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Sums of products of the degenerate Euler numbers

Abstract

The paper focuses on the degenerate Euler numbers, the degenerate Euler polynomials and the degenerate Bernoulli polynomials. By adopting the method of recurrences, two explicit expressions have been established for sums of products of the degenerate Euler polynomials and the degenerate Bernoulli polynomials. As a special case of the degenerate Euler polynomials, an expression can be obtained for j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) ε j 1 (λ) ε j 2 (λ) ε j N (λ).

MSC:11B68, 11B65, 11B73.

1 Introduction

The Bernoulli numbers B n are defined by

t e t 1 = n = 0 B n t n n ! .

The study of the Bernoulli numbers has a very long history. In fact, Euler had found that

k = 0 n ( n k ) B k B n k =(n1) B n n B n 1
(1.1)

for any n1.

In 1996, Dilcher [1] generalized (1.1) to the sums of products of N Bernoulli numbers:

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) B j 1 B j 2 B j N = N ( n N ) j = 0 N 1 ( 1 ) N 1 j s ( N , N j ) B n j n j ,
(1.2)

where ( n j 1 , j 2 , , j N ) are the multinomial coefficients defined by

( n j 1 , j 2 , , j N ) = n ! j 1 ! j 2 ! j N !

and s(n,k) are the Stirling numbers of the first kind (see, e.g., [2]). Furthermore, Dilcher also extended (1.2) to the Bernoulli polynomials.

On the other hand, in [3, 4], Carlitz defined the degenerate Bernoulli numbers β n (λ) as

t ( 1 + λ t ) 1 / λ 1 = n = 0 β n (λ) t n n ! .
(1.3)

Carlitz showed that β n (λ) is a polynomial in λ. And the explicit formula for β n (λ) was obtained by Howard [5]. Note that ( 1 + λ t ) 1 / λ tends to e t as λ0. So we have β n (0)= B n . Furthermore, we know that λ m β m (λ) tends to m! b n as λ+, where b n is the Bernoulli number of the second kind given by

t log ( 1 + t ) = n = 0 b n t n .

For more number-theoretical properties of the degenerate Bernoulli numbers, the readers may refer to [6, 7] and [8].

The analogues of (1.1) and (1.2) for b n have been given in [9] and [10]. Moreover, in [11], Zhang and Yang obtained a generalization of (1.2) for β n (λ):

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) β j 1 ( λ ) β j 2 ( λ ) β j N ( λ ) = N ( n N ) k = 0 N 1 ( 1 ) N 1 k σ N 1 , k ( λ , n ) β n k ( λ ) n k ,
(1.4)

where

σ N 1 , k (λ,n)= 1 i k < i k 1 < < i 2 < i 1 N 1 j = 1 k ( ( λ 1 ) i j ( n j ) λ ) .

The Euler numbers E n are another important kind of numbers, which are defined by

2 e t + e t = n = 0 E n t n n ! .

The Euler numbers have many similar properties as the Bernoulli numbers. For example, in the same paper, Dilcher also proved that

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( 2 n 2 j 1 , 2 j 2 , , 2 j N ) E 2 j 1 E 2 j 2 E 2 j N = 2 2 n + N 1 ( N 1 ) ! j = 0 [ ( N 1 ) / 2 ] ( k = 0 2 j ( N 1 k 2 j k ) s ( N , N k ) N 2 j k 2 2 j k ) E 2 n + N 1 2 j ( N 2 ) ,
(1.5)

where 2n>N and the Euler polynomials E n (x) are given by

2 e x t e t + 1 = n = 0 E n (x) t n n ! .

In [12], Wang obtained an explicit expression for sums of products of l Bernoulli polynomials and nl Euler polynomials.

In [3], Carlitz also defined the degenerate Euler numbers ε n (λ) by

2 ( 1 + λ t ) 1 / λ + ( 1 + λ t ) 1 / λ = n = 0 ε n (λ) t n n ! .

Motivated by (1.4), we shall establish a generalization of (1.5) for ε n (λ) in this note. We define a class of generalized Stirling-like polynomials of the first kind as follows:

τ m , k (λ,x,n)= 1 i k < i k 1 < < i 2 < i 1 m j = 1 k ( x + ( λ 1 ) i j ( n + m + 1 j ) λ ) .

In particular, here we set τ m , 0 (λ,x,n)=1.

Theorem 1.1 For 1Nn, we have

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) ε j 1 ( λ ) ε j 2 ( λ ) ε j N ( λ ) = 2 n + N 1 ( N 1 ) ! k = 0 N 1 ( 1 ) k τ N 1 , k ( λ 2 , N 2 , n ) ε n + N 1 k ( λ 2 , N 2 ) ,
(1.6)

where the degenerate Euler polynomials ε n (λ,x) are given by

2 ( 1 + λ t ) x / λ ( 1 + λ t ) 1 / λ + 1 = n = 0 ε n (λ,x) t n n ! .

