Cauchy numbers in type $B$

Alnour Altouma,1, Hasan Arslanb,2, Mariam Zaaroura,3

aGraduate School of Natural and Applied Sciences, Erciyes University, 38039,
Kayseri, Turkey
bDepartment of Mathematics, Faculty of Science, Erciyes University, 38039,
Kayseri, Turkey
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###### Abstract

In this paper, we will introduce the Cauchy numbers of both kinds in type B
and produce their corresponding exponential generating functions. Then we will
provide some identities involving Cauchy, Lah, and Stirling numbers in type B
through combinatorial methods.

Keywords: Cauchy numbers, Stirling numbers, Lah numbers, generating functions.

2020 Mathematics Subject Classification: 05A15, 05A19.

Cauchy numbers are defined as the integration of rising and falling factorials
[3]. These numbers can be classified into two categories: Cauchy numbers of
the first kind, denoted by $C_{n}$, and Cauchy numbers of the second kind,
denoted as $c_{n}$. The first kind is defined by

$C_{n}=\int_{0}^{1}(x)_{n}\,dx$

where $(x)_{n}:=x(x-1)(x-2)\cdots(x-n+1)$ represents the falling factorial.
The second kind is defined to be

$c_{n}=\int_{0}^{1}[x]_{n}\,dx$

where $[x]_{n}:=x(x+1)(x+2)(x+3)\cdots(x+n-1)$ represents the rising factorial
[3].

The ordinary generating function of any infinite sequence
$(a_{k})_{k\in\mathbb{N}}$ is defined by $g(x)=\sum_{k=0}^{\infty}a_{k}x^{k}$
and shortly denoted by $\mathcal{G}(a_{k})=g(x)$. The corresponding
exponential generating function to the sequence $(a_{k})_{k\in\mathbb{N}}$ has
the form $\hat{g}(x)=\sum_{k=0}^{\infty}\frac{a_{k}}{k!}x^{k}$ and briefly
denoted by $\mathcal{E}(a_{k})=\hat{g}(x)$. Due to [8], the inverse operator
of a formal power series $g(x)=\sum_{k=0}^{\infty}a_{k}x^{k}$ provides the
coefficient of $x^{k}$ such that $[x^{k}]g(x)=a_{k}$ for all $k\in\mathbb{N}$.

Riordan array is actually a direct coefficient derivation method and it was
first introduced by Shapiro in [11]. Thus, it is useful to derive the
exponential generating function of Cauchy numbers. Riordan array
$B=(b_{n,k})_{n,k\in\mathbb{N}}$, which is a lower triangular and infinite
matrix, is identified with a pair of formal power series such that
$B=\mathcal{R}(b_{n,k})=(b(x),c(x))$, where

$b_{n,k}=[x^{n}]b(x)(xc(x))^{k}$ (1)

for all $n\in\mathbb{N}$ where $\mathbb{N}$ denotes the set of natural
numbers. One of the fundamental properties of Riordan array is the summation
property which is given as follows:

$\sum_{k=0}^{n}b_{n,k}g_{k}=[x^{n}]b(x)g(xc(x))$ (2)

where $\mathcal{R}(b_{n,k})=(b(x),c(x))$ and $g(x)$ is the ordinary generating
function of the sequence $(g_{k})_{k\in\mathbb{N}}$. We will mainly use the
following expression instead of Eq. (2):

$\sum_{k=0}^{n}b_{n,k}g_{k}=[x^{n}]b(x)\left[g(y):y=xc(x)\right].$ (3)

It is well-known from [7] that,

$\mathcal{R}(\frac{k!}{n!}c(n,k))=\left(1,\frac{1}{x}ln\frac{1}{1-x}\right)~{}\textrm{and}~{}\mathcal{R}(\frac{k!}{n!}S(n,k))=\left(1,\frac{e^{x}-1}{x}\right).$

where $c(n,k)$ is the classical signless Stirling number of the first kind and
$S(n,k)$ is the classical Stirling numbers of the second kind.

