Some Formulas for a Family of Numbers Analogous to the Higher-Order Bernoulli

Numbers

Guo-Dong Liu Department of Mathematics

Huizhou University Huizhou 516007, Guangdong People’s Republic of China

gdliu@pub.huizhou.gd.cn

H. M. Srivastava^{1}

Department of Mathematics and Statistics University of Victoria Victoria, BC V8W 3R4

Canada

 harimsri@math.uvic.ca

Hai-Qing Wang Department of Mathematics

Huizhou University Huizhou 516007, Guangdong People’s Republic of China

     whq@hzu.edu.cn

Abstract

In this paper the authors establish several formulas and results for the D numbers D_{2n}^{(k)} and d_{2n}^{(k)}, which are analogous to the higher-order Bernoulli numbers. Some applications of these families of D numbers are also presented.

1Corresponding author.

1 Introduction

The main purpose of this paper is to prove several formulas and results for the D numbers D_{2n}^{(k)} and d_{2n}^{(k)}, which are (in a sense) analogous to the higher-order Bernoulli numbers. We also discuss some applications of the various results which are presented here for these D numbers. With a view to making our presentation as much self-contained as possible, and for the convenience of (and ready reference by) the interested reader, we have closely followed and chosen to freely reproduce here some basic definitions and preliminary results from such recent publications as (for example) [7] (and indeed also from the book [9]). We thus begin this paper by introducing the Bernoulli polynomials B_{n}^{(k)}(x) of order k and degree n, which may be defined (for any integer k) by means of the following generating function (see, for example, [2, 4, 8, 9, 13, 14, 15]):

(\frac{t}{e^{t} − 1})^{k} e^{xt} = ∑_{n=0}^{∞} B_{n}^{(k)}(x)\frac{t^{n}}{n!}  (|t| < 2π; k ∈ ℤ),  (1)

where ℤ denotes the set of integers. In partucular, the numbers B_{n}^{(k)} = B_{n}^{(k)}(0) are the Bernoulli numbers of order k and the numbers B_{n}^{(1)} = B_{n} are referred to as the ordinary Bernoulli numbers. By using the generating function (1), we can get

\frac{d}{dx}B_{n}^{(k)}(x) = nB_{n−1}^{(k)}(x),

B_{n}^{(k+1)}(x) = \frac{k − n}{k}B_{n}^{(k)}(x) + (x − k)\frac{n}{k}B_{n−1}^{(k)}(x)

and

B_{n}^{(k+1)}(x + 1) = \frac{nx}{k}B_{n−1}^{(k)}(x) − \frac{n − k}{k} B_{n}^{(k)}(x),

where

n ∈ ℕ (ℕ := {1, 2, 3, ⋯ } = ℕ_{0} \ {0}).

Specifically, the numbers B_{n}^{(n)} are called the Nörlund numbers (see [1, 3, 4]). A generating function for the numbers B_{n}^{(n−k)} is given by (see [9])

\frac{1}{1 + t} (\frac{t}{log(1 + t)})^{k+1} = ∑_{n=0}^{∞} B_{n}^{(n−k)} \frac{t^{n}}{n!}.

For many interesting applications of the numbers B_{n}^{(n)} and B_{n}^{(n−1)}, one may refer to [9] (see also [7]).

The so-called D numbers D_{2n}^{(k)} of the first kind are defined by means of the following generating function (see [5, 6, 7, 8, 9, 10]):

(t csc t)^{k} = ∑_{n=0}^{∞}(−1)^{n}D_{2n}^{(k)} \frac{t^{2n}}{(2n)!}  (|t| < π).  (2)

Indeed, by using (1) and (2), and by observing that

csc t = \frac{2i}{e^{it} − e^{−it}} ,

we easily find that

∑_{n=0}^{∞}(−1)^{n}D_{2n}^{(k)} \frac{t^{2n}}{(2n)!} = (\frac{2it}{e^{it} − e^{−it}} )^{k} = (\frac{2it}{e^{2it} − 1})^{k} e^{kit} = ∑_{n=0}^{∞} B_{n}^{(k)} (\frac{k}{2}) \frac{(2it)^{n}}{n!} .

Therefore, we have

D_{2n}^{(k)} = 4^{n}B_{2n}^{(k)} (\frac{k}{2}) .  (3)

Upon setting k = 1 and k = 2 in this last equation (3), if we note that

B_{2n}^{(1)} (\frac{1}{2}) = (2^{1−2n} − 1)B_{2n} and B_{2n}^{(2)}(1) = (1 − 2n)B_{2n},

we have

D_{2n}^{(1)} = (2 − 2^{2n})B_{2n} and D_{2n}^{(2)} = 4^{n}(1 − 2n)B_{2n}.

