1 Introduction

For r Z 0 , as is well known, the Bernoulli polynomials of order r are defined by the generating function to be

n = 0 B n ( r ) ( x ) n ! t n = ( t e t 1 ) r e x t (see [1–16]).
(1.1)

For kZ, the polylogarithm is defined by

Li k (x)= n = 1 x n n k .
(1.2)

Note that Li 1 (x)=log(1x).

The poly-Bernoulli polynomials are defined by the generating function to be

Li k ( 1 e t ) 1 e t e x t = n = 0 B n ( k ) (x) t n n ! (see [5, 8]).
(1.3)

When x=0, B n ( k ) = B n ( k ) (0) are called the poly-Bernoulli numbers (of index k).

For ν(0)R, the Hermite polynomials of order ν are given by the generating function to be

e ν t 2 2 e x t = n = 0 H n ( ν ) (x) t n n ! (see [6, 12, 13]).
(1.4)

When x=0, H n ( ν ) = H n ( ν ) (0) are called the Hermite numbers of order ν.

In this paper, we consider the Hermite and poly-Bernoulli mixed-type polynomials H B n ( ν , k ) (x) which are defined by the generating function to be

e ν t 2 2 Li k ( 1 e t ) 1 e t e x t = n = 0 H B n ( ν , k ) (x) t n n ! ,
(1.5)

where kZ and ν(0)R.

When x=0, H B n ( ν , k ) =H B n ( ν , k ) (0) are called the Hermite and poly-Bernoulli mixed-type numbers.

Let ℱ be the set of all formal power series in the variable t over ℂ as follows:

F= { f ( t ) = k = 0 a k t k k ! | a k C } .
(1.6)

Let P=C[x] and P denote the vector space of all linear functionals on ℙ.

L|p(x) denotes 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 ℂ. For f(t)F, let us define the linear functional on ℙ by setting

f ( t ) | x n = a n (n0).
(1.7)

Then, by (1.6) and (1.7), we get

t k | x n =n! δ n , k (n,k0),
(1.8)

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 ℱ. Henceforth, ℱ denotes both the algebra of formal power series in t and the vector space of all linear functionals on ℙ, and so an element f(t) of ℱ will be thought of as both a formal power series and a linear functional. We call ℱ the umbral algebra and the umbral calculus is the study of umbral algebra. The order O(f) of the power series f(t)0 is the smallest integer for which a k does not vanish. If O(f)=0, then f(t) is called an invertible series. If O(f)=1, then f(t) is called a delta series. For f(t),g(t)F, we have

f ( t ) g ( t ) | p ( x ) = f ( t ) | g ( t ) p ( x ) = g ( t ) | f ( t ) p ( x ) .
(1.9)

Let f(t)F and p(x)P. Then we have

f(t)= k = 0 f ( t ) | x k k ! t k ,p(x)= k = 0 t k | p ( x ) k ! x k (see [8, 9, 11, 13, 14]).
(1.10)

By (1.10), we get

p ( k ) (0)= t k | p ( x ) = 1 | p ( k ) ( x ) ,
(1.11)

where p ( k ) (0)= d k p ( x ) d x k | x = 0 .

From (1.11), we have

t k p(x)= p ( k ) (x)= d k p ( x ) d x k (see [8, 9, 13]).
(1.12)

By (1.12), we easily get

e y t p(x)=p(x+y), e y t | p ( x ) =p(y).
(1.13)

For O(f(t))=1, O(g(t))=0, there exists a unique sequence s n (x) of polynomials such that g(t)f ( t ) k | x n =n! δ n , k (n,k0).

The 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)).

