On the properties of k-Fibonacci and k-Lucas numbers

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International Journal of Advances in Applied Mathematics and Mechanics

On the properties of k-Fibonacci and k-Lucas numbers

Research Article

A. D. Godase^1,∗, M. B. Dhakne²

1Department of Mathematics, V. P. College,Vaijapur, India

2Department of Mathematics, Dr. B. A. M. University,Aurangabad, India

Received 07 July 2014; accepted (in revised version) 26 August 2014

Abstract: In this paper, some properties of k−Fibonacci and k −Lucas numbers are derived and proved by using matrices S=

 _k

21 2

k²+4 k2 2



and M = 0 1

k²+ 4 0

‹

. The identities we proved are not encountered in the k−Fibonacci and k−Lucas numbers literature.

MSC: 11B39• 11B83

Keywords: k−Fibonacci numbers • k −Lucas numbers • Fibonacci Matrix 2014 IJAAMM all rights reserved.c

1. Introduction

This paper represents an interesting investigation about some special relations between matrices and k−Fibonacci numbers,k−Lucas numbers.This investigation is valuable to obtain new k −Fibonacci ,k −Lucas identities by differ- ent methods.This paper contributes to k−Fibonacci,k −Lucas numbers literature,and encourage many researchers to investigate the properties of such number sequences.

Definition 1.1.

The k−Fibonacci sequence {Fk ,n}_n∈Nis defined as, F_{k ,0}= 0, Fk ,1= 1 and F_{k ,n+1}= k Fk ,n+ Fk ,n−1for n≥ 1 Definition 1.2.

The k−Lucas sequence {Lk ,n}_n∈Nis defined as, L_{k ,0}= 2, Lk ,1= k and L_{k ,n+1}= k Lk ,n+ Lk ,n−1for n≥ 1

2. Main theorems

Lemma 2.1.

If X is a square matrix with X²= k X + I , then Xⁿ= Fk ,nX+ F_{k ,n−1}I , for all n∈ Z Proof. If n= 0 then result is obivious,

If n= 1 then

(X )¹= Fk ,1X+ Fk ,0I

= 1X + I

= X

∗ Corresponding author.

E-mail address:ashokgodse2012@gmail.com

Hence result is true for n= 1 It can be shown by induction that,

Xⁿ= Fk ,nX+ F_{k ,n−1}I ,for all n∈ Z

Assume that, Xⁿ= Fk ,nX+ F_{k ,n−1}I , and prove that,Xⁿ⁺¹= F_{k ,n+1}X+ Fk ,nI , Consider,

F_{k ,n+1}X+ Fk ,nI= (Fk ,nX+ F_{k ,n−1}I)X + Fk ,nI

= (k X + I )Fk ,n+ X F_{k ,n−1}

= X²F_{k ,n}+ X Fk ,n−1= X (X Fk ,n+ Fk ,n−1)

= X (Xⁿ)

= Xⁿ⁺¹ Hence, Xⁿ⁺¹= Fk ,n+1X+ Fk ,nI ,

By Induction, Xⁿ= Fk ,nX+ F_{k ,n−1}I , for all n∈ Z

We now show that, X⁻⁽ⁿ⁾= F_{k ,−n}X+ F_{k ,−n−1}I , for all n∈ Z⁺ Let, Y = k I − X , then

Y²= (k I − X )²

= k²I− 2k X + X²

= k²I− 2k X + k X + I

= k²I− k X + I

= k(k I − X ) + X + I

= k Y + I Therefore, Y²= k Y + I , This shows that,

Yⁿ= Fk ,nY + Fk ,n−1I ,

i .e . (−X⁻¹)ⁿ = Fk ,n(k I − X ) + F_{k ,n−1}I(−1)ⁿX⁻ⁿ

= −Fk ,nX+ F_{k ,n+1}I X⁻ⁿ = (−1)ⁿ⁺¹F_{k ,n}X+ (−1)ⁿF_{k ,n+1}I

Since, F_{k ,−n}= (−1)ⁿ⁺¹F_{k ,n}, F_{k ,−n−1}= (−1)ⁿF_{k ,n+1}, therefore X⁻ⁿ= F_{k ,−n}X+ F_{k ,−n−1}I ,gives X⁻⁽ⁿ⁾= F_{k ,−n}X+ F_{k ,−n+1}I , for all n∈ Z⁺.

