# Higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials

## Abstract

In this paper, we consider higher-order Frobenius-Euler polynomials, associated with poly-Bernoulli polynomials, which are derived from polylogarithmic function. These polynomials are called higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials. The purpose of this paper is to give various identities of those polynomials arising from umbral calculus.

## 1 Introduction

For $λ∈C$ with $λ≠1$, the Frobenius-Euler polynomials of order α ($α∈R$) are defined by the generating function to be

$( 1 − λ e t − λ ) α e x t = ∑ n = 0 ∞ H n ( α ) (x|λ) t n n ! (see [1–5]).$
(1.1)

When $x=0$, $H n ( α ) (λ)= H n ( α ) (0|λ)$ are called the Frobenius-Euler numbers of order α. As is well known, the Bernoulli polynomials of order α are defined by the generating function to be

$( t e t − 1 ) α e x t = ∑ n = 0 ∞ B n ( α ) (x) t n n ! (see [6–8]).$
(1.2)

When $x=0$, $B n ( α ) = B n ( α ) (x)$ is called the n th Bernoulli number of order α. In the special case, $α=1$, $B n ( 1 ) (x)= B n (x)$ is called the n th Bernoulli polynomial. When $x=0$, $B n = B n (0)$ is called the n th ordinary Bernoulli number. Finally, we recall that the Euler polynomials of order α are given by

$( 2 e t + 1 ) α e x t = ∑ n = 0 ∞ E n ( α ) (x) t n n ! (see [9–13]).$
(1.3)

When $x=0$, $E n ( α ) = E n ( α ) (0)$ is called the n th Euler number of order α. In the special case, $α=1$, $E n ( 1 ) (x)= E n (x)$ is called the n th ordinary Euler polynomial. The classical polylogarithmic function $L i k (x)$ is defined by

$L i k (x)= ∑ n = 1 ∞ x n n k (k∈Z)(see ).$
(1.4)

As is known, poly-Bernoulli polynomials are defined by the generating function to be

(1.5)

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

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

Now, we use the notation $P=C[x]$. In this paper, $P ∗$ will be denoted by the vector space of all linear functionals on . Let us assume that $〈L|p(x)〉$ be the action of the linear functional L on the polynomial $p(x)$, and we remind 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 . The formal power series

$f(t)= ∑ k = 0 ∞ a k k ! t k ∈F$
(1.7)

defines a linear functional on by setting

(1.8)

From (1.7) and (1.8), we note that

$〈 t k | x n 〉 =n! δ n , k (see [14, 15]),$
(1.9)

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

Let us consider $f L (t)= ∑ k = 0 ∞ 〈 L | x n 〉 k ! t k$. Then we see that $〈 f L (t)| x n 〉=〈L| x n 〉$, and so $L= f L (t)$ as linear functionals. The map $L↦ f L (t)$ is a vector space isomorphism from $P ∗$ onto . Henceforth, will denote 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 (see ). We shall call the umbral algebra. The umbral calculus is the study of umbral algebra. The order $o(f(t))$ of a nonzero power series $f(t)$ is the smallest integer k, for which the coefficient of $t k$ does not vanish. A series $f(t)$ is called a delta series if $o(f(t))=1$, and an invertible series if $o(f(t))=0$. Let $f(t),g(t)∈F$. Then we have

$〈 f ( t ) g ( t ) | p ( x ) 〉 = 〈 f ( t ) | g ( t ) p ( x ) 〉 = 〈 g ( t ) | f ( t ) p ( x ) 〉 (see ).$
(1.10)

For $f(t),g(t)∈F$ with $o(f(t))=1$, $o(g(t))=0$, 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 $n,k≥0$. The sequence $S n (x)$ is called the Sheffer sequence for $(g(t),f(t))$, which is denoted by $S n (x)∼(g(t),f(t))$ (see [14, 15]). Let $f(t)∈F$ and $p(t)∈P$. Then we have

$f(t)= ∑ k = 0 ∞ 〈 f ( t ) | x k 〉 t k k ! ,p(x)= ∑ k = 0 ∞ 〈 t k | p ( x ) 〉 x k k ! .$
(1.11)