According to [7], the classical Cauchy numbers of the first kind have the
exponential generating function which is given by:

$\mathcal{E}(C_{n})=\sum_{k=0}^{\infty}C_{n}\frac{x^{n}}{n!}=\frac{x}{ln(1+x)}$

and these numbers are related to the Stirling numbers of the first kind
through the formula:

$C_{n}=\sum_{k=0}^{n}\frac{s(n,k)}{k+1}$

where $s(n,k):=(-1)^{n-k}c(n,k)$ is known as a Stirling number of the first
kind. The exponential generating function of Cauchy numbers of the second kind
has the following form (see [7]):

$\mathcal{E}(c_{n})=\sum_{k=0}^{\infty}c_{n}\frac{x^{n}}{n!}=\frac{x}{(1+x)ln(1+x)}$

and these numbers can be expressed in terms of signless Stirling numbers of
the first kind as follows:

$c_{n}=(-1)^{n}\sum_{k=0}^{n}\frac{c(n,k)}{k+1}.$

The rest of this paper is organized as follows: In section 2, we recall the
concept of Stirling numbers in type $B$. In section 3, we introduce the Cauchy
numbers of both kinds and Lah numbers in type B. Furthermore, we drive the
exponential generating functions of Cauchy numbers in type $B$ with the help
of the Riordan array. Finally, we generalize the Cauchy and Lah numbers into
$G_{m,n}$ type.

##  1 Preliminaries

The Stirling numbers of the second kind in type $B$ which is denoted by
$S_{B}(n,k)$ was defined first by Reiner in [9] by the following recurrence
relation:

$S_{B}(n,k)=S_{B}(n-1,k-1)+(2k+1)S_{B}(n-1,k),\quad 1\leq k<n$

with the initial conditions $S_{B}(n,n)=S_{B}(n,0)=1$ for all $n\geq 0$. The
signless Stirling numbers of the first kind in type $B$ are identified with
the recurrence relation

$c_{B}(n,k)=c_{B}(n-1,k-1)+(2n-1)c_{B}(n-1,k),\quad k\geq 0$

where $c_{B}(n,n)=c_{B}(1,0)=1$ and $c_{B}(n,k)=0$ if $k<0$ (see [10]). We
also note that the number $s_{B}(n,k):=(-1)^{n-k}c_{B}(n,k)$ is known as a
Stirling number of the first kind. The second kind of Stirling numbers in type
$B$ corresponds to the sequence oeis.org/A039755 in OEIS. One could see
Stirling numbers in type $B$ of the second kind for small values of $n$ and
$k$ in Table 1.

Table 1: Second kind Stirling numbers in type $B$

n | $S_{B}(n,0)$ | $S_{B}(n,1)$ | $S_{B}(n,2)$ | $S_{B}(n,3)$ | $S_{B}(n,4)$ | $S_{B}(n,5)$ | $S_{B}(n,6)$ | $S_{B}(n,7)$
---|---|---|---|---|---|---|---|---
0 | 1 |  |  |  |  |  |  |
1 | 1 | 1 |  |  |  |  |  |
2 | 1 | 4 | 1 |  |  |  |  |
3 | 1 | 13 | 9 | 1 |  |  |  |
4 | 1 | 40 | 58 | 16 | 1 |  |  |
5 | 1 | 121 | 330 | 170 | 25 | 1 |  |
6 | 1 | 364 | 1771 | 1520 | 395 | 36 | 1 |
7 | 1 | 1093 | 9219 | 12411 | 5075 | 791 | 49 | 1

The following table presents some Stirling numbers of the first kind
$c_{B}(n,k)$ which is associated with the sequence oeis.org/A039758 in OEIS.

