New approach to twisted q-Bernoulli polynomials

Abstract

By using the theory of basic hypergeometric series, we present some formulas for q-consecutive integers, and we find certain new identities for twisted q-Bernoulli polynomials and q-consecutive integers (Simsek in Adv. Stud. Contemp. Math. 16(2):251-278, 2008).

MSC: 11B68, 05A30.

1 Introduction

The classical Bernoulli polynomials $B_n(x)$ and the Euler polynomials $E_n(x)$ are usually defined by the generating functions

$$\begin{array}{c}\frac{t{e}^{xt}}{e^t-1}=\sum _{n=0}^{\mathrm{\infty }}{B}_n(x)\frac{t^n}{n!},|t|<2\pi \text{and}\hfill \\ \frac{2e^{xt}}{e^t+1}=\sum _{n=0}^{\mathrm{\infty }}{E}_n(x)\frac{t^n}{n!},|t|<\pi ,\hfill \end{array}$$

respectively. In addition, the Bernoulli numbers are given by $B_n:=B_n(0)$ for $n\ge 1$. Recently, the Bernoulli polynomials and Bernoulli numbers have gained considerable significance in the fields of physics and mathematics [1–4]. For example, Kim [3] defined a new q-analogy of the Bernoulli polynomials and Bernoulli numbers, and he deduced some important relations between them. Moreover, q-analogues have been investigated in the study of quantum groups and q-deformed superalgebras [1]. The connection here is similar, in that much the string theory is set in the language of Riemann surfaces, resulting in connections with elliptic curves, which in turn relate to q-series. A q-analogue is an identity for a q-series that returns a known result in the ‘bosonic’ limit (in contrast to the conventional ‘fermionic’ limit $q\to -1$) as $q\to 1$ (from inside the complex unit circle in most situations). In addition to the widely used q-series, we have q-numbers, q-factorials, and q-binomial coefficients. A q-number is obtained by observing ${lim}_{q\to 1}\frac{1-q^n}{1-q}=n$. Thus, we define a q-number as ${[n]}_q=\frac{1-q^n}{1-q}$. Accordingly, one can define the q-analogue of the factorial, namely, q-factorial, as

$$\begin{array}{rcl}{[n]}_q!& =& {[1]}_q{[2]}_q\cdots {[n-1]}_q{[n]}_q\\ =& \frac{1-q}{1-q}\frac{1-q^2}{1-q}\cdots \frac{1-q^n}{1-q}=1\cdot (1+q)\cdots (1+q+\cdots +q^{n-2})(1+q+\cdots +q^{n-1}).\end{array}$$

Using this notation, we can define the q-binomial coefficients, also known as Gaussian coefficients, by

$${\left[\begin{array}{c}n\\ k\end{array}\right]}_q=\frac{{[n]}_q!}{{[n-k]}_q!{[k]}_q!}.$$

Furthermore, the q-Bernoulli polynomials ${\beta }_{n,q}(x)$ and the q-Bernoulli numbers ${\beta }_{n,q}$ can be defined in terms of the generating function $F_t(x,q)$ as follows [5]:

$$F_t(x,q)=e^{\frac{1}{1-t}}\frac{t-1}{logt}-q\sum _{n=0}^{\mathrm{\infty }}{t}^{n+x}{e}^{{[x+n]}_{t}q}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_{n,t}(x)}{n!}{q}^n,|q|<1,|t|<1.$$

Kim [6] established an interesting relation between Bernoulli numbers and q-integers, that is,

$${\int }_0^n{\beta }_{l,q}d{[x]}_q=\frac{1}{l+1}({\beta }_{l+1,q}(k)-{\beta }_{l+1,q}).$$

In addition, Kim [[7], Theorem 1], Kim and Lee [[8], Lemma 2.1] derived the relations between the Euler polynomials $E_n^{(r)}(x)$ of order r using the alternating sum of powers of consecutive integers $T_k(n)$. Here, $T_k(n)={\sum }_{i=0}^n{(-1)}^r{l}^k$ and $E_n^{(r)}(x)$ is defined as

$${\left(\frac{2}{e^t+1}\right)}^r{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{E}_n^{(r)}(x)\frac{t^n}{n!}.$$

Simsek constructed twisted Bernoulli polynomials together with twisted Bernoulli numbers and obtained analytic properties of twisted L-functions [9, 10]. Further, he defined generating functions of the twisted q-Bernoulli numbers and polynomials [9]. In a complex case, the generating function of twisted q-Bernoulli numbers $f_{q,\omega }(t)$ and a q-analogue of the Hurwitz zeta function $f_{\omega }(t,x,q)$ are given by

$$\begin{array}{r}{f}_{q,\omega }(t)=\sum _{l=0}^{\mathrm{\infty }}{B}_{l,\omega }^{\ast }(q)\frac{t^l}{l!}\text{and}\\ f_{\omega }(t,x,q)=e^{-t{[x]}_q}{f}_{q,\omega }(t)=\sum _{l=0}^{\mathrm{\infty }}{B}_{l,\omega }^{\ast }(q,x)\frac{t^l}{l!},\end{array}$$

where $q\in \mathbb{C}$ with $|q|<1$, and ω is the r th root of 1. In a complex case, the generating function of twisted q-Bernoulli numbers $f_{q,\omega }(t)$ and a q-analogue of the Hurwitz zeta function $f_{\omega }(t,x,q)$ are given by

$$\begin{array}{r}{f}_{q,\omega }(t)=\sum _{l=0}^{\mathrm{\infty }}{B}_{l,\omega }^{\ast }(q)\frac{t^l}{l!}\text{and}\\ f_{\omega }(t,x,q)=e^{-t{[x]}_q}{f}_{q,\omega }(t)=\sum _{l=0}^{\mathrm{\infty }}{B}_{l,\omega }^{\ast }(q,x)\frac{t^l}{l!},\end{array}$$

where $q\in \mathbb{C}$ with $|q|<1$, and ω is the r th root of 1. Simsek [9] then derived the identities

$$B_{l,\omega }^{\ast }(q)={(-1)}^{l}l\sum _{n=0}^{\mathrm{\infty }}{\omega }^{-n}{q}^{-ln}{[n]}_q^{l-1},$$

(S1)

$${\zeta }_{\omega ,q}(1-l)=\frac{{(-1)}^{l+1}}{l}{B}_{l,\omega }^{\ast }(q),$$

(S2)

$$B_{l,\omega }^{\ast }(x,q)=-{(-1)}^{l}l\sum _{n=1}^{\mathrm{\infty }}{\omega }^{-n}{q}^{-ln}{[n+x]}_q^{l-1},$$

(S3)

$${\zeta }_{\omega ,q}(1-l,x)=\frac{{(-1)}^{l+1}}{l}{B}_{l,\omega }^{\ast }(x,q),$$

(S4)

where ${\zeta }_{\omega ,q}(s)={\sum }_{n=1}^{\mathrm{\infty }}\frac{{\omega }^{-n}{q}^{-n}}{{(q^{-n}{[n]}_q)}^s}$, ${\zeta }_{\omega ,q}(s,x)={\sum }_{n=0}^{\mathrm{\infty }}\frac{{\omega }^{-n}{q}^{-n}}{{(q^{-n}{[n+x]}_q)}^s}$, and x is a natural number. In this paper, we first study relations among q-consecutive integers, q-Bernoulli numbers, and q-Euler numbers.

In 1631, Faulhaber [11] evaluated the sums of powers of consecutive integers $1^k+\cdots +n^k$ up to $k=17$. Further, in 1993, Knuth [12] presented an insightful alternative account of Faulhaber’s work. Several mathematicians further considered the problems of q-analogues of such sums of powers [7–9, 13]. On the basis of Bernoulli’s concept, Kim derived a q-analogue of the sums of powers of consecutive integers, by setting

$$T_{l,q^h}(n)=\sum _{k=0}^{n-1}{[k]}_q^l{q}^{hk},$$

(1.1)

$$T_{l,t}=\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^l{t}^{k-1}$$

(1.2)

and

$$\begin{array}{rcl}{S}_{l,n}(q)& =& \sum _{k=1}^n\frac{1-q^{2k}}{1-q^2}{\left(\frac{1-q^k}{1-q}\right)}^{l-1}{q}^{\frac{(l+1)(n-k)}{2}}\\ =& \sum _{k=1}^n{[k]}_q^l\frac{1+q^k}{1+q}{q}^{\frac{(l+1)(n-k)}{2}}\\ =& \sum _{k=0}^{n-1}{[k+1]}_q^l\frac{1+q^{k+1}}{1+q}{q}^{\frac{(l+1)(n-k-1)}{2}}\end{array}$$

with $s\in \mathbb{C}$ and $|s|<1$.

In Section 2, we recall some necessary identities for basic hypergeometric series [14]. Further, we obtain a generalization of Proposition 2.1, and accordingly, we obtain q-consecutive integers for ${\sum }_{k=0}^{n-1}{[k]}_q{q}^k$. These new results are similar to the ones presented in some other studies [7–9] and [13].

