On Polycosecant Numbers

Masanobu Kaneko Faculty of Mathematics

Kyushu University Motooka 744, Nishi-ku

Fukuoka 819-0395

Japan

mkaneko@math.kyushu-u.ac.jp

Maneka Pallewatta Graduate School of Mathematics

Kyushu University Motooka 744, Nishi-ku

Fukuoka 819-0395

Japan

maneka.osh@gmail.com

Hirofumi Tsumura

Department of Mathematical Sciences

Tokyo Metropolitan University

1-1, Minami-Ohsawa Hachioji, Tokyo 192-0397

Japan

tsumura@tmu.ac.jp

Abstract

We introduce and study a “level two” generalization of the poly-Bernoulli numbers, which may also be regarded as a generalization of the cosecant numbers. We prove a recurrence relation, two exact formulas, and a duality relation for negative upper-index numbers.

1 Introduction

The first named author [7] defined the poly-Bernoulli numbers \underline{A099594} and later Arakawa and the first named author [2] studied a slightly modified version. They are, denoted by B_{n}^{(k)} and C_{n}^{(k)} respectively, defined by using generating series, as follows. For an integer k ∈ ℤ, let (B_{n}^{(k)})_{n≥0} and (C_{n}^{(k)})_{n≥0} be the sequences of rational numbers given respectively by

\frac{Li_{k}(1 − e^{−t})}{1 − e^{−t}} = ∑_{n=0}^{∞} B_{n}^{(k)} \frac{t^{n}}{n!}  (1)

and

\frac{Li_{k}(1 − e^{−t})}{e^{t} − 1} = ∑_{n=0}^{∞} C_{n}^{(k)} \frac{t^{n}}{n!},  (2)

where Li_{k}(z) is the polylogarithm function (or rational function when k ≤ 0) defined by

Li_{k}(z) = ∑_{m=1}^{∞} \frac{z^{m}}{m^{k}} (|z| < 1).  (3)

Since Li_{1}(z) = − log(1 − z), the generating functions on the left-hand sides of (1) and (2) when k = 1 become

\frac{te^{t}}{e^{t} − 1} and \frac{t}{e^{t} − 1}

respectively. Hence B_{n}^{(1)} and C_{n}^{(1)} represent the standard Bernoulli numbers \underline{A027641}, \underline{A027642}, the only difference being B_{1}^{(1)} = 1/2 and C_{1}^{(1)} = −1/2 and otherwise B_{n}^{(1)} = C_{n}^{(1)}.

Several properties of poly-Bernoulli numbers have been found including the following results:

B_{n}^{(k)} = (−1)^{n} ∑_{i=0}^{n} \frac{(−1)^{i}i! \begin{cases}n \\ i \end{cases}\}}{(i + 1)^{k}}, C_{n}^{(k)} = (−1)^{n} ∑_{i=0}^{n} \frac{(−1)^{i}i!\begin{cases}n + 1 \\ i + 1 \end{cases}\}}{(i + 1)^{k}},

where k is an integer, n a non-negative integer, and let \begin{cases}n \\ i \end{cases}\} denote the Stirling numbers of

the second kind \underline{A008277}. Moreover, their dualities

B_{n}^{(−k)} = B_{k}^{(−n)},  (4)
C_{n}^{(−k−1)} = C_{k}^{(−n−1)}  (5)

(k, n ∈ ℤ_{≥0}) are derived in [7, Theorems 1 and 2] or in [8, Section 2]. For combinatorial applications, see [3].

In this paper, we study the following ‘level 2’ analog of poly-Bernoulli numbers, denoted by D_{n}^{(k)}, which we call the polycosecant numbers. For each k ∈ ℤ, define D_{n}^{(k)} by

\frac{A_{k}(tanh(t/2))}{sinh t} = ∑_{n=0}^{∞} D_{n}^{(k)} \frac{t^{n}}{n!},  (6)

where A_{k}(z) is the series

A_{k}(z) = 2 ∑_{n=0}^{∞} \frac{z^{2n+1}}{(2n + 1)^{k}}  (7)

and tanh(z) and sinh(z) are the usual hyperbolic tangent and sine functions respectively. Since A_{k}(z), tanh(z) and sinh(z) are all odd functions, we immediately see that D_{2n+1}^{(k)} = 0 for all n ∈ ℤ_{≥0}. Note that A_{1}(z) = 2 tanh^{−1}(z), and thus

∑_{n=0}^{∞} D_{n}^{(1)} \frac{t^{n}}{n!} = \frac{t}{sinh t} = \frac{it}{sin(it)} (i = \sqrt{−1}).

Hence D_{n}^{(1)} is the cosecant number D_{n} which Nörlund [14, p. 27 (39) and p. 32 (52)] first introduced \underline{A001896}, \underline{A001897}. It should be noted that the terminology ‘cosecant number’ was not used by Nörlund. Apparently Kowalenko [11] was the first to use the terminology, but he adopted the name for D_{n}/n! instead of D_{n}. He gave many applications, and studied a generalization together with interesting number-theoretical applications [5, 11, 12, 13].

