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# Barnes-type Narumi polynomials

## Abstract

In this paper, we study the Barnes-type Narumi polynomials with umbral calculus viewpoint. From our study, we derive various identities of the Barnes-type Narumi polynomials.

MSC:05A19, 05A40, 11B68.

## 1 Introduction

As is well known, the Narumi polynomials of order α are defined by the generating function to be

$( t log ( 1 + t ) ) α ( 1 + t ) x = ∑ n = 0 ∞ N n ( α ) (x) t n n ! (see ).$
(1)

Let $r∈ Z > 0$. We consider the polynomials $N n (x| a 1 ,…, a r )$ and $N ˆ n (x| a 1 ,…, a r )$, respectively, called the Barnes-type Narumi polynomials of the first kind and those of the second kind and respectively given by

$∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( 1 + t ) x = ∑ n = 0 ∞ N n (x| a 1 ,…, a r ) t n n !$
(2)

and

$∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ( 1 + t ) a j ) ( 1 + t ) x = ∑ n = 0 ∞ N ˆ n (x| a 1 ,…, a r ) t n n ! ,$
(3)

where $a 1 , a 2 ,…, a r ≠0$.

When $x=0$,

$N n ( a 1 ,…, a r )= N n (0| a 1 ,…, a r )$

and

$N ˆ n ( a 1 ,…, a r )= N ˆ n (0| a 1 ,…, a r )$

are respectively called the Barnes-type Narumi numbers of the first kind and those of the second kind.

Note that

$N n ( x | 1 , … , 1 ⏟ r ) = N n ( r ) ( x ) , N ˆ n ( x | 1 , … , 1 ⏟ r ) = N ˆ n ( r ) ( x )$

and

$N ˆ n (x| 1 , … , 1 ⏟ r )= N n ( r ) (x−r).$

In the previous paper , $N n ( α ) (x)$ was denoted by $N n ( − α )$ and called the Narumi polynomial of order α.

The Bernoulli polynomials are defined by the generating function to be

$t e t − 1 e x t = ∑ n = 0 ∞ B n (x) t n n ! (see [3–6]).$
(4)

When $x=0$, $B n = B n (0)$ are called the Bernoulli numbers. In , it is known that the Cauchy numbers are given by

$t log ( 1 + t ) = ∑ n = 0 ∞ C n t n n ! .$
(5)

It is well known that the Stirling number of the first kind is given by

$( x ) n =x(x−1)⋯(x−n+1)= ∑ l = 0 ∞ S 1 (n,l) x l (n≥0)(see [1, 2, 7–11]).$
(6)

From (6), we have

$( log ( 1 + t ) ) n =n! ∑ l = n ∞ S 1 (l,n) t l l ! (n≥0).$
(7)

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

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

Let $P=C[x]$ and let $P ∗$ be the vector space of all linear functionals on . $〈L|p(x)〉$ denotes the action of the linear functional L on $p(x)$ which satisfies $〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉$, and $〈cL|p(x)〉=c〈L|p(x)〉$, where c is a complex constant. The linear functional $〈f(t)|⋅〉$ on is defined by $〈f(t)| x n 〉= a n$ ($n≥0$), where $f(t)∈F$. Thus, we note that

$〈 t k | x n 〉 =n! δ n , k (n,k≥0),$
(8)

where $δ n , k$ is the Kronecker symbol (see ).

For $f L (t)= ∑ k = 0 ∞ 〈 L | x k 〉 k ! t k$, we have $〈 f L (t)| x n 〉=〈L| x n 〉$. So, 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. The order $o(f(t))$ of a power series $f(t)≠0$ is the smallest integer 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 $g(t)$ is called an invertible series. Let $f(t),g(t)∈F$ with $o(f(t))=1$ and $o(g(t))=0$. Then 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 $s n (x)∼(g(t),f(t))$.

For $f(t),g(t)∈F$ and $p(x)∈P$, we have

$〈 f ( t ) g ( t ) | p ( x ) 〉 = 〈 f ( t ) | g ( t ) p ( x ) 〉 = 〈 g ( t ) | f ( t ) p ( x ) 〉$
(9)

and

$f ( t ) = ∑ k = 0 ∞ 〈 f ( t ) | x k 〉 t k k ! , p ( x ) = ∑ k = 0 ∞ 〈 t k | p ( x ) 〉 x k k ! .$
(10)

From (10), we can derive the following equation (11):

$t k p(x)= p ( k ) (x)= d k p ( x ) d x k , e y t p(x)=p(x+y)(see ).$
(11)

Let $s n (x)∼(g(t),f(t))$. Then the following will be used:

$d s n ( x ) d x = ∑ l = 0 n − 1 ( n l ) 〈 f ¯ ( t ) | x n − l 〉 s l (x),$
(12)

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

(13)
$f(t) s n (x)=n s n − 1 (x)(n≥1), s n (x)= ∑ j = 0 n 〈 g ( f ¯ ( t ) ) − 1 f ¯ ( t ) j | x n 〉 j ! x j ,$
(14)
(15)
(16)

and

$s n + 1 (x)= ( x − g ′ ( t ) g ( t ) ) 1 f ′ ( t ) s n (x)(n≥0)(see [1, 19]).$
(17)

