \!\!\left\{ k+1\atop k+1\right\} \!\!\right\} _q^{(r,s)}\\ {}&=\sum _{i_k=1}^{n-k}\bigl ([k+1]_q\bigr )^{(n-k-i_k)s}t_{i_k}\\ {}&=\sum _{i_{k-r+1}=0}^{n-k-1}\bigl ([k+1]_q\bigr )^{(n-k-1-i_{k-r+1})s}\sum _{i_{k-r}=0}^{i_{k-r+1}}\bigl ([k]_q\bigr )^{(i_{k-r+1}-i_{k-r})s}\\ {}&\qquad \times \sum _{i_{k-r-1}=0}^{i_{k-r}}\bigl ([k-1]_q\bigr )^{(i_{k-r}-i_{k-r-1})s}\cdots \sum _{i_1=0}^{i_2} \bigl ([r+1]_q\bigr )^{(i_{2}-i_{1})s}\bigl ([r]_q\bigr )^{i_{1} s}\,. \end{aligned}$$

By induction, the identity holds for any integer $k\ge r$. $\square$

The ordinary generating function of the q-Stirling numbers of the second kind with higher level is given as follows.

Theorem 4

For $k\ge r$, we have

$$\begin{aligned} \sum _{n=k}^\infty \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}x^n=x^k\prod _{\ell =r}^k\bigl (1-([\ell ]_q)^s x\bigr )^{-1}. \end{aligned}$$

Proof

Put

$$\begin{aligned} f_k(x)=\sum _{n=k}^\infty \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}x^n. \end{aligned}$$

Since $\left\{ \!\!\left\{ n\atop r\right\} \!\!\right\} _q^{(r,s)}=([r]_q)^{(n-r)s}$ ($n\ge 1$),

$$\begin{aligned} f_r(x)=\sum _{n=r}^\infty \left\{ \!\!\left\{ n\atop r\right\} \!\!\right\} _q^{(r,s)}x^n=\sum _{n=r}^\infty ([r]_q)^{(n-r)s}x^n=\frac{x^r}{1-([r]_q)^s x}. \end{aligned}$$

By using the recurrence relation (6) we have

$$\begin{aligned} f_k(x)&=\sum _{n=k}^\infty \left( \left\{ \!\!\left\{ n-1\atop k-1\right\} \!\!\right\} _q^{(r,s)}+\bigl ([k]_q\bigr )^s\left\{ \!\!\left\{ n-1\atop k\right\} \!\!\right\} _q^{(r,s)}\right) x^n\\&=x\sum _{n=k-1}^\infty \left\{ \!\!\left\{ n\atop k-1\right\} \!\!\right\} _q^{(r,s)}x^n+\bigl ([k]_q\bigr )^s x\sum _{n=k}^\infty \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}x^n\\&=x f_{k-1}(x)+\bigl ([k]_q\bigr )^s x f_k(x)\,. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} f_k(x)&=\sum _{n=k}^\infty \left( \left\{ \!\!\left\{ n-1\atop k-1\right\} \!\!\right\} _q^{(r,s)}+\bigl ([k]_q\bigr )^s\left\{ \!\!\left\{ n-1\atop k\right\} \!\!\right\} _q^{(r,s)}\right) x^n\\ {}&=x\sum _{n=k-1}^\infty \left\{ \!\!\left\{ n\atop k-1\right\} \!\!\right\} _q^{(r,s)}x^n+\bigl ([k]_q\bigr )^s x\sum _{n=k}^\infty \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}x^n\\ {}&=x f_{k-1}(x)+\bigl ([k]_q\bigr )^s x f_k(x)\,. \end{aligned}$$

$\square$

From Theorem 4, we have another expression of the q-Stirling numbers of the second kind with higher level.

Corollary 1

We have

$$\begin{aligned} \left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)}=\sum _{j=r}^k\frac{\bigl ([j]_q\bigr )^{s n}}{([j]_q)^{r s}\prod _{i=0,i\ne j}^k\bigl (\bigl ([j]_q\bigr )^{s}-\bigl ([i]_q\bigr )^{s}\bigr )}. \end{aligned}$$

We show identities which combine q-Stirling numbers with higher level and the ordinary q-Stirling numbers.

Theorem 5

For $j=1,2,\dots$, we have

$$\begin{aligned}{} & {} \sum _{\ell =1}^{n-1}(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}\left\{ \!\!\left\{ n-1+j-\ell \atop n-1\right\} \!\!\right\} _q^{(r,s)}\\{} & {} \quad =\sum _{i=1}^{j s}(-1)^i i\left[ \!\!\left[ n\atop n-i\right] \!\!\right] _q^{(r,1)}\left\{ \!\!\left\{ n-1+j s-i\atop n-1\right\} \!\!\right\} _q^{(r,1)}. \end{aligned}$$

Remark

When $r=1$ and $q\rightarrow 1$, the identity in Theorem 5 is the formula with Bernoulli polynomials $B_n(x)$ ( [7]):

$$\begin{aligned} \sum _{\ell =1}^{n-1}(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] ^{(1,s)}\left\{ \!\!\left\{ n-1+j-\ell \atop n-1\right\} \!\!\right\} ^{(1,s)}=\frac{B_{sj+1}(0)-B_{sj+1}(n)}{sj+1}, \end{aligned}$$

where $B_n(x)$ are defined by the exponential generating function

$$\begin{aligned} \frac{t e^{t x}}{e^t-1}=\sum _{n=0}^\infty B_n(x)\frac{t^n}{n!}. \end{aligned}$$

Proof of Theorem 5

From (1), for $|x|<([n-1]_q)^{-s}$ we get

$$\begin{aligned} F_1(x):=\prod _{i=r}^{n-1}\bigl (1-([i]_q)^s x\bigr )=\sum _{\ell =0}^n(-1)^\ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}x^\ell . \end{aligned}$$

