# Barnes-type Daehee polynomials

## Abstract

In this paper, we consider Barnes-type Daehee polynomials of the first kind and of the second kind. From the properties of the Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

MSC:05A15, 05A40, 11B68, 11B75, 65Q05.

## 1 Introduction

In this paper, we consider the polynomials $D n (x| a 1 ,…, a r )$ and $D ˆ n (x| a 1 ,…, a r )$ called the Barnes-type Daehee polynomials of the first kind and of the second kind, whose generating functions are given by

$∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) x = ∑ n = 0 ∞ D n (x| a 1 ,…, a r ) t n n ! ,$
(1)
$∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) x = ∑ n = 0 ∞ D ˆ n (x| a 1 ,…, a r ) t n n ! ,$
(2)

respectively, where $a 1 ,…, a r ≠0$. When $x=0$, $D n ( a 1 ,…, a r )= D n (0| a 1 ,…, a r )$ and $D ˆ n ( a 1 ,…, a r )= D ˆ n (0| a 1 ,…, a r )$ are called the Barnes-type Daehee numbers of the first kind and of the second kind, respectively.

Recall that the Daehee polynomials of the first kind and of the second kind of order r, denoted by $D n ( r ) (x)$ and $D ˆ n ( r ) (x)$, respectively, are given by the generating functions to be

$( ln ( 1 + t ) t ) r ( 1 + t ) x = ∑ n = 0 ∞ D n ( r ) ( x ) t n n ! , ( ( 1 + t ) ln ( 1 + t ) t ) r ( 1 + t ) x = ∑ n = 0 ∞ D ˆ n ( r ) ( x ) t n n ! ,$

respectively. If $a 1 =⋯= a r =1$, then $D n ( r ) (x)= D n (x| 1 , … , 1 ⏟ r )$ and $D ˆ n ( r ) (x)= D ˆ n (x| 1 , … , 1 ⏟ r )$. Daehee polynomials were defined by the second author  and have been investigated in .

In this paper, we consider Barnes-type Daehee polynomials of the first kind and of the second kind. From the properties of the Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

## 2 Umbral calculus

Let be the complex number field and let be the set of all formal power series in the variable t:

$F= { f ( t ) = ∑ k = 0 ∞ a k k ! t k | a k ∈ C } .$
(3)

Let $P=C[x]$ and let $P ∗$ be the vector space of all linear functionals on . $〈L|p(x)〉$ is the action of the linear functional L on the polynomial $p(x)$, and we recall that the vector space operations on $P ∗$ are defined by $〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉$, $〈cL|p(x)〉=c〈L|p(x)〉$, where c is a complex constant in . For $f(t)∈F$, let us define the linear functional on by setting

$〈 f ( t ) | x n 〉 = a n (n≥0).$
(4)

In particular,

$〈 t k | x n 〉 =n! δ n , k (n,k≥0),$
(5)

where $δ n , k$ is the Kronecker symbol.

For $f L (t)= ∑ k = 0 ∞ 〈 L | x k 〉 k ! t k$, we have $〈 f L (t)| x n 〉=〈L| x n 〉$. That is, $L= f L (t)$. The map $L↦ f L (t)$ is a vector space isomorphism from $P ∗$ onto . Henceforth, denotes both the algebra of formal power series in t and the vector space of all linear functionals on , and therefore an element $f(t)$ of will be thought of as both a formal power series and a linear functional. We call the umbral algebra and the umbral calculus is the study of umbral algebra. The order $O(f(t))$ of a power series $f(t)$ (≠0) is the smallest integer k for which the coefficient of $t k$ does not vanish. If $O(f(t))=1$, then $f(t)$ is called a delta series; if $O(f(t))=0$, then $f(t)$ is called an invertible series. For $f(t),g(t)∈F$ with $O(f(t))=1$ and $O(g(t))=0$, there exists a unique sequence $s n (x)$ ($deg s n (x)=n$) such that $〈g(t)f ( t ) k | s n (x)〉=n! δ n , k$, for $n,k≥0$. Such a sequence $s n (x)$ is called the Sheffer sequence for $(g(t),f(t))$, which is denoted by $s n (x)∼(g(t),f(t))$.

