- Email: [email protected]
Identities from the binomial transform
Journal of Number Theory 124 (2007) 142–150 www.elsevier.com/locate/jnt
Identities from the binomial transform Kwang-Wu Chen Department of Mathematics and Computer Science Education, Taipei Municipal University Education, No. 1, Ai-Kuo West Road, Taipei 100, Taiwan, ROC Received 31 May 2006 Available online 24 October 2006 Communicated by David Goss
Abstract
A sequence bn is the binomial transform of the sequence an if bn = nk=0 nk ak . We derive a general identity for such pairs of sequences. Various known identities are obtained as particular cases. © 2006 Elsevier Inc. All rights reserved. MSC: primary 05A19; secondary 11B65, 11B68 Keywords: Generalized Seidel matrix; Binomial transform; Bernoulli numbers
1. Introduction Given two sequences {an }n∈N and {bn }n∈N , where N = {0, 1, 2, . . .}. We say that the sequence bn is the binomial transform of the sequence an if bn =
n n k=0
k
ak .
In 1995 M. Kaneko [9] found the following recursion formula for the Bernoulli numbers Bn : k k+1 (k + j + 1)Bk+j = 0, j j =0
E-mail address: [email protected]. 0022-314X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2006.07.015
for k 1,
(1.1)
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
143
where the Bernoulli polynomials Bn (x) are defined by ∞
Bn (x)t n text = , et − 1 n! n=0
and the Bernoulli numbers Bn are defined by Bn = Bn (0). There are a lot of general formulae obtained by many authors [7,10,12,14]. All these general formulae are related to the binomial transform. In this paper, we obtain the following theorem. Theorem 1.1. Let s be an arbitrary integer and bn be the binomial transform of an . Then for all nonnegative integers m and n, m m (n + k)! an+k+s k (n + k + s)! k=0
=
n n (−1)n−k (m + k)! k=0
+
k
(m + k + s)!
bm+k+s
s−1 s−1−j s − 1 − j s − 1 j =0
i
i=0
j
(−1)n+1+i aj
. (s − 1)!(m + n + 1 + i) m+n+i n
In the next section we will derive some basic properties of the generalized Seidel matrices. Using these basic properties, we could give a detail proof of our main theorem in Section 3. In Section 4, we derive a general form of [12, Theorem 1.1]. All the identities mentioned in [12] are special cases of our theorem. This includes the general formula of Eq. (1.1). In the last section, we apply our main theorems to quite a few known sequences (e.g. the self-dual sequences, the Bernoulli polynomials, and the Stirling numbers of the second kind, etc.). 2. Generalized Seidel matrices Let α, β be any two nonzero complex numbers, and {an }n∈N be a sequence of complex numbers where N = {0, 1, 2, . . .}. We construct the generalized Seidel matrix (an,m )n,m∈N associated with {an }n∈N and a parameter pair (α, β) as follows [2,3]: 1. The first row (a0,n = an )n∈N of the matrix is called the initial sequence of the matrix. 2. Each entry an,m of the nth row is defined as an,m = αan−1,m + βan−1,m+1 ,
for n 1, m 0.
(2.1)
We denote this generalized Seidel matrix as GM(an ; α, β). The first column (an,0 )n∈N of this matrix is called the final sequence. From Eq. (2.1) the element an,m can be expressed as an,m =
n n k=0
k
α n−k β k a0,m+k .
If α = β = 1, then this matrix is the classical Seidel matrix [4–6].
(2.2)
144
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
Theorem 2.1. GM(an ; α, β) = t GM(an,0 ; −αβ −1 , β −1 ), where t M is the transpose of the matrix M. In particular, we have n m n n−k k m α β a0,m+k = (−α)m−k β −m an+k,0 . k k k=0
(2.3)
k=0
Proof. Let A(x) =
∞ a0,n x n n=0
n!
¯ be the exponential generating function (egf) of the initial sequence a0,n = an and A(x) the egf of the final sequence an,0 . From Eq. (2.2) we have ¯ A(x) = eαx A(βx)
(2.4)
or A(x) = e
−αx/β
x ¯ . A β
We rewrite Eq. (2.1) as an,m = −αβ −1 an,m−1 + β −1 an+1,m−1 ,
for n 0, m 1.
