Open Access

Identities involving harmonic and hyperharmonic numbers

Advances in Difference Equations20132013:235

https://doi.org/10.1186/1687-1847-2013-235

Received: 26 April 2013

Accepted: 22 July 2013

Published: 7 August 2013

Abstract

In this paper, we give some new and interesting identities involving harmonic and hyperharmonic numbers which are derived from the transfer formula for the associated sequences.

Keywords

Vector SpaceFormal Power SeriesLinear FunctionalSmall IntegerKronecker Symbol

1 Introduction

Let be the set of all formal power series in the variable t over C with
F = { f ( t ) = k = 0 a k k ! t k | a k C } .
(1)

Suppose that is the algebra of polynomials in the variable x over C and that P is the vector space of all linear functionals on . The action of the linear functional L on a polynomial p ( x ) is denoted by L | p ( x ) .

Let f ( t ) F . Then we consider a linear functional on by setting
f ( t ) | x n = a n ( n 0 )  (see [1, 2]) .
(2)
From (1) and (2), we note that
t k | x n = n ! δ n , k ( n , k 0 )  (see [1, 3–5]) ,
(3)

where δ n , k is the Kronecker symbol.

Let f L ( t ) = k = 0 L | x n k ! t k . Then we see that f L ( t ) | x n = L | x n . The map L f L ( t ) is a vector space isomorphism from P onto . Henceforth, is thought of as both a formal power series and a linear functional. We call the umbral algebra. The umbral calculus is the study of umbral algebra. The order O ( f ( t ) ) of the nonzero power series f ( t ) is the smallest integer k for which the coefficient of t k does not vanish. If O ( f ( t ) ) = 0 , then f ( t ) is called an invertible series. If O ( f ( t ) ) = 1 , then f ( t ) is called a delta series. Let O ( f ( t ) ) = 1 and O ( g ( t ) ) = 0 . Then there exists a unique sequence s n ( x ) of polynomials 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 [1, 3, 6]). If s n ( x ) ( 1 , f ( t ) ) , then s n ( x ) is called the associated sequence for f ( t ) . By (3), we easily see that e y t | p ( x ) = p ( y ) . 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 [1, 6, 7]) .
(4)
From (4), we note that
p ( k ) ( 0 ) = t k | p ( x ) , 1 | p ( k ) ( x ) = p ( k ) ( 0 ) .
(5)
By (5), we easily see that
t k p ( x ) = p ( k ) ( x ) = d k p ( x ) d x k ( k 0 )  (see [2, 3, 6, 7]) .
(6)
Let ϕ n ( x ) be exponential polynomials which are given by
k = 0 ϕ k ( x ) k ! t k = e x ( e t 1 ) (see [2, 6, 8]) .
(7)
Thus, by (7), we get
ϕ n ( x ) = k = 0 n S 2 ( n , k ) x k ( 1 , log ( 1 + t ) ) ,
(8)

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

The Stirling number of the first kind is defined by
( x ) n = x ( x 1 ) ( x n + 1 ) = k = 0 n S 1 ( n , k ) x k .
(9)
Thus, by (9), we get
S 1 ( n , k ) = 1 k ! t k | ( x ) n (see [2, 5]) .
(10)
Let p n ( x ) ( 1 , f ( t ) ) , q n ( x ) ( 1 , g ( t ) ) . Then the transfer formula for the associated sequences is given by
q n ( x ) = x ( f ( t ) g ( t ) ) n x 1 p n ( x ) (see [2, 8]) .
(11)

The n th harmonic number is H n = i = 1 n 1 i ( n 1 ) and H 0 = 0 .

In general, the hyperharmonic number H n ( r ) of order r is defined by
H n ( r ) = { 0 if  n 0  or  r < 0 , 1 n if  r = 0 , n 1 , i = 1 n H i ( r 1 ) if  r , n 1 (see [9, 10]) .
(12)
From (12), we note that H n ( 1 ) is the ordinary harmonic number H n . It is known that
H n ( r ) = ( n + r 1 r 1 ) ( H n + r 1 H r 1 ) (see [9, 10]) .
(13)
The generating functions of the harmonic and hyperharmonic numbers are given by
n = 1 H n t n = log ( 1 t ) 1 t
(14)
and
n = 1 H n ( r ) t n = log ( 1 t ) ( 1 t ) r , respectively .
(15)

The purpose of this paper is to give some new and interesting identities involving harmonic and hyperharmonic numbers which are derived from the transfer formula for the associated sequences.