In Section 3, we derive a formula for $S_{2,n}$ and $T_{2,q}(n)$ by using a property of basic hypergeometric series, such as

$$\sum _{k=0}^{\mathrm{\infty }}\frac{{(aq;q)}_k{(cq;q)}_k}{{(q;q)}_k{(bq;q)}_k}{t}^k=\frac{{(aq;q)}_{\mathrm{\infty }}{(ctq;q)}_{\mathrm{\infty }}}{{(bq;q)}_{\mathrm{\infty }}{(t;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(b/a;q)}_k{(t;q)}_k}{{(q;q)}_k{(ctq;q)}_k}{a}^k{q}^k.$$

The q-analogue Eulerian numbers are defined as [15]:

$$\begin{array}{r}{E}_{m,q}=\underset{N\to \mathrm{\infty }}{lim}\frac{1}{{[p^N]}_{-q}}\sum _{l=0}^{p^N-1}{[l]}_q^m{(-q)}^l\text{and}\\ E_{n,q}(x)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)q^{lx}{E}_{l,q}{[x]}_q^{n-l}.\end{array}$$

For these, we establish certain new identities by utilizing basic hypergeometric series, which differ from Bernoulli numbers and polynomials constructed by Kim et al. [16, 17] as follows:

$${(-1)}^{n+1}{q}^n{E}_{1,q}(n)+E_{1,q}=\{\begin{array}{ll}{[n]}_q-{[2]}_q^2\frac{{[2n]}_q}{{[4]}_q}& \text{if }n\text{ is even,}\\ {[\mathrm{\infty }]}_q(\frac{{[2n]}_q}{{[n]}_q}-\frac{{[2]}_q^2{[4n]}_q}{{[4]}_q{[2n]}_q})& \text{otherwise}\end{array}$$

and

$${(-1)}^{n+1}{q}^n{E}_{2,q}(n)+E_{2,q}=\{\begin{array}{ll}{[\mathrm{\infty }]}_q({[n]}_q-2\frac{{[2]}_q^2{[2n]}_q}{{[4]}_q}+\frac{{[2]}_q{[3]}_q{[3n]}_q}{{[6]}_q})& \text{if }n\text{ is even,}\\ {[\mathrm{\infty }]}_q^2(\frac{{[2n]}_q}{{[n]}_q}-2\frac{{[2]}_q^2{[4n]}_q}{{[4]}_q{[2n]}_q}+\frac{{[2]}_q{[3]}_q{[6n]}_q}{{[6]}_q{[3n]}_q})& \text{otherwise.}\end{array}$$

Here, we note that these are related to $T_{l,q^h}(n)$.

In Section 4, we deduce recursive formulas from Lemma 4.1 for basic hypergeometric series. More precisely, let $S_l(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{{(q^2;q)}_k^l}{{(q;q)}_k^l}{t}^k$. Then, we derive the recursive formulas

$$\begin{array}{rcl}{S}_{l+1}(t)& =& \frac{1}{1-q}(S_l(t)-q{S}_l(tq))\\ =& \frac{1}{{(1-q)}^l}\sum _{k=0}^{l+1}\left(\begin{array}{c}l+1\\ k\end{array}\right)S_0\left(t{q}^k\right){(-q)}^k.\end{array}$$

Using these identities, we obtain a formula for ${\sum }_{k=0}^{n-1}{[k]}^l{q}^k$, and we present relations between q-Bernoulli numbers and q-consecutive integers, which are related to (S1)-(S4). Lastly, the rank of partition is defined as the difference between its largest part and the number of its parts. The number of partitions of n with the rank r would be denoted by $P_r(n)$. We use the convention $P_0(0)=1$, $P_r(n)=0$ for $r\ne 0$, $n\le 0$ and $r=0$, $n<0$. Here, for the sake of convenience, we define

$$\begin{array}{c}{C}_1(t;q)=1,\hfill \\ C_l(t;q)=\frac{1}{1-q}(C_{l-1}(t;q)(1-t{q}^l)-q{C}_{l-1}(tq;q)(1-t)).\hfill \end{array}$$

Then, these are related to $P_r(n)$ by the following identity (Remark 4.13):

$$\sum _{l=0}^{\mathrm{\infty }}\frac{T_{l,\frac{q}{u}}}{C_l(u^{-1}q;q)}{u}^l{q}^{l+1}=\sum _{r=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{P}_r(n)u^r{q}^n$$

(1.3)

with ${|q|}^2<|u|<1$. Finally, we shall relate through Theorem 4.7 and Remark 4.14, q-Bernoulli polynomials with the third-order mock theta functions introduced by Ramanujan.

Throughout this paper, we adopt the following notations:

• ${[k]}_t=\frac{1-t^k}{1-t}$.

• ${[\mathrm{\infty }]}_t=\frac{1}{1-t}$.

• $S_{m,n}(q)={\sum }_{k=1}^n\frac{1-q^{2k}}{1-q^2}{(\frac{1-q^k}{1-q})}^{m-1}{q}^{\frac{m+1}{2}(n-k)}$.

• $T_{l,q^h}(n)={\sum }_{k=0}^{n-1}{[k]}_q^l{q}^{hk}$.

• ${(a;q)}_n=(1-a)(1-aq)(1-a{q}^2)\cdots (1-a{q}^{n-1})$.

• ${(a;q)}_{\mathrm{\infty }}={\prod }_{n=0}^{\mathrm{\infty }}(1-a{q}^n)$.

• ${(a;q)}_0=1$.

• ${\left[\begin{array}{c}m\\ k\end{array}\right]}_q=\{\begin{array}{ll}\frac{{(q;q)}_m}{{(q;q)}_k{(q;q)}_{m-k}}& \text{if }0\le k\le m,\\ 0& \text{otherwise.}\end{array}$

• ω: the r th root of unity.

• $F(a,b;t):=F(a,b;t:q)=1+{\sum }_{n=1}^{\mathrm{\infty }}\frac{(1-aq)(1-a{q}^2)\cdots (1-a{q}^n)}{(1-bq)(1-b{q}^2)\cdots (1-b{q}^n)}{t}^n={\sum }_{n=0}^{\mathrm{\infty }}\frac{{(aq;q)}_n}{{(bq;q)}_n}{t}^n$.

2 Identities of basic hypergeometric series and ${\sum }_{k=0}^{n-1}{[k]}_q{q}^k$

In this section, we investigate some identities of basic hypergeometric series. To this end, we refer to [14]. Now, we consider the series defined by

$$F(a,b;t:q)=1+\sum _{n=1}^{\mathrm{\infty }}\frac{(1-aq)(1-a{q}^2)\cdots (1-a{q}^n)}{(1-bq)(1-b{q}^2)\cdots (1-b{q}^n)}{t}^n.$$

(2.1)

Fine presented many interesting properties in his book; the following identity represents one such property:

$$F(a,b;t:q)=\frac{1-b}{1-t}+\frac{b-atq}{1-t}F(a,b;tq:q).$$

(2.2)

Throughout this paper, q denotes a fixed complex number of absolute value less than 1, so that we may write $q=exp(\pi i\tau )$, where τ is a complex number with a positive imaginary part. We use $q^c$ to denote $exp(c\pi i\tau )$. The partial product ${(aq;q)}_n$ converges for all values of a, as may be easily seen from the absolute convergence of $\sum q^n$. Hence, if b is not one of the values $q^{-1},q^{-2},\dots$ , the coefficients $\frac{{(aq;q)}_n}{{(bq;q)}_n}$ are bounded, and the series (2.1) converges for all t inside the unit circle, and represents an analytic function therein. Hence, the function on the right-hand of (2.2) is regular in the domain $|t|<{|q|}^{-1}$, except for a simple pole at $t=1$. Therefore, we obtain the continuation of F to a larger circle. Then, it is easy to apply (2.2) again to the continuation of F to the circle $|t|<{|q|}^{-2}$, and thus, we conclude that for $b\ne q^{-n}$, $n>1$, the only possible singularities of F occur at the points $t=q^{-n}$ ($n\ge 0$), which are simple poles in general. As a function of b, F is regular, except possibly at the simple poles $b=q^{-n}$ ($g\ge 1$), provided that b and t do not have one of the singular values mentioned above. First, we derive Theorem 2.2 by generalizing the following proposition.

Proposition 2.1 For the complex number q and t with $|q|<1$, we have

$$\sum _{m=0}^{\mathrm{\infty }}\frac{{(aq;q)}_{2m}{(bq;q)}_m}{{(aq;q)}_m{(q;q)}_m}{t}^m=\frac{{(btq;q)}_{\mathrm{\infty }}}{{(t;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(bq;q)}_k{(t;q)}_k}{{(q;q)}_k{(btq;q)}_{2k}}{(-at)}^k{q}^{\frac{3k^2+k}{2}}.$$

Proof Equation (25.96) in [14]. □

Theorem 2.2 For complex numbers q, t with $|q|<1$ and an integer $l\ge 0$, we get

$$\sum _{m=0}^{\mathrm{\infty }}\frac{{(aq;q)}_{(l+1)m}{(bq;q)}_m}{{(aq;q)}_{lm}{(q;q)}_m}{t}^m=\frac{{(btq;q)}_{\mathrm{\infty }}}{{(t;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(bq;q)}_k{(t;q)}_{lk}}{{(q;q)}_k{(btq;q)}_{(l+1)k}}{(-at)}^k{q}^{\frac{(2l+1)k^2+k}{2}}.$$

To prove this, we need some identities from [14].