Here, we give a table of D_{n}^{(k)} for small k and n. A table for k < 0 will be given in §3.

\begin{matrix}[figure: k; n] & 0 & 2 & 4 & 6 & 8 & 10 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & −\frac{1}{3} & \frac{7}{15} & −\frac{31}{21} & \frac{127}{15} & −\frac{2555}{33} \\ 2 & 1 & −\frac{4}{9} & \frac{176}{225} & −\frac{6464}{2205} & \frac{3328}{175} & −\frac{1037312}{5445} \\ 3 & 1 & −\frac{13}{27} & \frac{3103}{3375} & −\frac{859939}{231525} & \frac{12761501}{496125} & −\frac{63453851}{232925} \\ 4 & 1 & −\frac{40}{81} & \frac{49184}{50625} & −\frac{98447744}{24310125} & \frac{4519218688}{156279375} & −\frac{6868861044736}{21791298375} \\ 5 & 1 & −\frac{121}{243} & \frac{751927}{759375} & −\frac{10665916999}{2552563125} & \frac{1488186370469}{49228003125} & −\frac{25213417199300173}{75506848869375} \end{matrix}

Table 1: D_{n}^{(k)} (0 ≤ k ≤ 5, 0 ≤ n ≤ 10, even)

We should also mention that our D_{n}^{(k)} is (if slightly modified) a special case of a generalization of the poly-Bernoulli number which Sasaki [15, Definition 5] introduced. The numbers D_{n}^{(k)} are closely connected to the ‘multiple Hurwitz zeta functions’ and the ‘level 2’ multiple zeta functions. We shall explore these connections in an ongoing project, which will be stated in the forthcoming paper.

2 Recurrence and explicit formulas for polycosecant numbers

In this section, we obtain a recurrence and explicit formulas for polycosecant numbers. We begin by deriving a recurrence relation.

Proposition 1. For every integer k and n ≥ 0, the polycosecant numbers obey the recurrence relation of

D_{n}^{(k−1)} = ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \binom{n + 1}{2m + 1}D_{n−2m}^{(k)}.  (8)

Proof. First, differentiate (6), which yields

\frac{A_{k−1}(tanh(t/2))}{sinh t} = cosh t ∑_{n=0}^{∞} D_{n}^{(k)} \frac{t^{n}}{n!} + sinh t ∑_{n=1}^{∞} D_{n}^{(k)} \frac{t^{n−1}}{(n − 1)!}.

From the above result we have

∑_{n=0}^{∞} D_{n}^{(k−1)} \frac{t^{n}}{n!} = ∑_{m=0}^{∞} \frac{t^{2m}}{(2m)!} ∑_{n=0}^{∞} D_{n}^{(k)} \frac{t^{n}}{n!} + ∑_{m=0}^{∞} \frac{t^{2m+1}}{(2m + 1)!} ∑_{n=1}^{∞} D_{n}^{(k)} \frac{t^{n−1}}{(n − 1)!}

= ∑_{n=0}^{∞} ∑_{m=0}^{⌊ \frac{n}{2} ⌋} D_{n−2m}^{(k)} \frac{t^{n}}{(2m)!(n − 2m)!} + ∑_{n=1}^{∞} ∑_{m=0}^{⌊ \frac{n}{2} ⌋} D_{n−2m}^{(k)} \frac{t^{n}}{(2m + 1)!(n − 2m − 1)!}

= ∑_{n=0}^{∞} ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2m}D_{n−2m}^{(k)} \frac{t^{n}}{n!} + ∑_{n=1}^{∞} ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2m + 1}D_{n−2m}^{(k)} \frac{t^{n}}{n!}

= ∑_{n=0}^{∞} ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \binom{n + 1}{2m + 1}D_{n−2m}^{(k)} \frac{t^{n}}{n!}.

Since t is arbitrary, we can equate like powers of t on both sides, thereby obtaining the desired result.

Since A_{0}(tanh(t/2)) = sinh(t), we observe D_{0}^{(0)} = 1 and D_{n}^{(0)} = 0 for n ≥ 1. Hence equation (8) can be used to compute D_{n}^{(k)} for k < 0 recursively starting from D_{n}^{(0)}. For k > 0, we rewrite (8) as

(n + 1)D_{n}^{(k)} = D_{n}^{(k−1)} − ∑_{m=1}^{⌊ \frac{n}{2} ⌋} \binom{n + 1}{2m + 1}D_{n−2m}^{(k)}

in order to compute D_{n}^{(k)} recursively. We observe that D_{0}^{(k)} = 1 for all k ∈ ℤ.

We continue by presenting two formulas for polycosecant numbers. Before doing so, we require the following lemma.

Lemma 2. For n ≥ 1 we have,

x^{n} (\frac{d}{dx})^{n} = ∑_{m=1}^{n}(−1)^{n−m} [\begin{matrix}n \\ m \end{matrix}] (x\frac{d}{dx})^{m} .

Here, we let [\begin{matrix}n \\ m \end{matrix}] denote the Stirling numbers of the first kind.

Proof. This result can be proved in the same manner as [1, Proposition 2.6 (4)]. Hence we omit here.