2 Identities involving harmonic and hyperharmonic numbers

From (7) and (8), we note that
ϕ n ( x ) = j = 0 n S 2 ( n , j ) x j ( 1 , log ( 1 + t ) )
(16)
and
( 1 ) n ϕ n ( x ) ( 1 , log ( 1 t ) ) .
(17)
Let us assume that
q n ( x ) ( 1 , t ( 1 t ) r ) .
(18)
From (11), (18) and x n ( 1 , t ) , we note that
q n ( x ) = x ( t t ( 1 t ) r ) n x 1 x n = x ( 1 t ) r n x n 1 = x k = 0 n 1 ( r n k ) ( t ) k x n 1 = x k = 0 n 1 ( r n + k 1 k ) t k x n 1 = x k = 0 n 1 ( r n + k 1 k ) ( n 1 ) k x n 1 k = k = 1 n 1 ( r n + k 1 k ) ( n 1 ) k x n k = k = 1 n ( r n + n k 1 n k ) ( n 1 ) n k x k .
(19)
Now, we use the following fact:
n = 1 H n ( r ) t n = log ( 1 t ) ( 1 t ) r .
(20)
For n 1 , by (11), (17) and (18), we get
q n ( x ) = x ( log ( 1 t ) t ( 1 t ) r ) n x 1 ( 1 ) n ϕ n ( x ) = x ( l = 0 H l + 1 ( r ) t l ) n x 1 ( 1 ) n j = 1 n S 2 ( n , j ) ( x ) j = ( 1 ) n j = 1 n S 2 ( n , j ) ( 1 ) j x ( l = 0 H l + 1 ( r ) t l ) n x j 1 = ( 1 ) n j = 1 n S 2 ( n , j ) ( 1 ) j x ( l = 0 j 1 ( l 1 + + l n = l H l 1 + 1 ( r ) H l n + 1 ( r ) ) t l ) x j 1 = ( 1 ) n j = 1 n l = 0 j 1 l 1 + + l n = l S 2 ( n , j ) ( 1 ) j H l 1 + 1 ( r ) H l n + 1 ( r ) ( j 1 ) l x j l = ( 1 ) n j = 1 n k = 1 j l 1 + + l n = j k S 2 ( n , j ) ( 1 ) j H l 1 + 1 ( r ) H l n + 1 ( r ) ( j 1 ) j k x k = ( 1 ) n k = 1 n { j = k n l 1 + + l n = j k ( 1 ) j S 2 ( n , j ) H l 1 + 1 ( r ) H l n + 1 ( r ) ( j 1 ) j k } x k .
(21)

Therefore, by comparing coefficients on both sides of (19) and (20), we obtain the following theorem.