For $f(t),g(t)∈F$ and $p(x)∈P$, we have

$〈 f ( t ) g ( t ) | p ( x ) 〉 = 〈 f ( t ) | g ( t ) p ( x ) 〉 = 〈 g ( t ) | f ( t ) p ( x ) 〉$
(6)

and

$f(t)= ∑ k = 0 ∞ 〈 f ( t ) | x k 〉 t k k ! ,p(x)= ∑ k = 0 ∞ 〈 t k | p ( x ) 〉 x k k !$
(7)

, Theorem 2.2.5]. Thus, by (7), we get

$t k p(x)= p ( k ) (x)= d k p ( x ) d x k and e y t p(x)=p(x+y).$
(8)

Sheffer sequences are characterized by the generating function , Theorem 2.3.4].

Lemma 1 The sequence $s n (x)$ is Sheffer for $(g(t),f(t))$ if and only if

$1 g ( f ¯ ( t ) ) e y f ¯ ( t ) = ∑ k = 0 ∞ s k ( y ) k ! t k (y∈C),$

where $f ¯ (t)$ is the compositional inverse of $f(t)$.

For $s n (x)∼(g(t),f(t))$, we have the following equations , Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

$f(t) s n (x)=n s n − 1 (x)(n≥0),$
(9)
$s n (x)= ∑ j = 0 n 1 j ! 〈 g ( f ¯ ( t ) ) − 1 f ¯ ( t ) j | x n 〉 x j ,$
(10)
$s n (x+y)= ∑ j = 0 n ( n j ) s j (x) p n − j (y),$
(11)

where $p n (x)=g(t) s n (x)$.

Assume that $p n (x)∼(1,f(t))$ and $q n (x)∼(1,g(t))$. Then the transfer formula , Corollary 3.8.2] is given by

$q n (x)=x ( f ( t ) g ( t ) ) n x − 1 p n (x)(n≥1).$

For $s n (x)∼(g(t),f(t))$ and $r n (x)∼(h(t),l(t))$, assume that

$s n (x)= ∑ m = 0 n C n , m r m (x)(n≥0).$

Then we have , p.132]

$C n , m = 1 m ! 〈 h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n 〉 .$
(12)

## 3 Main results

We now note that $D n (x| a 1 ,…, a r )$ is the Sheffer sequence for

$g(t)= ∏ j = 1 r ( e a j t − 1 t ) andf(t)= e t −1.$

Therefore,

$D n (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( e a j t − 1 t ) , e t − 1 ) .$
(13)

$D ˆ n (x| a 1 ,…, a r )$ is the Sheffer sequence for

$g(t)= ∏ j = 1 r ( e a j t − 1 t e a j t ) andf(t)= e t −1.$

So,

$D ˆ n (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( e a j t − 1 t e a j t ) , e t − 1 ) .$
(14)

### 3.1 Explicit expressions

Recall that Barnes’ multiple Bernoulli polynomials $B n (x| a 1 ,…, a r )$ are defined by the generating function as

$t r ∏ j = 1 r ( e a j t − 1 ) e x t = ∑ n = 0 ∞ B n (x| a 1 ,…, a r ) t n n ! ,$
(15)

where $a 1 ,…, a r ≠0$ . Let $( n ) j =n(n−1)⋯(n−j+1)$ ($j≥1$) with $( n ) 0 =1$. The (signed) Stirling numbers of the first kind $S 1 (n,m)$ are defined by

$( x ) n = ∑ m = 0 n S 1 (n,m) x m .$

Theorem 1

$D n (x| a 1 ,…, a r )= ∑ m = 0 n S 1 (n,m) B m (x| a 1 ,…, a r )$
(16)
$D n ( x | a 1 , … , a r ) = ∑ j = 0 n ( ∑ l = j n ( n l ) S 1 ( l , j ) D n − l ( a 1 , … , a r ) ) x j$
(17)
$D n ( x | a 1 , … , a r ) = ∑ m = 0 n ( n m ) D n − m ( a 1 ,…, a r ) ( x ) m ,$
(18)
$D ˆ n (x| a 1 ,…, a r )= ∑ m = 0 n S 1 (n,m) B m (x+ a 1 +⋯+ a r | a 1 ,…, a r )$
(19)
$D ˆ n ( x | a 1 , … , a r ) = ∑ j = 0 n ( ∑ l = j n ( n l ) S 1 ( l , j ) D ˆ n − l ( a 1 , … , a r ) ) x j$
(20)
$D ˆ n ( x | a 1 , … , a r ) = ∑ m = 0 n ( n m ) D ˆ n − m ( a 1 ,…, a r ) ( x ) m .$
(21)