From (1.11), we note that

$p ( k ) (0)= 〈 t k | p ( x ) 〉 = 〈 1 | p ( k ) ( x ) 〉 .$
(1.12)

By (1.12), we get

$t k p(x)= p ( k ) (x)= d k p ( x ) d x k (see [14, 15]).$
(1.13)

From (1.13), we easily derive the following equation

$e y t p(x)=p(x+y), 〈 e y t | p ( x ) 〉 =p(y).$
(1.14)

For $p(x)∈P$, $f(t)∈F$, it is known that

$〈 f ( t ) | x p ( x ) 〉 = 〈 ∂ t f ( t ) | p ( x ) 〉 = 〈 f ′ ( t ) | p ( x ) 〉 (see ).$
(1.15)

Let $S n (x)∼(g(t),f(t))$. Then we have

(1.16)

where $f ¯ (t)$ is the compositional inverse of $f(t)$ with $f ¯ (f(t))=t$, and

$f(t) S n (x)=n S n − 1 (x)(see [14, 15]).$
(1.17)

The Stirling number of the second kind is defined by the generating function to be

$( e t − 1 ) m =m! ∑ l = m ∞ S 2 (l,m) t m m ! (m∈ Z ≥ 0 ).$
(1.18)

For $S n (x)∼(g(t),t)$, it is well known that

$S n + 1 (x)= ( x − g ′ ( t ) g ( t ) ) S n (x)(n≥0)(see [14, 15]).$
(1.19)

Let $S n (x)∼(g(t),f(t))$, $r n (x)∼(h(t),l(t))$. Then we have

$S n (x)= ∑ m = 0 n C n , m r m (x),$
(1.20)

where

$C n , m = 1 m ! 〈 h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n 〉 (see [14, 15]).$
(1.21)

In this paper, we study higher-order Frobeniuns-Euler polynomials associated with poly-Bernoulli polynomials, which are called higher-order Frobenius-Euler and poly-Beroulli mixed-type polynomials. The purpose of this paper is to give various identities of those polynomials arising from umbral calculus.

## 2 Higher-order Frobenius-Euler polynomials, associated poly-Bernoulli polynomials

Let us consider the polynomials $T n ( r , k ) (x|λ)$, called higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials, as follows:

$( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t e x t = ∑ n = 0 ∞ T n ( r , k ) (x|λ) t n n ! ,$
(2.1)

where $λ∈C$ with $λ≠1$, $r,k∈Z$.

When $x=0$, $T n ( r , k ) (λ)= T n ( r , k ) (0|λ)$ is called the n th higher-order Frobenius-Euler and poly-Bernoulli mixed type number.

From (1.16) and (2.1), we note that

$T n ( r , k ) (x|λ)∼ ( g r , k ( t ) = ( e t − λ 1 − λ ) r 1 − e − t L i k ( 1 − e − t ) , t ) .$
(2.2)

By (1.17) and (2.2), we get

$t T n ( r , k ) (x|λ)=n T n − 1 ( r , k ) (x|λ).$
(2.3)

From (2.1), we can easily derive the following equation

$T n ( r , k ) ( x | λ ) = ∑ l = 0 n ( n l ) H n − l ( r ) ( λ ) B l ( k ) ( x ) = ∑ l = 0 n ( n l ) H n − l ( r ) ( x | λ ) B l ( k ) .$
(2.4)

By (1.16) and (2.2), we get

$T n ( r , k ) (x|λ)= 1 g r , k ( t ) x n = ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t x n .$
(2.5)

In , it is known that

$L i k ( 1 − e − t ) 1 − e − t x n = ∑ m = 0 n 1 ( m + 1 ) k ∑ j = 0 m ( − 1 ) j ( m j ) ( x − j ) n .$
(2.6)

Thus, by (2.5) and (2.6), we get

$T n ( r , k ) ( x | λ ) = ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t x n = ∑ m = 0 ∞ 1 ( m + 1 ) k ∑ j = 0 m ( − 1 ) j ( m j ) ( 1 − λ e t − λ ) r ( x − j ) n = ∑ m = 0 n 1 ( m + 1 ) k ∑ j = 0 m ( − 1 ) j ( m j ) H n ( r ) ( x − j | λ ) .$
(2.7)

By (1.1), we easily see that

$H n ( r ) (x|λ)= ∑ l = 0 n ( n l ) H n − l ( r ) (λ) x l .$
(2.8)

Therefore, by (2.7) and (2.8), we obtain the following theorem.