Let us assume that $s n (x)∼(g(t),f(t))$ and $r n (x)∼(h(t),l(t))$. Then we have

$s n (x)= ∑ m = 0 n c n , m r m (x)(n≥0),$
(18)

where

$c n , m = 1 m ! 〈 h ( f ¯ ( t ) ) g ( f ¯ ( t ) ) l ( f ¯ ( t ) ) m | x n 〉 (see [1, 5]).$
(19)

From (2), (3) and (13), we note that

$N n (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( t e a j t − 1 ) , e t − 1 )$
(20)

and

$N ˆ n (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( t e a j t e a j t − 1 ) , e t − 1 ) .$
(21)

In this paper, we study the Barnes-type Narumi polynomials with umbral calculus viewpoint. From our study, we derive various identities of the Barnes-type Narumi polynomials.

## 2 Barnes-type Narumi polynomials

From (21), we note that

$∏ j = 1 r ( t e a j t − 1 ) N n (x| a 1 ,…, a r )∼ ( 1 , e t − 1 )$
(22)

and

$( x ) n ∼ ( 1 , e t − 1 ) .$
(23)

Thus, by (22) and (23), we get

$N n ( x | a 1 , … , a r ) = ∏ j = 1 r ( e a j t − 1 t ) ( x ) n = ∑ m = 0 n S 1 ( n , m ) ∏ j = 1 r ( e a j t − 1 t ) x m .$
(24)

Note that

$∏ j = 1 r ( e a j t − 1 t ) = ( ∑ l 1 = 0 ∞ a 1 l 1 + 1 ( l 1 + 1 ) ! t l 1 ) × ⋯ × ( ∑ l r = 0 ∞ a r l r + 1 ( l r + 1 ) ! t l r ) = ∑ l 1 , … , l r = 0 ∞ ∑ l 1 + ⋯ + l r = i a 1 l 1 + 1 ⋯ a r l r + 1 ( l 1 + 1 ) ! ⋯ ( l r + 1 ) ! t i .$
(25)

From (24) and (25), we have

$N n ( x | a 1 , … , a r ) = ∑ m = 0 ∞ S 1 ( n , m ) ∑ i = 0 m ∑ l 1 + ⋯ + l r = i a 1 l 1 + 1 ⋯ a r l r + 1 ( l 1 + 1 ) ! ⋯ ( l r + 1 ) ! t i x m = ∑ m = 0 n S 1 ( n , m ) ∑ i = 0 m i ! ( i + r ) ! ∑ l 1 + ⋯ + l r = i ( i + r l 1 + 1 , … , l r + 1 ) ( m i ) × a 1 l 1 + 1 ⋯ a r l r + 1 x m − i = ∑ i = 0 n { ∑ m = i n ∑ l 1 + ⋯ + l r = m − i S 1 ( n , m ) ( m − i ) ! ( m − i + r ) ! × ( m − i + r l 1 + 1 , … , l r + 1 ) ( m i ) a 1 l 1 + 1 ⋯ a r l r + 1 } x i .$
(26)

By (21), we see that

$∏ j = 1 r ( t e a j t e a j t − 1 ) N ˆ n (x| a 1 ,…, a r )∼ ( 1 , e t − 1 ) ,$
(27)

and we recall (23).

Thus, we have

$N ˆ n ( x | a 1 , … , a r ) = ∏ j = 1 r ( e a j t − 1 t e a j t ) ( x ) n = e − ∑ j = 1 r a j t ∏ j = 1 r ( e a j t − 1 t ) ( x ) n = e − ∑ j = 1 r a j t N n ( x | a 1 , … , a r ) = ∑ i = 0 n { ∑ m = i n ∑ l 1 + ⋯ + l r = m − i S 1 ( n , m ) ( m − i ) ! ( m − i + r ) ! × ( m − i + r l 1 + 1 , … , l r + 1 ) ( m i ) a 1 l 1 + 1 ⋯ a r l r + 1 } e − ∑ j = 1 r a j t x i = ∑ i = 0 n { ∑ m = i n ∑ l 1 + ⋯ + l r = m − i S 1 ( n , m ) ( m − i ) ! ( m − i + r ) ! × ( m − i + r l 1 + 1 , … , l r + 1 ) ( m i ) a 1 l 1 + 1 ⋯ a r l r + 1 } ( x − ∑ j = 1 r a j ) i$
(28)

or

$N ˆ n ( x | a 1 , … , a r ) = ∏ j = 1 r ( e a j t − 1 t e a j t ) ( x ) n = ∏ j = 1 r ( e − a j t − 1 − t ) ( x ) n = ∑ m = 0 n S 1 ( n , m ) ∑ i = 0 m ( − 1 ) i i ! ( i + r ) ! × ∑ l 1 + ⋯ + l r = i ( i + r i 1 + 1 , … , i r + 1 ) ( m i ) a 1 l 1 + 1 ⋯ a r l r + 1 x m − i = ∑ i = 0 n { ∑ m = i n ∑ l 1 + ⋯ + l r = m − i ( − 1 ) m − i S 1 ( n , m ) ( m − i ) ! ( m − i + r ) ! × ( m − i + r l 1 + 1 , … , l r + 1 ) ( m i ) a 1 l 1 + 1 ⋯ a r l r + 1 } x i .$
(29)

Therefore, by (26), (28) and (29), we obtain the following theorem.

