# Incompletek-Fibonacci and k-Lucas Numbers

Authors: José L. Ramírez
Journal: Chinese Journal of Mathematics
 (2013)

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## Abstract

We define the incompletek-Fibonacci and k-Lucas numbers; we study the recurrence relations and some properties of these numbers.


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## Body

## 1. Introduction

Fibonacci numbers and their generalizations have many interesting properties and applications to almost every field of science and art (e.g., see [1]). Fibonacci numbers Fn are defined by the recurrence relation
(1)F0=0,F1=1,Fn+1=Fn+Fn-1,n⩾1.There exist a lot of properties about Fibonacci numbers. In particular, there is a beautiful combinatorial identity to Fibonacci numbers [1]
(2)Fn=∑i=0⌊(n-1)/2⌋‍(n-i-1i).
From (2), Filipponi [2] introduced the incomplete Fibonacci numbers Fn(s) and the incomplete Lucas numbers Ln(s). They are defined by
(3)Fn(s)=∑j=0s‍(n-1-jj)(n=1,2,3,…;0≤s≤⌊n-12⌋),Ln(s)=∑j=0s‍nn-j(n-jj)(n=1,2,3,…;0≤s≤⌊n2⌋).Further in [3], generating functions of the incomplete Fibonacci and Lucas numbers are determined. In [4], Djordjević gave the incomplete generalized Fibonacci and Lucas numbers. In [5], Djordjević and Srivastava defined incomplete generalized Jacobsthal and Jacobsthal-Lucas numbers. In [6], the authors define the incomplete Fibonacci and Lucas p-numbers. Also the authors define the incomplete bivariate Fibonacci and Lucas p-polynomials in [7].On the other hand, many kinds of generalizations of Fibonacci numbers have been presented in the literature. In particular, a generalization is thek-Fibonacci Numbers.For any positive real numberk, the k-Fibonacci sequence, say {Fk,n}n∈ℕ, is defined recurrently by
(4)Fk,0=0,Fk,1=1,Fk,n+1=kFk,n+Fk,n-1,n⩾1.In [8], k-Fibonacci numbers were found by studying the recursive application of two geometrical transformations used in the four-triangle longest-edge (4TLE) partition. These numbers have been studied in several papers; see [8–14].For any positive real numberk, the k-Lucas sequence, say {Lk,n}n∈ℕ, is defined recurrently by
(5)Lk,0=2,Lk,1=k,Lk,n+1=kLk,n+Lk,n-1.Ifk=1, we have the classical Lucas numbers. Moreover, Lk,n=Fk,n-1+Fk,n+1,n⩾1; see [15].In [12], the explicit formula to k-Fibonacci numbers is
(6)Fk,n=∑i=0⌊(n-1)/2⌋‍(n-i-1i)kn-2i-1,
and the explicit formula of k-Lucas numbers is
(7)Lk,n(x)=∑i=0⌊n/2⌋‍nn-i(n-ii)kn-2i.From (6) and (7), we introduce the incomplete k-Fibonacci and k-Lucas numbers and we obtain new recurrent relations, new identities, and their generating functions.


## 2. The Incompletek-Fibonacci Numbers

Definition 1.
The incompletek-Fibonacci numbers are defined by
(8)Fk,nl=∑i=0l‍(n-1-ii)kn-2i-1,0≤l≤⌊n-12⌋.In Table1, some values of incomplete k-Fibonacci numbers are provided.Table 1
The numbersFk,nl, for 1⩽n⩽10.
n
∖
l
0
1
2
3
4
1
1
2
k
3
k
2
k
2
+
1
4
k
3
k
3
+
2
k
5
k
4
k
4
+
3
k
2
k
4
+
3
k
2
+
1
6
k
5
k
5
+
4
k
3
k
5
+
4
k
3
+
3
k
7
k
6
k
6
+
5
k
4
k
6
+
5
k
4
+
6
k
2
k
6
+
5
k
4
+
6
k
2
+
1
8
k
7
k
7
+
6
k
5
k
7
+
6
k
5
+
10
k
3
k
7
+
6
k
5
+
10
k
3
+
4
k
9
k
8
k
8
+
7
k
6
k
8
+
7
k
6
+
15
k
4
k
8
+
7
k
6
+
15
k
4
+
10
k
2
k
8
+
7
k
6
+
15
k
4
+
10
k
2
+
1
10
k
9
k
9
+
8
k
7
k
9
+
8
k
7
+
21
k
5
k
9
+
8
k
7
+
21
k
5
+
20
k
3
k
9
+
8
k
7
+
21
k
5
+
20
k
3
+
5
kWe note that(9)F1,n⌊(n-1)/2⌋=Fn.
For k=1, we get incomplete Fibonacci numbers [2].Some special cases of (8) are
(10)Fk,n0=kn-1;(n≥1),Fk,n1=kn-1+(n-2)kn-3;(n≥3)Fk,n2=kn-1+(n-2)kn-3+(n-4)(n-3)2kn-5;(n≥5)Fk,n⌊(n-1)/2⌋=Fk,n;(n≥1)Fk,n⌊(n-3)/2⌋={Fk,n-nk2(neven)Fk,n-1(nodd)(n≥3).