In fact, we shall prove a polynomial extension of (1.6) in the next section. In the third section, we also establish a generalization of (1.4) for the degenerate Bernoulli polynomial β n (λ,x) (see, e.g., [3]).

2 Degenerate Euler numbers and polynomials

Note that ε n (λ)= 2 n ε n (λ/2,1/2). So (1.6) is evidently a consequence of the following theorem.

Theorem 2.1 Let y= x 1 + x 2 ++ x N . Then for 1Nn, we have

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) ε j 1 ( λ , x 1 ) ε j 2 ( λ , x 2 ) ε j N ( λ , x N ) = 2 N 1 ( N 1 ) ! k = 0 N 1 ( 1 ) k τ N 1 , k ( λ , y , n ) ε n + N 1 k ( λ , y ) .
(2.1)

The degenerate Euler polynomials of order m are defined by

( 2 ( 1 + λ t ) 1 / λ + 1 ) m ( 1 + λ t ) x / λ = n = 0 ε n ( m ) (λ,x) t n n ! .
(2.2)

Clearly, ε n ( 1 ) (λ,x)= ε n (λ,x).

Lemma 2.1 For 1mn, we have

ε n ( m ) (λ,x)= 2 m 1 ( m 1 ) ! k = 0 m 1 ( 1 ) k τ m 1 , k (λ,x,n) ε n + m 1 k (λ,x).
(2.3)

Proof Observe that

2 m + 1 ( 1 + λ t ) x / λ ( ( 1 + λ t ) 1 / λ + 1 ) m + 1 = 2 ( 1 + λ t ) m d d t ( 2 m ( 1 + λ t ) x / λ ( ( 1 + λ t ) 1 / λ + 1 ) m ) 2 ( x m ) m 2 m ( 1 + λ t ) x / λ ( ( 1 + λ t ) 1 / λ + 1 ) m .

Comparing the coefficients of t n in both sides of the above equation, we have

ε n ( m + 1 ) (λ,x)= 2 m ( ε n + 1 ( m ) ( λ , x ) ( x + ( λ 1 ) m ( n + m ) λ ) ε n ( m ) ( λ , x ) ) .
(2.4)

Below we use induction on m to show (2.3). It is easy to see that (2.3) holds for m=1. Now let m>1 and assume that (2.3) holds for the smaller values of m. Then, by the induction hypothesis, we have

2 m ε n + 1 ( m ) ( λ , x ) 2 m ( x + ( λ 1 ) m ( n + m ) λ ) ε n ( m ) ( λ , x ) = 2 m m ! k = 0 m 1 ( 1 ) k τ m 1 , k ( λ , x , n + 1 ) ε n + m k ( λ , x ) 2 m ( x + ( λ 1 ) m ( n + m ) λ ) m ! k = 1 m ( 1 ) k 1 τ m 1 , k 1 ( λ , x , n ) ε n + m k ( λ , x ) .

It is easy to verify that

τ m , k (λ,x,n)= τ m 1 , k (λ,x,n+1)+ ( x + ( λ 1 ) m ( n + m ) λ ) τ m 1 , k 1 (λ,x,n)

for 1km1, and

τ m , m (λ,x,n)= ( x + ( λ 1 ) m ( n + m ) λ ) τ m 1 , m 1 (λ,x,n).

So by (2.4), we get

ε n ( m + 1 ) ( λ , x ) = 2 m m ! ( ε n + m ( λ , x ) + ( 1 ) m τ m , m ( λ , x , n ) ε n ( λ , x ) + k = 1 m 1 ( 1 ) k τ m , k ( λ , x , n ) ε n + m k ( λ , x ) ) = 2 m m ! k = 0 m ( 1 ) k τ m , k ( λ , x , n ) ε n + m k ( λ , x ) .

This concludes our proof. □

Let us turn to the proof of Theorem 2.1. Clearly,

2 ( 1 + λ t ) x 1 / λ ( 1 + λ t ) 1 / λ + 1 2 ( 1 + λ t ) x 2 / λ ( 1 + λ t ) 1 / λ + 1 2 ( 1 + λ t ) x N / λ ( 1 + λ t ) 1 / λ + 1 = 2 N ( 1 + λ t ) ( x 1 + x 2 + + x N ) / λ ( ( 1 + λ t ) 1 / λ + 1 ) N .

Hence we have

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) ε j 1 (λ, x 1 ) ε j 2 (λ, x 2 ) ε j N (λ, x N )= ε n ( N ) (λ,y).

Thus (2.1) immediately follows from (2.3).

3 Degenerate Bernoulli numbers and polynomials

The Bernoulli polynomials B n (x) are defined by

t e x t e t 1 = n = 0 B n (x) t n n ! .