Theorem 1 For n 1 , r 1 , 1 k n , we have
( r n + n k 1 n k ) ( n 1 ) n k = ( 1 ) n j = k n l 1 + + l n = j k S 2 ( n , j ) ( 1 ) j H l 1 + 1 ( r ) H l n + 1 ( r ) ( j 1 ) j k .
We recall the following equation:
( log ( 1 + t ) t ) n = l = 0 n ! ( l + n ) ! S 1 ( l + n , n ) t l .
(22)
For n 1 , from (11), (17) and (18), we have
q n ( x ) = x ( log ( 1 t ) t ( 1 t ) r ) n x 1 ( 1 ) n ϕ n ( x ) = x ( log ( 1 t ) t ) n ( 1 t ) r n x 1 ( 1 ) n ϕ n ( x ) = ( 1 ) n j = 1 n S 2 ( n , j ) ( 1 ) j x ( log ( 1 t ) t ) n ( 1 t ) r n x j 1 = ( 1 ) n j = 1 n S 2 ( n , j ) ( 1 ) j x ( log ( 1 t ) t ) n l = 0 j 1 ( r n + l 1 l ) ( j 1 ) l x j 1 l = ( 1 ) n j = 1 n S 2 ( n , j ) ( 1 ) j l = 0 j 1 ( r n + l 1 l ) ( j 1 ) l x m = 0 j 1 l n ! ( m + n ) ! × S 1 ( m + n , n ) ( t ) m x j 1 l = ( 1 ) n j = 1 n l = 0 j 1 m = 0 j 1 l ( 1 ) j + m ( r n + l 1 l ) n ! ( m + n ) ! ( j 1 ) ! ( j 1 l m ) ! × S 1 ( m + n , n ) S 2 ( n , j ) x j l m = ( 1 ) n k = 1 n { j = k n l = 0 j k ( 1 ) k + l ( r n + l 1 l ) n ! ( j l k + n ) ! ( j 1 ) ! ( k 1 ) ! × S 1 ( j l k + n , n ) S 2 ( n , j ) } x k .
(23)

Therefore, by (19) and (23), we obtain the following theorem.

Theorem 2 For r , n 1 , 1 k n , we have
( r n + n k 1 n k ) ( n 1 ) n k = ( 1 ) n j = k n l = 0 j k ( 1 ) k + l ( r n + l 1 l ) n ! ( j l k + n ) ! ( j 1 ) ! ( k 1 ) ! × S 1 ( j l k + n , n ) S 2 ( n , j ) .
Here we invoke the following identity:
n = 1 ( m = 1 n m H m ( r ) ) t n = t ( 1 r log ( 1 t ) ) ( 1 t ) r + 2 .
(24)
Let us consider the following associated sequence:
q n ( x ) ( 1 , t ( 1 t ) r + 2 ) .
(25)
For n 1 , by (19) and (25), we get
q n ( x ) = k = 1 n ( ( r + 3 ) n k 1 n k ) ( n 1 ) n k x k .
(26)
Let us assume that
p n ( x ) ( 1 , t ( 1 r log ( 1 t ) ) ) .
(27)
For n 1 , by (11), (27) and x n ( 1 , t ) , we get
p n ( x ) = 7 x ( t t ( 1 r log ( 1 t ) ) ) n x 1 x n = x ( 1 r log ( 1 t ) ) n x n 1 = x l = 0 ( n + l 1 l ) r l ( log ( 1 t ) ) l x n 1 = x l = 0 n 1 ( n + l 1 l ) r l j = 0 n 1 l l ! ( j + l ) ! S 1 ( j + l , l ) t j + l x n 1 = l = 0 n 1 j = 0 n 1 l l ! r l ( n + l 1 l ) ( n 1 j + l ) S 1 ( j + l , l ) x n j l = k = 1 n { l = 0 n k l ! r l ( n + l 1 l ) ( n 1 k 1 ) S 1 ( n k , l ) } x k .
(28)
For n 1 , from (11), (25) and (27), we can derive the following equation:
q n ( x ) = x ( t ( 1 r log ( 1 t ) ) t ( 1 t ) r + 2 ) n x 1 p n ( x ) = x ( j = 1 ( m = 1 j m H m ( r ) ) t j 1 ) n a = 1 n { l = 0 n a l ! r l ( n + l 1 l ) ( n 1 a 1 ) × S 1 ( n a , l ) } x a 1 = a = 1 n { l = 0 n a l ! r l ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) } × x [ j = 0 { j 1 + + j n = j ( m 1 = 1 j 1 + 1 m n = 1 j n + 1 m 1 m n H m 1 ( r ) H m n ( r ) ) } t j ] x a 1 = a = 1 n l = 0 n a k = 1 a j 1 + + j n = a k ( m 1 = 1 j 1 + 1 m n = 1 j n + 1 m 1 m n H m 1 ( r ) H m n ( r ) ) × l ! r l ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) ( a 1 ) a k x k = k = 1 n { a = k n l = 0 n a j 1 + + j n = a k ( m 1 = 1 j 1 + 1 m n = 1 j n + 1 m 1 m n H m 1 ( r ) H m n ( r ) ) × l ! r l ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) ( a 1 ) a k } x k .
(29)