Table 2: Fist kind signless Stirling numbers in type $B$

n | $c_{B}(n,0)$ | $c_{B}(n,1)$ | $c_{B}(n,2)$ | $c_{B}(n,3)$ | $c_{B}(n,4)$ | $c_{B}(n,5)$ | $c_{B}(n,6)$ | $c_{B}(n,7)$
---|---|---|---|---|---|---|---|---
0 | 1 |  |  |  |  |  |  |
1 | 1 | 1 |  |  |  |  |  |
2 | 3 | 4 | 1 |  |  |  |  |
3 | 15 | 23 | 9 | 1 |  |  |  |
4 | 105 | 176 | 86 | 16 | 1 |  |  |
5 | 945 | 1689 | 950 | 230 | 25 | 1 |  |
6 | 10395 | 19524 | 12139 | 3480 | 505 | 36 | 1 |
7 | 135135 | 264207 | 177331 | 57379 | 10045 | 973 | 49 | 1

The following theorem, which is provided by [2] and [10], expresses $x^{n}$ as
a sum of the terms involving both Stirling numbers of the second kind and
falling factorial in type $B$. The falling factorial in type $B$ is defined to
be $(x)_{n}^{B}=(x-1)(x-3)(x-5)\cdots(x-2n+1)$ with initial condition
$(x)^{B}_{0}=1$ (see [10]).

######  Theorem 1.1.

For any integer $n\geq 0$, we have

$x^{n}=\sum_{k=0}^{n}S_{B}(n,k)(x)_{k}^{B}.$

Taking into account Corollary 2.7 in [10], it is observed that the matrices
$[s_{B}(n,k)]_{n,k\geq 0}$ and $[S_{B}(n,k)]_{n,k\geq 0}$ are inverse of each
other. Therefore, it is easy to see that

$(x)_{n}^{B}=\sum_{k=0}^{n}s_{B}(n,k)x^{k}.$ (4)

The rising factorial in type $B$ is defined as
$[x]^{B}_{n}=(x+1)(x+3)(x+5)\cdots(x+2n-1)$ with the initial condition
$[x]^{B}_{0}=1$. It is well-known from part (c) of Theorem 2.1 in [10] that
for any nonnegative integer n

$[x]^{B}_{n}=\sum_{k=0}^{n}c_{B}(n,k)x^{k}.$ (5)

Conversely, for all $n\in\mathbb{N}$ the ordering powers $x^{n}$ can be easily
expressed as a linear combination of rising factorials $[x]_{n}^{B}$ as
follows:

$x^{n}=\sum_{k=0}^{n}S_{B}(n,k)(-1)^{n-k}[x]^{B}_{k}~{}~{}\textrm{for}~{}\textrm{all}~{}n\geq
0.$ (6)

##  2 Cauchy numbers in type $B$

In this section, we will introduce the notions of Cauchy and Lah numbers in
type $B$. These numbers will be defined using both falling and rising
factorials in type $B$. Additionally, we will explore some relationships
between Cauchy, Stirling, and Lah numbers.

######  Definition 2.1.

The type $B$ Cauchy numbers of the first kind are defined by the following
definite integral

$C_{n}^{B}=\int_{0}^{1}(x)^{B}_{n}\,dx.$

Table 3 records some values of the first kind of Cauchy numbers by giving a
few small $n$ values.

Table 3: Cauchy numbers of the first kind in type $B$ n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---|---
$C_{n}^{B}$ | 1 | -1/2 | 4/3 | -25/4 | 628/15 | -729/2 | 81994/21 | -1191619/24

######  Proposition 2.2.

The Cauchy numbers of the first kind hold for the following recurrence
relation

$C_{n+1}^{B}+(2n+1)C_{n}^{B}=\sum_{k=0}^{n}\frac{s_{B}(n,k)}{k+2}.$

###### Proof.

Due to the definition of the falling factorial of type $B$, we can write the
relation $x(x)_{n}^{B}=(x)_{n+1}^{B}+(2n+1)(x)_{n}^{B}$. Therefore, by Theorem
1.1 we get

$\displaystyle C_{n+1}^{B}$
$\displaystyle=\int_{0}^{1}(x)_{n+1}^{B}\,dx=\int_{0}^{1}(x(x)_{n}^{B}-(2n+1)(x)_{n}^{B})\,dx$
$\displaystyle=\int_{0}^{1}\sum_{k=0}^{n}s_{B}(n,k)x^{k+1}\,dx-(2n+1)\int_{0}^{1}(x)_{n}^{B}\,dx$
$\displaystyle=\sum_{k=0}^{n}\frac{s_{B}(n,k)}{k+2}-(2n+1)C_{n}^{B},$

as desired. ∎

######  Definition 2.3.

Cauchy numbers of the second kind in type $B$ are defined by definite integral
as below:

$c^{B}_{n}=\int_{0}^{1}[x]^{B}_{n}\,dx.$

Table 4 displays some special values for the second kind of Cauchy numbers of
type $B$.