Let p(x)P, f(t)F. Then we see that

f ( t ) | x p ( x ) = t f ( t ) | p ( x ) = d f ( t ) d t | p ( x ) .
(1.14)

For s n (x)(g(t),f(t)), we have the following equations:

h(t)= k = 0 h ( t ) | s k ( x ) k ! g(t)f ( t ) k ,p(x)= k = 0 g ( t ) f ( t ) k | p ( x ) k ! s k (x),
(1.15)

where h(t)F, p(x)P,

1 g ( f ¯ ( t ) ) e y f ¯ ( t ) = n = 0 s n (y) t n n ! ,
(1.16)

where f ¯ (t) is the compositional inverse for f(t) with f( f ¯ (t))=t,

s n (x+y)= k = 0 n ( n k ) s k (y) p n k (x),where  p n (x)=g(t) s n (x),
(1.17)
f(t) s n (x)=n s n 1 (x), s n + 1 (x)= ( x g ( t ) g ( t ) ) 1 f ( t ) s n (x),
(1.18)

and the conjugate representation is given by

s n (x)= j = 0 n 1 j ! g ( f ¯ ( t ) ) 1 f ¯ ( t ) j | x n x j .
(1.19)

For s n (x)(g(t),f(t)), r n (x)(h(t),l(t)), we have

s n (x)= m = 0 n C n , m r m (x),
(1.20)

where

C n , m = 1 m ! h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n (see [8, 9, 13]).
(1.21)

In this paper, we consider Hermite and poly-Bernoulli mixed-type polynomials and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give various identities associated with Bernoulli and Frobenius-Euler polynomials of higher order.

2 Hermite and poly-Bernoulli mixed-type polynomials

From (1.5) and (1.16), we note that

H B n ( ν , k ) (x) ( e ν t 2 2 1 e t Li k ( 1 e t ) , t ) ,
(2.1)

and, by (1.3), (1.4) and (1.16), we get

B n ( k ) (x) ( 1 e t Li k ( 1 e t ) , t ) ,
(2.2)
H n ( ν ) (x) ( e ν t 2 2 , t ) ,where n0.
(2.3)

From (1.18), (2.1), (2.2) and (2.3), we have

t B n ( k ) (x)=n B n 1 ( k ) (x),t H n ( ν ) (x)=n H n 1 ( ν ) (x),tH B n ( ν , k ) (x)=nH B n 1 ( ν , k ) (x).
(2.4)

By (1.5), (1.8) and (2.1), we get

H B n ( ν , k ) ( x ) = e ν t 2 2 Li k ( 1 e t ) 1 e t x n = e ν t 2 2 B n ( k ) ( x ) = m = 0 [ n 2 ] 1 m ! ( ν 2 ) m ( n ) 2 m B n 2 m ( k ) ( x ) = m = 0 [ n 2 ] ( n 2 m ) ( 2 m ) ! m ! ( ν 2 ) m B n 2 m ( k ) ( x ) .
(2.5)

Therefore, by (2.5), we obtain the following proposition.

Proposition 1 For n0, we have

H B n ( ν , k ) (x)= m = 0 [ n 2 ] ( n 2 m ) ( 2 m ) ! m ! ( ν 2 ) m B n 2 m ( k ) (x).

From (1.5), we can also derive

H B n ( ν , k ) ( x ) = Li k ( 1 e t ) 1 e t e ν t 2 2 x n = Li k ( 1 e t ) 1 e t H n ( ν ) ( x ) = m = 0 ( 1 e t ) m ( m + 1 ) k H n ( ν ) ( x ) = m = 0 n 1 ( m + 1 ) k j = 0 m ( m j ) ( 1 ) j e j t H n ( ν ) ( x ) = m = 0 n 1 ( m + 1 ) k j = 0 m ( m j ) ( 1 ) j H n ( ν ) ( x j ) .
(2.6)

Therefore, by (2.6), we obtain the following theorem.

Theorem 2 For n0, we have

H B n ( ν , k ) (x)= m = 0 n 1 ( m + 1 ) k j = 0 m ( m j ) ( 1 ) j H n ( ν ) (xj).

