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Poly-falling factorial sequences and poly-rising factorial sequences

  • Hye Kyung Kim EMAIL logo
Published/Copyright: December 31, 2021
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Open Mathematics
From the journal Open Mathematics Volume 19 Issue 1

Abstract

In this paper, we introduce generalizations of rising factorials and falling factorials, respectively, and study their relations with the well-known Stirling numbers, Lah numbers, and so on. The first stage is to define poly-falling factorial sequences in terms of the polyexponential functions, reducing them to falling factorials if k = 1 , necessitating a demonstration of the relations: between poly-falling factorial sequences and the Stirling numbers of the first and second kind, respectively; between poly-falling factorial sequences and the poly-Bell polynomials; between poly-falling factorial sequences and the poly-Bernoulli numbers; between poly-falling factorial sequences and poly-Genocchi numbers; and recurrence formula of these sequences. The later part of the paper deals with poly-rising factorial sequences in terms of the polyexponential functions, reducing them to rising factorial if k = 1 . We study some relations: between poly-falling factorial sequences and poly-rising factorial sequences; between poly-rising factorial sequences and the Stirling numbers of the first kind and the power of x ; and between poly-rising factorial sequences and Lah numbers and the poly-falling factorial sequences. We also derive recurrence formula of these sequences and reciprocal formula of the poly-falling factorial sequences.

Keywords: falling factorials; rising factorials; modified polyexponential functions; Stirling numbers of the first and second kind; poly-Bernoulli polynomials; poly-Genocchi polynomials
MSC 2010: 11B73; 11B83; 05A19

1 Introduction

In the study of falling factorials and rising factorials, the following properties produce important characteristics for special numbers such as Stirling numbers and Lah numbers: the nth falling factorial of x is expressed in terms of the Stirling numbers of the first kind and the power of x ; the nth power of x is expressed in terms of the Stirling numbers of the second kind S 2 ( n , l ) ; the falling factorials of x ; the nth rising factorial of x is expressed in terms of Lah numbers and falling factorials; and the nth falling factorial of x can also be expressed in terms of Lah numbers and rising factorials [1,2, 3,4,5, 6,7,8, 9,10]. Kim and Kim [11] introduced the polyexponential functions in the view of an inverse to the polylogarithm functions which were first studied by Hardy [12]. Many mathematicians have studied about “poly” for special polynomials and numbers arising from the polyexponential functions or the polylogarithm functions [4,8,10,13,14, 15,16]. Recently, the degenerate poly-Bell polynomials and the degenerate poly-Lah-Bell polynomials derived from the degenerate polyexponential functions were introduced in [17,18], respectively. In addition, it is briefly mentioned that the degenerate Daehee numbers of order k expressed the degenerate polyexponential functions in [19]. In this paper, we intend to study the above-mentioned properties by generalizing falling factorials and rising factorials, respectively, using the polyexponential functions. In Section 2, we define poly-falling factorial sequences arising from the polyexponential functions, reducing them to falling factorial if k = 1 . We demonstrate the relations: between poly-falling factorial sequences and the Stirling numbers of the first and second kind, respectively; between poly-falling factorial sequences and the poly-Bell polynomials; between poly-falling factorial sequences and the poly-Bernoulli numbers; between poly-falling factorial sequences and poly-Genocchi numbers; and recurrence formula of these sequences. In Section 3, we introduce poly-rising factorial sequences arising from the polyexponential functions, reducing them to rising factorial if k = 1 . To elaborate, we analyze the relationships: between poly-falling factorial sequences and poly-rising factorial sequences; between poly-rising factorial sequences of x and the Stirling numbers of the first kind and the power of x ; between poly-rising factorial sequences and Lah numbers the poly-falling factorial sequences; recurrence formula of these sequences; and reciprocal formula of the poly-falling factorial sequences.

First, definitions and preliminary properties required in this paper are introduced.

For x ∈ R , the falling factorials ( x ) n are defined by

(1) ( x ) n = x ( x − 1 ) ( x − 2 ) ⋯ ( x − n + 1 ) , ( n ≥ 1 ) and ( x ) 0 = 1

(see [1,2,7,20,21,22]).

For x ∈ R , the rising factorials ⟨ x ⟩ n are defined by

(2) ⟨ x ⟩ n = x ( x + 1 ) ( x + 2 ) ⋯ ( x + n − 1 ) , ( n ≥ 1 ) , and ⟨ x ⟩ 0 = 1

(see [2,5,6,20,21,22]).

It is well-known that for ∣ t ∣ < 1 ,

(3) ( 1 − t ) − m = ∑ l = 0 ∞ − m l ( − 1 ) l t l = ∑ l = 0 ∞ ⟨ m ⟩ l t l l !

(see [2,17,18]).

For n ≥ 0 , the Stirling numbers of the first kind S 1 ( n , l ) are the coefficients of x l in

(4) ( x ) n = ∑ l = 0 n S 1 ( n , l ) x l

(see [2,9,22]).

From (4), it is easy to see that

(5) 1 k ! ( log ( 1 + t ) ) k = ∑ n = k ∞ S 1 ( n , k ) t n n ! , ( ∣ t ∣ < 1 )

(see [2,9,22]).

In the inverse expression to (4), for n ≥ 0 , the nth power of x can be expressed in terms of the Stirling numbers of the second kind S 2 ( n , l ) as follows:

(6) x n = ∑ l = 0 n S 2 ( n , l ) ( x ) l ,

(see [2,7,21,22]).

From (6), it is easy to see that for ∣ t ∣ < 1 ,

(7) 1 k ! ( e t − 1 ) k = ∑ n = k ∞ S 2 ( n , k ) t n n ! , ( t ∈ C )

(see [2,7,9,22]).