On the other hand, if we put k = −1 in (2) and note that

sin t = ∑_{n=1}^{∞}(−1)^{n−1} \frac{t^{2n−1}}{(2n − 1)!}  and  cos t = ∑_{n=0}^{∞}(−1)^{n} \frac{t^{2n}}{(2n)!},

then it is easily seen that

D_{2n}^{(−1)} = \frac{1}{2n + 1} and D_{2n}^{(−2)} = \frac{4^{n}}{(n + 1)(2n + 1)}.

The D numbers D_{2n}^{(k)} satisfy the following recurrence relation (see [5]):

D_{2n}^{(k)} = \frac{(2n − k + 2)(2n − k + 1)}{(k − 2)(k − 1)}D_{2n}^{(k−2)} − \frac{2n(2n − 1)(k − 2)}{k − 1}D_{2n−2}^{(k−2)}.  (4)

In light of (4), we can immediately deduce the following known results (see [9]):

D_{2n}^{(2n+1)} = \frac{(−1)^{n}(2n)!}{4^{n}} \binom{2n}{n} and D_{2n}^{(2n+2)} = \frac{(−1)^{n}4^{n}}{2n + 1} (n!)^{2}

and

D_{2n}^{(2n+3)} = \frac{(−1)^{n}(2n)!}{2 · 4^{2n}} \binom{2n + 2}{n + 1} (1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + ⋯ + \frac{1}{(2n + 1)^{2}} ) .

Recently, Liu [7] derived the following exponential generating function for D_{2n}^{(2n−k)}:

∑_{n=0}^{∞} D_{2n}^{(2n−k)} \frac{t^{2n}}{(2n)!} = \frac{1}{\sqrt{1 + t^{2}}} (\frac{t}{log(t + \sqrt{1 + t^{2}})})^{k+1} .  (5)

Some of the important applications of the numbers D_{2n}^{(2n)} and D_{2n}^{(2n−1)} include (for example) each of the following known results:

∫0^{\frac{π}{2}} \frac{sin t}{t} dt = ∑_{n=0}^{∞} \frac{(−1)^{n}D_{2n}^{(2n)}}{(2n + 1)!} ,  (6)

∫0^{\frac{π}{2}} \frac{sin t}{t} dt = \frac{π}{2} ∑_{n=0}^{∞} \frac{(−1)^{n+1}D_{2n}^{(2n−1)}}{2^{2n}(2n − 1)(n!)^{2}}  (7)

and

\frac{2}{π} = ∑_{n=0}^{∞} \frac{(−1)^{n+1} D_{2n}^{(2n−1)}}{(2n − 1)(2n)!} .  (8)

The D numbers d_{2n}^{(k)} of the second kind may be defined by means of the following generating function:

(\frac{t}{log(t + \sqrt{1 + t^{2}})})^{k} = ∑_{n=0}^{∞} d_{2n}^{(k)}t^{2n}.  (9)

The numbers d_{n} = d_{n}^{(1)} are referred to as the ordinary D numbers of the second kind. Some consequences of the generating function (9) are given below:

d_{0} = 1, d_{2} = \frac{1}{6}, d_{4} = −\frac{17}{360}, d_{6} = \frac{367}{15120}, d_{8} = −\frac{195013}{27216000} and d_{10} = \frac{1295803}{252806400}.

Indeed, by using the generating function (9), we also find that

∑_{n=1}^{∞} 2nd_{2n}^{(k)}t^{2n−1} = k (\frac{t}{log(t + \sqrt{1 + t^{2}})})^{k−1} \frac{log(t + \sqrt{1 + t^{2}}) − \frac{t}{\sqrt{1+t^{2}}}}{[log(t + \sqrt{1 + t^{2}})]^{2}} ,

that is, that

∑_{n=1}^{∞} 2nd_{2n}^{(k)}t^{2n} = k ∑_{n=0}^{∞} d_{2n}^{(k)}t^{2n} − k ∑_{n=0}^{∞} D_{2n}^{(2n−k)} \frac{t^{2n}}{(2n)!}.  (10)

Consequently, we have (see [7, Theorem 4])

(2n)!d_{2n}^{(k)} = \frac{k}{k − 2n}D_{2n}^{(2n−k)}.  (11)

By applying (11), we obtain

(2n)!d_{2n}^{(2n+1)} = 1 and (2n)!d_{2n}^{(2n+2)} = \frac{4^{n}}{2n + 1},

and (2n)!d_{2n}^{(2n−1)} = (2n − 1)(2^{2n} − 2)B_{2n} and (2n)!d_{2n}^{(2n−2)} = 4^{n}(n − 1)(2n − 1)B_{2n}. We turn now to the central factorial numbers 𝔱(n, k) of the first kind, which are usually

defined by (see [11])

x (x + \frac{n}{2} − 1) (x + \frac{n}{2} − 2) ⋯ (x + \frac{n}{2} − n + 1) = ∑_{k=0}^{n} 𝔱(n, k)x^{k}  (12)

or, equivalently, by means of the following generating function:

