arXiv:1608.05925v1 [math.NT] 21 Aug 2016
Higher-order identities for balancing numbers Takao Komatsu School of Mathematics and Statistics Wuhan University Wuhan 430072 China
[email protected]
Prasanta Kumar Ray VSS University of Technology, Odisha Burla-768018, India
[email protected]
Abstract Let Bn be the n-thP balancing number. In this paper, we give some 2r−3 2r−3 P l explicit expressions of l=0 (−1) j1 +···+jr =n−2l Bj1 · · · Bjr and l j1 ,...,jr ≥1 P j1 +···+jr =n Bj1 · · · Bjr . We also consider the convolution identities j1 ,...,jr ≥1
with binomial coefficients: X
k1 +···+kr =n k1 ,...,kr ≥1
n Bk 1 · · · Bk r k1 , . . . , kr
This type can be generalized, so that Bn is a special case of the number un , where un = aun−1 + bun−2 (n ≥ 2) with u0 = 0 and u1 = 1.
1
Introduction
Higher-order convolutions for various types of numbers (or polynomials) have been studied, with or without binomial (or multinomial) coefficients, including Bernoulli, Euler, Genocchi, Cauchy, Stirling, and Fibonacci numbers
1
([1, 2, 3, 6, 7, 8, 10]). One typical one is due to Euler, given by n X n k=0
k
Bk Bn−k = −nBn−1 − (n − 1)Bn
(n ≥ 0) ,
where Bn are Bernoulli numbers, defined by ∞
X tn t = Bn et − 1 n=0 n!
(|t| < 2π) .
A positive integer x is called balancing number if 1 + 2 + · · · + (x − 1) = (x + 1) + · · · + (y − 1)
(1)
holds for some integer y ≥ x + 2. The problem of determining all balancing numbers leads to a Pell equation, whose solutions in x can be described by the recurrence Bn = 6Bn−1 − Bn−2 (n ≥ 2) with B0 = 0 and B1 = 1 (see [4, 5]). One of the most general extensions of balancing numbers is when (1) is being replaced by 1k + 2k + · · · + (x − 1)k = (x + 1)l + · · · + (y − 1)l ,
(2)
where the exponents k and l are given positive integers. In the work of Liptai et al. [12] effective and non-effective finiteness theorems on (2) are proved. In [9] a balancing problem of ordinary binomial coefficients is studied. Some more results can be seen in [13, 14, 15]. The generating function f (x) of balancing numbers Bn is given by ∞
X x f (x) := = Bn xn . 1 − 6x + x2 n=0 Then f (x) satisfies the relation: f (x)2 = or
x2 f ′ (x) 2 1−x
(1 − x2 )f (x)2 = x2 f ′ (x) .
2
(3) (4)
The left-hand side of (4) is ∞ X
(1 − x2 )
u=0 ∞ X n X
= (1 − x2 ) =
∞ X n X n=0 j=0
Bu xu
!
∞ X
Bv xv
v=0
!
Bj Bn−j xn
n=0 j=0
n
Bj Bn−j x −
∞ X n−2 X
Bj Bn−j−2xn .
n=2 j=0
The right-hand side of (4) is 2
x
∞ X
n−1
nBn x
=
∞ X n=0
n=1
(n − 1)Bn−1 xn .
Comparing the coefficients of both sides, we get (n − 1)Bn−1 = =
n X j=0
Bj Bn−j −
n−2 X
Bj Bn−j−2
j=0
n−1 X (Bj Bn−j − Bj−1 Bn−j−1) . j=1
Here, notice that B0 = 0. By changing n by n + 1, we get the following identity. Theorem 1. For n ≥ 1, we have nBn =
n X j=1
(Bj Bn−j+1 − Bj−1 Bn−j ) .
Differentiating both sides of (3) by x and dividing them by 2, we obtain f (x)f ′ (x) =
x x2 ′ f (x) + f ′′ (x) . (1 − x2 )2 2(1 − x2 )
3
(5)
By (3) and (5), we get x2 f (x)f ′ (x) 1 − x2 x3 x4 ′ = f (x) + f ′′ (x) (1 − x2 )3 2(1 − x2 )2
f (x)3 =
or
1 (1 − x2 )3 f (x)3 = x3 f ′ (x) + x4 (1 − x2 )f ′′ (x) . 2 The left-hand side of (7) is equal to equal to (1 − 3x2 + 3x4 − x6 ) =
3 X ∞ X l=0
∞ X n=0
3 (−1) l j n=2l
X
(7)
Bj1 Bj2 Bj3 xn
j1 +j2 +J3 =n j1 ,j2 ,j3 ≥0
X
l
(6)
Bj1 Bj2 Bj3 xn .