Theorem 1 For $n≥0$, we have

$N n ( x | a 1 , … , a r ) = ∑ i = 0 n { ∑ m = i n ∑ l 1 + ⋯ + l r = m − i S 1 ( n , m ) ( m − i ) ! ( m − i + r ) ! × ( m − i + r l 1 + 1 , … , l r + 1 ) ( m i ) a 1 l 1 + 1 ⋯ a r l r + 1 } x i$

and

$N ˆ n ( x | a 1 , … , a r ) = ∑ i = 0 n { ∑ m = i n ∑ l 1 + ⋯ + l r = m − i S 1 ( n , m ) ( m − i ) ! ( m − i + r ) ! × ( m − i + r l 1 + 1 , … , l r + 1 ) ( m i ) a 1 l 1 + 1 ⋯ a r l r + 1 } ( x − ∑ j = 1 r a j ) i = ∑ i = 0 n { ∑ m = i n ∑ l 1 + ⋯ + l r = m − i ( − 1 ) m − i S 1 ( n , m ) × ( m − i ) ! ( m − i + r ) ! ( m − i + r l 1 + 1 , … , l r + 1 ) ( m j ) a 1 l 1 + 1 ⋯ a r l r + 1 } x i .$

From (14), we can derive the following equation (30):

$N n (x| a 1 ,…, a r )= ∑ j = 0 n 1 j ! 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) j | x n 〉 x j ,$
(30)

where

$〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) j | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) | ( log ( 1 + t ) ) j x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) | j ! ∑ l = j ∞ S 1 ( l , j ) t l l ! x n 〉 = j ! ∑ l = j n ( n l ) S 1 ( l , j ) N n − l ( a 1 , … , a r ) .$
(31)

Thus, by (30) and (31), we obtain the following theorem.

Theorem 2 For $n≥0$, we have

$N n (x| a 1 ,…, a r )= ∑ j = 0 n { ∑ l = j n ( n l ) S 1 ( l , j ) N n − l ( a 1 , … , a r ) } x j .$

By the same methods as in (28), (29) and (30), we get

$N ˆ n ( x | a 1 , … , a r ) = ∑ j = 0 n { ∑ l = j n ( n l ) S 1 ( l , j ) N n − l ( a 1 , … , a r ) } ( x − ∑ i = 1 r a i ) j = ∑ j = 0 n { ∑ l = j n ( n l ) S 1 ( l , j ) N ˆ n − l ( a 1 , … , a r ) } x j .$
(32)

By (8), we get

$N n ( y | a 1 , … , a r ) = 〈 ∑ i = 0 ∞ N i ( y | a 1 , … , a r ) t i i ! | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( 1 + t ) y | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) | ∑ m = 0 ∞ ( y ) m t m m ! x m 〉 = ∑ m = 0 n ( y ) m ( n m ) 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) | x n − m 〉 = ∑ m = 0 n ( y ) m ( n m ) N n − m ( a 1 , … , a r )$
(33)

and

$N ˆ n ( y | a 1 , … , a r ) = 〈 ∑ i = 0 ∞ N ˆ i ( y | a 1 , … , a r ) t i i ! | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ( 1 + t ) a j ) ( 1 + t ) y | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ( 1 + t ) a j ) | ∑ m = 0 ∞ ( y ) m t m m ! x n 〉 = ∑ m = 0 n ( y ) m ( n m ) 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ( 1 + t ) a j ) | x n − m 〉 = ∑ m = 0 n ( n m ) ( y ) m N ˆ n − m ( a 1 , … , a r ) .$
(34)

Therefore, by (33) and (34), we obtain the following theorem.

Theorem 3 For $n≥0$, we have

$N n (x| a 1 ,…, a r )= ∑ m = 0 n ( n m ) N n − m ( a 1 ,…, a r ) ( x ) m$

and

$N ˆ n (x| a 1 ,…, a r )= ∑ m = 0 n ( n m ) N ˆ n − m ( a 1 ,…, a r ) ( x ) m .$

From (15), we note that

$N n (x+y| a 1 ,…, a r )= ∑ j = 0 n ( n j ) N j (x| a 1 ,…, a r ) ( y ) n − j$
(35)

and

$N ˆ n (x+y| a 1 ,…, a r )= ∑ j = 0 n ( n j ) N ˆ j (x| a 1 ,…, a r ) ( y ) n − j .$
(36)

By (14), we get

$( e t − 1 ) N n (x| a 1 ,…, a r )=n N n − 1 (x| a 1 ,…, a r )$
(37)

and

$( e t − 1 ) N n ( x | a 1 , … , a r ) = e t N n ( x | a 1 , … , a r ) − N n ( x | a 1 , … , a r ) = N n ( x + 1 | a 1 , … , a r ) − N ( x | a 1 , … , a r ) .$
(38)