From Theorem 4, for $|x|<([n-1]_q)^{-s}$ we get

$$\begin{aligned} F_2(x):=\prod _{i=r}^{n-1}\bigl (1-([i]_q)^s x\bigr )^{-1}=\sum _{\nu =0}^\infty \left\{ \!\!\left\{ n-1+\nu \atop n-1\right\} \!\!\right\} _q^{(r,s)} x^\nu . \end{aligned}$$

Note that $F_1(x)F_2(x)=1$. First,

$$\begin{aligned} \frac{d}{d x}\log F_2(x)&=\sum _{i=r}^{n-1}\frac{([i]_q)^s}{1-([i]_q)^s x}\nonumber \\&=-F_1'(x)F_2(x)\,. \end{aligned}$$

(7)

where

$$\begin{aligned} F_1'(x)=\sum _{\ell =1}^n(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}x^{\ell -1}. \end{aligned}$$

On the other hand, for $|x|<([n-1]_q)^{-s}$ we have

$$\begin{aligned} \frac{d}{d x}\log F_2(x)=\sum _{i=r}^{n-1}\frac{([i]_q)^s}{1-([i]_q)^s x}=\sum _{i=r}^{n-1}\sum _{j=1}^\infty ([i]_q)^{j s}x^{j-1}. \end{aligned}$$

(8)

Combining (7) and (8), we have

$$\begin{aligned}{} & {} \sum _{j=1}^\infty \sum _{i=r}^{n-1}([i]_q)^{j s}x^{j-1}\\{} & {} \quad =-\left( \sum _{\ell =1}^n(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}x^{\ell -1}\right) \left( \sum _{\nu =0}^\infty \left\{ \!\!\left\{ n-1+\nu \atop n-1\right\} \!\!\right\} _q^{(r,s)}x^\nu \right) . \end{aligned}$$

Comparing the coefficients on both sides, we obtain

$$\begin{aligned} \sum _{i=r}^{n-1}([i]_q)^{j s}=-\sum _{\ell =1}^{n-1}(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}\left\{ \!\!\left\{ n-1+j-\ell \atop n-1\right\} \!\!\right\} _q^{(r,s)}. \end{aligned}$$

Since a q-analogue of the sums of powers is given by

$$\begin{aligned} \sum _{j=r}^n\bigl ([j]_q\bigr )^k=-\sum _{i=1}^k(-1)^i i\left[ \!\!\left[ n+1\atop n+1-i\right] \!\!\right] _q^{(r,1)}\left\{ \!\!\left\{ n+k-i\atop n\right\} \!\!\right\} _q^{(r,1)} \end{aligned}$$

(see, e.g., [8]), we have we obtain the result. $\square$

From Theorem 5, we have the identity that $\left[ \!\!\left[ n\atop n-k\right] \!\!\right] _q^{(r,s)}$ is expressed in terms of $\left[ \!\!\left[ n\atop j\right] \!\!\right] _q^{(r,s)}$ ($n-k+1\le j\le n$) and the ordinary q-Stirling numbers.

Corollary 2

For $n\ge k\ge 0$, we have

$$\begin{aligned}{} & {} k\left[ \!\!\left[ n\atop n-k\right] \!\!\right] _q^{(r,s)}\\{} & {} \quad =\sum _{j=1}^k\sum _{i=1}^{j s}(-1)^{j+i}i\left[ \!\!\left[ n\atop n-k+j\right] \!\!\right] _q^{(r,s)}\left[ \!\!\left[ n\atop n-i\right] \!\!\right] _q^{(r,1)}\left\{ \!\!\left\{ n-1+j s-i\atop n-1\right\} \!\!\right\} _q^{(r,1)}. \end{aligned}$$

Remark

When $k=r=1$, from Corollary 2, we have the identity that $\left[ \!\!\left[ n\atop n-1\right] \!\!\right] _q^{(1,s)}$ is expressed in terms of the ordinary q-Stirling numbers:

$$\begin{aligned} \left[ \!\!\left[ n\atop n-1\right] \!\!\right] _q^{(1,s)}=\sum _{i=1}^{s}(-1)^{i+1}i\left[ \!\!\left[ n\atop n-i\right] \!\!\right] _q\left\{ \!\!\left\{ n-1+s-i\atop n-1\right\} \!\!\right\} _q. \end{aligned}$$

Proof of Corollary 2

By Theorem 5 and the orthogonal relation in Theorem 1, the right-hand side is equal to

$$\begin{aligned}&\sum _{j=1}^k(-1)^j \left[ \!\!\left[ n\atop n-k+j\right] \!\!\right] _q^{(r,s)}\sum _{\ell =1}^{n-1}(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}\left\{ \!\!\left\{ n-1+j-\ell \atop n-1\right\} \!\!\right\} _q^{(r,s)}\\&=\sum _{\ell =1}^{n-1}(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}\sum _{j=1}^k(-1)^j \left[ \!\!\left[ n\atop n-k+j\right] \!\!\right] _q^{(r,s)}\left\{ \!\!\left\{ n-1+j-\ell \atop n-1\right\} \!\!\right\} _q^{(r,s)}\\&=\sum _{\ell =1}^{n-1}(-1)^\ell \ell \left[ \!\!\left[ n\atop n-\ell \right] \!\!\right] _q^{(r,s)}\cdot (-1)^\ell \delta _{\ell ,k}\\&=k\left[ \!\!\left[ n\atop n-k\right] \!\!\right] _q^{(r,s)}\,. \end{aligned}$$