Lemma 2.3 (1) For a nonnegative integer N,

$${(tq;q)}_N=\sum _{k=0}^N{\left[\begin{array}{c}N\\ k\end{array}\right]}_q{(-t)}^k{q}^{\frac{k^2+k}{2}}.$$

It is an analogue of the binomial series, to which one can reduce termwise with $q=1$.

• (2)
  $$F(a,1;t:q)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(aq;q)}_n}{{(q;q)}_n}{t}^n=\frac{{(atq;q)}_{\mathrm{\infty }}}{{(t;q)}_{\mathrm{\infty }}}.$$
• (3)
  $${(aq;q)}_{\mathrm{\infty }}=\sum _{k=0}^{\mathrm{\infty }}\frac{{(-a)}^k}{{(q;q)}_k}{q}^{\frac{k^2+k}{2}}.$$

Proof Equations (6.23), (6.2), and (12.44) in [14], respectively. □

Proof of Theorem 2.2 We start with the left-hand side in our assertion:

$$\sum _{m=0}^{\mathrm{\infty }}\frac{{(aq;q)}_{(l+1)m}{(bq;q)}_m}{{(aq;q)}_{lm}{(q;q)}_m}{t}^m=\sum _{m=0}^{\mathrm{\infty }}{(a{q}^{lm+1};q)}_m\frac{{(bq;q)}_m}{{(q;q)}_m}{t}^m.$$

Replacing t by $a{q}^{lm}$ in Lemma 2.3(1), we claim that

$$\begin{array}{r}\sum _{m=0}^{\mathrm{\infty }}\frac{{(bq;q)}_m}{{(q;q)}_m}{t}^m\sum _{k=0}^m{\left[\begin{array}{c}m\\ k\end{array}\right]}_q{(-a{q}^{lm})}^k{q}^{\frac{k^2+k}{2}}\\ =\sum _{k=0}^{\mathrm{\infty }}{(-a)}^k{q}^{\frac{k^2+k}{2}}\sum _{m=k}^{\mathrm{\infty }}{\left[\begin{array}{c}m\\ k\end{array}\right]}_q{t}^m{q}^{klm}\frac{{(bq;q)}_m}{{(q;q)}_m}\\ =\sum _{k=0}^{\mathrm{\infty }}{(-a)}^k{q}^{\frac{k^2+k}{2}}\sum _{n=0}^{\mathrm{\infty }}{\left[\begin{array}{c}n+k\\ k\end{array}\right]}_q{t}^{n+k}{q}^{kl(n+k)}\frac{{(bq;q)}_{n+k}}{{(q;q)}_{n+k}}\\ =\sum _{k=0}^{\mathrm{\infty }}{(-a)}^k{q}^{\frac{k^2+k}{2}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(q;q)}_{n+k}}{{(q;q)}_k{(q;q)}_n}{t}^{n+k}{q}^{kl(n+k)}\frac{{(bq;q)}_{n+k}}{{(q;q)}_{n+k}}\\ =\sum _{k=0}^{\mathrm{\infty }}{(-at)}^k{q}^{\frac{(2l+1)k^2+k}{2}}\frac{{(bq;q)}_k}{{(q;q)}_k}\sum _{n=0}^{\mathrm{\infty }}\frac{{(b{q}^{k+1};q)}_n}{{(q;q)}_n}{t}^n{q}^{kln}.\end{array}$$

By substituting $b{q}^k$ and $t{q}^{kl}$ for a and t, respectively, in Lemma 2.3(2), we derive

$$\sum _{m=0}^{\mathrm{\infty }}\frac{{(aq;q)}_{(l+1)m}{(bq;q)}_m}{{(aq;q)}_{lm}{(q;q)}_m}{t}^m=\sum _{k=0}^{\mathrm{\infty }}\frac{{(bq;q)}_k}{{(q;q)}_k}{(-at)}^k{q}^{\frac{(2l+1)k^2+k}{2}}\frac{{(bt{q}^{(l+1)k+1};q)}_{\mathrm{\infty }}}{{(t{q}^{lk};q)}_{\mathrm{\infty }}}.$$

Since ${(a;q)}_n=\frac{{(a;q)}_{\mathrm{\infty }}}{{(a{q}^n;q)}_{\mathrm{\infty }}}$, it follows that the last can be written as

$$\frac{{(btq;q)}_{\mathrm{\infty }}}{{(t;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(bq;q)}_k{(t;q)}_{lk}}{{(q;q)}_k{(btq;q)}_{(l+1)k}}{(-at)}^k{q}^{\frac{(2l+1)k^2+k}{2}}.$$

Thus, we deduce the identity as desired. □

Next, we present alternative proofs of the following results of Kim [6] as an application of Theorem 2.2.

Corollary 2.4 (1)

$$\sum _{k=0}^{n-1}{[k]}_q{t}^k=\frac{1}{1-q}\sum _{k=0}^{n-1}(1-q^k)t^k=\frac{1}{1-q}(\frac{1-t^n}{1-t}-\frac{1-t^n{q}^n}{1-tq}).$$

• (2)
  $$T_{1,q}=\sum _{k=0}^{n-1}{[k]}_q{q}^k=q{\left[\begin{array}{c}n\\ 2\end{array}\right]}_q=\frac{1}{2}({[n]}_q^2-\frac{{[2n]}_q}{{[2]}_q}).$$

Note that it is exactly the same as (1.1).

• (3)
  $$\sum _{k=1}^n{[k]}_q{q}^{k-1}=\sum _{k=1}^n{[k]}_q{q}^{2n-2k}=\sum _{k=1}^n{[k]}_q\frac{1+q^k}{1+q}{q}^{n-k}={\left[\begin{array}{c}n+1\\ 2\end{array}\right]}_q.$$

These are the q-analogues of ${\sum }_{k=1}^{n}k=\left(\begin{array}{c}n+1\\ 2\end{array}\right)$.

Proof (1) Replacing both b and l by 0 in Theorem 2.2, we see that

$$\sum _{k=0}^{\mathrm{\infty }}\frac{{(aq;q)}_k}{{(q;q)}_k}{t}^k=\frac{1}{{(t;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}{\frac{1}{(q;q)}}_k{(-at)}^k{q}^{\frac{k^2+k}{2}}.$$

After substituting at for a in Lemma 2.3(3), if we apply it to the above, we get

$$\sum _{k=0}^{\mathrm{\infty }}\frac{{(aq;q)}_k}{{(q;q)}_k}{t}^k=\frac{{(atq;q)}_{\mathrm{\infty }}}{{(t;q)}_{\mathrm{\infty }}}.$$

Putting $a=q$ in the above, we get

$$\sum _{k=0}^{\mathrm{\infty }}\frac{{(q^2;q)}_k}{{(q;q)}_k}{t}^k=\frac{1}{(1-t)(1-tq)}.$$

However, by using the notation defined in Section 1, the left-hand side of the above can be written as

$$\sum _{k=0}^{\mathrm{\infty }}\frac{1-q^{k+1}}{1-q}{t}^k=\sum _{k=0}^{\mathrm{\infty }}{[k+1]}_q{t}^k.$$

Thus, from the calculations above and the fact that ${[0]}_q=1$, we derive

$$\begin{array}{rcl}\sum _{k=0}^{\mathrm{\infty }}{[k]}_q{t}^k& =& \sum _{k=0}^{\mathrm{\infty }}{[k+1]}_q{t}^{k+1}=\frac{t}{(1-t)(1-tq)}\\ =& \frac{1}{1-q}(\frac{1}{1-t}-\frac{1}{1-tq})=\frac{1}{1-q}\left(\sum _{k=0}^{\mathrm{\infty }}(1-q^k)t^k\right).\end{array}$$

By considering the exponent of t, we conclude that

$$\sum _{k=0}^{n-1}{[k]}_q{t}^k=\frac{1}{1-q}\sum _{k=0}^{n-1}(1-q^k)t^k=\frac{1}{1-q}(\frac{1-t^n}{1-t}-\frac{1-t^n{q}^n}{1-tq}).$$

• (2) If we put $t=q$ in (1), we have the first equality

  $$\begin{array}{rcl}{T}_{1,q}& =& \sum _{k=0}^{n-1}{[k]}_q{q}^k=\frac{1}{1-q}(\frac{1-q^n}{1-q}-\frac{1-q^{2n}}{1-q^2})\\ =& \frac{q(1-q^{n-1})(1-q^n)}{(1-q)(1-q^2)}=q{\left[\begin{array}{c}n\\ 2\end{array}\right]}_q.\end{array}$$

On the other hand, a direct calculation gives

$$\frac{1}{1-q}(\frac{1-q^n}{1-q}-\frac{1-q^{2n}}{1-q^2})={[\mathrm{\infty }]}_q({[n]}_q-\frac{{[2n]}_q}{{[2]}_q})=\frac{1}{2}({[n]}_q^2-\frac{{[2n]}_q}{{[2]}_q}).$$

Therefore, we establish the claim.