### 2.1. Some Recurrence Properties of the NumbersFk,nl

Proposition 2.
The recurrence relation of the incompletek-Fibonacci numbers Fk,nl is
(11)Fk,n+2l+1=kFk,n+1l+1+Fk,nl,0≤l≤n-22.
The relation (11) can be transformed into the nonhomogeneous recurrence relation
(12)Fk,n+2l=kFk,n+1l+Fk,nl-(n-1-ll)kn-1-2l.Proof.
Use Definition1 to rewrite the right-hand side of (11) as
(13)k∑i=0l+1‍(n-ii)kn-2i+∑i=0l‍(n-i-1i)kn-2i-1=∑i=0l+1‍(n-ii)kn-2i+1+∑i=1l+1‍(n-ii-1)kn-2i+1=kn-2i+1(∑i=0l+1‍[(n-ii)+(n-ii-1)])-kn+1(n-1)=∑i=0l+1‍(n-i+1i)kn-2i+1-0=Fk,n+2l.Proposition 3.
One has(14)∑i=0s‍(si)Fk,n+il+iki=Fk,n+2sl+s,0≤l≤n-s-12.Proof (by induction ons).
Sum (14) clearly holds for s=0 and s=1 (see (11)). Now suppose that the result is true for all j<s+1; we prove it for s+1:
(15)∑i=0s+1‍(s+1i)Fk,n+il+iki=∑i=0s+1‍[(si)+(si-1)]Fk,n+il+iki=∑i=0s+1‍(si)Fk,n+il+iki+∑i=0s+1‍(si-1)Fk,n+il+iki=Fk,n+2sl+s+(ss+1)Fk,n+s+1l+s+1ks+1+∑i=-1s‍(si)Fk,n+i+1l+i+1ki+1=Fk,n+2sl+s+0+∑i=0s‍(si)Fk,n+i+1l+i+1ki+1+(s-1)Fk,nl=Fk,n+2sl+s+k∑i=0s‍(si)Fk,n+i+1l+i+1ki+0=Fk,n+2sl+s+kFk,n+2s+1l+s+1=Fk,n+2s+2l+s+1.Proposition 4.
Forn≥2l+2,
(16)∑i=0s-1‍Fk,n+ilks-1-i=Fk,n+s+1l+1-ksFk,n+1l+1.Proof (by induction ons).
Sum (16) clearly holds for s=1 (see (11)). Now suppose that the result is true for all j<s. We prove it for s:
(17)∑i=0s‍Fk,n+ilks-i=k∑i=0s-1‍Fk,n+ilks-i-1+Fk,n+sl=k(Fk,n+s+1l+1-ksFk,n+1l+1)+Fk,n+sl=(kFk,n+s+1l+1+Fk,n+sl)-ks+1Fk,n+1l+1=Fk,n+s+2l+1-ks+1Fk,n+1l+1.Note that ifk, in (4), is a real variable, then Fk,n=Fx,n and they correspond to the Fibonacci polynomials defined by
(18)Fn+1(x)={1ifn=0xifn=1xFn(x)+Fn-1(x)ifn>1.Lemma 5.
One has(19)Fn′(x)=nFn+1(x)-xFn(x)+nFn-1(x)x2+4=nLn(x)-xFn(x)x2+4.See Proposition 13 of [12].Lemma 6.
One has(20)∑i=0⌊(n-1)/2⌋‍i(n-1-ii)kn-1-2i=((k2+4)n-4)Fk,n-nkLk,n2(k2+4).Proof.
From (6) we have that
(21)kFk,n=∑i=0⌊(n-1)/2⌋‍(n-1-ii)kn-2i.
By deriving into the previous equation (respect tok), it is obtained
(22)Fk,n+kFk,n′=∑i=0⌊(n-1)/2⌋‍(n-2i)(n-1-ii)kn-2i-1=nFk,n-2∑i=0⌊(n-1)/2⌋‍i(n-1-ii)kn-2i-1.
From Lemma5,
(23)Fk,n+k(nLk,n-kFk,nk2+4)=nFk,n-2∑i=0⌊(n-1)/2⌋‍i(n-1-ii)kn-2i-1.
From where, after some algebra (20) is obtained.Proposition 7.
One has(24)∑l=0⌊(n-1)/2⌋‍Fk,nl={4Fk,n+nkLk,n2(k2+4)(neven)(k2+8)Fk,n+nkLk,n2(k2+4)(nodd).Proof.
(25)
∑
l
=
0
⌊
(
n
-
1
)
/
2
⌋
‍
F
k
,
n
l
=
F
k
,
n
0
+
F
k
,
n
1
+
⋯
+
F
k
,
n
⌊
(
n
-
1
)
/
2
⌋
=
(
n
-
1
-
0
0
)
k
n
-
1
+
[
(
n
-
1
-
0
0
)
k
n
-
1
+
(
n
-
1
-
1
1
)
k
n
-
3
]
+
⋯
+
[
(
n
-
1
-
⌊
n
-
1
2
⌋
⌊
n
-
1
2
⌋
)
(
n
-
1
-
0
0
)
k
n
-
1
+
(
n
-
1
-
1
1
)
k
n
-
3
+
⋯
+
+
⋯
+
(
n
-
1
-
⌊
n
-
1
2
⌋
⌊
n
-
1
2
⌋
)
k
n
-
1
-
2
⌊
(
n
-
1
)
/
2
⌋
]
=
(
⌊
n
-
1
2
⌋
+
1
)
(
n
-
1
-
0
0
)
k
n
-
1
+
⌊
n
-
1
2
⌋
(
n
-
1
-
1
1
)
k
n
-
3
+
⋯
+
(
n
-
1
-
⌊
n
-
1
2
⌋
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-
1
2
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k
n
-
1
-
2
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(
n
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1
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2
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∑
i
=
0
⌊
(
n
-
1
)
/
2
⌋
‍
(
⌊
n
-
1
2
⌋
+
1
-
i
)
(
n
-
1
-
i
i
)
k
n
-
1
-
2
i
=
∑
i
=
0
⌊
(
n
-
1
)
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2
⌋
‍
(
⌊
n
-
1
2
⌋
+
1
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(
n
-
1
-
i
i
)
k
n
-
1
-
2
i
-
∑
i
=
0
⌊
(
n
-
1
)
/
2
⌋
‍
i
(
n
-
1
-
i
i
)
k
n
-
1
-
2
i
=
(
⌊
n
-
1
2
⌋
+
1
)
F
k
,
n
-
∑
i
=
0
⌊
(
n
-
1
)
/
2
⌋
‍
i
(
n
-
1
-
i
i
)
k
n
-
1
-
2
i
.
From Lemma 6, (24) is obtained.


