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SOME RESULTS ON GENERALIZED FIBONACCI
AND LUCAS NUMBERS AND DEDEKIND SUMS
Feng-Zhen Zhao
Dalian University of Technology, 116024 Dalian, China
Tianming Wang
Dalian University of Technology, 116024 Dalian, China
(Submitted December 2001–Final Revision September 2003)
ABSTRACT
In this paper, by using the reciprocity formula of Dedekind sums and the properties of generalized Fibonacci and Lucas numbers, the authors investigate Dedekind sums for generalized
Fibonacci and Lucas numbers and generalize some conclusions of other authors.
1. INTRODUCTION
Recently, Dedekind sums for Fibonacci numbers were investigated and some meaningful
results were obtained (see [7]). In this paper, the authors will consider Dedekind sums for
generalized Fibonacci and Lucas numbers.
The Binet forms of generalized Fibonacci and Lucas numbers are:
αn − β n
Un =
and Vn = αn + β n ,
α−β
√
√
where n ≥ 0, α = (p + ∆)/2, β = (p − ∆)/2, ∆ = p2 − 4q > 0, and p and q are integers with
pq 6= 0. Throughout this paper, we assume that p > 0. When n < 0, we define Un = (−1)n U−n
and Vn = (−1)n V−n . It is well known that {Un } and {Vn } satisfy the recurrence relation
Wn+2 = pWn+1 − qWn , n ≥ 0.
(1)
For p = −q = 1, {Un } and {Vn } are the classical Fibonacci and Lucas sequences {Fn } and
{Ln }, respectively.
The Dedekind sum S(h, t) is defined by
t X
a
ah
S(h, t) =
,
(2)
t
t
a=1
where t is a positive integer, h is an arbitrary integer, and
x − [x] − 1/2, if x is not an integer
((x)) =
0,
if x is an integer.
For various arithmetical properties of S(h, t), several articles have been written (see [1],
[4], and [6]). Regarding Dedekind sums and uniform distribution, Myerson [5] and Zheng
[8] have obtained some meaningful results. In [7], Zhang studied the distribution problem of
Dedekind sums for Fibonacci numbers Fn and obtained some interesting results. Zhang discussed
Pmthe mean value distribution of S(Fn , Fn+1 ) and presented a sharper asymptotic formula
for n=1 S(Fn , Fn+1 ). Inspired by Zhang’s results, we decided to investigate Dedekind sums
for Un and Vn . The principal purpose of this paper is to compute the values of S(Un , Un+1 ) and
S(Vn , Vn+1 ) for |q| = 1, by the reciprocity formula of Dedekind sums and properties of {Un }
250
SOME RESULTS ON GENERALIZED FIBONACCI AND LUCAS NUMBERS AND DEDEKIND SUMS
and {V
Pnm}. In the meantime, we present the sharper asymptotic formulas of
and n=1 S(Vn , Vn+1 ) when |q| = 1.
Pm
n=1
S(Un , Un+1 )
2. MAIN RESULTS
In this section, we state and prove the main results of this paper.
Theorem 1: Assume that q = −1 and m is a positive integer. Then
S(U2m , U2m+1 ) = 0,
S(U2m+1 , U2m+2 ) = −
S(V2m , V2m+1 ) = −
U2m + V2m−1
(p − 1)(p − 5)
1
+
−
, (2 6 | p),
12pV2m+1
24p
12p
S(V2m+1 , V2m+2 ) =
+
2m
X
S(Un , Un+1 ) =
n=1
U2m
(p − 1)(p − 2)
+
,
6pU2m+2
12p
S(Un , Un+1 ) =
n=1
p − 3 (p − 1)(p − 5)
1
+
−
, (2 6 | p),
6p
12
24p
n=1
S(Vn , Vn+1 ) =
(5)
(6)
m
(p − 3)m
U2m
1
+
−
−
2m+1
6α
12
24U2m+1
24α
U2m+1
(7)
(p − 3)(2m + 1)
U2m+1
1
m
+
−
−
2m+2
6α
24
24U2m+2
24α
U2m+2
∞
1
U2m
(p − 1)(p − 2) 1 X
1
1
+
−
+
+
+0
,
24α 12pU2m+2
24p
6 n=1 α2n U2n
α4m
2m
X
(4)
V2m+1
1
U2m
+
+
12pV2m+1
12V2m+2
12V2m+1 V2m+2
∞
1
1X
1
1
+
+
+0
,
24α 6 n=1 α2n U2n
α4m
2m+1
X
(3)
(8)
(p − 1)(p − 5)
m
(p − 3)m
+
+
48p
6α
12
∞
∞
∆−1 X
1
∆+1 X
1
1
− √
+ √
+0
, (2 6 | p),
α4m
12 ∆ n=1 α2n V2n
12 ∆ n=1 α2n+1 V2n+1
(9)
251
SOME RESULTS ON GENERALIZED FIBONACCI AND LUCAS NUMBERS AND DEDEKIND SUMS
2m+1
X
S(Vn , Vn+1 ) = −
n=1
+
(p − 1)(p − 5) (p − 3)(m + 1)
1
m
U2m
V2m+1
+
+
+
+
+
48p
12
6p 6α 12pV2m+1
12V2m+2
1
12V2m+1 V2m+2
∞
∆−1 X
1
− √
2n
12 ∆ n=1 α V2n
∞
∆+1 X
1
1
+ √
+0
, (2 6 | p).