From (37) and (38), we have

$N n (x+1| a 1 ,…, a r )− N n (x| a 1 ,…, a r )=n N n − 1 (x| a 1 ,…, a r ).$
(39)

By the same method as (39), we get

$N ˆ n (x+1| a 1 ,…, a r )− N ˆ n (x| a 1 ,…, a r )=n N ˆ n − 1 (x| a 1 ,…, a r ).$
(40)

Recall that $N n (x| a 1 ,…, a r )∼( ∏ j = 1 r ( t e a j t − 1 ), e t −1)$.

From (17), we can derive the following equation (41):

$N n + 1 (x| a 1 ,…, a r )=x N n (x−1| a 1 ,…, a r )− e − t g ′ ( t ) g ( t ) N n (x| a 1 ,…, a r ).$
(41)

Now, we observe that

$g ′ ( t ) g ( t ) = ( log g ( t ) ) ′ = ( r log t − ∑ j = 1 r log ( e a j t − 1 ) ) ′ = r t − ∑ j = 1 r a j e a j t e a j t − 1 = ∑ j = 1 r ∏ i ≠ j r ( e a i t − 1 ) { e a j t − 1 − t a j e a j t } t ∏ j = 1 r ( e a j t − 1 ) ,$
(42)

where

$r − ∑ j = 1 r a j t e a j t e a j t − 1 = ∑ j = 1 r ∏ i ≠ j ( e a i t − 1 ) { e a j t − 1 − a j t e a j t } ∏ j = 1 r ( e a j t − 1 ) = − 1 2 ( ∑ j = 1 r a 1 a 2 ⋯ a j − 1 a j 2 a j + 1 ⋯ a r ) t r + 1 + ⋯ ( a 1 a 2 ⋯ a r ) t r = − 1 2 ( ∑ j = 1 r a j ) t + ⋯$
(43)

has at least the order 1.

By (42) and (43), we get

$g ′ ( t ) g ( t ) N n ( x | a 1 , … , a r ) = r − ∑ j = 1 r a j t e a j t e a j t − 1 t ( ∑ i = 0 n { ∑ l = i n ( n l ) S 1 ( l , i ) N n − l ( a 1 , … , a r ) } x i ) = ∑ i = 0 n { ∑ l = i n ( n l ) S 1 ( l , i ) N n − l ( a 1 , … , a r ) } r − ∑ j = 1 r a j t e a j t e a j t − 1 t x i = ∑ i = 0 n { ∑ l = i n ( n l ) S 1 ( l , i ) N n − l ( a 1 , … , a r ) } ( r − ∑ j = 1 r a j t e a j t e a j t − 1 ) x i + 1 i + 1 = ∑ i = 0 n 1 i + 1 { ∑ l = i n ( n l ) S 1 ( l , i ) N n − l ( a 1 , … , a r ) } ( r x i + 1 − ∑ j = 1 r ∑ m = 0 ∞ B m ( − a j t ) m m ! x i + 1 ) = − ∑ i = 0 n 1 i + 1 { ∑ l = i n ( n l ) S 1 ( l , i ) N n − l ( a 1 , … , a r ) } × ∑ j = 1 r ∑ m = 1 i + 1 ( − 1 ) m ( i + 1 m ) B m a j m x i + 1 − m = − ∑ i = 0 n 1 i + 1 { ∑ l = i n ( n l ) S 1 ( l , i ) N n − l ( a 1 , … , a r ) } × ∑ j = 1 r ∑ m = 0 i ( − 1 ) i + 1 − m ( i + 1 m ) a j i + 1 − m B i + 1 − m x m .$
(44)

Therefore, by (41) and (44), we obtain the following theorem.

Theorem 4 For $n≥0$, we have

$N n + 1 ( x | a 1 , … , a r ) = x N n ( x − 1 | a 1 , … , a r ) + ∑ m = 0 n { ∑ i = m n ∑ l = i n ∑ j = 1 r 1 i + 1 ( n l ) ( i + 1 m ) S 1 ( l , i ) × B i + 1 − m ( − a j ) i + 1 − m N n − l ( a 1 , … , a r ) } ( x − 1 ) m .$

By the same method as the proof of Theorem 4, we get

$N ˆ n + 1 ( x | a 1 , … , a r ) = ( x − ∑ j = 1 r a j ) N ˆ n ( x − 1 | a 1 , … , a r ) + ∑ m = 0 n { ∑ i = m n ∑ l = i n ∑ j = 1 r 1 i + 1 ( n l ) ( i + 1 m ) S 1 ( l , i ) × B i + 1 − m ( − a j ) i + 1 − m N ˆ n − l ( a 1 , … , a r ) } ( x − 1 ) m .$
(45)

From (12) and (20), we can derive the following equation (46):

