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Several properties of hypergeometric Bernoulli numbers
Journal of Inequalities and Applications volume 2019, Article number: 113 (2019)
Abstract
In this paper, we give several characteristics of hypergeometric Bernoulli numbers, including several identities for hypergeometric Bernoulli numbers which the convergents of the continued fraction expansion of the generating function of the hypergeometric Bernoulli numbers entail. We show an analog of Kummer’s congruences in the classical Bernoulli numbers. We also give some determinant expressions of hypergeometric Bernoulli numbers and some relations between the hypergeometric and the classical Bernoulli numbers. By applying Trudi’s formula, we have some different expressions and inversion relations.
1 Introduction
Denote \({}_{1} F_{1}(a;b;z)\) be the confluent hypergeometric function defined by
with the rising factorial \((x)^{(n)}=x(x+1)\cdots (x+n-1)\) (\(n\ge 1\)) and \((x)^{(0)}=1\). For \(N\ge 1\), define hypergeometric Bernoulli numbers \(B_{N,n}\) [13,14,15,16, 18] by
When \(N=1\), \(B_{1,n}=B_{n}\) are the classical Bernoulli numbers, defined by
In addition, define hypergeometric Bernoulli polynomials \(B_{N,n}(z)\) [17] by the generating function
It is well known [25] that
Many kinds of generalizations of the Bernoulli numbers have been considered by many authors. For example, poly-Bernoulli numbers [20], multiple Bernoulli numbers [5], Apostol–Bernoulli numbers [23], multi-poly-Bernoulli numbers [10, 11], degenerate Bernoulli numbers [6], various types of q-Bernoulli numbers [4], Bernoulli–Carlitz numbers [3]. One of the advantages of hypergeometric numbers is the natural extension of determinant expressions of the numbers.
In [22], some determinant expressions of hypergeometric Cauchy numbers are considered. In this paper, first we show a similar determinant expression of hypergeometric Bernoulli numbers and their generalizations. Then we study some relations between the hypergeometric Bernoulli numbers and the classical Bernoulli numbers which include Kummer’s congruences. Furthermore, by applying Trudi’s formula, we also have some different expressions and inversion relations. Finally, we determine explicit forms of the convergents of the continued fraction expansion of the generating function of the hypergeometric Bernoulli numbers, from which several identities for hypergeometric Bernoulli numbers are given.
2 Some basic properties of hypergeometric Bernoulli numbers
In this section, we list some basic properties of hypergeometric Bernoulli numbers.
Proposition 1
Let \(N,n\ge 1\). We have
Remark
When \(N=1\) in (3), we have a famous identity for Bernoulli numbers,
In some literature (e.g., [20]), the Bernoulli numbers \(\mathfrak{B}_{n}\) are also defined by
Then we have
(see, e.g., [1]). Notice that \(B_{n}=(-1)^{n}\mathfrak{B} _{n}\) (\(n\ge 0\)).
By the identity (3), we get
with \(B_{N,0}=1\) (\(N\ge 1\)). So, we have the exact forms of \(B_{N,n}\) for small n one by one. For larger or general n, we need the identity (4).
In a later section about Trudi’s formula, we shall see a different expression of \(B_{N,n}\) in Corollary 2. Further, an inversion expression can be obtained:
where \(\binom{n}{t_{1}, \dots,t_{k}}= \frac{n!}{t_{1} !\cdots t_{k} !}\) are the multinomial coefficients.
Proof of Proposition 1.
From the definition in (1), we have
Hence, for \(n\ge 1\), we have (3).
The proof of (4) can be done by induction on n. Here, we shall prove directly by using the generating function. From the definition in (1), we have
The identity (4) immediately follows by comparing coefficients of both sides.
The identity (5) is a different expression of \(B_{N,n}\) with binomial coefficients. The proof is similar to that of (4) and is omitted. □
3 Analog of Kummer’s congruence
In this section, we show an analog of Kummer’s congruences in the classical Bernoulli numbers. Let p be a prime number, and \(\nu \geq 0\) be an integer. For a non-zero integer m, we define the p-adic order \({\mathrm{ord}}_{p}(m)\) \((\in \mathbb{Z})\) of m by \(m=up^{{\mathrm{ord}}_{p}(m)}\) where u is an integer satisfying \(p \nmid u\). If m and n are positive even integers with \(m\equiv n\ {(\operatorname{mod}(p-1)p^{\nu })}\) and \(m,n \not \equiv 0\ {(\operatorname{mod} p-1)}\), then we have
and this is called Kummer’s congruence [30, Corollary 5.14]. We get the similar congruence for hypergeometric Bernoulli numbers \(B_{N,n}\) for a special case that N is p-adically close enough to 1, that is, \({\mathrm{ord}}_{p}(N-1)\) is large enough compared to ν and the order of \(\prod_{k=0}^{ {\mathrm{max}} \{ m,n \}} (1+k)!\).
