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Identities on poly-Dedekind sums

Abstract

Dedekind sums occur in the transformation behavior of the logarithm of the Dedekind eta-function under substitutions from the modular group. In 1892, Dedekind showed a reciprocity relation for the Dedekind sums. Apostol generalized Dedekind sums by replacing the first Bernoulli function appearing in them by any Bernoulli functions and derived a reciprocity relation for the generalized Dedekind sums. In this paper, we consider the poly-Dedekind sums obtained from the Dedekind sums by replacing the first Bernoulli function by any type 2 poly-Bernoulli functions of arbitrary indices and prove a reciprocity relation for the poly-Dedekind sums.

Introduction

To give concise definition of the Dedekind sums, we introduce the notation

$$ ( (x) )= \textstyle\begin{cases} x-[x]-\frac{1}{2} & \text{if $x\notin \mathbb{Z}$,} \\ 0 & \text{if $x\in \mathbb{Z}$}, \end{cases}\displaystyle \quad (\text{see [1, 4]}), $$
(1)

where \([x]\) denotes the greatest integer not exceeding x.

It is well known that the Dedekind sums are defined by

$$ S(h,m)=\sum_{\mu =1}^{m} \biggl( \biggl( \frac{\mu }{m} \biggr) \biggr) \biggl( \biggl(\frac{h\mu }{m} \biggr) \biggr)\quad (\text{see [1, 4, 6--8, 11--13]}), $$
(2)

where h is any integer.

From (2) we note that

$$ S(h,m)=\sum_{\mu =1}^{m} \biggl( \frac{\mu }{m}-\frac{1}{2} \biggr) \biggl( \biggl(\frac{h\mu }{m} \biggr) \biggr)=\sum_{\mu =1}^{m} \frac{\mu }{m} \biggl( \biggl(\frac{h\mu }{m} \biggr) \biggr)\quad ( \text{see [7, 8]}). $$
(3)

As is well known, the Bernoulli polynomials are given by

$$ \frac{t}{e^{t}-1}e^{xt}=\sum_{n=0}^{\infty }B_{n}(x) \frac{t^{n}}{n!} \quad (\text{see [1--13]}). $$
(4)

When \(x=0\), \(B_{n}=B_{n}(0)\) (\(n\ge 0\)) are called the Bernoulli numbers.

From (4) we note that

$$ B_{n}(x)=\sum_{l=0}^{n} \binom{n}{l}B_{l}x^{n-l}=(B+x)^{n}\quad (n \ge 0)\ (\text{see [2--5, 7, 8]}), $$
(5)

with the usual convention about replacing \(B^{n}\) by \(B_{n}\).

We observe that

$$ \sum_{l=0}^{n-1}e^{lt}= \frac{t}{t(e^{t}-1)} \bigl(e^{nt}-1 \bigr)=\sum _{j=0}^{ \infty } \biggl(\frac{B_{j+1}(n)-B_{j+1}}{j+1} \biggr) \frac{t^{j}}{j!} \quad (n\in \mathbb{N}). $$
(6)

Thus by (6) we get

$$ \sum_{l=0}^{n-1}l^{j}= \frac{1}{j+1} \bigl(B_{j+1}(n)-B_{j+1} \bigr) \quad (n\in \mathbb{N}, j\ge 0). $$
(7)

Recently, Kim and Kim [5, 9] considered the polyexponential function of index k given by

$$ \operatorname{Ei}_{k}(x)=\sum_{n=1}^{\infty } \frac{x^{n}}{n^{k}(n-1)!} \quad (k \in \mathbb{Z}). $$
(8)

Note that \(\operatorname{Ei}_{1}(x)=e^{x}-1\).

In [5] the type 2 poly-Bernoulli polynomials of index k are defined in terms of the polyexponential function of index k as

$$ \frac{\operatorname{Ei}_{k}(\log (1+t))}{e^{t}-1}e^{xt}=\sum_{n=0}^{\infty }B_{n}^{(k)}(x) \frac{t^{n}}{n!}\quad (k\in \mathbb{Z}). $$
(9)

When \(x=0\), \(B_{n}^{(k)}=B_{n}^{(k)}(0)\) (\(n\ge 0\)) are called the type 2 poly-Bernoulli numbers of index k. Note that \(B_{n}^{(1)}(x)=B_{n}(x)\) are the Bernoulli polynomials.

The fractional part of x is denoted by

$$ \langle x\rangle =x-[x]. $$
(10)

The Bernoulli functions are defined by

$$ \overline{B}_{n}(x)=B_{n}\bigl(\langle x\rangle \bigr)\quad (n\ge 0)\ ( \text{see [1, 4, 11]}). $$
(11)

Thus by (3) and (11) we get

$$\begin{aligned} S(h,m) &= \sum_{\mu =1}^{m-1}\frac{\mu }{m} \biggl(\frac{h\mu }{m}- \biggl[\frac{h\mu }{m} \biggr]-\frac{1}{2} \biggr) \\ &= \sum_{\mu =1}^{m-1}\frac{\mu }{m} \overline{B}_{1} \biggl( \frac{h\mu }{m} \biggr) = \sum _{\mu =1}^{m-1}\overline{B}_{1} \biggl( \frac{\mu }{m} \biggr)\overline{B}_{1} \biggl(\frac{h\mu }{m} \biggr), \end{aligned}$$
(12)

where h, m are relatively prime positive integers.

We need the following lemma, which is well-known and easily shown.