Table 4: Cauchy numbers of the second kind in type $B$ n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---|---
$c_{n}^{B}$ | 1 | 3/2 | 16/3 | 119/4 | 3388/15 | 13013/6 | 528790/21 | 2742975/8

######  Proposition 2.4.

We have the following recurrence relation for the Cauchy numbers of the second
kind in type $B$:

$\quad c_{n+1}^{B}-(2n+1)c_{n}^{B}=\sum_{k=0}^{n}\frac{c_{B}(n,k)}{k+2}.$

###### Proof.

We can deduce the relation $[x]^{B}_{n+1}=x[x]^{B}_{n}+(2n+1)[x]^{B}_{n}$ from
the definition of the rising factorial of type $B$. Therefore, the desired
result can be easily seen from Eq. (5). ∎

######  Theorem 2.5.

For any positive integer $n$, we have the following formula

$\sum_{k=0}^{n}S_{B}(n,k)C_{k}^{B}=\frac{1}{n+1}=\sum_{k=0}^{n}S_{B}(n,k)(-1)^{n-k}c_{k}^{B}.$

###### Proof.

Considering Theorem 1.1 and the definition of the first kind of Cauchy numbers
in type $B$, we have

$\sum_{k=0}^{n}S_{B}(n,k)C_{k}^{B}=\sum_{k=0}^{n}S_{B}(n,k)\int_{0}^{1}(x)_{n}^{B}\,dx=\int_{0}^{1}x^{n}\,dx=\frac{1}{n+1}.$

Using Eq. (6) and the definition of the second kind of Cauchy numbers of type
B, then we get

$\sum_{k=0}^{n}S_{B}(n,k)(-1)^{n-k}c_{k}^{B}=\sum_{k=0}^{n}S_{B}(n,k)(-1)^{n-k}\int_{0}^{1}[x]_{k}^{B}\,dx=\int_{0}^{1}x^{n}\,dx=\frac{1}{n+1}.$

∎

###  2.1 The exponential generating functions of Cauchy numbers in type $B$

In order to derive the exponential generating functions for Cauchy numbers of
both kinds in type $B$, we will apply the Riordan array of the signless
Stirling numbers of the first kind in type $B$. Before going into a further
discussion of the exponential generating functions, we will give an important
relationship between Cauchy numbers of both kinds and Stirling numbers of the
first kind in type $B$.

######  Lemma 2.6.

For all $n\geq 0$, we have the following relations:

  * 1.

$C_{n}^{B}=\sum_{k=0}^{n}\frac{s_{B}(n,k)}{k+1}$,

  * 2.

$c_{n}^{B}=\sum_{k=0}^{n}\frac{c_{B}(n,k)}{k+1}$ .

###### Proof.

Considering Eq. (4), we then have

$\displaystyle C_{n}^{B}$ $\displaystyle=\int_{0}^{1}(x)_{n}^{B}\,dx$
$\displaystyle=\sum_{k=0}^{n}s_{B}(n,k)\int_{0}^{1}x^{k}\,dx$
$\displaystyle=\sum_{k=0}^{n}\frac{s_{B}(n,k)}{k+1}.$

From Eq. (5), we immediately obtain the second part as

$c_{n}^{B}=\int_{0}^{1}[x]_{n}^{B}\,dx=\sum_{k=0}^{n}c_{B}(n,k)\int_{0}^{1}x^{k}\,dx=\sum_{k=0}^{n}\frac{c_{B}(n,k)}{k+1}.$

∎

We are now in a position to give the exponential generating functions of
Cauchy numbers of both kinds in type $B$.

