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Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials

Abstract

In this paper, we consider Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials. From the properties of 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 T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r ) whose generating function is given by

j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t e x t = n = 0 T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r ) t n n ! ,
(1)

where r Z > , kZ, a 1 ,, a r 0, λ 1 ,, λ r 1 and

Li k (x)= m = 1 x m m k

is the k th polylogarithm function. T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r ) will be called Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials. When x=0, T n ( r , k ) ( a 1 ,, a r ; λ 1 ,, λ r )= T n ( r , k ) (0| a 1 ,, a r ; λ 1 ,, λ r ) will be called Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type numbers.

Recall that, for every integer k, the poly-Bernoulli polynomials B n ( k ) (x) are defined by the generating function as follows:

Li k ( 1 e t ) 1 e t e x t = n = 0 B n ( k ) (x) t n n !
(2)

([1], cf. [2]). Also, as a natural generalization of higher-order Frobenius-Euler polynomials, Barnes’ multiple Frobenius-Euler polynomials H n ( r ) (x| a 1 ,, a r ; λ 1 ,, λ r ) are defined by the generating function as follows:

j = 1 r ( 1 λ j e a j t λ j ) e x t = n = 0 H n ( r ) (x| a 1 ,, a r ; λ 1 ,, λ r ) t n n ! ,
(3)

where a 1 ,, a r 0. Note that 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., [3]).

In this paper, we consider Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials. 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 } .
(4)

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).
(5)

In particular,

t k | x n =n! δ n , k (n,k0),
(6)

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 )
(7)

and

f(t)= k = 0 f ( t ) | x k t k k ! ,p(x)= k = 0 t k | p ( x ) x k k !
(8)

[[4], Theorem 2.2.5]. Thus, by (8), 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).
(9)

Sheffer sequences are characterized in the generating function [[4], 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 [[4], Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

f(t) s n (x)=n s n 1 (x)(n0),
(10)
s n (x)= j = 0 n 1 j ! g ( f ¯ ( t ) ) 1 f ¯ ( t ) j | x n x j ,
(11)
s n (x+y)= j = 0 n ( n j ) s j (x) p n j (y),
(12)

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 [[4], 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 [[4], p.132]

C n , m = 1 m ! h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n .
(13)

3 Main results

We now note that B n ( k ) (x), H n ( r ) (x| a 1 ,, a r ; λ 1 ,, λ r ) and T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r ) are the Appell sequences for

g k ( t ) = 1 e t Li k ( 1 e t ) , g r ( t ) = j = 1 r ( e a j t λ j 1 λ j ) , g r , k ( t ) = j = 1 r ( e a j t λ j 1 λ j ) 1 e t Li k ( 1 e t ) .

So,

B n ( k ) (x) ( 1 e t Li k ( 1 e t ) , t ) ,
(14)
H n ( r ) (x| a 1 ,, a r ; λ 1 ,, λ r ) ( j = 1 r ( e a j t λ j 1 λ j ) , t ) ,
(15)
T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r ) ( j = 1 r ( e a j t λ j 1 λ j ) 1 e t Li k ( 1 e t ) , t ) .
(16)

In particular, we have

t B n ( k ) (x)= d d x B n ( k ) (x)=n B n 1 ( k ) (x),
(17)
t H n ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = d d x H n ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = n H n 1 ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) ,
(18)
t T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = d d x T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = n T n 1 ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) .
(19)

Notice that

d d x Li k (x)= 1 x Li k 1 (x).

3.1 Explicit expressions

Write H n ( r ) ( a 1 ,, a r ; λ 1 ,, λ r ):= H n ( r ) (0| a 1 ,, a r ; λ 1 ,, λ r ). Let ( n ) j =n(n1)(nj+1) (j1) with ( n ) 0 =1.