By (1.5), we get

H B n ( ν , k ) ( x ) = e ν t 2 2 B n ( k ) ( x ) = l = 0 1 l ! ( ν 2 ) l t 2 l B n ( k ) ( x ) = l = 0 [ n 2 ] 1 l ! ( ν 2 ) l m = 0 n 1 ( m + 1 ) k j = 0 m ( 1 ) j ( m j ) t 2 l ( x j ) n = l = 0 [ n 2 ] j = 0 n { m = j n ( n 2 l ) ( 2 l ) ! l ! ( ν 2 ) l ( 1 ) j ( m j ) ( m + 1 ) k } ( x j ) n 2 l .
(2.7)

Therefore, by (2.7), we obtain the following theorem.

Theorem 3 For n0, we have

H B n ( ν , k ) (x)= l = 0 [ n 2 ] j = 0 n { m = j n ( n 2 l ) ( 2 l ) ! l ! ( ν 2 ) l ( 1 ) j ( m j ) ( m + 1 ) k } ( x j ) n 2 l .

By (2.6), we get

H B n ( ν , k ) ( x ) = m = 0 n ( 1 e t ) m ( m + 1 ) k H n ( ν ) ( x ) = m = 0 n 1 ( m + 1 ) k a = 0 n m m ! ( a + m ) ! ( 1 ) a S 2 ( a + m , m ) ( n ) a + m H n a m ( ν ) ( x ) = m = 0 n a = 0 n m ( 1 ) n a m m ! ( m + 1 ) k ( n n a ) S 2 ( n a , m ) H a ( ν ) ( x ) = ( 1 ) n a = 0 n { m = 0 n a ( 1 ) m + a m ! ( m + 1 ) k ( n a ) S 2 ( n a , m ) } H a ( ν ) ( x ) ,
(2.8)

where S 2 (n,m) is the Stirling number of the second kind.

Therefore, by (2.8), we obtain the following theorem.

Theorem 4 For n0, we have

H B n ( ν , k ) (x)= ( 1 ) n a = 0 n { m = 0 n a ( 1 ) a + m m ! ( m + 1 ) k ( n a ) S 2 ( n a , m ) } H a ( ν ) (x).

From (1.19) and (2.1), we have

H B n ( ν , k ) ( x ) = j = 0 n ( n j ) e ν t 2 2 Li k ( 1 e t ) 1 e t | x n j x j = j = 0 n ( n j ) e ν t 2 2 | B n j ( k ) ( x ) x j = j = 0 n ( n j ) l = 0 [ n j 2 ] ( ν 2 ) l l ! ( n j ) 2 l 1 | B n j 2 l ( k ) ( x ) x j = j = 0 n ( n j ) l = 0 [ n j 2 ] 1 l ! ( ν 2 ) l ( n j ) 2 l B n j 2 l ( k ) x j = j = 0 n { l = 0 [ n j 2 ] ( n j ) ( n j 2 l ) ( 2 l ) ! l ! ( ν 2 ) l B n j 2 l ( k ) } x j .
(2.9)

Therefore, by (2.9), we obtain the following theorem.

Theorem 5 For n0, we have

H B n ( ν , k ) (x)= j = 0 n { l = 0 [ n j 2 ] ( n j ) ( n j 2 l ) ( 2 l ) ! l ! ( ν 2 ) l B n j 2 l ( k ) } x j .

Remark By (1.17) and (2.1), we easily get

H B n ( ν , k ) (x+y)= j = 0 n ( n j ) H B j ( ν , k ) (x) y n j .
(2.10)

We note that

H B n ( ν , k ) (x) ( g ( t ) = e ν t 2 2 1 e t Li k ( 1 e t ) , f ( t ) = t ) .
(2.11)

From (1.18) and (2.11), we have

H B n + 1 ( ν , k ) (x)= ( x g ( t ) g ( t ) ) H B n ( ν , k ) (x).
(2.12)