Lemma 1

Let n be a nonnegative integer, and let d be a positive integer. Then we have:

  1. (a)

    \(\sum_{i=0}^{d-1} B_{n} (\frac{x+i}{d} )=d^{1-n}B_{n}(x)\),

  2. (b)

    \(\sum_{i=0}^{d-1} \overline{B}_{n} (\frac{x+i}{d} )=d^{1-n} \overline{B}_{n}(x)\), and

  3. (c)

    \(\sum_{i=0}^{d-1} B_{n} (\frac{\langle x \rangle +i}{d} )= \sum_{i=0}^{d-1} \overline{B}_{n} (\frac{x+i}{d} )\) for all real x.

Dedekind showed that the quantity \(S(h,m) = \sum_{\mu =1}^{m-1}\frac{\mu }{m}\overline{B}_{1} ( \frac{h\mu }{m} )\) occurs in the transformation behavior of the logarithm of the Dedekind eta-function under substitutions from the modular group. In 1892, he showed the following reciprocity relation for Dedekind sums:

$$ S(h,m)+S(m,h)=\frac{1}{12} \biggl(\frac{h}{m}+\frac{1}{hm}+ \frac{m}{h} \biggr)-\frac{1}{4} $$

if h and m are relatively prime positive integers.

Apostol [1] considered the generalized Dedekind sums given by

$$ S_{p}(h,m)= \sum_{\mu =1}^{m-1} \frac{\mu }{m}\overline{B}_{p} \biggl( \frac{h\mu }{m} \biggr) $$
(13)

and showed that they satisfy the reciprocity relation

$$\begin{aligned} &(p+1) \bigl(hm^{p}S_{p}(h,m) + mh^{p}S_{p}(m,h) \bigr) \\ &\quad =pB_{p+1}+\sum_{s=0}^{p+1} \binom{p+1}{s}(-1)^{s}B_{s}B_{p+1-s}h^{s}m^{p+1-s}. \end{aligned}$$

In this paper, we consider the poly-Dedekind sums defined by

$$ S_{p}^{(k)}(h,m)= \sum_{\mu =1}^{m-1} \frac{\mu }{m}\overline{B}_{p}^{(k)} \biggl( \frac{h\mu }{m} \biggr), $$

where \(B_{p}^{(k)}(x)\) are the type 2 poly-Bernoulli polynomials of index k (see (9)), and \(\overline{B}_{p}^{(k)}(x)=B_{p}^{(k)}(\langle x \rangle )\) are the type 2 poly-Bernoulli functions of index k. Note that \(S_{p}^{(1)}(h,m)=S_{p}(h,m)\). We show the following reciprocity relation for the poly-Dedekind sums (see Theorem 10):

$$\begin{aligned} &hm^{p}S_{p}^{(k)}(h,m)+mh^{p}S_{p}^{(k)}(m,h) \\ &\quad =\sum_{\mu =0}^{m-1}\sum _{j=0}^{p}\sum_{\nu =0}^{h-1} \sum_{l=1}^{p-j+1} \frac{(mh)^{j-1}\binom{p}{j}S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \bigl((\mu h)m^{p-j}+(m \nu ) h^{p-j} \bigr)\overline{B}_{j} \biggl(\frac{\nu }{h}+\frac{\mu }{m} \biggr). \end{aligned}$$

For \(k=1\), this reciprocity relation for the poly-Dedekind sums reduces to that for the generalized Dedekind sums given by (see Corollary 11)

$$\begin{aligned} &hm^{p}S_{p}(h,m)+mh^{p}S_{p}(m,h) \\ &\quad = \sum_{\mu =0}^{m-1}\sum _{\nu =0}^{h-1}(mh)^{p-1}(\mu h+m\nu ) \overline{B}_{p} \biggl(\frac{\nu }{h}+\frac{\mu }{m} \biggr). \end{aligned}$$

In Sect. 2, we derive various facts about the type 2 poly-Bernoulli polynomials, which will be needed in the next section. In Sect. 3, we define the poly-Dedekind sums and demonstrate a reciprocity relation for them.

On type 2 poly-Bernoulli polynomials

Note that by (9)

$$\begin{aligned} \frac{\operatorname{Ei}_{k}(\log (1+t))}{e^{t}-1}e^{xt} &= \sum_{l=0}^{ \infty }B_{l}^{(k)} \frac{t^{l}}{l!}\sum_{m=0}^{\infty } \frac{x^{m}}{m!}t^{m} \\ &= \sum_{n=0}^{\infty } \Biggl(\sum _{l=0}^{n}\binom{n}{l}B_{l}^{(k)}x^{n-l} \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(14)

Thus by (14) we get

$$\begin{aligned} B_{n}^{(k)}(x) &= \sum_{l=0}^{n} \binom{n}{l}B_{l}^{(k)}x^{n-l} \quad (n\ge 0). \end{aligned}$$
(15)

By (15) we get

$$ \frac{d}{dx}B_{n}^{(k)}(x)=nB_{n-1}^{(k)}(x) \quad (n\ge 1). $$
(16)

From (9) we have

$$\begin{aligned} \operatorname{Ei}_{k} \bigl(\log (1+t) \bigr) &= \sum _{l=0}^{\infty }B_{l}^{(k)} \frac{t^{l}}{l!} \bigl(e^{t}-1 \bigr) \\ &= \sum_{n=0}^{\infty } \bigl(B_{n}^{(k)}(1)-B_{n}^{(k)} \bigr) \frac{t^{n}}{n!} = \sum_{n=1}^{\infty } \bigl(B_{n}^{(k)}(1)-B_{n}^{(k)} \bigr) \frac{t^{n}}{n!}. \end{aligned}$$
(17)

