# Some Properties of Hyperfibonacci and Hyperlucas Numbers o and Feng-Zhen Zhao ```1
Journal of Integer Sequences, Vol. 13 (2010),
Article 10.8.8
2
3
47
6
23 11
Some Properties of Hyperfibonacci
and Hyperlucas Numbers
Ning-Ning Cao1 and Feng-Zhen Zhao1
School of Mathematical Sciences
Dalian University of Technology
Dalian 116024
China
caoning0928@163.com
fengzhenzhao@yahoo.com.cn
Abstract
In this paper, we discuss the properties of hyperfibonacci numbers and hyperlucas
numbers. We derive some identities for hyperfibonacci and hyperlucas numbers by the
method of coefficients. Furthermore, we give asymptotic expansions of certain sums
involving hyperfibonacci and hyperlucas numbers by Darboux’s method.
1
Introduction
(r)
Dil and Mező  introduced the definitions of “hyperfibonacci” numbers Fn and “hyperlu(r)
cas” numbers Ln
Fn(r) =
n
X
(r−1)
Fk
,
with Fn(0) = Fn ,
(r)
F0 = 0,
(r)
F1 = 1,
k=0
L(r)
n
=
n
X
(r−1)
Lk
,
with L(0)
n = Ln ,
(r)
L0 = 2,
(r)
L1 = 2r + 1,
k=0
1
This work supported by the Science Research Foundation of Dalian University of Technology (2008).
1
where r is a positive integer, and Fn and Ln are Fibonacci and Lucas numbers, respectively.
(r)
(r)
The generating functions of Fn and Ln are as follows:
∞
X
n=0
∞
X
Fn(r) tn =
n
L(r)
=
n t
n=0
(r)
t
,
(1 − t − t2 )(1 − t)r
2−t
.
(1 − t − t2 )(1 − t)r
(r)
The first few values of Fn and Ln are as follows:
Fn(1) : 0, 1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, . . . ;
Fn(2) : 0, 1, 3, 7, 14, 26, 46, 79, 133, 221, 364, 596, 972, 1581, 2567, 4163, 6746, . . . ;
L(1)
n : 2, 3, 6, 10, 17, 28, 46, 75, 122, 198, 321, 520, 842, 1363, 2206, 3570, 5777, . . . ;
L(2)
n : 2, 5, 11, 21, 38, 66, 112, 187, 309, 507, 828, 1348, 2190, 3553, 5759, 9329, 15106, . . . .
(r)
(r)
There are some elementary identities for Fn and Ln when r = 1, 2. For example,
Fn(1) = Fn+2 − 1,
Fn(2) = Fn+4 − n − 3
n
X
=
(n − k)Fk ,
k=0
L(1)
n
L(2)
n
=
=
=
=
Ln+2 − 1
Fn + Fn+2 − 1,
4(Fn+1 − 1) + 3Fn − n
Ln+3 − (n + 4).
(r)
(r)
For the above values and elementary identities of Fn and Ln , see  (A000071, A001924,
A001610, A023548).
Hyperfibonacci numbers and hyperlucas numbers are useful, and Dil and Mező  derived
some equalities for Fibonacci and Lucas numbers by applying them. Hence, hyperfibonacci
numbers and hyperlucas numbers deserve to be investigated. In this paper, we discuss
(r)
(r)
(r)
(r)
properties of Fn and Ln . We establish some identities for Fn and Ln . Furthermore, we
(r)
(r)
give asymptotic expansions of certain sums related to Fn and √
Ln .
For convenience, we first recall some notation. Let α = (1 + 5)/2. It is well known that
Fn =
αn − (−1)n α−n
√
,
5
Ln = αn + (−1)n α−n ,
and Fn and Ln satisfy the following recurrence relation
Wn+1 = Wn + Wn−1 ,
2
n ≥ 1.
(1)
n
is defined by
As usual, the binomial coefficient m

n!

, if n ≥ m;
n
= m!(n − m)!

m
0,
if n &lt; m;
where n and m are nonnegative integers. Throughout, [z n ]f (z) denotes the coefficient of z n
in f (z), where
∞
X
f (z) =
fn z n .
n=0
If f (t) and g(t) are formal power series, the following relations hold 
[tn ](af (t) + bg(t)) = a[tn ]f (t) + b[tn ]g(t),
[tn ]tf (t) = [tn−1 ]f (t),
n
X
n
[t ]f (t)g(t) =
[y k ]f (y)[tn−k ]g(t).