## 2.1. Some Recurrence Properties of the NumbersFk,nl

Proposition 2.
The recurrence relation of the incompletek-Fibonacci numbers Fk,nl is
(11)Fk,n+2l+1=kFk,n+1l+1+Fk,nl,0≤l≤n-22.
The relation (11) can be transformed into the nonhomogeneous recurrence relation
(12)Fk,n+2l=kFk,n+1l+Fk,nl-(n-1-ll)kn-1-2l.Proof.
Use Definition1 to rewrite the right-hand side of (11) as
(13)k∑i=0l+1‍(n-ii)kn-2i+∑i=0l‍(n-i-1i)kn-2i-1=∑i=0l+1‍(n-ii)kn-2i+1+∑i=1l+1‍(n-ii-1)kn-2i+1=kn-2i+1(∑i=0l+1‍[(n-ii)+(n-ii-1)])-kn+1(n-1)=∑i=0l+1‍(n-i+1i)kn-2i+1-0=Fk,n+2l.Proposition 3.
One has(14)∑i=0s‍(si)Fk,n+il+iki=Fk,n+2sl+s,0≤l≤n-s-12.Proof (by induction ons).
Sum (14) clearly holds for s=0 and s=1 (see (11)). Now suppose that the result is true for all j<s+1; we prove it for s+1:
(15)∑i=0s+1‍(s+1i)Fk,n+il+iki=∑i=0s+1‍[(si)+(si-1)]Fk,n+il+iki=∑i=0s+1‍(si)Fk,n+il+iki+∑i=0s+1‍(si-1)Fk,n+il+iki=Fk,n+2sl+s+(ss+1)Fk,n+s+1l+s+1ks+1+∑i=-1s‍(si)Fk,n+i+1l+i+1ki+1=Fk,n+2sl+s+0+∑i=0s‍(si)Fk,n+i+1l+i+1ki+1+(s-1)Fk,nl=Fk,n+2sl+s+k∑i=0s‍(si)Fk,n+i+1l+i+1ki+0=Fk,n+2sl+s+kFk,n+2s+1l+s+1=Fk,n+2s+2l+s+1.Proposition 4.
Forn≥2l+2,
(16)∑i=0s-1‍Fk,n+ilks-1-i=Fk,n+s+1l+1-ksFk,n+1l+1.Proof (by induction ons).
Sum (16) clearly holds for s=1 (see (11)). Now suppose that the result is true for all j<s. We prove it for s:
(17)∑i=0s‍Fk,n+ilks-i=k∑i=0s-1‍Fk,n+ilks-i-1+Fk,n+sl=k(Fk,n+s+1l+1-ksFk,n+1l+1)+Fk,n+sl=(kFk,n+s+1l+1+Fk,n+sl)-ks+1Fk,n+1l+1=Fk,n+s+2l+1-ks+1Fk,n+1l+1.Note that ifk, in (4), is a real variable, then Fk,n=Fx,n and they correspond to the Fibonacci polynomials defined by
(18)Fn+1(x)={1ifn=0xifn=1xFn(x)+Fn-1(x)ifn>1.Lemma 5.
One has(19)Fn′(x)=nFn+1(x)-xFn(x)+nFn-1(x)x2+4=nLn(x)-xFn(x)x2+4.See Proposition 13 of [12].Lemma 6.
One has(20)∑i=0⌊(n-1)/2⌋‍i(n-1-ii)kn-1-2i=((k2+4)n-4)Fk,n-nkLk,n2(k2+4).Proof.
From (6) we have that
(21)kFk,n=∑i=0⌊(n-1)/2⌋‍(n-1-ii)kn-2i.
By deriving into the previous equation (respect tok), it is obtained
(22)Fk,n+kFk,n′=∑i=0⌊(n-1)/2⌋‍(n-2i)(n-1-ii)kn-2i-1=nFk,n-2∑i=0⌊(n-1)/2⌋‍i(n-1-ii)kn-2i-1.
From Lemma5,
(23)Fk,n+k(nLk,n-kFk,nk2+4)=nFk,n-2∑i=0⌊(n-1)/2⌋‍i(n-1-ii)kn-2i-1.
From where, after some algebra (20) is obtained.Proposition 7.
One has(24)∑l=0⌊(n-1)/2⌋‍Fk,nl={4Fk,n+nkLk,n2(k2+4)(neven)(k2+8)Fk,n+nkLk,n2(k2+4)(nodd).Proof.
(25)
∑
l
=
0
⌊
(
n
-
1
)
/
2
⌋
‍
F
k
,
n
l
=
F
k
,
n
0
+
F
k
,
n
1
+
⋯
+
F
k
,
n
⌊
(
n
-
1
)
/
2
⌋
=
(
n
-
1
-
0
0
)
k
n
-
1
+
[
(
n
-
1
-
0
0
)
k
n
-
1
+
(
n
-
1
-
1
1
)
k
n
-
3
]
+
⋯
+
[
(
n
-
1
-
⌊
n
-
1
2
⌋
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n
-
1
2
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)
(
n
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1
-
0
0
)
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n
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+
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n
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1
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1
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n
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+
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+
+
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+
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n
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]
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n
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1
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+
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n
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n
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+
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n
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+
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+
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1
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2
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1
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n
-
1
)
/
2
⌋
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(
⌊
n
-
1
2
⌋
+
1
-
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n
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1
-
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i
)
k
n
-
1
-
2
i
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i
=
0
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(
n
-
1
)
/
2
⌋
‍
(
⌊
n
-
1
2
⌋
+
1
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(
n
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1
-
i
i
)
k
n
-
1
-
2
i
-
∑
i
=
0
⌊
(
n
-
1
)
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2
⌋
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i
(
n
-
1
-
i
i
)
k
n
-
1
-
2
i
=
(
⌊
n
-
1
2
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+
1
)
F
k
,
n
-
∑
i
=
0
⌊
(
n
-
1
)
/
2
⌋
‍
i
(
n
-
1
-
i
i
)
k
n
-
1
-
2
i
.
From Lemma 6, (24) is obtained.