The Bell polynomials bel n ( x ) = ∑ k = 0 n S 2 ( n , k ) x n are natural extensions of the Bell numbers which are the number of ways to partition a set with n elements into nonempty subsets. It is well known that the generating function of the Bell polynomials is given by

e x ( e t − 1 ) = ∑ n = 0 ∞ bel n ( x ) t n n ! , ( t ∈ C )

(see [2,3,10,23,24]).

The unsigned Lah number L ( n , j ) counts the number of ways a set of n elements can be partitioned into j nonempty linearly ordered subsets and has an explicit formula

(8) L ( n , j ) = n − 1 j − 1 n ! j ! , ( j ≥ 0 )

(see [2,5,6,8,18,25,26]).

From (1), the generating function of L ( n , k ) is given by

(9) 1 j ! t 1 − t j = ∑ n = j ∞ L ( n , j ) t n n ! , ( j ≥ 0 ) , ( t ∈ C ⧹ { 1 } )

(see [2,5,6,8,25,26]).

We observe that

(10) 1 1 − t x = 1 + t 1 − t x = ∑ j = 0 ∞ x j t 1 − t j = ∑ n = 0 ∞ ∑ j = 0 n L ( n , j ) ( x ) j t n n ! .

On the other hand, we get

(11) 1 1 − t x = ( 1 − t ) − x = ∑ n = 0 ∞ ⟨ x ⟩ n t n n ! .

From (10) and (11), we have

(12) ⟨ x ⟩ n = ∑ j = 0 n L ( n , j ) ( x ) j

(see [2,5]). Replacing x by − x , we get easily

(13) ( x ) n = ∑ j = 0 n ( − 1 ) n − j L ( n , j ) ⟨ x ⟩ j

(see [5,6]).

Kim and Kim introduced the modified polyexponential functions as

(14) Ei k ( x ) = ∑ n = 1 ∞ x n ( n − 1 ) ! n k , ( k ∈ Z )

(see [11]). When k = 1 , we see that Ei 1 ( x ) = e x − 1 .

The poly-Bernoulli polynomials are defined by

(15) Ei k ( log ( 1 + t ) ) e t − 1 e x t = ∑ n = 0 ∞ B n ( k ) ( x ) t n n !

(see [14,16]). When x = 0 , B n ( k ) = B n ( k ) ( 0 ) , which are called the poly-Bernoulli numbers [14,16].

Since Ei 1 ( log ( 1 + t ) ) = t , B n ( 1 ) ( x ) are the Bernoulli polynomials [1,5,13,19,22].

The poly-Genocchi polynomials are given by

(16) 2 Ei k ( log ( 1 + t ) ) e t + 1 e x t = ∑ n = 0 ∞ G n ( k ) ( x ) t n n !

(see [27]), and G 0 ( k ) ( x ) = 0 . When x = 0 , G n ( k ) = G n ( k ) ( 0 ) , which are called the poly-Genocchi numbers.

When k = 1 , G n ( 1 ) ( x ) are the Genocchi polynomials [4,27,28,29].

We introduced the poly-Bell polynomials by

(17) 1 + Ei k ( x ( e t − 1 ) ) = ∑ n = 0 ∞ bel n ( k ) ( x ) t n n !

(see [17]), and bel 0 ( k ) ( x ) = 1 (see [17]).

When k = 1 , from (17), we note that

(18) 1 + Ei 1 ( x ( e t − 1 ) ) = 1 + ∑ n = 1 ∞ ( x ( e t − 1 ) ) n ( n − 1 ) ! n = exp ( x ( e t − 1 ) ) = ∑ n = 0 ∞ bel n ( x ) t n n ! .

Combining (17) with (18), we have the Bell polynomials

bel n ( 1 ) ( x ) = bel n ( x )

(see [2,3,22]).

As the multivariate version of the Stirling numbers S ( n , k ) of the second kind, the generation function of the incomplete Bell polynomials B n , k ( x 1 , x 2 , … , x n − k + 1 ) is given by

(19) 1 k ! ∑ h = 1 ∞ x h t h h ! k = ∑ n = k ∞ B n , k ( α 1 , α 2 , … , α n − k + 1 ) t n n !

(see [2,23,25,30]).

From (19), we obtain

(20) B n , k ( α 1 , α 2 , … , α n − k + 1 ) = ∑ h 1 + 2 h 2 + ⋯ + n h n = n n ! h 1 ! h 2 ! ⋯ h n − k + 1 ! α 1 1 h 1 α 2 2 ! h 2 ⋯ α n − k + 1 ( n − k + 1 ) ! h n − k + 1

(see [2,23,25,31]), where the sum over all nonnegative integers h 1 , h 2 , … , h n satisfying h 1 + h 2 + ⋯ + h n − k + 1 = k and h 1 + 2 h 2 + ⋯ + ( n − k + 1 ) h n − k + 1 = n .

2 Poly-falling factorial sequences

In this section, we define the poly-falling factorial sequences by using the polyexponential functions. We also give some relations between them and special numbers, and derive recurrence formula of these sequences.

For n ∈ N ∪ { 0 } and x ∈ R , we consider the poly-falling factorial sequences ( x ) n ( k ) , which are arising from the polyexponential functions to be

(21) 1 + Ei k ( x log ( 1 + t ) ) = ∑ n = 0 ∞ ( x ) n ( k ) t n n ! and ( x ) 0 ( k ) = 1 ( x ≠ 0 , ∣ t ∣ < 1 ) .

Note that Ei 1 ( x ) = e x − 1 .

When k = 1 , since Ei 1 ( x ) = e x − 1 , we note that

(22) 1 + Ei 1 ( x log ( 1 + t ) ) = 1 + e x log ( 1 + t ) − 1 = ( 1 + t ) x = ∑ n = 0 ∞ ( x ) n t n n ! .

Therefore, by (22), we have ( x ) n ( 1 ) = ( x ) n .