1 +j2 +J3 =n−2l j1 ,j2 ,j3 ≥1
The right-hand side of (7) is x3 =
∞ X
nBn xn−1 +
n=1 ∞ X n=2
∞ ∞ x6 X x4 X n(n − 1)Bn xn−2 − n(n − 1)Bn xn−2 2 n=2 2 n=2
∞ X (n − 1)(n − 2) (n − 4)(n − 5) n Bn−2 x − Bn−4 xn . 2 2 n=4
Comparing the coefficients of both sides, we get the following result. Theorem 2. For n ≥ 4, we have 3 X X n−4 n−1 l 3 Bn−4 . Bn−2 − Bj Bj Bj = (−1) 2 2 l j +j +J =n−2l 1 2 3 l=0 1
2 3 j1 ,j2 ,j3 ≥1
In this paper, P we give some explicit expression P2r−3 P of a more general case l 2r−3 j1 +···+jr =n Bj · · · Bjr . We j1 +···+jr =n−2l Bj1 · · · Bjr and 1 l=0 (−1) l j1 ,...,jr ≥1
j1 ,...,jr ≥1
also consider a different type for more general numbers: X n u k1 · · · u kr , k , . . . , k 1 r k +···+kr =n 1 k1 ,...,kr ≥1
4
where un = aun−1 + bun−2 (n ≥ 2) with u0 = 0 and u1 = 1. When a = 6 and b = −1, this is reduced to the convolution identity for balancing numbers. When a = b = 1, this is reduced to the convolution identity for Fibonacci numbers. The corresponding identities for Luca-balancing numbers are also given.
2
Main results
First, as a general case of (3) and (6), we can have the following. Lemma 1. For r ≥ 2, we have Pk−1 k r−2 2r−k+2j−2 r−2 2r−2 (r−1) X x f (x) j=0 j k−j−1 x f (x)r = + f (r−k−1) (x) . 2 r−1 2 )r+k−1 (r − 1)!(1 − x ) k(r − k − 2)!(1 − x k=1 (8) Proof. The proof is done by induction. It is trivial to see that the identity holds for r = 2. Suppose that the identity holds for some r. Differentiating
5
both sides by x, we obtain rf (x)r−1 f ′ (x) =
x2r−2 f (r) (x) (2r − 2)x2r−3 f (r−1) (x) + (r − 1)!(1 − x2 )r−1 (r − 1)!(1 − x2 )r Pk−1 k r−2 2r−k−2+2j r−2 X j=0 j k−j−1 x + f (r−k) (x) 2 )r+k−1 k(r − k − 2)!(1 − x k=1 r−2 2r−k−3+2j Pk−1 k r−2 X j=0 (2r − k − 2 + 2j) j k−j−1 x + f (r−k−1) (x) 2 r+k k(r − k − 2)!(1 − x ) k=1 r−2 2r−k−1+2j P k−1 k r−2 X j=0 (3k − 2j) j k−j−1 x f (r−k−1) (x) + 2 )r+k k(r − k − 2)!(1 − x k=1
x2r−2 f (r) (x) 2x2r−3 f (r−1) (x) + (r − 1)!(1 − x2 )r−1 (r − 2)!(1 − x2 )r Pk−1 k r−2 2r−k−2+2j r−2 X j=0 j k−j−1 x f (r−k) (x) + 2 r+k−1 k(r − k − 2)!(1 − x ) k=1 r−2 2r−k−2+2j P k−2 k−1 r−1 X x j=0 (2r − k − 1 + 2j) j k−j−2 + f (r−k) (x) (k − 1)(r − k − 1)!(1 − x2 )r+k−1 k=2 r−2 2r−k−2+2j Pk−1 k−1 r−1 X j=1 (3k − 2j − 1) j−1 k−j−1 x f (r−k) (x) + 2 )r+k−1 (k − 1)(r − k − 1)!(1 − x k=2 Pk−1 k r−1 2r−k−2+2j r−1 X x2r−2 f (r) (x) j=0 j k−j−1 x = +r f (r−k) (x) . 2 r−1 2 )r+k−1 (r − 1)!(1 − x ) k(r − k − 1)!(1 − x k=1 =
Here, we used the relations 2 1 r + = (r − 2)! (r − 3)! (r − 2)!