## 3. The Incompletek-Lucas Numbers

Definition 8.
The incompletek-Lucas numbers are defined by
(26)Lk,nl=∑i=0l‍nn-i(n-ii)kn-2i,0≤l≤⌊n2⌋.In Table2, some numbers of incomplete k-Lucas numbers are provided.Table 2
The numbersLk,nl, for 1⩽n⩽9.
n
∖
l
0
1
2
3
4
1
k
2
k
2
k
2
+
2
3
k
3
k
3
+
3
k
4
k
4
k
4
+
4
k
2
k
4
+
4
k
2
+
2
5
k
5
k
5
+
5
k
3
k
5
+
5
k
3
+
5
k
6
k
6
k
6
+
6
k
4
k
6
+
6
k
4
+
9
k
2
k
6
+
6
k
4
+
9
k
2
+
2
7
k
7
k
7
+
7
k
5
k
7
+
7
k
5
+
14
k
3
k
7
+
7
k
5
+
14
k
3
+
7
k
8
k
8
k
8
+
8
k
6
k
8
+
8
k
6
+
20
k
4
k
8
+
8
k
6
+
20
k
4
+
16
k
2
k
8
+
8
k
6
+
20
k
4
+
16
k
2
+
2
9
k
9
k
9
+
9
k
7
k
9
+
9
k
7
+
27
k
5
k
9
+
9
k
7
+
27
k
5
+
30
k
3
k
9
+
9
k
7
+
27
k
5
+
30
k
3
+
9
kWe note that(27)L1,n⌊n/2⌋=Ln.Some special cases of (26) are
(28)Lk,n0=kn;(n≥1),Lk,n1=kn+nkn-2;(n≥2),Lk,n2=kn+nkn-2+n(n-3)2kn-4;(n≥4),Lk,n⌊n/2⌋=Lk,n;(n≥1),Lk,n⌊(n-2)/2⌋={Lk,n-2(neven)Lk,n-nk(nodd)(n≥2).