Theorem 1

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = l = 0 n ( n l ) B l ( k ) ( x ) H n l ( r ) ( a 1 , , a r ; λ 1 , , λ r )
(20)
= l = 0 n ( n l ) B n l ( k ) H l ( r ) (x| a 1 ,, a r ; λ 1 ,, λ r )
(21)
= l = 0 n m = 0 n j = 0 m ( 1 ) j ( m j ) ( n l ) 1 ( m + 1 ) k H n l ( r ) ( a 1 ,, a r ; λ 1 ,, λ r ) ( x j ) l
(22)
= l = 0 n ( j = l n m = 0 n j ( 1 ) n m j ( n j ) ( j l ) × m ! ( m + 1 ) k S 2 ( n j , m ) H j l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) ) x l
(23)
= j = 0 n ( n j ) T n j ( r , k ) ( a 1 ,, a r ; λ 1 ,, λ r ) x j .
(24)

Proof By (1), (2) and (3), we have

T n ( r , k ) ( y | a 1 , , a r ; λ 1 , , λ r ) = i = 0 T i ( r , k ) ( y | a 1 , , a r , λ 1 , , λ r ) t i i ! | x n = j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t e y t | x n = j = 1 r ( 1 λ j e a j t λ j ) | Li k ( 1 e t ) 1 e t e y t x n = j = 1 r ( 1 λ j e a j t λ j ) | l = 0 B l ( k ) ( y ) t l l ! x n = l = 0 n ( n l ) B l ( k ) ( y ) j = 1 r ( 1 λ j e a j t λ j ) | x n l = l = 0 n ( n l ) B l ( k ) ( y ) i = 0 H i ( r ) ( a 1 , , a r ; λ 1 , , λ r ) t i i ! | x n l = l = 0 n ( n l ) B l ( k ) ( y ) H n l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) .

So, we get (20).

We also have

T n ( r , k ) ( y | a 1 , , a r ; λ 1 , , λ r ) = i = 0 T i ( r , k ) ( y | a 1 , , a r ; λ 1 , , λ r ) t i i ! | x n = j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t e y t | x n = Li k ( 1 e t ) 1 e t | j = 1 r ( 1 λ j e a j t λ j ) e y t x n = Li k ( 1 e t ) 1 e t | l = 0 H l ( r ) ( y | a 1 , , a r ; λ 1 , , λ r ) t l l ! x n = l = 0 n ( n l ) H l ( r ) ( y | a 1 , , a r ; λ 1 , , λ r ) Li k ( 1 e t ) 1 e t | x n l = l = 0 n ( n l ) H l ( r ) ( y | a 1 , , a r ; λ 1 , , λ r ) i = 0 B i ( k ) t i i ! | x n l = l = 0 n ( n l ) H l ( r ) ( y | a 1 , , a r ; λ 1 , , λ r ) B n l ( k ) .

Thus, we get (21).

In [5] we obtained that

Li k ( 1 e t ) 1 e t x n = m = 0 n 1 ( m + 1 ) k j = 0 m ( 1 ) j ( m j ) ( x j ) n .

So,

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t x n = m = 0 n 1 ( m + 1 ) k j = 0 m ( 1 ) j ( m j ) j = 1 r ( 1 λ j e a j t λ j ) ( x j ) n = m = 0 n 1 ( m + 1 ) k j = 0 m ( 1 ) j ( m j ) l = 0 n ( n l ) H n l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) ( x j ) l = l = 0 n m = 0 n j = 0 m ( 1 ) j ( m j ) ( n l ) 1 ( m + 1 ) k H n l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) ( x j ) l ,

which is identity (22).

In [5] we obtained that

Li k ( 1 e t ) 1 e t x n = j = 0 n ( m = 0 n j ( 1 ) n m j ( m + 1 ) k ( n j ) m ! S 2 ( n j , m ) ) x j ,

where S 2 (l,m) are the Stirling numbers of the second kind, defined by

( e t 1 ) m =m! l = m S 2 (l,m) t l l ! .