(2)
(3)
(4)
k=0
The above relations (2)–(4) will be used later on.
Now we recall the notation of the binomial transform of a sequence, the inverse binomial
transform of a sequence and Euler-Seidel infinite
[1, 5]. Let {ak }∞
k=0 be a sequence.
Pn matrix
n
The binomial transform
of {ak } is given by k=0 k ak , the inverse binomial transform of
P
{ak } is given by nk=0 nk (−1)k ak , and the Euler-Seidel infinite matrix corresponding to the
sequence {ak } is determined by the following formulas
a
n = an
(n ≥ 0),
[k−1]
[k−1]
a[k]
+ an+1
n = an
(n ≥ 0,
k ≥ 0),
[k]

where an is the element at the (k + 1)th row and the (n + 1)th column. The sequences {an }
[n]
and {a0 } satisfy 
n X
n 
[n]
a ,
(5)
a0 =
k k
k=0
n X
n
[k]

(−1)n−k a0 .
(6)
an =
k
k=0

[n]
Let a(t) and ā(t) be the generating function of {an } and {a0 }, respectively,
a(t) =
∞
X
n
a
n t ,
ā(t) =
n=0
∞
X
[n]
a0 tn .
n=0
The functions a(t) and ā(t) satisfy 
1
t
ā(t) =
,
a
1−t 1−t
1
t
a(t) =
.
ā
t+1 t+1
3
(7)
(8)
2
Main Results
(r)
(r)
In this section, we derive some identities for Fn and Ln . Later, we give asymptotic
(r)
(r)
expansions of certain sums involving Fn and Ln .
Various identities involving Fibonacci and Lucas numbers were established. The following
sums were investigated [3, 7, 8]
X
X
Fj1 Fj2 &middot; &middot; &middot; Fjk ,
L j1 L j2 &middot; &middot; &middot; L jk .
j1 +j2 +&middot;&middot;&middot;+jk =n
j1 +j2 +&middot;&middot;&middot;+jk =n
For example,
(n − 1)Ln + 2Fn−1
,
5
X
Fj1 Fj2 =
X
Lj1 Lj2 = (n + 1)Ln + 2Fn+1 .
j1 +j2 =n
j1 +j2 =n
(r)
(r)
Now we derive some identities for Fn and Ln . Denote
X
X
(r) (r)
(r)
An,k,r =
Fj1 Fj2 &middot; &middot; &middot; Fjk , Bn,k,r =
j1 +j2 +&middot;&middot;&middot;+jk =n
j1 +j2 +&middot;&middot;&middot;+jk =n
(r)
(r)
(r)
L j1 L j2 &middot; &middot; &middot; L jk .
These sums are interesting because they can help us to find some new convolution properties.
For An,k,r and Bn,k,r , we have
Theorem 1. Let k, n ≥ 1 and r ≥ 1 be positive integers. For An,k,r and Bn,k,r , we have
(n + 1)Ln+4 − 2Fn+1
,
5
(5n + 9)Ln+4 + 4Ln+6
= n + 9 − 10Fn+4 − 2Fn+1 +
,
5
n
X
(r)
=
An,k,r Fj ,
An,2,1 = n + 5 − 2Fn+4 +
Bn,2,1
An,k+1,r
Bn,k+1,r =
j=0
n
X
(r)
Bn,k,r Lj .
Fr (t) =
(10)
(11)
(12)
j=0
Proof. Let
(9)
t
,
(1 − t − t2 )(1 − t)r
Lr (t) =
4
2−t
.
(1 − t − t2 )(1 − t)r
Clearly,
α2 − α−2
1
α
α−1
√
F1 (t) =
−
−
−1
t−1
t−α
t+α
5
X
∞
∞
X
=
Fn+2 tn −
tn t,
n=0
n
[t ]F12 (t),
[tn ]L21 (t)
[tn ]F12 (t) −
n=0
An,2,1 =
Bn,2,1 =
=
4[tn+1 ]F12 (t) + 4[tn+2 ]F12 (t)
= An,2,1 − 4An+1,2,1 + 4An+2,2,1 .