Proof Since

$∏ j = 1 r ( e a j t − 1 t ) D n (x| a 1 ,…, a r )∼ ( 1 , e t − 1 )$
(22)

and

$( x ) n ∼ ( 1 , e t − 1 ) ,$
(23)

we have

$D n ( x | a 1 , … , a r ) = ∏ j = 1 r ( t e a j t − 1 ) ( x ) n = ∑ m = 0 n S 1 ( n , m ) ∏ j = 1 r ( t e a j t − 1 ) x m = ∑ m = 0 n S 1 ( n , m ) B m ( x | a 1 , … , a r ) .$

So, we get (16).

Similarly, by

$∏ j = 1 r ( e a j t − 1 t e a j t ) D ˆ n (x| a 1 ,…, a r )∼ ( 1 , e t − 1 )$
(24)

and (23), we have

$D ˆ n ( x | a 1 , … , a r ) = ∏ j = 1 r ( t e a j t e a j t − 1 ) ( x ) n = ∑ m = 0 n S 1 ( n , m ) ∏ j = 1 r ( t e a j t e a j t − 1 ) x m = ∑ m = 0 n S 1 ( n , m ) e ( a 1 + ⋯ + a r ) t ∏ j = 1 r ( t e a j t − 1 ) x m = ∑ m = 0 n S 1 ( n , m ) B m ( x + a 1 + ⋯ + a r | a 1 , … , a r ) .$

Therefore, we get (19).

By (10) with (13), we get

$D n (x| a 1 ,…, a r )= ∑ j = 0 n 1 j ! 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) j | x n 〉 x j .$

Since

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) j | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ( ln ( 1 + t ) ) j x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | j ! ∑ l = j ∞ S 1 ( l , j ) t l l ! x n 〉 = j ! ∑ l = j n ( n l ) S 1 ( l , j ) 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − l 〉 = j ! ∑ l = j n ( n l ) S 1 ( l , j ) 〈 ∑ i = 0 ∞ D i ( a 1 , … , a r ) t i i ! | x n − l 〉 = j ! ∑ l = j n ( n l ) S 1 ( l , j ) D n − l ( a 1 , … , a r ) ,$

we obtain (17).

Similarly, by (10) with (14), we get

$D ˆ n (x| a 1 ,…, a r )= ∑ j = 0 n 1 j ! 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) j | x n 〉 x j .$

Since

$〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) j | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ( ln ( 1 + t ) ) j x n 〉 = j ! ∑ l = j n ( n l ) S 1 ( l , j ) 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − l 〉 = j ! ∑ l = j n ( n l ) S 1 ( l , j ) 〈 ∑ i = 0 ∞ D ˆ i ( a 1 , … , a r ) t i i ! | x n − l 〉 = j ! ∑ l = j n ( n l ) S 1 ( l , j ) D ˆ n − l ( a 1 , … , a r ) ,$

we obtain (20).

Next, we obtain

$D n ( y | a 1 , … , a r ) = 〈 ∑ i = 0 ∞ D i ( y | a 1 , … , a r ) t i i ! | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ∑ m = 0 ∞ ( y ) m t m m ! x n 〉 = ∑ m = 0 n ( y ) m ( n m ) 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − m 〉 = ∑ m = 0 n ( n m ) D n − m ( a 1 , … , a r ) ( y ) m .$

Thus, we get the identity (18).

Similarly,

$D ˆ n ( y | a 1 , … , a r ) = 〈 ∑ i = 0 ∞ D ˆ i ( y | a 1 , … , a r ) t i i ! | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ∑ m = 0 ∞ ( y ) m t m m ! x n 〉 = ∑ m = 0 n ( y ) m ( n m ) 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − m 〉 = ∑ m = 0 n ( n m ) D ˆ n − m ( a 1 , … , a r ) ( y ) m .$

Thus, we get the identity (21). □

### 3.2 Sheffer identity

Theorem 2

$D n (x+y| a 1 ,…, a r )= ∑ j = 0 n ( n j ) D j (x| a 1 ,…, a r ) ( y ) n − j ,$
(25)
$D ˆ n (x+y| a 1 ,…, a r )= ∑ j = 0 n ( n j ) D ˆ j (x| a 1 ,…, a r ) ( y ) n − j .$
(26)

Proof By (13) with

$p n ( x ) = ∏ j = 1 r ( e a j t − 1 t ) D n ( x | a 1 , … , a r ) = ( x ) n ∼ ( 1 , e t − 1 ) ,$

using (11), we have (25).