Now, we observe that

g ( t ) g ( t ) = ( log ( g ( t ) ) ) = ( log e ν t 2 2 + log ( 1 e t ) log ( Li k ( 1 e t ) ) ) = ν t + e t 1 e t ( 1 Li k 1 ( 1 e t ) Li k ( 1 e t ) ) .
(2.13)

By (2.12) and (2.13), we get

H B n + 1 ( ν , k ) ( x ) = x H B n ( ν , k ) ( x ) g ( t ) g ( t ) H B n ( ν , k ) ( x ) = x H B n ( ν , k ) ( x ) ν n H B n 1 ( ν , k ) ( x ) e ν t 2 2 t e t 1 Li k ( 1 e t ) Li k 1 ( 1 e t ) t ( 1 e t ) x n .
(2.14)

It is easy to show that

Li k ( 1 e t ) Li k 1 ( 1 e t ) 1 e t = m = 2 ( 1 m k 1 m k 1 ) ( 1 e t ) m 1 = ( 1 2 k 1 2 k 1 ) t + .
(2.15)

Thus, by (2.15), we get

Li k ( 1 e t ) Li k 1 ( 1 e t ) t ( 1 e t ) x n = Li k ( 1 e t ) Li k 1 ( 1 e t ) 1 e t x n + 1 n + 1 .
(2.16)

From (2.16), we can derive

e ν t 2 2 t e t 1 Li k ( 1 e t ) Li k 1 ( 1 e t ) t ( 1 e t ) x n = 1 n + 1 ( l = 0 B l l ! t l ) ( H B n + 1 ( ν , k ) ( x ) H B n + 1 ( ν , k 1 ) ( x ) ) = 1 n + 1 l = 0 n + 1 B l l ! t l ( H B n + 1 ( ν , k ) ( x ) H B n + 1 ( ν , k 1 ) ( x ) ) = 1 n + 1 l = 0 n + 1 ( n + 1 l ) B l ( H B n + 1 l ( ν , k ) ( x ) H B n + 1 l ( ν , k 1 ) ( x ) ) .
(2.17)

Therefore, by (2.14) and (2.17), we obtain the following theorem.

Theorem 6 For n0, we have

H B n + 1 ( ν , k ) ( x ) = x H B n ( ν , k ) ( x ) ν n H B n 1 ( ν , k ) ( x ) 1 n + 1 l = 0 n + 1 ( n + 1 l ) B l { H B n + 1 l ( ν , k ) ( x ) H B n + 1 l ( ν , k 1 ) ( x ) } .
(2.18)

Let us take t on the both sides of (2.18). Then we have

( n + 1 ) H B n ( ν , k ) ( x ) = ( x t + 1 ) H B n ( ν , k ) ( x ) ν n ( n 1 ) H B n 2 ( ν , k ) ( x ) 1 n + 1 l = 0 n + 1 ( n + 1 l ) ( n + 1 l ) B l { H B n l ( ν , k ) ( x ) H B n l ( ν , k 1 ) ( x ) } = n x H B n 1 ( ν , k ) ( x ) + H B n ( ν , k ) ( x ) ν n ( n 1 ) H B n 2 ( ν , k ) ( x ) l = 0 n ( n l ) B l ( H B n l ( ν , k ) ( x ) H B n l ( ν , k 1 ) ( x ) ) ,
(2.19)

where n3.

Thus, by (2.19), we obtain the following theorem.

Theorem 7 For n3, we have

l = 0 n ( n l ) B l H B n l ( ν , k 1 ) ( x ) = ( n + 1 ) H B n ( ν , k ) ( x ) n ( x + 1 2 ) H B n 1 ( ν , k ) ( x ) + n ( n 1 ) ( ν + 1 12 ) H B n 2 ( ν , k ) ( x ) + l = 0 n 3 ( n l ) B n l H B l ( ν , k ) ( x ) .

