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Barnes-type Daehee of the second kind and poly-Cauchy of the second kind mixed-type polynomials

Abstract

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

MSC:05A19, 05A40, 11B68, 11B75.

1 Introduction

In this paper, we consider the polynomials D ˆ n ( k ) (x| a 1 ,, a r ) called the Barnes-type Daehee of the second kind and poly-Cauchy of the second kind mixed-type polynomials, whose generating function is given by

j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) x = n = 0 D ˆ n ( k ) (x| a 1 ,, a r ) t n n ! ,
(1)

where a 1 ,, a r 0. Here, Lif k (x) (kZ) is the polyfactorial function [1] defined by

Lif k (x)= m = 0 x m m ! ( m + 1 ) k .

When x=0, D ˆ n ( k ) ( a 1 ,, a r )= D ˆ n ( k ) (0| a 1 ,, a r ) is called the Barnes-type Daehee of the second kind and poly-Cauchy of the second kind mixed-type number.

Recall that the Barnes-type Daehee polynomials of the second kind, denoted by D ˆ n (x| a 1 ,, a r ), are given by the generating function to be

j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ( 1 + t ) x = n = 0 D ˆ n (x| a 1 ,, a r ) t n n ! .

If a 1 == a r =1, then D ˆ n ( r ) (x)= D ˆ n (x| 1 , , 1 r ) are the Daehee polynomials of the second kind of order r. Daehee polynomials were defined by the second author [2] and were investigated in [3, 4].

The poly-Cauchy polynomials of the second kind, denoted by c ˆ n ( k ) (x) [5, 6], are given by the generating function as follows:

Lif k ( ln ( 1 + t ) ) ( 1 + t ) x = n = 0 c ˆ n ( k ) (x) t n n ! .

In this paper, we introduce the mixed-type polynomials: Barnes-type Daehee polynomials of the second kind and poly-Cauchy polynomials of the second kind. From the properties of 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 } .
(2)

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)=cL|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 (n0).
(3)

In particular,

t k | x n =n! δ n , k (n,k0),
(4)

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 so 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,k0. 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 )
(5)

and

f ( t ) = k = 0 f ( t ) | x k t k k ! , p ( x ) = k = 0 t k | p ( x ) x k k !
(6)

[7, Theorem 2.2.5]. Thus, by (6), 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).
(7)

Sheffer sequences are characterized in the generating function [7, 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 (yC),

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

For s n (x)(g(t),f(t)), we have the following equations [7, Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

f(t) s n (x)=n s n 1 (x)(n0),
(8)
s n (x)= j = 0 n 1 j ! g ( f ¯ ( t ) ) 1 f ¯ ( t ) j | x n x j ,
(9)
s n (x+y)= j = 0 n ( n j ) s j (x) p n j (y),
(10)

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 [7, Corollary 3.8.2] is given by

q n (x)=x ( f ( t ) g ( t ) ) n x 1 p n (x)(n1).

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)(n0).

Then we have [7, p.132]

C n , m = 1 m ! h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n .
(11)

3 Main results

From definition (1), D ˆ n ( k ) (x| a 1 ,, a r ) is the Sheffer sequence for the pair

g(t)= j = 1 r ( e a j t 1 t e a j t ) 1 Lif k ( t ) andf(t)= e t 1.

So,

D ˆ n ( k ) (x| a 1 ,, a r ) ( j = 1 r ( e a j t 1 t e a j t ) 1 Lif k ( t ) , e t 1 ) .
(12)

3.1 Explicit expressions

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

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 ! ,
(13)

where a 1 ,, a r 0 [8, 9]. Let ( n ) j =n(n1)(nj+1) (j1) 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 ˆ n ( k ) (x| a 1 ,, a r )= m = 0 n l = 0 m S 1 (n,m) ( 1 ) m l ( m l ) ( m l + 1 ) k B l (x+ a 1 ++ a r | a 1 ,, a r )
(14)
= j = 0 n l = 0 n j i = 0 l ( n l ) ( l i ) S 1 (nl,j) c ˆ i ( k ) D ˆ l i ( a 1 ,, a r ) x j
(15)
= l = 0 n ( n l ) D ˆ n l ( a 1 ,, a r ) c ˆ l ( k ) (x)
(16)
= l = 0 n ( n l ) c ˆ n l ( k ) D ˆ l (x| a 1 ,, a r ).
(17)