By (14) with

$p n ( x ) = ∏ j = 1 r ( e a j t − 1 t e a j t ) D ˆ n ( x | a 1 , … , a r ) = ( x ) n ∼ ( 1 , e t − 1 ) ,$

using (11), we have (26). □

### 3.3 Difference relations

Theorem 3

$D n (x+1| a 1 ,…, a r )− D n (x| a 1 ,…, a r )=n D n − 1 (x| a 1 ,…, a r ),$
(27)
$D ˆ n (x+1| a 1 ,…, a r )− D ˆ n (x| a 1 ,…, a r )=n D ˆ n − 1 (x| a 1 ,…, a r ).$
(28)

Proof By (9) with (13), we get

$( e t − 1 ) D n (x| a 1 ,…, a r )=n D n − 1 (x| a 1 ,…, a r ).$

By (8), we have (27).

Similarly, by (9) with (14), we get

$( e t − 1 ) D ˆ n (x| a 1 ,…, a r )=n D ˆ n − 1 (x| a 1 ,…, a r ).$

By (8), we have (28). □

### 3.4 Recurrence

Theorem 4

$D n + 1 ( x | a 1 , … , a r ) = x D n ( x − 1 | a 1 , … , a r ) D n + 1 ( x | a 1 , … , a r ) = − ∑ m = 0 n ( ∑ i = m n ∑ l = i n ∑ j = 1 r 1 i + 1 ( n l ) ( i + 1 m ) S 1 ( l , i ) D n + 1 ( x | a 1 , … , a r ) = × B i + 1 − m ( − a j ) i + 1 − m D n − l ( a 1 , … , a r ) ) ( x − 1 ) m ,$
(29)
$D ˆ n + 1 ( x | a 1 , … , a r ) = ( x + ∑ j = 1 r a j ) D ˆ n ( x − 1 | a 1 , … , a r ) D ˆ n + 1 ( x | a 1 , … , a r ) = − ∑ m = 0 n ( ∑ i = m n ∑ l = i n ∑ j = 1 r 1 i + 1 ( n l ) ( i + 1 m ) S 1 ( l , i ) D ˆ n + 1 ( x | a 1 , … , a r ) = × B i + 1 − m ( − a j ) i + 1 − m D ˆ n − l ( a 1 , … , a r ) ) ( x − 1 ) m ,$
(30)

where $B n$ is the nth ordinary Bernoulli number.

Proof By applying

$s n + 1 (x)= ( x − g ′ ( t ) g ( t ) ) 1 f ′ ( t ) s n (x)$
(31)

, Corollary 3.7.2] with (13), we get

$D n + 1 (x| a 1 ,…, a r )=x D n (x−1| a 1 ,…, a r )− e − t g ′ ( t ) g ( t ) D n (x| a 1 ,…, a r ).$

Now,

$g ′ ( t ) g ( t ) = ( ln g ( t ) ) ′ = ( ∑ j = 1 r ln ( e a j t − 1 ) − r ln t ) ′ = ∑ j = 1 r a j e a j t e a j t − 1 − r t = ∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) ( a j t e a j t − e a j t + 1 ) t ∏ j = 1 r ( e a j t − 1 ) .$

Since

$∑ j = 1 r a j t e a j t e a j t − 1 − r = ∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) ( a j t e a j t − e a j t + 1 ) ∏ j = 1 r ( e a j t − 1 ) = 1 2 ( ∑ j = 1 r a 1 ⋯ a j − 1 a j 2 a j + 1 ⋯ a r ) t r + 1 + ⋯ ( a 1 ⋯ a r ) t r + ⋯ = 1 2 ( ∑ j = 1 r a j ) t + ⋯$

is a series with order ≥1, by (17) we have

$D n + 1 ( x | a 1 , … , a r ) = x D n ( x − 1 | a 1 , … , a r ) − e − t ∑ j = 1 r a j t e a j t e a j t − 1 − r t D n ( x | a 1 , … , a r ) = x D n ( x − 1 | a 1 , … , a r ) − e − t ∑ j = 1 r a j t e a j t e a j t − 1 − r t ( ∑ i = 0 n ∑ l = i n ( n l ) S 1 ( l , i ) D n − l ( a 1 , … , a r ) x i ) = x D n ( x − 1 | a 1 , … , a r ) − ∑ i = 0 n ∑ l = i n ( n l ) S 1 ( l , i ) D n − l ( a 1 , … , a r ) e − t ( ∑ j = 1 r a j t e a j t e a j t − 1 − r ) x i + 1 i + 1 .$