5
Then we get
X
(1) (1)
Fj1 Fj2 = [tn ]F12 (t)
j1 +j2 =n
1 (α2 − α−2 )2
α2
α−2
2α(α2 − α−2 )
= [t ]
+
+
−
5
(t − 1)2
(t − α−1 )2 (t + α)2 (t − 1)(t − α−1 )
−1
2
−2
2α (α − α )
2
−
+
(t − 1)(t + α)
(t − α−1 )(t + α)
2
α4
α−4
1 (α − α−2 )2
+
+
= [tn ]
5
(t − 1)2
(1 − αt)2 (1 + α−1 t)2
2(α2 − α−2 )(α3 + α−3 )
1
α
4
2
−2
+
− 2 α (α − α ) +
−1
1−t
α+α
1 − αt
−1
α
1
+2 α−4 (α2 − α−2 ) −
−1
α+α
1 + α−1 t
∞
∞
X
X
2
−2 2
n
n 1
(n + 1)t +
[αn+4 + (−1)n α−n−4 ](n + 1)tn
= [t ] (α − α )
5
n=0
n=0
n
2
−2
3
−3
+2(α − α )(α + α )
∞
X
tn
n=0
X
∞
α
αn tn
α + α−1 n=0
X
∞
α−1
−4
2
−2
n −n n
+2 α (α − α ) −
(−1) α t
α + α−1 n=0
X
∞
∞
∞
X
X
αn+4 + (−1)n α−n−4
n
= [t ]
(n + 1)tn +
(n + 1)tn + 2F3
tn
5
n=0
n=0
n=0
−2 α4 (α2 − α−2 ) +
∞
X
2 2
−2
− (α − α )
[αn+4 − (−1)n α−n−4 ]tn
5
n=0
∞
X
2
n+1
n −n−1 n
−
[α
+ (−1) α
]t
5(α + α−1 ) n=0
= [tn ]
∞
X
(n + 1)tn + [tn ]
n=0
−[tn ]2
∞
X
n=0
∞
X
(n + 1)Ln+4
n=0
Fn+4 tn − [tn ]
∞
2X
5
5
tn + [tn ]4
∞
X
tn
n=0
Fn+1 tn .
n=0
By applying (3) and the definitions of Fn and Ln , we have (9). Naturally, we deduce that
4(n + 3)Ln+6 − 4(n + 2)Ln+5 + (n + 1)Ln+4
− 2Fn+1 .
5
By using (1), we prove that (10) holds. By means of (4), we can show that (11) and (12)
hold.
Bn,2,1 = n + 9 − 10Fn+4 +
6
It is interesting that we can get congruence relations from (9) and (10)
(n + 1)Ln+4 − 2Fn+1
An,2,1 ≡ n − 2Fn+4 +
(mod 5),
5
(5n + 9)Ln+4 + 4Ln+6
Bn,2,1 ≡ n − 10Fn+4 − 2Fn+1 +
(mod 9).
5
When k or r gets large, it is difficult to compute the closed forms of An,k,r and Bn,k,r .
However, we can give their asymptotic values. Now we recall a lemma .
P
Lemma 2. Assume that f (t) = n≥0 an tn is an analytic function for |t| &lt; r and with a
finite number of algebraic singularities on the circle |t| = r. α1 , α2 , &middot; &middot; &middot; , αl are singularities
of order ω, where ω is the highest order of all singularities. Then
an = (n
ω−1
/Γ(ω)) &times;
X
l
gk (αk )αk−n
+ o(r
−n
k=1
) ,
(13)
where Γ(ω) is the gamma function, and
gk (αk ) = lim (1 − (t/αk ))ω f (t).
t→αk
Theorem 3. Suppose that k and r are fixed positive integers. For An,k,r and Bn,k,r , when
n → ∞,
An,k,r
Bn,k,r
nk−1
=
(k − 1)!
nk−1
=
(k − 1)!
Proof. Let
k
α
α2 + 1
2α2 − α
α2 + 1
(1 + α) α + o(α )) ,
k
kr
n
n
(14)
(1 + α) α + o(α )) .
kr
n
n
(15)
tk
fk,r (t) =
.
(1 − t − t2 )k (1 − t)kr
We know that fk,r (t) is analytic for |t| &lt; 1/α, and with one algebraic singularity on the
circle |t| = 1/α. The order of 1/α is k. One can compute that
k
lim (1 − αt) fk,r (t) =
t→1/α
α
2
α +1
k
(1 + α)kr .
By using Lemma 2, we can prove that (14) holds. Using the same method, we can prove
that (15) holds.