Proof Since

j = 1 r ( e a j t 1 t e a j t ) 1 Lif k ( t ) D ˆ n ( k ) (x| a 1 ,, a r ) ( 1 , e t 1 )
(18)

and

( x ) n ( 1 , e t 1 ) ,
(19)

we have

D ˆ n ( k ) ( x | a 1 , , a r ) = j = 1 r ( t e a j t e a j t 1 ) Lif k ( t ) ( x ) n = m = 0 n S 1 ( n , m ) j = 1 r ( t e a j t e a j t 1 ) Lif k ( t ) x m = m = 0 n S 1 ( n , m ) j = 1 r ( t e a j t e a j t 1 ) l = 0 m ( 1 ) l t l l ! ( l + 1 ) k x m = m = 0 n S 1 ( n , m ) j = 1 r ( t e a j t e a j t 1 ) l = 0 m ( 1 ) l ( m ) l l ! ( l + 1 ) k x m l = m = 0 n S 1 ( n , m ) l = 0 m ( 1 ) l ( m ) l l ! ( l + 1 ) k j = 1 r ( t e a j t e a j t 1 ) x m l = m = 0 n l = 0 m S 1 ( n , m ) ( 1 ) l ( m l ) ( l + 1 ) k B m l ( x + a 1 + + a r | a 1 , , a r ) = m = 0 n l = 0 m S 1 ( n , m ) ( 1 ) m l ( m l ) ( m l + 1 ) k B l ( x + a 1 + + a r | a 1 , , a r ) .

So, we get (14).

By (9) with (12), we get

g ( f ¯ ( t ) ) 1 f ¯ ( t ) j | x n = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) j | x n = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) | l = 0 j ! ( l + j ) ! S 1 ( l + j , j ) t l + j x n = l = 0 n j j ! ( l + j ) ! S 1 ( l + j , j ) ( n ) l + j j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) | x n l j = l = 0 n j j ! ( n l + j ) S 1 ( l + j , j ) j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) | i = 0 n l j c ˆ i ( k ) t i i ! x n l j = l = 0 n j j ! ( n l + j ) S 1 ( l + j , j ) i = 0 n l j c ˆ i ( k ) ( n l j ) i i ! j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) | x n l j i = l = 0 n j j ! ( n l + j ) S 1 ( l + j , j ) i = 0 n l j c ˆ i ( k ) ( n l j ) i i ! m = 0 D ˆ m ( a 1 , , a r ) t m m ! | x n l j i = l = 0 n j i = 0 n l j j ! ( n l + j ) ( n l j i ) S 1 ( l + j , j ) c ˆ i ( k ) D ˆ n l j i ( a 1 , , a r ) = l = 0 n j i = 0 l j ! ( n l ) ( l i ) S 1 ( n l , j ) c ˆ i ( k ) D ˆ l i ( a 1 , , a r ) .

Thus, we obtain

D ˆ n ( k ) (x| a 1 ,, a r )= j = 0 n l = 0 n j i = 0 l ( n l ) ( l i ) S 1 (nl,j) c ˆ i ( k ) D ˆ l i ( a 1 ,, a r ) x j ,

which is identity (15).

Next,

D ˆ n ( k ) ( y | a 1 , , a r ) = i = 0 D ˆ i ( k ) ( y | a 1 , , a r ) t i i ! | x n = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) | Lif k ( ln ( 1 + t ) ) ( 1 + t ) y x n = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) | l = 0 n c ˆ l ( k ) ( y ) t l l ! x n = l = 0 n ( n l ) c ˆ l ( k ) ( y ) j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) | x n l = l = 0 n ( n l ) c ˆ l ( k ) ( y ) i = 0 D ˆ i ( a 1 , , a r ) t i i ! | x n l = l = 0 n ( n l ) c ˆ l ( k ) ( y ) D ˆ n l ( a 1 , , a r ) .

Thus, we obtain (16).