Theorem 1

For n ∈ N , k ∈ Z , and x ∈ R , we have

( x ) n ( k ) = ∑ d = 1 n 1 d k − 1 S 1 ( n , d ) x d .

Proof

First, from (5), (14), and (21), we observe that

(23) 1 + Ei k ( x log ( 1 + t ) ) = 1 + ∑ d = 1 ∞ x d ( log ( 1 + t ) ) d ( d − 1 ) ! d k = 1 + ∑ d = 1 ∞ x d d k − 1 1 d ! ( log ( 1 + t ) ) d = 1 + ∑ d = 1 ∞ x d d k − 1 ∑ n = d ∞ S 1 ( n , d ) t n n ! = 1 + ∑ n = 1 ∞ ∑ d = 1 n 1 d k − 1 S 1 ( n , d ) x d t n n ! .

By comparing the coefficients of both sides of (23), we get the desired result.□

Theorem 2

For n ∈ N , k ∈ Z , and x ∈ R , we have

n k − 1 ∑ d = 1 n S 2 ( n , d ) ( x ) d ( k ) = x n .

Proof

Substituting t by e t − 1 in (21), the left-hand side of (21) is

(24) 1 + Ei k ( x t ) = 1 + ∑ n = 1 ∞ ( x t ) n ( n − 1 ) ! n k = 1 + ∑ n = 1 ∞ x n n k − 1 t n n ! .

On the other hand, from (7), the right-hand side of (21) is

(25) ∑ d = 0 ∞ ( x ) d ( k ) ( e t − 1 ) d d ! = 1 + ∑ d = 1 ∞ ( x ) d ( k ) ∑ n = d ∞ S 2 ( n , d ) t n n ! = 1 + ∑ n = 1 ∞ ∑ d = 1 n ( x ) d ( k ) S 2 ( n , d ) t n n ! .

Combining (24) with (25), we have the desired result.□

In Theorems 1 and 2, when k = 1 , we note that

( x ) n = ∑ d = 0 n S 1 ( n , d ) x d and ∑ d = 0 n S 2 ( n , d ) ( x ) d = x n .

Theorem 3

For n ∈ N , k ∈ Z , and x ∈ R , we have

bel n ( k ) ( x ) = ∑ m = 1 n ∑ d = 1 m S 2 ( n , m ) S 2 ( m , d ) ( x ) d ( k ) .

Proof

Substituting t by e ( e t − 1 ) − 1 in (21), from (7), we observe that

(26) 1 + Ei k ( x ( e t − 1 ) ) = 1 + ∑ d = 1 ∞ ( x ) d ( k ) ( e ( e t − 1 ) − 1 ) d d ! = 1 + ∑ d = 1 ∞ ( x ) d ( k ) ∑ m = d ∞ S 2 ( m , d ) ( e t − 1 ) m m ! = 1 + ∑ m = 1 ∞ ∑ d = 1 m ( x ) d ( k ) S 2 ( m , d ) ∑ n = m ∞ S 2 ( n , m ) t n n ! = 1 + ∑ n = 1 ∞ ∑ m = 1 n ∑ d = 1 m S 2 ( n , m ) S 2 ( m , d ) ( x ) d ( k ) t n n ! .

By comparing the coefficients of (17) and (26), for n ≥ 1 we get

(27) bel n ( k ) ( x ) = ∑ m = 1 n ∑ d = 1 m S 2 ( n , m ) S 2 ( m , d ) ( x ) d ( k ) .

□

From (4) and (5), we observe that

(28) x n = ∑ l = 0 n S 2 ( n , l ) ( x ) l = ∑ d = 0 n S 2 ( n , d ) ∑ j = 0 d S 1 ( d , j ) x j = ∑ d = 0 n ∑ j = 0 d S 2 ( n , d ) S 1 ( d , j ) x j .

From (28), we have

(29) ∑ d = 0 n ∑ j = 0 d S 2 ( n , d ) S 1 ( d , j ) = δ n , j .

In Theorem 3, for n ≥ 1 when k = 1 , by using (4), (29), and Theorem 1, we note that

(30) bel n ( x ) = ∑ m = 1 n ∑ d = 1 m S 2 ( n , m ) S 2 ( m , d ) ( x ) d = ∑ m = 1 n ∑ d = 1 m ∑ j = 0 d S 2 ( n , m ) S 2 ( m , d ) S 1 ( d , j ) x j = ∑ j = 1 n S 2 ( n , j ) x j = ∑ j = 0 n S 2 ( n , j ) x j .

Theorem 4

For n ∈ N , k ∈ Z , and x ∈ R , we have

( x ) n ( k ) = ∑ d = 1 n ∑ m = d n S 1 ( n , m ) S 1 ( m , d ) bel d ( k ) ( x ) .

Proof

First, we observe that

(31) ( log ( log ( 1 + t ) + 1 ) ) d d ! = ∑ m = d ∞ S 1 ( m , d ) ( log ( 1 + t ) ) m m ! = ∑ m = d ∞ S 1 ( m , d ) ∑ n = m ∞ S 1 ( n , m ) t n n ! = ∑ n = d ∞ ∑ m = d n S 1 ( n , m ) S 1 ( m , d ) t n n ! .

Substituting t by log ( log ( 1 + t ) + 1 ) in (17), from (5), we observe that

(32) 1 + Ei k ( x log ( 1 + t ) ) = ∑ d = 0 ∞ bel d ( k ) ( x ) ( log ( log ( 1 + t ) + 1 ) ) d d ! = 1 + ∑ d = 1 ∞ bel d ( k ) ( x ) ∑ n = d ∞ ∑ m = d n S 1 ( n , m ) S 1 ( m , d ) t n n ! = 1 + ∑ n = 1 ∞ ∑ d = 1 n ∑ m = d n S 1 ( n , m ) S 1 ( m , d ) bel d ( k ) ( x ) t n n ! .