$〈 f ¯ ( t ) | x n − l 〉 = 〈 log ( 1 + t ) | x n − l 〉 = 〈 ∑ m = 1 ∞ ( − 1 ) m − 1 m t m | x n − l 〉 = ( − 1 ) n − l − 1 (n−l−1)!.$
(46)

Thus, by (46), we get

$d d x N n ( x | a 1 , … , a r ) = ∑ l = 0 n − 1 ( n l ) ( − 1 ) n − l − 1 ( n − l − 1 ) ! N l ( x | a 1 , … , a r ) = n ! ∑ l = 0 n − 1 ( − 1 ) n − l − 1 l ! ( n − l ) N l ( x | a 1 , … , a r ) .$
(47)

By the same method as (47), we get

$d d x N ˆ n (x| a 1 ,…, a r )=n! ∑ l = 0 n − 1 ( − 1 ) n − l − 1 l ! ( n − l ) N ˆ l (x| a 1 ,…, a r ).$
(48)

From (8), we note that, for $n≥1$,

$N n ( y | a 1 , … , a r ) = 〈 ∑ i = 0 ∞ N i ( y | a 1 , … , a r ) t i i ! | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( 1 + t ) y | x n 〉 = 〈 ∂ t ( ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( 1 + t ) y ) | x n − 1 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( ∂ t ( 1 + t ) y ) | x n − 1 〉 + 〈 ( ∂ t ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ) ( 1 + t ) y | x n − 1 〉 = y N n − 1 ( y − 1 | a 1 , … , a r ) + 〈 ( ∂ t ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ) ( 1 + t ) y | x n − 1 〉 .$
(49)

Now, we observe that

$∂ t ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) = ∑ j = 1 r ∏ i ≠ j ( ( 1 + t ) a i − 1 log ( 1 + t ) ) a j ( 1 + t ) a j − 1 log ( 1 + t ) − ( ( 1 + t ) a j − 1 ) 1 1 + t ( log ( 1 + t ) ) 2 = 1 1 + t ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ∑ j = 1 r { a j ( 1 + t ) a j ( 1 + t ) a j − 1 − 1 log ( 1 + t ) } = 1 1 + t ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ∑ j = 1 r { a j t ( 1 + t ) a j ( 1 + t ) a j − 1 − t log ( 1 + t ) } t ,$
(50)

where

$∑ j = 1 r { a j t ( 1 + t ) a j ( 1 + t ) a j − 1 − t log ( 1 + t ) } = 1 2 ( ∑ j = 1 r a j ) t+⋯$
(51)

is a series with order greater than or equal to 1.

By (50) and (51), we get

$〈 ( ∂ t ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ) ( 1 + t ) y | x n − 1 〉 = 〈 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) y 1 + t | ∑ j = 1 r { a j t ( 1 + t ) a j ( 1 + t ) a j − 1 − t log ( 1 + t ) } t x n − 1 〉 = 1 n 〈 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) y − 1 | ∑ j = 1 r { a j t ( 1 + t ) a j ( 1 + t ) a j − 1 − t log ( 1 + t ) } x n 〉 = 1 n { ∑ j = 1 r a j 〈 log ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) y + a j − 1 | t log ( 1 + t ) x n 〉 − r 〈 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) y − 1 | t log ( 1 + t ) x n 〉 } = 1 n ∑ j = 1 r a j ∑ l = 0 n ( n l ) C l 〈 log ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) y + a j − 1 | x n − l 〉 − r n ∑ l = 0 n ( n l ) C l 〈 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) y − 1 | x n − l 〉 = 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j C l N n − l ( y + a j − 1 | a 1 , … , a ˆ j , … , a r ) − r n ∑ l = 0 n ( n l ) C l N n − l ( y − 1 | a 1 , … , a r ) ,$
(52)

where $a ˆ j$ means that $a j$ is omitted.

Therefore, by (49) and (52), we obtain the following theorem.

Theorem 5 For $n≥1$, we have

$N n ( x | a 1 , … , a r ) = x N n − 1 ( x − 1 | a 1 , … , a r ) + 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j C l N n − l ( x + a j − 1 | a 1 , … , a ˆ j , … , a r ) − r n ∑ l = 0 r ( n l ) C l N n − l ( x − 1 | a 1 , … , a r ) ,$

where $C n$ are the Cauchy numbers with the generating function given by

$t log ( 1 + t ) = ∑ n = 0 ∞ C n t n n ! .$

By the same method as the proof of Theorem 5, we get

$N ˆ n ( x | a 1 , … , a r ) = ( x − ∑ j = 1 r a j ) N ˆ n − 1 ( x − 1 | a 1 , … , a r ) − r n ∑ l = 0 n ( n l ) C l N ˆ n − l ( x − 1 | a 1 , … , a r ) − 1 n ∑ j = 1 r ∑ l = 0 n ( n l ) a j C l N ˆ n − l ( x − 1 | a 1 , … , a ˆ j , … , a r ) .$
(53)

Now we compute the following formula (54) in two different ways:

$〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m | x n 〉 .$
(54)