Since

$e − t ( ∑ j = 1 r a j t e a j t e a j t − 1 − r ) x i + 1 = e − t ( ∑ j = 1 r ∑ m = 0 ∞ ( − 1 ) m B m a j m m ! t m − r ) x i + 1 = e − t ( ∑ j = 1 r ∑ m = 0 i + 1 ( i + 1 m ) B m ( − a j ) m x i + 1 − m − r x i + 1 ) = ∑ j = 1 r ∑ m = 1 i + 1 ( i + 1 m ) B m ( − a j ) m ( x − 1 ) i + 1 − m = ∑ j = 1 r ∑ m = 0 i ( i + 1 m ) B i + 1 − m ( − a j ) i + 1 − m ( x − 1 ) m ,$
(32)

we have

$D n + 1 ( x | a 1 , … , a r ) = x D n ( x − 1 | a 1 , … , a r ) − ∑ i = 0 n ∑ l = i n ∑ j = 1 r ∑ m = 0 i 1 i + 1 ( n l ) ( i + 1 m ) S 1 ( l , i ) × B i + 1 − m ( − a j ) i + 1 − m D n − l ( a 1 , … , a r ) ( x − 1 ) m = x D n ( x − 1 | a 1 , … , a r ) − ∑ m = 0 n ( ∑ i = m n ∑ l = i n ∑ j = 1 r 1 i + 1 ( n l ) ( i + 1 m ) S 1 ( l , i ) × B i + 1 − m ( − a j ) i + 1 − m D n − l ( a 1 , … , a r ) ) ( x − 1 ) m ,$

which is the identity (29).

Next, by applying (31) with (14), we get

$D ˆ n + 1 (x| a 1 ,…, a r )=x D ˆ n (x−1| a 1 ,…, a r )− e − t g ′ ( t ) g ( t ) D ˆ n (x| a 1 ,…, a r ).$

Now,

$g ′ ( t ) g ( t ) = ( ln g ( t ) ) ′ = ( ∑ j = 1 r ln ( e a j t − 1 ) − r ln t − ( ∑ j = 1 r a j ) t ) ′ = ∑ j = 1 r a j e a j t e a j t − 1 − r t − ∑ j = 1 r a j .$

By (20) we have

$D ˆ n + 1 ( x | a 1 , … , a r ) = ( x + ∑ j = 1 r a j ) D ˆ n ( x − 1 | a 1 , … , a r ) − e − t ∑ j = 1 r a j t e a j t e a j t − 1 − r t D ˆ n ( x | a 1 , … , a r ) = ( x + ∑ j = 1 r a j ) D ˆ n ( x − 1 | a 1 , … , a r ) − e − t ∑ j = 1 r a j t e a j t e a j t − 1 − r t ( ∑ i = 0 n ∑ l = i n ( n l ) S 1 ( l , i ) D ˆ n − l ( a 1 , … , a r ) x i ) = ( x + ∑ j = 1 r a j ) D ˆ n ( x − 1 | a 1 , … , a r ) − ∑ i = 0 n ∑ l = i n ( n l ) S 1 ( l , i ) D ˆ n − l ( a 1 , … , a r ) e − t ( ∑ j = 1 r a j t e a j t e a j t − 1 − r ) x i + 1 i + 1 .$

By (32), we have the identity (30). □

### 3.5 Differentiation

Theorem 5

$d d x D n (x| a 1 ,…, a r )=n! ∑ l = 0 n − 1 ( − 1 ) n − l − 1 l ! ( n − l ) D l (x| a 1 ,…, a r ),$
(33)
$d d x D ˆ n (x| a 1 ,…, a r )=n! ∑ l = 0 n − 1 ( − 1 ) n − l − 1 l ! ( n − l ) D ˆ l (x| a 1 ,…, a r ).$
(34)

