On poly-Bell numbers and polynomials

TL;DR

This paper introduces a new family of poly-Bell numbers and polynomials related to Bell numbers via the confluent hypergeometric function, exploring their properties and applications in combinatorics and analysis.

Contribution

It constructs and analyzes poly-Bell numbers and polynomials, providing explicit formulas, generating functions, and integral representations, expanding the theory of Bell-related special functions.

Findings

Derived explicit formulas and generating functions for poly-Bell numbers.

Established recurrence relations and integral representations.

Connected poly-Bell numbers to combinatorial sums and special functions.

Abstract

This paper aims to construct a new family of numbers and polynomials which are related to the Bell numbers and polynomials by means of the confluent hypergeometric function. We give various properties of these numbers and polynomials (generating functions, explicit formulas, integral representations, recurrence relations, probabilistic representation,...). We also derive some combinatorial sums including the generalized Bernoulli polynomials, lower incomplete gamma function, generalized Bell polynomials. Finally, by applying Cauchy formula for repeated integration, we introduce poly-Bell numbers and polynomials.

Equations

$$g_p\left(z\right)=\left(p+1\right)!\sum _{n\ge 0}\frac{n!}{\left(n+p+1\right)!}{\left(1-e^z\right)}^n,$$

$$f_p\left(z\right)=p!\sum _{n\ge 0}\frac{{\left(e^z-1\right)}^n}{\left(n+p\right)!}.$$

$$\left\{\begin{tabular}[c]{ll}$x^{\underline{0}}=1,$&\\ $x^{\underline{n}}=x\left(x-1\right)\cdots\left(x-n+1\right),$&for $n>0$\end{tabular}\right.$$

$$x^{\underline{n}}=\sum _{k=0}^{n}s\left(n,k\right)x^k,$$

$$s\left(n+1,k\right)=s\left(n,k-1\right)-ns\left(n,k\right)\left(1\le k\le n\right).$$

$$x^n=\sum _{k=0}^n\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}{x}^{\underline{k}}.$$

$${\varphi }_n=\sum _{k=0}^n\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}.$$

$${\varphi }_n\left(x\right)=\sum _{k=0}^n\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}{x}^k$$

$$\sum _{n\ge k}s\left(n,k\right)\frac{z^n}{n!}$$

$$\sum _{n\ge k}\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}\frac{z^n}{n!}$$

$$\sum _{n\ge 0}{\varphi }_n\left(x\right)\frac{z^n}{n!}=\exp \left(x\left(e^z-1\right)\right).$$

$$f_{p}\left(z\right):={\displaystyle\sum\limits_{n\geq 0}}\mathcal{B}_{n,p}\frac{z^{n}}{n!}=\text{ }_{1}F_{1}\left(\begin{array}[c]{c}1\\ p+1\end{array};e^{z}-1\right)={\displaystyle\sum\limits_{n\geq 0}}\binom{n+p}{p}^{-1}\frac{\left(e^{z}-1\right)^{n}}{n!},$$

$$\sum _{n\ge 0}\frac{a^{\overline{n}}}{c^{\overline{n}}}\frac{z^n}{n!}.$$

$$\sum _{n\ge 0}{\mathcal{B}}_{n,1}\frac{z^n}{n!}$$

$$\sum _{n\ge 0}{\mathcal{B}}_{n,2}\frac{z^n}{n!}$$

$$\sum _{n\ge 0}{\mathcal{B}}_{n,3}\frac{z^n}{n!}$$

$${\mathcal{B}}_{n,p}=\sum _{k=0}^n{\left(\genfrac{}{}{0ex}{}{k+p}{k}\right)}^{-1}\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}.$$

$$\sum _{n\ge 0}\sum _{k=0}^n{\left(\genfrac{}{}{0ex}{}{k+p}{k}\right)}^{-1}\left\{\genfrac{}{}{0ex}{}{n}{k}\right\}\frac{z^n}{n!}$$

$$=\sum _{k\ge 0}\frac{p!k!}{\left(k+p\right)!}\frac{{\left(e^z-1\right)}^k}{k!}$$