• (3) It follows from (2) that

  $$\sum _{k=1}^n{[k]}_q{q}^{k-1}={\left[\begin{array}{c}n+1\\ 2\end{array}\right]}_q.$$

  (2.3)

Moreover, if we replace $n-1$ by n in (1) and multiply both sides by $q^{2n}$, we obtain

$$q^{2n}\sum _{k=1}^n{[k]}_q{t}^k=\frac{q^{2n}}{1-q}({[n+1]}_t-{[n+1]}_{tq}).$$

Observe that the identity above with $t=q^{-2}$ turns out to be a Warnaar’s identity [13]:

$$\sum _{k=1}^n{[k]}_q{q}^{2n-2k}={\left[\begin{array}{c}n+1\\ 2\end{array}\right]}_q.$$

(2.4)

Finally, on the basis of geometric series, we get

$$q^n\sum _{k=0}^n\frac{1-q^{2k}}{1-q^2}{t}^k=\frac{q^n}{1-q^2}(\frac{1-t^{n+1}}{1-t}-\frac{1-{(t{q}^2)}^{n+1}}{1-t{q}^2}).$$

When $t=q^{-1}$ in the above, we have

$$\begin{array}{rcl}\sum _{k=0}^n\frac{1-q^{2k}}{1-q^2}{q}^{n-k}& =& \frac{q^n}{1-q^2}(\frac{1-q^{-n-1}}{1-q^{-1}}-\frac{1-q^{n+1}}{1-q})\\ =& \frac{(1-q^n)(1-q^{n+1})}{(1-q)(1-q^2)}={\left[\begin{array}{c}n+1\\ 2\end{array}\right]}_q.\end{array}$$

Note that this formula was also derived by several mathematicians such as Schlosser [17] and Warnaar [13]. Since ${[0]}_q=0$,

$$\sum _{k=1}^n\frac{1+q^k}{1+q}{[k]}_q{q}^{n-k}=\sum _{k=0}^n\frac{(1-q^k)(1+q^k)}{(1-q)(1+q)}{q}^{n-k}={\left[\begin{array}{c}n+1\\ 2\end{array}\right]}_q.$$

(2.5)

Thus, by combining (2.3), (2.4), and (2.5), we reach the conclusion. □

3 q-Consecutive and q-analogue of Eulerian numbers $E_{1,q}$and $E_{2,q}$

We have studied the infinite sum with linear coefficients for q-numbers in the previous section. In this section, we consider the sum with quadratic coefficients, i.e., the following equation.

Lemma 3.1

$$\begin{array}{rcl}\sum _{k=0}^{n-1}{[k]}_q^2{t}^k& =& \frac{1}{{(1-q)}^2}\sum _{k=0}^{n-1}(1-2q^k+q^{2k})t^k\\ =& \frac{1}{{(1-q)}^2}(\frac{1-t^n}{1-t}-2\frac{1-t^n{q}^n}{1-tq}+\frac{1-t^n{q}^{2n}}{1-t{q}^2}).\end{array}$$

Theorem 3.2 As a finite sum for q, we get

$$\begin{array}{rcl}\sum _{k=0}^{n-1}{[k]}_q^2{q}^{k+1}& =& {[\mathrm{\infty }]}_q({[\mathrm{\infty }]}_q-{[1]}_q)({[n]}_q-2\frac{{[2n]}_q}{{[2]}_q}+\frac{{[3n]}_q}{{[3]}_q})\\ =& \frac{1}{3}({[n]}_q^3-\frac{{[3n]}_q}{{[3]}_q})-\frac{1}{2}({[n]}_q^2-\frac{{[2n]}_q}{{[2]}_q}).\end{array}$$

Proof Putting $t=q$ in Lemma 3.1, we derive

$$\begin{array}{rcl}\sum _{k=0}^{n-1}{[k]}_q^2{q}^k& =& \frac{1}{{(1-q)}^2}(\frac{1-q^n}{1-q}-2\frac{1-q^{2n}}{1-q^2}+\frac{1-q^{3n}}{1-q^3})\\ =& {[\mathrm{\infty }]}_q^2({[n]}_q-2\frac{{[2n]}_q}{{[2]}_q}+\frac{{[3n]}_q}{{[3]}_q}).\end{array}$$

Since ${[\mathrm{\infty }]}_q-{[1]}_q=\frac{1}{1-q}-\frac{1-q}{1-q}=\frac{q}{1-q}$, we get the first equality.

By routine calculations, we have

$$\begin{array}{r}\frac{q}{{(1-q)}^2}(\frac{1-q^n}{1-q}-2\frac{1-q^{2n}}{1-q^2}+\frac{1-q^{3n}}{1-q^3})\\ =\frac{q^2(1-q^{n-1})(1-q^n)}{(1-q)(1-q^2)(1-q^3)}\{q^2(1-q^{n-1})+(1-q^n)\}\\ =\frac{q^2{[n-1]}_q{[n]}_q}{{[2]}_q{[3]}_q}(q^2{[n-1]}_q+{[n]}_q).\end{array}$$

Moreover, $\frac{1}{3}({[n]}_q^3-\frac{{[3n]}_q}{{[3]}_q})-\frac{1}{2}({[n]}_q^2-\frac{{[2n]}_q}{{[2]}_q})$ is also equal to

$$\frac{q^2{[n-1]}_q{[n]}_q}{{[2]}_q{[3]}_q}(q^2{[n-1]}_q^2+{[n]}_q)$$

by the definition of ${[n]}_q$. Therefore, the last equality follows. □

Theorem 3.3 Let $S_l(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{{(q^2;q)}_k^l}{{(q;q)}_k^l}{t}^k$. For $l=2$, we obtain

$$S_{2,n}(q)=\frac{(1-q^{n+\frac{1}{2}})(1-q^n)(1-q^{n+1})}{(1-q^{\frac{3}{2}})(1-q)(1-q^2)}.$$

The other cases $l>2$ will be studied in greater detail in the next section. This was previously proved by Schlosser [17]by using Bailey’s terminating very-well-poised balanced ${}_{10}\varphi _9$ transformation.

Proof By definition in Section 1, we see that

$$\begin{array}{rcl}{S}_{2,n}(q)& =& \sum _{k=1}^n\frac{1-q^{2k}}{1-q}\left(\frac{1-q^k}{1-q}\right)q^{\frac{3n-3k}{2}}\\ =& \sum _{k=1}^n\frac{1+q^k}{1+q}{\left(\frac{1-q^k}{1-q}\right)}^2{q}^{\frac{3n-3k}{2}}\\ =& \frac{q^{\frac{3}{2}n}}{1+q}(\sum _{k=1}^n{[k]}_q^2{q}^{-\frac{3}{2}k}+\sum _{k=1}^n{[k]}_q^2{q}^{-\frac{1}{2}k}).\end{array}$$

From Lemma 3.1, we know that

$$\sum _{k=0}^n{[k]}_q^2{t}^k=\frac{1}{{(1-q)}^2}(\frac{1-t^{n+1}}{1-t}-2\frac{1-t^{n+1}{q}^{n+1}}{1-tq}+\frac{1-t^{n+1}{q}^{2n+2}}{1-t{q}^2}).$$

Then, the sum of formulas after setting $t=q^{-\frac{3}{2}}$ and $t=q^{-\frac{1}{2}}$ shows that our corollary is true. □

Carlitz [18] constructed a q-analogue of Eulerian numbers. On the other hand, Kim considered the following functions [19]:

$$H_q(t)={[2]}_q{e}^{\frac{t}{1-q}}\sum _{j=0}^{\mathrm{\infty }}\frac{{(-1)}^j}{1+q^{j+1}}{\left(\frac{1}{1-q}\right)}^j\frac{t^j}{j!}=\sum _{k=0}^{\mathrm{\infty }}{E}_{k,q}\frac{t^k}{k!}$$

and

$$H_q(x,t)={[2]}_q{e}^{\frac{t}{1-q}}\sum _{j=0}^{\mathrm{\infty }}\frac{{(-1)}^j{q}^{jx}}{1+q^{j+1}}{\left(\frac{1}{1-q}\right)}^j\frac{t^j}{j!}=\sum _{k=0}^{\mathrm{\infty }}{E}_{k,q}(x)\frac{t^k}{k!}.$$

For $m,n\in \mathbb{N}$, he showed that [[19], Proposition 2]

$$\sum _{k=0}^{n-1}{(-1)}^k{[k]}_q^l{q}^k=\frac{1}{{[2]}_q}({(-1)}^{n+1}{q}^n{E}_{l,q}(n)+E_{l,q}).$$

(3.1)

A similar result is in [[8], Lemma 2.1]. Thus, we get the results for $l=1$ and 2 as follows.