Finally, we obtain that

D ˆ n ( k ) ( y | a 1 , , a r ) = i = 0 D ˆ i ( k ) ( y | a 1 , , a r ) t i i ! | x n = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y | x n = Lif k ( ln ( 1 + t ) ) | j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ( 1 + t ) y x n = Lif k ( ln ( 1 + t ) ) | l = 0 n D ˆ l ( y | a 1 , , a r ) t l l ! x n = l = 0 n D ˆ l ( y | a 1 , , a r ) ( n l ) Lif k ( ln ( 1 + t ) ) | x n l = l = 0 n D ˆ l ( y | a 1 , , a r ) ( n l ) i = 0 c ˆ i ( k ) t i i ! | x n l = l = 0 n ( n l ) D ˆ l ( y | a 1 , , a r ) c ˆ n l ( k ) .

Thus, we get identity (17). □

3.2 Sheffer identity

Theorem 2

D ˆ n ( k ) (x+y| a 1 ,, a r )= j = 0 n ( n j ) D ˆ j ( k ) (x| a 1 ,, a r ) ( y ) n j .
(20)

Proof By (12) with

p n ( x ) = j = 1 r ( e a j t 1 t e a j t ) 1 Lif k ( t ) D ˆ n ( k ) ( x | a 1 , , a r ) = ( x ) n ( 1 , e t 1 ) ,

using (10), we have (20). □

3.3 Difference relations

Theorem 3

D ˆ n ( k ) (x+1| a 1 ,, a r ) D ˆ n ( k ) (x| a 1 ,, a r )=n D ˆ n 1 ( k ) (x| a 1 ,, a r ).
(21)

Proof By (8) with (12), we get

( e t 1 ) D ˆ n ( k ) (x| a 1 ,, a r )=n D ˆ n 1 ( k ) (x| a 1 ,, a r ).

By (7), we have (21). □

3.4 Recurrence

Theorem 4

D ˆ n + 1 ( k ) ( x | a 1 , , a r ) = x D ˆ n ( k ) ( x 1 | a 1 , , a r ) m = 0 n j = 1 r l = 0 m i = 0 l ( 1 ) l i ( m + 1 l ) ( l i ) ( m + 1 ) ( l i + 1 ) k S 1 ( n , m ) × ( a j ) m + 1 l B m + 1 l B i ( x + a 1 + + a r 1 | a 1 , , a r ) + j = 1 r a j m = 0 n i = 0 m ( 1 ) m i ( m i ) ( m i + 1 ) k S 1 ( n , m ) × B i ( x + a 1 + + a r 1 | a 1 , , a r ) m = 0 n l = 0 m + 1 ( 1 ) m + 1 l ( m l ) ( m + 2 l ) k S 1 ( n , m ) × B l ( x + a 1 + + a r 1 | a 1 , , a r ) ,
(22)

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)
(23)

[7, Corollary 3.7.2] with (12), we get

D ˆ n + 1 ( k ) (x| a 1 ,, a r )=x D ˆ n ( k ) (x1| a 1 ,, a r ) e t g ( t ) g ( t ) D ˆ n ( k ) (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 ln Lif k ( t ) ) = j = 1 r a j e a j t e a j t 1 r t j = 1 r a j + Lif k ( t ) Lif k ( 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 ) j = 1 r a j + Lif k ( t ) Lif k ( t ) .

Observe that

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. Since

D ˆ n ( k ) ( x | a 1 , , a r ) = j = 1 r ( t e a j t e a j t 1 ) Lif k ( t ) ( x ) n = m = 0 n S 1 ( n , m ) j = 1 r ( t e a j t e a j t 1 ) Lif k ( t ) x m ,

we have

g ( t ) g ( t ) D ˆ n ( k ) ( x | a 1 , , a r ) = m = 0 n S 1 ( n , m ) Lif k ( t ) ( j = 1 r t e a j t e a j t 1 ) g ( t ) g ( t ) x m = m = 0 n S 1 ( n , m ) Lif k ( t ) ( j = 1 r t e a j t e a j t 1 ) × 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 ) x m j = 1 r a j m = 0 n S 1 ( n , m ) Lif k ( t ) ( j = 1 r t e a j t e a j t 1 ) x m + m = 0 n S 1 ( n , m ) ( j = 1 r t e a j t e a j t 1 ) Lif k ( t ) x m .
(24)