Combining (21) with (32), for n ≥ 1 , we have the desired result.□

Theorem 5

For n ∈ N and k ∈ Z , we have

( x ) 1 ( k − 1 ) = − ( x ) 1 ( k )

and

( x ) n ( k − 1 ) = ∑ j = 1 n n j ( − 1 ) j + 1 ( j − 1 ) ! ( ( x ) n − j + 1 ( k ) + ( n − j ) ( x ) n − j ( k ) ) , ( n ≥ 2 ) .

Proof

Differentiating with respect to t in (21), the left-hand side of (21) is

(33) ∂ ∂ t ( 1 + Ei k ( x log ( 1 + t ) ) ) = ∑ n = 1 ∞ x n ( log ( 1 + t ) ) n − 1 ( n − 1 ) ! n k − 1 1 1 + t = 1 ( 1 + t ) log ( 1 + t ) ∑ n = 1 ∞ ( x log ( 1 + t ) ) n ( n − 1 ) ! n k − 1 = 1 ( 1 + t ) log ( 1 + t ) Ei k − 1 ( x log ( 1 + t ) ) = 1 ( 1 + t ) log ( 1 + t ) ∑ n = 1 ∞ ( x ) n ( k − 1 ) t n n ! .

On the other hand, the right-hand side of (21) is

(34) ∂ ∂ t ∑ n = 0 ∞ ( x ) n ( k ) t n n ! = ∑ n = 1 ∞ ( x ) n ( k ) t n − 1 ( n − 1 ) ! = ∑ n = 0 ∞ ( x ) n + 1 ( k ) t n n ! .

By (33) and (34), we get

(35) ∑ n = 1 ∞ ( x ) n ( k − 1 ) t n n ! = ( 1 + t ) log ( 1 + t ) ∑ m = 0 ∞ ( x ) m + 1 ( k ) t m m ! = ∑ j = 1 ∞ ( − 1 ) j + 1 j t j ∑ m = 0 ∞ ( x ) m + 1 ( k ) t m m ! + ∑ j = 1 ∞ ( − 1 ) j + 1 j t j ∑ m = 0 ∞ ( x ) m + 1 ( k ) t m + 1 m ! = ∑ j = 1 ∞ ( − 1 ) j + 1 ( j − 1 ) ! t j j ! ∑ m = 0 ∞ ( x ) m + 1 ( k ) t m m ! + ∑ j = 1 ∞ ( − 1 ) j + 1 ( j − 1 ) ! t j j ! ∑ m = 1 ∞ ( x ) m ( k ) m t m m ! = ∑ n = 1 ∞ ∑ j = 1 n n j ( − 1 ) j + 1 ( j − 1 ) ! ( x ) n − j + 1 ( k ) t n n ! + ∑ n = 2 ∞ ∑ j = 1 n n j ( − 1 ) j + 1 ( j − 1 ) ! ( n − j ) ( x ) n − j ( k ) t n n ! = − ( x ) 1 ( k ) + ∑ n = 2 ∞ ∑ j = 1 n n j ( − 1 ) j + 1 ( j − 1 ) ! ( ( x ) n − j + 1 ( k ) + ( n − j ) ( x ) n − j ( k ) ) t n n ! .

By comparing the coefficients of both sides of (35), we get the desired result.□

Theorem 6

For n ∈ N and k ∈ Z , we have

( 1 ) n ( k ) = ∑ m = 0 n n m B m ( k ) ,

where B n ( k ) are the poly-Bernoulli numbers.

Proof

From (15), we get

(36) Ei k ( log ( 1 + t ) ) = ( e t − 1 ) ∑ m = 0 ∞ B m ( k ) t m m ! = ∑ j = 1 ∞ t j j ! ∑ m = 0 ∞ B m ( k ) t m m ! = ∑ n = 1 ∞ ∑ m = 0 n n m B m ( k ) t n n ! .

On the other hand, when x = 1 in (21), we observe that

(37) 1 + Ei k ( log ( 1 + t ) ) = ∑ n = 0 ∞ ( 1 ) n ( k ) t n n ! = 1 + ∑ n = 1 ∞ ( 1 ) n ( k ) t n n ! .

By (36) and (37), for n ≥ 1 , we attain the desired result.□

Corollary 7

For n ∈ N and k ∈ Z , we have

( 1 ) n ( k ) = B n ( k ) ( 1 ) − B n ( k ) .

Proof

From (15), we observe that

(38) ∑ n = 0 ∞ B n ( k ) ( x ) t n n ! = 2 Ei k ( log ( 1 + t ) ) e t + 1 e x t = ∑ j = 0 ∞ B j ( k ) t j j ! ∑ m = 0 ∞ x m t m m ! = ∑ n = 0 ∞ ∑ j = 0 n n j B j ( k ) x n − j t n n ! .

By comparing the coefficients of both sides of (38), we get

(39) B n ( k ) ( x ) = ∑ j = 0 n n j x n − j B j ( k ) .

From (15) and (39), we have

(40) Ei k ( log ( 1 + t ) ) = ( e t − 1 ) ∑ d = 0 ∞ B d ( k ) t d d ! = ∑ n = 0 ∞ ∑ d = 0 n n d B d ( k ) − B n ( k ) t n n ! = ∑ n = 1 ∞ ( B n ( k ) ( 1 ) − B n ( k ) ) t n n ! .

Combining (37) with (40), we have the desired result.□

Theorem 8

For n ∈ N and k ∈ Z , we have

( 1 ) n ( k ) = 1 2 ∑ d = 1 n − 1 n d G d ( k ) + 2 G n ( k ) ,

where G d ( k ) are the poly-Genocchi numbers.