$$=\sum _{k\ge 0}\frac{{\left(1\right)}^{\overline{k}}}{{\left(p+1\right)}^{\overline{k}}}\frac{{\left(e^z-1\right)}^k}{k!}$$

$$\displaystyle=\text{ }_{1}F_{1}\left(\begin{array}[c]{c}1\\ p+1\end{array};e^{z}-1\right)$$

$$=\sum _{n\ge 0}{\mathcal{B}}_{n,p}\frac{z^n}{n!}.$$

$${\mathcal{B}}_{n,1}$$

$$=\underset{0}{\overset{1}{\int }}{\varphi }_n\left(t\right)dt$$

$${\mathcal{B}}_{n,1}=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{{\varphi }_{k+1}{B}_{n-k}}{k+1},$$

$$\sum _{n\ge 0}{B}_n\frac{z^n}{n!}=\frac{z}{e^z-1}.$$

$$\sum _{n\ge 0}{\mathcal{B}}_{n,1}\frac{z^n}{n!}$$

$$=\sum _{n\ge 0}\left(\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{{\varphi }_{k+1}{B}_{n-k}}{k+1}\right)\frac{z^n}{n!}.$$

$$\sum _{p\ge 0}\sum _{n\ge p}{\mathcal{B}}_{n,p}\frac{z^n}{n!}\frac{y^p}{p!}=\frac{\left(e^z-1\right)\exp \left(e^z-1\right)}{e^z-1-y}.$$

$$\sum _{p\ge 0}\sum _{n\ge p}{\mathcal{B}}_{n,p}\frac{z^n}{n!}\frac{y^p}{p!}$$

$$=\sum _{n\ge 0}\frac{{\left(e^z-1\right)}^n}{n!}\sum _{p\ge 0}{\left(\frac{y}{e^z-1}\right)}^p$$

$$=\frac{\left(e^z-1\right)\exp \left(e^z-1\right)}{e^z-1-y}.$$

$$\sum _{n\ge k}{\left\{\genfrac{}{}{0ex}{}{n+r}{k+r}\right\}}_r\frac{z^n}{n!}=\frac{1}{k!}{e}^{rz}{\left(e^z-1\right)}^k.$$

$$\sum _{k=0}^{m}s\left(m,k\right){\mathcal{B}}_{n+k,p}=\sum _{k=0}^n{\left\{\genfrac{}{}{0ex}{}{n+m}{k+m}\right\}}_m{\left(\genfrac{}{}{0ex}{}{m+k+p}{p}\right)}^{-1}.$$

$${}_{1}F_{1}\left(\begin{array}[c]{c}a\\ b\end{array};z\right)=\frac{\Gamma\left(b\right)}{\Gamma\left(b-a\right)\Gamma\left(a\right)}{\displaystyle\int\limits_{0}^{1}}e^{zt}t^{a-1}\left(1-t\right)^{b-a-1}dt,$$

$$\sum _{n\ge 0}{\mathcal{B}}_{n,p}\frac{z^n}{n!}=p\underset{0}{\overset{1}{\int }}\exp (\left(e^z-1\right)t){\left(1-t\right)}^{p-1}dt.$$

Full text

Keywords: Bell numbers and polynomials, Bernoulli polynomials, generating function, probabilistic representation, Stirling numbers.

Mathematics Subject Classification: 11B73, 33C15, 11B68, 60C05.

1. Introduction

Recently, Rahmani in [17] published a paper on the function

[TABLE]

which generalized the generating function of Bernoulli numbers and gave an interesting algorithm for computing Bernoulli numbers and polynomials. Our main goal in this paper is to consider a class of numbers outcome from the following generating function which generalize Bell numbers

[TABLE]

and we give some of its properties.