Theorem 3. For k ∈ ℤ and n ≥ 0, the following results hold.

(1)

D_{n}^{(k)} = 4 ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \frac{1}{(2m + 1)^{k+1}} ∑_{p=1}^{2m+1} ∑_{q=0}^{n−2m}(2^{p+q+1} − 1)\binom{n}{q} [\begin{matrix}2m + 1 \\ p \end{matrix}] \begin{cases}n − q \\ 2m \end{cases}\} \frac{B_{p+q+1}}{p + q + 1},

where B_{n} (= C_{n}^{(1)}) are the Bernoulli numbers, and

(2)

D_{n}^{(k)} = ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \frac{1}{(2m + 1)^{k+1}} ∑_{p=2m}^{n} \frac{(−1)^{p}(p + 1)!}{2^{p}} \binom{p}{2m} \begin{cases}n + 1 \\ p + 1 \end{cases}\} .

Proof. We may express (6) as

∑_{n=0}^{∞} D_{n}^{(k)} \frac{t^{n}}{n!} = \frac{A_{k}(tanh(t/2))}{sinh t}

= 2 ∑_{m=0}^{∞} \frac{(tanh(t/2))^{2m+1}}{(2m + 1)^{k}} \frac{1}{sinh t}

= 4 ∑_{m=0}^{∞} \frac{1}{(2m + 1)^{k}} \frac{e^{t}(e^{t} − 1)^{2m}}{(e^{t} + 1)^{2m+2}} .  (9)

Since

\frac{1}{(x + 1)^{n+1}} = \frac{(−1)^{n}}{n!} (\frac{d}{dx})^{n} \frac{1}{x + 1},  (10)

we see that by setting x = e^{t} and introducing Lemma 2,

\frac{e^{nt}}{(e^{t} + 1)^{n+1}} = \frac{1}{n!} ∑_{p=1}^{n}(−1)^{p} [\begin{matrix}n \\ p \end{matrix}] (\frac{d}{dt})^{p} \frac{1}{e^{t} + 1}.  (11)

Moreover, from the generating functions

\frac{t}{e^{t} − 1} = ∑_{q=0}^{∞} B_{q} \frac{t^{q}}{q!},

and

\frac{1}{e^{t} + 1} = \frac{1}{e^{t} − 1} − \frac{2}{e^{2t} − 1},

we find that

\frac{1}{e^{t} + 1} = ∑_{q=0}^{∞}(1 − 2^{q})B_{q} \frac{t^{q−1}}{q!} .

Taking the p-th derivative on both sides yields

(\frac{d}{dt})^{p} (\frac{1}{e^{t} + 1}) = ∑_{q=p+1}^{∞}(1 − 2^{q})\frac{B_{q}}{q} \frac{t^{q−p−1}}{(q − p − 1)!} = ∑_{q=0}^{∞}(1 − 2^{p+q+1})\frac{B_{p+q+1}}{p + q + 1} \frac{t^{q}}{q!}.

Now we substitute the above result into (11) to obtain

\frac{e^{nt}}{(e^{t} + 1)^{n+1}} = \frac{1}{n!} ∑_{p=1}^{n}(−1)^{p} [\begin{matrix}n \\ p \end{matrix}] ∑_{q=0}^{∞}(1 − 2^{p+q+1})\frac{B_{p+q+1}}{p + q + 1} \frac{t^{q}}{q!}

= \frac{1}{n!} ∑_{q=0}^{∞} ∑_{p=1}^{n}(−1)^{p} [\begin{matrix}n \\ p \end{matrix}] (1 − 2^{p+q+1})\frac{B_{p+q+1}}{p + q + 1} \frac{t^{q}}{q!}.

Hence we obtain

\frac{e^{t}}{(e^{t} + 1)^{2m+2}} = \frac{e^{−(2m+1)t}}{(e^{−t} + 1)^{2m+2}}

= \frac{1}{(2m + 1)!} ∑_{q=0}^{∞} ∑_{p=1}^{2m+1}(−1)^{p+q} [\begin{matrix}2m + 1 \\ p \end{matrix}] (1 − 2^{p+q+1})\frac{B_{p+q+1}}{p + q + 1} \frac{t^{q}}{q!}.

With the aid of the generating function given by [1, Proposition 2.6 (7)] and noting that

\begin{cases}s \\ 2m \end{cases}\} = 0 if s < 2m and

(e^{t} − 1)^{2m} = (2m)! ∑_{s=0}^{∞} \begin{cases}s \\ 2m \end{cases}\} \frac{t^{s}}{s!},

we arrive at

\frac{e^{t}(e^{t} − 1)^{2m}}{(e^{t} + 1)^{2m+2}}

= \frac{1}{2m + 1} ∑_{q=0}^{∞} ∑_{s=0}^{∞} ∑_{p=1}^{2m+1}(−1)^{p+q}(1 − 2^{p+q+1}) [\begin{matrix}2m + 1 \\ p \end{matrix}] \begin{cases}s \\ 2m \end{cases}\} \frac{B_{p+q+1}}{p + q + 1} \frac{t^{q+s}}{q!s!}

= \frac{1}{2m + 1} ∑_{n=0}^{∞} ∑_{q=0}^{n} ∑_{p=1}^{2m+1}(−1)^{p+q}(1 − 2^{p+q+1})\binom{n}{q} [\begin{matrix}2m + 1 \\ p \end{matrix}] \begin{cases}n − q \\ 2m \end{cases}\} \frac{B_{p+q+1}}{p + q + 1} \frac{t^{n}}{n!}.