Thus,

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = j = 0 n ( m = 0 n j ( 1 ) n m j ( m + 1 ) k ( n j ) m ! S 2 ( n j , m ) ) j = 1 r ( 1 λ j e a j t λ j ) x j = j = 0 n ( m = 0 n j ( 1 ) n m j ( m + 1 ) k ( n j ) m ! S 2 ( n j , m ) ) H j ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = j = 0 n ( m = 0 n j ( 1 ) n m j ( m + 1 ) k ( n j ) m ! S 2 ( n j , m ) ) l = 0 j ( j l ) H j l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) x l = l = 0 n ( j = l n m = 0 n j ( 1 ) n m j ( n j ) ( j l ) m ! ( m + 1 ) k S 2 ( n j , m ) H j l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) ) x l ,

which is identity (23).

By (11) with (16), we have

g ( f ¯ ( t ) ) 1 f ¯ ( t ) j | x n = j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t t j | x n = ( n ) j j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | x n j = ( n ) j i = 0 T i ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) t i i ! | x n j = ( n ) j T n j ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) .

Thus, we get (24). □

3.2 Sheffer identity

Theorem 2

T n ( r , k ) (x+y| a 1 ,, a r ; λ 1 ,, λ r )= j = 0 n ( n j ) T j ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r ) y n j .
(25)

Proof By (16) with

p n ( x ) = j = 1 r ( e a j t λ j 1 λ j ) 1 e t Li k ( 1 e t ) T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = x n ( 1 , t ) ,

using (12), we have (25). □

3.3 Recurrence

Theorem 3

T n + 1 ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = x T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) j = 1 r a j 1 λ j T n ( r + 1 , k ) ( x + a j | a 1 , , a r , a j ; λ 1 , , λ r , λ j ) 1 n + 1 l = 0 n + 1 ( n + 1 l ) B n + 1 l × ( T l ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) T l ( r , k 1 ) ( x | a 1 , , a r ; λ 1 , , λ r ) ) ,
(26)

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)

[[4], Corollary 3.7.2] with (16), we get

T n + 1 ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r )= ( x g r , k ( t ) g r , k ( t ) ) T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r ).

Now,

g r , k ( t ) g r , k ( t ) = ( ln g r , k ( t ) ) = ( j = 1 r ln ( e a j t λ j ) j = 1 r ln ( 1 λ j ) + ln ( 1 e t ) ln Li k ( 1 e t ) ) = j = 1 r a j e a j t e a j t λ j + e t 1 e t ( 1 Li k 1 ( 1 e t ) Li k ( 1 e t ) ) = j = 1 r a j e a j t e a j t λ j + t e t 1 Li k ( 1 e t ) Li k 1 ( 1 e t ) t Li k ( 1 e t ) .

Since

T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r )= i = 1 r ( 1 λ i e a i t λ i ) Li k ( 1 e t ) 1 e t x n ,

we have

T n + 1 ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = x T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) j = 1 r a j e a j t 1 λ j 1 λ j e a j t λ j i = 1 r ( 1 λ i e a i t λ i ) Li k ( 1 e t ) 1 e t x n t e t 1 i = 1 r ( 1 λ i e a i t λ i ) Li k ( 1 e t ) Li k 1 ( 1 e t ) t ( 1 e t ) x n .

Since

Li k ( 1 e t ) Li k 1 ( 1 e t ) 1 e t = ( 1 2 k 1 2 k 1 ) t+

is a delta series, we get

Li k ( 1 e t ) Li k 1 ( 1 e t ) t ( 1 e t ) x n = 1 n + 1 Li k ( 1 e t ) Li k 1 ( 1 e t ) 1 e t x n + 1 .