α4m
12 ∆ n=1 α2n+1 V2n+1
(10)
Proof: We can show that (Um , Um+1 ) = 1 (m ≥ 1) by induction. From the reciprocity
formula of Dedekind sums (see [1]), we obtain
2
2
Um
+ Um+1
+1 1
S(Um , Um+1 ) + S(Um+1 , Um ) =
− .
12Um Um+1
4
By using (1) and (2), we get S(Um+1 , Um ) = S(Um−1 , Um ). Thus,
1
S(Um , Um+1 ) + S(Um−1 , Um ) =
12
=
Um−1
Um
1
+
+
Um
Um+1
Um Um+1
−
1
p
+
4 12
1
Um−1
Um−2
1
p
1
−
−
− +
+
12Um Um+1
12pUm+1
12pUm
4 12 6p
(11)
so that
1
Um−2
1
p
1
Um−1
S(Um , Um+1 ) +
=
− S(Um−1 , Um ) +
− +
+
12pUm+1
12Um Um+1
12pUm
4 12 6p
1
Um−3
1
−
+ S(Um−2 , Um−1 ) +
.
=
12Um Um+1
12Um−1 Um
12pUm−1
Hence,
S(U2m , U2m+1 ) +
U2m−1
1
1
1
=
−
+
12pU2m+1
12U2m U2m+1
12U2m−1 U2m
12U2m−2 U2m−1
1
U0
1
p
1
− ··· +
− S(U1 , U2 ) +
− +
+ ,
12U2 U3
12pU2
4 12 6p
and
S(U2m+1 , U2m+2 ) +
U2m
1
1
1
=
−
+
12pU2m+2
12U2m+1 U2m+2
12U2m U2m+1
12U2m−1 U2m
U0
− · · · + S(U1 , U2 ) +
.
12pU2
252
SOME RESULTS ON GENERALIZED FIBONACCI AND LUCAS NUMBERS AND DEDEKIND SUMS
From the definition of Un , we can prove that
1
αm Um
+
1
αm+1 U
=
m+1
1
.
Um Um+1
(12)
On the other hand, S(U1 , U2 ) = S(1, p) = (p−1)(p−2)
and U0 = 0. Therefore, we have (3) and
12p
(4).
Using the same method for getting (3-4), and the formula S(V0 , V1 ) = S(2, p) =
(p−1)(p−5)
24p
(2 6 | p) in the meantime, we can show that (5) and (6) hold.
Summing both sides of (11), we obtain
m
X
[S(Un , Un+1 ) + S(Un−1 , Un )] = 2
n=1
m
X
S(Un , Un+1 ) − S(Um , Um+1 )
n=1
m 1 X
Un
Un−1
1
m pm
=
+
+
−
+
12 n=1 Un+1
Un
Un Un+1
4
12
=
m
m
1 X Un
1 X
1
Um
m pm
+
−
−
+
.
6 n=1 Un+1
12 n=1 Un Un+1
12Um+1
4
12
It follows from the definition of Un that
Pm
Un
n=1 Un+1
=
1
α
Pm
n=1
Un+1 −(−1/α)n
.
Un+1
Then
m
X
m
m
S(Um , Um+1 )
1 X Un
1 X
1
Um
(p − 3)m
S(Un , Un+1 ) =
+
+
−
+
2
12 n=1 Un+1
24 n=1 Un Un+1
24Um+1
24
n=1
=
m
m
S(Um , Um+1 )
m
(p − 3)m
1 X (−1)n+1
Um
1 X
1
+
+
+
−
+
.
n+1
2
12α
24
12 n=1 α
Un+1
24Um+1
24 n=1 Un Un+1
By (12), we have
m
X
m
S(Um , Um+1 )
m
(p − 3)m
Um
1 X (−1)n+1
S(Un , Un+1 ) =
+
+
−
+
2
12α
24
24Um+1
12 n=1 αn+1 Un+1
n=1
m 1 X
1
1
+
+ n+1
.