We need the following lemma in order to prove the main result.
Lemma 1
Let p be a prime number. For \(N\geq 1 \) and \(n\geq 0\), we have
where \(t={\mathrm{ord}}_{p}(N-1)\).
Proof
In the case \(n=0\), the assertion is trivial. Assume that the result is true up to \(n-1\). By Proposition 1, we have
□
From this lemma, we have the following corollary.
Corollary 1
Let p be a prime number, and \(N,n\geq 1, \nu \geq 0\) be integers with \(n\not \equiv 0 {(\operatorname{mod} p-1)}\). If \({\mathrm{ord}}_{p}(N-1)\geq \nu +1+ {\mathrm{ord}}_{p} ( \prod_{k=0}^{n} (1+k)! ) +{\mathrm{ord}} _{p}(n)\), then we have
Furthermore, by using (6), we have the following proposition.
Proposition 2
Let p be a prime number, and \(\nu \geq 0\) be an integer. If m and n are positive even integers with \(m\geq n\), \(m\equiv n\ {(\operatorname{mod} (p-1)p^{\nu })}\) and \(m,n\not \equiv 0\ {(\operatorname{mod} (p-1))}\), and \({\mathrm{ord}}_{p}(N-1)\geq \nu +1 +{\mathrm{ord}}_{p} ( \prod_{k=0}^{m} (1+k)! )+ {\mathrm{max}} \{ {\mathrm{ord}} _{p}(m), {\mathrm{ord}}_{p}(n) \}\), then we have
Note that \({\mathrm{ord}}_{p} ( \prod_{k=0}^{m} (1+k)!) +{\mathrm{max}} \{ {\mathrm{ord}}_{p} (m), {\mathrm{ord}}_{p}(m) \}=0\) if \(p>m+1\) in Proposition 2.
Example 1
Consider the case \(p=5, m=6, n=2\). For any integer N satisfying
we have
Example 2
Consider the case \(p=5, m=22, n=2\). For any integer N satisfying
we have
5 A relation between \(B_{N,n}\) and \(B_{N-1,n}\)
In this section, we show the following relation between \(B_{N,n}\) and \(B_{N-1,n}\).
Proposition 3
For \(N\geq 2\) and \(n\geq 1\), we have
Example 3
-
(i)
\({ B_{N,1}=\frac{N}{N+1} B_{N-1,1}}\),
-
(ii)
\({ B_{N,2}=\frac{N}{N+2} \{ B_{N-1,2}+\frac{N}{N+1} B_{N-1,1} B _{N-1,2} \} }\),
-
(iii)
$$\begin{aligned} B_{N,3} ={} & \frac{N}{N+3} \biggl\{ B_{N-1,3}+ \frac{N}{N+1} B_{N-1,1}B _{N-1,3}+\frac{3N}{N+2} B_{N-1,2}^{2} \\ & {} +\frac{3N^{2}}{(N+1)(N+2)}B_{N-1,1}B_{N-1,2}^{2} \biggr\} . \end{aligned}$$
By using Proposition 3 for \(N=2\) and \(B_{n}=0\) for odd \(n\geq 3\), the numbers \(B_{2,n} (0\leq n\leq 4)\) are explicitly given by the classical Bernoulli numbers \(B_{n}\) (cf. [15, §9]),
Lemma 2
For \(N\geq 2\) and \(n\geq 1\), we have
Proof
From the derivative of (1), we have
By \(B_{N-1,0}=1\), we have
Therefore, we have
for \(n\geq 1\). By \(B_{N-1,1}=-\frac{1}{N}\), the assertion follows. □
Proof of Proposition 3
We give the proof by induction for n. In the case \(n=1\), the assertion means \(B_{N,1}=\frac{N}{N+1} B_{N-1,1}\), and this equality follows from \(B_{N,1}=-\frac{1}{N+1}\) and \(B_{N-1,1}=-\frac{1}{N}\). Assume that the assertion holds up to \(n-1\). By Lemma 2, we have
□
6 Multiple hypergeometric Bernoulli numbers
In this section, we define a more general hypergeometric Bernoulli number and give the properties.
For positive integers N and r, define the higher order hypergeometric Bernoulli numbers \(B_{N,n}^{(r)}\) [18, 26] by the generating function
The higher order hypergeometric Bernoulli polynomials \(B_{N,n}^{(r)}(x)\) are studied in [17], we have \(B_{N,n}^{(r)}=B_{N,n}^{(r)}(0)\).
From the definition (8), we have
Hence, as a generalization of Proposition 1 (3), for \(n\ge 1\), we have the following.