On the other hand,

$$\begin{aligned} \operatorname{Ei}_{k} \bigl(\log (1+t) \bigr) &= \sum _{m=1}^{\infty } \frac{ (\log (1+t) )^{m}}{m^{k}(m-1)!} = \sum _{m=1}^{\infty } \frac{1}{m^{k-1}}\frac{1}{m!} \bigl( \log (1+t) \bigr)^{m} \\ &= \sum_{m=1}^{\infty }\frac{1}{m^{k-1}}\sum _{n=m}^{\infty }S_{1}(n,m) \frac{t^{n}}{n!} \\ &= \sum_{n=1}^{\infty } \Biggl(\sum _{m=1}^{n}\frac{1}{m^{k-1}}S_{1}(n,m) \Biggr)\frac{t^{n}}{n!}, \end{aligned}$$
(18)

where \(S_{1}(n,m)\) are the Stirling numbers of the first kind.

Therefore by (17) and (18) we obtain the following theorem.

Theorem 2

For \(n\ge 1\), we have

$$\begin{aligned}& B_{n}^{(k)}(1)-B_{n}^{(k)}=\sum _{m=1}^{n}S_{1}(n,m)\frac{1}{m^{k-1}} \quad (k\in \mathbb{Z}). \end{aligned}$$

By Theorem 2 we get

$$\begin{aligned}& B_{n}^{(1)}-B_{n}^{(1)}=\delta _{1,n}, \quad\quad B_{0}^{(k)}=1, \quad\quad B_{1}^{(k)}=-1+ \frac{1}{2^{k}},\ldots, \end{aligned}$$

where \(\delta _{n,k}\) is the Kronecker symbol.

With (16) in mind, we now compute

$$\begin{aligned} \biggl(\frac{d}{dx} \biggr)^{s-1} \bigl(xB_{p}^{(k)}(x) \bigr) \bigg| _{x=1}&= \sum_{l=0}^{s-1} \binom{s-1}{l} \biggl( \biggl(\frac{d}{dx} \biggr)^{l}x \biggr) \biggl( \biggl(\frac{d}{dx} \biggr)^{s-1-l}B_{p}^{(k)}(x) \biggr) \bigg| _{x=1} \\ &= \biggl(\frac{d}{dx} \biggr)^{s-1}B_{p}^{(k)}(x) \bigg| _{x=1}+ \binom{s-1}{1} \biggl(\frac{d}{dx} \biggr)^{s-2}B_{p}^{(k)}(x) \bigg| _{x=1} \\ &= \frac{s!}{p+1}\binom{p+1}{s}B_{p-s+1}^{(k)}(1) + \frac{(s-1)s!}{(p+1)(p+2)}\binom{p+2}{s}B_{p-s+2}^{(k)}(1). \end{aligned}$$
(19)

On the other hand, by (15) we get

$$\begin{aligned} \biggl(\frac{d}{dx} \biggr)^{s-1} \bigl(xB_{p}^{(k)}(x) \bigr) \bigg| _{x=1} &= \sum_{\nu =0}^{p} \binom{p}{\nu }B_{\nu }^{(k)} \biggl( \biggl( \frac{d}{dx} \biggr)^{s-1}x^{p-\nu +1} \biggr) \bigg| _{x=1} \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu }B_{\nu }^{(k)}(p- \nu +1)\cdots (p- \nu -s+3) \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu } \frac{s!B_{\nu }^{(k)}}{p-\nu +2} \binom{p-\nu +2}{s}. \end{aligned}$$
(20)

Therefore by (19) and (20) we obtain the following theorem.

Theorem 3

For \(s,p\in \mathbb{N}\), we have

$$\begin{aligned}& \sum_{\nu =0}^{p}\binom{p}{\nu } \binom{p-\nu +2}{s} \frac{B_{\nu }^{(k)}}{p-\nu +2}=\binom{p+1}{s} \frac{B_{p-s+1}^{(k)}(1)}{p+1}+ \frac{s-1}{p+1}\binom{p+2}{s} \frac{B_{p-s+2}^{(k)}(1)}{p+2}. \end{aligned}$$

Now we observe that

$$\begin{aligned} \sum_{\nu =0}^{p}\binom{p}{\nu } \binom{p-\nu +2}{s} \frac{B_{\nu }^{(k)}}{p-\nu +2}&=\sum_{\nu =0}^{p-s+2} \frac{\binom{p}{\nu }\binom{p-\nu +2}{s}}{p-\nu +2} B_{\nu }^{(k)} \\ &=\sum_{\nu =0}^{p-s+1} \frac{\binom{p}{\nu }\binom{p-\nu +2}{s}}{p-\nu +2}B_{\nu }^{(k)}+ \frac{1}{s}\binom{p}{s-2}B_{p-s+2}^{(k)}. \end{aligned}$$
(21)

Therefore by Theorem 3 and (21) we obtain the following corollary.