######  Theorem 2.7.

For the Cauchy numbers of the second kind, we have

$\mathcal{E}(c_{n}^{B})=\sum_{n\geq
0}c_{n}^{B}~{}\frac{x^{n}}{n!}=\frac{1-\sqrt{1-2x}}{(2x-1)ln\sqrt{1-2x}}.$

###### Proof.

We can write $\sum_{n\geq
0}c_{B}(n,k)~{}\frac{x^{n}}{n!}=\frac{1}{k!\sqrt{1-2x}}\left(ln\frac{1}{\sqrt{1-2x}}\right)^{k}$
from Theorem 4.1 (c) in [10]. Therefore, we can deduce from Eq. (1) that

$\mathcal{R}(\frac{k!}{n!}c_{B}(n,k))=\left(\frac{1}{\sqrt{1-2x}},\frac{1}{x}ln\frac{1}{\sqrt{1-2x}}\right).$
(7)

Using the second part of Lemma 2.6, we can get
$\frac{c_{n}^{B}}{n!}=\sum_{k=0}^{n}\frac{k!}{n!}c_{B}(n,k)\frac{1}{(k+1)!}$.
Since the ordinary generating function of $\frac{1}{(k+1)!}$ is equal to
$\frac{e^{x}-1}{x}$ and considering Eq. (3) and Eq. (7) , we then extract

$\frac{c_{n}^{B}}{n!}=[x^{n}]\frac{1}{\sqrt{1-2x}}\left[\frac{e^{y}-1}{y}:y=ln\frac{1}{\sqrt{1-2x}}\right],$

as desired. ∎

######  Theorem 2.8.

For the Cauchy numbers of the first kind, we have

$\mathcal{E}((-1)^{(n-1)}C_{n}^{B})=\sum_{n\geq
0}(-1)^{(n-1)}C_{n}^{B}~{}\frac{x^{n}}{n!}=\frac{1-\sqrt{1-2x}}{\sqrt{1-2x}~{}~{}ln\sqrt{1-2x}}.$

###### Proof.

Using the first part of Lemma 2.6, we can write
$\frac{C_{n}^{B}}{n!}=(-1)^{n-1}\sum_{k=0}^{n}\frac{k!}{n!}c_{B}(n,k)\frac{(-1)^{k+1}}{(k+1)!}$.
Since the ordinary generating function of $\frac{(-1)^{k+1}}{(k+1)!}$ is equal
to $\frac{e^{-x}-1}{x}$ and taking into consideration Eq. (3) and Eq. (7), we
then derive the desired formula by means of the following relation:

$\frac{C_{n}^{B}}{n!}=[x^{n}]\frac{1}{\sqrt{1-2x}}\left[\frac{e^{-y}-1}{y}:y=ln\frac{1}{\sqrt{1-2x}}\right].$

∎

One can illustrate Theorem 2.7 and Theorem 2.8 by considering Table 4 and 3,
respectively.

###  2.2 Lah numbers in type $B$

The classical Lah numbers, which were discovered by Ivo Lah in 1954 (see [5]),
are defined by binomial coefficient as

$L(n,k)=\frac{n!}{k!}\binom{n-1}{k-1},$

and were also defined by means of the Stirling numbers as follows (see [6]):

$L(n,k)=\sum_{j=k}^{n}c(n,j)S(j,k)$

where $c(n,j)$ and $S(j,k)$ are the classical Stirling numbers of the first
and second kind, respectively. The recurrence relation of $L(n,k)$ is given by

$L(n,k)=L(n-1,k-1)+(n-1+k)L(n-1,k)~{}~{}\textrm{for}~{}\textrm{all}~{}~{}n,k\in\mathbb{N}.$

It is well-known from [6] that the exponential generating function of $L(n,k)$
is stated as

$\mathcal{E}(L(n,k))=\sum_{n\geq
0}L(n,k)~{}\frac{x^{n}}{n!}=\frac{1}{k!}(\frac{x}{1-x})^{k}.$

Thus, we can conclude that Riordon array
$\mathcal{R}(\frac{k!}{n!}L(n,k))=\left(1,\frac{1}{1-x}\right).$