Theorem 3.4 (1)

$${(-1)}^{n+1}{q}^n{E}_{1,q}(n)+E_{1,q}=\{\begin{array}{ll}{[n]}_q-{[2]}_q^2\frac{{[2n]}_q}{{[4]}_q}& \mathit{\text{if }}n\mathit{\text{ is even,}}\\ {[\mathrm{\infty }]}_q(\frac{{[2n]}_q}{{[n]}_q}-\frac{{[2]}_q^2{[4n]}_q}{{[4]}_q{[2n]}_q})& \mathit{\text{otherwise.}}\end{array}$$

(2)

$$\begin{array}{r}{(-1)}^{n+1}{q}^n{E}_{2,q}(n)+E_{2,q}\\ =\{\begin{array}{ll}{[\mathrm{\infty }]}_q({[n]}_q-2\frac{{[2]}_q^2{[2n]}_q}{{[4]}_q}+\frac{{[2]}_q{[3]}_q{[3n]}_q}{{[6]}_q})& \mathit{\text{if }}n\mathit{\text{ is even,}}\\ {[\mathrm{\infty }]}_q^2(\frac{{[2n]}_q}{{[n]}_q}-2\frac{{[2]}_q^2{[4n]}_q}{{[4]}_q{[2n]}_q}+\frac{{[2]}_q{[3]}_q{[6n]}_q}{{[6]}_q{[3n]}_q})& \mathit{\text{otherwise.}}\end{array}\end{array}$$

Proof Replacing t by −t in Corollary 2.4(1), we see that

$$\sum _{k=0}^{n-1}{(-1)}^k{[k]}_q{t}^k=\frac{1}{1-q}(\frac{1-{(-t)}^n}{1+t}-\frac{1-{(-tq)}^n}{1+tq}).$$

If we let $t=q$, it becomes

$$\sum _{k=0}^{n-1}{(-1)}^k{[k]}_q{q}^k=\frac{1}{1-q}(\frac{1-{(-q)}^n}{1+q}-\frac{1-{(-1)}^n{q}^{2n}}{1+q^2}).$$

(3.2)

Therefore,

$$\sum _{k=0}^{n-1}{(-1)}^k{[k]}_q{q}^k=\{\begin{array}{ll}\frac{{[n]}_q}{{[2]}_q}-\frac{{[2n]}_q{[2]}_q}{{[4]}_q}& \text{if }n\text{ is even,}\\ {[\mathrm{\infty }]}_q(\frac{{[2n]}_q}{{[2]}_q{[n]}_q}-\frac{{[2]}_q{[4n]}_q}{{[4]}_q{[2n]}_q})& \text{otherwise.}\end{array}$$

Comparing this with (3.1), we can prove (1).

As for (2), if we substitute −q with t in Lemma 3.1, it turns out that

$$\sum _{k=0}^{n-1}{(-1)}^k{[k]}_q^2{q}^k=\frac{1}{{(1-q)}^2}(\frac{1-{(-q)}^n}{1+q}-2\frac{1-{(-q^2)}^n}{1+q^2}+\frac{1-{(-q^3)}^n}{1+q^3}).$$

For an even integer n, the above becomes

$$\begin{array}{r}\frac{1}{1-q}(\frac{1-q^n}{1-q^2}-2\frac{(1-q^{2n})(1-q^2)}{(1-q)(1-q^4)}+\frac{(1-qf{o}^{3n})(1-q^3)}{(1-q)(1-q^6)})\\ ={[\mathrm{\infty }]}_q(\frac{{[n]}_q}{{[2]}_q}-2\frac{{[2n]}_q{[2]}_q}{{[4]}_q}+\frac{{[3n]}_q{[3]}_q}{{[6]}_q}).\end{array}$$

Similarly, for an odd integer n, we get the result

$${[\mathrm{\infty }]}_q^2(\frac{{[2n]}_q}{{[2]}_q{[n]}_q}-2\frac{{[2]}_q{[4n]}_q}{{[4]}_q{[2n]}_q}+\frac{{[3]}_q{[6n]}_q}{{[6]}_q{[3n]}_q})$$

as desired. □

Remark 3.5 In the proof above, as in the case of (3.2), we can obtain an equation by plugging −q into t:

$$\sum _{k=0}^{n-1}{[k]}_q{q}^k=\frac{1}{1-q}(\frac{1-q^n}{1-q}-\frac{1-q^{2n}}{1-q^2}).$$

(3.3)

Further, we have $2{\sum }_{k=0}^{n-1}{[2k]}_q{q}^{2k}$ by adding (3.2) and (3.3). Indeed,

$$\begin{array}{r}\sum _{k=0}^{2n-1}{(-1)}^k{[k]}_q{q}^k+\sum _{k=0}^{2n-1}{[k]}_q{q}^k\\ =\frac{1}{1-q}(\frac{1-q^{2n}}{1+q}-\frac{1-q^{4n}}{1+q^2})+\frac{1}{1-q}(\frac{1-q^{2n}}{1-q}-\frac{1-q^{4n}}{1-q^2})\\ =\frac{2}{1-q}(\frac{1-q^{2n}}{1-q^2}-\frac{1-q^{4n}}{1-q^4}).\end{array}$$

Note that this can be written as $2{[\mathrm{\infty }]}_q(\frac{{[2n]}_q}{{[2]}_q}-\frac{{[4n]}_q}{{[4]}_q})$ in terms of q-number notation. Alternatively, it may be factorized and expressed as

$$\frac{2q^2(1-q^{2n-2})(1-q^{2n})}{(1-q)(1-q^4)}=\frac{2q^2{[2n-2]}_q{[2n]}_q}{{[4]}_q},$$

from which we derive

$$\sum _{k=0}^{n-1}{[2k]}_q{q}^{2k}={[\mathrm{\infty }]}_q(\frac{{[2n]}_q}{{[2]}_q}-\frac{{[4n]}_q}{{[4]}_q})=\frac{q^2{[2n-2]}_q{[2n]}_q}{{[4]}_q}.$$

Remark 3.6 In [[15], Theorem 1], Kim derived a summation formula for $E_{m,q}$,

$$E_{m,q}={[2]}_q{\left(\frac{1}{1-q}\right)}^m\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right){(-1)}^k\frac{1}{1+q^{k+1}}.$$

We also see one for $E_{k,q}(x)$,

$$E_{k,q}(x)={[2]}_q\sum _{m=1}^{\mathrm{\infty }}{(-1)}^n{q}^n{[n+x]}_q^k$$

for any positive integer k from [[20], Proposition 1]. The proof of Theorem 3.4 is obtained without the help of the summations above.

4 Difference equation and q-consecutive integer

As mentioned in Theorem 3.3, in this section, we study ${\sum }_{k=0}^{\mathrm{\infty }}\frac{{(q^2;q)}_k^l}{{(q;q)}_k^l}{t}^k$ for more general cases l and its similar sum with q-binomial coefficients. In addition, we show the relations between these and twisted q-Bernoulli numbers. To this end, we need the following lemma.

Lemma 4.1 Given a sequence $A_k$ ($k\ge 0$) for which $g(t)={\sum }_{k=0}^{\mathrm{\infty }}{A}_k{t}^k$ converges,

$$\sum _{k=0}^{\mathrm{\infty }}\frac{{(aq;q)}_k}{{(bq;q)}_k}{A}_k{t}^k=\frac{{(aq;q)}_{\mathrm{\infty }}}{{(bq;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(b/a;q)}_k}{{(q;q)}_k}{a}^k{q}^{k}g\left(t{q}^k\right).$$

Proof See Section 20, [14]. □

Using this lemma, we generalize the identities considered in the previous two sections.

Proposition 4.2 For a positive integer l, we have the identities

$$\begin{array}{rcl}\sum _{k=0}^{n-1}{[k]}_q^l{q}^k& =& \frac{q}{{(1-q)}^l}(\frac{1-q^{n-1}}{1-q}-lq\frac{1-q^{2(n-1)}}{1-q^2}+\cdots +{(-1)}^l{q}^l\frac{1-q^{(l+1)(n-1)}}{1-q^{l+1}})\\ =& \frac{q}{{(1-q)}^l}\sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)\frac{1-q^{(k+1)(n-1)}}{1-q^{k+1}}{(-q)}^k.\end{array}$$

If l is 1 (respectively, 2), this would be the result of Corollary  2.4(1) (respectively, Lemma  3.1).