(r)
(r)
In addition, we give asymptotic expansions of other sums for Fn and Ln .
7
Theorem 4. Let n be a positive integer. When n → ∞,
n X
α(1 + α)r
n
(r)
Fk
=
+ o((2 − α)−n ),
2
n
k
(α + 1)(2 − α)
k=0
n
X n (r)
(1 + α)r
Lk =
+ o((2 − α)−n ),
n
(2
−
α)
k
k=0
n X
−αn+1 (2 − α)r
n
(r)
+ o((−α)n ),
(−1)k Fk
=
2+1
α
k
k=0
n X
n
(r)
(−1)k Lk = (2 − α)r αn + o((−α)n ).
k
k=0
(16)
(17)
(18)
(19)
Proof. We only give the proofs of (16) and (18). The proofs of (17) and (19) are similar to
(r)
those of (16) and (18), respectively, and they are omitted here. Let Fk be the first row,
(r)
and we get the Euler-Seidel infinite matrix A. Then let Fk be the first column, and we get
[k]
[k]
Euler-Seidel infinite matrix B. The elements of A and B are denoted by an and bn . By
using (5) and (6), we obtain
n X
n
(r)
[n]
Fk ,
a0 =
k
k=0
n X
n
(r)

(−1)n−k Fk .
bn =
k
k=0
By means of (7) and (8), we get
n X
n
k=0
k
(r)
Fk
[n]
= a0
= [tn ]
n X
n
k=0
k
(r)
(−1)n−k Fk
t(1 − t)r
,
(t2 − 3t + 1)(1 − 2t)r
= b
n
= [tn ]
−t(1 + t)r
.
t2 − t − 1
t(1 − t)r
is analytic for |t| &lt; 2 − α with one
We know that the function ā(t) = 2
(t − 3t + 1)(1 − 2t)r
−t(1 + t)r
is analytic
algebraic singularity α1 = 2 − α on the circle |t| = 2 − α, and a(t) = 2
t −t−1
for |t| &lt; α−1 with one algebraic singularity β1 = −α−1 on the circle |t| = α−1 . The orders of
α1 and β1 are 1. We can compute that
t
α(1 + α)r
lim (1 −
)ā(t) =
,
t→2−α
α2 + 1
2 − α
α(2 − α)r
t
.
lim−1 1 + −1 a(t) = − 2
t→−α
α
α +1
8
By using Lemma 2, we prove that (16) and (18) hold.
Theorem 5. Suppose that m and r are fixed positive integers. When n → ∞,
n
X
αn+1 (1 + α)r+m
j+m−1
(r)
=
+ o(αn ),
Fn−j
2+1
j
α
j=0
n
X
j+m−1
(r)
= (1 + α)r+m αn + o(αn ).
Ln−j
j
j=0
(20)
(21)
Proof. We can verify that
n
n
X
X
t
j+m−1
1
(r)
=
[tn−j ]
Fn−j
[tj ]
2
r
j
(1 − t − t )(1 − t)
(1 − t)m
j=0
j=0
= [tn ]
n
X
(r)
Ln−j
j=0
t
(1 − t −
t2 )(1
− t)m+r
,
2−t
j+m−1
.
= [tn ]
(1 − t − t2 )(1 − t)m+r
j
By using Lemma 2, we have (20) and (21).
In the final of this section, we compare the accurate values with the asymptotic values.
In Theorem 4, let
n X
α(1 + α)r
n
(r)
Xn =
Fk , Yn = 2
.
n
k
(α
+
1)(2
−
α)
k=0
n
Xn
50 9.2737156629317196 &times; 1020
100 7.345448671565505 &times; 1041
150 5.818115698360039 &times; 1062
Yn
9.2737269219308562 &times; 1020
7.3454486715782857 &times; 1041
5.8181156983601641 &times; 1062
|Xn − Yn |/Xn
1.2141 &times; 10−6
1.7399 &times; 10−12
2.1352 &times; 10−14
Table 1: some values of Xn and Yn
From the above table, we find that the value of |Xn − Yn |/Xn gets smaller and smaller
with the increasing of n.