Proof We shall use

$d d x s n (x)= ∑ l = 0 n − 1 ( n l ) 〈 f ¯ ( t ) | x n − l 〉 s l (x)$

(cf. , Theorem 2.3.12]). Since

$〈 f ¯ ( t ) | x n − l 〉 = 〈 ln ( 1 + t ) | x n − l 〉 = 〈 ∑ m = 1 ∞ ( − 1 ) m − 1 t m m | x n − l 〉 = ∑ m = 1 n − l ( − 1 ) m − 1 m 〈 t m | x n − l 〉 = ∑ m = 1 n − l ( − 1 ) m − 1 m ( n − l ) ! δ m , n − l = ( − 1 ) n − l − 1 ( n − l − 1 ) ! ,$

with (13), we have

$d d x D n ( x | a 1 , … , a r ) = ∑ l = 0 n − 1 ( n l ) ( − 1 ) n − l − 1 ( n − l − 1 ) ! D l ( x | a 1 , … , a r ) = n ! ∑ l = 0 n − 1 ( − 1 ) n − l − 1 l ! ( n − l ) D l ( x | a 1 , … , a r ) ,$

which is the identity (33). Similarly, with (14), we have the identity (34). □

### 3.6 More relations

The classical Cauchy numbers $c n$ are defined by

$t ln ( 1 + t ) = ∑ n = 0 ∞ c n t n n !$

(see e.g. [9, 10]).

Theorem 6

$D n ( x | a 1 , … , a r ) = x D n − 1 ( x − 1 | a 1 , … , a r ) D n ( x | a 1 , … , a r ) = + r n ∑ l = 0 n ( n l ) c l D n − l ( x − 1 | a 1 , … , a r ) D n ( x | a 1 , … , a r ) = − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D n − l ( x + a j − 1 | a 1 , … , a r , a j ) ,$
(35)
$D ˆ n ( x | a 1 , … , a r ) = ( x + ∑ j = 1 r a j ) D ˆ n − 1 ( x − 1 | a 1 , … , a r ) D ˆ n ( x | a 1 , … , a r ) = + r n ∑ l = 0 n ( n l ) c l D ˆ n − l ( x − 1 | a 1 , … , a r ) D ˆ n ( x | a 1 , … , a r ) = − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D ˆ n − l ( x − 1 | a 1 , … , a r , a j ) .$
(36)

Proof For $n≥1$, we have

$D n ( y | a 1 , … , a r ) = 〈 ∑ l = 0 ∞ D l ( y | a 1 , … , a r ) t l l ! | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y | x n 〉 = 〈 ∂ t ( ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y ) | x n − 1 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ∂ t ( 1 + t ) y ) | x n − 1 〉 + 〈 ( ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) ( 1 + t ) y | x n − 1 〉 = y D n − 1 ( y − 1 | a 1 , … , a r ) + 〈 ( ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) ( 1 + t ) y | x n − 1 〉 .$

Observe that

$∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) = ∑ j = 1 r ∏ i ≠ j ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) 1 1 + t ( ( 1 + t ) a j − 1 ) − ln ( 1 + t ) ( a j ( 1 + t ) a j − 1 ) ( ( 1 + t ) a j − 1 ) 2 = 1 1 + t ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ∑ j = 1 r ( 1 ln ( 1 + t ) − a j ( 1 + t ) a j ( 1 + t ) a j − 1 ) = 1 1 + t ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) t .$

Since

$∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) =− 1 2 ( ∑ j = 1 r a j ) t+⋯$

is a series with order ≥1, we have

$〈 ( ∂ t ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ) ( 1 + t ) y | x n − 1 〉 = 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) t x n − 1 〉 = 1 n ∑ j = 1 r 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) x n 〉 = r n 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | t ln ( 1 + t ) x n 〉 − 1 n ∑ j = 1 r a j 〈 ln ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y + a j − 1 | t ln ( 1 + t ) x n 〉 = r n 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | ∑ l = 0 ∞ c l t l l ! x n 〉 − 1 n ∑ j = 1 r a j 〈 ln ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y + a j − 1 | ∑ l = 0 ∞ c l t l l ! x n 〉 = r n ∑ l = 0 n c l ( n l ) 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | x n − l 〉 − 1 n ∑ j = 1 r a j ∑ l = 0 n c l ( n l ) 〈 ln ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y + a j − 1 | x n − l 〉 = r n ∑ l = 0 n ( n l ) c l D n − l ( y − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D n − l ( y + a j − 1 | a 1 , … , a r , a j ) .$

Therefore, we obtain

$D n ( x | a 1 , … , a r ) = x D n − 1 ( x − 1 | a 1 , … , a r ) + r n ∑ l = 0 n ( n l ) c l D n − l ( x − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D n − l ( x + a j − 1 | a 1 , … , a r , a j ) ,$

which is the identity (35).