### 3.1. Some Recurrence Properties of the NumbersLk,nl

Proposition 9.
One has(29)Lk,nl=Fk,n-1l-1+Fk,n+1l,0≤l≤⌊n2⌋.Proof.
By (8), rewrite the right-hand side of (29) as
(30)∑i=0l-1‍(n-2-ii)kn-2-2i+∑i=0l‍(n-ii)kn-2i=∑i=1l‍(n-1-ii-1)kn-2i+∑i=0l‍(n-ii)kn-2i=∑i=0l‍[(n-1-ii-1)+(n-ii)]kn-2i-(n-1-1)=∑i=0l‍nn-i(n-ii)kn-2i+0=Lk,nl.Proposition 10.
The recurrence relation of the incompletek-Lucas numbers Lk,nl is
(31)Lk,n+2l+1=kLk,n+1l+1+Lk,nl,0≤l≤⌊n2⌋.
The relation (31) can be transformed into the nonhomogeneous recurrence relation
(32)Lk,n+2l=kLk,n+1l+Lk,nl-nn-l(n-ll)kn-2l.Proof.
Using (29) and (11), we write
(33)Lk,n+2l+1=Fk,n+1l+Fk,n+3l+1=kFk,nl+Fk,n-1l-1+kFk,n+2l+1+Fk,n+1l=k(Fk,nl+Fk,n+2l+1)+Fk,n-1l-1+Fk,n+1l=kLk,n+1l+1+Lk,nl.Proposition 11.
One has(34)kLk,nl=Fk,n+2l-Fk,n-2l-2,0≤l≤⌊n-12⌋.Proof.
By (29),
(35)Fk,n+2l=Lk,n+1l-Fk,nl-1,Fk,n-2l-2=Lk,n-1l-1-Fk,nl-1,
whence, from (31),
(36)Fk,n+2l-Fk,n-2l-2=Lk,n+1l-Lk,n-1l-1=kLk,nl.Proposition 12.
One has(37)∑i=0s‍(si)Lk,n+il+iki=Lk,n+2sl+s,0≤l≤n-s2.Proof.
Using (29) and (14), we write
(38)∑i=0s‍(si)Lk,n+il+iki=∑i=0s‍(si)[Fk,n+i-1l+i-1+Fk,n+i+1l+i]ki=∑i=0s‍(si)Fk,n+i-1l+i-1ki+∑i=0s‍(si)Fk,n+i+1l+iki=Fk,n-1+2sl-1+s+Fk,n+1+2sl+s=Lk,n+2sl+s.Proposition 13.
Forn≥2l+1,
(39)∑i=0s-1‍Lk,n+ilks-1-i=Lk,n+s+1l+1-ksLk,n+1l+1.The proof can be done by using (31) and induction on s.Lemma 14.
One has(40)∑i=0⌊n/2⌋‍inn-i(n-ii)kn-2i=n2[Lk,n-kFk,n].The proof is similar to Lemma6.Proposition 15.
One has(41)∑l=0⌊n/2⌋‍Lk,nl={Lk,n+nk2Fk,n(neven)12(Lk,n+nkFk,n)(nodd).Proof.
An argument analogous to that of the proof of Proposition7 yields
(42)∑l=0⌊n/2⌋‍Lk,nl=(⌊n2⌋+1)Lk,n-∑i=0⌊n/2⌋‍inn-i(n-ii)kn-2i.
From Lemma14, (41) is obtained.