Substituting the above result into (9), we have

∑_{n=0}^{∞} D_{n}^{(k)} \frac{t^{n}}{n!}

= 4 ∑_{m=0}^{∞} \frac{1}{(2m + 1)^{k+1}} ∑_{n=0}^{∞} ∑_{q=0}^{n} ∑_{p=1}^{2m+1}(−1)^{p+q}(1 − 2^{p+q+1})

× \binom{n}{q} [\begin{matrix}2m + 1 \\ p \end{matrix}] \begin{cases}n − q \\ 2m \end{cases}\} \frac{B_{p+q+1}}{p + q + 1} \frac{t^{n}}{n!}

= 4 ∑_{n=0}^{∞} ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \frac{1}{(2m + 1)^{k+1}} ∑_{p=1}^{2m+1} ∑_{q=0}^{n−2m}(2^{p+q+1} − 1)\binom{n}{q} [\begin{matrix}2m + 1 \\ p \end{matrix}] \begin{cases}n − q \\ 2m \end{cases}\} \frac{B_{p+q+1}}{p + q + 1} \frac{t^{n}}{n!}.

In obtaining the above result, we have used B_{p+q+1} = 0 for p + q ≥ 1 and even, while

\begin{cases}n − q \\ 2m \end{cases}\} = 0 for n − q < 2m. By equating like powers, we arrive at the first result in the

theorem.

To prove the second result, we require a formula from [4] for the higher order tangent numbers, T_{n,m}, whose generating function is

\frac{tan^{m} t}{m!} = ∑_{n=m}^{∞} T_{n,m} \frac{t^{n}}{n!}.  (12)

The formula is

T_{n,m} = \frac{i^{n−m}}{m!} ∑_{p=m}^{n}(−2)^{n−p}p!\binom{p − 1}{m − 1} \begin{cases}n \\ p \end{cases}\} .  (13)

From (6),

∑_{n=0}^{∞} D_{n}^{(k)} \frac{t^{n}}{n!} = \frac{A_{k}(tanh(t/2))}{sinh t} = \frac{d}{dt}A_{k+1}(tanh(t/2))

= 2\frac{d}{dt} ∑_{m=0}^{∞} \frac{(tanh(t/2))^{2m+1}}{(2m + 1)^{k+1}} .  (14)

By using tanh t = −i tan(it) and equations (12) and (13), we can write

(tanh(t/2))^{m} = (−i)^{m}m! ∑_{n=m}^{∞} T_{n,m} \frac{i^{n}}{2^{n}} \frac{t^{n}}{n!}

= (−i)^{m}(−1)^{\frac{n−m}{2}} ∑_{n=m}^{∞} ∑_{p=m}^{n}(−2)^{n−p}p!\binom{p − 1}{m − 1} \begin{cases}n \\ p \end{cases}\} \frac{i^{n}}{2^{n}} \frac{t^{n}}{n!}

= (−1)^{m} ∑_{n=m}^{∞} ∑_{p=m}^{n}(−1)^{p} \frac{p!}{2^{p}} \binom{p − 1}{m − 1} \begin{cases}n \\ p \end{cases}\} \frac{t^{n}}{n!}.

Therefore, we find that

∑_{n=0}^{∞} D_{n}^{(k)} \frac{t^{n}}{n!} = ∑_{m=0}^{∞} \frac{1}{(2m + 1)^{k+1}} ∑_{n=2m+1}^{∞} ∑_{p=2m+1}^{n}(−1)^{p+1} \frac{p!}{2^{p−1}} \binom{p − 1}{2m} \begin{cases}n \\ p \end{cases}\} \frac{t^{n−1}}{(n − 1)!}

= ∑_{m=0}^{∞} \frac{1}{(2m + 1)^{k+1}} ∑_{n=2m}^{∞} ∑_{p=2m}^{n}(−1)^{p} \frac{(p + 1)!}{2^{p}} \binom{p}{2m} \begin{cases}n + 1 \\ p + 1 \end{cases}\} \frac{t^{n}}{n!}

= ∑_{n=0}^{∞} ∑_{m=0}^{⌊ \frac{n}{2} ⌋} \frac{1}{(2m + 1)^{k+1}} ∑_{p=2m}^{n} \frac{(−1)^{p}(p + 1)!}{2^{p}} \binom{p}{2m} \begin{cases}n + 1 \\ p + 1 \end{cases}\} \frac{t^{n}}{n!}.

By equating like powers of t, we arrive at the second result in the theorem.

3 Duality

We now turn our attention to the duality property of the polycosecant numbers. We shall present two different proofs using the same generating function. The first proof is based on a closed symmetric formula for the generating function, while the second is more indirect and complicated. However, we have decided to include the latter proof since it reveals fascinating results, especially regarding hyperbolic trigonometric functions.