$\square$

5 Finite differences

The method of finite differences [13] is useful to deal with elementary sequences, that is, if the sequence can be simplified after taking the difference. Consider the sequences $\{a_k\}_{k\ge 0}$ and $\{b_k\}_{k\ge 0}$ such that $a_k$ is the finite difference of $b_k$, that is, $a_k=b_{k+1}-b_k$. For $n\ge r$, let

$$\begin{aligned} g_n=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} a_k\quad \hbox {and}\quad h_n=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} b_k. \end{aligned}$$

Theorem 6

For $n\ge r$, we have

$$\begin{aligned} g_n=h_{n+1}-\bigl (([n]_q)^s+1\bigr )h_n \end{aligned}$$

and

$$\begin{aligned} h_n=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\left( b_r+\sum _{k=r}^{n-1}\frac{g_{k}}{\prod _{\ell =r}^{k}\bigl (([\ell ]_q)^s+1\bigr )}\right) . \end{aligned}$$

Proof

By the recurrence relation in (5), we have for $n\ge r$

$$\begin{aligned} h_{n+1}-([n]_q)^s h_n&=\sum _{k=r}^{n+1}\left[ \!\!\left[ n+1\atop k\right] \!\!\right] _s b_k-([n]_q)^s\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} b_k\\&=\sum _{k=r+1}^{n+1}\left[ \!\!\left[ n\atop k-1\right] \!\!\right] _s b_k=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} b_{k+1}\,. \end{aligned}$$

Hence,

$$\begin{aligned} h_{n+1}-\bigl (([n]_q)^s+1\bigr )h_n&=\sum _{k=r}^{n}\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} b_{k+1}-\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} b_k\\&=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} a_{k}=g_n\,. \end{aligned}$$

Next, summing up the equations

$$\begin{aligned} h_{n}&=\bigl (([n-1]_q)^s+1\bigr )h_{n-1}+g_{n-1}\\ \prod _{j=n-1}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot h_{n-1}&=\prod _{j=n-2}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot h_{n-2}+\prod _{j=n-1}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot g_{n-2}\\ \prod _{j=n-2}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot h_{n-2}&=\prod _{j=n-3}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot h_{n-3}+\prod _{j=n-2}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot g_{n-3}\\ \dots \\ \prod _{j=r+1}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot h_{r+1}&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot h_{r}+\prod _{j=r+1}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot g_{r}\,, \end{aligned}$$

we have

$$\begin{aligned} h_n=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\left( h_r+\sum _{k=r}^{n-1}\frac{g_{k}}{\prod _{\ell =r}^{k}\bigl (([\ell ]_q)^s+1\bigr )}\right) . \end{aligned}$$

Finally, $h_0=b_0$. $\square$

Proposition 1

For $n\ge 1$,

$$\begin{aligned} \sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr ). \end{aligned}$$

Remark

When $r=s=1$, Proposition 1 is reduced to [13, Identity 33]

$$\begin{aligned} \sum _{k=0}^n\left[ n\atop k\right] =\sum _{k=1}^n\left[ n\atop k\right] =n!=\prod _{j=1}^{n-1}\bigl (j+1\bigr )=\prod _{j=0}^{n-1}\bigl (j+1\bigr ). \end{aligned}$$

Proof of Proposition 1

If $b_k=1$ ($k\ge r$), then $a_k=0$ ($k\ge r$). So, $g_n=0$ ($n\ge r$). From Theorem 6, we get the result. $\square$

Theorem 7

For $n\ge m\ge r$,

$$\begin{aligned} \sum _{k=m}^n\left[ \!\!\left[ k\atop m\right] \!\!\right] _q^{(r,s)}\frac{([n]_q!)^s}{([k]_q!)^s}=\left[ \!\!\left[ n+1\atop m+1\right] \!\!\right] _q^{(r,s)}. \end{aligned}$$

Remark

When $r=s=1$ and $q\rightarrow 1$, Theorem 7 is reduced to [13, Identity 34]

$$\begin{aligned} \sum _{k=m}^n\left[ k\atop m\right] \frac{n!}{k!}=\left[ n+1\atop m+1\right] . \end{aligned}$$

Proof of Theorem 7

When $n=m$, it is clear that both sides are equal to 1. Suppose that the identity holds for some n. Then for $n+1$,

$$\begin{aligned} \sum _{k=0}^{n+1}\left[ \!\!\left[ k\atop m\right] \!\!\right] _q^{(r,s)}\frac{\bigl ([n+1]_q!\bigr )^s}{([k]_q!)^s}&=([n+1]_q)^s\left[ n+1\atop m+1\right] _q^{(r,s)}+\left[ \!\!\left[ n+1\atop m\right] \!\!\right] _q^{(r,s)}\frac{\bigl ([n+1]_q!\bigr )^s}{\bigl ([n+1]_q!\bigr )^s}\\&=\left[ \!\!\left[ n+2\atop m+1\right] \!\!\right] _q^{(r,s)}\,. \end{aligned}$$

By induction on n, the identity is proved. $\square$

Proposition 2

For $n\ge r\ge 1$,

$$\begin{aligned} \sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} k=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot \left( r+\sum _{\ell =r}^{n-1}\frac{1}{([\ell ]_q)^s+1}\right) . \end{aligned}$$

Remark

When $r=s=1$ and $q\rightarrow 1$, Proposition 2 is reduced to [13, Identity 35]

$$\begin{aligned} \sum _{k=0}^n\left[ n\atop k\right] k=\sum _{k=1}^n\left[ n\atop k\right] k=\left[ n+1\atop 2\right] . \end{aligned}$$

Note that the case $r=0$ is included in the case $r=1$ in Proposition 2.