Therefore, by

t e t 1 x n + 1 = B n + 1 (x)= l = 0 n + 1 ( n + 1 l ) B n + 1 l x l ,

we obtain

T n + 1 ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = x T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) j = 1 r a j 1 λ j T n ( r + 1 , k ) ( x + a j | a 1 , , a r , a j ; λ 1 , , λ r , λ j ) 1 n + 1 i = 1 r ( 1 λ i e a i t λ i ) Li k ( 1 e t ) Li k 1 ( 1 e t ) 1 e t t e t 1 x n + 1 = x T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) j = 1 r a j 1 λ j T n ( r + 1 , k ) ( x + a j | a 1 , , a r , a j ; λ 1 , , λ r , λ j ) 1 n + 1 l = 0 n + 1 ( n + 1 l ) B n + 1 l × ( T l ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) T l ( r , k 1 ) ( x | a 1 , , a r ; λ 1 , , λ r ) ) ,

which is identity (26). □

3.4 A more relation

Theorem 4

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = x T n 1 ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) j = 1 r a j 1 λ j T n 1 ( r + 1 , k ) ( x + a j | a 1 , , a r , a j ; λ 1 , , λ r , λ j ) 1 n l = 0 n ( n l ) B n l ( T l ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) T l ( r , k 1 ) ( x | a 1 , , a r ; λ 1 , , λ r ) ) .
(27)

Proof For n1, we have

T n ( r , k ) ( y | a 1 , , a r ; λ 1 , , λ r ) = l = 0 T l ( r , k ) ( y | a 1 , , a r ; λ 1 , , λ r ) t l l ! | x n = j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t e y t | x n = t ( j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t e y t ) | x n 1 = ( t j = 1 r ( 1 λ j e a j t λ j ) ) Li k ( 1 e t ) 1 e t e y t | x n 1 + j = 1 r ( 1 λ j e a j t λ j ) ( t Li k ( 1 e t ) 1 e t ) e y t | x n 1 + j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t ( t e y t ) | x n 1 = y T n 1 ( r , k ) ( y | a 1 , , a r ; λ 1 , , λ r ) + ( t j = 1 r ( 1 λ j e a j t λ j ) ) Li k ( 1 e t ) 1 e t e y t | x n 1 + j = 1 r ( 1 λ j e a j t λ j ) ( t Li k ( 1 e t ) 1 e t ) e y t | x n 1 .

Observe that

t j = 1 r ( 1 λ j e a j t λ j ) = j = 1 r ( 1 λ j ) t ( 1 j = 1 r ( e a j t λ j ) ) = j = 1 r ( 1 λ j ) ( j = 1 r ( e a j t λ j ) ) ( j = 1 r ( e a j t λ j ) ) 2 = j = 1 r ( 1 λ j ) j = 1 r a j e a j t i j ( e a i t λ i ) ( j = 1 r ( e a j t λ j ) ) 2 = j = 1 r a j e a j t e a j t λ j i = 1 r ( 1 λ i e a i t λ i ) = j = 1 r a j e a j t 1 λ j 1 λ j e a j t λ j i = 1 r ( 1 λ i e a i t λ i ) .

Thus,

( t j = 1 r ( 1 λ j e a j t λ j ) ) Li k ( 1 e t ) 1 e t e y t | x n 1 = j = 1 r a j 1 λ j 1 λ j e a j t λ j i = 1 r ( 1 λ i e a i t λ i ) Li k ( 1 e t ) 1 e t e ( y + a j ) t | x n 1 = j = 1 r a j 1 λ j T n 1 ( r + 1 , k ) ( y + a j | a 1 , , a r , a j ; λ 1 , , λ r , λ j ) .

Since

t ( Li k ( 1 e t ) 1 e t ) = e t ( Li k 1 ( 1 e t ) Li k ( 1 e t ) ) ( 1 e t ) 2 = t e t 1 Li k 1 ( 1 e t ) Li k ( 1 e t ) t ( 1 e t )