Theorem 2.1 For $r,k∈Z$, $n≥0$, we have

$T n ( r , k ) ( x | λ ) = ∑ m = 0 n 1 ( m + 1 ) k ∑ j = 0 m ( − 1 ) j ( m j ) ∑ l = 0 n ( n l ) H n − l ( r ) ( λ ) ( x − j ) l = ∑ l = 0 n { ( n l ) H n − l ( r ) ( λ ) ∑ m = 0 ∞ 1 ( m + 1 ) k ∑ j = 0 m ( − 1 ) j ( m j ) } ( x − j ) l .$

In , it is known that

$L i k ( 1 − e − t ) 1 − e − t x n = ∑ j = 0 n { ∑ m = 0 n − j ( − 1 ) n − m − j ( m + 1 ) k ( n j ) m ! S 2 ( n − j , m ) } x j .$
(2.9)

By (2.5) and (2.9), we get

$T n ( r , k ) ( x | λ ) = ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t x n = ∑ j = 0 n { ∑ m = 0 n − j ( − 1 ) n − m − j ( m + 1 ) k ( n j ) m ! S 2 ( n − j , m ) } ( 1 − λ e t − λ ) r x j = ∑ j = 0 n { ∑ m = 0 n − j ( − 1 ) n − m − j ( m + 1 ) k ( n j ) m ! S 2 ( n − j , m ) } H j ( r ) ( x | λ ) .$
(2.10)

Therefore, by (2.8) and (2.10), we obtain the following theorem.

Theorem 2.2 For $r,k∈Z$, $n∈ Z ≥ 0$, we have

$T n ( r , k ) (x|λ)= ∑ l = 0 n { ∑ j = l n ∑ m = 0 n − j ( − 1 ) n − m − j ( n j ) ( j l ) m ! ( m + 1 ) k H j − l ( r ) ( λ ) S 2 ( n − j , m ) } x l .$

From (1.19) and (2.2), we have

$T n + 1 ( r , k ) (x|λ)= ( x − g r , k ′ ( t ) g r , k ( t ) ) T n ( r , k ) (x|λ).$
(2.11)

Now, we note that

$g r , k ′ ( t ) g r , k ( t ) = ( log g r , k ( t ) ) ′ = ( r log ( e t − λ ) − r log ( 1 − λ ) + log ( 1 − e − t ) − log L i k ( 1 − e t ) ) ′ = r + r λ e t λ + ( t e t − 1 ) L i k ( 1 − e − t ) − L i k − 1 ( 1 − e − t ) t L i k ( 1 − e − t ) .$
(2.12)

By (2.11) and (2.12), we get

$T n + 1 ( r , k ) ( x | λ ) = x T n ( r , k ) ( x | λ ) − r T n ( r , k ) ( x | λ ) − r λ 1 − λ ( 1 − λ e t − λ ) r + 1 L i k ( 1 − e − t ) 1 − e − t x n − ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) − L i k − 1 ( 1 − e − t ) t ( 1 − e − t ) ( t e t − 1 ) x n = ( x − r ) T n ( r , k ) ( x | λ ) − r λ 1 − λ T n ( r + 1 , k ) ( x | λ ) − ∑ l = 0 n ( n l ) B n − l ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) − L i k − 1 ( 1 − e − t ) t ( 1 − e − t ) x l .$
(2.13)