Proof of Proposition 2

If $b_k=k$ ($k\ge 0$), then $a_k=1$ ($k\ge 0$). So, by Proposition 1, for $n\ge 0$

$$\begin{aligned} g_n=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr ). \end{aligned}$$

From Theorem 6,

$$\begin{aligned} h_n&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot \left( r+\sum _{k=r}^{n-1}\frac{\prod _{j=r}^{k-1}\bigl (([j]_q)^s+1\bigr )}{\prod _{j=r}^{k}\bigl (([j]_q)^s+1\bigr )}\right) \\&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot \left( r+\sum _{k=r}^{n-1}\frac{1}{([k]_q)^s+1}\right) \,. \end{aligned}$$

$\square$

By using the results in Proposition 1 and Proposition 2, we can obtain the transform of the square.

Proposition 3

For $n\ge 1$,

$$\begin{aligned}{} & {} \sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} k^2=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\\{} & {} \times \left( r^2+(2 r+1)\sum _{k=r}^{n-1}\frac{1}{([k]_q)^s+1}+2\sum _{r\le i_1<i_2\le n-1}\frac{1}{\bigl ([i_1]_q)^s+1\bigr )\bigl ([i_2]_q)^s+1\bigr )}\right) . \end{aligned}$$

Remark

When $r=s=1$ and $q\rightarrow 1$, Proposition 3 is reduced to

$$\begin{aligned} \sum _{k=0}^n\left[ n\atop k\right] k^2=\sum _{k=1}^n\left[ n\atop k\right] k^2&=n!\left( H_n+2\sum _{k=1}^n\frac{H_{k-1}}{k}\right) \\&=n!(H_n+H_n^2-H_n^{(2)})\\&=\left[ n+1\atop 2\right] +2\left[ n+1\atop 3\right] \,, \end{aligned}$$

where

$$\begin{aligned} H_n^{(\mu )}:=\sum _{k=1}^n\frac{1}{k^\mu } \end{aligned}$$

are the higher order harmonic numbers.

Proof of Proposition 3

When $b_k=k^2$, since $a_k=2 k+1$, we get from Proposition 1 and Proposition 2,

$$\begin{aligned} g_n&=2\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} k+\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}\\&=2\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot \left( r+\sum _{\ell =r}^{n-1}\frac{1}{([\ell ]_q)^s+1}\right) +\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\\&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot \left( 2 r+1+2\sum _{\ell =r}^{n-1}\frac{1}{([\ell ]_q)^s+1}\right) \,. \end{aligned}$$

Hence, by Theorem 6,

$$\begin{aligned} h_n&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot \left( r^2+\sum _{k=r}^{n-1}\frac{\prod _{i=r}^{k-1}\bigl (([i]_q)^s+1\bigr )}{\prod _{\ell =r}^{k}\bigl (([\ell ]_q)^s+1\bigr )}\left( 2 r+1+\sum _{\nu =r}^{k-1}\frac{2}{([\nu ]_q)^s+1}\right) \right) \\&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\\&\quad \times \biggl (r^2+(2 r+1)\sum _{k=0}^{n-1}\frac{1}{([k]_q)^s+1}+2\sum _{k=r}^{n-1}\sum _{\nu =r}^{k-1}\frac{1}{\bigl (([k]_q)^s+1\bigr )\bigl (([\nu ]_q)^s+1\bigr )}\biggr )\,. \end{aligned}$$

$\square$

By repeating a similar process, we can obtain the transform of the power

$$\begin{aligned} \sum _{k=0}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} k^m \end{aligned}$$

for $m=3,4,\dots$ one by one. Nevertheless, the general result can be obtained by using Proposition 4 below.

The transform of the falling factorial

$$\begin{aligned} (k)_m=k(k-1)\cdots (k-m+1)\quad (m\ge 1) \end{aligned}$$

with $(k)_0=1$ is given as follows.

Proposition 4

For positive integers n and m,

$$\begin{aligned}&\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}(k)_m=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\\&\quad \times \sum _{l=0}^m\frac{m!}{(m-l)!}(r)_{m-l}\sum _{r\le i_1<i_2<\cdots <i_l\le n-1}\frac{1}{\bigl (([i_1]_q)^s+1\bigr )\bigl (([i_2]_q)^s+1\bigr )\cdots \bigl (([i_l]_q)^s+1\bigr )}\,. \end{aligned}$$

Remark

When $r=s=1$ and $q\rightarrow 1$, Proposition 4 is reduced to

$$\begin{aligned} \sum _{k=0}^n\left[ n\atop k\right] (k)_m&=n!\sum _{0\le i_1<i_2<\cdots <i_m\le n-1}\frac{m!}{(i_1+1)(i_2+1)\cdots (i_m+1)}\\&=\left[ n+1\atop m+1\right] m! \end{aligned}$$

( [13, Identity 36,Identity 42]).

Proof of Proposition 4

From Proposition 2, the identity is valid for $m=1$. Suppose that the identity is valid up to $m-1$. When $b_k=(k)_m$, by $a_k=m(k)_{m-1}$ we get

$$\begin{aligned} g_n=m\sum _{k=0}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}(k)_{m-1}. \end{aligned}$$