6
(k = 1)
and 2r − k − 1 + 2j k − 1 r−2 r−2 r−k−1 k + k−j−1 k−j−2 j j k k−1 3k − 2j − 1 k − 1 r−2 + k−j −1 j−1 k−1 r k r−1 = (k ≥ 2) . k−j−1 k j Together with (3), we get f (x)r+1 =
x2 f (x)r−1 f ′ (x) 1 − x2
! Pk−1 k r−1 2r−k−2+2j r−1 X x2r−2 f (r) (x) j=0 j k−j−1 x + f (r−k) (x) r!(1 − x2 )r−1 k=1 k(r − k − 1)!(1 − x2 )r+k−1 Pk−1 k r−1 2r−k+2j r−1 X x2r f (r) (x) j=0 j k−j−1 x = + f (r−k) (x) . 2 r 2 )r+k r!(1 − x ) k(r − k − 1)!(1 − x k=1 x2 = 1 − x2
In general, we can state the following. Theorem 3. Let r ≥ 2. Then for n ≥ 3r − 5, we have 2r−3 X
2r − 3 (−1) l j l=0
=
r−1 X
l
X
1 +···+jr =n−2l j1 ,...,jr ≥1
(−1)
k−1 n
k=1
Bj1 · · · Bjr
− 2k − r + 3 n − 2k + 1 n − k − 2r + 3 Bn−2k−r+3 . k−1 r−k−1 r−1
Proof. By Lemma 1 we get 2 2r−3
(1 − x )
2r−2 (r−1) f (x) 2 r−2 x
r
f (x) = (1 − x )
+
r−2 X k=1
2 r−k−2
(1 − x )
(r − 1)! Pk−1 k r−2 j=0
j
k−j−1
x2r−k−2+2j
k(r − k − 2)!
7
f (r−k−1) (x) . (9)
Since B0 = 0, the left-hand side of (9) is equal to 2 2r−3
(1 − x ) =
2r−3 ∞ XX l=0
∞ X n=0
X
j1 +···+jr =n j1 ,...,jr ≥0
Bj1 · · · Bjr xn
2r − 3 (−1) l j n=2l l
X
1 +···+jr =n−2l j1 ,...,jr ≥1
Bj1 · · · Bjr xn .
On the other hand, x2r−2 f (r−1) (x) (r − 1)! ∞ r−2 X n! r − 2 2i x2r−2 X Bn xn−r+1 x = (r − 1)! (n − r + 1)! i n=r−1 i=0 ∞ r−2 X X r−2 (n − r − 2i + 1)! 1 Bn−r−2i+1 xn . = i (r − 1)! i=0 (n − 2r − 2i + 2)! n=2r+2i−2
(1 − x2 )r−2
For i = r − 2, we have ∞ (−1)r−2 X (n − 3r + 5)! Bn−3r+5 xn (r − 1)! n=4r−6 (n − 4r + 6)! ∞ X n − 3r + 5 n − 3r + 4 r−2 = (−1) Bn−3r+5 xn , r − 2 r − 1 n=3r−5
which yields the term for k = r − 1 on the right-hand side of the identity in Theorem 3. Notice that ′ γ = 0 (γ ′ < γ) . γ
8
The second term of the right-hand side of (9) is r−2 r−k−2 k−1 X X X 1 r−2 k 2i i r−k−2 x (−1) x2r−k−2+2j i k − j − 1 j k(r − k − 2)! j=0 k=1 i=0 ×
∞ X
n! Bn xn−r+k+1 (n − r + k + 1)! n=r−k−1
r−3 r−i−3 X X r−i−3 X
(−1)i r−k−3 = i (k + 1)(r − k − 3)!(n − 2r + k − 2i − 2j + 3)! i=0 j=0 k=j X k+1 r−2 (n − r − 2i − 2j + 1)!Bn−r−2i−2j+1 xn × k−j j n=2r+2i+2j−k−3 r−3 X r−2 r−i−3 X X r−k−3 (−1)i = i (k + 1)(r − k − 3)!(n − 2r + k − 2κ + 5)! i=0 κ=i+1 k=κ−i−1 X r−2 k+1 (n − r − 2κ + 3)!Bn−r−2κ+3xn . × κ−i−1 k−κ+i+1 n=2r+2κ−k−5
Together with the first term of the right-hand side of (9) we can prove that (−1)k−1 r − 2 (n − r − 2k + 3)! (r − 1)! k − 1 (n − 2r − 2k + 4)! k−1 r−i−3 X X (−1)k−1 + (l + 1)(r − l − 3)!(n − 2r + l − 2k + 5)! i=0 l=k−i−1 r−2 l+1 r−l−3 (n − r − 2k + 3)! × k−i−1 l−k+i+1 i n − k − 2r + 3 k−1 n − 2k − r + 3 n − 2k + 1 . (10) = (−1) k−1 r−k−1 r−1 Then the proof is done.