Here is a table of D_{2n}^{(−2k−1)} for small k and n.

Theorem 4. For n, k ∈ ℤ_{≥0}, the polycosecant numbers possess the duality property of

D_{2n}^{(−2k−1)} = D_{2k}^{(−2n−1)}.  (15)

\begin{matrix}[figure: k; n] & 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 13 & 121 & 1093 & 9841 & 88573 \\ 2 & 1 & 121 & 4081 & 111721 & 2880481 & 72799321 \\ 3 & 1 & 1093 & 111721 & 7256173 & 403087441 & 20966597653 \\ 4 & 1 & 9841 & 2880481 & 403087441 & 42931692481 & 4032800405041 \\ 5 & 1 & 88573 & 72799321 & 20966597653 & 4032800405041 & 638704166793133 \end{matrix} ; Table 2: D_{2n}^{(−2k−1)} (0 ≤ k ≤ 5, 0 ≤ n ≤ 5)

First proof. We show that the generating function of D_{2n}^{(−2k−1)},

F (x, y) := ∑_{n=0}^{∞} ∑_{k=0}^{∞} D_{2n}^{(−2k−1)} \frac{x^{2n}}{(2n)!} \frac{y^{2k}}{(2k)!},  (16)

is symmetric in x and y. This is ensured by the following closed formula for F (x, y). Proposition 5. Let

G(x, y) = \frac{e^{x+y}}{(1 + e^{x} + e^{y} − e^{x+y})^{2}} .

Then one finds

F (x, y) = G(x, y) + G(x, −y) + G(−x, y) + G(−x, −y).

Proof. We first compute the generating function of all D_{n}^{(−k)},

f(x, y) = ∑_{n=0}^{∞} ∑_{k=0}^{∞} D_{n}^{(−k)} \frac{x^{n}}{n!} \frac{y^{k}}{k!} .  (17)

We claim that the formula

f(x, y) = \frac{e^{x}(e^{y} − 1)}{1 + e^{x} + e^{y} − e^{x+y}} + \frac{e^{−x}(e^{y} − 1)}{1 + e^{−x} + e^{y} − e^{−x+y}}  (18)

holds. To prove this, we first observe that, by definition,

f(x, y) = ∑_{k=0}^{∞} \frac{A_{−k}(tanh(x/2))}{sinh x} \frac{y^{k}}{k!}

= \frac{2}{sinh x} ∑_{k=0}^{∞} ∑_{n=0}^{∞}(2n + 1)^{k}(tanh(x/2))^{2n+1} \frac{y^{k}}{k!} .

Noting that

2 ∑_{n=0}^{∞}(2n + 1)^{k}t^{2n+1} = 2 (t\frac{d}{dt})^{k} \frac{t}{1 − t^{2}} = (t\frac{d}{dt})^{k} (\frac{1}{1 − t} − \frac{1}{1 + t}) ,  (19)

and by the standard formula (cf., e.g., [1, Proposition 2.6 (4)])

(t\frac{d}{dt})^{k} = ∑_{m=1}^{k} \begin{cases}k \\ m \end{cases}\} t^{m} (\frac{d}{dt})^{m} ,

we find that the right-hand side of (19) becomes

∑_{m=1}^{k} \begin{cases}k \\ m \end{cases}\} t^{m} (\frac{d}{dt})^{m} (\frac{1}{1 − t} − \frac{1}{1 + t})

= ∑_{m=1}^{k} \begin{cases}k \\ m \end{cases}\} m! (\frac{t^{m}}{(1 − t)^{m+1}} − \frac{(−t)^{m}}{(1 + t)^{m+1}} ) .

Therefore, by setting t = tanh(x/2) and noting t/(1−t) = (e^{x}−1)/2, −t/(1+t) = (e^{−x}−1)/2, (sinh x)(1 − t) = e^{−x}(e^{x} − 1), (sinh x)(1 + t) = e^{x} − 1, we arrive at

f(x, y) = \frac{1}{sinh x} ∑_{k=0}^{∞} ∑_{m=1}^{k} \begin{cases}k \\ m \end{cases}\} m! (\frac{t^{m}}{(1 − t)^{m+1}} − \frac{(−t)^{m}}{(1 + t)^{m+1}} ) \frac{y^{k}}{k!}

= ∑_{k=0}^{∞} ∑_{m=1}^{k} \begin{cases}k \\ m \end{cases}\} m!(\frac{e^{x}}{e^{x} − 1} (\frac{e^{x} − 1}{2})^{m} − \frac{1}{e^{x} − 1} (\frac{e^{−x} − 1}{2})^{m} )\frac{y^{k}}{k!}

= ∑_{m=1}^{∞}(e^{y} − 1)^{m}(\frac{e^{x}}{e^{x} − 1} (\frac{e^{x} − 1}{2})^{m} − \frac{1}{e^{x} − 1} (\frac{e^{−x} − 1}{2})^{m} )

= \frac{e^{x}}{e^{x} − 1} · \frac{(e^{y} − 1)(e^{x} − 1)}{2 − (e^{y} − 1)(e^{x} − 1)} − \frac{1}{e^{x} − 1} · \frac{(e^{y} − 1)(e^{−x} − 1)}{2 − (e^{y} − 1)(e^{−x} − 1)}

= \frac{e^{x}(e^{y} − 1)}{1 + e^{x} + e^{y} − e^{x+y}} + \frac{e^{−x}(e^{y} − 1)}{1 + e^{−x} + e^{y} − e^{−x+y}} .