[2 log (\frac{x}{2} + \sqrt{1 + \frac{x^{2}}{4}} )]^{k} = k! ∑_{n=k}^{∞} 𝔱(n, k)\frac{x^{n}}{n!} .  (13)

By appealing to (12) or (13), we can show that

𝔱(n, k) = 𝔱(n − 2, k − 2) − \frac{1}{4}(n − 2)^{2} 𝔱(n − 2, k)  (14)

and that

𝔱(n, 0) = δ_{n,0} (n ∈ ℕ_{0} := ℕ ∪ {0}) and 𝔱(n, n) = 1 (n ∈ ℕ)

and

𝔱(n, k) = 0 (n + k odd) and 𝔱(n, k) = 0 (k > n or k < 0),

where δ_{m,n} denotes the Kronecker symbol.

Next, by making use of (12), we obtain

(x^{2} − 1^{2})(x^{2} − 3^{2}) ⋯ [x^{2} − (2n − 1)^{2}] = ∑_{k=0}^{n} 4^{n−k}𝔱(2n + 1, 2k + 1)x^{2k}  (15)

and

x^{2}(x^{2} − 1^{2})(x^{2} − 2^{2}) ⋯ [x^{2} − (n − 1)^{2}] = ∑_{k=0}^{n} 𝔱(2n, 2k)x^{2k}.  (16)

By applying (15) and (16), we find for n ∈ ℕ_{0} that

𝔱(2n + 1, 1) = (−\frac{1}{4})^{n} · 1^{2} · 3^{2} ⋯ (2n − 1)^{2} and 𝔱(2n + 2, 2) = (−1)^{n}(n!)^{2}

and

𝔱(2n + 2, 4) = (−1)^{n+1}(n!)^{2} (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + ⋯ + \frac{1}{n^{2}} ) (n ∈ ℕ).

By using (4) and (14), we have

D_{2n−2k}^{(2n+1)} = \frac{4^{n−k}}{\binom{2n}{2k}} 𝔱(2n + 1, 2k + 1) (n ≧ k ≧ 0)  (17)

and

D_{2n−2k}^{(2n)} = \frac{4^{n−k}}{\binom{2n − 1}{2k − 1}} 𝔱(2n, 2k) (n ≧ k ≧ 1).  (18)

In Sections 2 and 3 of this paper, we shall state and prove several formulas and results for the D numbers D_{2n}^{(k)} and d_{2n}^{(k)}, respectively. Then, in Section 4, we discuss some applications of these families of D numbers.

2 Formulas and Results Involving the Numbers D_{2n}^{(k)}

Theorem 1. Let n ∈ ℕ. Then

D_{2n}^{(2n)} = ∫_{0}^{1}(x^{2} − 1^{2})(x^{2} − 3^{2}) ⋯ [x^{2} − (2n − 1)^{2}]dx.  (19)

Proof. By applying (15) and (17) and noting that (see [5])

D_{2k}^{(−1)} = \frac{1}{2k + 1} and D_{2n}^{(k)} = ∑_{j=0}^{n} \binom{2n}{2j}D_{2n−2j}^{(k−l)}D_{2j}^{(l)},

we get

D_{2n}^{(2n)} = ∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n+1)}D_{2k}^{(−1)} = ∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n+1)}D_{2k}^{(−1)} ∫_{0}^{1}(2k + 1)x^{2k} dx

= ∫_{0}^{1} ∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n+1)}x^{2k} dx = ∫_{0}^{1} ∑_{k=0}^{n} 4^{n−k}𝔱(2n + 1, 2k + 1)x^{2k} dx

= ∫_{0}^{1}(x^{2} − 1^{2})(x^{2} − 3^{2}) ⋯ [x^{2} − (2n − 1)^{2}]dx,

which completes the proof of the assertion (19) of Theorem 1.