Then we can express an,m as a linear combination of an,0 . m m an,m = (−α)m−k β −m an+k,0 . k k=0
Let bm,n be the (m, n)-entry of GM(an,0 ; −αβ −1 , β −1 ). Equation (2.2) implies m m−k −1 k m bm,n = −αβ −1 β an+k,0 k k=0
=
m m k=0
Therefore bm,n = an,m .
k
(−α)m−k β −m an+k,0 .
2
3. Proofs of main theorems Theorem 3.1. If the sequence bn is the binomial transform of the sequence an , then for all nonnegative integers
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
m
m
k n+k+s s k=0
an+k+s =
n
n k m+k+s (−1)n−k bm+k+s s
k=0
s−1 s−1−j
+
145
j =0
s −1−j i
i=0
(−1)n+1+i saj s −1 . j (m + n + 1 + i) · m+n+i n
Proof. We use induction on the number s. For s = 0, the sequence bn can be viewed as the final sequence of the general Seidel matrix GM(an ; 1, 1). Applying Eq. (2.3), we have m m
k
k=0
an+k =
n n k=0
k
(−1)n−k bm+k .
Assume that the theorem is true for the number s. Then for all nonnegative integers m and n, we have m
m
k n+k+s s k=0
an+k+s =
n
n k m+k+s (−1)n−k bm+k+s s
k=0
+
s−1 s−1−j s − 1 − j s − 1 j =0
i
i=0
j
(−1)n+1+i saj . (m + n + 1 + i) · m+n+i n
(3.1)
Let an# = an+1 /(n + 1) and bn# be the binomial transform of an# . Then bn#
=
n n k=0
k
ak#
=
n n ak+1 k=0
k k+1
=
bn+1 a0 − . n+1 n+1
By the inductive hypothesis, Eq. (3.1) holds for these two sequences an# and bn# . Changing an# , bn# to an , bn and multiplying (s + 1) into the resulted identity, we get the following identity: m
m
an+k+s+1 n+k+s +1 n
k n+k+s+1 s+1
k=0
=
n
k m+k+s+1 s+1 k=0
+
n−k
(−1)
k=0
s−1 s−1−j j =0
i=0
bm+k+s+1 +
n n
s −1−j i
(−1)n−k+1 a0 (s + 1)! k (m + k + 1)(m + k + 2) · · · (m + k + s + 1)
(−1)n+1+i (s + 1)aj +1 s . j + 1 (m + n + 1 + i) m+n+i n
Since Eq. (5.41) of [8] n n (−1)k k=0
k x +k
=
n! , x(x + 1) · · · (x + n)
(3.2)
146
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
the third factor of Eq. (3.2) becomes s s (−1)n+1+i (s + 1)a0 . i (m + n + 1 + i) m+n+i n
i=0
Changing the dummy variable j to k − 1, the fourth factor of Eq. (3.2) becomes s s−k s −k s (−1)n+1+i (s + 1)ak . i k (m + n + 1 + i) m+n+i n
k=1 i=0
Now we collect the above information and rewrite Eq. (3.2): m
m
k n+k+s+1 s+1 k=0
an+k+s+1 =
n
n k m+k+s+1 (−1)n−k bm+k+s+1
k=0
+
s+1
s−j s s−j s j =0 i=0
This completes our proof.
i
(−1)n+1+i (s + 1)aj . j (m + n + 1 + i) · m+n+i n
2
Theorem 3.2. If bn is the binomial transform of the sequence an , then for all nonnegative integers m, n, s: m m n+k k=0
k
s
an+k−s =
n n m+k (−1)n−k bm+k−s . k s k=0
Proof. The proof is similar to the proof of Theorem 3.1, but we use an# = nan−1 and bn# = nbn−1 . 2 4. Dual sequences Lemma 4.1. Let bn (α1 , β1 ) be the final sequence of GM(an ; α1 , β1 ) and cn (α1 , β1 , α2 , β2 ) be the final sequence of GM(bn (α1 , β1 ); α2 , β2 ). Then the sequence cn (α1 , β1 , α2 , β2 ) is also the final sequence of GM(an ; α1 β2 + α2 , β1 β2 ). Proof. Let Θ(z), Ψ (z), and Φ(z) be the exponential generating functions of an , bn (α1 , β1 ), and cn (α1 , β1 , α2 , β2 ), respectively. Applying Eq. (2.4) we have Ψ (z) = eα1 z Θ(β1 z)
(4.1)
Φ(z) = eα2 z Ψ (β2 z).