3
Examples
When r = 2 and r = 3, Theorem 3 is reduced to Theorem 1 and Theorem 2, respectively. When r = 4, 5, 6 in Theorem 3, we get the following Corollaries as examples. 9
Corollary 1. For n ≥ 7, we have 5 X l=0
5 (−1) l j l
X
Bj1 Bj2 Bj3 Bj4
1 +j2 +j3 +j4 =n−2l j1 ,j2 ,j3 ,j4 ≥1
(n − 3)(n − 5)(n − 7) n−1 n−7 Bn−3 − = Bn−7 . Bn−5 + 3 3 3 Corollary 2. For n ≥ 10, we have 7 X
7 (−1) l j l=0 l
X
Bj1 Bj2 Bj3 Bj4 Bj5
1 +j2 +j3 +j4 +j5 =n−2l j1 ,j2 ,j3 ,j4 ,j5 ≥1
(n − 3)(n − 4)(n − 6)(n − 9) n−1 Bn−4 − = Bn−6 4 8 n − 10 (n − 5)(n − 8)(n − 10)(n − 11) Bn−10 . Bn−6 − + 8 4
Corollary 3. For n ≥ 13, we have 9 X
9 (−1) l j l=0 l
X
1 +···+j6 =n−2l j1 ,...,j6 ≥1
Bj1 · · · Bj6
(n − 3)(n − 4)(n − 5)(n − 7)(n − 11) n−1 Bn−5 − = Bn−7 5 30 (n − 5)(n − 6)(n − 9)(n − 12)(n − 13) + Bn−9 20 (n − 7)(n − 11)(n − 13)(n − 14)(n − 15) n − 13 − Bn−13 . Bn−11 + 5 30
4
Another result
In this section, we shall give an expression of 10
P
j1 +···+jr =n j1 ,...,jr ≥1
Bj1 · · · Bjr .
The left-hand side of (3) is equal to ! ∞ ! ∞ X n ∞ X X X u v Bj Bn−j xn . Bu x Bv x = u=0
n=0 j=0
v=0
The right-hand side of (3) is equal to ! ! ∞ ∞ X X mBm xm−1 x2j x2 m=1
j=0
=x
∞ X j=0
=x
∞ X
1 + (−1)j j x 2
!
∞ X
mBm xm
m=1
n X 1 + (−1)m (n − m)Bn−m 2 m=0
n=0 ∞ n−1 X X
!
!
xn
1 + (−1)m = (n − m − 1)Bn−m−1 2 n=1 m=0 n−1 ⌊ ∞ 2 ⌋ X X (n − 2m − 1)Bn−2m−1 xn . = n=1
!
xn
m=0
Comparing the coefficients of both sides, we have the following. Theorem 4. For n ≥ 2, we have n−1 X j=1
Bj Bn−j =
n−1 ⌊X 2 ⌋
m=0
(n − 2m − 1)Bn−2m−1 .
In general, we have the following. Theorem 5. For n ≥ r ≥ 2, we have X Bj1 · · · Bjr j1 +···+jr =n j1 ,...,jr ≥1
⌋ ⌊ n−r+1 2 X n − m − 1 m + r − 2 n − 2m − r + 1 Bn−2m−r+1 . = r−2 r−2 r−1 m=0 11
Proof. The left-hand side of (8) in Lemma 1 is equal to ∞ X n=0
X
j1 +···+jr =n j1 ,...,jr ≥1
Bj1 · · · Bjr xn .
The first term on the right-hand side of (8) in Lemma 1 is equal to x2r−2 f (r−1) (x) (r − 1)!(1 − x2 )r−1 ∞ ∞ x2r−2 X i + r − 2 2i X (m + r − 1)! x = Bm+r−1 xm r−2 (r − 1)! i=0 m! m=0 ∞
x2r−2 X 1 (k + 2r − 4)!! 1 + (−1)k k = x (r − 1)! (r − 2)!2r−2 k!! 2 k=0
×
∞ X
(m + r − 1)! Bm+r−1 xm m! m=0 ∞
n
x2r−2 X X 1 (n − m + 2r − 4)!! = r−2 (r − 1)! n=0 m=0 (r − 2)!2 (n − m)!!
1 + (−1)n−m (m + r − 1)! Bm+r−1 xn 2 m! ∞ n−2r+2 X X (n − m − 2)!! 1 = r−2 (r − 1)!(r − 2)!2 (n − m − 2r + 2)!! n=2r−2 m=0 ×
1 + (−1)n−m (m + r − 1)! Bm+r−1 xn 2 m! n−r+1 ∞ X (n − m + r − 3)!! X 1 = (r − 1)!(r − 2)!2r−2 n=2r−2 m=r−1 (n − m − r + 1)!! ×
1 + (−1)n−m−r+1 m! × Bm xn . 2 (m − r + 1)!
12
Concerning the second term, we have Pk−1 k r−2 2r−k−2+2j j=0 j k−j−1 x f (r−k−1) (x) (1 − x2 )r+k−1 ∞ k−1 X X i + r + k − 2 2i r−2 k 2r−k−2+2j x x = r + k − 2 k − j − 1 j i=0 j=0 ∞ X (m + r − k − 1)! Bm+r−k−1 xm m! m=0 ∞ k−1 X X k r − 2 1 2r−k−2+2j x = (r + k − 2)!2r+k−2 k−j−1 j j=0 l=0
×
∞ (l + 2r + 2k − 4)!! 1 + (−1)l l X (m + r − k − 1)! × x Bm+r−k−1 xm l!! 2 m! m=0 n ∞ X k−1 X X 1 r−2 k 2r−k−2+2j x = k−j−1 j (r + k − 2)!2r+k−2 n=0 m=0 j=0
× =
(n − m + 2r + 2k − 4)!! 1 + (−1)n−m (m + r − k − 1)! Bm+r−k−1 xn (n − m)!! 2 m! n−2r+k+2−2j ∞ k−1 X X X r−2 k 1
(r + k − 2)!2r+k−2
j=0
j
k−j−1
(n − m + 3k − 2 − 2j)!! 1 + (−1) (n − m − 2r + k + 2 − 2j)!! 2 Since
n=2r−k−2+2j
n−m+k
(m + r − k − 1)! Bm+r−k−1 xn . m!