Corollary 4

For \(s,p\in \mathbb{N}\), we have

$$\begin{aligned} &\sum_{\nu =0}^{p-s+1}\binom{p}{\nu } \binom{p-\nu +2}{s} \frac{B_{\nu }^{(k)}}{p-\nu +2} \\ &\quad =\binom{p+1}{s}\frac{B_{p-s+1}^{(k)}(1)}{p+1}+\frac{s-1}{p+1} \binom{p+2}{s} \frac{B_{p-s+2}^{(k)}(1)}{p+2}-\frac{1}{s} \binom{p}{s-2}B_{p-s+2}^{(k)}. \end{aligned}$$

From (16) we have

$$\begin{aligned} \int _{0}^{1}xB_{p}^{(k)}(x) \,dx &= \biggl[x \frac{B_{p+1}^{(k)}(x)}{p+1} \biggr]_{0}^{1}- \frac{1}{p+1} \int _{0}^{1}B_{p+1}^{(k)}(x)\,dx \\ &= \frac{B_{p+1}^{(k)}(1)}{p+1}-\frac{1}{p+1} \biggl[\frac{1}{p+2}B_{p+2}^{(k)}(x) \biggr]_{0}^{1} \\ &= \frac{B_{p+1}^{(k)}(1)}{p+1}-\frac{B_{p+2}^{(k)}(1)}{(p+1)(p+2)}+ \frac{B_{p+2}^{(k)}}{(p+1)(p+2)}. \end{aligned}$$
(22)

On the other hand, by (15) we get

$$\begin{aligned} \int _{0}^{1}xB_{p}^{(k)}(x) \,dx &= \sum_{s=0}^{p}\binom{p}{s}B_{s}^{(k)} \int _{0}^{1}x^{p-s+1}\,dx \\ &= \sum_{s=0}^{p}\binom{p}{s}B_{s}^{(k)} \frac{1}{p+2-s}. \end{aligned}$$
(23)

Therefore by (22) and (23) we obtain the following theorem.

Theorem 5

For \(p\in \mathbb{N}\), we have

$$\begin{aligned}& \sum_{s=0}^{p}\binom{p}{s}B_{s}^{(k)} \frac{1}{p+2-s}= \frac{B_{p+1}^{(k)}(1)}{p+1}-\frac{B_{p+2}^{(k)}(1)}{(p+1)(p+2)}+ \frac{B_{p+2}^{(k)}}{(p+1)(p+2)}. \end{aligned}$$

Poly-Dedekind sums

Apostol considered the generalized Dedekind sums given by

$$ S_{p}(h,m)=\sum_{\mu =1}^{m-1}(\mu /m)\overline{B}_{p}(h\mu /m) \quad (h,m,p\in \mathbb{N}), $$
(24)

where \(\overline{B}_{p}(h\mu /m)=B_{p} (\langle h\mu /m\rangle )\).

Note that, for any relatively prime positive integers h, m, we have

$$\begin{aligned} S_{1}(h,m) &= \sum_{\mu =1}^{m-1}( \mu /m)\overline{B}_{1}(h\mu /m) \\ &= \sum_{\mu =1}^{m-1}( (\mu /m) ) ( (h\mu /m) ) = S(h,m). \end{aligned}$$

In this section, we consider the poly-Dedekind sums given by

$$ S_{p}^{(k)}(h,m) = \sum_{\mu =1}^{m-1} (\mu /m )\overline{B}_{p}^{(k)} (h\mu /m ), $$
(25)

where \(h,m,p\in \mathbb{N}\), \(k\in \mathbb{Z}\), and \(\overline{B}_{p}^{(k)}(x)=B_{p}^{(k)}(\langle x \rangle )\) are the type 2 poly-Bernoulli functions of index k.

Note that

$$\begin{aligned}& S_{p}^{(1)}(h,m) = \sum_{\mu =1}^{m-1} (\mu /m )\overline{B}_{p} (h\mu /m )=S_{p}(h,m). \end{aligned}$$

Assume now that \(h=1\). Then we have

$$\begin{aligned} S_{p}^{(k)}(1,m) &= \sum_{\mu =1}^{m-1} (\mu /m ) \overline{B}_{p}^{(k)} (\mu /m ) \\ &= \sum_{\mu =1}^{m-1} (\mu /m)\sum _{\nu =0}^{p}\binom{p}{\nu }B_{ \nu }^{(k)} (\mu /m )^{p-\nu } \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu }B_{\nu }^{(k)}m^{-(p-\nu +1)} \sum_{ \mu =1}^{m-1}\mu ^{p+1-\nu } \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu }B_{\nu }^{(k)}m^{-(p+1-\nu )} \frac{1}{p+2-\nu } \bigl(B_{p+2-\nu }(m)-B_{p+2-\nu } \bigr). \end{aligned}$$
(26)

From (5) we have

$$\begin{aligned} B_{p+2-\nu }(m)-B_{p+2-\nu } &= \sum_{i=0}^{p+2-\nu } \binom{p+2-\nu }{i}B_{i}m^{p+2-\nu -i}-B_{p+2-\nu } \\ &= \sum_{i=0}^{p+1-\nu } \binom{p+2-\nu }{i} B_{i}m^{p+2-\nu -i}. \end{aligned}$$
(27)

By (26) and (27) we get

$$\begin{aligned} S_{p}^{(k)}(1,m) &= \sum_{\nu =0}^{p} \binom{p}{\nu }B_{\nu }^{(k)}m^{-(p+1- \nu )}\frac{1}{p+2-\nu } \sum_{i=0}^{p+1-\nu }\binom{p+2-\nu }{i}B_{i}m^{p+2- \nu -i} \\ &= \frac{1}{m^{p}}\sum_{\nu =0}^{p} \binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }\sum_{i=0}^{p+1-\nu } \binom{p+2-\nu }{i}B_{i}m^{p+1-i}. \end{aligned}$$
(28)