The present paper is organized as follows. We first introduce in Section $2$, our notations and definitions. Then we present in Section $3$ some properties related to the $p$-Bell numbers. The poly-Bell numbers is introduced in Section $4$. The $p$-Bell and poly-Bell polynomials are presented in Section $5$. Finally, the probabilistic representation of $p$-Bell polynomials is given in Section $6.$

2. Preliminaries

As usual [10], the falling factorial $x^{\underline{n}}$ $\left(x\in\mathbb{C}\right)$ is defined by

[TABLE]

and the rising factorial denoted by $x^{\overline{n}}$, is defined by $x^{\overline{n}}=x\left(x+1\right)\cdots\left(x+n-1\right)$ with $x^{\overline{0}}=1$. The (signed) Stirling numbers of the first kind $s\left(n,k\right)$ are the coefficients in the expansion

[TABLE]

and satisfy the recurrence relation given by

[TABLE]

The Stirling numbers of the second kind, denoted $\genfrac{\{}{\}}{0.0pt}{}{n}{k}$ are the coefficients in the expansion

[TABLE]

The Stirling numbers of the second kind $\genfrac{\{}{\}}{0.0pt}{}{n}{k}$ count the number of ways to partition a set of $n$ elements into exactly $k$ nonempty subsets. The number of all partitions is the Bell number $\phi_{n},$ thus

[TABLE]

The polynomials

[TABLE]

are called single-variable Bell polynomials or exponential polynomials.

The exponential generating functions are respectively

[TABLE]

and

[TABLE]

3. The $p$-Bell numbers

In this section, we introduce and study a new generalization of the Bell number which we call the $p$-Bell numbers. For every integer $p\geq 0$, we define a sequence of rational numbers $\mathcal{B}_{n,p}$ $\left(n\geq 0\right)$ by

[TABLE]

where $\mathcal{B}_{n,0}:=\phi_{n}$ denotes the classical Bell numbers and {}_{1}F_{1}\left(\begin{array}[c]{c}a\\ c\end{array};z\right) denotes the Kummer confluent hypergeometric function [1] is defined by

[TABLE]

The first three exponential generating functions are:

[TABLE]

3.1. Generating functions and Explicit formulas

We start with an explicit formula for $p$-Bell numbers $\mathcal{B}_{n,p}$ involving Stirling numbers of the second kind.

Theorem 1**.**

For $p\geq 0$, we have

[TABLE]

Proof.

By using (4), we obtain

[TABLE]

Comparing the coefficients of $\frac{z^{n}}{n!}$, on both sides, we arrive at the result (9). ∎

In particular, for $p=1$ we have

[TABLE]

and we can write $\mathcal{B}_{n,1}$ also in the following form.

Proposition 1**.**

For $n\geq 0$, we have

[TABLE]

where $B_{n}$ denotes the $n$-th Bernoulli number [10], which is defined by means of the following generating function

[TABLE]

Proof.

We can rewrite $\left(\ref{BN2}\right)$ as

[TABLE]

Equating the coefficients of $\frac{z^{n}}{n!}$, we get the desired result. ∎

Remark 1**.**

We note that the formula (10) is a particular case of Ramanujan’s identity [3, Example 4, p. 51].

Theorem 2**.**

The double exponential generating function of $\mathcal{B}_{n,p}$ is given by

[TABLE]

Proof.

[TABLE]

∎

In order to establish some properties of $\mathcal{B}_{n,p}$, recall that the $r$-Stirling numbers of the second kind [5] $\genfrac{\{}{\}}{0.0pt}{}{n}{k}_{r}$ counts the number of partitions of a set of $n$ objects into exactly k nonempty, disjoint subsets, such that the first $r$ elements are in distinct subsets. The exponential generating function is given by

[TABLE]

Now, by means of the generalized Stirling transform [15], we obtain the following result.

Theorem 3**.**

For $n,m\geq 0$, we have

[TABLE]

In the next, we give some integral representations involving the exponential polynomials. It follows from the general theory of hypergeometric functions that the confluent hypergeometric function has an integral representation [1]

[TABLE]

where $\Gamma$ denotes the gamma function.

Theorem 4**.**

For $p\geq 1$, we have

[TABLE]

Corollary 1**.**

For $p\geq 1$, we have

[TABLE]

Proof.