######  Definition 2.9.

The Lah numbers in type $B$ may be defined by

$L_{B}(n,k)=\binom{n}{k}^{2}2^{n-k}(n-k)!$

and the type $B$ Lah numbers can be expressed as a linear combination of
Stirling numbers:

$L_{B}(n,k)=\sum_{j=k}^{n}c_{B}(n,j)S_{B}(j,k)~{}~{}\textrm{for}~{}\textrm{all}~{}~{}n,k\in\mathbb{N}.$

The recurrence relation of $L_{B}(n,k)$ is defined by

$L_{B}(n,k)=L_{B}(n-1,k-1)+2(n+k)L_{B}(n-1,k)~{}~{}\textrm{for}~{}\textrm{all}~{}~{}n,k\in\mathbb{N},$
(8)

with the initial conditions $L_{B}(n,0)=2^{n}n!$, $L_{B}(n,n)=1$ and
$L_{B}(n,k)=0$ if $k<0$.

######  Theorem 2.10.

The exponential generating function of $L_{B}(n,k)$ is given by

$\mathcal{E}(L_{B}(n,k))=\sum_{n\geq
0}L_{B}(n,k)~{}\frac{x^{n}}{n!}=\frac{x^{k}}{(1-2x)^{k+1}k!}.$ (9)

###### Proof.

We argue by induction on $k$. If $k=0$, then we get

$\sum_{n\geq 0}L_{B}(n,0)~{}\frac{x^{n}}{n!}=\sum_{n\geq
0}2^{n}n!~{}\frac{x^{n}}{n!}=\frac{1}{1-2x}.$

Now let $\mathcal{E}(L_{B}(n,k)):=f_{k}(x)$. Thus we can write
$f_{k}(x)=\sum_{n\geq 0}L_{B}(n,k)~{}\frac{x^{n}}{n!}$. Considering Eq. (8),
we conclude that

$\displaystyle f_{k}(x)$ $\displaystyle:=\sum_{n\geq
k}L_{B}(n,k)~{}\frac{x^{n}}{n!}$ $\displaystyle=\sum_{n\geq
k}L_{B}(n-1,k-1)~{}\frac{x^{n}}{n!}+\sum_{n\geq
k+1}2(n+k)L_{B}(n-1,k)~{}\frac{x^{n}}{n!}$ $\displaystyle=\sum_{n\geq
k-1}L_{B}(n,k-1)~{}\frac{x^{n+1}}{(n+1)!}+\sum_{n\geq
k}2(n+1+k)L_{B}(n,k)~{}\frac{x^{n+1}}{(n+1)!}$ $\displaystyle=\sum_{n\geq
k-1}L_{B}(n,k-1)~{}\frac{x^{n+1}}{(n+1)!}+2\sum_{n\geq
k}L_{B}(n,k)~{}\frac{x^{n+1}}{n!}+2k\sum_{n\geq
k}L_{B}(n,k)~{}\frac{x^{n+1}}{(n+1)!}$ $\displaystyle=\sum_{n\geq
k-1}L_{B}(n,k-1)~{}\frac{x^{n+1}}{(n+1)!}+2xf_{k}(x)+2k\sum_{n\geq
k}L_{B}(n,k)~{}\frac{x^{n+1}}{(n+1)!}$

hence

$(1-2x)f_{k}(x)=\sum_{n\geq
k-1}L_{B}(n,k-1)~{}\frac{x^{n+1}}{(n+1)!}+2k\sum_{n\geq
k}L_{B}(n,k)~{}\frac{x^{n+1}}{(n+1)!}.$

If we take the derivative of both sides of the previous equation with respect
to $x$, then we get

$(1-2x)f^{\prime}_{k}(x)-2f_{k}(x)=f_{k-1}(x)+2kf_{k}(x)$

hence

$(1-2x)f^{\prime}_{k}(x)-2(k+1)f_{k}(x)=f_{k-1}(x).$

Multiplying both sides of the above equality by $(1-2x)^{k}$, then we obtain
by the induction hypothesis for $f_{k-1}(x)$ that