## 3.1. Some Recurrence Properties of the NumbersLk,nl

Proposition 9.
One has(29)Lk,nl=Fk,n-1l-1+Fk,n+1l,0≤l≤⌊n2⌋.Proof.
By (8), rewrite the right-hand side of (29) as
(30)∑i=0l-1‍(n-2-ii)kn-2-2i+∑i=0l‍(n-ii)kn-2i=∑i=1l‍(n-1-ii-1)kn-2i+∑i=0l‍(n-ii)kn-2i=∑i=0l‍[(n-1-ii-1)+(n-ii)]kn-2i-(n-1-1)=∑i=0l‍nn-i(n-ii)kn-2i+0=Lk,nl.Proposition 10.
The recurrence relation of the incompletek-Lucas numbers Lk,nl is
(31)Lk,n+2l+1=kLk,n+1l+1+Lk,nl,0≤l≤⌊n2⌋.
The relation (31) can be transformed into the nonhomogeneous recurrence relation
(32)Lk,n+2l=kLk,n+1l+Lk,nl-nn-l(n-ll)kn-2l.Proof.
Using (29) and (11), we write
(33)Lk,n+2l+1=Fk,n+1l+Fk,n+3l+1=kFk,nl+Fk,n-1l-1+kFk,n+2l+1+Fk,n+1l=k(Fk,nl+Fk,n+2l+1)+Fk,n-1l-1+Fk,n+1l=kLk,n+1l+1+Lk,nl.Proposition 11.
One has(34)kLk,nl=Fk,n+2l-Fk,n-2l-2,0≤l≤⌊n-12⌋.Proof.
By (29),
(35)Fk,n+2l=Lk,n+1l-Fk,nl-1,Fk,n-2l-2=Lk,n-1l-1-Fk,nl-1,
whence, from (31),
(36)Fk,n+2l-Fk,n-2l-2=Lk,n+1l-Lk,n-1l-1=kLk,nl.Proposition 12.
One has(37)∑i=0s‍(si)Lk,n+il+iki=Lk,n+2sl+s,0≤l≤n-s2.Proof.
Using (29) and (14), we write
(38)∑i=0s‍(si)Lk,n+il+iki=∑i=0s‍(si)[Fk,n+i-1l+i-1+Fk,n+i+1l+i]ki=∑i=0s‍(si)Fk,n+i-1l+i-1ki+∑i=0s‍(si)Fk,n+i+1l+iki=Fk,n-1+2sl-1+s+Fk,n+1+2sl+s=Lk,n+2sl+s.Proposition 13.
Forn≥2l+1,
(39)∑i=0s-1‍Lk,n+ilks-1-i=Lk,n+s+1l+1-ksLk,n+1l+1.The proof can be done by using (31) and induction on s.Lemma 14.
One has(40)∑i=0⌊n/2⌋‍inn-i(n-ii)kn-2i=n2[Lk,n-kFk,n].The proof is similar to Lemma6.Proposition 15.
One has(41)∑l=0⌊n/2⌋‍Lk,nl={Lk,n+nk2Fk,n(neven)12(Lk,n+nkFk,n)(nodd).Proof.
An argument analogous to that of the proof of Proposition7 yields
(42)∑l=0⌊n/2⌋‍Lk,nl=(⌊n2⌋+1)Lk,n-∑i=0⌊n/2⌋‍inn-i(n-ii)kn-2i.
From Lemma14, (41) is obtained.