By (1.5) and (1.8), we get

H B n ( ν , k ) ( y ) = e ν t 2 2 Li k ( 1 e t ) 1 e t e y t | x n = t ( e ν t 2 2 Li k ( 1 e t ) 1 e t e y t ) | x n 1 = ( t e ν t 2 2 ) Li k ( 1 e t ) 1 e t e y t | x n 1 + e ν t 2 2 ( t Li k ( 1 e t ) 1 e t ) e y t | x n 1 + e ν t 2 2 Li k ( 1 e t ) 1 e t ( t e y t ) | x n 1 = ν ( n 1 ) e ν t 2 2 Li k ( 1 e t ) 1 e t e y t | x n 2 + y e ν t 2 2 Li k ( 1 e t ) 1 e t e y t | x n 1 + e ν t 2 2 ( t Li k ( 1 e t ) 1 e t ) e y t | x n 1 = ν ( n 1 ) H B n 2 ( ν , k ) ( y ) + y H B n 1 ( ν , k ) ( y ) + e ν t 2 2 ( t Li k ( 1 e t ) 1 e t ) e y t | x n 1 .
(2.20)

Now, we observe that

t ( Li k ( 1 e t ) 1 e t ) = Li k 1 ( 1 e t ) Li k ( 1 e t ) ( 1 e t ) 2 e t .
(2.21)

From (2.21), we have

e ν t 2 2 ( t Li k ( 1 e t ) 1 e t ) e y t | x n 1 = e ν t 2 2 ( Li k 1 ( 1 e t ) Li k ( 1 e t ) ( 1 e t ) 2 ) e t e y t | 1 n t x n = 1 n e ν t 2 2 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 e ν t 2 2 Li k 1 ( 1 e t ) Li k ( 1 e t ) 1 e t e y t | B n ( x ) = 1 n l = 0 n ( n l ) B l e ν t 2 2 Li k 1 ( 1 e t ) Li k ( 1 e t ) 1 e t e y t | x n l = 1 n l = 0 n ( n l ) B l { H B n l ( ν , k 1 ) ( y ) H B n l ( ν , k ) ( y ) } ,
(2.22)

where B n are the ordinary Bernoulli numbers which are defined by the generating function to be

t e t 1 = n = 0 B n n ! t n .

Therefore, by (2.20) and (2.22), we obtain the following theorem.

Theorem 8 For n2, we have

H B n ( ν , k ) ( x ) = ν ( n 1 ) H B n 2 ( ν , k ) ( x ) + x H B n 1 ( ν , k ) ( x ) + 1 n l = 0 n ( n l ) B l ( H B n l ( ν , k 1 ) ( x ) H B n l ( ν , k ) ( x ) ) .

Now, we compute

e ν t 2 2 Li k ( 1 e t ) | x n + 1

in two different ways.

On the one hand,

e ν t 2 2 Li k ( 1 e t ) | x n + 1 = e ν t 2 2 Li k ( 1 e t ) 1 e t ( 1 e t ) | x n + 1 = e ν t 2 2 Li k ( 1 e t ) 1 e t | ( 1 e t ) x n + 1 = e ν t 2 2 Li k ( 1 e t ) 1 e t | x n + 1 ( x 1 ) n + 1 = m = 0 n ( 1 ) n m ( n + 1 m ) e ν t 2 2 Li k ( 1 e t ) 1 e t | x m = m = 0 n ( 1 ) n m ( n + 1 m ) H B m ( ν , k ) .
(2.23)

On the other hand,

e ν t 2 2 Li k ( 1 e t ) | x n + 1 = Li k ( 1 e t ) | e ν t 2 2 x n + 1 = 0 t ( Li k ( 1 e s ) ) d s | e ν t 2 2 x n + 1 = 0 t e s Li k 1 ( 1 e s ) 1 e s d s | e ν t 2 2 x n + 1 = l = 0 ( m = 0 l ( 1 ) l m ( l m ) B m ( k 1 ) t l + 1 ( l + 1 ) ! ) | H n + 1 ( ν ) ( x ) = l = 0 n m = 0 l ( 1 ) l m ( l m ) B m ( k 1 ) 1 ( l + 1 ) ! t l + 1 | H n + 1 ( ν ) ( x ) = l = 0 n m = 0 l ( 1 ) l m ( l m ) ( n + 1 l + 1 ) B m ( k 1 ) H n l ( ν ) .
(2.24)

Therefore, by (2.23) and (2.24), we obtain the following theorem.

