# Hermite and poly-Bernoulli mixed-type polynomials

## Abstract

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 Stirling numbers, Bernoulli and Frobenius-Euler polynomials of higher order.

## 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 $k∈Z$, the polylogarithm is defined by

$Li k (x)= ∑ n = 1 ∞ x n n k .$
(1.2)

Note that $Li 1 (x)=−log(1−x)$.

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 $k∈Z$ 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 be the set of all formal power series in the variable t over 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 .

$〈L|p(x)〉$ 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 $〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉$, $〈cL|p(x)〉=c〈L|p(x)〉$, where c is a complex constant in . For $f(t)∈F$, let us define the linear functional on by setting

$〈 f ( t ) | x n 〉 = a n (n≥0).$
(1.7)

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

$〈 t k | x n 〉 =n! δ n , k (n,k≥0),$
(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 . 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 $O(f)$ 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 $〈g(t)f ( t ) k | x n 〉=n! δ n , k$ ($n,k≥0$).

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 $f( f ¯ (t))=t$,

(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)
(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 $n≥0$, 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 $n≥0$, we have

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

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 $n≥0$, 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 $n≥0$, 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 $n≥0$, 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 $n≥0$, 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 $n≥3$.

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

Theorem 7 For $n≥3$, 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 $n≥2$, 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 $n≥0$, 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

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

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## Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786).

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Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kim, D.S., Kim, T. Hermite and poly-Bernoulli mixed-type polynomials. Adv Differ Equ 2013, 343 (2013). https://doi.org/10.1186/1687-1847-2013-343 