Now we assume that \(p\ge 3\) is an odd positive integer, so that \(B_{p}=0\). Then we have

$$\begin{aligned} m^{p}S_{p}^{(k)}(1,m) &= \sum _{\nu =0}^{p}\binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }\sum _{i=0}^{p+1-\nu }\binom{p+2-\nu }{i}B_{i}m^{p+1-i} \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }m^{p+1}+ \sum_{\nu =0}^{p} \binom{p}{\nu }\frac{B_{\nu }^{(k)}}{p+2-\nu }\sum_{i=1}^{p+1- \nu } \binom{p+2-\nu }{i}B_{i}m^{p+1-i} \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }m^{p+1}+ \sum_{i=1}^{p+1} \sum_{\nu =0}^{p+1-i}\binom{p}{\nu } \binom{p+2-\nu }{i}\frac{B_{\nu }^{(k)}}{p+2-\nu }B_{i}m^{p+1-i} \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }m^{p+1}+ \sum_{i=1}^{p-1} \sum_{\nu =0}^{p+1-i}\binom{p}{\nu } \binom{p+2-\nu }{i}\frac{B_{\nu }^{(k)}}{p+2-\nu } B_{i} m^{p+1-i} \\ &\quad {} +\frac{1}{p+2}\binom{p+2}{p+1}B_{p+1}+\sum _{\nu =0}^{1} \binom{p}{\nu }\binom{p+2-\nu }{p} \frac{B_{\nu }^{(k)}}{p+2-\nu }B_{p}m \\ &= \sum_{\nu =0}^{p}\binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }m^{p+1}+ \sum_{i=1}^{p-1} \sum_{\nu =0}^{p+1-i}\binom{p}{\nu } \frac{\binom{p+2-\nu }{i} }{p+2-\nu }B_{\nu }^{(k)}B_{i}m^{p+1-i}+B_{p+1}. \end{aligned}$$
(29)

Therefore by (29) we obtain the following proposition.

Proposition 6

Let \(p\ge 3\) be an odd positive integer. Then we have

$$\begin{aligned}& \begin{aligned} m^{p}S_{p}^{(k)}(1,m)= {}&\sum _{\nu =0}^{p}\binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }m^{p+1}+ \sum_{i=1}^{p-1}\sum _{\nu =0}^{p+1-i} \binom{p}{\nu } \binom{p+2-\nu }{i} \frac{B_{\nu }^{(k)}}{p+2-\nu }B_{i}m^{p+1-i}\\ &{}+B_{p+1}. \end{aligned} \end{aligned}$$

We still assume that \(p\ge 3\) is an odd positive integer, so that \(B_{p}=0\). Then from Corollary 4, Theorem 5, and Proposition 6 we note that

$$\begin{aligned} &m^{p}S_{p}^{(k)}(1,m) \\ &\quad = \sum_{\nu =0}^{p}\binom{p}{\nu } \frac{B_{\nu }^{(k)}}{p+2-\nu }m^{p+1}+ \sum_{i=1}^{p-1} \sum_{\nu =0}^{p+1-i}\binom{p}{\nu } \binom{p+2-\nu }{i}\frac{B_{\nu }^{(k)}}{p+2-\nu }B_{i}m^{p+1-i}+B_{p+1} \\ &\quad = \biggl(\frac{B_{p+1}^{(k)}(1)}{p+1}- \frac{B_{p+2}^{(k)}(1)}{(p+1)(p+2)}+\frac{B_{p+2}^{(k)}}{(p+1)(p+2)} \biggr)m^{p+1}+B_{p+1} \\ &\quad\quad{} +\sum_{i=1}^{p-1} \biggl( \binom{p+1}{i}\frac{B_{p+1-i}^{(k)}(1)}{p+1}+ \frac{(i-1)}{(p+1)(p+2)}\binom{p+2}{i}B_{p+2-i}^{(k)}(1) \\ &\quad\quad{} - \binom{p}{i-2}\frac{1}{i}B_{p+2-i}^{(k)} \biggr)B_{i}m^{p+1-i}. \end{aligned}$$
(30)

To proceed further, we note that \(\binom{p}{i-2}\frac{p+1}{i}=\frac{1}{p+2}\binom{p+2}{i}(i-1)\) for \(i \ge 1\) and that \(B_{1}^{(k)}(1)-B_{1}^{(k)}=1\) by Theorem 2. Then from (30) we see that

$$\begin{aligned} (p+1)m^{p}S_{p}^{(k)}(1,m)&= \biggl(B_{p+1}^{(k)}(1)- \frac{B_{p+2}^{(k)}(1)}{p+2}+\frac{B_{p+2}^{(k)}}{p+2} \biggr)m^{p+1} \\ &\quad {} +\sum_{i=1}^{p-1} \binom{p+1}{i}B_{i}B_{p+1-i}^{(k)}(1)m^{p+1-i}+(p+1)B_{p+1} \\ &\quad {} +\frac{1}{p+2}\sum_{i=1}^{p-1} \binom{p+2}{i}(i-1)B_{i}B_{p+2-i}^{(k)}(1)m^{p+1-i} \\ &\quad {}- \sum_{i=1}^{p-1}\binom{p}{i-2} \frac{(p+1)}{i}B^{(k)}_{p+2-i}B_{i}m^{p+1-i} \\ &=m^{p+1}B_{p+1}^{(k)}(1)+\sum _{i=1}^{p-1}\binom{p+1}{i}B_{i}m^{p+1-i}B_{p+1-i}^{(k)}(1)+B_{p+1} \\ &\quad {} +\frac{1}{p+2}(-1)m^{p+1} \bigl(B_{p+2}^{(k)}(1)-B_{p+2}^{(k)} \bigr) \\ &\quad {}+ \frac{1}{p+2}\sum_{i=1}^{p-1} \binom{p+2}{i}(i-1)B_{i}m^{p+1-i} \bigl(B_{p+2-i}^{(k)}(1)-B_{p+2-i}^{(k)} \bigr) \\ &\quad {} +pB_{p+1} \\ &= \sum_{i=0}^{p+1}\binom{p+1}{i}B_{i}m^{p+1-i}B_{p+1-i}^{(k)}(1) \\ &\quad {}+ \frac{1}{p+2}\sum_{i=0}^{p+1} \binom{p+2}{i}(i-1)B_{i}m^{p+1-i} \bigl(B_{p+2-i}^{(k)}(1)-B_{p+2-i}^{(k)} \bigr). \end{aligned}$$
(31)