Theorem 2. Let n ∈ ℕ. Then

D_{2n}^{(2n−1)} = (1 − 2n) ∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n)}.  (20)

Proof. By using (5), (18) and (13), we have

∑_{n=0}^{∞} ∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n)} \frac{t^{2n}}{(2n)!} = ∑_{n=0}^{∞} D_{2n}^{(2n)} \frac{t^{2n}}{(2n)!} + ∑_{n=1}^{∞} ∑_{k=1}^{n} \frac{n4^{n−k}}{k} 𝔱(2n, 2k)\frac{t^{2n}}{(2n)!}

= \frac{t}{\sqrt{1 + t^{2}} log (t + \sqrt{1 + t^{2}})} + ∑_{k=1}^{∞} \frac{t}{2k} ∑_{n=k}^{∞} \frac{d}{dt} \{4^{n−k} 𝔱(2n, 2k) \frac{t^{2n}}{(2n)!}\}

= \frac{t}{\sqrt{1 + t^{2}} log (t + \sqrt{1 + t^{2}})} + ∑_{k=1}^{∞} \frac{t}{(2k) · (2k)!} \frac{d}{dt} \{[log (t + \sqrt{1 + t^{2}})]^{2k}\}

= \frac{t}{\sqrt{1 + t^{2}} log (t + \sqrt{1 + t^{2}})} ∑_{k=0}^{∞} \frac{[log (t + \sqrt{1 + t^{2}})]^{2k}}{(2k)!}

= \frac{t}{log (t + \sqrt{1 + t^{2}})} = ∑_{n=0}^{∞} d_{2n}t^{2n},

which, in view of (11), yields

∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n)} = (2n)!d_{2n} = \frac{1}{1 − 2n}D_{2n}^{(2n−1)}.

This completes the proof of the assertion (20) of Theorem 2. Theorem 3. Let n ∈ ℕ. Then

D_{2n}^{(2n−1)} = (1 − 2n) ∫_{0}^{1} x^{2}(x^{2} − 2^{2}) ⋯ [x^{2} − (2n − 2)^{2}]dx.  (21)

Proof. By using (16), (18) and (20), we have

∫_{0}^{1} x^{2}(x^{2} − 2^{2}) ⋯ [x^{2} − (2n − 2)^{2}]dx

= 2 ∫0^{\frac{1}{2}} 4x^{2}(4x^{2} − 2^{2}) ⋯ [4x^{2} − (2n − 2)^{2}]dx

= 2^{2n+1} ∫0^{\frac{1}{2}} ∑_{k=0}^{n} 𝔱(2n, 2k)x^{2k} dx = ∑_{k=0}^{n} \frac{4^{n−k}}{2k + 1}𝔱(2n, 2k)

= ∑_{k=0}^{n} \frac{\binom{2n−1}{2k−1}}{2k + 1}D_{2n−2k}^{(2n)} = \frac{1}{n} ∑_{k=0}^{n} \frac{k}{2k + 1}\binom{2n}{2k}D_{2n−2k}^{(2n)}

= \frac{1}{2n} ∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n)} − \frac{1}{2n} ∑_{k=0}^{n} \frac{1}{2k + 1}\binom{2n}{2k}D_{2n−2k}^{(2n)}

= \frac{1}{2n(1 − 2n)}D_{2n}^{(2n−1)} − \frac{1}{2n} ∑_{k=0}^{n} \binom{2n}{2k}D_{2n−2k}^{(2n)}D_{2k}^{(−1)}

= \frac{1}{2n(1 − 2n)}D_{2n}^{(2n−1)} − \frac{1}{2n}D_{2n}^{(2n−1)} = \frac{1}{1 − 2n}D_{2n}^{(2n−1)},

which completes the proof of the assertion (21) of Theorem 3.

3 Formulas and Results Involving the Numbers d_{2n}^{(k)}

Theorem 4. Let n, k ∈ ℕ. Then

k(k + 1)d_{2n}^{(k+2)} = (2n − k)(2n − k − 1)d_{2n}^{(k)} + (2n − k − 2)^{2}d_{2n−2}^{(k)}.  (22)

Proof. By making use of (9), we find that

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

= \frac{k(k + 1)}{1 + t^{2}} (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+2} + (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+1} \frac{kt}{(1 + t^{2})^{3/2}} . (23)

On the other hand, we also have

∑_{n=1}^{∞}(2n − k − 2)d_{2n−2}^{(k)}t^{2n−k−3} = \frac{d}{dt} (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k}

= −\frac{k}{\sqrt{1 + t^{2}}} (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+1} .

Therefore, we get

∑_{n=1}^{∞}(2n − k − 2)^{2}d_{2n−2}^{(k)}t^{2n−k−3} = −\frac{d}{dt} \frac{kt}{\sqrt{1 + t^{2}}} (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+1}

= \frac{k(k + 1)t}{1 + t^{2}} (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+2} − (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+1} \frac{k}{(1 + t^{2})^{3/2}} ,

that is,

∑_{n=1}^{∞}(2n − k − 2)^{2}d_{2n−2}^{(k)}t^{2n−k−2} = \frac{k(k + 1)t^{2}}{1 + t^{2}} (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+2}

− (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+1} \frac{kt}{(1 + t^{2})^{3/2}} .  (24)

Now, by using (23) and (24), we obtain

∑_{n=0}^{∞}(2n − k)(2n − k − 1)d_{2n}^{(k)}t^{2n−k−2} + ∑_{n=1}^{∞}(2n − k − 2)^{2}d_{2n−2}^{(k)}t^{2n−k−2}

= k(k + 1) (\frac{1}{log (t + \sqrt{1 + t^{2}})})^{k+2} = k(k + 1) ∑_{n=0}^{∞} d_{2n}^{(k+2)}t^{2n−k−2}.  (25)

Finally, by comparing the coefficients of t^{2n−k−2} on both sides of (25), we are led easily to (22). This completes the proof of Theorem 4.