(n − m + r + 2k − 3 − 2j)!! =0 (n − m − r + k + 3 − 2j)!!
13
m=0
if m = n − 2r + k + 2 − 2j (j = 1, 2, . . . , k − 2), this is equal to X ∞ n−2r+k+2 k−1 X X 1 r−2 k k − j − 1 n=2r−k−2 m=0 (r + k − 2)!2r+k−2 j=0 j
(n − m + 3k − 2 − 2j)!! 1 + (−1)n−m+k (m + r − k − 1)! Bm+r−k−1 xn (n − m + k)!! 2 m! ∞ n−r+1 k−1 X X X 1 r−2 k
× =
(r + k − 2)!2r+k−2 n=2r−k−2 m=r−k−1 j=0
×
j
k−j−1
m! (n − m + r + 2k − 3 − 2j)!! 1 + (−1)n−m−r+1 Bm xn . (n − m − r + k + 3 − 2j)!! 2 (m − r + k + 1)!
This is also equal to ∞ n−r+1 k−1 X X X r−2 k 1 k−j −1 (r + k − 2)!2r+k−2 n=2r−k−2 m=r−k−1 j=0 j × =
(n − m + r + 2k − 3 − 2j)!! 1 + (−1)n−m−r+1 m! Bm xn (n − m − r + k + 3 − 2j)!! 2 (m − r + k + 1)! ∞ n−r+1 X X (n − m + r − 1)!! r + k − 2 1
(r + k − 2)!2r+k−2 ×
(n − m − r + 1)!!
n=2r−k−2 m=r−k−1 n−m−r+1
(n − m + r − 3)!! 1 + (−1) (n − m + r − 2k − 1)!! 2
14
k−1
m! Bm xn . (m − r + k + 1)!
Therefore, the right-hand side of the relation in Theorem 4 is ∞ n−r+1 X X (n − m + r − 3)!! 1 + (−1)n−m−r+1 1 (r − 1)!(r − 2)!2r−2 n=2r−2 m=r−1 (n − m − r + 1)!! 2
m! Bm xn (m − r + 1)! r−1 X 1
×
∞ X 1 + k(r − k − 2)! (r + k − 2)!2r+k−2 n=2r−k−2 k=1 n−r+1 X (n − m + r − 1)!! r + k − 2 (n − m + r − 3)!! k−1 (n − m − r + 1)!! (n − m + r − 2k − 1)!! m=r−k−1
× =
m! 1 + (−1)n−m−r+1 Bm xn 2 (m − r + k + 1)! ∞ n−r+1 X X (n − m + r − 3)!! 1 + (−1)n−m−r+1 1
(r − 1)!(r − 2)!2r−2 n=2r−2 m=r−1 (n − m − r + 1)!! m! Bm xn (m − r + 1)! r−2 ∞ X X 1
2
× +
n=r−1
(r −
1)!2r−2
r−2 X
1 (n − m + r − 1)!! k k!(r − k − 2)!2 (n − m − r + 1)!! m=1 k=r−m−1
(n − m + r − 3)!! 1 + (−1)n−m−r+1 m! Bm xn (n − m + r − 2k − 1)!! 2 (m − r + k + 1)! n−r+1 r−2 ∞ X X X 1 (n − m + r − 1)!! 1
× +
n=r−1
×
(r − 1)!2r−2 m=r−1 k=1 k!(r − k − 2)!2k (n − m − r + 1)!!
(n − m + r − 3)!! 1 + (−1)n−m−r+1 m! Bm xn . (n − m + r − 2k − 1)!! 2 (m − r + k + 1)!
15
Since for 1 ≤ m ≤ r − 2 we have r−2 X 1 1 (n − m + r − 1)!! r−2 k (r − 1)!2 k!(r − k − 2)!2 (n − m − r + 1)!! k=r−m−1
(n − m + r − 3)!! m! (n − m + r − 2k − 1)!! (m − r + k + 1)! 1 (n + m + r − 3)!! (n − m + r − 3)!! = m 2r−4 (r − 1)!(r − 2)!2 (n + m − r + 1)!! (n − m − r + 1)!! ×
and for r − 1 ≤ m ≤ n − r + 1 we have m! (n − m + r − 3)!! 1 r−2 (r − 1)!(r − 2)!2 (n − m − r + 1)!! (m − r + 1)! r−2 X 1 (n − m + r − 1)!! 1 + (r − 1)!2r−2 k=r−m−1 k!(r − k − 2)!2k (n − m − r + 1)!!