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

Theorem 3 For n , r 1 , 1 k n , we have
( ( r + 3 ) n k 1 n k ) ( n 1 ) n k = a = k n l = 0 n a j 1 + + j n = a k ( m 1 = 1 j 1 + 1 m n = 1 j n + 1 m 1 m n H m 1 ( r ) H m n ( r ) ) l ! r l × ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) ( a 1 ) n k .
Here we use the following identity:
n = 1 n H n ( r ) t n = t ( 1 r log ( 1 t ) ) ( 1 t ) r + 1 .
(30)
Let us consider the following associated sequence:
q n ( x ) ( 1 , t ( 1 t ) r + 1 ) .
(31)
For n 1 , from (19) and (31), we have
q n ( x ) = k = 1 n ( ( r + 2 ) n k 1 n k ) ( n 1 ) n k x k .
(32)
Let us assume that
p n ( x ) ( 1 , t ( 1 r log ( 1 t ) ) ) .
(33)
Then, from (28) and (33), we note that, for n 1 ,
p n ( x ) = k = 1 n { l = 0 n k l ! r l ( n + l 1 l ) ( n 1 k 1 ) S 1 ( n k , l ) } x k .
(34)
For n 1 , by (11), (32) and (33), we get
q n ( x ) = x ( t ( 1 r log ( 1 t ) ) t ( 1 t ) r + 1 ) n x 1 p n ( x ) = x ( j = 1 j H j ( r ) t j 1 ) n x 1 a = 1 n { l = 0 n a l ! r l ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) } x a = a = 1 n { l = 0 n a l ! r l ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) } × x j = 0 a 1 ( j 1 + + j n = j ( j 1 + 1 ) ( j n + 1 ) H j 1 + 1 ( r ) H j n + 1 ( r ) ) t j x a 1 = a = 1 n l = 0 n a j = 0 a 1 ( j 1 + + j n = j ( j 1 + 1 ) ( j n + 1 ) H j 1 + 1 ( r ) H j n + 1 ( r ) ) l ! r l × ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) ( a 1 ) j x a j = k = 1 n { a = k n l = 0 n a ( j 1 + + j n = a k ( j 1 + 1 ) ( j n + 1 ) H j 1 + 1 ( r ) H j n + 1 ( r ) ) l ! r l × ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) ( a 1 ) a k } x k .
(35)

Therefore, by (32) and (35), we obtain the following theorem.