Remark 5. Upon setting k = 2n − 2 in Theorem 4, if we make use of (11), we immediately obtain the following result:

(2n)!d_{2n}^{(2n)} = 4^{n}B_{2n} (n ∈ ℕ_{0}).  (26)

A generalization of the above result (26) and other analogous results can be found in the recent work by Liu [7, Corollary 1 and Theorem 4].

Theorem 6. Let n, k ∈ ℕ and n ≧ k + 1. Then

(2k)!(2n − 2k − 1)!d_{2n}^{(2k+1)} = ∑_{j=0}^{k}(2n − 1 − 2j)!σ(n, k, j)d_{2n−2j},  (27)

where

σ(n, k, j) = ∑_{v_{1},··· ,v_{k−j+1}∈ℕ_{0}} ; (v_{1}+···+v_{k−j+1}=j)(2n − 2j − 1)^{2v_{1}}(2n − 2j − 3)^{2v_{2}} ⋯ (2n − 2k − 1)^{2v_{k−j+1}}.

Proof. We prove the assertion (27) of Theorem 6 by using the principle of mathematical induction. Indeed, when k = 1, (27) is true by virtue of (22). Suppose now that (27) is true for some natural number k ∈ ℕ \ {1}. Then, by the superposition of (22), we have

(2k + 1)(2k + 2)d_{2n}^{(2k+3)}

= (2n − 2k − 1)(2n − 2k − 2)d_{2n}^{(2k+1)} + (2n − 2k − 3)^{2}d_{2n−2}^{(2k+1)}

= \frac{(2n − 2k − 1)(2n − 2k − 2)}{(2k)!(2n − 2k − 1)!} ∑_{j=0}^{k}(2n − 1 − 2j)!σ(n, k, j)d_{2n−2j}

+ \frac{(2n − 2k − 3)^{2}}{(2k)!(2n − 2k − 3)!} ∑_{j=0}^{k}(2n − 3 − 2j)!σ(n − 1, k, j)d_{2n−2−2j}

= \frac{1}{(2k)!(2n − 2k − 3)!} ∑_{j=0}^{k}(2n − 1 − 2j)!σ(n, k, j)d_{2n−2j}

+ \frac{(2n − 2k − 3)^{2}}{(2k)!(2n − 2k − 3)!} ∑_{j=1}^{k+1}(2n − 2 − 2j)!σ(n − 1, k, j − 1)d_{2n−2j}.  (28)

In light of this last result (28), and by noting that

σ(n, k + 1, 0) = σ(n, k, 0),
σ(n, k + 1, k + 1) = (2n − 2k − 3)^{2}σ(n − 1, k, k)

and

σ(n, k + 1, j) = σ(n, k, j) + (2n − 2k − 3)^{2}σ(n − 1, k, j − 1),

we find that

(2k + 2)!(2n − 2k − 3)!d_{2n}^{(2k+3)}

= ∑_{j=0}^{k}(2n − 1 − 2j)!σ(n, k, j)d_{2n−2j}

+ (2n − 2k − 3)^{2} ∑_{j=1}^{k+1}(2n − 1 − 2j)!σ(n − 1, k, j − 1)d_{2n−2j}

= (2n − 1)!σ(n, k, 0)d_{2n} + (2n − 2k − 3)^{2}(2n − 3 − 2k)!σ(n − 1, k, k)d_{2n−2k−2}

+ ∑_{j=1}^{k}(2n − 1 − 2j)! [σ(n, k, j) + (2n − 2k − 3)^{2}σ(n − 1, k, j − 1)] d_{2n−2j}

= (2n − 1)!σ(n, k + 1, 0)d_{2n} + (2n − 3 − 2k)!σ(n, k + 1, k + 1)d_{2n−2k−2}

+ ∑_{j=1}^{k}(2n − 1 − 2j)!σ(n, k + 1, j)d_{2n−2j}

= ∑_{j=0}^{k+1}(2n − 1 − 2j)!σ(n, k + 1, j)d_{2n−2j},

which shows that (27) is also true for the natural number k + 1. Thus, by the principle of mathematical induction, (27) holds true for all k ∈ ℕ. This completes the proof of Theorem 6.