Theorem 9 For n0, we have

m = 0 n ( 1 ) n m ( n + 1 m ) H B m ( ν , k ) = m = 0 n l = m n ( 1 ) l m ( l m ) ( n + 1 l + 1 ) B m ( k 1 ) H n l ( ν ) .

Let us consider the following two Sheffer sequences:

H B n ( ν , k ) (x) ( e ν t 2 2 1 e t Li k ( 1 e t ) , t )
(2.25)

and

B n ( r ) (x) ( ( e t 1 t ) r , t ) (r Z 0 ).
(2.26)

Let us assume that

H B n ( ν , k ) (x)= m = 0 n C n , m B m ( r ) (x).
(2.27)

Then, by (1.20) and (1.21), we get

C n , m = 1 m ! ( e t 1 t ) r t m | e ν t 2 2 Li k ( 1 e t ) 1 e t x n = 1 m ! ( e t 1 t ) r | t m H B n ( ν , k ) ( x ) = 1 m ! ( n ) m ( e t 1 t ) r | H B n m ( ν , k ) ( x ) = ( n m ) l = 0 r ! ( l + r ) ! S 2 ( l + r , r ) t l | H B n m ( ν , k ) ( x ) = ( n m ) l = 0 n m ( n m ) l r ! ( l + r ) ! S 2 ( l + r , r ) H B n m l ( ν , k ) = ( n m ) l = 0 n m ( n m l ) ( l + r r ) S 2 ( l + r , r ) H B n m l ( ν , k ) .
(2.28)

Therefore, by (2.27) and (2.28), we obtain the following theorem.

Theorem 10 For n,r Z 0 , we have

H B n ( ν , k ) (x)= m = 0 n { ( n m ) l = 0 n m ( n m l ) ( l + r r ) S 2 ( l + r , r ) H B n m l ( ν , k ) } B m ( r ) (x).

For λ(1)C, r Z 0 , the Frobenius-Euler polynomials of order r are defined by the generating function to be

( 1 λ e t λ ) r e x t = n = 0 H n ( r ) (x|λ) t n n ! (see [1, 4, 7, 9, 10]).
(2.29)

From (1.16) and (2.29), we note that

H n ( r ) (x|λ) ( ( e t λ 1 λ ) r , t ) .
(2.30)

Let us assume that

H B n ( ν , k ) (x)= m = 0 n C n , m H m ( r ) (x|λ).
(2.31)

By (1.21), we get

C n , m = 1 m ! ( e t λ 1 λ ) r t m | e ν t 2 2 Li k ( 1 e t ) 1 e t x n = ( n ) m m ! ( 1 λ ) r l = 0 r ( r l ) ( λ ) r l e l t | H B n m ( ν , k ) ( x ) = ( n m ) 1 ( 1 λ ) r l = 0 r ( r l ) ( λ ) r l 1 | e l t H B n m ( ν , k ) ( x ) = ( n m ) ( 1 λ ) r l = 0 r ( r l ) ( λ ) r l H B n m ( ν , k ) ( l ) .
(2.32)

Therefore, by (2.31) and (2.32), we obtain the following theorem.

Theorem 11 For n,r Z 0 , we have

H B n ( ν , k ) (x)= 1 ( 1 λ ) r m = 0 n ( n m ) { l = 0 r ( r l ) ( λ ) r l H B n m ( ν , k ) ( l ) } H m ( r ) (x|λ).