Proposition 4
We have
By using Proposition 4 or
with \(B_{N,0}^{(r)}=1\) (\(N\ge 1\)), some values of \(B_{N,n}^{(r)}\) (\(0 \le n\le 4\)) are explicitly given by the following:
As a generalization of Proposition 1 (4), we have an explicit expression of \(B_{N,n}^{(r)}\).
Proposition 5
For \(N,n\ge 1\), we have
where
We shall introduce the Hasse–Teichmüller derivative in order to prove Proposition 5 easily. Let \(\mathbb{F}\) be a field of any characteristic, \(\mathbb{F}[[z]]\) the ring of formal power series in one variable z, and \(\mathbb{F}((z))\) the field of Laurent series in z. Let n be a nonnegative integer. We define the Hasse–Teichmüller derivative \(H^{(n)}\) of order n by
for \(\sum_{m=R}^{\infty } c_{m} z^{m}\in \mathbb{F}((z))\), where R is an integer and \(c_{m}\in \mathbb{F}\) for any \(m\geq R\). Note that \(\binom{m}{n}=0\) if \(m< n\).
The Hasse–Teichmüller derivatives satisfy the product rule [27], the quotient rule [8] and the chain rule [12]. One of the product rules can be described as follows.
Lemma 3
For \(f_{i}\in \mathbb{F}[[z]]\) \((i=1,\dots,k)\) with \(k\ge 2\) and for \(n\ge 1\), we have
The quotient rules can be described as follows.
Lemma 4
For \(f\in \mathbb{F}[[z]]\backslash \{0\}\) and \(n\ge 1\), we have
Proof of Proposition 5
Put \(h(x)= (f(x) )^{r}\), where
Since
by the product rule of the Hasse–Teichmüller derivative in Lemma 3, we get
Hence, by the quotient rule of the Hasse–Teichmüller derivative in Lemma 4 (11), we have
□
Now, we can also show a determinant expression of \(B_{N,n}^{(r)}\).
Theorem 2
For \(N,n\ge 1\), we have
where \(M_{r}(e)\) are given in (10).
Remark
When \(r=1\) in Theorem 2, we have the result in Theorem 1.
Proof
For simplicity, put \(A_{N,n}^{(r)}=(-1)^{n} B_{N,n}^{(r)}/n!\). Then we shall prove that for any \(n\ge 1\),
When \(n=1\), (13) is valid because
Assume that (13) is valid up to \(n-1\). Notice that by (9), we have
Thus, by expanding the first row of the right-hand side of (13), it is equal to
Note that \(A_{N,1}^{(r)}=M_{r}(1)\) and \(A_{N,0}^{(r)}=1\). □
7 A relation between \(B_{N,n}^{(r)}\) and \(B_{N,n}\)
In this section, we show the following relation between \(B_{N,n}^{(r)}\) and \(B_{N,n}\).
Lemma 5
For \(r,N\geq 1\) and \(n\geq 0\), we have
Proof
From the definition (1) and (8), we get the assertion. □
We give some examples for Lemma 5. Note that \(B_{N,n}=B_{N,n}^{(1)}\) by the definition.
Example 4
-
(i)
\(B_{N,0}^{(r)}=B_{N,0}^{r}\ (r\geq 1)\),
-
(ii)
\(B_{N,1}^{(r)}=rB_{N,1} B_{N,0}^{r-1}\ (r \geq 1)\),
-
(iii)
$$\begin{aligned} B_{N,2}^{(r)}& = \sum_{{n_{1},\ldots,n_{r} \geq 0 \atop n_{1}+\cdots +n_{r}=2}} \frac{2}{n _{1}! \cdots n_{r}!} B_{N,n_{1}} \cdots B_{N,n_{r}} \\ & = \sum_{ {n_{1},\ldots,n_{r} \in \{ 0,2 \} \atop n_{1}+\cdots +n_{r}=2}} \frac{2}{n _{1}! \cdots n_{r}!} B_{N,n_{1}} \cdots B_{N,n_{r}} + \sum_{{n_{1},\ldots,n_{r} \in \{ 0,1 \} \atop n_{1}+\cdots +n_{r}=2}} \frac{2}{n_{1}! \cdots n_{r}!} B_{N,n_{1}} \cdots B_{N,n_{r}} \\ &= r B_{N,2}B_{N,0}^{N-1}+r(r-1)B_{N,1}^{2}B_{N,0}^{r-2}\quad (r \geq 2). \end{aligned}$$
8 Applications by the Trudi’s formula and inversion expressions
We can obtain different explicit expressions for the numbers \(B_{N,n}^{(r)}\), \(B_{N,n}\) and \(B_{n}\) by using the Trudi’s formula. We also show some inversion formulas. The following relation is known as Trudi’s formula [24, Vol. 3, p. 214], [28] and the case \(a_{0}=1\) of this formula is known as Brioschi’s formula [2], [24, Vol. 3, pp. 208–209].