Remark 7. Upon setting k = 1, 2, 3 in (27), we can immediately deduce

2!d_{2n}^{(3)} = (2n − 1)(2n − 2)d_{2n} + (2n − 3)^{2}d_{2n−2},

4!d_{2n}^{(5)} = (2n − 1)(2n − 2)(2n − 3)(2n − 4)d_{2n}
+ (2n − 3)(2n − 4)[(2n − 3)^{2} + (2n − 5)^{2}]d_{2n−2} + (2n − 5)^{4}d_{2n−4}

and

6!d_{2n}^{(7)} = (2n − 1)(2n − 2)(2n − 3)(2n − 4)(2n − 5)(2n − 6)d_{2n}
+ (2n − 3)(2n − 4)(2n − 5)(2n − 6)[(2n − 3)^{2} + (2n − 5)^{2}
+ (2n − 7)^{2}]d_{2n−2} + (2n − 5)(2n − 6)[(2n − 5)^{4}
+ (2n − 5)^{2}(2n − 7)^{2} + (2n − 7)^{4}]d_{2n−4} + (2n − 7)^{6}d_{2n−6}.

Theorem 8. Let n, k ∈ ℕ and n ≧ k. Then

(2k − 1)!(2n − 2k)!d_{2n}^{(2k)} = ∑_{j=0}^{k−1}(2n − 2 − 2j)!τ(n, k, j)d_{2n−2j}^{(2)},  (29)

where

τ(n, k, j) = ∑_{v_{1},··· ,v_{k−j}∈ℕ_{0}} ; (v_{1}+···+v_{k−j}=j)(2n − 2j − 2)^{2v_{1}}(2n − 2j − 4)^{2v_{2}} ⋯ (2n − 2k)^{2v_{k−j}}.

Proof. We prove the assertion (29) of Theorem 8 by using the principle of mathematical induction. In fact, when k = 1 and k = 2, (29) is true by (22). Suppose now that (29) is true for some natural number k ∈ ℕ \ {1}. Then, by the superposition of (22), we have

2k(2k + 1)d_{2n}^{(2k+2)}

= (2n − 2k)(2n − 2k − 1)d_{2n}^{(2k)} + (2n − 2k − 2)^{2}d_{2n−2}^{(2k)}

= \frac{(2n − 2k)(2n − 2k − 1)}{(2k − 1)!(2n − 2k)!} ∑_{j=0}^{k−1}(2n − 2 − 2j)!τ(n, k, j)d_{2n−2j}^{(2)}

+ \frac{(2n − 2k − 2)^{2}}{(2k − 1)!(2n − 2k − 2)!} ∑_{j=0}^{k−1}(2n − 4 − 2j)!τ(n − 1, k, j)d_{2n−2−2j}^{(2)}

= \frac{1}{(2k − 1)!(2n − 2k − 2)!} ∑_{j=0}^{k−1}(2n − 2 − 2j)!τ(n, k, j)d_{2n−2j}^{(2)}

+ \frac{(2n − 2k − 2)^{2}}{(2k − 1)!(2n − 2k − 2)!} ∑_{j=1}^{k}(2n − 2 − 2j)!τ(n − 1, k, j − 1)d_{2n−2j}^{(2)}.  (30)

By using (30), and noting that

τ(n, k + 1, 0) = τ(n, k, 0),
τ(n, k + 1, k) = (2n − 2k − 2)^{2}τ(n − 1, k, k − 1)

and

τ(n, k + 1, j) = τ(n, k, j) + (2n − 2k − 2)^{2}τ(n − 1, k, j − 1),

(2k + 1)!(2n − 2k − 2)!d_{2n}^{(2k+2)}

= ∑_{j=0}^{k−1}(2n − 2 − 2j)!τ(n, k, j)d_{2n−2j}^{(2)}

+ (2n − 2k − 2)^{2} ∑_{j=1}^{k}(2n − 2 − 2j)!τ(n − 1, k, j − 1)d_{2n−2j}^{(2)}

= (2n − 2)!τ(n, k, 0)d_{2n}^{(2)} + (2n − 2k − 2)^{2}(2n − 2 − 2k)!τ(n − 1, k, k − 1)d_{2n−2k}^{(2)}

+ ∑_{j=1}^{k−1}(2n − 2 − 2j)! [τ(n, k, j) + (2n − 2k − 2)^{2}τ(n − 1, k, j − 1)] d_{2n−2j}^{(2)}

= (2n − 2)!τ(n, k + 1, 0)d_{2n}^{(2)} + (2n − 2 − 2k)!τ(n, k + 1, k)d_{2n−2k}^{(2)}

+ ∑_{j=1}^{k−1}(2n − 2 − 2j)! [τ(n, k, j) + (2n − 2k − 2)^{2}τ(n − 1, k, j − 1)] d_{2n−2j}^{(2)}

= ∑_{j=0}^{k}(2n − 2 − 2j)!τ(n, k + 1, j)d_{2n−2j}^{(2)},

which shows that (29) holds true also for the natural number k +1. This evidently completes the proof of Theorem 8 by the principle of mathematical induction on k ∈ ℕ.