$(1-2x)^{k+1}f^{\prime}_{k}(x)-2(k+1)(1-2x)^{k}f_{k}(x)=\frac{x^{k-1}}{(k-1)!}.$

Thus we have

$((1-2x)f_{k}(x))^{\prime}=\frac{x^{k-1}}{(k-1)!}.$

If we solve the initial value problem
$((1-2x)f_{k}(x))^{\prime}=\frac{x^{k-1}}{(k-1)!},~{}~{}f_{k}(0)=0$, we then
obtain

$f_{k}(x)=\frac{x^{k}}{(1-2x)^{k+1}k!},$

as desired. ∎

It is straightforward to check that Eq. (9) gives the Riordon array of
$\frac{k!}{n!}L_{B}(n,k)$ as

$\mathcal{R}(\frac{k!}{n!}L_{B}(n,k))=\left(\frac{1}{1-2x},\frac{1}{1-2x}\right).$

Due to [4], the classical Lah-Bell numbers $L(n)$ are defined as

$L(n)=\sum_{k=0}^{n}L(n,k)$

and the exponential generating function of classical Lah-Bell numbers is given
by

$\mathcal{E}(L(n))=\sum_{n=0}^{\infty}L(n)\frac{x^{n}}{n!}=e^{\frac{x}{1-x}}.$

Therefore, we can define Lah-Bell numbers in type $B$ as

$LB(n)=\sum_{k=0}^{n}L_{B}(n,k).$

From Eq. (9), we can immediately deduce the exponential generating function of
$LB(n)$ in the following form:

$\mathcal{E}(LB(n))=\sum_{n=0}^{\infty}LB(n)\frac{x^{n}}{n!}=(\frac{1}{1-2x})e^{\frac{x}{1-2x}}.$

In Table 5, one could see both Lah numbers and Lah-Bell numbers of type $B$
for small values of $n$ and $k$.

Table 5: Lah numbers and Lah-Bell numbers $LB(n)$ of type $B$

n | $LB(n)$ | $L_{B}(n,0)$ | $L_{B}(n,1)$ | $L_{B}(n,2)$ | $L_{B}(n,3)$ | $L_{B}(n,4)$ | $L_{B}(n,5)$ | $L_{B}(n,6)$ | $L_{B}(n,7)$
---|---|---|---|---|---|---|---|---|---
0 | 1 | 1 |  |  |  |  |  |  |
1 | 3 | 2 | 1 |  |  |  |  |  |
2 | 17 | 8 | 8 | 1 |  |  |  |  |
3 | 139 | 48 | 72 | 18 | 1 |  |  |  |
4 | 1473 | 384 | 768 | 288 | 32 | 1 |  |  |
5 | 19091 | 3840 | 9600 | 4800 | 800 | 50 | 1 |  |
6 | 291793 | 46080 | 138240 | 86400 | 19200 | 1800 | 72 | 1 |
7 | 5129307 | 645120 | 2257920 | 1693440 | 470400 | 58800 | 3528 | 98 | 1

We observe that Lah numbers of type $B$ can be expressed by the falling
factorial as follows:

$\displaystyle L_{B}(n,k)=\binom{n}{k}(2n+1)_{n-k}^{B}$ (10)

where $(2n+1)_{n-k}^{B}$ is falling factorial of type $B$. As a consequence of
Eq. (10), we can give the next result.

######  Lemma 2.11.

Let $n\geq 0$. Thus we have

  * 1.

$[x]^{B}_{n}=(x+2n)_{n}^{B}=\sum_{k=0}^{n}\binom{n}{k}(2n+1)_{n-k}^{B}(x)_{k}^{B}=\sum_{k=0}^{n}L_{B}(n,k)(x)_{k}^{B}$,

  * 2.

$(x)^{B}_{n}=\sum_{k=0}^{n}(-1)^{n-k}L_{B}(n,k)[x]_{k}^{B}$.

