Open Access

Hermite and poly-Bernoulli mixed-type polynomials

Advances in Difference Equations20132013:343

https://doi.org/10.1186/1687-1847-2013-343

Received: 23 September 2013

Accepted: 7 November 2013

Published: 27 November 2013

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 ) , c L | 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 ,
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  n 0 .
(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 ) , t H B n ( ν , k ) ( x ) = n H 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 ) .

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics, Sogang University
(2)
Department of Mathematics, Kwangwoon University

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© Kim and Kim; licensee Springer. 2013

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