Remark 9. By setting k = 2, 3, 4 in (29), we can immediately deduce

3!d_{2n}^{(4)} = (2n − 2)(2n − 3)d_{2n}^{(2)} + (2n − 4)^{2}d_{2n−2}^{(2)},
5!d_{2n}^{(6)} = (2n − 2)(2n − 3)(2n − 4)(2n − 5)d_{2n}^{(2)}
+ (2n − 4)(2n − 5)[(2n − 4)^{2} + (2n − 6)^{2}]d_{2n−2}^{(2)} + (2n − 6)^{4}d_{2n−4}^{(2)}

and

7!d_{2n}^{(8)} = (2n − 2)(2n − 3)(2n − 4)(2n − 5)(2n − 6)(2n − 7)d_{2n}^{(2)}

+ (2n − 4)(2n − 5)(2n − 6)(2n − 7)[(2n − 4)^{2} + (2n − 6)^{2} + (2n − 8)^{2}]d_{2n−2}^{(2)}

+ (2n − 6)(2n − 7)[(2n − 6)^{4} + (2n − 6)^{2}(2n − 8)^{2} + (2n − 8)^{4}]d_{2n−4}^{(2)}
+ (2n − 8)^{6}d_{2n−6}^{(2)}.

4 A Set of Applications

We first give the following application of the results presented in the preceding sections.

Theorem 10. Let k ∈ ℤ. Then

∫0^{\frac{π}{2}} (\frac{sin t}{t})^{k+1} dt = ∑_{n=0}^{∞} \frac{(−1)^{n}D_{2n}^{(2n−k)}}{(2n + 1)!} .  (31)

Proof. By using (5), we find that

∑_{n=0}^{∞}(−1)^{n}D_{2n}^{(2n−k)} \frac{x^{2n}}{(2n)!} = \frac{1}{\sqrt{1 − x^{2}}} (\frac{ix}{log(ix + \sqrt{1 − x^{2}})})^{k+1} .

Therefore, we have

∑_{n=0}^{∞} \frac{(−1)^{n}D_{2n}^{(2n−k)}}{(2n + 1)!} = ∫_{0}^{1} \frac{1}{\sqrt{1 − x^{2}}} (\frac{ix}{log(ix + \sqrt{1 − x^{2}})})^{k+1} dx

= ∫0^{\frac{π}{2}} \frac{1}{cos θ} (\frac{i sin θ}{log(i sin θ + cos θ)})^{k+1} d(sin θ) = ∫0^{\frac{π}{2}} (\frac{sin θ}{θ})^{k+1} dθ,

which obviously completes the proof of Theorem 10.

Remark 11. Setting k = 0 in (31), we immediately obtain (6). Moreover, if we set k = −1 and k = −2 in (31) and note that

D_{2n}^{(2n+1)} = \frac{(−1)^{n}(2n)!}{4^{n}} \binom{2n}{n} and D_{2n}^{(2n+2)} = \frac{(−1)^{n}4^{n}}{2n + 1} (n!)^{2},

we get

∑_{n=0}^{∞} \frac{1}{(2n + 1)4^{n}} \binom{2n}{n} = \frac{π}{2}

and

∫0^{\frac{π}{2}} \frac{t}{sin t} dt = ∑_{n=0}^{∞} \frac{4^{n}}{(2n + 1)^{2}\binom{2n}{n}},

The sum on the right-hand side of this last result can be expressed as a Clausenian hypergeometric series as follows:

3F_{2} (\frac{1}{2}, 1, 1; \frac{3}{2}, \frac{3}{2}; 1) .

Moreover, by using a computer algebra system, the integral on the left-hand side can be found to be twice the Catalan constant G which is defined by (see, for details, [14, p. 43 et seq.])

G := \frac{1}{2} ∫_{0}^{1} K(κ)dκ = ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n + 1)^{2}} ≐ 0.91596 55941 77219 015 ⋯,

where K(κ) is the complete elliptic integral of the first kind, given by

K(κ) := ∫0^{\frac{π}{2}} \frac{dt}{\sqrt{1 − κ^{2} sin^{2} t}} (|κ| < 1).

In fact, the above observation can be verified analytically by setting τ = tan (\frac{t}{2}) and dτ = \frac{1}{2} sec^{2} (\frac{t}{2}) dt in the following known result [13, p. 110, Equation 2.3(37)]:

∫_{0}^{1} \frac{arctan τ}{τ} dτ = G.