This proves the identity (18). From (18) we see that f(x, y) is even in x, and so we have

\frac{f(x, y) − f(x, −y)}{2} = ∑_{n=0}^{∞} ∑_{k=0}^{∞} D_{2n}^{(−2k−1)} \frac{x^{2n}}{(2n)!} \frac{y^{2k+1}}{(2k + 1)!}.

Our generating function F (x, y) is the derivative of this relation with respect to y, and Proposition 5 follows from a straightforward calculation, and by the symmetry of F (x, y) in x and y, Theorem 4 is proved.

Before presenting the second proof of Theorem 4, we require several lemmas.

Lemma 6. The function F (x, y) defined by (16) can be expressed as

F (x, y) = 2 ∑_{n=0}^{∞} \frac{∂}{∂x} (tanh^{2n+1}(x/2)) cosh((2n + 1)y).

Proof. By (6) and (16), we have

F (x, y) = 2 ∑_{k=0}^{∞} \frac{A_{−2k−1}(tanh(x/2))}{sinh(x)} \frac{y^{2k}}{(2k)!}

= \frac{2}{sinh(x)} ∑_{k=0}^{∞} ∑_{n=0}^{∞}(2n + 1)^{2k+1} tanh^{2n+1}(x/2)\frac{y^{2k}}{(2k)!}

= \frac{2}{sinh(x)} ∑_{n=0}^{∞}(2n + 1) tanh^{2n+1}(x/2) cosh((2n + 1)y)

= \frac{1}{sinh(x/2) cosh(x/2)} ∑_{n=0}^{∞}(2n + 1) tanh^{2n}(x/2)\frac{sinh(x/2)}{cosh(x/2)} cosh((2n + 1)y)

= 2 ∑_{n=0}^{∞} \frac{∂}{∂x} (tanh^{2n+1}(x/2)) cosh((2n + 1)y).

This completes the proof.

We write

F (x, y) = ∑_{m=0}^{∞} g_{m}(x)\frac{y^{2m}}{(2m)!} = ∑_{m=0}^{∞} h_{m}(y)\frac{x^{2m}}{(2m)!}.

If one can show that g_{m}(x) = h_{m}(x) for all m ≥ 0, then the second proof will be complete.

First, we consider g_{m}(x). Expanding cosh((2n + 1)y) in Lemma 6 and equating like powers of y, we obtain

g_{m}(x) = (\frac{∂}{∂y} )^{2m} F (x, y) |_{y=0} = 2\frac{d}{dx} ∑_{n=0}^{∞}(2n + 1)^{2m} tanh^{2n+1}(x/2).

Next we note that

∑_{n=0}^{∞}(2n + 1)^{2m}t^{2n+1} = (t\frac{d}{dt})^{2m} ∑_{n=0}^{∞} t^{2n+1} = (t\frac{d}{dt})^{2m} \frac{t}{1 − t^{2}} .  (20)

Setting t = tanh(x/2) and noting

dt = \frac{1}{2} \frac{1}{cosh^{2}(x/2)}dx, \frac{t}{1 − t^{2}} = \frac{tanh(x/2)}{1 − tanh^{2}(x/2)} = \frac{1}{2} sinh x,

we have

t\frac{d}{dt} = tanh(x/2) · 2 cosh^{2}(x/2)\frac{d}{dx} = sinh x \frac{d}{dx}.

Therefore we obtain

g_{m}(x) = \frac{d}{dx} (sinh x \frac{d}{dx})^{2m} sinh x.  (21)

We can explicitly write down the right-hand side by using the following lemma. For m ∈ ℤ_{≥0}, we define sequences (a_{i}^{(m)})_{0≤i≤m} ⊂ ℚ inductively by

\begin{matrix}a_{0}^{(0)} = 1, \\ a_{i}^{(m)} = \frac{1}{2} (i(2i − 1)a_{i−1}^{(m−1)} − (2i + 1)^{2}a_{i}^{(m−1)} + (i + 1)(2i + 3)a_{i+1}^{(m−1)}) (m ≥ 1), \end{matrix} (22)

where we formally interpret a_{i}^{(m)} = 0 for i < 0 or i > m.

Lemma 7. For m ∈ ℤ_{≥0},

(sinh x \frac{d}{dx})^{2m} sinh x = ∑_{i=0}^{m} a_{i}^{(m)} sinh((2i + 1)x).  (23)

Proof. We give the proof by induction on m. For m = 0, the identity trivially holds. We assume

(sinh x \frac{d}{dx})^{2(m−1)} sinh x = ∑_{i=0}^{m−1} a_{i}^{(m−1)} sinh((2i + 1)x).