(n − m + r − 3)!! m! (n − m + r − 2k − 1)!! (m − r + k + 1)! (n + m + r − 3)!! (n − m + r − 3)!! 1 m, = 2r−4 (r − 1)!(r − 2)!2 (n + m − r + 1)!! (n − m − r + 1)!! ×
16
By comparing the coefficients, we have X Bj1 · · · Bjr j1 +···+jr =n j1 ,...,jr ≥1
=
=
=
1 (r − 1)!(r − 2)!22r−4 n−r+1 X (n + m + r − 3)!!(n − m + r − 3)!! 1 + (−1)n−m−r+1 mFm × (n + m − r + 1)!!(n − m − r + 1)!! 2 m=0 1 (r − 1)!(r − 2)!22r−4 n−r+1 X (2n − m − 2)!!(m + 2r − 4)!! × (n − m − r + 1)Bn−m−r+1 (2n − m − 2r + 2)!!m!! m=0
1 (r − 1)!(r − 2)!22r−4 ⌊ n−r+1 ⌋ 2 X (2n − 2m − 2)!!(2m + 2r − 4)!! (n − 2m − r + 1)Bn−2m−r+1 × (2n − 2m − 2r + 2)!!(2m)!! m=0
⌊ n−r+1 ⌋ 2 X n − m − 1 m + r − 2 n − 2m − r + 1 = Bn−2m−r+1 . r−2 r−2 r−1 m=0
5
Some other generating functions
Another kinds of the generating functions of balancing numbers and Lucasbalancing numbers are given by ∞
eαt − eβt X tn √ = Bn b(t) := n! 4 2 n=0 and
∞
eαt + eβt X tn = Cn c(t) := 2 n! n=0
because they satisfy the differential equation y ′′ − 6y ′ + y = 0. 17
Since b′ (t) = 3b(t) + c(t) and c′ (t) = 8b(t) + 3c(t), we have for n ≥ 0 Bn+1 = 3Bn + Cn and Cn+1 = 8Bn + 3Cn . Since c(t)2 =
e2αt + e2βt e6t + , 4 2
we have ∞ X n ∞ ∞ X 1 X (2t)n 1 X (6t)n n Ck Cn−k = Cn + , k 2 n=0 n! 2 n=0 n! n=0 k=0
yielding
Similarly, by
n X 2n Cn + 6n n Ck Cn−k = k 2 k=0
b(t)2 = we have
n X n k=0
k
(n ≥ 0) .
e2αt + e2βt e6t − , 32 16
Bk Bn−k =
2n Cn − 6n 16
(n ≥ 0) .
Since by 2α + β = α + 6 and α + 2β = β + 6 e3αt − e3βt 3(e(2α+β)t − e(α+2β)t ) √ √ − (4 2)3 (4 2)3 eαt − eβt 3 1 e3αt − e3βt √ √ − e6t = 32 4 2 32 4 2 ∞ ∞ ∞ 1 X (3t)n 3 X (6t)i X tk = − Bn Bk 32 n=0 n! 32 i=0 i! k! k=0 ∞ ∞ n 3 X X n n−k tn 1 X n tn 6 Bk , 3 Bn − = 32 n=0 n! 32 n=0 k=0 k n!
b(t)3 =
18
we have X
k1 +k2 +k3 =n k1 ,k2 ,k3 ≥1
1 n Bk1 Bk2 Bk3 = k1 , k2 , k3 32
! n X n 6n−k Bk . 3n Bn − 3 k k=0
Notice that B0 = 0. Similarly, we can obtain that X
k1 +k2 +k3 =n k1 ,k2 ,k3 ≥0
1 n Ck1 Ck2 Ck3 = k1 , k2 , k3 4
3n Cn + 3
n X n k=0
k
6n−k Ck
!
.