Theorem 4 For n , r 1 , 1 k n , we have
( ( r + 2 ) n k 1 n k ) ( n 1 ) n k = a = k n l = 0 n a ( j 1 + + j n = a k ( j 1 + 1 ) ( j n + 1 ) H j 1 + 1 ( r ) H j n + 1 ( r ) ) l ! r l × ( n + l 1 l ) ( n 1 a 1 ) S 1 ( n a , l ) ( a 1 ) a k .
Now, we utilize the following identity:
n = 1 ( n + 1 ) H n t n = t log ( 1 t ) ( 1 t ) 2 .
(36)
Let us consider the following associated sequence:
q n ( x ) ( 1 , t ( 1 t ) 2 ) .
(37)
For n 1 , from (19) and (37), we have
q n ( x ) = k = 1 n ( 3 n k 1 n k ) ( n 1 ) n k x k .
(38)
Let us assume that
p n ( x ) ( 1 , t log ( 1 t ) ) .
(39)
We observe that
t log ( 1 t ) = t + n = 1 t n n = 2 t + n = 2 t n n .
(40)
From (11), (39), (40) and x n ( 1 , t ) , we can derive the following equation:
p n ( x ) = x ( t 2 ( t + n = 2 t n n ) ) n x 1 x n = 2 n x ( 1 + n = 2 t n 1 2 n ) n x n 1 = 2 n x l = 0 ( n l ) ( n = 2 t n 1 2 n ) l x n 1 = 2 n x l = 0 n 1 ( 1 ) l ( n + l 1 l ) × m = 0 n 1 l m 1 + + m l = m 1 2 l ( m 1 + 2 ) ( m l + 2 ) t m + l x n 1 = 2 n l = 0 n 1 m = 0 n 1 l m 1 + + m l = m ( 1 2 ) l ( n + l 1 l ) ( n 1 ) m + l ( m 1 + 2 ) ( m l + 2 ) x n l m = 2 n k = 1 n { l = 0 n k m 1 + + m l = n l k ( 1 2 ) l ( n + l 1 l ) ( n 1 ) n k ( m 1 + 2 ) ( m l + 2 ) } x k .
(41)
For n 1 , by (11), (37), (39) and (41), we get
q n ( x ) = x ( t log ( 1 t ) t ( 1 t ) 2 ) n x 1 p n ( x ) = x ( j = 0 ( j + 2 ) H j + 1 t j ) n x 1 2 n a = 1 n { l = 0 n a m 1 + + m l = n l a ( 1 2 ) l × ( n + l 1 l ) ( n 1 ) n a ( m 1 + 2 ) ( m l + 2 ) } x a = 2 n a = 1 n { l = 0 n a m 1 + + m l = n l a ( 1 2 ) l ( n + l 1 l ) × ( n 1 ) n a ( m 1 + 2 ) ( m l + 2 ) } x j = 0 a 1 ( j 1 + + j n = j ( j 1 + 2 ) ( j n + 2 ) × H j 1 + 1 H j n + 1 ) ( a 1 ) j x a 1 j = 2 n a = 1 n l = 0 n a k = 1 a m 1 + + m l = n l a j 1 + + j n = a k ( 1 2 ) l ( n + l 1 l ) × ( n 1 ) n a ( a 1 ) a k ( m 1 + 2 ) ( m l + 2 ) ( j 1 + 2 ) ( j n + 2 ) H j 1 + 1 H j n + 1 x k = 2 n k = 1 n { a = k n l = 0 n a m 1 + + m l = n l a j 1 + + j n = a k ( 1 2 ) l ( n + l 1 l ) × ( n 1 ) n a ( a 1 ) a k ( m 1 + 2 ) ( m l + 2 ) ( j 1 + 2 ) ( j n + 2 ) H j 1 + 1 H j n + 1 } x k .
(42)

Therefore, by (38) and (42), we obtain the following theorem.