Using

cosh(kx) sinh(x) = \frac{1}{2} (sinh((k + 1)x) − sinh((k − 1)x)) ,

we have

(sinh x \frac{d}{dx})^{2m−1} sinh x = \frac{1}{2} ∑_{i=0}^{m−1}(2i + 1)a_{i}^{(m−1)} (sinh((2i + 2)x) − sinh(2ix)) ,

and

(sinh x \frac{d}{dx})^{2m} sinh x

= ∑_{i=0}^{m−1}(2i + 1)a_{i}^{(m−1)}(\frac{i + 1}{2} (sinh((2i + 3)x) − sinh((2i + 1)x))

− \frac{i}{2} (sinh((2i + 1)x) − sinh((2i − 1)x)) )

= \frac{1}{2} ∑_{i=1}^{m} i(2i − 1)a_{i−1}^{(m−1)} sinh((2i + 1)x)

− \frac{1}{2} ∑_{i=0}^{m−1}(2i + 1)^{2}a_{i}^{(m−1)} sinh((2i + 1)x)

+ \frac{1}{2} ∑_{i=0}^{m−2}(i + 1)(2i + 3)a_{i+1}^{(m−1)} sinh((2i + 1)x).

Thus we observe that the coefficients of sinh((2i + 1)x) are in accordance with (22), thereby completing this proof by induction.

Using this lemma, we obtain

g_{m}(x) = ∑_{i=0}^{m}(2i + 1)a_{i}^{(m)} cosh((2i + 1)x).  (24)

Secondly, we compute h_{m}(y). Again by using Lemma 6, we have

h_{m}(y) = (\frac{∂}{∂x})^{2m} F (x, y) |_{x=0}

= 2 ∑_{n=0}^{∞} (\frac{d}{dx})^{2m+1} (tanh^{2n+1}(x/2)) cosh((2n + 1)y) |_{x=0}

= 2 ∑_{n=0}^{m} (\frac{d}{dx})^{2m+1} tanh^{2n+1}(x/2) |_{x=0} · cosh((2n + 1)y)  (25)

because

tanh^{2n+1}(x/2) = \frac{x^{2n+1}}{2^{2n+1}} + O(x^{2n+2}) (x → 0).

We write down the right-hand side of (25) by using the following lemma.

Lemma 8. For n, l ∈ ℤ_{≥0}, there exist sequences (b_{j}^{(n,l)})_{0≤j≤l} ⊂ ℚ such that

(\frac{d}{dx})^{l} tanh^{2n+1}(x/2) = ∑_{j=0}^{l} b_{j}^{(n,l)} tanh^{2n+1−l+2j}(x/2),  (26)

where b_{j}^{(n,l)} = 0 if 2n + 1 − l + 2j < 0. In particular,

(\frac{d}{dx})^{2m+1} tanh^{2n+1}(x/2) |_{x=0} = b_{m−n}^{(n,2m+1)}.  (27)

Proof. For each n, we can immediately obtain the form (26) by induction on l, using the relation

\frac{d}{dx} tanh^{2n+1}(x/2) = \frac{2n + 1}{2} (tanh^{2n}(x/2) − tanh^{2n+2}(x/2)) .

Combining Lemma 8 and (25), we obtain

h_{m}(y) = 2 ∑_{n=0}^{m} b_{m−n}^{(n,2m+1)} cosh((2n + 1)y).  (28)

Now we are going to show 2b_{m−n}^{(n,2m+1)} = (2i + 1)a_{i}^{(m)}, which implies g_{m}(x) = h_{m}(x). For m, n ∈ ℤ_{≥0} with n ≤ m, set b̃_{n}^{(m)} = 2b_{m−n}^{(n,2m+1)}. Then, by (27), we have b̃_{0}^{(0)} = 1. Furthermore the following lemma holds.

Lemma 9. For m ∈ ℤ_{≥1}, the b̃_{n}^{(m)} satisfy the recurrence relation given by

b̃_{n}^{(m)} = \frac{2n + 1}{2} (nb̃_{n−1}^{(m−1)} − (2n + 1)b̃_{n}^{(m−1)} + (n + 1)b̃_{n+1}^{(m−1)}) (n ≤ m),  (29)

where we interpret b_{i}^{(k)} = 0 for i < 0 or i > k.

Proof. It follows from (26) that

(\frac{d}{dx})^{2m+1} tanh^{2n+1}(x/2) = ∑_{j=0}^{2m+1} b_{j}^{(n,2m+1)} tanh^{2n−2m+2j}(x/2).  (30)

Differentiating twice and using (26), we see that the left-hand side is equal to

(\frac{d}{dx})^{2m} (\frac{2n + 1}{2} tanh^{2n}(x/2) − tanh^{2n+2}(x/2))

= \frac{2n + 1}{2} (\frac{d}{dx})^{2m−1} (n tanh^{2n−1}(x/2) − (2n + 1) tanh^{2n+1}(x/2) + (n + 1) tanh^{2n+3}(x/2))

= \frac{2n + 1}{2}(n ∑_{j=0}^{2m−1} b_{j}^{(n−1,2m−1)} tanh^{2n−2m+2j}(x/2)

− (2n + 1) ∑_{j=0}^{2m−1} b_{j}^{(n,2m−1)} tanh^{2n−2m+2+2j}(x/2)

+ (n + 1) ∑_{j=0}^{2m−1} b_{j}^{(n+1,2m−1)} tanh^{2n−2m+4+2j}(x/2)).