Hence proof.

Corollary 2.1.

Let, M = k 1

1 0

‹

, then Mⁿ= F_{k ,n+1}F_{k ,n} F_{k ,n}F_{k ,n−1}

‹

Proof. Since,

M²= kM + I = Fk ,nM+ Fk ,n−1I Using Lemma2.1

= k F_{k ,n} F_{k ,n}

F_{k ,n} 0

‹

+ F_{k ,n−1} 0

0 F_{k ,n−1}

‹

= F_{k ,n+1} F_{k ,n}

F_{k ,n} F_{k ,n−1}

‹ , for all n∈ Z

Hence proof.

Corollary 2.2.

Let, S=

 _k

2 k²+4 k 2 2

k 2



, then Sⁿ=

 _L

k ,n 2

(k²+4)Fk ,n Fk ,n 2

2

Lk ,n 2



, for every n∈ Z

Lemma 2.2.

L²_{k ,n}− (k²+ 4)F_{k ,n}² = 4(−1)ⁿ,for all n∈ Z

101

Proof. Since, d e t(S) = −1, d e t (Sⁿ) = [d e t (S)]ⁿ= (−1)ⁿ, Moreover since,

Sⁿ=

 _{Lk ,n}

Fk ,n2 2

(k²+4)Fk ,n L2k ,n

2

 , We get

d e t(Sⁿ) =L²_{k ,n}

4 −(k²+ 4)F_{k ,n}²

4 ,

Thus it follows that L²_{k ,n}− (k²+ 4)F_{k ,n}² = 4(−1)ⁿ,for all n∈ Z Hence proof.

Lemma 2.3.

2L_{k ,n+m}= Lk ,nL_{k ,m}+ (k²+ 4)Fk ,nF_{k ,m} and 2F_{k ,n+m}= Fk ,nL_{k ,m}+ Lk ,nF_{k ,m} for all n , m∈ Z

Proof. Since, S^n+m= Sⁿ· S^m

=

 _{Lk ,n}

F_{k ,n}2 2

(k²+4)Fk ,n L2k ,n

2

 .

 _{Lk ,m}

F_{k ,m}2 2

(k²+4)Fk ,m Lk ,m2

2



=

‚ Lk ,nLk ,m+(k²+4)Fk ,nFk ,m Lk ,nFk ,m+F4 k ,nLk ,m

4

(k²+4)[Lk ,nFk ,m+Fk ,nLk ,m Lk ,nLk ,m+(k4²+4)Fk ,nFk ,m

4

Œ

But,

S^n+m=

 _{Lk ,n+m}

Fk ,n+m2 2

(k²+4)Fk ,n+m Lk ,n+m2

2

 , Gives,

2L_{k ,n+m}= Lk ,nL_{k ,m}+ (k²+ 4)Fk ,nF_{k ,m} and

2F_{k ,n+m}= Fk ,nL_{k ,m}+ Lk ,nF_{k ,m} for all n , m∈ Z

Hence proof.

Lemma 2.4.

2(−1)^mL_{k ,n−m}= Lk ,nL_{k ,m}− (k²+ 4)Fk ,nF_{k ,m} and

2(−1)^mF_{k ,n−m}= Fk ,nL_{k ,m}− Lk ,nF_{k ,m} for all n , m∈ Z .

Proof. Since, S^n−m= Sⁿ· S^−m

= Sⁿ· [S^m]⁻¹

= Sⁿ· (−1)^m

 _{Lk ,m}

−F2k ,m 2

−(k²+4)Fk ,m Lk ,m2

2



= (−1)^m

 _{Lk ,n}

Fk ,n2 2

(k²+4)Fk ,n L2k ,n

2

 .