3
Remarks
(r)
(r)
Consider the sequences {un } and {vn } defined by
∞
X
n
u(r)
n z =
n=0
∞
X
n=0
vn(r) z n =
z
,
(1 − pz − z 2 )(1 − z)r
2 − pz
,
(1 − pz − z 2 )(1 − z)r
9
(r)
(r)
(r)
(r)
(r)
where p &gt; 0. It is clear that un = Fn and vn = Ln when p = 1. The conclusions of Fn
(r)
(r)
(r)
and Ln can be generalized to the cases of {un } and {vn }. For example, put
X
X
(r) (r)
(r)
(r) (r)
(r)
Un,k,r =
uj1 uj2 &middot; &middot; &middot; ujk , Vn,k,r =
v j1 v j2 &middot; &middot; &middot; v jk .
j1 +j2 +&middot;&middot;&middot;+jk =n
j1 +j2 +&middot;&middot;&middot;+jk =n
Then we have
n−1
2 (0)
(0)
(0)
(0)
(vn(0) + 2vn+1 + vn+2 ) − 3 (un+1 + 2u(0)
n + un−1 )
+ 4)
p
np + p + 4
2
(0)
un−1 +
,
+ 2
p(p + 4)
p3
n
X
(r)
=
Un,k,r uj .
Un,2,1 =
Un,k+1,r
p2 (p2
(22)
(23)
j=0
We can verify that
t2
Un,2,1 = [tn ]
(1 − pt − t2 )2 (1 − t)2
2
1
t2
1
1
1
t2
n
= [t ]
+
−
∆(1 + τ )2 (t + τ )2 ∆ 1 − τ −1 1 + τ (1 − t)2
2 1
t2
2
1
1
1
1
t
1
+
−
−
+
∆ (1 − τ −1 )2 (t − τ −1 )2 ∆(1 + τ ) 1 − τ −1 1 + τ τ + 1 t + τ
1−t
2
2t
1
1
−1
√
−
+
−1
−1
∆(1 + τ )(1 − τ ) t + τ
t−τ
∆
2
2t
1
1
1
1
1
+
−
+
∆(1 − τ −1 ) 1 − τ −1 1 + τ
t − τ −1 1 − t 1 − τ −1
∞
∞
1 X
1 X
(0)
(0)
n
n+2 (0)
= [t ] 2
(n + 1)tn+2
(n + 1)t (vn+2 + 2vn+3 + vn+4 ) + 2
p ∆ n=0
p n=0
∞
∞
∞
2 X n+2 (0)
2 X n+2 (0)
4 + 2p X n+2
(0)
(0)
− 3
,
t (un+3 + 2un+2 + un−1 ) +
t
t un+1 +
p n=0
p∆ n=0
p3 n=0
√
where ∆ = p2 + 4, τ = p+2 ∆ . Hence (22) holds. The proof of (23) is similar to that of (11),
(0)
and it is omitted here. The identities about {vn } can be found in the same way.
4
Acknowledgments
The authors would like to thank the anonymous referees for their criticism and useful suggestions.
10
References
 A. Dil and I. Mező, A symmetric algorithm for hyperharmonic and Fibonacci numbers,
Appl. Math. Comput. 206 (2008), 942–951.
 D. Merlini, R. Sprugnoli and M. C. Verri, The method of coefficients, Amer. Math.
Monthly 114 (2007), 40–57.
 N. Robbins, Some convolution-type and combinatorial identities pertaining to binary
linear recurrences, Fibonacci Quart. 29 (1991), 249–255.
 N. J. A. Sloane, The On–Line Encyclopedia of Integer Sequences. Published electronically at http://www.research.att.com/~njas/sequences, 2010.
 M. Z. Spivey, Combinatorial sums and finite difference, Discrete Math. 307 (2007),
3130–3146.
 G. Szegő, Orthogonal Polynomials, Amer. Math. Soc. Colloq. Publ., Vol. 23, rev. ed.,
1959.
 Wenpeng Zhang, Some identities involving the Fibonacci numbers, Fibonacci Quart. 35
(1997), 225–228.
 Fengzhen Zhao and Tianming Wang, Generalizations of some identities involving the
Fibonacci numbers, Fibonacci Quart. 39 (2001), 165–167.
2000 Mathematics Subject Classification: Primary 11B37, 11B39; Secondary 05A16, 05A19.
Keywords: Fibonacci numbers, Lucas numbers, generating function, asymptotic expansion,
Darboux’s method.
(Concerned with sequences A000048, A000071, A001610, A001924, and A023548.)
Received June 6 2010; revised version received September 8 2010. Published in Journal of
Integer Sequences, September 10 2010. Corrected, March 26 2011.