## 4. Generating Functions of the Incompletek-Fibonacci and k-Lucas Number

In this section, we give the generating functions of incompletek-Fibonacci and k-Lucas numbers.Lemma 16 (see [3, page 592]).
Let{sn}n=0∞ be a complex sequence satisfying the following nonhomogeneous recurrence relation:
(43)sn=asn-1+bsn-2+rn(n>1),
where a and b are complex numbers and {rn} is a given complex sequence. Then, the generating function U(t) of the sequence {sn} is
(44)U(t)=G(t)+s0-r0+(s1-s0a-r1)t1-at-bt2,
where G(t) denotes the generating function of {rn}.Theorem 17.
The generating function of the incompletek-Fibonacci numbers Fk,nl is given by
(45)Rk,l(x)=∑i=0∞‍Fk,ilti=t2l+1[Fk,2l+1+(Fk,2l+2-kFk,2l+1)t-t2(1-kt)l+1]×[1-kt-t2]-1.Proof.
Letl be a fixed positive integer. From (8) and (12), Fk,nl=0 for 0≤n<2l+1, Fk,2l+1l=Fk,2l+1, and Fk,2l+2l=Fk,2l+2, and
(46)Fk,nl=kFk,n-1l+Fk,n-2l-(n-3-ll)kn-3-2l.
Now let(47)s0=Fk,2l+1l,s1=Fk,2l+2l,sn=Fk,n+2l+1l.
Also let(48)r0=r1=0,rn=(n+l-2n-2)kn-2.
The generating function of the sequence {rn} is G(t)=t2/(1-kt)l+1 (see [16, page 355]). Thus, from Lemma 16, we get the generating function Rk,l(x) of sequence {sn}.Theorem 18.
The generating function of the incompletek-Lucas numbers Lk,nl is given by
(49)Sk,l(x)=∑i=0∞‍Lk,ilti=t2l[Lk,2l+(Lk,2l+1-kLk,2l)t-t2(2-t)(1-kt)l+1]×[1-kt-t2]-1.Proof.
The proof of this theorem is similar to the proof of Theorem17. Let l be a fixed positive integer. From (26) and (32), Lk,nl=0 for 0≤n<2l, Lk,2ll=Lk,2l, and Lk,2l+1l=Lk,2l+1, and
(50)Lk,nl=kLk,n-1l+Lk,n-2l-n-2n-2-l(n-2-ln-2-2l)kn-2-2l.
Now let(51)s0=Lk,2ll,s1=Lk,2l+1l,sn=Lk,n+2ll.
Also let(52)r0=r1=0,rn=(n+2l-2n+l-2)kn+2l-2.
The generating function of the sequence {rn} is G(t)=t2(2-t)/(1-kt)l+1 (see [16, page 355]). Thus, from Lemma 16, we get the generating function Sk,l(x) of sequence {sn}.


## 5. Conclusion

In this paper, we introduce incompletek-Fibonacci and k-Lucas numbers, and we obtain new identities. In [17], the authors introduced the h(x)-Fibonacci polynomials. That generalizes Catalan’s Fibonacci polynomials and the k-Fibonacci numbers. Let h(x) be a polynomial with real coefficients. The h(x)-Fibonacci polynomials {Fh,n(x)}n∈ℕ are defined by the recurrence relation
(53)Fh,0(x)=0,Fh,1(x)=1,Fh,n+1(x)=h(x)Fh,n(x)+Fh,n-1(x),n⩾1.It would be interesting to study a definition of incompleteh(x)-Fibonacci polynomials and research their properties.



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*Source: 107145-2013-10-30.xml*