Therefore by (31) we obtain the following theorem.

Theorem 7

For \(m\in \mathbb{N}\) and any odd positive integer \(p \ge 3\), we have

$$\begin{aligned} &(p+1)m^{p}S_{p}^{(k)}(1,m) \\ &\quad = \sum_{i=0}^{p+1}\binom{p+1}{i}B_{i}m^{p+1-i}B_{p+1-i}^{(k)}(1) \\ &\quad\quad{} + \frac{1}{p+2}\sum_{i=0}^{p+1} \binom{p+2}{i}(i-1)B_{i}m^{p+1-i} \bigl(B_{p+2-i}^{(k)}(1)-B_{p+2-i}^{(k)} \bigr). \end{aligned}$$

Now we employ the notation

$$\begin{aligned}& B_{n}(x)=(B+x)^{n},\quad\quad B_{n}^{(k)}(x)= \bigl(B^{(k)}+x \bigr)^{n} \quad (n\ge 0). \end{aligned}$$

Assume that h, m are relatively prime positive integers. Then we see that

$$\begin{aligned} &m^{p}\sum_{\mu =0}^{m-1}\sum _{s=0}^{p+1}\binom{p+1}{s}h^{s}B_{s}^{(k)} (\mu /m )B_{p+1-s} \bigl(h-[h\mu /m] \bigr) \\ &\quad =m^{p} \sum_{\mu =0}^{m-1} \sum_{s=0}^{p+1}\binom{p+1}{s}h^{s} \bigl(B^{(k)}+ \mu m^{-1} \bigr)^{s} \bigl(B+h-[h \mu /m] \bigr)^{p+1-s} \\ &\quad =m^{p}\sum_{\mu =0}^{m-1} \bigl(hB^{(k)}+h\mu m^{-1}+B+h-[h\mu /m] \bigr)^{p+1} \\ &\quad =m^{p}\sum_{\mu =0}^{m-1} \biggl(hB^{(k)}+h+B+\frac{1}{2}+ \frac{h\mu }{m} -[h\mu /m]- \frac{1}{2} \biggr)^{p+1} \\ &\quad =m^{p}\sum_{\mu =0}^{m-1} \biggl(hB^{(k)}+h+B+\frac{1}{2}+ \overline{B}_{1} (h\mu /m ) \biggr)^{p+1}. \end{aligned}$$
(32)

Now, as the index μ ranges over the values \(\mu =0,1,2,\ldots,m-1\), the product ranges over a complete residue system modulo m, and due to the periodicity of \(\overline{B}_{1}(x)\), the term \(\overline{B}_{1}(h\mu/m)\) may be replaced by \(\overline{B}_{1}(\mu /m)\) without altering the sum over μ. Thus the sum (32) is equal to

$$\begin{aligned} & m^{p}\sum_{m=0}^{m-1} \biggl(hB^{(k)}+h+B+\frac{1}{2}+\overline{B}_{1} \biggl(\frac{\mu }{m} \biggr) \biggr)^{p+1} \\ &\quad = m^{p}\sum_{m=0}^{m-1} \biggl(h \bigl(B^{(k)}+1 \bigr)+B+\frac{\mu }{m} \biggr) )^{p+1} \\ &\quad = m^{p}\sum_{\mu =0}^{m-1} \sum_{s=0}^{p+1}\binom{p+1}{s} \biggl(B+ \frac{\mu }{m} \biggr)^{s}h^{p+1-s} \bigl(B^{(k)}+1 \bigr)^{p+1-s} \\ &\quad = m^{p}\sum_{\mu =0}^{m-1} \sum_{s=0}^{p+1}\binom{p+1}{s}B_{s} \biggl(\frac{\mu }{m} \biggr)h^{p+1-s}B^{(k)}_{p+1-s}(1) \\ &\quad = \sum_{s=0}^{p+1} \binom{p+1}{s}m^{s-1}\sum_{\mu =0}^{m-1}B_{s} \biggl(\frac{\mu }{m} \biggr) (mh)^{p+1-s}B_{p+1-s}^{(k)}(1) \\ &\quad = \sum_{s=0}^{p+1} \binom{p+1}{s}B_{s}(mh)^{p+1-s}B_{p+1-s}^{(k)}(1), \end{aligned}$$
(33)

where we used the fact (a) in Lemma 1.

Therefore we obtain the following theorem.