Proof

By (16) and G 0 ( k ) = 0 , we observe that

(41) 2 Ei k ( log ( 1 + t ) ) = ( e t + 1 ) ∑ d = 0 ∞ G d ( k ) t d d ! = ∑ j = 0 ∞ t j j ! ∑ d = 0 ∞ G d ( k ) t d d ! + ∑ d = 0 ∞ G d ( k ) t d d ! = ∑ n = 1 ∞ ∑ d = 0 n n d G d ( k ) t n n ! + ∑ n = 1 ∞ G n ( k ) t n n ! = ∑ n = 1 ∞ ∑ d = 1 n − 1 n d G d ( k ) + 2 G n ( k ) t n n ! .

From (37) and (41), we get what we want.□

3 Poly-rising factorial sequences

In this section, we define poly-rising factorial sequences by using the polyexponential functions and give relations between poly-falling factorial sequences and poly-rising factorial sequences, and special numbers. We also study recurrence formula of these sequences and a reciprocal formula of the poly-falling factorial sequences.

Now, we consider the poly-rising factorial sequences ⟨ x ⟩ n ( k ) , which are arising from the polyexponential functions to be

(42) 1 + Ei k ( − x log ( 1 − t ) ) = ∑ n = 0 ∞ ⟨ x ⟩ n ( k ) t n n ! and ⟨ x ⟩ 0 ( k ) = 1 ( x ≠ 0 , ∣ t ∣ < 1 ) .

When k = 1 , since Ei 1 ( x ) = e x − 1 , we note that

(43) 1 + Ei 1 ( − x log ( 1 − t ) ) = 1 + ∑ n = 1 ∞ ( − x log ( 1 + t ) ) n n ! = e − x log ( 1 − t ) = ( 1 − t ) − x = ∑ n = 0 ∞ ⟨ x ⟩ n t n n ! .

Therefore, we have ⟨ x ⟩ n ( 1 ) = ⟨ x ⟩ n .

Theorem 9

For n ∈ N ∪ { 0 } and k ∈ Z

⟨ x ⟩ n ( k ) = ( − 1 ) n ( − x ) n ( k ) and ( x ) n ( k ) = ( − 1 ) n ⟨ − x ⟩ n ( k ) .

Proof

From (21) and (42), we observe that

(44) ∑ n = 0 ∞ ⟨ x ⟩ n ( k ) t n n ! = 1 + Ei k ( − x log ( 1 − t ) ) = 1 + ∑ n = 1 ∞ ( − x log ( 1 + ( − t ) ) ) n ( n − 1 ) ! n k = ∑ n = 0 ∞ ( − x ) n ( k ) ( − 1 ) n t n n ! = ∑ n = 0 ∞ ( − 1 ) n ( − x ) n ( k ) t n n !

and

(45) ∑ n = 0 ∞ ( x ) n ( k ) t n n ! = 1 + Ei k ( x log ( 1 + t ) ) = 1 + ∑ n = 1 ∞ ( − ( − x ) log ( 1 − ( − t ) ) ) n ( n − 1 ) ! n k = ∑ n = 0 ∞ ⟨ − x ⟩ n ( k ) ( − 1 ) n t n n ! = ∑ n = 0 ∞ ( − 1 ) n ⟨ − x ⟩ n ( k ) t n n ! .

By comparing the coefficients of both sides of (44) and (45), respectively, we have the desired result.□

When k = 1 , we note that ⟨ x ⟩ n = ( − 1 ) n ( − x ) n .

From Theorems 6, 7, 8, and 9, we observe the following identities, respectively.

Corollary 10

For n ∈ N and k ∈ Z , we have

  1. ⟨ − 1 ⟩ n ( k ) = ∑ m = 0 n ( − 1 ) n n m B m ( k ) ,

  2. ⟨ − 1 ⟩ n ( k ) = ( − 1 ) n ( B n ( k ) ( 1 ) − B n ( k ) ) ,

  3. ⟨ − 1 ⟩ n ( k ) = 1 2 ( − 1 ) n ∑ d = 1 n − 1 d n G d ( k ) + 2 G n ( k ) ,

where B n ( k ) are poly-Bernoulli numbers and G n ( k ) are poly-Genocchi numbers.

Next two theorems are relations of poly-rising factorial sequences and powers of x .

Theorem 11

For n ∈ N and k ∈ Z , we have

⟨ x ⟩ n ( k ) = ∑ d = 1 n ( − 1 ) n + d d k − 1 S 1 ( n , d ) x d .

Proof

First, from (5) and (14), we observe that

(46) 1 + Ei k ( − x log ( 1 − t ) ) = 1 + ∑ d = 1 ∞ ( − 1 ) d x d ( log ( 1 − t ) ) d ( d − 1 ) ! d k = 1 + ∑ d = 1 ∞ ( − 1 ) d x d d k − 1 1 d ! ( log ( 1 − t ) ) d = 1 + ∑ d = 1 ∞ ( − 1 ) d x d d k − 1 ∑ n = d ∞ S 1 ( n , d ) ( − 1 ) n t n n ! = 1 + ∑ n = 1 ∞ ∑ d = 1 n ( − 1 ) n + d d k − 1 S 1 ( n , d ) x d t n n ! .

Combining (42) with (46), we get what we want.□

Theorem 12

For n ∈ N and k ∈ Z , we have

n k − 1 ∑ d = 0 n ( − 1 ) n + d S 2 ( n , d ) ⟨ x ⟩ d ( k ) = x n .

Proof

Substituting t by 1 − e − t in (42), the left-hand side of (42) is

(47) 1 + Ei k ( x t ) = 1 + ∑ n = 1 ∞ ( x t ) n ( n − 1 ) ! n k = 1 + ∑ n = 1 ∞ x n n k − 1 t n n ! .

On the other hand, from (7), the right-hand side (42) is

(48) ∑ d = 0 ∞ ⟨ x ⟩ d ( k ) ( 1 − e − t ) d d ! = ∑ d = 0 ∞ ( − 1 ) d ⟨ x ⟩ d ( k ) ∑ n = d ∞ S 2 ( n , d ) ( − 1 ) n t n n ! = 1 + ∑ n = 1 ∞ ∑ d = 0 n ( − 1 ) n + d S 2 ( n , d ) ⟨ x ⟩ d ( k ) t n n ! .