Since

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 ) x m = 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 ) x m + 1 m + 1 = 1 m + 1 j = 1 r ( a j t e a j t e a j t 1 1 ) x m + 1 = 1 m + 1 j = 1 r ( l = 0 ( a j ) l B l l ! t l 1 ) x m + 1 = 1 m + 1 j = 1 r l = 0 m ( m + 1 l ) ( a j ) m + 1 l B m + 1 l x l ,

the first term in (24) is

m = 0 n S 1 ( n , m ) m + 1 j = 1 r l = 0 m ( m + 1 l ) ( a j ) m + 1 l B m + 1 l Lif k ( t ) ( j = 1 r t e a j t e a j t 1 ) x l = m = 0 n S 1 ( n , m ) m + 1 j = 1 r l = 0 m ( m + 1 l ) ( a j ) m + 1 l B m + 1 l × i = 0 l ( 1 ) i t i i ! ( i + 1 ) k B l ( x + a 1 + + a r | a 1 , , a r ) = m = 0 n j = 1 r l = 0 m i = 0 l ( 1 ) l i ( m + 1 l ) ( l i ) ( m + 1 ) ( l i + 1 ) k S 1 ( n , m ) ( a j ) m + 1 l B m + 1 l × B i ( x + a 1 + + a r | a 1 , , a r ) .

Since

Lif k 1 (t) Lif k (t)= ( 1 2 k 1 2 k 1 ) t+,
(25)

the second term in (24) is

j = 1 r a j m = 0 n S 1 ( n , m ) Lif k ( t ) B m ( x + a 1 + + a r | a 1 , , a r ) = j = 1 r a j m = 0 n S 1 ( n , m ) i = 0 m ( 1 ) i t i i ! ( i + 1 ) k B m ( x + a 1 + + a r | a 1 , , a r ) = j = 1 r a j m = 0 n i = 0 m ( 1 ) m i ( m i ) ( m i + 1 ) k S 1 ( n , m ) B i ( x + a 1 + + a r | a 1 , , a r ) .

The third term in (24) is

m = 0 n S 1 ( n , m ) Lif k 1 ( t ) Lif k ( t ) t B m ( x + a 1 + + a r | a 1 , , a r ) = m = 0 n S 1 ( n , m ) ( Lif k 1 ( t ) Lif k ( t ) ) B m + 1 ( x + a 1 + + a r | a 1 , , a r ) m + 1 = m = 0 n S 1 ( n , m ) m + 1 ( l = 0 m + 1 ( 1 ) l t l l ! ( l + 1 ) k 1 B m + 1 ( x + a 1 + + a r | a 1 , , a r ) l = 0 m + 1 ( 1 ) l t l l ! ( l + 1 ) k B m + 1 ( x + a 1 + + a r | a 1 , , a r ) ) = m = 0 n l = 0 m + 1 ( 1 ) m + 1 l ( m l ) ( m + 2 l ) k S 1 ( n , m ) B l ( x + a 1 + + a r | a 1 , , a r ) .

Thus we have identity (22). □

3.5 Differentiation

Theorem 5

d d x D ˆ n ( k ) (x| a 1 ,, a r )=n! l = 0 n 1 ( 1 ) n l 1 l ! ( n l ) D ˆ l ( k ) (x| a 1 ,, a r ).
(26)

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. [7, 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 = ( 1 ) n l 1 ( n l 1 ) ! ,

with (12), we have

d d x D ˆ n ( k ) ( x | a 1 , , a r ) = l = 0 n 1 ( n l ) ( 1 ) n l 1 ( n l 1 ) ! D ˆ l ( k ) ( x | a 1 , , a r ) = n ! l = 0 n 1 ( 1 ) n l 1 l ! ( n l ) D ˆ l ( k ) ( x | a 1 , , a r ) ,

which is identity (26). □

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. [1, 10]).