(4.2)
and
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
147
Substituting Eq. (4.1) into Eq. (4.2), we have Φ(z) = e(α1 β2 +α2 )z Θ(β1 β2 z). Hence cn (α1 , β1 , α2 , β2 ) is also the final sequence of GM(an ; α1 β2 + α2 , β1 β2 ).
2
The dual sequence an∗ of an is defined by an∗
n n
=
k=0
k
(−1)k ak .
So an∗ is the final sequence of GM(an ; 1, −1). Z.-W. Sun [12] defined two related polynomial sequences as An (x) =
n n k=0
k
(−1)k ak x n−k
and A∗n (x) =
n n k=0
k
(−1)k ak∗ x n−k .
In our viewpoint, An (x) is the final sequence of GM(an ; x, −1) and A∗n (x) is the final sequence of GM(an∗ ; x, −1). Now we give a generalization of Theorem 1.1 in [12]. Theorem 4.2. Let m, n, s be nonnegative integers and x + y + z = 1. Then we have m
m k n+k+s x m−k An+k+s (y) s
k=0
=
n
k m+k+s (−1)n+m+s x n−k A∗m+k+s (z)
k=0
+
n s
s−1 s−1−j j =0
i=0
s −1−j i
s − 1 (−1)n+1+i x m+n+s−j sAj (y) , j (m + n + 1 + i) · m+n+i
(4.3)
n
and m n m n + k m−k n m+k x (−1)m+n+s x n−k A∗m+k−s (z). An+k−s (y) = k s k s k=0
(4.4)
k=0
Proof. Since x + y + z = 1, A∗n (z) = A∗n (1 − x − y) is the final sequence of GM(an∗ ; 1 − x − y, −1). an∗ is the final sequence of GM(an ; 1, −1). Applying Lemma 4.1, A∗n (z) is thus the final sequence of GM(an ; −x − y, 1).
148
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
Now we consider the generalized Seidel matrix GM(An (y); −x, −1). Its final sequence is un =
n n (−x)n−k (−1)k Ak (y). k k=0
An (y) is the final sequence of GM(an ; y, −1). We apply Lemma 4.1, then the sequence un is the final sequence of GM(an ; −y − x, 1). Hence un = A∗n (z). Therefore A∗n (z)/(−x)n is the binomial transform of An (y)/x n . Now we can apply Theorems 3.1 and 3.2. Then we get the desired identities. 2 Equation (1.4) of [12] is just the case s = 1 in Eq. (4.3). Equations (1.5) and (1.6) of [12] are the cases s = 0 and s = 1 in Eq. (4.4), respectively. Hence all the identities mentioned in [12] are special cases of Theorem 4.2. 5. More applications The self-dual sequence an∗ = an is of particular interest [11–13]. It is known that [11]
2
−n
,
n + 2m − 1 −1 , m
−1 x dx, (−1) n n
nFn−1 ,
Ln ,
(−1)n Bn
0
are all self-dual sequences, where Fn , Ln , and Bn are the Fibonacci numbers, Lucas numbers, and Bernoulli numbers, respectively. If an is a self-dual sequence, then an is the binomial transform of (−1)n an . Now we apply Theorems 3.1 and 3.2 to a self-dual sequence and we set m = n, to obtain the following. Theorem 5.1. Let an be a self-dual sequence. Then for n ∈ N, s, t ∈ Z+ with t odd, n
n (−1)k kn+k+s 1 − (−1)s an+k+s
k=0
=
s s−1 s−1−j j =0
i=0
s −1−j i
s(−1)i+j aj s−1 , j (2n + 1 + i) 2n+i
(5.1)
n
and n n n+k k=0
k
t
(−1)k an+k−t = 0.