 _{Lk ,m}

−F2k ,m 2

−(k²+4)Fk ,m Lk ,m2

2



=

‚ Lk ,nLk ,m−(k²+4)Fk ,nFk ,m Lk ,nFk ,m−F4 k ,nLk ,m

4

(k²+4)[Lk ,nFk ,m−Fk ,nLk ,m 4

Lk ,nLk ,m−(k²+4)Fk ,nFk ,m 4

Œ

But,

S^n−m=

 _L

k ,n−m F_{k ,n−m}2 2

(k²+4)Fk ,n−m L_{k ,n−m}2

2

 ,

Gives,

2(−1)^mL_{k ,n−m}= Lk ,nL_{k ,m}− (k²+ 4)Fk ,nF_{k ,m} and

2(−1)^mF_{k ,n−m}= Fk ,nL_{k ,m}− Lk ,nF_{k ,m} for all n , m∈ Z .

Lemma 2.5.

(−1)^mL_{k ,n−m}+ L_{k ,n+m}= Lk ,nL_{k ,m} and

(−1)^mF_{k ,n−m}+ F_{k ,n+m}= Fk ,nL_{k ,m} for all n , m∈ Z .

Proof. By definition of the matrix Sⁿ, it can be seen that

S^n+m+ (−1)^mS^n−m=

‚ _L

k ,n+m+(−1)^mL_{k ,n−m} Fk ,n+m+(−1)2 ^mF_{k ,n−m}

2

(k²+4)[Fk ,n+m+(−1)^mF_{k ,n−m} L_{k ,n+m}+(−1)2 ^mL_{k ,n−m}

2

Œ

On the other hand,

S^n+m+ (−1)^mS^n−m= SⁿS^m+ (−1)^mSⁿS^−m

= Sⁿ[S^m+ (−1)^mS^−m]

=

 _{Lk ,n}

Fk ,n2 2

(k²+4)Fk ,n Lk ,n2

2

 .

 _{Lk ,m}

Fk ,m2 2

(k²+4)Fk ,m Lk ,m2

2

 + (−1)^m

 _{Lk ,m}

−F2k ,m 2

−(k²+4)Fk ,m Lk ,m2

2



=

 _{Lk ,n}

Fk ,n2 2

(k²+4)Fk ,n L2k ,n

2

 . L_{k ,m}

0 0 L_{k ,m}

‹

=

 _{Lk ,mLk ,n}

Fk ,n2Lk ,m 2

(k²+4)Fk ,nLk ,m Lk ,m2Lk ,n

2



Gives,

(−1)^mL_{k ,n−m}+ Lk ,n+m= Lk ,nLk ,m

and

(−1)^mF_{k ,n−m}+ Fk ,n+m= Fk ,nL_{k ,m} for all n , m∈ Z .

Lemma 2.6.

8F_{k ,x+y +z}= Lk ,xL_{k ,y}F_{k ,z}+ Fk ,xL_{k ,y}L_{k ,z}+ Lk ,xF_{k ,y}L_{k ,z}+ (k²+ 4)Fk ,xF_{k ,y}F_{k ,z} and

8Lk ,x+y +z= Lk ,xLk ,yLk ,z+ (k²+ 4)[Lk ,xFk ,yFk ,z+ Fk ,xLk ,yFk ,z+ Fk ,xFk ,yLk ,z

for all x , y , z∈ Z .

103

Proof. By definition of the matrix Sⁿ,it can be seen that

S^{x+y +z}=

‚ _{Lk ,x+y +z}

Fk ,x+y +z2 2

(k²+4)Fk ,x+y +z L_{k ,x+y +z}2

2

Œ

On the other hand, S^{x+y +z}= S^x+yS^z

=

‚ _{Lk ,x+y}

F_{k ,x+y}2 2

(k²+4)Fk ,x+y L_{k ,x+y}2

2

Π.