Proof Let $S_l(t)$ be a series defined by ${\sum }_{k=0}^{\mathrm{\infty }}\frac{{(q^2;q)}_k^l}{{(q;q)}_k^l}{t}^k$ for a nonnegative integer l. By setting $a=q$, $b=1$, $A_k=\frac{{(q^2;q)}_k^l}{{(q;q)}_k^l}$, and $g(t)=S_l(t)$ in Lemma 4.1, we derive the following recursive formula:

$$\begin{array}{rl}{S}_{l+1}(t)& =\sum _{k=0}^{\mathrm{\infty }}\frac{{(q^2;q)}_k}{{(q;q)}_k}\frac{{(q^2;q)}_k^l}{{(q;q)}_k^l}{t}^k=\frac{{(q^2;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(1/q;q)}_k}{{(q;q)}_k}{q}^{2k}{S}_l\left(t{q}^k\right)\\ =\frac{1}{1-q}(S_l(t)+\frac{1-\frac{1}{q}}{1-q}{q}^2{S}_l(tq))=\frac{1}{1-q}(S_l(t)-q{S}_l(tq)).\end{array}$$

(4.1)

Multiplying both sides by t, we get

$$\begin{array}{c}t{S}_1(t)=\frac{t}{1-q}(S_0(t)-q{S}_0(tq)),\hfill \\ t{S}_2(t)=\frac{t}{1-q}(S_1(t)-q{S}_1(tq))=\frac{t}{{(1-q)}^2}(S_0(t)-2q{S}_0(tq)+q^2{S}_0\left(t{q}^2\right)).\hfill \end{array}$$

Further, by induction,

$$t{S}_l(t)=\frac{t}{{(1-q)}^l}\sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)S_0\left(t{q}^k\right){(-q)}^k.$$

Considering $t{S}_0(t)=t{\sum }_{k=0}^{\mathrm{\infty }}{t}^k=\frac{t}{1-t}$, we are able to rewrite the above as

$$\begin{array}{r}\sum _{k=0}^{\mathrm{\infty }}{[k]}_q{t}^k=\frac{t}{1-q}(\frac{1}{1-t}-\frac{q}{1-tq}),\\ \sum _{k=0}^{\mathrm{\infty }}{[k]}_q^2{t}^k=\frac{t}{{(1-q)}^2}(\frac{1}{1-t}-\frac{2q}{1-tq}+\frac{q^2}{1-t{q}^2}),\\ \dots ,\\ \sum _{k=0}^{\mathrm{\infty }}{[k]}_q^l{t}^k=\frac{t}{{(1-q)}^l}(\frac{1}{1-t}-\frac{lq}{1-tq}+\cdots +\frac{{(-q)}^l}{1-t{q}^l})=\frac{t}{{(1-q)}^l}\sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)\frac{{(-q)}^k}{1-t{q}^k}.\end{array}$$

(4.2)

If we take a finite sum from the above, we get

$$\begin{array}{rl}\sum _{k=0}^{n-1}{[k]}_q^l{t}^k& =\frac{t}{{(1-q)}^l}(\sum _{k=0}^{n-2}{t}^k-lq\sum _{k=0}^{n-2}{t}^k{q}^k+\cdots +{(-q)}^l\sum _{k=0}^{n-2}{t}^k{q}^{lk})\\ =\frac{t}{{(1-q)}^l}(\frac{1-t^{n-1}}{1-t}-lq\frac{1-t^{n-1}{q}^{n-1}}{1-tq}+\cdots +{(-q)}^l\frac{1-t^{n-1}{q}^{l(n-1)}}{1-t{q}^l})\\ =\frac{t}{{(1-q)}^l}\sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right)\frac{1-{(t{q}^k)}^{n-1}}{1-t{q}^k}{(-q)}^k.\end{array}$$

(4.3)

Therefore, if we let $t=q$, we are done, which amounts to recovering Corollary 2.4, Lemma 3.1, and Theorem 3.2. □

Remark 4.3 In Section 1, we mentioned Kim’s relation about q-Bernoulli polynomials and q-consecutive integers, from which we obtain some identities for ${\int }_0^n{\beta }_{l,q}d{[x]}_q$, namely

$$\begin{array}{rcl}{\int }_0^n{\beta }_{l,q}d{[x]}_q& =& \frac{1}{l+1}({\beta }_{l+1,q}(n)-{\beta }_{l+1,q})=T_{l,q}(n)\\ =& \sum _{k=0}^{n-1}{[k]}_q^l{q}^k=\frac{1}{{(1-q)}^l}\sum _{k=0}^l{(-1)}^k\left(\begin{array}{c}l\\ k\end{array}\right)\frac{1-q^{(k+1)(n-1)}}{1-q^{k+1}}{q}^{k+1}\\ =& {[\mathrm{\infty }]}_q^l\sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right){(-1)}^k{[n-1]}_{q^{k+1}}{q}^{k+1}.\end{array}$$

Next, we would like to consider the sum $T_{l,t}$ from (4.2) when $l=1$,

$$T_{1,t}=\sum _{k=0}^{\mathrm{\infty }}{[k]}_q{t}^{k-1}=\frac{1}{(1-t)(1-tq)}.$$

By the same argument as that in the proof of Proposition 4.2, we have more general identities for $T_{l,t}$:

$$\begin{array}{r}{T}_{2,t}=\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^2{t}^{k-1}=\frac{1+tq}{(1-t)(1-tq)(1-t{q}^2)}=:g(t),\\ T_{3,t}=\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^3{t}^{k-1}=\frac{1}{1-q}\{g(t)-qg(tq)\}\\ \phantom{T_{3,t}}=\frac{1}{1-q}(\frac{1+tq}{(1-t)(1-tq)(1-t{q}^2)}-\frac{q(1+t{q}^2)}{(1-tq)(1-t{q}^2)(1-t{q}^3)})\\ \phantom{T_{3,t}}=\frac{1+2tq+2t{q}^2+t^2{q}^3}{(1-t)(1-tq)(1-t{q}^2)(1-t{q}^3)}\text{and}\\ T_{4,t}=\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^4{t}^{k-1}=\frac{(1+t{q}^2)(1+3tq+4t{q}^2+3t{q}^3+t^2{q}^4)}{(1-t)(1-tq)(1-t{q}^2)(1-t{q}^3)(1-t{q}^4)}.\end{array}$$

All the denominators on the right-hand side are factorized as $l+1$ terms. However, the numerators are somewhat complex. Therefore, we recursively define a sequence $C_l(t;q)$ with $l\ge 1$ as follows:

$$\begin{array}{c}{C}_0(t;q)=1,\hfill \\ C_l(t;q)=\frac{1}{1-q}\{(1-t{q}^l)C_{l-1}(t;q)-q(1-t)C_{l-1}(tq;q)\}.\hfill \end{array}$$

Then, we get the following theorem.

Theorem 4.4 (1) The infinite sum $T_{l,t}$ is expressed as a quotient of $C_l(t;q)$ by $l+1$ products, precisely speaking,

$$T_{l,t}=\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^l{t}^{k-1}=\frac{C_l(t;q)}{{(t;q)}_{l+1}}.$$

• (2) Since $C_l(q;q)=C_{l-1}(q;q){[l+1]}_q-q{C}_{l-1}(q^2;q)$, we have

  $$\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^l{q}^{k-1}=\frac{C_l(q;q)}{{(q;q)}_{l+1}}.$$

Replacing t by $\frac{1}{t}$ in Theorem 4.4(1), we can deduce one of Simsek’s relations [[9], Proposition 3.1].

Theorem 4.5 The generating function (complex cases) of twisted q-Bernoulli numbers is given by

$$f_{q,\omega }(t)=\sum _{k=0}^{\mathrm{\infty }}{B}_{k,\omega }^{\ast }(q)\frac{t^k}{k!},$$

where ω is the rth root of unity and

$$B_{l,\omega }^{\ast }(q)=\{\begin{array}{ll}0& \mathit{\text{if }}l=0,\\ \frac{\omega q}{1-\omega q}& \mathit{\text{if }}l=1,\\ \frac{l{\omega }^{l-1}{q}^{\frac{l(l-1)}{2}}}{{(\omega q;q)}_l}{C}_{l-1}({\omega }^{-1}{q}^{-l};q)& \mathit{\text{if }}l\ge 2.\end{array}$$

Proof If we recall (S1) from Section 1, we get

$$\begin{array}{r}{B}_{0,\omega }^{\ast }(q)=0,\\ B_{1,\omega }^{\ast }(q)=-\sum _{k=0}^{\mathrm{\infty }}{\omega }^{-k}{q}^{-k}=\frac{\omega q}{1-\omega q},\\ B_{2,\omega }^{\ast }(q)=2\sum _{k=0}^{\mathrm{\infty }}{\omega }^{-k}{q}^{-2k}{[k]}_q=\frac{2}{w{q}^2}\sum _{k=0}^{\mathrm{\infty }}{[k]}_q{w}^{-k+1}{q}^{-2k+2}\\ \phantom{B_{2,\omega }^{\ast }(q)}=\frac{2}{\omega q^2}\sum _{k=1}^{\mathrm{\infty }}{[k]}_q{\omega }^{-k+1}{q}^{-2k+2}=\frac{2}{\omega q^2}\sum _{k=0}^{\mathrm{\infty }}\frac{{(q^2;q)}_k}{{(q;q)}_k}{\omega }^{-k}{q}^{-2k}\\ \phantom{B_{2,\omega }^{\ast }(q)}=\frac{2}{\omega q^2}F(q,1;{\omega }^{-1}{q}^{-2}:q).\end{array}$$