(5.2)
Substituting the self-dual sequence (−1)n Bn into Theorem 5.1, we have the following. Corollary 5.2. Let n ∈ N, and t, s ∈ Z+ . Then n
n
k n+k+s Bn+k+s s k=0
=
s−1 s−1−j s − 1 − j s − 1 j =0
i=0
i
j
s(−1)i+n+s Bj , 2(2n + 1 + i) 2n+i n
(5.3)
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
149
and n 2n n n+k n k Bn+k−s = Bk−s = 0. k s k−n s
(5.4)
k=n
k=0
Since Eq. (23.1.7) of [1] Bn (x + y) =
n n k=0
k
Bk (x)y n−k ,
(5.5)
we know that Bn (x + y)/y n is the binomial transform of Bn (x)/y n . Applying Theorems 3.1 and 3.2, we get the following. Theorem 5.3. Let m, n, s be nonnegative integers. Then we have m
m k n+k+s y m−k Bn+k+s (x) s
k=0
=
n
n k m+k+s (−y)n−k Bm+k+s (x + y)
k=0
+
s
s−1 s−1−j j =0
i=0
s −1−j i
s − 1 (−1)n+1+i sy m+n+s−j Bj (x) , j (m + n + 1 + i) m+n+i n
(5.6)
and m m n+k k=0
k
s
y m−k Bn+k−s (x) =
n n m+k k=0
k
s
(−y)n−k Bm+k−s (x + y).
(5.7)
There are many polynomial sequences satisfied the relation equation (5.5), for example, the Euler polynomials En (x), the Genocchi polynomials Gn (x), the polynomial sequences An (x) and A∗n (x) which are defined in Section 4, etc. Thus the analogue of Theorem 5.3 to these polynomial sequences also hold. The Stirling numbers of the second kind S(n, m) are defined by ∞ xn (ex − 1)m = S(n, m) . m! n! n=m
Since we have the well-known identity n n k=0
k
k!S(q, k) = nq ,
it can be seen that nq is the binomial transform of n!S(q, n). Applying Theorems 3.1 and 3.2, we get the following.
150
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
Theorem 5.4. Let m, n, s be nonnegative integers. Then we have m m k=0
k
(n + k)!S(q, n + k + s)
n n (m + k + s)q (m + k)!(−1)n−k = k (m + k + s)! k=0
+
s−1 s−1−j s − 1 − j j =0
i=0
i
(−1)n+1+i S(q, j )
, (s − 1 − j )!(m + n + 1 + i) m+n+i n
(5.8)
and m n n (m + k − s)q (m + k)!(−1)n−k . (n + k)!S(q, n + k − s) = k (m + k − s)! k=0
(5.9)
k=0
References [1] M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover Publications, New York, 1972. [2] K.-W. Chen, A summation on Bernoulli numbers, J. Number Theory 111 (2005) 372–391, doi:10.1016/ j.jnt.2004.08.011. [3] K.-W. Chen, Applications related to the generalized Seidel matrix, Ars Combin. 77 (2005), 12 pp. [4] D. Dumont, Matrices d’Euler–Seidel, in: Sém. Lothar. Combin., 5-éme session, 1981, vol. 182/s-04, Publ. IRMA, Strasbourg, 1982, pp. 59–78. [5] D. Dumont, Further triangles of Seidel–Arnold type and continued fractions related to Euler and Springer numbers, Adv. in Appl. Math. 16 (3) (1995) 275–296, doi:10.1006/aama.1995.1014. [6] D. Dumont, J. Zeng, Further results on the Euler and Genocchi numbers, Aequationes Math. 47 (1994) 31–42. [7] I.M. Gessel, Applications of the classical umbra calculus, Algebra Universalis 49 (2003) 397–434. [8] R.L. Graham, D.E. Knuth, O. Patashnik, Concrete Mathematics, Addison–Wesley, New York, 1994. [9] M. Kaneko, A recurrence formula for the Bernoulli numbers, Proc. Japan Acad. Ser. A Math. Sci. 71 (1995) 192– 193. [10] H. Momiyama, A new recurrence formula for Bernoulli numbers, Fibonacci Quart. 39 (3) (2001) 285–288. [11] Z.-H. Sun, Invariant sequences under binomial transformation, Fibonacci Quart. 39 (4) (2001) 324–333. [12] Z.-W. Sun, Combinatorial identities in dual sequences, European J. Combin. 24 (6) (2003) 709–718. [13] Y. Wang, Self-inverse sequences related to a binomial inverse pair, Fibonacci Quart. 43 (1) (2005) 46–52. [14] K.-J. Wu, Z.-W. Sun, H. Pan, Some identities for Bernoulli and Euler polynomials, Fibonacci Quart. 42 (4) (2004) 295–299.