On the one hand,

$〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m | x n 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) | m ! ∑ l = m ∞ S 1 ( l , m ) t l l ! x n 〉 = m ! ∑ l = m n ( n l ) S 1 ( l , m ) 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) | x n − l 〉 = m ! ∑ l = m n ( n l ) S 1 ( l , m ) N n − l ( a 1 , … , a r ) = m ! ∑ l = 0 n − m ( n l ) S 1 ( n − l , m ) N l ( a 1 , … , a r ) .$
(55)

On the other hand,

$〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m | x n 〉 = 〈 ∂ t ( ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m ) | x n − 1 〉 = 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( ∂ t ( ( log ( 1 + t ) ) m ) ) | x n − 1 〉 + 〈 ( ∂ t ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ) ( log ( 1 + t ) ) m | x n − 1 〉 .$
(56)

Note that

$〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( ∂ t ( ( log ( 1 + t ) ) m ) ) | x n − 1 〉 = m 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) 1 1 + t | log ( 1 + t ) m − 1 x n − 1 〉 = m 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( 1 + t ) − 1 | ( m − 1 ) ! ∑ l = m − 1 ∞ S 1 ( l , m − 1 ) t l l ! x n − 1 〉 = m ! ∑ l = m − 1 n − 1 ( n − 1 l ) S 1 ( l , m − 1 ) 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( 1 + t ) − 1 | x n − 1 − l 〉 = m ! ∑ l = m − 1 n − 1 ( n − 1 l ) S 1 ( l , m − 1 ) N n − 1 − l ( − 1 | a 1 , … , a r ) = m ! ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − 1 − l , m − 1 ) N l ( − 1 | a 1 , … , a r )$
(57)

and

$〈 ( ∂ t ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ) ( log ( 1 + t ) ) m | x n − 1 〉 = 〈 ( ∂ t ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ) | m ! ∑ l = m ∞ S 1 ( l , m ) t l l ! x n − 1 〉 = m ! ∑ l = m n − 1 ( n − 1 l ) S 1 ( l , m ) × 〈 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) − 1 | ∑ j = 1 r { a j t ( 1 + t ) a j ( 1 + t ) a j − 1 − t log ( 1 + t ) } t x n − 1 − l 〉 = m ! ∑ l = m n − 1 ( n − 1 l ) S 1 ( l , m ) n − l × 〈 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) − 1 | ∑ j = 1 r { a j t ( 1 + t ) a j ( 1 + t ) a j − 1 − t log ( 1 + t ) } x n − l 〉 = m ! n ∑ l = m n − 1 ( n l ) S 1 ( l , m ) × { ∑ j = 1 r a j 〈 log ( 1 + t ) ( 1 + t ) a j − 1 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) a j − 1 | t log ( 1 + t ) x n − l 〉 − r 〈 ∏ i = 1 r ( ( 1 + t ) a i − 1 log ( 1 + t ) ) ( 1 + t ) − 1 | t log ( 1 + t ) x n − l 〉 } = m ! n ∑ l = m n − 1 ( n l ) S 1 ( l , m ) { ∑ j = 1 r a j ∑ k = 0 n − l C k ( n − l k ) N n − l − k ( a j − 1 | a 1 , … , a ˆ j , … , a r ) − r ∑ k = 0 n − l C k ( n − l k ) N n − l − k ( − 1 | a 1 , … , a r ) } .$
(58)

Therefore, by (55), (56), (57) and (58), we obtain the following theorem.

Theorem 6 For $n−1≥m≥1$, we have

$∑ l = 0 n − m ( n l ) S 1 ( n − l , m ) N l ( a 1 , … , a r ) = ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) N l ( − 1 | a 1 , … , a r ) + 1 n ∑ l = m n − 1 ∑ k = 0 n − l ∑ j = 1 r ( n l ) ( n − l k ) a j C n − l − k S 1 ( l , m ) N k ( a j − 1 | a 1 , … , a ˆ j , … , a r ) − r n ∑ l = m n − 1 ∑ k = 0 n − l ( n l ) ( n − l k ) C n − l − k S 1 ( l , m ) N k ( − 1 | a 1 , … , a r ) .$

By the same method as the proof of Theorem 6, we get

$∑ l = 0 n − m ( n l ) S 1 ( n − l , m ) N ˆ l ( a 1 , … , a r ) = ∑ l = 0 n − m ( n − 1 l ) S 1 ( n − l − 1 , m − 1 ) N ˆ l ( − 1 | a 1 , … , a r ) + 1 n ∑ j = 1 r ∑ l = m n − 1 ∑ k = 0 n − l ( n l ) ( n − l k ) a j C n − l − k S 1 ( l , m ) N ˆ k ( − 1 | a 1 , … , a ˆ j , … , a r ) − r n ∑ l = m n − 1 ∑ k = 0 n − l ( n l ) ( n − l k ) C n − l − k S 1 ( l , m ) N ˆ k ( − 1 | a 1 , … , a r ) − ∑ j = 1 r a j ∑ l = 0 n − m − 1 ( n − 1 l ) S 1 ( n − l − 1 , m ) N ˆ k ( − 1 | a 1 , … , a r ) ,$
(59)

where $n−1≥m≥1$.