Let r ≥ 1. If r is odd, then r αt e − eβt r √ b(t) = 4 2 r−1 2 X 1 j r (e((r−j)α+jβ)t − e(jα+(r−j)β)t ) (−1) = √ j (4 2)r j=0 r−1
2 X 1 j r = √ e6jt (e(r−2j)αt − e(r−2j)βt ) (−1) r j (4 2) j=0
r−1 k X ∞ ∞ 2 i X X (r − 2j)t r (6jt) 1 Bk (−1)j = √ k! j i=0 i! k=0 (4 2)r−1 j=0 r−1
1 = √ (4 2)r−1
X ∞ X n tn n r n−k k (6j) (r − 2j) Bk . (−1) k n! j n=0 j=0
2 X
j
k=0
Therefore, we get X
k1 +···+kr =n k1 ,...,kr ≥1
n Bk1 · · · Bkr k1 , . . . , kr
r−1 X n 2 X n 1 j r (6j)n−k (r − 2j)k Bk . (−1) = √ r−1 k j (4 2) j=0 k=0
19
Similarly, we get X
k1 +···+kr =n k1 ,...,kr ≥0
n Ck1 · · · Ckr k1 , . . . , kr
r−1 X n 2 1 X n r = r−1 (6j)n−k (r − 2j)k Ck . k j 2 j=0 k=0
If r is even, then r −1 2 1 X r j r b(t) = √ (e((r−j)α+jβ)t + e(jα+(r−j)β)t ) (−1) r j (4 2) j=0 r r ( r α+ r β)t +(−1) 2 r e 2 2 2r −1 2 1 X j r e6jt (e(r−2j)αt + e(r−2j)βt ) = √ (−1) r j (4 2) j=0 r r +(−1) 2 r e3rt 2r k −1 X ∞ ∞ 2 X (r − 2j)t 1 X (6jt)i j r = √ ·2 Ck (−1) k! j i=0 i! (4 2)r j=0 k=0 ! X ∞ r (3r)n n r +(−1) 2 r t n! 2 r n=0 −1 X ∞ X n 2 tn n 1 X j r (6j)n−k (r − 2j)k Ck (−1) = √ 2 j n=0 k=0 k n! (4 2)r j=0 ! X ∞ n r t r (3r)n . +(−1) 2 r n! 2 n=0
20
Therefore, we get X k1 +···+kr =n k1 ,...,kr ≥1
n Bk1 · · · Bkr k1 , . . . , kr
r −1 X n 2 n 1 X j r (6j)n−k (r − 2j)k Ck = √ (−1) 2 r k j (4 2) j=0 k=0 r r n 2 +(−1) r (3r) .
2
Similarly, we get X k1 +···+kr =n k1 ,...,kr ≥0
n Ck1 · · · Ckr k1 , . . . , kr
=
6
r −1 n 2 X r X n
1 2 2r j=0
j
k=0
k
r (6j)n−k (r − 2j)k Ck + r (3r)n . 2
More general cases
Let {un }n≥0 and {vn }n≥0 be integer sequences, satisfying the same recurrence relation: un = aun−1 + bun−2 (n ≥ 2) and vn = avn−1 + bvn−2 (n ≥ 2) with initial values u0 , u1 , v0 and v1 . If the general terms are given by un = where
αn − β n α−β
and vn = αn + β n
(n ≥ 0)
√ a2 + 4b a − a2 + 4b and β = , α= 2 2 we can set u0 = 0, u1 = 1, v = 0 = 2 and v1 = a. Then, the generating functions of un and vn are given by a+
√
∞
X tn eαt − eβt u(t) := √ = un a2 + 4b n=0 n!
αt
βt
and v(t) := e + e
=
∞ X n=0
vn
tn n!
respectively, because they satisfy the differential equation y ′′ − ay ′ − by = 0. Our main results can be stated as follows. 21
Theorem 6. If r is odd, then X
k1 +···+kr =n k1 ,...,kr ≥1
n u k1 · · · u kr k1 , . . . , kr
r−1 n X X 1 n 2 j r = √ (aj)n−k (r − 2j)k uk (11) (−1) j ( a2 + 4b)r−1 k=0 k j=0
and X
k1 +···+kr =n k1 ,...,kr ≥0
n vk1 · · · vkr k1 , . . . , kr
=
r−1 n X X n 2 r
k=0
k
j=0
j
(aj)n−k (r − 2j)k vk . (12)
If r is even, then X
k1 +···+kr =n k1 ,...,kr ≥1
n u k1 · · · u kr k1 , . . . , kr
r −1 n X 2 X n 1 j r (aj)n−k (r − 2j)k vk = √ (−1) j ( a2 + 4b)r k=0 k j=0 r ar n r 2 +(−1) r (13) 2 2 and X
k1 +···+kr =n k1 ,...,kr ≥0
n vk1 · · · vkr k1 , . . . , kr
=
r −1 n X X n 2 r
k=0
k
j=0
j
(aj)
22
n−k
ar n r . (14) (r − 2j) vk + r 2 2 k
Proof. If r is odd, then by α + β = a r αt e − eβt r u(t) = √ a2 + 4b r−1 2 X 1 j r = √ (e((r−j)α+jβ)t − e(jα+(r−j)β)t ) (−1) 2 r j ( a + 4b) j=0 r−1
2 X 1 j r eajt (e(r−2j)αt − e(r−2j)βt ) (−1) = √ j ( a2 + 4b)r j=0