Next, for $n≥1$ we have

$D ˆ n ( y | a 1 , … , a r ) = 〈 ∑ l = 0 ∞ D ˆ l ( y | a 1 , … , a r ) t l l ! | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y | x n 〉 = 〈 ∂ t ( ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) y ) | x n − 1 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ∂ t ( 1 + t ) y ) | x n − 1 〉 + 〈 ( ∂ t ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) ( 1 + t ) y | x n − 1 〉 = y D ˆ n − 1 ( y − 1 | a 1 , … , a r ) + 〈 ( ∂ t ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) ( 1 + t ) y | x n − 1 〉 .$

Observe that

$∂ t ∏ j = 1 r ( ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ) = ∂ t ( ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ∏ j = 1 r ( 1 + t ) a j ) = ( ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) ∏ j = 1 r ( 1 + t ) a j + ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ∂ t ∏ j = 1 r ( 1 + t ) a j ) = 1 1 + t ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) t + 1 1 + t ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ∑ j = 1 r a j .$

Thus, we have

$〈 ( ∂ t ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ) ( 1 + t ) y | x n − 1 〉 = 〈 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) t x n − 1 〉 + ( ∑ j = 1 r a j ) 〈 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | x n − 1 〉 = ( ∑ j = 1 r a j ) D ˆ n − 1 ( y − 1 | a 1 , … , a r ) + 1 n 〈 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) x n 〉 = ( ∑ j = 1 r a j ) D ˆ n − 1 ( y − 1 | a 1 , … , a r ) + r n 〈 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | t ln ( 1 + t ) x n 〉 − 1 n ∑ j = 1 r a j 〈 ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | t ln ( 1 + t ) x n 〉 = ( ∑ j = 1 r a j ) D ˆ n − 1 ( y − 1 | a 1 , … , a r ) + r n 〈 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | ∑ l = 0 ∞ c l t l l ! x n 〉 − 1 n ∑ j = 1 r a j 〈 ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | ∑ l = 0 ∞ c l t l l ! x n 〉 = ( ∑ j = 1 r a j ) D ˆ n − 1 ( y − 1 | a 1 , … , a r ) + r n ∑ l = 0 n c l ( n l ) 〈 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | x n − l 〉 − 1 n ∑ j = 1 r a j ∑ l = 0 n c l ( n l ) 〈 ( 1 + t ) a j ln ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ( 1 + t ) a i ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) y − 1 | x n − l 〉 = ( ∑ j = 1 r a j ) D ˆ n − 1 ( y − 1 | a 1 , … , a r ) + r n ∑ l = 0 n ( n l ) c l D ˆ n − l ( y − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D ˆ n − l ( y − 1 | a 1 , … , a r , a j ) .$

Therefore, we obtain

$D ˆ n ( x | a 1 , … , a r ) = ( x + ∑ j = 1 r a j ) D ˆ n − 1 ( x − 1 | a 1 , … , a r ) + r n ∑ l = 0 n ( n l ) c l D ˆ n − l ( x − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j c l D ˆ n − l ( x − 1 | a 1 , … , a r , a j ) ,$