Theorem 8

For \(m,n,h\in \mathbb{N}\) with \((h,m)=1\) and any positive odd integer \(p\ge 3\), we have

$$\begin{aligned} & \sum_{s=0}^{p+1}\binom{p+1}{s}B_{s}B_{p+1-s}^{(k)}(1) (mh)^{p+1-s} \\ &\quad =m^{p} \sum_{\mu =0}^{m-1} \sum_{s=0}^{p+1}\binom{p+1}{s}h^{s}B_{s}^{(k)} (\mu /m )B_{p+1-s} \biggl(h- \biggl[\frac{h\mu }{m} \biggr] \biggr). \end{aligned}$$

Now we observe that

$$\begin{aligned} \sum_{n=0}^{\infty }B_{n}^{(k)}(x) \frac{t^{n}}{n!} &= \frac{\operatorname{Ei}_{k} (\log (1+t) )}{e^{t}-1}e^{xt}= \frac{\operatorname{Ei}_{k} (\log (1+t) )}{e^{dt}-1}\sum _{i=0}^{d-1}e^{(i+x)t} \\ &= \frac{\operatorname{Ei}_{k} (\log (1+t) )}{dt}\sum_{i=0}^{d-1}e^{(i+x)t} \frac{dt}{e^{dt}-1} \\ &= \sum_{j=0}^{\infty }d^{j-1}\sum _{i=0}^{d-1}B_{j} \biggl( \frac{x+i}{d} \biggr)\frac{t^{j}}{j!}\frac{1}{t}\sum _{l=1}^{\infty } \frac{ (\log (1+t) )^{l}}{(l-1)!l^{k}} \\ &= \sum_{j=0}^{\infty }d^{j-1}\sum _{i=0}^{d-1}B_{j} \biggl( \frac{x+i}{d} \biggr)\frac{t^{j}}{j!}\frac{1}{t}\sum _{l=1}^{\infty } \frac{1}{l^{k-1}}\sum _{m=l}^{\infty }S_{1}(m,l)\frac{t^{m}}{m!} \\ &= \sum_{j=0}^{\infty }d^{j-1}\sum _{i=0}^{d-1}B_{j} \biggl( \frac{x+i}{d} \biggr)\frac{t^{j}}{j!}\frac{1}{t}\sum _{m=1}^{\infty } \sum_{l=1}^{m} \frac{S_{1}(m,l)}{l^{k-1}}\frac{t^{m}}{m!} \\ &= \sum_{j=0}^{\infty }d^{j-1}\sum _{i=0}^{d-1}B_{j} \biggl( \frac{x+i}{d} \biggr)\frac{t^{j}}{j!}\sum_{m=0}^{\infty } \sum_{l=1}^{m+1} \frac{S_{1}(m+1,l)}{l^{k-1}(m+1)} \frac{t^{m}}{m!} \\ &= \sum_{n=0}^{\infty } \Biggl(\sum _{j=0}^{n}\sum_{i=0}^{d-1} \sum_{l=1}^{n-j+1} \binom{n}{j}d^{j-1}B_{j} \biggl(\frac{x+i}{d} \biggr) \frac{S_{1}(n-j+1,l)}{(n-j+1)l^{k-1}} \Biggr)\frac{t^{n}}{n!}, \end{aligned}$$
(34)

where d is a positive integer.

Therefore by comparing the coefficients on both sides of (34) we obtain the following theorem.

Theorem 9

For \(k\in \mathbb{Z}\), \(d\in \mathbb{N}\), and \(n\ge 0\), we have

$$\begin{aligned}& B_{n}^{(k)}(x)=\sum_{j=0}^{n} \sum_{i=0}^{d-1}\sum _{l=1}^{n-j+1} \binom{n}{j}d^{j-1}B_{j} \biggl(\frac{x+i}{d} \biggr) \frac{S_{1}(n-j+1,l)}{(n-j+1)l^{k-1}}. \end{aligned}$$

From (25), using Theorem 9 and (c) in Lemma 1, we see that

$$\begin{aligned} & hm^{p}S_{p}^{(k)}(h,m)+mh^{p}S_{p}^{(k)}(m,h) \\ &\quad =hm^{p}\sum_{\mu =0}^{m-1} \frac{\mu }{m}\overline{B}_{p}^{(k)} \biggl( \frac{h\mu }{m} \biggr)+mh^{p}\sum_{\nu =0}^{h-1} \frac{\nu }{h} \overline{B}_{p}^{(k)} \biggl( \frac{m\nu }{h} \biggr) \\ &\quad =hm^{p}\sum_{\mu =0}^{m-1} \frac{\mu }{m}\sum_{j=0}^{p}h^{j-1} \binom{p}{j}\sum_{\nu =0}^{h-1}\sum _{l=1}^{p-j+1} \frac{S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \overline{B}_{j} \biggl( \frac{\mu }{m}+\frac{\nu }{h} \biggr) \\ &\quad\quad {} +mh^{p}\sum_{\nu =0}^{h-1} \frac{\nu }{h}\sum_{j=0}^{p}m^{j-1} \binom{p}{j}\sum_{\mu =0}^{m-1}\sum _{l=1}^{p-j+1} \frac{S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \overline{B}_{j} \biggl( \frac{\nu }{h}+\frac{\mu }{m} \biggr) \\ &\quad =\sum_{\mu =0}^{m-1}\frac{\mu }{m} \sum_{j=0}^{p}m^{p-j}(mh)^{j} \binom{p}{j}\sum_{\nu =0}^{h-1}\sum _{l=1}^{p-j+1}\overline{B}_{j} \biggl(\frac{\mu }{m}+\frac{\nu }{h} \biggr) \frac{S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \\ &\quad\quad {} +\sum_{\nu =0}^{h-1} \frac{\nu }{h}\sum_{j=0}^{p}h^{p-j}(mh)^{j} \binom{p}{j}\sum_{\mu =0}^{m-1}\sum _{l=1}^{p-j+1}\overline{B}_{j} \biggl(\frac{\nu }{h}+\frac{\mu }{m} \biggr) \frac{S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \\ &\quad =\sum_{\mu =0}^{m-1}\sum _{j=0}^{p}\sum_{\nu =0}^{h-1} \sum_{l=1}^{p-j+1}( \mu h) (mh)^{-1}m^{p-j}(mh)^{j} \binom{p}{j}\overline{B}_{j} \biggl( \frac{\mu }{m}+ \frac{\nu }{h} \biggr) \frac{S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \\ &\quad\quad {} +\sum_{\mu =0}^{m-1}\sum _{j=0}^{p}\sum_{\nu =0}^{h-1} \sum_{l=1}^{p-j+1}(m \nu ) (mh)^{-1}h^{p-j}(mh)^{j} \binom{p}{j}\overline{B}_{j} \biggl( \frac{\nu }{h}+ \frac{\mu }{m} \biggr) \frac{S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \\ &\quad =\sum_{\mu =0}^{m-1}\sum _{j=0}^{p}\sum_{\nu =0}^{h-1} \sum_{l=1}^{p-j+1} \frac{(mh)^{j-1}\binom{p}{j}S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \bigl((\mu h)m^{p-j}+(m \nu ) h^{p-j} \bigr)\overline{B}_{j} \biggl(\frac{\nu }{h}+\frac{\mu }{m} \biggr). \end{aligned}$$
(35)