and the fact that

Li k 1 ( 1 e t ) Li k ( 1 e t ) 1 e t = ( 1 2 k 1 1 2 k ) t+

is a delta series, we have

j = 1 r ( 1 λ j e a j t λ j ) ( t Li k ( 1 e t ) 1 e t ) e y t | x n 1 = j = 1 r ( 1 λ j e a j t λ j ) t e t 1 Li k 1 ( 1 e t ) Li k ( 1 e t ) t ( 1 e t ) e y t | x n 1 = j = 1 r ( 1 λ j e a j t λ j ) t e t 1 Li k 1 ( 1 e t ) Li k ( 1 e t 1 e t e y t | x n n = 1 n j = 1 r ( 1 λ j e a j t λ j ) Li k 1 ( 1 e t ) Li k ( 1 e t ) 1 e t e y t | t e t 1 x n = 1 n l = 0 n ( n l ) B n l j = 1 r ( 1 λ j e a j t λ j ) Li k 1 ( 1 e t ) Li k ( 1 e t ) 1 e t e y t | x l = 1 n l = 0 n ( n l ) B n l ( T l ( r , k 1 ) ( y | a 1 , , a r ; λ 1 , , λ r ) T l ( r , k ) ( y | a 1 , , a r ; λ 1 , , λ r ) ) .

Therefore, we obtain the desired result. □

Remark After n is replaced by n+1, identity (27) becomes the recurrence formula (26).

3.5 Relations with poly-Bernoulli numbers and Barnes’ multiple Bernoulli numbers

Theorem 5

m = 0 n ( 1 ) n m ( n + 1 m ) T m ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) = l = 0 n m = 0 l ( 1 ) l m ( l m ) ( n + 1 l + 1 ) B m ( k 1 ) H n l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) .
(28)

Proof We shall compute

j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) | x n + 1

in two different ways. On the one hand,

j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) | x n + 1 = j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | ( 1 e t ) x n + 1 = j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | x n + 1 ( x 1 ) n + 1 = m = 0 n ( n + 1 m ) ( 1 ) n m j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | x m = m = 0 n ( n + 1 m ) ( 1 ) n m T m ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) .

On the other hand,

j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) | x n + 1 = Li k ( 1 e t ) | j = 1 r ( 1 λ j e a j t λ j ) x n + 1 = Li k ( 1 e t ) | H n + 1 ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = 0 t ( Li k ( 1 e s ) ) d s | H n + 1 ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = 0 t e s Li k 1 ( 1 e s ) 1 e s d s | H n + 1 ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = 0 t ( j = 0 ( s ) j j ! ) ( m = 0 B m ( k 1 ) m ! s m ) d s | H n + 1 ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = l = 0 ( m = 0 l ( 1 ) l m ( l m ) B m ( k 1 ) ) 1 l ! 0 t s l d s | H n + 1 ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = l = 0 n m = 0 l ( 1 ) l m ( l m ) B m ( k 1 ) ( l + 1 ) ! t l + 1 | H n + 1 ( r ) ( x | a 1 , , a r ; λ 1 , , λ r ) = l = 0 n m = 0 l ( 1 ) l m ( l m ) B m ( k 1 ) ( l + 1 ) ! ( n + 1 ) l + 1 H n l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) = l = 0 n m = 0 l ( 1 ) l m ( l m ) ( n + 1 l + 1 ) B m ( k 1 ) H n l ( r ) ( a 1 , , a r ; λ 1 , , λ r ) .

Here, H n l ( r ) ( a 1 ,, a r ; λ 1 ,, λ r )= H n l ( r ) (0| a 1 ,, a r ; λ 1 ,, λ r ). Thus, we get (28). □

3.6 Relations with the Stirling numbers of the second kind and the falling factorials

Theorem 6

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = m = 0 n ( l = m n ( n l ) S 2 ( l , m ) T n l ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) ) ( x ) m .
(29)

Proof For (16) and ( x ) n (1, e t 1), assume that

T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r )= m = 0 n C n , m ( x ) m .

By (13), we have

C n , m = 1 m ! 1 j = 1 r ( e a j t λ j 1 λ j ) 1 e t Li k ( 1 e t ) ( e t 1 ) m | x n = 1 m ! j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | ( e t 1 ) m x n = 1 m ! j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | m ! l = m n S 2 ( l , m ) t l l ! x n = l = m n ( n l ) S 2 ( l , m ) j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | x n l = l = m n ( n l ) S 2 ( l , m ) T n l ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) .