Theorem 6

D ˆ n ( k ) ( x | a 1 , , a r ) = ( x + j = 1 r a j ) D ˆ n 1 ( k ) ( x 1 | a 1 , , a r ) + 1 n l = 0 n ( n l ) c l D ˆ n l ( k 1 ) ( x 1 | a 1 , , a r ) + r 1 n l = 0 n ( n l ) c l D ˆ n l ( k ) ( x 1 | a 1 , , a r ) 1 n j = 1 r l = 0 n ( n l ) a j c l D ˆ n l ( k ) ( x 1 | a 1 , , a r , a j ) .
(27)

Proof For n1, we have

D ˆ n ( k ) ( y | a 1 , , a r ) = l = 0 D ˆ l ( k ) ( y | a 1 , , a r ) t l l ! | x n = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y | x n = t ( j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y ) | x n 1 = ( t j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y | x n 1 + j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ( t Lif k ( ln ( 1 + t ) ) ) ( 1 + t ) y | x n 1 + j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( t ( 1 + t ) y ) | x n 1 .

The third term is

y j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y 1 | x n 1 = y D n 1 ( k ) ( y 1 | a 1 , , a r ) .

By (25), the second term is

j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k 1 ( ln ( 1 + t ) ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) ln ( 1 + t ) ( 1 + t ) y | x n 1 = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k 1 ( ln ( 1 + t ) ) Lif k ( ln ( 1 + t ) ) t × ( 1 + t ) y 1 | t ln ( 1 + t ) x n 1 = 1 n l = 0 n 1 ( n l ) c l ( j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k 1 ( ln ( 1 + t ) ) ( 1 + t ) y 1 | x n l j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y 1 | x n l ) = 1 n l = 0 n 1 ( n l ) c l ( D ˆ n l ( k 1 ) ( y 1 | a 1 , , a r ) D ˆ n l ( k ) ( y 1 | a 1 , , a r ) ) .

Observe that

t j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) = 1 1 + t j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) j = 1 r ( t ln ( 1 + t ) a j t ( 1 + t ) a j ( 1 + t ) a j 1 ) t + ( j = 1 r a j ) 1 1 + t j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ,

with

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+

a series with order ≥1.

Now, the first term is

j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 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 j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) y 1 | x n 1 = ( j = 1 r a j ) D ˆ n 1 ( k ) ( y 1 | a 1 , , a r ) + r n j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 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 + t ) a j 1 j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) × ( 1 + t ) y 1 | t ln ( 1 + t ) x n = ( j = 1 r a j ) D ˆ n 1 ( k ) ( y 1 | a 1 , , a r ) + r n l = 0 n ( n l ) c l D ˆ n l ( k ) ( y 1 | a 1 , , a r ) 1 n i = 1 r a j l = 0 n ( n l ) c l D ˆ n l ( k ) ( y 1 | a 1 , , a r , a j ) .

Altogether, we obtain

D ˆ n ( k ) ( x | a 1 , , a r ) = ( x + j = 1 r a j ) D ˆ n 1 ( k ) ( x 1 | a 1 , , a r ) + 1 n l = 0 n 1 ( n l ) c l ( D ˆ n l ( k 1 ) ( x 1 | a 1 , , a r ) D ˆ n l ( k ) ( x 1 | a 1 , , a r ) ) + r n l = 0 n ( n l ) c l D ˆ n l ( k ) ( x 1 | a 1 , , a r ) 1 n j = 1 r l = 0 n ( n l ) a j c l D ˆ n l ( k ) ( x 1 | a 1 , , a r , a j ) ,

from which identity (27) follows. □

3.7 A relation including the Stirling numbers of the first kind

Theorem 7 For n1m1, we have

m l = 0 n m ( n l ) S 1 ( n l , m ) D ˆ l ( k ) ( a 1 , , a r ) = m j = 1 r a j l = 0 n m 1 ( n 1 l ) S 1 ( n l 1 , m ) D ˆ l ( k ) ( 1 | a 1 , , a r ) + m r n l = 0 n m i = 0 l ( n l ) ( l i ) S 1 ( n l , m ) c l i D ˆ i ( k ) ( 1 | a 1 , , a r ) m n l = 0 n m i = 0 l j = 1 r ( n l ) ( l i ) S 1 ( n l , m ) a j c l i D ˆ i ( k ) ( 1 | a 1 , , a r , a j ) + l = 0 n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k 1 ) ( 1 | a 1 , , a r ) + ( m 1 ) l = 0 n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k ) ( 1 | a 1 , , a r ) .
(28)