We have

[TABLE]

Comparing the coefficients of $\frac{z^{n}}{n!}$, on both sides, we arrive at the result (13). ∎

The integral representation $\left(\ref{za}\right)$ can be also proved by using a well-known result [19]

[TABLE]

and $\left(\ref{Exp11}\right)$, we write

[TABLE]

Now, using the well-known Dobinski’s formula for the Bell polynomials [8]

[TABLE]

we get, the Dobinski’s formula for the $p$-Bell numbers.

Corollary 2**.**

The $p$-Bell numbers $\mathcal{B}_{n,p}$ can also be written as

[TABLE]

Proof.

From $\left(\ref{za}\right)$ and $\left(\ref{zaa}\right),$ we obtain

[TABLE]

and (15) follows from (11). ∎

We present here another expression for the generating function of $\mathcal{B}_{n,p}$ involving the lower incomplete gamma function. Recall that the lower incomplete gamma function $\gamma\left(s,z\right)$ is defined as:

[TABLE]

Theorem 5**.**

For $p\geq 1$, the exponential generating function for $p$-Bell numbers is given by

[TABLE]

Proof.

This can be achieved by a simple change of variable $t=1-y/(e^{z}-1)$ in (12). ∎

Theorem 6**.**

The exponential generating function for $p$-Bell numbers is given by

[TABLE]

Proof.

After an integration by parts, we can rewrite (12) as

[TABLE]

Now, applying (17) inductively we get the desired result. ∎

Next, we will show the relationship between the $p$-Bell numbers, Bell numbers and the generalized Bernoulli numbers. Recall that the generalized Bernoulli polynomials $B_{n}^{\left(\alpha\right)}(x)$ of degree $n$ in $x$ are defined by the exponential generating function [4, 18]

[TABLE]

for arbitrary parameter $\alpha$. In particular $B_{n}^{\left(\alpha\right)}(0):=B_{n}^{\left(\alpha\right)}$ denotes the generalized Bernoulli numbers of order $\alpha$, and $B_{n}^{\left(1\right)}(x)$ denotes the classical Bernoulli polynomials.

Corollary 3**.**

For $n,p\geq 0$, we have

[TABLE]

Proof.

From (16), we write

[TABLE]

Comparing the coefficients of $\frac{z^{n}}{n!},$ on both sides, we reach the result (19). ∎

The following lemma will be useful for the proof of next theorem.

Lemma 1**.**

For $p\geq 0$, we have

[TABLE]

and

[TABLE]

Proof.

By induction on $p$. ∎

Theorem 7**.**

For $p\geq 1$, we have

[TABLE]

Proof.

According to the Kummer transformation [1], we can write

[TABLE]

By above Lemma we obtain the result. ∎

The next result give us the integral representation of Cesàro type [7].

Theorem 8**.**

For $n\geq 1$, we have

[TABLE]

where $\operatorname{Im}\left(z\right)$, denotes the imaginary part of $z$.

Proof.

By using the integral identity:

[TABLE]

and (9) we get (20). ∎

3.2. Recurrence relations

Now we give a recurrence relation for the $p$-Bell numbers $\mathcal{B}_{n,p}.$ The proof is based on (12).

Theorem 9**.**

The $p$-Bell numbers satisfies the following recurrence relation

[TABLE]

with the initial sequence $\mathcal{B}_{0,p}=1.$

Proof.

By differentiation (12) with respect to $z$, we obtain

[TABLE]

or equivalently

[TABLE]

After some rearrangement, we get

[TABLE]

The conclusion follows by comparing coefficients of $\frac{z^{n}}{n!}.$ ∎

By using (21), we list the $p$-Bell numbers $\mathcal{B}_{n,p}$ for $0\leq n\leq 6$ and $0\leq p\leq 3$

[TABLE]

When $p=0,$ the formula (21) becomes

Corollary 4**.**

We have

[TABLE]

The next result presents a different types of recurrence for $p$-Bell numbers $\mathcal{B}_{n,p}$. We need the following lemma in order to prove the Theorem 10.

Lemma 2**.**

We have

[TABLE]

Proof.