It is easy to show that

$L i k ( 1 − e − t ) − L i k − 1 ( 1 − e − t ) 1 − e − t = 1 1 − e − t ∑ n = 1 ∞ { ( 1 − e − t ) n n k − ( 1 − e − t ) n n k − 1 } = ( 1 − e − t 2 k − 1 − e − t 2 k − 1 ) + ⋯ = ( 1 2 k − 1 2 k − 1 ) t + ⋯ .$
(2.14)

For any delta series $f(t)$, we have

$f ( t ) t x n =f(t) 1 n + 1 x n + 1 .$
(2.15)

Thus, by (2.13), (2.14) and (2.15), we get

$T n + 1 ( r , k ) ( x | λ ) = ( x − r ) T n ( r , k ) ( x | λ ) − r λ 1 − λ T n ( r + 1 , k ) ( x | λ ) − ∑ l = 0 n ( n l ) B n − l 1 l + 1 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) − L i k − 1 ( 1 − e − t ) 1 − e − t x l + 1 = ( x − r ) T n ( r , k ) ( x | λ ) − r λ 1 − λ T n ( r + 1 , k ) ( x | λ ) − ∑ l = 0 n ( n l ) l + 1 B n − l { T l + 1 ( r , k ) ( x | λ ) − T l + 1 ( r , k − 1 ) ( x | λ ) } = ( x − r ) T n ( r , k ) ( x | λ ) − r λ 1 − λ T n ( r + 1 , k ) ( x | λ ) − 1 n + 1 ∑ l = 1 n + 1 ( n + 1 l ) B n + 1 − l { T l ( r , k ) ( x | λ ) − T l ( r , k − 1 ) ( x | λ ) } = ( x − r ) T n ( r , k ) ( x | λ ) − r λ 1 − λ T n ( r + 1 , k ) ( x | λ ) − 1 n + 1 ∑ l = 0 n + 1 ( n + 1 l ) B n + 1 − l { T l ( r , k ) ( x | λ ) − T l ( r , k − 1 ) ( x | λ ) } = ( x − r ) T n ( r , k ) ( x | λ ) − r λ 1 − λ T n ( r + 1 , k ) ( x | λ ) − 1 n + 1 ∑ l = 0 n + 1 ( n + 1 l ) B l { T n + 1 − l ( r , k ) ( x | λ ) − T n + 1 − l ( r , k − 1 ) ( x | λ ) } .$
(2.16)

Therefore, by (2.16), we obtain the following theorem.

Theorem 2.3 For $r,k∈Z$, $n∈ Z ≥ 0$, we have

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

Remark 1 If $r=0$, then we have

$∑ n = 0 ∞ B n ( k ) (x) t n n ! = L i k ( 1 − e − t ) ( 1 − e − t ) e x t = ∑ n = 0 ∞ T n ( 0 , k ) (x|λ) t n n ! .$
(2.17)

Thus, by (2.17), we get $B n ( k ) (x)= T n ( 0 , k ) (x|λ)$.

From (2.4), we have

$t x T n ( r , k ) ( x | λ ) = t ( x ∑ l = 0 n ( n l ) H n − l ( r ) ( λ ) B l ( k ) ( x ) ) = ∑ l = 0 n ( n l ) H n − l ( r ) ( λ ) { l x B l − 1 ( k ) ( x ) + B l ( k ) ( x ) } = n x ∑ l = 0 n − 1 ( n − 1 l ) H n − 1 − l ( r ) ( λ ) B l ( k ) ( x ) + ∑ l = 0 n ( n l ) H n − l ( r ) ( λ ) B l ( k ) ( x ) = n x T n − 1 ( r , k ) ( x | λ ) + T n ( r , k ) ( x | λ ) .$
(2.18)

Applying t on both sides of Theorem 2.3, we get

$( n + 1 ) T n ( r , k ) ( x | λ ) = n x T n − 1 ( r , k ) ( x | λ ) + T n ( r , k ) ( x | λ ) − r n T n − 1 ( r , k ) ( x | λ ) − r n λ 1 − λ T n − 1 ( r + 1 , k ) ( x | λ ) − 1 n + 1 ∑ l = 0 n + 1 ( n + 1 l ) B l { ( n + 1 − l ) T n − l ( r , k ) ( x | λ ) − ( n + 1 − l ) T n − l ( r , k − 1 ) ( x | λ ) } .$
(2.19)