###### Proof.

It is clear from the definitions of the rising and the falling factorials of
type $B$ that $[x]^{B}_{n}=(x+2n)_{n}^{B}$. Eq. (10) yields to the equality

$\sum_{k=0}^{n}L_{B}(n,k)(x)_{k}^{B}=\sum_{k=0}^{n}\binom{n}{k}(2n+1)_{n-k}^{B}(x)_{k}^{B}.$

The signless Stirling numbers of the first kind in type $B$ actually appear as
coefficients in the expression of the rising factorial of type $B$ in terms of
ordinary powers $x^{n}$. Therefore, based on Eq. (5) and Theorem 1.1, we
conclude the following result:

$\displaystyle[x]^{B}_{n}$
$\displaystyle=\sum_{j=0}^{n}c_{B}(n,j)x^{j}=\sum_{j=0}^{n}c_{B}(n,j)\sum_{k=0}^{j}S_{B}(j,k)(x)_{k}^{B}$
$\displaystyle=\sum_{k=0}^{n}\left(\sum_{j=k}^{n}c_{B}(n,j)S_{B}(j,k)\right)(x)_{k}^{B}=\sum_{k=0}^{n}L_{B}(n,k)(x)_{k}^{B}.$

Considering the above facts together, we obtain the first part of the lemma.
Similarly, the second part of the lemma can be easily proven by applying Eq.
(4) and Eq. (6). ∎

######  Remark 2.12.

Lemma 2.11 has the following additional interpretation. The set $V=K[x]$ be a
vector space of all polynomials in the indeterminate $x$ with coefficients in
the field $K$. The sets $B_{1}=\\{1,[x]_{1}^{B},[x]_{2}^{B},\cdots\\}$ and
$B_{2}=\\{1,(x)_{1}^{B},(x)_{2}^{B},\cdots\\}$ are both bases of $V$. Then the
first part of Lemma 2.11 asserts that the infinite matrix
$\textbf{L}=[L_{B}(n,k)]_{k,n\in\mathbb{N}}$ is the transition matrix between
the basis $B_{2}$ and the basis $B_{1}$. Conversely, the second part of Lemma
2.11 asserts that the infinite matrix
$\textbf{M}=[(-1)^{n-k}L_{B}(n,k)]_{k,n\in\mathbb{N}}$ is the transition
matrix between the basis $B_{1}$ and the basis $B_{2}$. Therefore, the matrix
M is the inverse to the matrix L.

The following theorem provides an important relationship between Cauchy
numbers of both kinds in type $B$ by using Lah numbers $L_{B}(n,k)$.

######  Theorem 2.13.

Let $n\geq 0$, then

$\displaystyle c_{n}^{B}=\sum_{k=0}^{n}L_{B}(n,k)C_{k}^{B}\quad and\quad
C_{n}^{B}=\sum_{k=0}^{n}(-1)^{n-k}L_{B}(n,k)c_{k}^{B}$

where $L_{B}(n,k)$ is of type $B$ Lah numbers.

###### Proof.

Consider the first part of Lemma 2.11. We obtain by integrating of
$[x]^{B}_{n}=\sum_{k=0}^{n}L_{B}(n,k)(x)_{k}^{B}$ from $0$ to $1$ on both
sides that $c_{n}^{B}=\sum_{k=0}^{n}L_{B}(n,k)C_{k}^{B}$. Using the second
part of Lemma 2.11, it can be easily proved that
$C_{n}^{B}=\sum_{k=0}^{n}(-1)^{n-k}L_{B}(n,k)c_{k}^{B}$ in a similar way. ∎

######  Remark 2.14.

Lemma 2.11 and Theorem 2.13 can both be thought of as a Lah inversion in type
$B$ in the sense of 3.38 Corollary (iii) on page 96 of [1].

##  3 Future Directions

Two kinds of incomplete Cauchy numbers of type $B$, which are generalizations
of the Cauchy numbers of type $B$, may be investigated in future studies by
introducing the restricted Stirling numbers of type $B$. In addition, the
identities associated with Cauchy numbers of both kinds in type $B$ can be
generalized into colored type $G_{m,n}$ as a future work. Furthermore, new
Cauchy numbers may be studied by considering the Stirling numbers of the
second kind in type $D$.

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