By Theorem 6 and the induction hypothesis, we have

$$\begin{aligned}&\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}(k)_m\\ {}&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\cdot \left( (r)_m+\sum _{k=r}^{n-1}\frac{m}{\prod _{\ell =r}^{k}\bigl (([\ell ]_q)^s+1\bigr )}\sum _{\nu =r}^k\left[ \!\!\left[ k\atop \nu \right] \!\!\right] _s(\nu )_{m-1}\right) \\ {}&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\biggl ((r)_m\\ {}&\qquad +\sum _{k=r}^{n-1}\frac{m}{([k]_q)^s+1}\sum _{l=0}^{m-1}\frac{(m-1)!}{(m-l-1)!}(r)_{m-l-1}\\ {}&\qquad \qquad \times \sum _{r\le i_1<\cdots<i_l\le k-1}\frac{1}{\bigl (([i_1]_q)^s+1\bigr )\cdots \bigl (([i_l]_q)^s+1\bigr )}\biggr )\\ {}&=\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\biggl ((r)_m\\ {}&\quad +\sum _{l=1}^{m}\frac{m!}{(m-l)!}(r)_{m-l}\\ {}&\qquad \times \sum _{r\le i_1<\cdots<i_{l-1}<k\le k-1}\frac{1}{\bigl (([i_1]_q)^s+1\bigr ) \cdots \bigl (([i_{l-1}]_q)^s+1\bigr )\bigl (([k]_q)^s+1\bigr )}\biggr )\,. \end{aligned}$$

$\square$

Now, by using the relation $k^m=\sum _{h=0}^m\left\{ m\atop h\right\} (k)_h$ to Proposition 4, we can obtain the transform of the power.

Proposition 5

For $n\ge r\ge 1$,

$$\begin{aligned}&\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} k^m =\prod _{j=r}^{n-1}\bigl (([j]_q)^s+1\bigr )\\ {}&\quad \times \sum _{h=0}^m\sum _{l=0}^m\left\{ m\atop h\right\} \frac{m!}{(m-l)!}(k)_h(r)_{m-l}\\ {}&\quad \times \sum _{r\le i_1<i_2<\cdots <i_l\le n-1}\frac{1}{\bigl (([i_1]_q)^s+1\bigr )\bigl (([i_2]_q)^s+1\bigr )\cdots \bigl (([i_l]_q)^s+1\bigr )}\,. \end{aligned}$$

Remark

When $r=s=1$ and $q\rightarrow 1$, Proposition 3 is reduced to [13, Identity 37]

$$\begin{aligned} \sum _{k=0}^n\left[ n\atop k\right] k^m=\sum _{\ell =0}^m\left[ n+1\atop \ell +1\right] \left\{ m\atop \ell \right\} \ell !. \end{aligned}$$

Next, for $n\ge r$, let

$$\begin{aligned} g_n'=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}(-1)^{n-k}a_k\quad \hbox {and}\quad h_n'=\sum _{k=r}^n\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}(-1)^{n-k}b_k. \end{aligned}$$

Then, similarly to Theorem 6, we have the following.

Theorem 8

For $n\ge r\ge 1$, we have

$$\begin{aligned} g_n'=h_{n+1}'+\bigl (([n]_q)^s-1\bigr )h_n' \end{aligned}$$

and

$$\begin{aligned} h_n'=(-1)^n\prod _{j=r+1}^{n-1}\bigl (([j]_q)^s-1\bigr )\cdot \left( (-1)^r\bigl (([r]_q)^s-1\bigr )b_r'+\sum _{k=r}^{n-1}\frac{(-1)^{k+1}g_{k}'}{\prod _{\ell =r+1}^{k}\bigl (([\ell ]_q)^s-1\bigr )}\right) \end{aligned}$$

with $h_r'=b_r$.

Proposition 6

For $n\ge r\ge 1$

$$\begin{aligned} \sum _{k=r}^n(-1)^{n-k}\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)}=(-1)^{n-r}\prod _{j=r}^{n-1}\bigl (([j]_q)^s-1\bigr ). \end{aligned}$$

Remark

When $r=s=1$ and $q\rightarrow 1$, Proposition 6 is reduced to [13, Identity 40]

$$\begin{aligned} \sum _{k=0}^n(-1)^{n-k}\left[ n\atop k\right] ={\left\{ \begin{array}{ll} 1&{}\hbox { if}\ n=0,1;\\ 0&{}\hbox { if}\ n\ge 2. \end{array}\right. } \end{aligned}$$

Proof of Proposition 6

When $b_k=1$, we see that $a_k=g_k=0$. The result is clear from Theorem 8. $\square$

Proposition 7

For $n\ge r\ge 1$

$$\begin{aligned} {}&{} \sum _{k=r}^n(-1)^{n-k}\left[ \!\!\left[ n\atop k\right] \!\!\right] _q^{(r,s)} k\\{}&{} =(-1)^{n-r}\prod _{j=r+1}^{n-1}\bigl (([j]_q)^s-1\bigr )\left( r\bigl (([r]_q)^s-1\bigr ) -\sum _{k=r+1}^{n-1}\frac{1}{([k]_q)^s-1}\right) . \end{aligned}$$

Remark

When $r=s=1$ and $q\rightarrow 1$, Proposition 7 is reduced to [13, Identity 41]

$$\begin{aligned} \sum _{k=0}^n(-1)^{n-k}\left[ n\atop k\right] k={\left\{ \begin{array}{ll} 0&{}\hbox { if}\ n=0;\\ 1&{}\hbox { if}\ n=1;\\ (-1)^n(n-2)!&{}\hbox { if}\ n\ge 2. \end{array}\right. } \end{aligned}$$

Proof of Proposition 7

When $b_k=k$, we see that $a_k=1$. By Proposition 6,

$$\begin{aligned} g_k'=(-1)^{k-r}\prod _{j=r}^{k-1}\bigl (([j]_q)^s-1\bigr ). \end{aligned}$$

The result is obtained from Theorem 8. $\square$

The Stirling transform by the second kind

$$\begin{aligned} b_n=\sum _{k=r}^n\left\{ \!\!\left\{ n\atop k\right\} \!\!\right\} _q^{(r,s)} a_k \end{aligned}$$

is not suitably achieved by the method of finite differences. When $r=s=1$ and $q\rightarrow 1$, if $a_n=1$, then $b_n$ becomes the Bell number $\textrm{Bell}_n$. Then

$$\begin{aligned} \sum _{k=0}^n\left\{ n\atop k\right\} k=\textrm{Bell}_{n+1}-\textrm{Bell}_n. \end{aligned}$$

6 A q-multiple zeta function

In this section, set $r=1$. For simplicity, we write

$$\begin{aligned} \left[ \!\!\left[ n\atop m\right] \!\!\right] _q^{(s)}=\left[ \!\!\left[ n\atop m\right] \!\!\right] _q^{(1,s)}. \end{aligned}$$

Then we give some applications from the q-version of Stirling numbers of the first kind with higher level.