Lemma 6
For a positive integer m, we have
where \(\binom{t_{1}+\cdots + t_{m}}{t_{1}, \dots,t_{m}}=\frac{(t_{1}+ \cdots + t_{m})!}{t_{1} !\cdots t_{m} !}\) are the multinomial coefficients.
In addition, there exists the following inversion formula (see, e.g., [21]), which is based upon the relation
Lemma 7
If \(\{\alpha _{n}\}_{n\geq 0}\) is a sequence defined by \(\alpha _{0}=1\) and
Moreover, if
From Trudi’s formula, it is possible to give the combinatorial expression
By applying these lemmas to Theorem 2, we obtain an explicit expression for the generalized hypergeometric Bernoulli numbers \(B_{N,n}^{(r)}\).
Theorem 3
For \(n\geq 1\),
where \(M_{r}(e)\) are given in (10). Moreover,
and
When \(r=1\) in Theorem 3, we have an explicit expression for the numbers \(B_{N,n}\).
Corollary 2
For \(n\geq 1\),
and
When \(r=N=1\) in Theorem 3, we have a different expression of the classical Bernoulli numbers.
Corollary 3
We have for \(n\ge 1\),
and
9 Continued fractions of hypergeometric Bernoulli numbers
In the final section, we show several identities for hypergeometric Bernoulli numbers which the convergents of the continued fraction expansion of the generating function of hypergeometric Bernoulli numbers entail.
In [1, 19] by studying the convergents of the continued fraction of
some identities of Bernoulli numbers are obtained. In this section, the nth convergent of the generating function of hypergeometric Bernoulli numbers is explicitly given. As an application, we give some identities of hypergeometric Bernoulli numbers in terms of binomial coefficients.
The generating function on the left-hand side of (1) can be expanded as a continued fraction
(cf. [29, (91.2)]). Its nth convergent \(P_{n}(x)/Q _{n}(x)\) (\(n\ge 0\)) is given by the recurrence relation
with initial values
where, for \(n\ge 1\), \(a_{n}(x)=N+n\), \(b_{2n}(x)=n x\) and \(b_{2 n+1}(x)=-(N+n)x\).
We have explicit expressions of both the numerator and the denominator of the nth convergent of (14).
Theorem 4
For \(n\ge 1\), we have
and
Remark
Here we use the convenient values
and recognize the empty product as 1. Otherwise, we should write \(Q_{2 n}(x)\) as
If we use the unsigned Stirling numbers of the first kind, \([{n\atop k} ]\), of which the generating function is given by
[9, (7.50)], we can express the products as
or
Proof of Theorem 4
The proof is done by induction on n. It is easy to see that for \(n=0\), we have \(P_{0}(x)=Q_{0}(x)=1\), and for \(n=1\), we have \(P_{1}(x)=(N+1)-x\) and \(Q_{1}(x)=N+1\). Assume that the results hold up to \(n-1(\ge 2)\). Then by using the recurrence relation in (16)
Since
we get
Next,
Since
we get
Concerning \(Q_{n}(x)\),
Since \(N+2 n=(N+2 n-j)+j\),
and \((2 n-j-1)_{k}+k(2 n-j-1)_{k-1}=(2 n-j)_{k}\), we obtain
Similarly,
Since
and
we get
Since
we have
□
9.1 Some more identities of hypergeometric Bernoulli numbers
Since \(P_{2 n-1}(x)\), \(P_{2 n}(x)\) and \(Q_{2 n}(x)\) are polynomials with degree n and \(Q_{2 n-1}(x)\) is a polynomial with degree \(n-1\), by the approximation property of the continued fraction, we have the following.
Lemma 8
Let \(P_{n}(x)/Q_{n}(x)\) denote the nth convergent of the continued fraction expansion of (14). Then for \(n\ge 0\),
By this approximation property, the coefficients \(x^{j}\) (\(0\le j \le n\)) of
are nullified. By Theorem 4,
and
Therefore,
Similarly, since
and
we have
Theorem 5
We have
and
In particular, when \(N=1\), we have the relations for the classical Bernoulli numbers.
Corollary 4
We have
and
Remark
Since
we can write (19) as
Since
we can write (20) as
Here the empty summation is recognized as 0, as usual.
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Aoki, M., Komatsu, T. & Panda, G.K. Several properties of hypergeometric Bernoulli numbers. J Inequal Appl 2019, 113 (2019). https://doi.org/10.1186/s13660-019-2066-y
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DOI: https://doi.org/10.1186/s13660-019-2066-y