Theorem 5 For n 1 , 1 k n , we have
( 3 n k 1 n k ) ( n 1 ) n k = 2 n a = k n l = 0 n a m 1 + + m l = n l a j 1 + + j n = a k ( 1 2 ) l ( n + l 1 l ) × ( n 1 ) n a ( a 1 ) a k ( m 1 + 2 ) ( m l + 2 ) ( j 1 + 2 ) ( j n + 2 ) H j 1 + 1 H j n + 1 .
Now, we recall the following identity:
n = 1 n 2 H n t n = t { 1 + 2 t ( 1 + t ) log ( 1 t ) } ( 1 t ) 3 .
(43)
Let us consider the following associated sequence:
q n ( x ) ( 1 , t ( 1 t ) 3 ) .
(44)
For n 1 , from (19) and (44), we can derive the following equation:
q n ( x ) = k = 1 n ( 4 n k 1 n k ) ( n 1 ) n k x k .
(45)
Let us assume that
p n ( x ) ( 1 , t { 1 + 2 t ( 1 + t ) log ( 1 t ) } ) .
(46)
We observe that
1 + 2 t ( 1 + t ) log ( 1 t ) = 1 + 2 t + ( 1 + t ) j = 1 t j j = 1 + 2 t + t + j = 2 t j j + j = 1 t j + 1 j = 1 + 3 t + j = 0 t j + 2 j + 2 + j = 0 t j + 2 j + 1 = 1 + 3 t + j = 0 2 j + 3 ( j + 2 ) ( j + 1 ) t j + 2 .
(47)
For n 1 , by (11), (46), (47) and x n ( 1 , t ) , we get
p n ( x ) = x ( t t { 1 + 2 t ( 1 + t ) log ( 1 t ) } ) n x 1 x n = x ( 1 + 3 t + j = 0 2 j + 3 ( j + 1 ) ( j + 2 ) t j + 2 ) n x n 1 = x l = 0 n 1 ( 1 ) l ( n + l 1 l ) ( 3 + j = 0 2 j + 3 ( j + 1 ) ( j + 2 ) t j + 1 ) l t l x n 1 = l = 0 n 1 a = 0 n 1 l k = 1 n a l j 1 + + j a = n a k l ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a ( n 1 ) n k × ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) x k = k = 1 n { l = 0 n k a = 0 n k l j 1 + + j a = n a k l ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a ( n a ) n k × ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) } x k .
(48)
For n 1 , from (11), (44), (46) and (48), we have
q n ( x ) = x ( t ( 1 + 2 t ( 1 + t ) log ( 1 t ) ) t ( 1 t ) 3 ) n x 1 p n ( x ) = m = 1 n l = 0 n m a = 0 n m l j 1 + + j a = n a m l ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a ( n 1 ) n m × ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) x b = 0 m 1 b 1 + + b n = b ( i = 1 n ( b i + 1 ) 2 H b i + 1 ) t b x m 1 = m = 1 n l = 0 n m a = 0 n m l j 1 + + j a = n a m l ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a ( n 1 ) n m × ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) b = 0 m 1 b 1 + + b n = b i = 1 n ( b i + 1 ) 2 H b i + 1 ( m 1 ) b x m b = k = 1 n { m = k n l = 0 n m a = 0 n m l j 1 + + j a = n a m l b 1 + + b n = m k ( 1 ) l ( n + l 1 l ) ( l a ) × 3 l a ( n 1 ) n m ( m 1 ) m k ( i = 1 a ( 2 j i + 3 ) i = 1 n ( b i + 1 ) 2 H b i + 1 i = 1 a ( j i + 1 ) ( j i + 2 ) ) } x k .
(49)

Therefore, by (45) and (49), we obtain the following theorem.

Theorem 6 For n 1 , 1 k n , we have
( 4 n k 1 n k ) ( n 1 ) n k = m = k n l = 0 n m a = 0 n m l j 1 + + j a = n a m l b 1 + + b n = m k ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a × ( n 1 ) n m ( m 1 ) m k ( i = 1 a ( 2 j i + 3 ) i = 1 n ( b i + 1 ) 2 H b i + 1 i = 1 a ( j i + 1 ) ( j i + 2 ) ) .
Here we invoke the following identity:
b = 1 ( c = 1 b c 2 H c ) t b = t { 1 + 2 t ( 1 + t ) log ( 1 t ) } ( 1 t ) 4 .
(50)
Let us consider the following associated sequence:
q n ( x ) ( 1 , t ( 1 t ) 4 ) .
(51)
From (19) and (51), we note that
q n ( x ) = k = 1 n ( 5 n k 1 n k ) ( n 1 ) n k x k .
(52)
Let us assume that
p n ( x ) ( 1 , t ( 1 + 2 t ( 1 + t ) log ( 1 t ) ) ) .
(53)
For n 1 , from (48) and (49), we have
p n ( x ) = k = 1 n { l = 0 n k a = 0 n k l j 1 + + j a = n a k l ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a ( n 1 ) n k × ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) } x k .
(54)
For n 1 , from (11), (51), (53) and (50), we can derive the following identity:
q n ( x ) = x ( t { 1 + 2 t ( 1 + t ) log ( 1 t ) } t ( 1 t ) 4 ) n x 1 p n ( x ) = x ( b = 0 ( c = 1 b + 1 c 2 H c ) t b ) n x 1 p n ( x ) = x b = 0 b 1 + + b n = b { c 1 = 1 b 1 + 1 c n = 1 b n + 1 c 1 2 c n 2 H c 1 H c n } t b × m = 1 n { l = 0 n m a = 0 n m l j 1 + + j a = n a m l ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a × ( n 1 ) n m ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) } x m 1 = m = 1 n l = 0 n m a = 0 n m l j 1 + + j a = n a m l ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a ( n 1 ) n m × ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) b = 0 m 1 b 1 + + b n = b { c 1 = 1 b 1 + 1 c n = 1 b n + 1 c 1 2 c n 2 H c 1 H c n } × ( m 1 ) b x m b = k = 1 n { m = k n l = 0 n m a = 0 n m l j 1 + + j a = n a m l b 1 + + b n = m k ( 1 ) l ( n + l 1 l ) × ( l a ) 3 l a ( n 1 ) n m ( m 1 ) m k ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) × c 1 = 1 b 1 + 1 c n = 1 b n + 1 i = 1 n c i 2 H c i } x k .
(55)

Therefore, by (52) and (55), we obtain the following theorem.