which is the identity (36). □

### 3.7 Relations including the Stirling numbers of the first kind

Theorem 7 For $n−1≥m≥1$, we have

$∑ l = 0 n − m ( n l ) S 1 ( n − l , m ) D l ( a 1 , … , a r ) = ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D l ( − 1 | a 1 , … , a r ) + 1 n ∑ l = 0 n − m − 1 ( n l + 1 ) S 1 ( n − l − 1 , m ) × ( r ∑ i = 0 l + 1 ( l + 1 i ) c i D l + 1 − i ( − 1 | a 1 , … , a r ) − ∑ j = 1 r ∑ i = 0 l + 1 ( l + 1 i ) a j c i D l + 1 − i ( a j − 1 | a 1 , … , a r , a j ) ) ,$
(37)
$∑ l = 0 n − m ( n l ) S 1 ( n − l , m ) D ˆ l ( a 1 , … , a r ) = ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D ˆ l ( − 1 | a 1 , … , a r ) + 1 n ∑ l = 0 n − m − 1 ( n l + 1 ) S 1 ( n − l − 1 , m ) × ( r ∑ i = 0 l + 1 ( l + 1 i ) c i D ˆ l + 1 − i ( − 1 | a 1 , … , a r ) − ∑ j = 1 r ∑ i = 0 l + 1 ( l + 1 i ) a j c i D ˆ l + 1 − i ( − 1 | a 1 , … , a r , a j ) ) + ∑ l = 0 n − m − 1 ( n − 1 l ) S 1 ( n − l − 1 , m ) ∑ j = 1 r a j D ˆ l ( − 1 | a 1 , … , a r ) .$
(38)

Proof We shall compute

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) m | x n 〉$

in two different ways. On the one hand,

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) m | x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ( ln ( 1 + t ) ) m x n 〉 = 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ∑ l = 0 ∞ m ! ( l + m ) ! S 1 ( l + m , m ) t l + m x n 〉 = ∑ l = 0 n − m m ! ( l + m ) ! S 1 ( l + m , m ) ( n ) l + m 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − l − m 〉 = ∑ l = 0 n − m m ! ( n l + m ) S 1 ( l + m , m ) D n − l − m ( a 1 , … , a r ) = ∑ l = 0 n − m m ! ( n l ) S 1 ( n − l , m ) D l ( a 1 , … , a r ) .$

On the other hand,

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) m | x n 〉 = 〈 ∂ t ( ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( ln ( 1 + t ) ) m ) | x n − 1 〉 = 〈 ( ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) ( ln ( 1 + t ) ) m | x n − 1 〉 + 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ∂ t ( ( ln ( 1 + t ) ) m ) | x n − 1 〉 .$
(39)

The second term of (39) is equal to

$〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ∂ t ( ( ln ( 1 + t ) ) m ) | x n − 1 〉 = m 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) − 1 | ( ln ( 1 + t ) ) m − 1 x n − 1 〉 = m 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) − 1 | ∑ l = 0 n − m ( m − 1 ) ! ( l + m − 1 ) ! S 1 ( l + m − 1 , m − 1 ) t l + m − 1 x n − 1 〉 = m ∑ l = 0 n − m ( m − 1 ) ! ( l + m − 1 ) ! S 1 ( l + m − 1 , m − 1 ) ( n − 1 ) l + m − 1 × 〈 ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ( 1 + t ) − 1 | x n − l − m 〉 = m ! ∑ l = 0 n − m ( n − 1 l + m − 1 ) S 1 ( l + m − 1 , m − 1 ) D n − l − m ( − 1 | a 1 , … , a r ) = m ! ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) D l ( − 1 | a 1 , … , a r ) .$

The first term of (39) is equal to

$〈 ( ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) ) ( ln ( 1 + t ) ) m | x n − 1 〉 = 〈 ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ( ln ( 1 + t ) ) m x n − 1 〉 = 〈 ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | ∑ l = 0 n − m − 1 m ! ( l + m ) ! S 1 ( l + m , m ) t l + m x n − 1 〉 = ∑ l = 0 n − m − 1 m ! ( l + m ) ! S 1 ( l + m , m ) ( n − 1 ) l + m 〈 ∂ t ∏ j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j − 1 ) | x n − l − m − 1 〉 = ∑ l = 0 n − m − 1 m ! ( n − 1 l + m ) S 1 ( l + m , m ) × 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) − 1 | ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) t x n − l − m − 1 〉 = m ! ∑ l = 0 n − m − 1 1 n − l − m ( n − 1 l + m ) S 1 ( l + m , m ) × 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) − 1 | ∑ j = 1 r ( t ln ( 1 + t ) − a j t ( 1 + t ) a j ( 1 + t ) a j − 1 ) x n − l − m 〉 = m ! n ∑ l = 0 n − m − 1 ( n l + 1 ) S 1 ( n − 1 − l , m ) ( r 〈 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a i − 1 ) ( 1 + t ) − 1 | t ln ( 1 + t ) x l + 1 〉 − ( ∑ j = 1 r a j ) 〈 ln ( 1 + t ) ( 1 + t ) a j − 1 ( 1 + t ) a j − 1 ∏ i = 1 r ( ln ( 1 + t ) ( 1 + t ) a$