Let us consider the following two Sheffer sequences:

$N n (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( t e a j t − 1 ) , e t − 1 )$
(60)

and (23).

We let

$N n (x| a 1 ,…, a r )= ∑ m = 0 n C n , m ( x ) m .$
(61)

From (18) and (19), we note that

$C n , m = 1 m ! 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) t m | x n 〉 = ( n m ) 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) | x n − m 〉 = ( n m ) N n − m ( a 1 , … , a r ) .$
(62)

Therefore, by (61) and (62), we obtain the following theorem.

Theorem 7 For $n≥0$, we have

$N n (x| a 1 ,…, a r )= ∑ m = 0 n ( n m ) N n − m ( a 1 ,…, a r ) ( x ) m .$

By the same method as the proof of Theorem 7, we get

$N ˆ n (x| a 1 ,…, a r )= ∑ m = 0 n ( n m ) N ˆ n − m ( a 1 ,…, a r ) ( x ) m .$
(63)

For

$N n (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( t e a j t − 1 ) , e t − 1 )$

and

let us assume that

$N n (x| a 1 ,…, a r )= ∑ m = 0 n C n , m H m ( s ) (x|λ),$
(64)

where $H m ( s ) (x|λ)$ are the Frobenius-Euler polynomials of order s defined by the generating function as

$( 1 − λ e t − λ ) s e x t = ∑ n = 0 ∞ H n ( s ) (x|λ) t n n ! .$

From (18) and (19), we note that

$C n , m = 1 m ! ( 1 − λ ) s 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m ( 1 − λ + t ) s | x n 〉 = 1 m ! ( 1 − λ ) s 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m | ∑ j = 0 min { s , n } ( s j ) ( 1 − λ ) s − j t j x n 〉 = 1 m ! ( 1 − λ ) s ∑ j = 0 n − m ( s j ) ( 1 − λ ) s − j ( n ) j × 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 t ) ) ( log ( 1 + t ) ) m | x n − j 〉 = 1 m ! ( 1 − λ ) s ∑ j = 0 n − m ( s j ) ( 1 − λ ) s − j ( n ) j m ! × ∑ l = 0 n − j − m ( n − j l ) S 1 ( n − j − l , m ) N l ( a 1 , … , a r ) = ∑ j = 0 n − m ∑ l = 0 n − m − j ( s j ) ( n − j l ) × ( n ) j ( 1 − λ ) − j S 1 ( n − j − l , m ) N l ( a 1 , … , a r ) .$
(65)

Therefore, by (64) and (65), we obtain the following theorem.

Theorem 8 For $n≥0$, we have

$N n ( x | a 1 , … , a r ) = ∑ m = 0 n { ∑ j = 0 n − m ∑ l = 0 n − m − j ( s j ) ( n − j l ) ( n ) j ( 1 − λ ) − j × S 1 ( n − j − l , m ) N l ( a 1 , … , a r ) } H m ( s ) ( x | λ ) .$

By the same method as the proof of Theorem 8, we get

$N ˆ n ( x | a 1 , … , a r ) = ∑ m = 0 n { ∑ j = 0 n − m ∑ l = 0 n − m − j ( s j ) ( n − j l ) ( n ) j × ( 1 − λ ) − j S 1 ( n − j − l , m ) N ˆ l ( a 1 , … , a r ) } H m ( s ) ( x | λ ) .$
(66)

Now, we consider the following two Sheffer sequences:

$N n (x| a 1 ,…, a r )∼ ( ∏ j = 1 r ( t e a j t − 1 ) , e t − 1 )$
(67)

and

$B n ( s ) (x)∼ ( ( e t − 1 t ) s , e t − 1 ) ,$

where $B n ( s ) (x)$ are the Bernoulli polynomials of order s given by the generating function as

$( t e t − 1 ) s e x t = ∑ n = 0 ∞ B n ( s ) (x) t n n ! .$

Let us assume that

$N n (x| a 1 ,…, a r )= ∑ m = 0 n C n , m B m ( s ) (x).$
(68)

By (18) and (19), we get

$C n , m = 1 m ! 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m ( t log ( 1 + t ) ) s | x n 〉 = 1 m ! 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m | ( t log ( 1 + t ) ) s x n 〉 = 1 m ! 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m | ∑ k = 0 ∞ C k ( s ) t k k ! x n 〉 = 1 m ! ∑ k = 0 n − m ( n k ) C k ( s ) 〈 ∏ j = 1 r ( ( 1 + t ) a j − 1 log ( 1 + t ) ) ( log ( 1 + t ) ) m | x n − k 〉 = 1 m ! ∑ k = 0 n − m ( n k ) C k ( s ) m ! ∑ l = 0 n − m − k ( n − k l ) S 1 ( n − l − k , m ) N l ( a 1 , … , a r ) = ∑ k = 0 n − m ∑ l = 0 n − m − k ( n k ) ( n − k l ) C k ( s ) S 1 ( n − k − l , m ) N l ( a 1 , … , a r ) ,$
(69)

where $C k ( s )$ are the Cauchy numbers of the first kind of order s defined by the generating function as

$( t log ( 1 + t ) ) s = ∑ n = 0 ∞ C n ( s ) t n n ! .$

Therefore, by (68) and (69), we obtain the following theorem.