Thus, by (2.19), we have

$( n + 1 ) T n ( r , k ) ( x | λ ) + n ( r − 1 2 − x ) T n − 1 ( r , k ) ( x | λ ) + ∑ l = 0 n − 2 ( n l ) B n − l T l ( r , k ) ( x | λ ) = − r λ n 1 − λ T n − 1 ( r + 1 , k ) ( x | λ ) + ∑ l = 0 n ( n l ) B n − l T l ( r , k − 1 ) ( x | λ ) .$
(2.20)

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

Theorem 2.4 For $r,k∈Z$, $n∈Z$ with $n≥2$, we have

$( n + 1 ) T n ( r , k ) ( x | λ ) + n ( r − 1 2 − x ) T n − 1 ( r , k ) ( x | λ ) + ∑ l = 0 n − 2 ( n l ) B n − l T l ( r , k ) ( x | λ ) = − r λ n 1 − λ T n − 1 ( r + 1 , k ) ( x | λ ) + ∑ l = 0 n ( n l ) B n − l T l ( r , k − 1 ) ( x | λ ) .$

From (1.14) and (2.5), we note that

$T n ( r , k ) ( y | λ ) = 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t e y t | x n 〉 = 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t e y t | x x n − 1 〉 .$
(2.21)

By (1.15) and (2.21), we get

$T n ( r , k ) ( y | λ ) = 〈 ∂ t ( ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t e y t ) | x n − 1 〉 = 〈 ( ∂ t ( 1 − λ e t − λ ) r ) L i k ( 1 − e − t ) 1 − e − t e y t | x n − 1 〉 + 〈 ( 1 − λ e t − λ ) r ( ∂ t L i k ( 1 − e − t ) 1 − e − t ) e y t | x n − 1 〉 + 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t ∂ t e y t | x n − 1 〉 .$
(2.22)

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

Theorem 2.5 For $r,k∈Z$, $n≥1$, we have

$T n ( r , k ) ( x | λ ) = ( x − r ) T n − 1 ( r , k ) ( x | λ ) − r λ 1 − λ T n − 1 ( r + 1 , k ) ( x | λ ) + ∑ l = 0 n − 1 { ( − 1 ) n − 1 − l ( n − 1 l ) ∑ m = 0 n − 1 − l ( − 1 ) m ( m + 1 ) ! ( m + 2 ) k S 2 ( n − 1 − l , m ) } H l ( r ) ( x − 1 | λ ) .$

Now, we compute $〈 ( 1 − λ e t − λ ) r L i k (1− e − t )| x n + 1 〉$ in two different ways.

On the one hand,

$〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) | x n + 1 〉 = 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | ( 1 − e − t ) x n + 1 〉 = 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | x n + 1 − ( x − 1 ) n + 1 〉 = ∑ m = 0 n ( n + 1 m ) ( − 1 ) n − m 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | x m 〉 = ∑ m = 0 n ( n + 1 m ) ( − 1 ) n − m 〈 1 | T m ( r , k ) ( x | λ ) 〉 = ∑ m = 0 n ( n + 1 m ) ( − 1 ) n − m T m ( r , k ) ( λ ) .$
(2.23)

On the other hand, we get

$〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) | x n + 1 〉 = 〈 L i k ( 1 − e − t ) | ( 1 − λ e t − λ ) r x n + 1 〉 = 〈 ∫ 0 t ( L i k ( 1 − e − s ) ) ′ d s | H n + 1 ( r ) ( x | λ ) 〉 = 〈 ∫ 0 t e − s L i k ( 1 − e − s ) ( 1 − e − s ) d s | H n + 1 ( r ) ( x | λ ) 〉 = ∑ l = 0 n ( ∑ m = 0 l ( l m ) ( − 1 ) l − m B m ( k − 1 ) ) 1 l ! 〈 ∫ 0 t s l d s | H n + 1 ( r ) ( x | λ ) 〉 = ∑ l = 0 n ∑ m = 0 l ( l m ) ( − 1 ) l − m B m ( k − 1 ) ( l + 1 ) ! 〈 t l + 1 | H n + 1 ( r ) ( x | λ ) 〉 = ∑ l = 0 n ∑ m = 0 l ( l m ) ( n + 1 l + 1 ) ( − 1 ) l − m B m ( k − 1 ) H n − l ( r ) ( λ ) .$
(2.24)