Several different types of q-multiple zeta functions have been studied by many researchers. For example, the functions

$$\begin{aligned} \zeta [s_1,\dots ,s_m]&:=\sum _{1\le i_1<\dots<i_{m}}\frac{q^{(s_1-1)i_1+\cdots +(s_m-1)i_m}}{\bigl ([i_1]_q\bigr )^{s_1}\dots \bigl ([i_{m}]_q\bigr )^{s_m}} \quad {\mathrm{[1]}}\\ \bar{\zeta }[s_1,\dots ,s_m]&:=\sum _{1\le i_1<\dots<i_{m}}\frac{q^{s_1 i_1+\cdots +s_m i_m}}{(1-q^{i_1})^{s_1}\dots (1-q^{i_m})^{s_m}}\quad {\mathrm{[16]}}\\ \bar{\mathfrak z}_q[s_1,\dots ,s_m]&:= \sum _{0<i_1<\dots <i_m}\frac{q^{i_m}}{(1-q^{i_1})^{s_1}\dots (1-q^{i_m})^{s_m}}\quad {\mathrm{[10]}}\,. \end{aligned}$$

One can consider a little different q-multiple zeta function

$$\begin{aligned} \zeta _q(s_1,\dots ,s_m):=\sum _{1\le i_1<\dots <i_{m}}\frac{1}{\bigl ([i_1]_q\bigr )^{s_1}\dots \bigl ([i_{m}]_q\bigr )^{s_m}}. \end{aligned}$$

In [11], the function

$$\begin{aligned} \frac{\zeta _q(s_1,\dots ,s_m)}{(1-q)^{s_1+\cdots +s_m}}=\sum _{1\le i_1<\dots <i_{m}}\frac{1}{(1-q^{i_1})^{s_1}\dots (1-q^{i_m})^{s_m}} \end{aligned}$$

is proposed. A more general function is introduced in [15]. In this section, we consider the function

$$\begin{aligned} \mathfrak Z_n(q;m,s)&:=\sum _{1\le i_1<\dots <i_{m}\le n-1}\frac{1}{(1-q^{i_1})^s\cdots (1-q^{i_m})^s}\\&=\frac{1}{(1-q)^{s m}\bigl ([n-1]_q!\bigr )^s}\left[ \!\!\left[ n\atop m+1\right] \!\!\right] _q^{(s)}\,. \end{aligned}$$

From the first formula of Theorem 2, we get

$$\begin{aligned} \zeta _q(\underbrace{s,\dots ,s}_m)=\lim _{n\rightarrow \infty }\frac{1}{\bigl ([n-1]_q!\bigr )^s}\left[ \!\!\left[ n\atop m+1\right] \!\!\right] _q^{(s)}. \end{aligned}$$

When $s=1$, we have the following.

Theorem 9

For $n\ge m+1$, we have

$$\begin{aligned} \mathfrak Z_n(\zeta _n;m,1)=\frac{1}{m+1}\left( {\begin{array}{c}n-1\\ m\end{array}}\right) , \end{aligned}$$

where $\zeta _n=e^{2\pi \sqrt{-1}/n}$ is the n-th primitive root of unity. For $m\ge 1$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{[n-1]_{\zeta _n}!}\left[ \!\!\left[ n\atop m+1\right] \!\!\right] _{\zeta _n}^{(1)}=\frac{(-2\pi \sqrt{-1})^m}{(m+1)!}. \end{aligned}$$

Proof

Since

$$\begin{aligned} \sum _{k=1}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) x^l&=(x+1)^n-1\\&=x(x+1-\zeta _n)\dots (x+1-\zeta _n^2)\cdots (x+1-\zeta _n^{n-1})\,, \end{aligned}$$

for $k=1,2,\dots ,n-1$, we get

$$\begin{aligned} \sum _{1\le i_1<\dots <i_{k}\le n-1}(1-\zeta _n^{i_1})\cdots (1-\zeta _n^{i_{k}})=\left( {\begin{array}{c}n\\ k\end{array}}\right) . \end{aligned}$$

(9)

Thus, by the definition in (1), we have

$$\begin{aligned} \left[ \!\!\left[ n\atop k\right] \!\!\right] _{\zeta _n}^{(1)}&=\sum _{1\le i_1<\dots <i_{n-k}\le n-1}[i_1]_{\zeta _n}\cdots [i_{n-k}]_{\zeta _n}\\&=\frac{1}{(1-\zeta _{n})^{n-k}}\left( {\begin{array}{c}n\\ k\end{array}}\right) \,. \end{aligned}$$

Hence, by (9) with $k=1$

$$\begin{aligned} \mathfrak Z_n(\zeta _n;m,1)&=\frac{1}{(1-\zeta _n)^{m}[n-1]_{\zeta _n}!}\left[ \!\!\left[ n\atop m+1\right] \!\!\right] _{\zeta _n}^{(1)}\\&=\frac{1-\zeta _n)^{n-1}}{(1-\zeta _n)^{m}\prod _{i=1}^{n-1}(1-\zeta _n^i)}\frac{1}{(1-\zeta _{n})^{n-m-1}}\left( {\begin{array}{c}n\\ m+1\end{array}}\right) \\&=\frac{1}{m+1}\left( {\begin{array}{c}n-1\\ m\end{array}}\right) \,. \end{aligned}$$