r−1 k X ∞ ∞ 2 X (r − 2j)t 1 (ajt)i X j r = √ uk (−1) k! j i=0 i! k=0 ( a2 + 4b)r−1 j=0 r−1
X ∞ X n 2 X tn n 1 j r (aj)n−k (r − 2j)k uk . (−1) = √ n! j n=0 k=0 k ( a2 + 4b)r−1 j=0 Therefore, we get X
k1 +···+kr =n k1 ,...,kr ≥1
n u k1 · · · u kr k1 , . . . , kr
r−1 X n 2 X n 1 j r (aj)n−k (r − 2j)k uk . (−1) = √ 2 r−1 k j ( a + 4b) j=0 k=0
Similarly, we get X
k1 +···+kr =n k1 ,...,kr ≥0
r−1 X n 2 X n n r vk1 · · · vkr = (aj)n−k (r − 2j)k vk . k1 , . . . , kr k j j=0 k=0
23
If r is even, then r −1 2 X 1 r j r u(t) = √ (e((r−j)α+jβ)t + e(jα+(r−j)β)t ) (−1) j ( a2 + 4b)r j=0 r r ( r α+ r β)t 2 2 2 +(−1) r e 2 r −1 2 X 1 j r eajt (e(r−2j)αt + e(r−2j)βt ) = √ (−1) j ( a2 + 4b)r j=0 r art/2 r +(−1) 2 r e 2 r k −1 X ∞ ∞ 2 i X X (r − 2j)t 1 r (ajt) j (−1) = √ · vk k! j i=0 i! ( a2 + 4b)r j=0 k=0 ! X ∞ r ar n tn r +(−1) 2 r 2 n! 2 n=0 r −1 X ∞ X n 2 X 1 tn n j r = √ (aj)n−k (r − 2j)k vk (−1) n! j n=0 k=0 k ( a2 + 4b)r j=0 ! X ∞ ar n tn r r . +(−1) 2 r 2 n! 2 n=0 Therefore, we get X
k1 +···+kr =n k1 ,...,kr ≥1
n u k1 · · · u kr k1 , . . . , kr
r −1 X n 2 X 1 n r j (−1) = √ (aj)n−k (r − 2j)k vk 2 r j k=0 k ( a + 4b) j=0 r ar n r +(−1) 2 r . 2 2
24
Similarly, we get X
k1 +···+kr =n k1 ,...,kr ≥0
n vk1 · · · vkr k1 , . . . , kr
r
−1 n 2 X r ar n r X n n−k k (aj) (r − 2j) vk + r . = 2 j k=0 k 2 j=0
7
Examples
When a = 6 and b = −1, then Bn = un are balancing numbers and Cn = vn /2 are Lucas-balancing numbers. Thus, Theorem 6 can be reduced as follows. Corollary 4. If r is odd, then X
k1 +···+kr =n k1 ,...,kr ≥1
n Bk1 · · · Bkr k1 , . . . , kr
r−1 n X X 1 n 2 j r = √ (6j)n−k (r − 2j)k Bk (−1) j (4 2)r−1 k=0 k j=0
and X
k1 +···+kr =n k1 ,...,kr ≥0
n Ck1 · · · Ckr k1 , . . . , kr
r−1 n 2 r 1 X n X (6j)n−k (r − 2j)k Ck . = r−1 j k 2 j=0 k=0
25
If r is even, then X k1 +···+kr =n k1 ,...,kr ≥1
n Bk1 · · · Bkr k1 , . . . , kr
r
n 2 −1 1 X n X j r = √ (6j)n−k (r − 2j)k Ck 2 (−1) r k j (4 2) j=0 k=0 r r n r +(−1) 2 r 2 2 and X
k1 +···+kr =n k1 ,...,kr ≥0
n Ck1 · · · Ckr k1 , . . . , kr
r −1 n n X 2 X 6r n 1 r r = r 2 . (6j)n−k (r − 2j)k Ck + r k j=0 j 2 2 2 k=0
When a = b = 1, then Fn = un are Fibonacci numbers and Ln = vn are Lucas numbers. Thus, Theorem 6 can be reduced as follows. Corollary 5. If r is odd, then X n Fk1 · · · Fkr k1 , . . . , kr k +···+kr =n 1 k1 ,...,kr ≥1
r−1 n X X 1 n 2 j r = √ j n−k (r − 2j)k Fk (−1) r−1 k j=0 j ( 5) k=0
and X
k1 +···+kr =n k1 ,...,kr ≥0
n Lk1 · · · Lkr k1 , . . . , kr
r−1 n X X n 2 r n−k = j (r − 2j)k Lk . k j j=0 k=0
26
If r is even, then X
k1 +···+kr =n k1 ,...,kr ≥1
n Fk1 · · · Fkr k1 , . . . , kr
r n 2 −1 1 X n X j r = √ j n−k (r − 2j)k Lk (−1) r j k j=0 ( 5) k=0 r r n r +(−1) 2 r 2 2
and X
k1 +···+kr =n k1 ,...,kr ≥0
r −1 n X X r n n 2 r n r n−k k . Lk1 · · · Lkr = j (r−2j) Lk + r 2 k j=0 j k1 , . . . , kr 2
k=0
When r = 2 and a = b = 1, we have n X 2n Ln − 2 n Fk Fn−k = 5 k n=0 and
n X n Lk Ln−k = 2n Ln + 2 . k n=0
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