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Identities involving harmonic and hyperharmonic numbers

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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.

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 (n0) (see [1, 2]).
(2)

From (1) and (2), we note that

t k | x n =n! δ n , k (n,k0) (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,k0. 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 (k0) (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(x1)(xn+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 (n1) 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 n1, 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 n1, r1, 1kn, 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 n1, 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,n1, 1kn, 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 n1, 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 n1, 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 n1, 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,r1, 1kn, 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 n1, 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 n1,

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 n1, 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,r1, 1kn, 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 n1, 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

tlog(1t)=t+ n = 1 t n n =2t+ 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 n1, 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 n1, 1kn, 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 n1, 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 n1, 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 n1, 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 n1, 1kn, 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 n1, 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 n1, 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 n1, 1kn, 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(2n+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 (n1).
(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(1t)=3+6t+ n = 1 4 n + 1 n ( n + 1 ) t n + 1 .
(60)

For n1, 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 n1, 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]).

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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).

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Correspondence to Taekyun Kim.

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Keywords

  • Vector Space
  • Formal Power Series
  • Linear Functional
  • Small Integer
  • Kronecker Symbol