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

Theorem 2.6 For $r,k∈Z$, $n∈ Z ≥ 0$, we have

$∑ m = 0 n ( n + 1 m ) ( − 1 ) n − m T m ( r , k ) ( λ ) = ∑ l = 0 n ∑ m = 0 l ( − 1 ) l − m ( l m ) ( n + 1 l + 1 ) B m ( k − 1 ) H n − l ( r ) ( λ ) .$

Now, we consider the following two Sheffer sequences:

$T n ( r , k ) ( x | λ ) ∼ ( ( e t − λ 1 − λ ) r 1 − e − t L i k ( 1 − e − t ) , t ) , B ( s ) ∼ ( ( e t − 1 t ) s , t ) ,$
(2.25)

where $s∈ Z ≥ 0$, $r,k∈Z$ and $λ∈C$ with $λ≠1$. Let us assume that

$T n ( r , k ) (x|λ)= ∑ m = 0 n C n ⋅ m B m ( s ) (x).$
(2.26)

By (1.21) and (2.26), we get

$C n , m = 1 m ! 〈 ( e t − 1 t ) s ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t t m | x n 〉 = 1 m ! 〈 ( e t − 1 t ) s ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | t m x n 〉 = ( n m ) 〈 ( e t − 1 t ) s ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | x n − m 〉 = ( n m ) ∑ l = 0 n − m s ! ( l + s ) ! S 2 ( l + s , s ) 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | t l x n − m 〉 = ( n m ) ∑ l = 0 n − m s ! l ! ( l + s ) ! ( n − m ) l l ! S 2 ( l + s , s ) 〈 1 | T n − m − l ( r , k ) ( x | λ ) 〉 = ( n m ) ∑ l = 0 n − m ( n − m l ) ( s + l l ) S 2 ( l + s , s ) T n − m − l ( r , k ) ( λ ) .$
(2.27)

Therefore, by (2.26) and (2.27), we obtain the following theorem.

Theorem 2.7 For $r,k∈Z$, $s∈ Z ≥ 0$, we have

$T n ( r , k ) (x|λ)= ∑ m = 0 n { ( n m ) ∑ l = 0 n − m ( n − m l ) ( s + l l ) S 2 ( l + s , s ) T n − m − l ( r , k ) ( λ ) } B m ( s ) (x).$

From (1.3) and (2.1), we note that

$T n ( r , k ) ( x | λ ) ∼ ( ( e t − λ 1 − λ ) r 1 − e − t L i k ( 1 − e − t ) , t ) , E n ( r , s ) ( x ) ∼ ( ( e t + 1 2 ) s , t ) ,$
(2.28)

where $r,k∈Z$, $s∈ Z ≥ 0$.

By the same method, we get

$T n ( r , k ) (x|λ)= 1 2 s ∑ m = 0 n { ( n m ) ∑ j = 0 s ( s j ) T n − m ( r , k ) ( j ) } E m ( s ) (x).$
(2.29)

From (1.1) and (2.1), we note that

$T n ( r , k ) ( x | λ ) ∼ ( ( e t − λ 1 − λ ) r 1 − e − t L i k ( 1 − e − t ) , t ) , H n ( s ) ( x | μ ) ∼ ( ( e t − μ 1 − μ ) s , t ) ,$
(2.30)

where $r,k∈Z$, and $λ,μ∈C$ with $λ≠1$, $μ≠1$, $s∈ Z ≥ 0$.