From (47) and (48), we attain the desired result.□

To prove the next theorem, we observe that

(49) 1 1 − t x = ( 1 − t ) − x = e − x log ( 1 − t ) = ∑ d = 0 ∞ ( − 1 ) d x d ( log ( 1 − t ) ) d d ! = ∑ n = 0 ∞ ∑ d = 0 n ( − 1 ) n + d S 1 ( n , d ) x d t n n ! .

From (10) and (49), we obtain

(50) ∑ j = 0 n L ( n , j ) ( x ) j = ∑ d = 0 n ( − 1 ) n + d S 1 ( n , d ) x d .

From (6) and (50), we note that

(51) ∑ j = 0 n L ( n , j ) ( x ) j = ∑ d = 0 n ( − 1 ) n + d S 1 ( n , d ) x d = ∑ d = 0 n ( − 1 ) n + d S 1 ( n , d ) ∑ j = 0 d S 2 ( d , j ) ( x ) j = ∑ j = 0 n ∑ d = j n ( − 1 ) n + d S 1 ( n , d ) S 2 ( d , j ) ( x ) j .

By comparing the coefficients of both sides of (51), we obtain

(52) L ( n , j ) = ∑ d = j n ( − 1 ) n + d S 1 ( n , d ) S 2 ( d , j ) .

Theorem 13

For n ∈ N and k ∈ Z , we have

( x ) n ( k ) = ∑ v = 0 n ( − 1 ) v − n L ( n , v ) ⟨ x ⟩ v ( k ) and ⟨ x ⟩ n ( k ) = ∑ v = 0 n L ( n , v ) ( x ) v ( k ) .

Proof

By (52), Theorems 1 and 12, we obtain

(53) ( x ) n ( k ) = ∑ d = 1 n 1 d k − 1 S 1 ( n , d ) x d = ∑ d = 1 n 1 d k − 1 S 1 ( n , d ) ∑ v = 0 d d k − 1 ( − 1 ) v + d S 2 ( d , v ) ⟨ x ⟩ v ( k ) = ∑ v = 1 n ∑ d = 1 n ( − 1 ) v − n ( − 1 ) n + d S 1 ( n , d ) S 2 ( d , v ) ⟨ x ⟩ v ( k ) = ∑ v = 1 n ( − 1 ) v − n L ( n , v ) ⟨ x ⟩ v ( k ) .

By n ≥ 1 , L ( n , 0 ) = 0 , and (53), we obtain

(54) ( x ) n ( k ) = ∑ v = 0 n ( − 1 ) v − n L ( n , v ) ⟨ x ⟩ v ( k ) .

Thus, substituting x by − x in (54) and by using Theorem 9, we arrive at the result.□

In Theorem 13, when k = 1 , we note that

( x ) n = ∑ v = 0 n ( − 1 ) v − n L ( n , v ) ⟨ x ⟩ v and ⟨ x ⟩ n = ∑ v = 0 n L ( n , v ) ( x ) v .

Theorem 14

For n ∈ N and k ∈ Z , we have

⟨ x ⟩ 1 ( k − 1 ) = − ⟨ x ⟩ 1 ( k )

and

⟨ x ⟩ n ( k − 1 ) = ∑ j = 1 n n j ( j − 1 ) ! ( ⟨ x ⟩ n − j + 1 ( k ) − ( n − j ) ⟨ x ⟩ n − j ( k ) ) ( n ≥ 2 ) .

Proof

Differentiating with respect to t in (42), the left-hand side of (42) is

(55) ∂ ∂ t ( 1 + Ei k ( − x log ( 1 − t ) ) ) = ∑ n = 1 ∞ ( − x ) n ( log ( 1 − t ) ) n − 1 ( n − 1 ) ! n k − 1 − 1 1 − t = 1 ( t − 1 ) log ( 1 − t ) ∑ n = 1 ∞ ( − x log ( 1 − t ) ) n ( n − 1 ) ! n k − 1 = 1 ( t − 1 ) log ( 1 − t ) Ei k − 1 ( − x log ( 1 − t ) ) = 1 ( t − 1 ) log ( 1 − t ) ∑ n = 1 ∞ ⟨ x ⟩ n ( k − 1 ) t n n ! .

On the other hand, the right-hand side of (42) is

(56) ∂ ∂ t ∑ n = 0 ∞ ⟨ x ⟩ n ( k ) t n n ! = ∑ n = 1 ∞ ⟨ x ⟩ n ( k ) t n − 1 ( n − 1 ) ! = ∑ n = 0 ∞ ⟨ x ⟩ n + 1 ( k ) t n n ! .

By (56) and (57), we obtain

(57) ∑ n = 1 ∞ ⟨ x ⟩ n ( k − 1 ) t n n ! = ( t − 1 ) log ( 1 − t ) ∑ m = 0 ∞ ⟨ x ⟩ m + 1 ( k ) t m m ! = − ∑ j = 1 ∞ 1 j t j ∑ m = 0 ∞ ⟨ x ⟩ m + 1 ( k ) t m m ! + ∑ j = 1 ∞ 1 j t j ∑ m = 0 ∞ ⟨ x ⟩ m + 1 ( k ) t m + 1 m ! = − ∑ j = 1 ∞ j ! j t j j ! ∑ m = 0 ∞ ⟨ x ⟩ m + 1 ( k ) t m m ! + ∑ j = 1 ∞ j ! j t j j ! ∑ m = 1 ∞ ⟨ x ⟩ m ( k ) t m ( m − 1 ) ! = − ∑ n = 1 ∞ ∑ j = 1 n n j ( j − 1 ) ! ⟨ x ⟩ n − j + 1 ( k ) t n n ! + ∑ n = 2 ∞ ∑ j = 1 n n j ( j − 1 ) ! ( n − j ) ⟨ x ⟩ n − j ( k ) t n n ! = − ⟨ x ⟩ 1 ( k ) + ∑ n = 2 ∞ ∑ j = 1 n n j ( j − 1 ) ! ( ⟨ x ⟩ n − j + 1 ( k ) − ( n − j ) ⟨ x ⟩ n − j ( k ) ) t n n ! .