Identities from the binomial transform Kwang-Wu Chen Department of Mathematics and Computer Science Education, Taipei Municipal University Education, No. 1, Ai-Kuo West Road, Taipei 100, Taiwan, ROC Received 31 May 2006 Available online 24 October 2006 Communicated by David Goss
Abstract
A sequence bn is the binomial transform of the sequence an if bn = nk=0 nk ak . We derive a general identity for such pairs of sequences. Various known identities are obtained as particular cases. © 2006 Elsevier Inc. All rights reserved. MSC: primary 05A19; secondary 11B65, 11B68 Keywords: Generalized Seidel matrix; Binomial transform; Bernoulli numbers
1. Introduction Given two sequences {an }n∈N and {bn }n∈N , where N = {0, 1, 2, . . .}. We say that the sequence bn is the binomial transform of the sequence an if bn =
n n k=0
k
ak .
In 1995 M. Kaneko [9] found the following recursion formula for the Bernoulli numbers Bn : k k+1 (k + j + 1)Bk+j = 0, j j =0
E-mail address: [email protected]. 0022-314X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2006.07.015
for k 1,
(1.1)
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
143
where the Bernoulli polynomials Bn (x) are defined by ∞
Bn (x)t n text = , et − 1 n! n=0
and the Bernoulli numbers Bn are defined by Bn = Bn (0). There are a lot of general formulae obtained by many authors [7,10,12,14]. All these general formulae are related to the binomial transform. In this paper, we obtain the following theorem. Theorem 1.1. Let s be an arbitrary integer and bn be the binomial transform of an . Then for all nonnegative integers m and n, m m (n + k)! an+k+s k (n + k + s)! k=0
=
n n (−1)n−k (m + k)! k=0
+
k
(m + k + s)!
bm+k+s
s−1 s−1−j s − 1 − j s − 1 j =0
i
i=0
j
(−1)n+1+i aj
. (s − 1)!(m + n + 1 + i) m+n+i n
In the next section we will derive some basic properties of the generalized Seidel matrices. Using these basic properties, we could give a detail proof of our main theorem in Section 3. In Section 4, we derive a general form of [12, Theorem 1.1]. All the identities mentioned in [12] are special cases of our theorem. This includes the general formula of Eq. (1.1). In the last section, we apply our main theorems to quite a few known sequences (e.g. the self-dual sequences, the Bernoulli polynomials, and the Stirling numbers of the second kind, etc.). 2. Generalized Seidel matrices Let α, β be any two nonzero complex numbers, and {an }n∈N be a sequence of complex numbers where N = {0, 1, 2, . . .}. We construct the generalized Seidel matrix (an,m )n,m∈N associated with {an }n∈N and a parameter pair (α, β) as follows [2,3]: 1. The first row (a0,n = an )n∈N of the matrix is called the initial sequence of the matrix. 2. Each entry an,m of the nth row is defined as an,m = αan−1,m + βan−1,m+1 ,
for n 1, m 0.
(2.1)
We denote this generalized Seidel matrix as GM(an ; α, β). The first column (an,0 )n∈N of this matrix is called the final sequence. From Eq. (2.1) the element an,m can be expressed as an,m =
n n k=0
k
α n−k β k a0,m+k .
If α = β = 1, then this matrix is the classical Seidel matrix [4–6].
(2.2)
144
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
Theorem 2.1. GM(an ; α, β) = t GM(an,0 ; −αβ −1 , β −1 ), where t M is the transpose of the matrix M. In particular, we have n m n n−k k m α β a0,m+k = (−α)m−k β −m an+k,0 . k k k=0
(2.3)
k=0
Proof. Let A(x) =
∞ a0,n x n n=0
n!
¯ be the exponential generating function (egf) of the initial sequence a0,n = an and A(x) the egf of the final sequence an,0 . From Eq. (2.2) we have ¯ A(x) = eαx A(βx)
(2.4)
or A(x) = e
−αx/β
x ¯ . A β
We rewrite Eq. (2.1) as an,m = −αβ −1 an,m−1 + β −1 an+1,m−1 ,
for n 0, m 1.