Proof We shall compute

j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n

in two different ways. On the one hand, it is

j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) | ( ln ( 1 + t ) ) m x n = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) | l = 0 m ! ( l + m ) ! S 1 ( l + m , m ) t l + m x n = l = 0 n m m ! ( n l ) S 1 ( n l , m ) D ˆ l ( k ) ( a 1 , , a r ) .

On the other hand, it is

t ( j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m ) | x n 1 = ( t j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n 1 + j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ( t Lif k ( ln ( 1 + t ) ) ) ( ln ( 1 + t ) ) m | x n 1 + j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( t ( ln ( 1 + t ) ) m ) | x n 1 .
(29)

The third term of (29) is equal to

m j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) 1 | ( ln ( 1 + t ) ) m 1 x n 1 = m j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 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 ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k ) ( 1 | a 1 , , a r ) .

The second term of (29) is equal to

j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) ( Lif k 1 ( ln ( 1 + t ) ) Lif k 1 ( ln ( 1 + t ) ) ( 1 + t ) ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n 1 = j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k 1 ( ln ( 1 + t ) ) ( 1 + t ) 1 | ( ln ( 1 + t ) ) m 1 x n 1 j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) 1 | ( ln ( 1 + t ) ) m 1 x n 1 = ( m 1 ) ! l = 0 n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k 1 ) ( 1 | a 1 , , a r ) ( m 1 ) ! l = 0 n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k ) ( 1 | a 1 , , a r ) .

The first term of (29) is equal to

1 n j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) 1 × j = 1 r ( t ln ( 1 + t ) a j t ( 1 + t ) a j ( 1 + t ) a j 1 ) | ( ln ( 1 + t ) ) m x n + j = 1 r a j j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) 1 | ( ln ( 1 + t ) ) m x n 1 = m ! j = 1 r a j l = 0 n m 1 ( n 1 l + m ) S 1 ( l + m , m ) D ˆ n l m 1 ( k ) ( 1 | a 1 , , a r ) + m ! n l = 0 n m ( n l + m ) S 1 ( l + m , m ) × ( r j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( 1 + t ) 1 | t ln ( 1 + t ) x n l m j = 1 r a j ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) × Lif k ( ln ( 1 + t ) ) ( 1 + t ) 1 | t ln ( 1 + t ) x n l m ) = m ! j = 1 r a j l = 0 n m 1 ( n 1 l ) S 1 ( n l 1 , m ) D ˆ l ( k ) ( 1 | a 1 , , a r ) + m ! n l = 0 n m ( n l ) S 1 ( n l , m ) ( r i = 0 l ( l i ) c i D ˆ l i ( k ) ( 1 | a 1 , , a r ) j = 1 r i = 0 l ( l i ) a j c i D ˆ l i ( k ) ( 1 | a 1 , , a r , a j ) ) .

Therefore, we get, for n1m1,

m ! l = 0 n m ( n l ) S 1 ( n l , m ) D ˆ l ( k ) ( a 1 , , a r ) = m ! j = 1 r a j l = 0 n m 1 ( n 1 l ) S 1 ( n l 1 , m ) D ˆ l ( k ) ( 1 | a 1 , , a r ) + m ! r n l = 0 n m i = 0 l ( n l ) ( l i ) S 1 ( n l , m ) c i D ˆ l i ( k ) ( 1 | a 1 , , a r ) m ! 1 n l = 0 n m i = 0 l j = 1 r ( n l ) ( l i ) S 1 ( n l , m ) a j c i D ˆ l i ( k ) ( 1 | a 1 , , a r , a j ) + ( m 1 ) ! l = 0 n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k 1 ) ( 1 | a 1 , , a r ) ( m 1 ) ! l = 0 n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k ) ( 1 | a 1 , , a r ) + m ! l = 0 n m ( n 1 l ) S 1 ( n l 1 , m 1 ) D ˆ l ( k ) ( 1 | a 1 , , a r ) .