By comparing the coefficients of both sides of (58), we get what we want.□

Now, we consider the reciprocal of the poly-falling factorial sequences given by

(58) 1 1 + Ei k ( x log ( 1 + t ) ) = ∑ n = 0 ∞ R n ( k ) ( x ) t n n ! .

In (59), when k = 1 , by (1), (2), (11), and (12), we observe that

(59) ∑ n = 0 ∞ R n ( 1 ) ( x ) t n n ! = 1 1 + Ei 1 ( x log ( 1 + t ) ) = 1 e x log ( 1 + t ) = ( 1 + t ) − x = ∑ n = 0 ∞ ( − 1 ) n ⟨ x ⟩ n t n n ! = ∑ n = 0 ∞ ( − 1 ) n ∑ j = 0 n L ( n , j ) ( x ) j t n n ! .

From (59), we obtain

(60) R n ( 1 ) ( x ) = ( − 1 ) n ⟨ x ⟩ n = ( − x ) n = ( − 1 ) n ∑ j = 0 n L ( n , j ) ( x ) j .

To give an explicit reciprocal formula of power series of the poly-falling factorial sequence ( x ) n ( k ) , when k = 1 , we get the following theorem.

Theorem 15

Assume that

∑ n = 0 ∞ ( x ) n ( 1 ) t n n ! ∑ m = 0 ∞ R m ( 1 ) ( x ) t m m ! = 1 with ( x ) 0 ( k ) = 1 .

Then we have

R n ( 1 ) ( x ) = ( − 1 ) n ⟨ x ⟩ n = ( − x ) n = ∑ j = 0 n ( − 1 ) n L ( n , j ) ( x ) j .

The following theorem can be obtained simply by using Theorem 5 of [23].

Theorem 16

Assume that

∑ n = 0 ∞ ( x ) n ( k ) t n n ! ∑ m = 0 ∞ R m ( k ) ( x ) t m m ! = 1 with ( x ) 0 ( k ) = 1 .

Then we have

R 0 ( k ) ( x ) = 1 and R n ( k ) ( x ) = ∑ d = 1 n B n , d ( ( x ) 1 ( k ) , ( x ) 2 ( k ) , … , ( x ) n − d + 1 ( k ) ) ( − 1 ) d d ! = ∑ d = 1 n ∑ h 1 + 2 h 2 + ⋯ + n h n = n n ! h 1 ! h 2 ! ⋯ h n − d + 1 ! x 1 ! h 1 ∑ j = 1 2 1 j k − 1 S 1 ( 2 , j ) x j 2 ! h 2 ⋯ ∑ j = 1 n − d + 1 1 j k − 1 S 1 ( n − d + 1 , j ) x j ( n − d + 1 ) ! h n − d + 1 .

Proof

We observe that

(61) ∑ d = 0 ∞ ( x ) d ( k ) t d d ! ∑ m = 0 ∞ R m ( k ) ( x ) t m m ! = ∑ n = 0 ∞ ∑ d = 0 n n d ( x ) d ( k ) R n − d ( k ) ( x ) t n n ! = 1 .

From (61), we have

(62) R 0 ( k ) ( x ) = 1 and ∑ d = 0 n n d ( x ) d ( k ) R n − d ( k ) ( x ) = 0 , ( n ≥ 1 ) .

From (62), we obtain

(63) R n ( k ) ( x ) = − ∑ d = 1 n n d ( x ) d ( k ) R n − d ( k ) ( x ) .

When n = 1 , we have

(64) R 1 ( k ) ( x ) = − ( x ) 1 ( k ) R 0 ( k ) ( x ) = − ( x ) 1 ( k ) .

When n = 2 , we obtain

(65) R 2 ( k ) ( x ) = − 2 1 ( x ) 2 ( k ) R 1 ( k ) ( x ) − ( x ) 2 ( k ) = 2 ( ( x ) 1 ( k ) ) 2 − ( x ) 2 ( k ) .

On the other hand, from (19), we observe that

B 2 , 1 ( x 1 , x 2 ) = ∑ h 1 + h 2 = 1 h 1 + 2 h 2 = 2 2 ! h 1 ! h 2 ! x 1 1 ! h 1 x 2 2 ! h 2 = 2 ! 0 ! 1 ! x 1 1 ! 0 x 2 2 ! 1 = x 2

and

(66) B 2 , 2 ( x 1 ) = ∑ h 1 = 2 2 ! 2 ! x 1 1 ! 2 = x 1 2 .

By (66), we have

(67) ∑ d = 1 2 B 2 , d ( x 1 , x 2 , … , x 2 − k + 1 ) ( − 1 ) d d ! = − B 2 , 1 ( x 1 , x 2 ) + 2 ! B 2 , 2 ( x 1 ) = − x 2 + 2 ( x 1 ) 2 .

From (65) and (67), we obtain

(68) R 2 ( k ) ( x ) = − ( x ) 2 ( k ) + 2 ( ( x ) 1 ( k ) ) 2 = ∑ d = 1 n B 2 , d ( ( x ) 1 ( k ) , ( x ) 2 ( k ) ) ( − 1 ) d d ! .