Then we can express an,m as a linear combination of an,0 . m m an,m = (−α)m−k β −m an+k,0 . k k=0
Let bm,n be the (m, n)-entry of GM(an,0 ; −αβ −1 , β −1 ). Equation (2.2) implies m m−k −1 k m bm,n = −αβ −1 β an+k,0 k k=0
=
m m k=0
Therefore bm,n = an,m .
k
(−α)m−k β −m an+k,0 .
2
3. Proofs of main theorems Theorem 3.1. If the sequence bn is the binomial transform of the sequence an , then for all nonnegative integers
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
m
m
k n+k+s s k=0
an+k+s =
n
n k m+k+s (−1)n−k bm+k+s s
k=0
s−1 s−1−j
+
145
j =0
s −1−j i
i=0
(−1)n+1+i saj s −1 . j (m + n + 1 + i) · m+n+i n
Proof. We use induction on the number s. For s = 0, the sequence bn can be viewed as the final sequence of the general Seidel matrix GM(an ; 1, 1). Applying Eq. (2.3), we have m m
k
k=0
an+k =
n n k=0
k
(−1)n−k bm+k .
Assume that the theorem is true for the number s. Then for all nonnegative integers m and n, we have m
m
k n+k+s s k=0
an+k+s =
n
n k m+k+s (−1)n−k bm+k+s s
k=0
+
s−1 s−1−j s − 1 − j s − 1 j =0
i
i=0
j
(−1)n+1+i saj . (m + n + 1 + i) · m+n+i n
(3.1)
Let an# = an+1 /(n + 1) and bn# be the binomial transform of an# . Then bn#
=
n n k=0
k
ak#
=
n n ak+1 k=0
k k+1
=
bn+1 a0 − . n+1 n+1
By the inductive hypothesis, Eq. (3.1) holds for these two sequences an# and bn# . Changing an# , bn# to an , bn and multiplying (s + 1) into the resulted identity, we get the following identity: m
m
an+k+s+1 n+k+s +1 n
k n+k+s+1 s+1
k=0
=
n
k m+k+s+1 s+1 k=0
+
n−k
(−1)
k=0
s−1 s−1−j j =0
i=0
bm+k+s+1 +
n n
s −1−j i
(−1)n−k+1 a0 (s + 1)! k (m + k + 1)(m + k + 2) · · · (m + k + s + 1)
(−1)n+1+i (s + 1)aj +1 s . j + 1 (m + n + 1 + i) m+n+i n
Since Eq. (5.41) of [8] n n (−1)k k=0
k x +k
=
n! , x(x + 1) · · · (x + n)
(3.2)
146
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
the third factor of Eq. (3.2) becomes s s (−1)n+1+i (s + 1)a0 . i (m + n + 1 + i) m+n+i n
i=0
Changing the dummy variable j to k − 1, the fourth factor of Eq. (3.2) becomes s s−k s −k s (−1)n+1+i (s + 1)ak . i k (m + n + 1 + i) m+n+i n
k=1 i=0
Now we collect the above information and rewrite Eq. (3.2): m
m
k n+k+s+1 s+1 k=0
an+k+s+1 =
n
n k m+k+s+1 (−1)n−k bm+k+s+1
k=0
+
s+1
s−j s s−j s j =0 i=0
This completes our proof.
i
(−1)n+1+i (s + 1)aj . j (m + n + 1 + i) · m+n+i n
2
Theorem 3.2. If bn is the binomial transform of the sequence an , then for all nonnegative integers m, n, s: m m n+k k=0
k
s
an+k−s =
n n m+k (−1)n−k bm+k−s . k s k=0
Proof. The proof is similar to the proof of Theorem 3.1, but we use an# = nan−1 and bn# = nbn−1 . 2 4. Dual sequences Lemma 4.1. Let bn (α1 , β1 ) be the final sequence of GM(an ; α1 , β1 ) and cn (α1 , β1 , α2 , β2 ) be the final sequence of GM(bn (α1 , β1 ); α2 , β2 ). Then the sequence cn (α1 , β1 , α2 , β2 ) is also the final sequence of GM(an ; α1 β2 + α2 , β1 β2 ). Proof. Let Θ(z), Ψ (z), and Φ(z) be the exponential generating functions of an , bn (α1 , β1 ), and cn (α1 , β1 , α2 , β2 ), respectively. Applying Eq. (2.4) we have Ψ (z) = eα1 z Θ(β1 z)
(4.1)
Φ(z) = eα2 z Ψ (β2 z).