Therefore we obtain the following reciprocity relation.

Theorem 10

For \(m,h,p\in \mathbb{N}\) and \(k\in \mathbb{Z}\), we have

$$\begin{aligned} &hm^{p}S_{p}^{(k)}(h,m)+mh^{p}S_{p}^{(k)}(m,h) \\ &\quad =\sum_{\mu =0}^{m-1}\sum _{j=0}^{p}\sum_{\nu =0}^{h-1} \sum_{l=1}^{p-j+1} \frac{(mh)^{j-1}\binom{p}{j}S_{1}(p-j+1,l)}{(p-j+1)l^{k-1}} \bigl((\mu h)m^{p-j}+(m \nu ) h^{p-j} \bigr)\overline{B}_{j} \biggl(\frac{\nu }{h}+\frac{\mu }{m} \biggr). \end{aligned}$$

In the case \(k=1\), we obtain the following reciprocity relation for the generalized Dedekind sum defined by Apostol.

Corollary 11

For \(m,h,p\in \mathbb{N}\), we have

$$\begin{aligned} &hm^{p}S_{p}(h,m)+mh^{p}S_{p}(m,h) \\ &\quad = \sum_{\mu =0}^{m-1}\sum _{\nu =0}^{h-1}(mh)^{p-1}(\mu h+m\nu ) \overline{B}_{p} \biggl(\frac{\nu }{h}+\frac{\mu }{m} \biggr) \\ &\quad = (mh)^{p} \sum_{\mu =0}^{m-1}\sum _{\nu =0}^{h-1}(mh)^{-1}(\mu h+m \nu ) \overline{B}_{p} \biggl(\frac{\nu }{h}+\frac{\mu }{m} \biggr). \end{aligned}$$

Conclusion

The Dedekind sums are defined by

$$ S(h,m)=\sum_{\mu =1}^{m} \biggl( \biggl( \frac{\mu }{m} \biggr) \biggr) \biggl( \biggl(\frac{h\mu }{m} \biggr) \biggr) \quad ( \text{see [1, 4, 6--8, 11--13]}). $$

In 1952, Apostol considered the generalized Dedekind sums and introduced interesting and important identities and theorems related to his generalized Dedekind sums. These Dedekind sums are a field studied by various researchers. Recently, the modified Hardy polyexponential function of index k is introduced by

$$ \operatorname{Ei}_{k}(x)=\sum_{n=1}^{\infty } \frac{x^{n}}{n^{k}(n-1)!}, \quad (k \in \mathbb{Z})\ (\text{see [5, 9]}). $$

In [5] the type 2 poly-Bernoulli polynomials of index k are defined in terms of the polyexponential function of index k by

$$ \frac{\operatorname{Ei}_{k}(\log (1+t))}{e^{t}-1}e^{xt}=\sum_{n=0}^{\infty }B_{n}^{(k)}(x) \frac{t^{n}}{n!}\quad (k\in \mathbb{Z}). $$

In this paper, we thought of the poly-Dedekind sums from the perspective of the Apostol generalized Dedekind sums. That is, we considered the poly-Dedekind sums derived from the type 2 poly-Bernoulli functions and polynomials.

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Acknowledgements

The authors thank Jangjeon Institute for Mathematical Science for the support of this research.

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TK and DSK conceived of the framework and structured the whole paper; TK and DSK wrote the paper; L-CJ and HL checked the results of the paper; DSK and TK completed the revision of the paper. All authors have read and agreed to the published version of the manuscript.

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Correspondence to Lee-Chae Jang.

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Kim, T., Kim, D.S., Lee, H. et al. Identities on poly-Dedekind sums. Adv Differ Equ 2020, 563 (2020). https://doi.org/10.1186/s13662-020-03024-x

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MSC

  • 11F20
  • 11B68
  • 11B83

Keywords

  • Poly-Dedekind sum
  • Polyexponential function
  • Type 2 poly-Bernoulli polynomial