If we let x → 0, the above result goes to

\frac{2n + 1}{2}(nb_{m−n}^{(n−1,2m−1)} − (2n + 1)b_{m−n−1}^{(n,2m−1)} + (n + 1)b_{m−n−2}^{(n+1,2m−1)})

= \frac{2n + 1}{4}(nb̃_{n−1}^{(m−1)} − (2n + 1)b̃_{n}^{(m−1)} + (n + 1)b̃_{n+1}^{(m−1)}).

On the other-hand, the right-hand side of equation (30) tends to b_{m−n}^{(n,2m+1)} = b̃_{n}^{(m)}/2 as x → 0. Thus we obtain (29).

Second proof of Theorem 4. For (a_{i}^{(m)})_{0≤i≤m} defined by (22), set ã_{i}^{(m)} = (2i + 1)a_{i}^{(m)}. Then (22) can be written as ã_{0}^{(0)} = 1 and

ã_{i}^{(m)} = \frac{2i + 1}{2} (iã_{i−1}^{(m−1)} − (2i + 1)^{2}ã_{i}^{(m−1)} + (i + 1)ã_{i+1}^{(m−1)})

which has exactly the same form as the recurrence relation (29) for b̃_{n}^{(m)}. Therefore one concludes ã_{n}^{(m)} = b̃_{n}^{(m)}. Comparing (24) and (28), we obtain g_{m}(x) = h_{m}(x). Thus we complete our second proof of Theorem 4.

4 Acknowledgments

The authors wish to express their sincere gratitude to the referee for valuable suggestions and comments. This work was supported by Japan Society for the Promotion of Science, Grant-in-Aid for Scientific Research (S) 16H06336 (M. Kaneko), and (C) 18K03218 (H. Tsumura).

References

[1] T. Arakawa, T. Ibukiyama, and M. Kaneko, Bernoulli Numbers and Zeta Functions, Springer, 2014.

[2] T. Arakawa and M. Kaneko, Multiple zeta values, poly-Bernoulli numbers, and related zeta functions, Nagoya Math. J. 153 (1999), 189–209.

[3] B. Bényi and P. Hajnal, Combinatorial properties of poly-Bernoulli relatives, Integers 17 (2017), No. A31.

[4] D. Cvijovic, Higher-order tangent and secant numbers, Comp. and Math. with Appl. 62 (2011), 1879–1886.

[5] C. M. da Fonseca, M. L. Glasser, and V. Kowalenko, Generalized cosecant numbers and trigonometric inverse power sums, Appl. Anal. Discrete Math. (2018) 12, 70–109.

[6] K. Imatomi, M. Kaneko, and E. Takeda, Multi-poly-Bernoulli numbers and finite multiple zeta values, J. Integer Sequences 17 (2014), Article 14.4.5.

[7] M. Kaneko, Poly-Bernoulli numbers, J. Théor. Nombres Bordeaux 9 (1997), 199–206.

[8] M. Kaneko, Poly-Bernoulli numbers and related zeta functions, Algebraic and Analytic Aspects of Zeta Functions and L-functions, MSJ Mem. 21, 73–85, Math. Soc. Japan, 2010.

[9] M. Kaneko and H. Tsumura, Multi-poly-Bernoulli numbers and related zeta functions, Nagoya Math. J. 232 (2018), 19–54.

[10] M. Kaneko and H. Tsumura, Zeta functions connecting multiple zeta values and poly-Bernoulli numbers, in H. Mishou, T. Nakamura, M. Suzuki and Y. Umegaki, eds., Various Aspects of Multiple Zeta Functions, Adv. Stud. Pure Math. 84, Math. Soc. Japan, 2020, pp. 181–204.

[11] V. Kowalenko, Applications of the cosecant and related numbers, Acta Appl. Math. 114 (2011), 15–134.

[12] V. Kowalenko, The Partition Method for a Power Series Expansion: Theory and Applications, Academic Press/Elsevier, 2017.

[13] V. Kowalenko, Generalized cosecant numbers and the Hurwitz zeta function, Integers 19 (2019), A52.

[14] N. E. Nörlund, Vorlesungen über Differenzenrechnung, Springer-Verlag, 1924.

[15] Y. Sasaki, On generalized poly-Bernoulli numbers and related L-functions, J. Number Theory 132 (2012), 156–170.

2010 Mathematics Subject Classification: Primary 11B68, Secondary 11M32, 11M99. Keywords: poly-Bernoulli number, multiple zeta value, multiple zeta function, polylogarithm.

(Concerned with sequences \underline{A001896}, \underline{A001897}, \underline{A008277}, \underline{A027641}, \underline{A027642}, and \underline{A099594}.)

Received August 1 2019; revised versions received April 12 2020; May 12 2020; May 18 2020. Published in Journal of Integer Sequences, June 9 2020.

Return to Journal of Integer Sequences home page.