Since we know from [[14], (6.2)] that

$$F(a,1;t:q)=\frac{{(atq:q)}_{\mathrm{\infty }}}{{(t:q)}_{\mathrm{\infty }}},$$

by setting q and t to be a and ${\omega }^{-1}{q}^{-2}$, respectively, we obtain

$$B_{2,\omega }^{\ast }(q)=\frac{2\omega q}{(1-\omega q)(1-\omega q^2)}.$$

When l is greater than 2, we get

$$\begin{array}{rcl}{B}_{l,\omega }^{\ast }(q)& =& {(-1)}^{l}l\sum _{k=0}^{\mathrm{\infty }}{\omega }^{-k}{q}^{-lk}{[k]}_q^{l-1}=\frac{{(-1)}^{l}l}{\omega q^l}{T}_{l-1,{\omega }^{-1}{q}^{-l}}\\ =& \frac{{(-1)}^{l}l}{\omega q^l}\frac{C_{l-1}({\omega }^{-1}{q}^{-l};q)}{{({\omega }^{-1}{q}^{-l};q)}_l}=\frac{l{\omega }^{l-1}{q}^{\frac{l(l-1)}{2}}}{{(\omega q;q)}_l}{C}_{l-1}({\omega }^{-1}{q}^{-l};q).\end{array}$$

 □

By (S2) and Theorem 4.5, we get a corollary.

Corollary 4.6 If l is an integer greater than 1, we have

$${\zeta }_{\omega ,q}(1-l)={(-1)}^{l+1}\frac{{\omega }^{l-1}{q}^{\frac{l(l-1)}{2}}}{{(\omega q;q)}_l}{C}_{l-1}({\omega }^{-1}{q}^{-l};q).$$

Theorem 4.7 For any integer x,

$$B_{l,\omega }^{\ast }(x,q)=\{\begin{array}{ll}\frac{-\omega q}{1-\omega q}& \mathit{\text{if }}l=1,\\ \frac{2(1-\omega q-q^x+\omega q^{x+2})}{\omega q^2(1-q)(1-\omega q)(1-\omega q^2)}& \mathit{\text{if }}l=2,\\ {\omega }^x{q}^{lx}(-\frac{l{\omega }^l{q}^{\frac{l(l-1)}{2}}}{{(\omega q;q)}_l}{C}_l({\omega }^{-1}{q}^{-l};q)\\ +\frac{{(-1)}^{l}l}{{(1-q)}^{l-1}}{\sum }_{k=0}^{l-1}\left(\begin{array}{c}l-1\\ k\end{array}\right){[x]}_{\omega q^{l-k}}{\omega }^{-x}{q}^{-lx}{(-q^x)}^k)& \mathit{\text{if }}l>2.\end{array}$$

Proof It follows from (3.6) in [9] that

$$\begin{array}{rcl}{B}_{l,\omega }^{\ast }(x,q)& =& {(-1)}^{l+1}l\sum _{k=1}^{\mathrm{\infty }}{\omega }^{-k}{q}^{-kl}{[k+x]}_q^{l-1}\\ =& {\omega }^x{q}^{lx}({(-1)}^{l+1}l\sum _{k=1}^{\mathrm{\infty }}{\omega }^{-k}{q}^{-kl}{[k]}_q^{l-1}-{(-1)}^{l+1}l\sum _{k=1}^x{\omega }^{-k}{q}^{-kl}{[k]}_q^{l-1})\end{array}$$

with $x\in \mathbb{N}$. By direct calculation, we get the identities

$$\begin{array}{r}{B}_{1,\omega }^{\ast }(x,q)=\sum _{k=1}^{\mathrm{\infty }}{\omega }^{-k}{q}^{-k}=-\frac{\omega q}{1-\omega q},\\ B_{2,\omega }^{\ast }(x,q)=-2\sum _{k=1}^{\mathrm{\infty }}{\omega }^{-k}{q}^{-2k}\frac{1-q^{k+x}}{1-q}=\frac{2(1-\omega q-q^x+\omega q^{x+2})}{\omega q^2(1-q)(1-\omega q)(1-\omega q^2)},\end{array}$$

and when $l>2$, we derive from Proposition 4.2 and Theorem 4.5 that

$$B_{l,\omega }^{\ast }(x,q)={\omega }^x{q}^x(-B_{l,\omega }^{\ast }(q)+{(-1)}^{l}l\sum _{k=1}^x\frac{1}{{\omega }^k{q}^{lk}}{[k]}_q^{l-1}).$$

Substituting $l-1$, $x+1$, and ${\omega }^{-1}{q}^{-l}$ for l, n, and t in (4.3), respectively, we establish the last identities. □

As its immediate corollary, we have the following.

Corollary 4.8

$${\zeta }_{\omega ,q}(1-l,x)=\{\begin{array}{ll}-\frac{\omega q}{1-\omega q}& \mathit{\text{if }}l=1,\\ -\frac{1-\omega q-q^x+\omega q^{x+2}}{\omega q^2(1-q)(1-\omega q)(1-\omega q^2)}& \mathit{\text{if }}l=2,\\ {\omega }^x{q}^{lx}({(-1)}^l\frac{{\omega }^l{q}^{\frac{l(l-1)}{2}}}{{(\omega q;q)}_l}{C}_l({\omega }^{-1}{q}^{-l};q)\\ -\frac{1}{{(1-q)}^{l-1}}{\sum }_{k=0}^{l-1}\left(\begin{array}{c}l-1\\ k\end{array}\right){[x]}_{\omega q^{l-k}}{\omega }^{-x}{q}^{-lx}{(-q^x)}^k)& \mathit{\text{if }}l>2.\end{array}$$

Moreover, we can deduce the following corollary, which is analogous to Theorem 4.4.

Corollary 4.9

$$\sum _{k=1}^{\mathrm{\infty }}{[k]}_q^l\frac{1+q^k}{1+q}{t}^{k-1}=\frac{1}{1+q}(\frac{C_l(t;q)}{{(t;q)}_{l+1}}+q\frac{C_l(tq;q)}{{(tq;q)}_{l+1}}).$$

Proposition 4.10 For a nonnegative integer n,

$$\text{(1)}\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{c}k+n\\ n\end{array}\right]}_q{t}^k=\frac{1}{{(t;q)}_{n+1}},$$

and when $n=2$, we get the following by considering the summation from 0 to $n-1$ in the above:

$$\text{(2)}\sum _{k=0}^{n-1}{\left[\begin{array}{c}k+2\\ 2\end{array}\right]}_q{q}^k={\left[\begin{array}{c}n+2\\ 3\end{array}\right]}_q.$$

Furthermore, we obtain

$$\text{(3)}\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{c}k+2\\ 2\end{array}\right]}_q^2{q}^k=\frac{1+q^2+2q^3+q^4+q^6}{(1-q)(1-q^2)(1-q^3)(1-q^4)(1-q^5)}.$$

Proof By Lemma 2.3(2), we get

$$\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{c}k+n\\ n\end{array}\right]}_q{t}^k=\sum _{k=0}^{\mathrm{\infty }}\frac{{(q^{n+1};q)}_k}{{(q;q)}_k}{t}^k=\frac{1}{{(t;q)}_{n+1}}.$$

As for the second, we set $n=2$. Then, it follows from Lemma 4.1 that

$$\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{c}k+2\\ 2\end{array}\right]}_q{t}^k=\frac{{(q^3;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(q^{-2};q)}_k}{{(q;q)}_k}{q}^{3k}\sum _{l=0}^{\mathrm{\infty }}{t}^l{q}^{lk}.$$

Considering that the exponent of t is less than n only in the above, we have

$$\begin{array}{rcl}\sum _{k=0}^{n-1}{\left[\begin{array}{c}k+2\\ 2\end{array}\right]}_q{t}^k& =& \frac{{(q^3;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(q^{-2};q)}_k}{{(q;q)}_k}{q}^{3k}\sum _{l=0}^{n-1}{t}^l{q}^{2l}\\ =& \frac{1}{(1-q)(1-q^2)}(\sum _{l=0}^{n-1}{t}^l+\frac{(1-q^{-2})}{1-q}{q}^3\sum _{l=0}^{n-1}{t}^l{q}^l\\ +\frac{(1-q^{-2})(1-q^{-1})}{(1-q)(1-q^2)}{q}^6\sum _{l=0}^{n-1}{t}^l{q}^{2l}).\end{array}$$