Theorem 12. Let k ∈ ℤ. Then

∫0^{\frac{π}{2}} (\frac{sin t}{t})^{k} cos t dt = ∑_{n=0}^{∞} \frac{(−1)^{n+1}kD_{2n}^{(2n−k)}}{(2n − k)(2n + 1)!}.  (32)

Proof. By using (9) and (11), we find that

∑_{n=0}^{∞} \frac{(−1)^{n+1}kD_{2n}^{(2n−k)}}{(2n − k)(2n)!} x^{2n} = ∑_{n=0}^{∞}(−1)^{n}d_{2n}^{(k)}x^{2n} = (\frac{ix}{log(ix + \sqrt{1 − x^{2}})})^{k} .

Therefore, we have

∑_{n=0}^{∞} \frac{(−1)^{n+1}kD_{2n}^{(2n−k)}}{(2n − k)(2n + 1)!} = ∫_{0}^{1} (\frac{ix}{log(ix + \sqrt{1 − x^{2}})})^{k} dx

= ∫0^{\frac{π}{2}} (\frac{i sin t}{log(i sin t + cos t)})^{k} d(sin t) = ∫0^{\frac{π}{2}} (\frac{sin t}{t})^{k} cos t dt,

which evidently completes the proof of Theorem 12.

Remark 13. By setting k = 1 and k = −1 in (32), we have

∫_{0}^{π} \frac{sin t}{t} dt = 2 ∑_{n=0}^{∞} \frac{(−1)^{n+1}D_{2n}^{(2n−1)}}{(2n − 1)(2n + 1)!}

and

∫0^{\frac{π}{2}} t cot t dt = ∑_{n=0}^{∞} \frac{1}{4^{n} (2n + 1)^{2}} \binom{2n}{n}

or, equivalently,

∫_{0}^{1} \frac{arcsin τ}{τ} dτ = ∑_{n=0}^{∞} \frac{1}{4^{n} (2n + 1)^{2}} \binom{2n}{n}

by setting sin t = τ and cos t dt = dτ. The sum on the right-hand side of this last result can be expressed in a closed form via a Clausenian hypergeometric series as follows:

∫_{0}^{1} \frac{arcsin τ}{τ} dτ = ∑_{n=0}^{∞} \frac{1}{4^{n} (2n + 1)^{2}} \binom{2n}{n} = _{3}F_{2} (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; \frac{3}{2}, \frac{3}{2}; 1) = \frac{π}{2} ln 2.

∑_{n=0}^{∞} \frac{(−1)^{n+1}k}{(2n − k)(2n)!}D_{2n}^{(2n−k)} = (\frac{2}{π} )^{k} ,  (33)

provided that each member of (33) exists.

Proof. By applying (9) and (11), we obtain

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

= (\frac{ix}{log(ix + \sqrt{1 − x^{2}})})^{k} = (\frac{1}{∑_{n=0}^{∞} \frac{1}{(2n+1)4^{n}} \binom{2n}{n}x^{2n}} )^{k} .

Therefore, we get

∑_{n=0}^{∞} \frac{(−1)^{n+1}k}{(2n − k)(2n)!}D_{2n}^{(2n−k)} = (\frac{1}{∑_{n=0}^{∞} \frac{1}{(2n+1)4^{n}} \binom{2n}{n}} )^{k} = (\frac{2}{π} )^{k} .

This completes the proof of Theorem 14.

Remark 15. Upon setting k = 1 in (33), we immediately obtain (8). If, on the other hand, we put k = −2 in (33) and note that

D_{2n}^{(2n+2)} = \frac{(−1)^{n}4^{n}}{2n + 1} (n!)^{2},

then we find that

∑_{n=0}^{∞} \frac{4^{n}(n!)^{2}}{(2n + 2)!} = \frac{π^{2}}{8} .

This last identity follows also from a known power-series expansion for the function (arcsin x)^{2}.

5 Acknowledgements

This work was supported by the Guangdong Provincial Natural Science Foundation of the People’s Republic of China (Grant No. 8151601501000002).

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2010 Mathematics Subject Classification: Primary 11B68, 11B83; Secondary 05A19. Keywords: Bernoulli polynomial, Bernoulli number, D number, D number of the second kind, Nörlund number; central factorial number of the first kind, Catalan’s constant, Clausenian hypergeometric series.

(Concerned with sequences \underline{A000367}, \underline{A002445}, \underline{A002657}, \underline{A002790}, \underline{A027641}, and \underline{A027642}.)

Received October 23 2013; revised versions received February 15 2014; February 21 2014. Published in Journal of Integer Sequences, February 22 2014.

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