 _{Lk ,z}

F2k ,z 2

(k²+4)Fk ,z Lk ,z2

2



=

‚ _L

k ,x+yL_{k ,z}+(k²+4)Fk ,x+yF_{k ,z} Lk ,zF_{k ,x+y}+F4 k ,zL_{k ,x+y}

4

(k²+4)[Lk ,x+yFk ,z+Fk ,x+yLk ,z L_{k ,x+y}Lk ,z+(k4²+4)Fk ,x+yFk ,z

4

Œ

Using,

2L_{k ,x+y}= Lk ,xL_{k ,y}+ (k²+ 4)Fk ,xF_{k ,y} 2F_{k ,x+y}= Lk ,yF_{k ,x}+ (k²+ 4)Fk ,yL_{k ,x} Gives,

8F_{k ,x+y +z}= Lk ,xLk ,yFk ,z+ Fk ,xLk ,yLk ,z+ Lk ,xFk ,yLk ,z+ (k²+ 4)Fk ,xFk ,yFk ,z

and

8L_{k ,x+y +z}= Lk ,xL_{k ,y}L_{k ,z}+ (k²+ 4)[Lk ,xF_{k ,y}F_{k ,z}+ Fk ,xL_{k ,y}F_{k ,z}+ Fk ,xF_{k ,y}L_{k ,z} for all x , y , z ∈ Z .

Theorem 2.1.

L_{k ,x+y}² − (k²+ 4)(−1)^{x+y +1}F_{k ,z−x}L_{k ,x+y}F_{k ,y+z}− (k²+ 4)(−1)^x+zF_{k ,y+z}² = (−1)^y+zL_{k ,z}² _−x for all x , y , z∈ Z .

Proof. Consider matrix multiplication given below.

That is,

 _{Lk ,x}

Fk ,z2 2

(k²+4)Fk ,x L2k ,z

2

 . L_{k ,y}

F_{k ,y}

‹

= L_{k ,x+y} F_{k ,y+z}

‹

Now,

d e t

 _{Lk ,x}

Fk ,z2 2

(k²+4)Fk ,x L2k ,z

2



=L_{k ,x}L_{k ,z}− (k²+ 4)Fk ,xF_{k ,z} 4

=(−1)^xL_{k ,z−x} 2

= Q 6= 0 Therefore we can write

 L_{k ,y} F_{k ,y}

‹

=

 _{Lk ,x}

Fk ,z2 2

(k²+4)Fk ,x L2k ,z

2

−1

· L_{k ,x+y} F_{k ,y+z}

‹

= 1 Q

 _{Lk ,z}

−F2k ,z 2

−(k²+4)Fk ,x Lk ,x2

2



. L_{k ,x+y} F_{k ,y+z}

‹

Gives,

L_{k ,y}=(−1)^x[Lk ,zL_{k ,x+y}− (k²+ 4)Fk ,xF_{k ,y+z}] L_{k ,z−x}

and

F_{k ,y}=(−1)^x[Lk ,xF_{k ,z+y}− Fk ,zL_{k ,y+x}] L_{k ,z−x}

Since,

L_{k ,y}² − (k²+ 4)F_{k ,y}² = 4(−1)^y We get,

[Lk ,zL_{k ,x+y}− (k²+ 4)Fk ,xF_{k ,y+z}]²− (k²+ 4)²[Lk ,xF_{k ,z+y}− Fk ,zL_{k ,y+x}]²= 4(−1)^yL²_{k ,z−x} Using Lemma2.4and Lemma2.6,

€L_{k ,z}² L²_{k ,x+y}− 2(k²+ 4)Lk ,zF_{k ,x+y}F_{k ,y+z}+ (k²+ 4)²F_{k ,x}² F_{k ,y+z}² Š

−(k²+ 4)(L_{k ,x}² F_{k ,y+z}² − 2Lk ,xF_{k ,z}F_{k ,y+z}L_{k ,x+y}+ F_{k ,z}² L²_{k ,x+y}) = 4(−1)^yL_{k ,z}² _−x Gives,

L_{k ,x+y}² − (k²+ 4)(−1)^{x+y +1}F_{k ,z−x}L_{k ,x+y}F_{k ,y+z}− (k²+ 4)(−1)^x+zF_{k ,y+z}² = (−1)^y+zL_{k ,z}² _−x for all x , y , z ∈ Z .