(4.2)
and
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
147
Substituting Eq. (4.1) into Eq. (4.2), we have Φ(z) = e(α1 β2 +α2 )z Θ(β1 β2 z). Hence cn (α1 , β1 , α2 , β2 ) is also the final sequence of GM(an ; α1 β2 + α2 , β1 β2 ).
2
The dual sequence an∗ of an is defined by an∗
n n
=
k=0
k
(−1)k ak .
So an∗ is the final sequence of GM(an ; 1, −1). Z.-W. Sun [12] defined two related polynomial sequences as An (x) =
n n k=0
k
(−1)k ak x n−k
and A∗n (x) =
n n k=0
k
(−1)k ak∗ x n−k .
In our viewpoint, An (x) is the final sequence of GM(an ; x, −1) and A∗n (x) is the final sequence of GM(an∗ ; x, −1). Now we give a generalization of Theorem 1.1 in [12]. Theorem 4.2. Let m, n, s be nonnegative integers and x + y + z = 1. Then we have m
m k n+k+s x m−k An+k+s (y) s
k=0
=
n
k m+k+s (−1)n+m+s x n−k A∗m+k+s (z)
k=0
+
n s
s−1 s−1−j j =0
i=0
s −1−j i
s − 1 (−1)n+1+i x m+n+s−j sAj (y) , j (m + n + 1 + i) · m+n+i
(4.3)
n
and m n m n + k m−k n m+k x (−1)m+n+s x n−k A∗m+k−s (z). An+k−s (y) = k s k s k=0
(4.4)
k=0
Proof. Since x + y + z = 1, A∗n (z) = A∗n (1 − x − y) is the final sequence of GM(an∗ ; 1 − x − y, −1). an∗ is the final sequence of GM(an ; 1, −1). Applying Lemma 4.1, A∗n (z) is thus the final sequence of GM(an ; −x − y, 1).
148
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
Now we consider the generalized Seidel matrix GM(An (y); −x, −1). Its final sequence is un =
n n (−x)n−k (−1)k Ak (y). k k=0
An (y) is the final sequence of GM(an ; y, −1). We apply Lemma 4.1, then the sequence un is the final sequence of GM(an ; −y − x, 1). Hence un = A∗n (z). Therefore A∗n (z)/(−x)n is the binomial transform of An (y)/x n . Now we can apply Theorems 3.1 and 3.2. Then we get the desired identities. 2 Equation (1.4) of [12] is just the case s = 1 in Eq. (4.3). Equations (1.5) and (1.6) of [12] are the cases s = 0 and s = 1 in Eq. (4.4), respectively. Hence all the identities mentioned in [12] are special cases of Theorem 4.2. 5. More applications The self-dual sequence an∗ = an is of particular interest [11–13]. It is known that [11]
2
−n
,
n + 2m − 1 −1 , m
−1 x dx, (−1) n n
nFn−1 ,
Ln ,
(−1)n Bn
0
are all self-dual sequences, where Fn , Ln , and Bn are the Fibonacci numbers, Lucas numbers, and Bernoulli numbers, respectively. If an is a self-dual sequence, then an is the binomial transform of (−1)n an . Now we apply Theorems 3.1 and 3.2 to a self-dual sequence and we set m = n, to obtain the following. Theorem 5.1. Let an be a self-dual sequence. Then for n ∈ N, s, t ∈ Z+ with t odd, n
n (−1)k kn+k+s 1 − (−1)s an+k+s
k=0
=
s s−1 s−1−j j =0
i=0
s −1−j i
s(−1)i+j aj s−1 , j (2n + 1 + i) 2n+i
(5.1)
n
and n n n+k k=0
k
t
(−1)k an+k−t = 0.