When n = 3 ,

(69) R 3 ( k ) ( x ) = − 3 1 ( x ) 1 ( k ) R 2 ( k ) ( x ) − 3 2 ( x ) 2 ( k ) R 1 ( k ) ( x ) − ( x ) 3 ( k ) = − 3 ( x ) 1 ( k ) ( 2 ( ( x ) 1 ( k ) ) 2 − ( x ) 2 ( k ) ) − 3 ( x ) 2 ( k ) ( − ( x ) 1 ( k ) ) − ( x ) 3 ( k ) = − 6 ( ( x ) 1 ( k ) ) 3 + 6 ( x ) 2 ( k ) ( x ) 1 ( k ) − ( x ) 3 ( k ) .

On the other hand, from (19), we observe that

(70) B 3 , 1 ( x 1 , x 2 , x 3 ) = ∑ h 1 + h 2 + h 3 = 1 h 1 + 2 h 2 + 3 h 3 = 3 2 ! h 1 ! h 2 ! 3 ! h 1 ! h 2 ! h 3 ! x 1 1 ! h 1 x 2 2 ! h 2 x 3 3 ! h 3 = 3 ! 0 ! 0 ! 1 ! x 3 3 ! = x 3 , B 3 , 2 ( x 1 , x 2 ) = 3 x 1 x 2 , and B 3 , 3 ( x 1 ) = x 1 3 .

By (70), we have

(71) ∑ d = 1 3 B 3 , d ( x 1 , x 2 , … , x 3 − k + 1 ) ( − 1 ) d d ! = − B 3 , 1 ( x 1 , x 2 , x 3 ) + 2 ! B 3 , 2 ( x 1 , x 2 ) − 3 ! B 3 , 3 ( x 1 ) = − x 3 + 6 x 1 x 2 − 3 ! ( x 1 ) 3 .

From (69) and (71), we obtain

(72) R 3 ( k ) ( x ) = − 6 ( ( x ) 1 ( k ) ) 3 + 6 ( x ) 2 ( k ) ( x ) 1 ( k ) − ( x ) 3 ( k ) = ∑ d = 1 n B 3 , d ( ( x ) 1 ( k ) , ( x ) 2 ( k ) , ( x ) 3 ( k ) ) ( − 1 ) d d ! .

Continuing this process, we have

(73) R n ( k ) ( x ) = ∑ d = 1 n B n , d ( ( x ) 1 ( k ) , ( x ) 2 ( k ) , … , ( x ) n − d + 1 ( k ) ) ( − 1 ) d d ! .

By comparing (20), (73), and Theorem 1, we obtain the desired result.□

4 Conclusion

To conclude, note that in Section 2 we introduced poly-falling factorial sequences in terms of the polyexponential functions, reducing them to rising factorial if k = 1 . We demonstrated the following: the nth poly-falling factorial sequences of x were expressed in terms of the Stirling numbers of the first kind and the power of x ; the nth power of x were expressed in terms of the Stirling numbers of the second kind and poly-falling factorial sequences of x ; the poly-Bell polynomials were represented in terms of the poly-falling factorial sequences; the poly-falling poly-falling factorial sequences were represented in terms of the poly-Bell polynomials; recurrence formula; and the poly-falling factorial sequences were expressed in terms of the poly-Bernoulli numbers and poly-Genocchi numbers, respectively. In addition, in Section 3 we introduced poly-rising factorial sequences in terms of the polyexponential functions, reducing them to rising factorial if k = 1 . Thus, these relations between poly-falling factorial sequences and poly-rising factorial sequences were expressed as: the nth poly-rising factorial sequences of x were expressed in terms of the Stirling numbers of the first kind and the power of x ; the nth power of x were expressed in terms of the Stirling numbers of the second kind and poly-rising factorial sequences; the poly-falling factorial sequences are represented in terms of Lah numbers and the poly-rising factorial sequences; the poly-rising factorial sequences were represented in terms of Lah numbers and the poly-falling factorial sequences; a recurrence formula; and a reciprocal formula of the poly-falling factorial sequences.

In conclusion, in [8], one of the generalizations of Lah numbers, multi-Lah number was studied using the polylogarithms. From a similar point of view, we may generalize the results of the poly-falling factorial sequences and the poly-rising factorial sequences, respectively. There are various methods for studying special polynomials and numbers, including generating functions, combinatorial methods, umbral calculus, differential equations, and probability theory. The next academic project for our continuing research would be to examine the application of “poly” versions of certain special polynomials and numbers in the domains of physics, science, and engineering as well as mathematics of course.

Acknowledgements

The author would like to thank Jangjeon Institute for Mathematical Science for the support of this research.

  1. Funding information: This work was supported by the Basic Science Research Program, the National Research Foundation of Korea (NRF-2021R1F1A1050151).

  2. Conflict of interest: The author states no conflict of interest.

  3. Data availability statement: This paper does not use data and materials.

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Received: 2021-10-31
Accepted: 2021-12-01
Published Online: 2021-12-31

© 2021 Hye Kyung Kim, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  110. Epi-quasi normality
  111. Derivative and higher-order Cauchy integral formula of matrix functions
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  113. Solutions to a multi-phase model of sea ice growth
  114. Existence and simulation of positive solutions for m-point fractional differential equations with derivative terms
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  134. General decay rate for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping
  135. On hyponormality on a weighted annulus
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  137. Convergence results on Picard-Krasnoselskii hybrid iterative process in CAT(0) spaces
  138. Special Issue on Boundary Value Problems and their Applications on Biosciences and Engineering (Part I)
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  141. Existence of random attractors and the upper semicontinuity for small random perturbations of 2D Navier-Stokes equations with linear damping
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  143. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  144. On the mixed fractional quantum and Hadamard derivatives for impulsive boundary value problems
  145. The Lp dual Minkowski problem about 0 < p < 1 and q > 0
https://doi.org/10.1515/math-2021-0128
Keywords for this article
falling factorials; rising factorials; modified polyexponential functions; Stirling numbers of the first and second kind; poly-Bernoulli polynomials; poly-Genocchi polynomials
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