Next, note that $(1-\zeta _n)(n-c)\rightarrow -2\pi \sqrt{-1}$ ($n\rightarrow \infty$) for a fixed real number c because $\sin x/x\rightarrow 1$ and $(1-\cos x)/x\rightarrow 0$ ($x\rightarrow 0$). Then, we can get

$$\begin{aligned} \frac{1}{[n-1]_{\zeta _n}!}\left[ \!\!\left[ n\atop m+1\right] \!\!\right] _{\zeta _n}^{(1)}&=(1-\zeta _n)^m\mathfrak Z_n(\zeta _n;m,1)\\&=\frac{(1-\zeta _n)^m}{m+1}\left( {\begin{array}{c}n-1\\ m\end{array}}\right) \\&\rightarrow \frac{(-2\pi \sqrt{-1})^m}{(m+1)!}\quad (n\rightarrow \infty )\,. \end{aligned}$$

$\square$

When $s\ge 2$, some of the specific values are given by the following.

Proposition 8

$$\begin{aligned} \mathfrak Z_n(\zeta _n;1,2)&=-\frac{(n-1)(n-5)}{12}\quad (n\ge 2)\,,\\ \mathfrak Z_n(\zeta _n;1,3)&=-\frac{(n-1)(n-3)}{8}\quad (n\ge 2)\,,\\ \mathfrak Z_n(\zeta _n;1,4)&=\frac{(n-1)(n^3+n^2-109n+251)}{720}\quad (n\ge 2)\,,\\ \mathfrak Z_n(\zeta _n;2,2)&=\frac{(n-1)(n-2)(n^2-12 n+47)}{360}\quad (n\ge 3)\,. \end{aligned}$$

Proof

Since

$$\begin{aligned} \frac{1}{(1-\zeta _n^k)^2}+\frac{1}{(1-\zeta _n^{n-k})^2}&=1-\frac{2}{2-\zeta ^k-\zeta ^{n-k}}=1-\frac{1}{1-\cos \frac{2 k\pi }{n}}\\&=1-\frac{1}{2\sin ^2\frac{k\pi }{n}}\,, \end{aligned}$$

when n is odd with $n\ge 3$, we have

$$\begin{aligned} \sum _{i=1}^{n-1}\frac{1}{(1-\zeta _n^i)^2}&=\frac{n-1}{2}-\sum _{k=1}^{\frac{n-1}{2}}\frac{1}{2\sin ^2\frac{k\pi }{n}}\\&=\frac{n-1}{2}-\frac{(n+1)(n-1)}{12}\\&=-\frac{(n-1)(n-5)}{12}\,. \end{aligned}$$

When n is even, we have

$$\begin{aligned} \sum _{i=1}^{n-1}\frac{1}{(1-\zeta _n^i)^2}&=\frac{n}{2}-1-\sum _{k=1}^{\frac{n}{2}-1}\frac{1}{2\sin ^2\frac{k\pi }{n}}+\frac{1}{\bigl (1-(-1)\bigr )^2}\\&=\frac{n}{2}-1-\frac{(n+1)(n-1)}{12}+\frac{1}{4}\\&=-\frac{(n-1)(n-5)}{12}\,. \end{aligned}$$

Since

$$\begin{aligned} \frac{1}{(1-\zeta _n^k)^3}+\frac{1}{(1-\zeta _n^{n-k})^3}&=1-\frac{3}{2-\zeta ^k-\zeta ^{n-k}}=1-\frac{3}{2(1-\cos \frac{2 k\pi }{n})}\\&=1-\frac{3}{4\sin ^2\frac{k\pi }{n}}\,, \end{aligned}$$

when n is odd with $n\ge 3$, we have

$$\begin{aligned} \sum _{i=1}^{n-1}\frac{1}{(1-\zeta _n^i)^3}&=\frac{n-1}{2}-\sum _{k=1}^{\frac{n-1}{2}}\frac{3}{4\sin ^2\frac{k\pi }{n}}\\&=\frac{n-1}{2}-\frac{(n+1)(n-1)}{8}\\&=-\frac{(n-1)(n-3)}{8}\,. \end{aligned}$$

When n is even, we have

$$\begin{aligned} \sum _{i=1}^{n-1}\frac{1}{(1-\zeta _n^i)^3}&=\frac{n}{2}-1-\sum _{k=1}^{\frac{n}{2}-1}\frac{3}{4\sin ^2\frac{k\pi }{n}}+\frac{1}{\bigl (1-(-1)\bigr )^3}\\&=\frac{n}{2}-1-\frac{(n+1)(n-1)}{8}+\frac{1}{8}\\&=-\frac{(n-1)(n-3)}{8}\,. \end{aligned}$$

Since

$$\begin{aligned} \frac{1}{(1-\zeta _n^k)^4}+\frac{1}{(1-\zeta _n^{n-k})^4}&=1-\frac{4}{2-\zeta ^k-\zeta ^{n-k}}+\frac{2}{(2-\zeta ^k-\zeta ^{n-k})^2}\\&=1-\frac{1}{\sin ^2\frac{k\pi }{n}}+\frac{1}{8\sin ^4\frac{k\pi }{n}}\,, \end{aligned}$$

when n is odd with $n\ge 3$, we have

$$\begin{aligned} \sum _{i=1}^{n-1}\frac{1}{(1-\zeta _n^i)^4}&=\frac{n-1}{2}-\sum _{k=1}^{\frac{n-1}{2}}\frac{1}{\sin ^2\frac{k\pi }{n}}+\sum _{k=1}^{\frac{n-1}{2}}\frac{1}{8\sin ^4\frac{k\pi }{n}}\\&=\frac{n-1}{2}-\frac{(n+1)(n-1)}{6}+\frac{1}{8}\frac{(n+1)(n-1)(n^2+11)}{90}\\&=\frac{(n-1)(n^2+n^2-109 n+251)}{720}\,. \end{aligned}$$

It is similar when n is even.