Thus, we get identity (29). □

3.7 Relations with the Stirling numbers of the second kind and the rising factorials

Theorem 7

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = m = 0 n ( l = m n ( n l ) S 2 ( l , m ) T n l ( r , k ) ( m | a 1 , , a r ; λ 1 , , λ r ) ) ( x ) ( m ) .
(30)

Proof For (16) and ( x ) ( n ) =x(x+1)(x+n1)(1,1 e t ), assume that T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r )= m = 0 n C n , m ( x ) ( m ) . By (13), we have

C n , m = 1 m ! 1 j = 1 r ( e a j t λ j 1 λ j ) 1 e t Li k ( 1 e t ) ( 1 e t ) m | x n = 1 m ! j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t e m t | ( e t 1 ) m x n = l = m n ( n l ) S 2 ( l , m ) e m t | j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t x n l = l = m n ( n l ) S 2 ( l , m ) e m t | T n l ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = l = m n ( n l ) S 2 ( l , m ) T n l ( r , k ) ( m | a 1 , , a r ; λ 1 , , λ r ) .

Thus, we get identity (30). □

3.8 Relations with higher-order Frobenius-Euler polynomials

Theorem 8

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = m = 0 n ( ( n m ) ( 1 λ ) s l = 0 s ( s l ) ( λ ) s l T n m ( r , k ) ( l | a 1 , , a r ; λ 1 , , λ r ) ) H m ( s ) ( x | λ ) .
(31)

Proof For (16) and

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

assume that T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r )= m = 0 n C n , m H m ( s ) (x|λ). By (13), we have

C n , m = 1 m ! ( e t λ 1 λ ) s j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t t m | x n = 1 m ! ( 1 λ ) s l = 0 s ( s l ) ( λ ) s l e l t j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | t m x n = ( n m ) ( 1 λ ) s l = 0 s ( s l ) ( λ ) s l e l t | j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t x n m = ( n m ) ( 1 λ ) s l = 0 s ( s l ) ( λ ) s l T n m ( r , k ) ( l | a 1 , , a r ; λ 1 , , λ r ) .

Thus, we get identity (31). □

3.9 Relations 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., [[4], Section 2.2]).

Theorem 9

T n ( r , k ) ( x | a 1 , , a r ; λ 1 , , λ r ) = m = 0 n ( n m ) ( l = 0 n m ( n m l ) ( l + s l ) S 2 ( l + s , s ) T n m l ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) ) B m ( s ) ( x ) .
(32)

Proof For (16) and

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

assume that T n ( r , k ) (x| a 1 ,, a r ; λ 1 ,, λ r )= m = 0 n C n , m B m ( s ) (x). By (13), we have

C n , m = 1 m ! ( e t 1 t ) s j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t t m | x n = ( n m ) j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | ( e t 1 t ) s x n m = ( n m ) j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | l = 0 n m s ! ( l + s ) ! S 2 ( l + s , s ) t l x n m = ( n m ) l = 0 n m s ! ( l + s ) ! S 2 ( l + s , s ) ( n m ) l j = 1 r ( 1 λ j e a j t λ j ) Li k ( 1 e t ) 1 e t | x n m l = ( n m ) l = 0 n m s ! ( l + s ) ! S 2 ( l + s , s ) ( n m ) l T n m l ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) = ( n m ) l = 0 n m ( n m l ) ( l + s l ) S 2 ( l + s , s ) T n m l ( r , k ) ( a 1 , , a r ; λ 1 , , λ r ) .

Thus, we get identity (32). □

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Acknowledgements

The authors would like to thank the referees for their valuable comments and suggestions. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786). The fourth author was supported in part by the Grant-in-Aid for Scientific research (C) (No. 22540005), the Japan Society for the Promotion of Science.

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Kim, D.S., Kim, T., Seo, JJ. et al. Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials. Adv Differ Equ 2014, 92 (2014). https://doi.org/10.1186/1687-1847-2014-92

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Keywords

  • Formal Power Series
  • Bernoulli Number
  • Bernoulli Polynomial
  • Stirling Number
  • Polylogarithm Function