Theorem 7 For n 1 , 1 k n , we have
( 5 n k 1 n k ) ( n 1 ) n k = m = k n l = 0 n m a = 0 n m l j 1 + + j a = n a m l b 1 + + b n = m k ( 1 ) l ( n + l 1 l ) ( l a ) 3 l a × ( n 1 ) n m ( m 1 ) m k ( i = 1 a ( 2 j i + 3 ) i = 1 a ( j i + 1 ) ( j i + 2 ) ) c 1 = 1 b 1 + 1 c n = 1 b n + 1 i = 1 n c i 2 H c i .
Here we use the following identity:
n = 1 n ( 2 n + 1 ) H n t n = t { 3 ( 1 + t ) ( t + 3 ) log ( 1 t ) } ( 1 t ) 3 .
(56)
Let us consider the following associated sequence:
q n ( x ) ( 1 , t ( 1 t ) 3 ) .
(57)
By (19) and (57), we get
q n ( x ) = k = 1 n ( 4 n k 1 n k ) ( n 1 ) n k x k ( n 1 ) .
(58)
Let us assume that
p n ( x ) ( 1 , t { 3 ( 1 + t ) ( t + 3 ) log ( 1 t ) } ) .
(59)
We see that
3 ( 1 + t ) ( t + 3 ) log ( 1 t ) = 3 + 6 t + n = 1 4 n + 1 n ( n + 1 ) t n + 1 .
(60)
For n 1 , from (11), (59), (60) and x n ( 1 , t ) , we have
p n ( x ) = x ( t t { 3 ( 1 + t ) ( t + 3 ) log ( 1 t ) } ) n x 1 x n = x ( 3 ( 1 + t ) ( t + 3 ) log ( 1 t ) ) n x n 1 = x ( 3 + 6 t + j = 1 4 j + 1 j ( j + 1 ) t j + 1 ) n x n 1 .
(61)
From (61), by the same method of (48), we get
p n ( x ) = 3 n k = 1 n { l = 0 n k a = 0 n k l j 1 + + j a = n a l k ( 1 ) l ( n + l 1 l ) ( l a ) 2 l a × ( n 1 ) n k ( i = 1 a ( 4 j i + 5 ) 3 ( j i + 1 ) ( j i + 2 ) ) } x k .
(62)
For n 1 , by (11), (56), (57), (59) and (62), we get
q n ( x ) = x ( t { 3 ( 1 + t ) ( t + 3 ) log ( 1 t ) } t ( 1 t ) 3 ) n x 1 p n ( x ) = x ( b = 0 ( b + 1 ) ( 2 b + 3 ) H b + 1 t b ) n x 1 p n ( x ) = x b = 0 ( b 1 + + b n = b ( i = 1 b ( b i + 1 ) ( 2 b i + 3 ) H b i + 1 ) t b ) × 3 n m = 1 n { l = 0 n m a = 0 n m l j 1 + + j a = n a l m ( 1 ) l ( n + l 1 l ) ( l a ) 2 l a × ( n 1 ) n m i = 1 a ( 4 j i + 5 ) 3 ( j i + 1 ) ( j i + 2 ) } x m 1 = 3 n m = 1 n l = 0 n m a = 0 n m l j 1 + + j a = n a l m ( 1 ) l ( n + l 1 l ) ( l a ) 2 l a ( n 1 ) n m × ( i = 1 a ( 4 j i + 5 ) 3 ( j i + 1 ) ( j i + 2 ) ) b = 0 m 1 b 1 + + b n = b ( i = 1 n ( b i + 1 ) ( 2 b i + 3 ) H b i + 1 ) × ( m 1 ) b x m b .
(63)
By the same method, we can derive the following identity from (63):
q n ( x ) = 3 n k = 1 n { m = k n l = 0 n m a = 0 n m l j 1 + + j a = n a l m b 1 + + b n = m k ( 1 ) l × ( n + l 1 l ) ( l a ) 2 l a ( n 1 ) n m ( m 1 ) m k ( i = 1 a ( 4 j i + 5 ) 3 ( j i + 1 ) ( j i + 2 ) ) × i = 1 n ( b i + 1 ) ( 2 b i + 3 ) H b i + 1 } x k .
(64)
By comparing coefficients on both sides of (58) and (64), we get
( 4 n k 1 n k ) ( n 1 ) n k = 3 n m = k n l = 0 n m a = 0 n m l j 1 + + j a = n a l m b 1 + + b n = m k ( 1 ) l ( n + l 1 l ) ( l a ) × 2 l a ( n 1 ) n m ( m 1 ) m k ( i = 1 a ( 4 j i + 5 ) 3 ( j i + 1 ) ( j i + 2 ) ) × ( i = 1 n ( b i + 1 ) ( 2 b i + 3 ) H b i + 1 ) .
(65)