This comes directly from $\left(\ref{R5}\right).$ ∎

Theorem 10**.**

The $p$-Bell numbers satisfy the following recurrence relation

[TABLE]

with the initial sequence $\mathcal{B}_{0,p}=1$ and the final sequence $\mathcal{B}_{n,0}=\phi_{n}$ .

Proof.

By $\left(\ref{C1}\right)$ with $z:=e^{z}-1$, we have

[TABLE]

Equating the coefficients of $\frac{z^{n}}{n!}$ and using (21), we get the result. ∎

In the next, we propose an algorithm, which is based on a three-term recurrence relation, for calculating the $p$-Bell numbers. Let consider the following sequence $\mathcal{Z}_{n,m}\left(p\right)$ with two indices by

[TABLE]

From Theorem 3, we can write

[TABLE]

with

[TABLE]

Theorem 11**.**

The $\mathcal{Z}_{n,m}\left(p\right)$ satisfy the following three-term recurrence relation:

[TABLE]

with the initial sequence given by $\mathcal{Z}_{0,m}\left(p\right)=1.$

Proof.

From (1) and (23), we have

[TABLE]

which is obviously equivalent to (24). ∎

Remark 2**.**

Note that for $p=0$, the recurrence relation (24) reduces to a known formula in [15].

4. Poly-Bell numbers

It is well known that the Cauchy formula for repeated integration [14] is given by

[TABLE]

where $\mathcal{D}^{-n}$ denotes the $n$-fold integral and $f$ be a continuous function on the real line. Comparing the formula (25) with the formula (13) we obtain:

Theorem 12**.**

For $p\geq 1$, we have

[TABLE]

The iterated integral expression (26) motivate a definition of poly-Bell numbers. Define the poly-Bell numbers $\mathbb{B}_{n}^{\left(p\right)}$as follows:

[TABLE]

For $\mathbb{B}_{n}^{\left(p\right)}$ the generating function is given by

[TABLE]

and can be extended for $p$ negative as

[TABLE]

An explicit formula of the poly-Bell numbers at a negative upper index $p$ can be obtained by a direct use of (27).

Theorem 13**.**

The poly-Bell numbers $\mathbb{B}_{n}^{\left(-p\right)}$is positive and we have

[TABLE]

Corollary 5**.**

For $p\geq 0$, we have

[TABLE]

Proof.

From (2), we have

[TABLE]

which reduces to $\mathbb{B}_{n}^{\left(-p\right)}$ by setting $x=1$. On the other hand, from (5), we have

[TABLE]

Equating coefficients of $\frac{z^{n}}{n!}$ and setting $x=1,$ we get the result. ∎

A natural question that arises is: the poly-Bernoulli numbers have the duality property [2, 12]. What about the poly-Bell numbers?

Theorem 14**.**

The double generating function of $\mathbb{B}_{n}^{\left(-p\right)}$ is given by

[TABLE]

By the above theorem, we can answer negatively the question, the duality property does not hold for the poly-Bell numbers.

Proof.

The double generating function of $\mathbb{B}_{n}^{\left(-p\right)}$ can be obtained by using (27) and (4)

[TABLE]

∎

As a consequence of Theorem 14, we obtain the generating function of poly-Bell numbers $\mathbb{B}_{n}^{\left(-p\right)}$ at negative upper index $p$.

Corollary 6**.**

We have

[TABLE]

Proof.

It follows immediately from

[TABLE]

∎

Here is a small table of the numbers $\mathbb{B}_{n}^{\left(p\right)}\left(-4\leq p\leq-1,0\leq n\leq 9\right).$

[TABLE]

5. The $p$-Bell polynomials

For $p\geq 0,$ let us consider the $p$-Bell polynomials $\mathcal{B}_{n,p}\left(x\right)$ is defined by means of the following generating function

[TABLE]

Now, using (29) we can easily get the explicit formula for the $p$-Bell polynomials.