Theorem 9 For $n≥0$, we have

$N n ( x | a 1 , … , a r ) = ∑ m = 0 n { ∑ k = 0 n − m ∑ l = 0 n − m − k ( n k ) ( n − k l ) C k ( s ) × S 1 ( n − k − l , m ) N l ( a 1 , … , a r ) } B m ( s ) ( x ) .$

By the same method as the proof of Theorem 9, we get

$N ˆ n ( x | a 1 , … , a r ) = ∑ m = 0 n { ∑ k = 0 n − m ∑ l = 0 n − m − k ( n k ) ( n − k l ) C k ( s ) × S 1 ( n − k − l , m ) N ˆ l ( a 1 , … , a r ) } B m ( s ) ( x ) .$

Recently, several authors have studied umbral calculus (see [15, 718, 20]).

## References

1. 1.

Roman S: The Umbral Calculus. Dover, New York; 2005.

2. 2.

Kim T, Kim DS: Some identities involving associated sequences of special polynomials. J. Comput. Anal. Appl. 2014, 16: 626-642.

3. 3.

Kim DS, Kim T, Lee S-H, Rim S-H: Umbral calculus and Euler polynomials. Ars Comb. 2013, 112: 293-306.

4. 4.

Kim DS, Kim T: Some identities of Bernoulli and Euler polynomials arising from umbral calculus. Adv. Stud. Contemp. Math. (Kyungshang) 2013, 23(1):159-171.

5. 5.

Kim DS, Kim T, Dolgy DV, Rim S-H: Some new identities of Bernoulli, Euler and Hermite polynomials arising from umbral calculus. Adv. Differ. Equ. 2013., 2013: Article ID 73

6. 6.

Kim T, Mansour T, Rim S-H, Lee S-H: Apostol-Euler polynomials arising from umbral calculus. Adv. Differ. Equ. 2013., 2013: Article ID 301

7. 7.

Kim DS, Kim T: Higher-order Cauchy of the first kind and poly-Cauchy of the first kind mixed type polynomials. Adv. Stud. Contemp. Math. (Kyungshang) 2013, 23(21):621-636.

8. 8.

Kim T, Kim DS, Mansour T, Rim S-H, Schork M: Umbral calculus and Sheffer sequences of polynomials. J. Math. Phys. 2013., 54(8): Article ID 083504

9. 9.

Kim T: Identities involving Laguerre polynomials derived from umbral calculus. Russ. J. Math. Phys. 2014, 21(1):36-45. 10.1134/S1061920814010038

10. 10.

Lu D-Q, Xiang Q-M, Luo C-H: Some results for Apostol-type polynomials associated with umbral algebra. Adv. Differ. Equ. 2013., 2013: Article ID 201

11. 11.

Maldonado M, Prada J, Senosiain MJ: Appell bases on sequence spaces. J. Nonlinear Math. Phys. 2011, 18: 189-194. suppl. 1 10.1142/S1402925111001362

12. 12.

Araci S, Kong X, Acikgoz M, Şen E: A new approach to multivariate q -Euler polynomials using the umbral calculus. J. Integer Seq. 2014., 17(1): Article ID 14.1.2

13. 13.

Dere R, Simsek Y: Applications of umbral algebra to some special polynomials. Adv. Stud. Contemp. Math. (Kyungshang) 2012, 22(3):433-438.

14. 14.

Ernst T: Examples of a q -umbral calculus. Adv. Stud. Contemp. Math. (Kyungshang) 2008, 16(1):1-22.

15. 15.

Fang Q, Wang T: Umbral calculus and invariant sequences. Ars Comb. 2011, 101: 257-264.

16. 16.

Gessel IM: Applications of the classical umbral calculus. Algebra Univers. 2003, 49(4):397-434. Dedicated to the memory of Gian-Carlo Rota 10.1007/s00012-003-1813-5

17. 17.

Kim DS, Kim T: Some identities of Frobenius-Euler polynomials arising from umbral calculus. Adv. Differ. Equ. 2012., 2012: Article ID 196

18. 18.

Kim DS, Kim T, Ryoo CS: Sheffer sequences for the powers of Sheffer pairs under umbral composition. Adv. Stud. Contemp. Math. (Kyungshang) 2013, 23(2):275-285.

19. 19.

Ryoo CS, Song H, Agarwal RP: On the roots of the q -analogue of Euler-Barnes’ polynomials. Adv. Stud. Contemp. Math. 2004, 9(2):153-163.

20. 20.

Taylor BD: Umbral presentations for polynomial sequences. Comput. Math. Appl. 2001, 41(9):1085-1098. Umbral calculus and its applications (Cortona, 1998) 10.1016/S0898-1221(01)00083-9

## Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MOE) (No. 2012R1A1A2003786). The authors would like to thank the referees for their valuable comments.

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