Remark Recently, several authors have studied the q-extension of harmonic and hyperharmonic numbers (see [1113]).

Declarations

Acknowledgements

The authors express their sincere gratitude to the referees for their valuable suggestions and comments. 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, Seoul, Republic of Korea
(2)
Department of Mathematics, Kwangwoon University, Seoul, Republic of Korea

References

  1. Roman S: More on the umbral calculus, with emphasis on the q -umbral calculus. J. Math. Anal. Appl. 1985, 107: 222-254. 10.1016/0022-247X(85)90367-1MathSciNetView ArticleGoogle Scholar
  2. Roman S: The Umbral Calculus. Dover, New York; 2005.Google Scholar
  3. Araci S, Acikgoz M, Kilicman A: Extended p -adic q -invariant integrals on Z p associated with applications of umbral calculus. Adv. Differ. Equ. 2013., 2013: Article ID 96Google Scholar
  4. Kim DS, Kim T, Lee S-H, Rim S-H: Some identities of Bernoulli, Euler and Abel polynomials arising from umbral calculus. Adv. Differ. Equ. 2013., 2013: Article ID 15Google Scholar
  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 73Google Scholar
  6. Kim DS, Kim T, Lee S-H, Rim S-H: A note on the higher-order Frobenius-Euler polynomials and Sheffer sequences. Adv. Differ. Equ. 2013., 2013: Article ID 41Google Scholar
  7. Dere R, Simsek Y: Applications of umbral algebra to some special polynomials. Adv. Stud. Contemp. Math. 2012, 22(3):433-438.MathSciNetGoogle Scholar
  8. Kim DS, Kim T: Some identities of Frobenius-Euler polynomials arising from umbral calculus. Adv. Differ. Equ. 2012., 2012: Article ID 196Google Scholar
  9. Dil A, Kurt V: Polynomials related to harmonic numbers and evaluation of harmonic number series. Integers 2012., 12: Article ID 38Google Scholar
  10. Dil A, Kurt V: Polynomials related to harmonic numbers and evaluation of harmonic number series II. Appl. Anal. Discrete Math. 2011, 5: 212-229. 10.2298/AADM110615015DMathSciNetView ArticleGoogle Scholar
  11. Mansour T, Shattuck M, Song C: q -Analogs of identities involving harmonic numbers and binomial coefficients. Appl. Appl. Math. 2012, 7(1):22-36.MathSciNetGoogle Scholar
  12. Mansour T: Identities on harmonic and q -harmonic number sums. Afr. Math. 2012, 23: 135-143. 10.1007/s13370-011-0023-0View ArticleGoogle Scholar
  13. Mansour T, Shattuck M: A q -analog of the hyperharmonic numbers. Afr. Math. 2012. 10.1007/s13370-012-0106-6Google Scholar

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