Let us assume that

$T n ( r , k ) (x|λ)= ∑ m = 0 n C n , m H m ( s ) (x|μ).$
(2.31)

By (1.21) and (2.31), we get

$C n , m = 1 m ! 〈 ( e t − μ 1 − μ ) s ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t t m | x n 〉 = ( n m ) ( 1 − μ ) s 〈 ( e t − μ ) s | ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t x n − m 〉 = ( n m ) ( 1 − μ ) s ∑ j = 0 s ( s j ) ( − μ ) s − j 〈 e j t | T n − m ( r , k ) ( x | λ ) 〉 = ( n m ) ( 1 − μ ) s ∑ j = 0 s ( s j ) ( − μ ) s − j T n − m ( r , k ) ( j | λ ) .$
(2.32)

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

Theorem 2.8 For $r,k∈Z$, $s∈ Z ≥ 0$, we have

$T n ( r , k ) (x|λ)= 1 ( 1 − μ ) s ∑ m = 0 n { ( n m ) ∑ j = 0 s ( s j ) ( − μ ) s − j T n − m ( r , k ) ( j | λ ) } H m ( s ) (x|μ).$

It is known that

$T n ( r , k ) ( x | λ ) ∼ ( ( e t − λ 1 − λ ) r 1 − e − t L i k ( 1 − e − t ) , t ) , ( x ) n ∼ ( 1 , e t − 1 ) .$
(2.33)

Let

$T n ( r , k ) (x|λ)= ∑ m = 0 n C n , m ( x ) m .$
(2.34)

Then, by (1.21) and (2.34), we get

$C n , m = 1 m ! 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t ( e t − 1 ) m | x n 〉 = ∑ l = 0 ∞ S 2 ( l + m , m ) ( l + m ) ! 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | t m + l x n 〉 = ∑ l = 0 n − m S 2 ( l + m , m ) ( l + m ) ! ( n ) m + l 〈 1 | ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t x n − m − l 〉 = ∑ l = 0 n − m ( n l + m ) S 2 ( l + m , m ) T n − m − l ( r , k ) ( λ ) .$
(2.35)

Therefore, by (2.34) and (2.35), we obtain the following theorem.

Theorem 2.9 For $r,k∈Z$, we have

$T n ( r , k ) (x|λ)= ∑ m = 0 n { ∑ l = 0 n − m ( n l + m ) S 2 ( l + m , m ) T n − m − l ( r , k ) ( λ ) } ( x ) m .$

Finally, we consider the following two Sheffer sequences:

$T n ( r , k ) ( x | λ ) ∼ ( ( e t − λ 1 − λ ) r 1 − e − t L i k ( 1 − e − t ) , t ) , x [ n ] ∼ ( 1 , 1 − e − t ) ,$
(2.36)

where $x [ n ] =x(x+1)⋯(x+n−1)$.

Let us assume that

$T n ( r , k ) (x|λ)= ∑ m = 0 n C n , m x [ m ] .$
(2.37)

Then, by (1.21) and (2.37), we get

$C n , m = 1 m ! 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t ( 1 − e − t ) m | x n 〉 = ∑ l = 0 ∞ ( − 1 ) l S 2 ( l + m , m ) ( l + m ) ! 〈 ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t | t m + l x n 〉 = ∑ l = 0 n − m ( − 1 ) l S 2 ( l + m , m ) ( l + m ) ! ( n ) m + l 〈 1 | ( 1 − λ e t − λ ) r L i k ( 1 − e − t ) 1 − e − t x n − m − l 〉 = ∑ l = 0 n − m ( − 1 ) l ( n l + m ) S 2 ( l + m , m ) T n − m − l ( r , k ) ( λ ) .$
(2.38)

Therefore, by (2.37) and (2.38), we obtain the following theorem.

Theorem 2.10 For $r,k∈Z$, $n≥0$, we have

$T n ( r , k ) (x|λ)= ∑ m = 0 n { ∑ l = 0 n − m ( − 1 ) l ( n l + m ) S 2 ( l + m , m ) T n − m − l ( r , k ) ( λ ) } x [ m ] .$

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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.

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

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Kim, D.S., Kim, T. Higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials. Adv Differ Equ 2013, 251 (2013). https://doi.org/10.1186/1687-1847-2013-251 