24 n=1 αn Un
α
Un+1
Pm
P∞
1
Due to n=1 αn1Un = n=1 αn1Un + 0 α2m
, the equalities (7-8) hold.
Similarly, we can prove that (9) and (10) hold.
Theorem 2: Suppose that q = 1 and m is a positive integer. Then
S(Um , Um+1 ) =
Um
m(p − 3)
+
,
6Um+1
12
253
(13)
SOME RESULTS ON GENERALIZED FIBONACCI AND LUCAS NUMBERS AND DEDEKIND SUMS
p−6
Vm
(p − 3)m
1
S(Vm , Vm+1 ) =
+
+
+ √
24
12Vm+1
12
12 ∆
1
1
− m+1
2 α
Vm+1
(2 6 | p),
∞
(p − 3)m(m + 1)
m
1X
1
1
S(Un , Un+1 ) =
+
−
+0
,
n+1 U
2m
24
6α
6
α
α
n+1
n=1
n=1
m
X
m
X
S(Vn , Vn+1 ) =
n=1
(14)
(15)
m
m
[2p − 9 + (p − 3)m]m
+
+ √
24
12α 24 ∆
∞
∆−1 X
1
1
+ √
+0
.
α2m
12 ∆ n=1 αn+1 Vn+1
(16)
Proof: One can verify that (Um , Um+1 ) = 1(m ≥ 1) by induction. So,
2
2
Um
+ Um+1
+1 1
S(Um , Um+1 ) + S(Um+1 , Um ) =
− .
12Um Um+1
4
It follows from (1-2) and ((−x)) = − ((x)) that
S(Um+1 , Um ) = −S(Um−1 , Um ).
Thus, the following identity holds:
1
S(Um , Um+1 ) − S(Um−1 , Um ) =
12
Um
Um−1
1
−
+
Um+1
Um
Um Um+1
+
p
1
− .
12 4
Summing both sides of (17) and noticing that S(U0 , U1 ) = 0, we get
m
Um
(p − 3)m
1 X
1
S(Um , Um+1 ) =
+
+
.
12Um+1
12
12 n=1 Un Un+1
From [2] or [3], we know that
Pm
1
n=1 Un Un+1
S(Um , Um+1 ) =
By the definition of Un , we have
By (13), we have
m
X
n=1
=
αm Um+1 −1
αm+1 Um+1 .
Hence,
αm+1 Um + αm Um+1 − 1 (p − 3)m
+
.
12αm+1 Um+1
12
αm+1 Um +αm Um+1 −1
αm+1 Um
S(Un , Un+1 ) =
=
2Um
Um+1 .
Hence, equality (13) holds.
m
(p − 3)m(m + 1) 1 X Un
+
.
24
6 n=1 Un+1
From the definition of Un , we can deduce that
αUn
Un+1 − β n
1
=
=1− n
.
Un+1
Un+1
α Un+1
254
(17)
SOME RESULTS ON GENERALIZED FIBONACCI AND LUCAS NUMBERS AND DEDEKIND SUMS
Therefore, equality (15) holds.
Similarly, we can show that (14) and (16) hold.
We note that (7) and (8) generalize Zhang’s results (see [7]):
√
m
X
( 5 − 1)2 m
22m
S(Fn , Fn+1 ) = −
+ C(m) + 0 √
,
48
( 5 + 1)2m
n=1
where
C(m) =



1
12
P∞
1
12
P∞
1
n=1 F2n F2n+1
1
n=1 F2n+1 F2n+2
(−2)n+1
√
n=1 ( 5+1)n+1 Fn+1 ,
P∞
1
12
+
+
1
12
(−2)n+1
√
n=1 ( 5+1)n+1 Fn+1 ,
P∞
if m is an even number;
if m is an odd number,
(in [7],
C(m) =



1
12
P∞
1
12
P∞
1
n=1 F2n F2n+1
+
1
n=1 F2n+1 F2n+2
1
12
+
n+1
√ 2
n=1 ( 5+1)n+1 Fn ,
P∞
1
12
n+1
√ 2
n=1 ( 5+1)n+1 Fn ,
P∞
if m is an even number;
if m is an odd number,
in which there exist printing errors, 2n+1 and Fn should be (−2)n+1 and Fn+1 , respectively).
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[4] L. J. Mordell. “The Reciprocity Formula for Dedekind Sums.” Amer. J. Math. 73 (1951):
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AMS Classification Numbers: 11B39, 11B37
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255