(5.2)
Substituting the self-dual sequence (−1)n Bn into Theorem 5.1, we have the following. Corollary 5.2. Let n ∈ N, and t, s ∈ Z+ . Then n
n
k n+k+s Bn+k+s s k=0
=
s−1 s−1−j s − 1 − j s − 1 j =0
i=0
i
j
s(−1)i+n+s Bj , 2(2n + 1 + i) 2n+i n
(5.3)
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
149
and n 2n n n+k n k Bn+k−s = Bk−s = 0. k s k−n s
(5.4)
k=n
k=0
Since Eq. (23.1.7) of [1] Bn (x + y) =
n n k=0
k
Bk (x)y n−k ,
(5.5)
we know that Bn (x + y)/y n is the binomial transform of Bn (x)/y n . Applying Theorems 3.1 and 3.2, we get the following. Theorem 5.3. Let m, n, s be nonnegative integers. Then we have m
m k n+k+s y m−k Bn+k+s (x) s
k=0
=
n
n k m+k+s (−y)n−k Bm+k+s (x + y)
k=0
+
s
s−1 s−1−j j =0
i=0
s −1−j i
s − 1 (−1)n+1+i sy m+n+s−j Bj (x) , j (m + n + 1 + i) m+n+i n
(5.6)
and m m n+k k=0
k
s
y m−k Bn+k−s (x) =
n n m+k k=0
k
s
(−y)n−k Bm+k−s (x + y).
(5.7)
There are many polynomial sequences satisfied the relation equation (5.5), for example, the Euler polynomials En (x), the Genocchi polynomials Gn (x), the polynomial sequences An (x) and A∗n (x) which are defined in Section 4, etc. Thus the analogue of Theorem 5.3 to these polynomial sequences also hold. The Stirling numbers of the second kind S(n, m) are defined by ∞ xn (ex − 1)m = S(n, m) . m! n! n=m
Since we have the well-known identity n n k=0
k
k!S(q, k) = nq ,
it can be seen that nq is the binomial transform of n!S(q, n). Applying Theorems 3.1 and 3.2, we get the following.
150
K.-W. Chen / Journal of Number Theory 124 (2007) 142–150
Theorem 5.4. Let m, n, s be nonnegative integers. Then we have m m k=0
k
(n + k)!S(q, n + k + s)
n n (m + k + s)q (m + k)!(−1)n−k = k (m + k + s)! k=0
+
s−1 s−1−j s − 1 − j j =0
i=0
i
(−1)n+1+i S(q, j )
, (s − 1 − j )!(m + n + 1 + i) m+n+i n
(5.8)
and m n n (m + k − s)q (m + k)!(−1)n−k . (n + k)!S(q, n + k − s) = k (m + k − s)! k=0
(5.9)
k=0
References [1] M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover Publications, New York, 1972. [2] K.-W. Chen, A summation on Bernoulli numbers, J. Number Theory 111 (2005) 372–391, doi:10.1016/ j.jnt.2004.08.011. [3] K.-W. Chen, Applications related to the generalized Seidel matrix, Ars Combin. 77 (2005), 12 pp. [4] D. Dumont, Matrices d’Euler–Seidel, in: Sém. Lothar. Combin., 5-éme session, 1981, vol. 182/s-04, Publ. IRMA, Strasbourg, 1982, pp. 59–78. [5] D. Dumont, Further triangles of Seidel–Arnold type and continued fractions related to Euler and Springer numbers, Adv. in Appl. Math. 16 (3) (1995) 275–296, doi:10.1006/aama.1995.1014. [6] D. Dumont, J. Zeng, Further results on the Euler and Genocchi numbers, Aequationes Math. 47 (1994) 31–42. [7] I.M. Gessel, Applications of the classical umbra calculus, Algebra Universalis 49 (2003) 397–434. [8] R.L. Graham, D.E. Knuth, O. Patashnik, Concrete Mathematics, Addison–Wesley, New York, 1994. [9] M. Kaneko, A recurrence formula for the Bernoulli numbers, Proc. Japan Acad. Ser. A Math. Sci. 71 (1995) 192– 193. [10] H. Momiyama, A new recurrence formula for Bernoulli numbers, Fibonacci Quart. 39 (3) (2001) 285–288. [11] Z.-H. Sun, Invariant sequences under binomial transformation, Fibonacci Quart. 39 (4) (2001) 324–333. [12] Z.-W. Sun, Combinatorial identities in dual sequences, European J. Combin. 24 (6) (2003) 709–718. [13] Y. Wang, Self-inverse sequences related to a binomial inverse pair, Fibonacci Quart. 43 (1) (2005) 46–52. [14] K.-J. Wu, Z.-W. Sun, H. Pan, Some identities for Bernoulli and Euler polynomials, Fibonacci Quart. 42 (4) (2004) 295–299.