Then, putting $t=q$ we get

$$\begin{array}{r}\frac{1}{(1-q)(1-q^2)}(\frac{1-q^n}{1-q}-\frac{q(1+q)(1-q^{2n})}{1-q^2}+\frac{q^3(1-q^{3n})}{1-q^3})\\ =\frac{(1-q^n)(1-q^{n+1})(1-q^{n+2})}{(1-q)(1-q^2)(1-q^3)}={\left[\begin{array}{c}n+2\\ 3\end{array}\right]}_q.\end{array}$$

In order to show (3), let $g(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{{(q^3;q)}_k}{{(q;q)}_k}{t}^k=\frac{1}{{(t;q)}_3}$. Then, we derive

$$\begin{array}{rcl}\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{c}k+2\\ 2\end{array}\right]}_q^2{t}^k& =& \sum _{k=0}^{\mathrm{\infty }}\frac{{(q^3;q)}_k}{{(q;q)}_k}\frac{{(q^3;q)}_k}{{(q;q)}_k}{t}^k\\ =& \frac{{(q^3;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(q^{-2};q)}_k}{{(q;q)}_k}{q}^{3k}g\left(t{q}^k\right)\\ =& \frac{1}{(1-q)(1-q^2)}(g(t)+\frac{1-q^{-2}}{1-q}{q}^{3}g(tq)+\frac{(1-q^{-2})(1-q^{-1})}{(1-q)(1-q^2)}{q}^{6}g\left(t{q}^2\right))\\ =& \frac{1}{(1-q)(1-q^2)}(\frac{1}{{(t;q)}_3}-\frac{q(1+q)}{{(tq;q)}_3}+\frac{q^3}{{(t{q}^2;q)}_3})\\ =& \frac{1+tq+2t{q}^2+t{q}^3+t^2{q}^4}{(1-t)(1-tq)(1-t{q}^2)(1-t{q}^3)(1-t{q}^4)}.\end{array}$$

Thus, by substituting q for t, we conclude (3). □

Proposition 4.11 For a nonnegative integer l,

$$\begin{array}{c}\text{(1)}\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^l{t}^{k-1}=C_l(t;q)\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{c}l+k\\ k\end{array}\right]}_q{t}^k,\hfill \\ \text{(2)}\sum _{k=0}^{\mathrm{\infty }}{[k]}_q^l{q}^{k-1}=C_l(q;q)\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{c}l+k\\ k\end{array}\right]}_q{q}^k.\hfill \end{array}$$

Proof We see from (6.22) in [14] that ${\sum }_{k=0}^{\mathrm{\infty }}{\left(\begin{array}{c}l+k\\ k\end{array}\right)}_q{t}^k=\frac{1}{{(t;q)}_{l+1}}$. Thus, the proposition follows from Theorem 4.4(2). □

Henceforth, we concentrate on $S_{l,n}(t)$ introduced in Section 1.

Theorem 4.12 For a complex number s with $|s|<1$ and positive integers m and n,

$$S_{l,n}(q)=\frac{q^{\frac{(l+1)n}{2}}}{(1+q){(1-q)}^l}\sum _{m=0}^l\left(\begin{array}{c}l\\ m\end{array}\right){(-q)}^m(\sum _{k=0}^{n-1}{q}^{(m-\frac{l+1}{2})k}-q\sum _{k=0}^{n-1}\left(q^{(m+1-\frac{l+1}{2})k}\right)).$$

Here, we consider $\left(\begin{array}{c}0\\ 0\end{array}\right)$ as 1.

Proof For fixed n, we consider ${\sum }_{k=0}^{n-1}{[k+1]}_q^l\frac{1+q^{k+1}}{1+q}{t}^k$, and we denote it by $g_l(t)$ so that $g_0(t)=\frac{1}{1+q}({\sum }_{k=0}^{n-1}{t}^k-q{\sum }_{k=0}^{n-1}{t}^k{q}^k)$. By adopting the arguments used in Lemma 4.1, we obtain the following recursion

$$\begin{array}{rcl}{g}_{l+1}(t)& =& \sum _{k=0}^{n-1}\frac{1-q^{k+1}}{1-q}{[k+1]}_q^l\frac{1+q^{k+1}}{1+q}{t}^k\\ =& \sum _{k=0}^n\frac{{(q^2;q)}_k}{{(q;q)}_k}{[k+1]}_q^l\frac{1+q^{k+1}}{1+q}{t}^k\\ =& \frac{{(q^2;q)}_{\mathrm{\infty }}}{{(q;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(\frac{1}{q};q)}_k}{{(q;q)}_k}{q}^{2k}{g}_l\left(t{q}^k\right)\\ =& \frac{1}{1-q}(g_l(t)-q{g}_l(tq)).\end{array}$$

Since the above is true for all adjacent integers, we obtain

$$g_l(t)=\frac{1}{{(1-q)}^l}\sum _{k=0}^l\left(\begin{array}{c}l\\ k\end{array}\right){(-q)}^k{g}_0\left(t{q}^k\right).$$

Multiplying both sides by $q^{\frac{(l+1)n}{2}}$ and replacing t by $q^{-\frac{l+1}{2}}$, we complete the proof. □

Remark 4.13 In Section 1, we mentioned that the generating function for the rank of partition can be written as $T_{l,t}$ and $C_l(t;q)$:

$$\sum _{l=0}^{\mathrm{\infty }}\frac{T_{l,\frac{q}{u}}}{C_l(u^{-1}q;q)}{u}^l{q}^{l+1}=\sum _{r=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{P}_r(n)u^r{q}^n.$$

By letting $u=-1$ on the right-hand side, we get

$$\begin{array}{rl}\sum _{r=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{P}_r(n){(-1)}^r=& \sum _{n=1}^{\mathrm{\infty }}(\sum _{r\text{: even}}{P}_r(n)-\sum _{r\text{: odd}}{P}_r(n))q^r\\ =& \sum _{n=1}^{\mathrm{\infty }}\frac{q^{n^2}}{{(-q;q)}_n^2}.\end{array}$$

Further, it gives rise to a third-order mock theta function $f(q)=1+{\sum }_{n=1}^{\mathrm{\infty }}\alpha (n)q^n=1+{\sum }_{n=1}^{\mathrm{\infty }}\frac{q^{n^2}}{{(-q;q)}_n^2}$, where the explicit formula of $\alpha (n)$ was conjectured by both Andrews [21] and Dragonette [22], and later proved by Bringmann and Ono [23]. Let $N_e(n)$ (resp., $N_o(n)$) be the infinite sum ${\sum }_{r\text{: even}}{P}_r(n)$ (resp., ${\sum }_{r\text{: odd}}{P}_r(n)$). Then, we can find the coefficients of the infinite sum ${\sum }_{n=0}^{\mathrm{\infty }}\frac{T_{n,-q}{(-1)}^n}{C_n(-q;q)}{q}^{n+1}$, because the formula for the partition function $p(n)$ is already known and $\alpha (n)=N_e(n)-N_o(n)$.

Remark 4.14 In Ramanujan’s lost notebook, there are 4 third-order mock theta functions [[24], p.345]:

$$\begin{array}{c}f(q)=\sum _{k=0}^{\mathrm{\infty }}\frac{q^{k^2}}{{(-q;q)}_k^2},\hfill \\ \varphi (q)=1+\frac{q}{1+q^2}+\frac{q^4}{(1+q^2)(1+q^4)}+\frac{q^9}{(1+q^2)(1+q^4)(1+q^6)}+\cdots ,\hfill \\ \psi (q)=\frac{q}{1-q}+\frac{q^4}{(1-q)(1-q^3)}+\frac{q^9}{(1-q)(1-q^3)(1-q^5)}+\cdots ,\hfill \\ \chi (q)=1+\frac{q}{1-q+q^2}+\frac{q^4}{(1-q+q^2)(1-q^2+q^4)}+\cdots .\hfill \end{array}$$

By utilizing our notations, we interpret them as follows.

From Theorem 4.4, Proposition 4.10, Proposition 4.11, and the definition of the mock theta functions [[14], pp.55-57], we are able to connect q-consecutive integers with these mock theta functions, namely

$$\begin{array}{c}1+\sum _{l=0}^{\mathrm{\infty }}\frac{T_{l,q}}{C_l(q;q)}{q}^{l+1}=\frac{1}{{(q;q)}_{\mathrm{\infty }}}=\sum _{k=0}^{\mathrm{\infty }}p(k)q^k,\hfill \\ 1+\sum _{l=0}^{\mathrm{\infty }}\frac{T_{l,-q}{(-1)}^l}{C_l(-q;q)}{q}^{l+1}=f(q),\hfill \\ 1+\sum _{l=0}^{\mathrm{\infty }}\frac{T_{l,-iq}}{C_l(-iq;q)}{(i)}^l{q}^{l+1}=\varphi (q),\hfill \\ 1+\sum _{l=0}^{\mathrm{\infty }}\frac{T_{l,-\omega q}}{C_l(-\omega q;q)}{(-{\omega }^2)}^l{q}^{l+1}=\chi (q),\hfill \end{array}$$

with $i=e^{2\pi i/4}$ and $\omega =e^{2\pi i/3}$.