Dividing both sides by (m1)!, we get (28). □

3.8 A relation with the falling factorials

Theorem 8

D ˆ n ( k ) (x| a 1 ,, a r )= m = 0 n ( n m ) D ˆ n m ( k ) ( a 1 ,, a r ) ( x ) m .
(30)

Proof For (12) and (19), assume that D ˆ n ( k ) (x| a 1 ,, a r )= m = 0 n C n , m ( x ) m . By (11), we have

C n , m = 1 m ! j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) | t m x n = ( n m ) D ˆ n m ( k ) ( a 1 , , a r ) .

Thus, we get identity (30). □

3.9 A relation with higher-order Frobenius-Euler polynomials

For λC with λ1, the Frobenius-Euler polynomials of order r, H n ( r ) (x|λ) are defined by the generating function

( 1 λ e t λ ) r e x t = n = 0 H n ( r ) (x|λ) t n n !

(see e.g. [11]).

Theorem 9

D ˆ n ( k ) ( x | a 1 , , a r ) = m = 0 n ( j = 0 n m l = 0 n m j ( s j ) ( n j l ) ( n ) j × ( 1 λ ) j S 1 ( n j l , m ) D ˆ l ( k ) ( a 1 , , a r ) ) H m ( s ) ( x | λ ) .
(31)

Proof For (12) and

H n ( s ) (x|λ) ( ( e t λ 1 λ ) s , t ) ,
(32)

assume that D ˆ n ( k ) (x| a 1 ,, a r )= m = 0 n C n , m H m ( s ) (x|λ). By (11), similarly to the proof of (28), we have

C n , m = 1 m ! ( 1 λ ) s j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m ( 1 λ + t ) s | x n = 1 m ! ( 1 λ ) s j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) × ( ln ( 1 + t ) ) m | i = 0 min { s , n } ( s i ) ( 1 λ ) s i t i x n = 1 m ! ( 1 λ ) s i = 0 n m ( s i ) ( 1 λ ) s i ( n ) i × j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | x n i = 1 m ! ( 1 λ ) s i = 0 n m ( s i ) ( 1 λ ) s i ( n ) i l = 0 n m i m ! ( n i l ) S 1 ( n i l , m ) D ˆ l ( k ) ( a 1 , , a r ) = i = 0 n m l = 0 n m i ( s i ) ( n i l ) ( n ) i ( 1 λ ) i S 1 ( n i l , m ) D ˆ l ( k ) ( a 1 , , a r ) .

Thus, we get identity (31). □

3.10 A relation with higher-order Bernoulli polynomials

Bernoulli polynomials B n ( r ) (x) of order r are defined by

( t e t 1 ) r e x t = n = 0 B n ( r ) ( x ) n ! t n

(see e.g. [7, Section 2.2]). In addition, the Cauchy numbers of the first kind C n ( r ) of order r are defined by

( t ln ( 1 + t ) ) r = n = 0 C n ( r ) n ! t n

(see e.g. [[12], (2.1)], [[13], (6)]).

Theorem 10

D ˆ n ( k ) ( x | a 1 , , a r ) = m = 0 n ( i = 0 n m l = 0 n m i ( n i ) ( n i l ) C i ( s ) S 1 ( n i l , m ) D ˆ l ( k ) ( a 1 , , a r ) ) B m ( s ) ( x ) .
(33)

Proof For (12) and

B n ( s ) (x) ( ( e t 1 t ) s , t ) ,
(34)

assume that D ˆ n ( k ) (x| a 1 ,, a r )= m = 0 n C n , m B m ( s ) (x). By (11), similarly to the proof of (28), we have

C n , m = 1 m ! j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | ( t ln ( 1 + t ) ) s x n = 1 m ! j = 1 r ( ln ( 1 + t ) ( 1 + t ) a j ( 1 + t ) a j 1 ) Lif k ( ln ( 1 + t ) ) ( ln ( 1 + t ) ) m | i = 0 C