Clearly, B n (0)= B n . In [1], Dilcher proved that

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) B j 1 ( x 1 ) B j 2 ( x 2 ) B j N ( x N ) = N ( n N ) j = 0 N 1 ( 1 ) N 1 j ( k = 0 j ( n 1 k j k ) s ( N , N k ) y j k ) B n j ( y ) n j ,
(3.1)

where y= x 1 + x 2 ++ x N .

The degenerate Bernoulli polynomials β n (λ,x) are defined by

t ( 1 + λ t ) x / λ ( 1 + λ t ) 1 / λ 1 = n = 0 β n (λ,x) t n n ! .
(3.2)

In this section, we shall give a generalization of (3.1) for β n (λ,x).

Define σ m , k (λ,x,n) by

σ m , k (λ,x,n)= 1 i k < i k 1 < < i 2 < i 1 m j = 1 k ( x + ( λ 1 ) i j ( n j ) λ ) .

In particular, we set σ m , 0 (λ,x,n)=1.

Theorem 3.1 Let y= x 1 + x 2 ++ x N . Then for 1Nn, we have

j 1 + j 2 + + j N = n j 1 , j 2 , , j N 0 ( n j 1 , j 2 , , j N ) β j 1 ( λ , x 1 ) β j 2 ( λ , x 2 ) β j N ( λ , x N ) = N ( n N ) k = 0 N 1 ( 1 ) N 1 k σ N 1 , k ( λ , y , n ) β n k ( λ , y ) n k .
(3.3)

Proof Similarly as in the proof of Theorem 2.1, it suffices to show that

β n ( m ) (λ,x)=m ( n m ) k = 0 m 1 ( 1 ) m 1 k σ m 1 , k (λ,x,n) β n k ( λ , x ) n k
(3.4)

for 1mn, where the degenerate Bernoulli polynomials of order m are defined by

( t ( 1 + λ t ) 1 / λ 1 ) m ( 1 + λ t ) x / λ = n = 0 β n ( m ) (λ,x) t n n ! .
(3.5)

We shall use induction on m. Clearly (3.4) holds for m=1. Let m1 and assume that (3.4) holds for m. Note that

t m + 1 ( 1 + λ t ) x / λ ( ( 1 + λ t ) 1 / λ 1 ) m + 1 = ( ( 1 + λ t ) + ( x m ) t m ) t m ( 1 + λ t ) x / λ ( ( 1 + λ t ) 1 / λ 1 ) m ( 1 + λ t ) t m d d t ( t m ( 1 + λ t ) x / λ ( ( 1 + λ t ) 1 / λ 1 ) m ) .

Comparing the coefficients of t n n ! in both sides of the above equation, we get

β n ( m + 1 ) (λ,x)= m n m β n ( m ) (λ,x)+ n m ( x + ( λ 1 ) m ( n 1 ) λ ) β n 1 ( m ) (λ,x).
(3.6)

Applying (3.4) for β n ( m ) (λ,x) and β n 1 ( m ) (λ,x), we have

m n m β n ( m ) ( λ , x ) + n m ( x + ( λ 1 ) m ( n 1 ) λ ) β n 1 ( m ) ( λ , x ) n 1 = ( m + 1 ) ( n m + 1 ) ( k = 0 m 1 ( 1 ) m 1 k σ m 1 , k ( λ , x , n ) β n k ( λ , x ) n k + ( x + ( λ 1 ) m ( n 1 ) λ ) k = 1 m ( 1 ) m k σ m 1 , k 1 ( λ , x , n 1 ) β n k ( λ , x ) n k ) .

It is not difficult to check that

σ m , k (λ,x,n)= σ m 1 , k (λ,x,n)+ ( x + ( λ 1 ) m ( n 1 ) λ ) σ m 1 , k 1 (λ,x,n1)

for 1km1, and

σ m , m (λ,x,n)= ( x + ( λ 1 ) m ( n 1 ) λ ) σ m 1 , m 1 (λ,x,n1).

It follows from (3.6) that

β n ( m + 1 ) ( λ , x ) = ( m + 1 ) ( n m + 1 ) ( ( 1 ) m β n ( λ , x ) n + σ m , m ( λ , x , n ) β n m ( λ , x ) n m ) + ( m + 1 ) ( n m + 1 ) k = 1 m 1 ( 1 ) m k σ m , k ( λ , x , n ) β n k ( λ , x ) n k = ( m + 1 ) ( n m + 1 ) k = 0 m ( 1 ) m k σ m , k ( λ , x , n ) β n k ( λ , x ) n k .

All proofs thus are done. □

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Acknowledgements

The authors are grateful to two anonymous referees for their helpful suggestions. The second author is supported by National Natural Science Foundation of China (Grant No.11271185). The second author is the corresponding author.

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Wu, M., Pan, H. Sums of products of the degenerate Euler numbers. Adv Differ Equ 2014, 40 (2014). https://doi.org/10.1186/1687-1847-2014-40

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Keywords

  • degenerate Euler number
  • degenerate Euler polynomial
  • degenerate Bernoulli polynomial