[TABLE]

For every integer $p,$ we define also the following class of a polynomials $\mathbb{B}_{n}^{\left(p\right)}\left(x\right),$ which we call poly-Bell polynomials, by

[TABLE]

The first few $p$-Bell polynomials are

[TABLE]

and

[TABLE]

We present here another expression for the explicit formula of $\mathcal{B}_{n,p}\left(x\right)$ involving weighted Stirling numbers $S_{n}^{k}\left(x\right)$ of the second kind [6]. Recall that $S_{n}^{k}\left(x\right)$ are defined by

[TABLE]

where $\Delta$ denotes the forward difference operator. These numbers are related to $r$-Stirling numbers of the second kind$\genfrac{\{}{\}}{0.0pt}{}{n}{k}_{r}$ and Whitney numbers of the second kind [16] $W_{m,r}\left(n,k\right)$ by

[TABLE]

respectively.

Theorem 15**.**

For $p\geq 0$, we have

[TABLE]

Proof.

From (31) and the generating function of $S_{n}^{k}\left(x\right)$

[TABLE]

we have

[TABLE]

This evidently completes the proof of the theorem. ∎

In particular, we have

[TABLE]

and

[TABLE]

Now, we want to generalize Theorem 11 to the polynomials case. Let consider the polynomials $\mathcal{Z}_{n,m}\left(z;p\right)$ which is defined by

[TABLE]

with $\mathcal{Z}_{0,m}\left(x;p\right)=1$ and $\mathcal{Z}_{n,0}\left(x;p\right)=\mathcal{B}_{n,p}\left(x\right).$

Theorem 16**.**

The polynomials $\mathcal{Z}_{n,m}\left(x;p\right)$ satisfy the following three-term recurrence relation:

[TABLE]

with the initial sequence given by

[TABLE]

Proof.

From (32) and (24), we have

[TABLE]

Thus

[TABLE]

which is equivalent to (33). ∎

The next theorem establishes the integral representation for $p$-Bell polynomials. From (29) and (12), we have

[TABLE]

Since

[TABLE]

we get the following theorem.

Theorem 17**.**

The $p$-Bell polynomials $\mathcal{B}_{n,p}\left(x\right)$ satisfy the following integral representation

[TABLE]

Note that the polynomials $G_{n,x}\left(t\right)$ are a particular case of the more general polynomials considered by Corcino et al. in [9]. Now using the properties of $G_{n,x}\left(t\right),$ we get the following corollaries.

Corollary 7** (Dobinski’s formula).**

For $p\geq 1$, we have

[TABLE]

Proof.

From [9, Formula (1.18)], we obtain

[TABLE]

∎

Corollary 8** (Recurrence relations ).**

For $p\geq 0$, we have

[TABLE]

Proof.

From [9, Formula (2.9)], we obtain

[TABLE]

After some manipulation we get (35). ∎

Note that when $p=0,$ (35) gives

[TABLE]

6. Probabilistic representation

We consider a Poisson distributed random variable $Z$ with intensity $\lambda>0$. Suppose further that $\lambda$ is a random variable a beta distribution $\lambda\leadsto\operatorname*{beta}(1,p),$ the probability density function, for $0\leq t\leq 1$, and the parameter $p>0$ as follows:

[TABLE]

because

[TABLE]

where $\beta$ denotes beta function.

Then, the resultant distribution is a particular case of the beta-Poisson distribution [11, 13]. The probability mass function of $Z$ is given by

[TABLE]

thus, the probability mass function of a weighted Poisson distribution. The moment generating function of the beta-Poisson distribution is given by

[TABLE]

where $\mathbb{E}$ is the mathematical expectation.

It is well known that the higher moments $\mathbb{E}\left[Z^{n}\right]$ of the Poisson distribution are $\phi_{n}(t)$. In the next paragraph, we show that $\mathcal{B}_{n,p}(x)$ can be viewed as the $n$th moment of a random variable $x+Z$ where $Z$ follows the beta-Poisson law.

Theorem 18**.**

Let $Z$ be a random variable follows the beta-Poisson law, we have

[TABLE]

Proof.

From (34) and [9, p. 13, formula $\left(4,17\right)$], we have

[TABLE]

on the other hand, we have

[TABLE]

from which we obtain the result by comparison. ∎