In general,

$$\begin{aligned} \frac{1}{(1-\zeta _n^k)^s}+\frac{1}{(1-\zeta _n^{n-k})^s}&=\sum _{j=0}^{\left\lfloor \frac{s}{2}\right\rfloor }(-1)^j\frac{s}{s-j}\left( {\begin{array}{c}s-j\\ j\end{array}}\right) \frac{1}{(2-\zeta ^k-\zeta ^{n-k})^j}\\&=\sum _{j=0}^{\left\lfloor \frac{s}{2}\right\rfloor }(-1)^j\frac{s}{s-j}\left( {\begin{array}{c}s-j\\ j\end{array}}\right) \frac{1}{\bigl (4\sin \frac{k\pi }{n}\bigr )^j}\,. \end{aligned}$$

Here, the coefficients $\frac{n}{n-j}\left( {\begin{array}{c}n-j\\ j\end{array}}\right)$ appear in those of the term $x^{n-2 j}$ of Lucas polynomial $L_n(x)$ ( [12, A034807]), defined by

$$\begin{aligned} L_0(x)=2,\quad L_1(x)=x,\quad L_n(x)=x L_{n-1}(x)+L_{n-2}(x)\quad (n\ge 2). \end{aligned}$$

Indeed,

$$\begin{aligned} L_n(x)&=\left( \frac{x+\sqrt{x^2+4}}{2}\right) ^n+\left( \frac{x-\sqrt{x^2+4}}{2}\right) ^n\\&=\frac{2}{2^{n}}\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\left( {\begin{array}{c}n\\ 2 k\end{array}}\right) x^{n-2 k}(x^2+4)^k\\&=\frac{1}{2^{n-1}}\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\left( {\begin{array}{c}n\\ 2 k\end{array}}\right) \sum _{j=0}^k\left( {\begin{array}{c}k\\ j\end{array}}\right) x^{2(k-j)}4^j\\&=\sum _{j=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\left( \frac{1}{2^{n-1}}\sum _{k=j}^{\left\lfloor \frac{n}{2}\right\rfloor }\left( {\begin{array}{c}n\\ 2 k\end{array}}\right) \left( {\begin{array}{c}k\\ j\end{array}}\right) 4^j\right) x^{n-2 j}\\&=\sum _{j=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\frac{n}{n-j}\left( {\begin{array}{c}n-j\\ j\end{array}}\right) x^{n-2 j}\,. \end{aligned}$$

Thus, when n is odd, we have

$$\begin{aligned} \sum _{i=1}^{n-1}\frac{1}{(1-\zeta _n^i)^s}=\sum _{k=1}^{\frac{n-1}{2}}\sum _{j=0}^{\left\lfloor \frac{s}{2}\right\rfloor }(-1)^j\frac{s}{s-j}\left( {\begin{array}{c}s-j\\ j\end{array}}\right) \frac{1}{\bigl (4\sin \frac{k\pi }{n}\bigr )^j}. \end{aligned}$$

When n is even, we have

$$\begin{aligned} \sum _{i=1}^{n-1}\frac{1}{(1-\zeta _n^i)^s}=\sum _{k=1}^{\frac{n}{2}-1}\sum _{j=0}^{\left\lfloor \frac{s}{2}\right\rfloor }(-1)^j\frac{s}{s-j}\left( {\begin{array}{c}s-j\\ j\end{array}}\right) \frac{1}{\bigl (4\sin \frac{k\pi }{n}\bigr )^j}+\frac{1}{2^s}. \end{aligned}$$

On the other hand, from the formulae for $\sum _{i=1}^{n-1}(1-\zeta _n^i)^{-s}$, we have

$$\begin{aligned} \mathfrak Z_n(\zeta _n;2,1)&=\frac{1}{2}\left( \bigl (\mathfrak Z_n(\zeta _n;1,1)\bigr )^2-\mathfrak Z_n(\zeta _n;1,2)\right) \\&=\frac{1}{2}\left( \left( \frac{n-1}{2}\right) ^2+\frac{(n-1)(n-5)}{12}\right) \\&=\frac{(n-1)(n-2)}{6}\,.\\ \mathfrak Z_n(\zeta _n;2,2)&=\frac{1}{2}\left( \bigl (\mathfrak Z_n(\zeta _n;1,2)\bigr )^2-\mathfrak Z_n(\zeta _n;1,4)\right) \\&=\frac{1}{2}\left( \left( -\frac{(n-1)(n-5)}{12}\right) ^2-\frac{(n-1)(n^3+n^2-109 n+251)}{720}\right) \\&=\frac{(n-1)(n-2)(n^2-12 n+47)}{360}\,.\\ \mathfrak Z_n(\zeta _n;3,1)&=\frac{1}{6}\left( \bigl (\mathfrak Z_n(\zeta _n;1,1)\bigr )^3-3\mathfrak Z_n(\zeta _n;1,1)\mathfrak Z_n(\zeta _n;1,2)+2\mathfrak Z_n(\zeta _n;1,3)\right) \\&=\frac{1}{6}\left( \left( \frac{n-1}{2}\right) ^3+3\frac{n-1}{2}\frac{(n-1)(n-5)}{12}-2\frac{(n-1)(n-3)}{8}\right) \\&=\frac{(n-1)(n-2)(n-3)}{6}\,. \end{aligned}$$

$\square$