Theorem 2.2.

L_{k ,x+y}² − (−1)^x+zL_{k ,z−x}L_{k ,x+y}L_{k ,y+z}+ (−1)^x+zL²_{k ,y+z}= (−1)^{y+z +1}(k²+ 4)F_{k ,z−x}² for all x , y , z∈ Z , x 6= z .

Proof. Consider matrix multiplication,

‚ _L

k ,x L2k ,z 2

(k²+4)Fk ,x (k²+4)F2k ,z

2

Œ . L_{k ,y}

F_{k ,y}

‹

= L_{k ,x+y} L_{k ,y+z}

‹

Now, d e t

‚ _L

k ,x L2k ,z 2

(k²+4)Fk ,x (k²+4)F2 k ,z

2

Œ

=(k²+ 4)(−1)^xF_{k ,z−x} 2

= P 6= 0, ( if x 6= z ) Therefore for x6= z , we can write

 L_{k ,y} F_{k ,y}

‹

=

‚ _L

k ,x L2k ,z 2

(k²+4)Fk ,x (k²+4)F2 k ,z

2

Œ−1

. L_{k ,x+y} L_{k ,y+z}

‹

= 1 P

 _(k2+4)Fk ,z

−L2k ,z 2

−(k²+4)Fk ,x L2k ,x

2



. L_{k ,x+y} L_{k ,y+z}

‹

Gives,

L_{k ,y}=(−1)^x[Fk ,zL_{k ,x+y}− Fk ,xL_{k ,y+z}] F_{k ,z−x}

and

F_{k ,y}=(−1)^x[Lk ,xL_{k ,z+y}− Lk ,zL_{k ,y+x}] (k²+ 4)Fk ,z−x

Since,

L_{k ,y}² − (k²+ 4)F_{k ,y}² = 4(−1)^y We get,

(k²+ 4)[Fk ,zL_{k ,x+y}− Fk ,xL_{k ,y+z}]²− [Lk ,xL_{k ,z+y}− Lk ,zL_{k ,y+x}]²= 4(k²+ 4)(−1)^yF_{k ,z−x}² Using Lemma2.4and Lemma2.6, We obtain

L_{k ,x+y}² − (−1)^x+zL_{k ,z−x}L_{k ,x+y}L_{k ,y+z}+ (−1)^x+zL²_{k ,y+z}= (−1)^{y+z +1}(k²+ 4)F_{k ,z−x}² for all x , y , z ∈ Z , x 6= z .

105

Theorem 2.3.

F_{k ,x+y}² − Lk ,x−zF_{k ,x+y}F_{k ,y+z}+ (−1)^x+zF_{k ,y}² _+z= (−1)^y+zF_{k ,z−x}² for all x , y , z∈ Z , x 6= z .

Proof. Consider matrix multiplication,

 _{Fk ,x}

Fk ,z2 2

Fk ,x L2k ,z 2

 . L_{k ,y}

F_{k ,y}

‹

= F_{k ,x+y} F_{k ,y+z}

‹

Now, d e t

 _{Fk ,x}

F2k ,z 2

Fk ,x L2k ,z 2



=(−1)^zF_{k ,x−z} 2

= R 6= 0, ( if x 6= z ) Therefore for x6= z , we get,

 L_{k ,y} F_{k ,y}

‹

=

 _{Fk ,x}

F2k ,z 2

Fk ,x L2k ,z 2

−1

. F_{k ,x+y} F_{k ,y+z}

‹

= 1 R

 _{Lk ,z}

−F2k ,z 2

−Lk ,x Fk ,x2

2



. F_{k ,x+y} F_{k ,y+z}

‹

Gives,

L_{k ,y}=(−1)^z[Lk ,zF_{k ,x+y}− Lk ,xF_{k ,y+z}] F_{k ,x−z}

and

F_{k ,y}=(−1)^z[Fk ,xF_{k ,z+y}− Fk ,zF_{k ,y+x}] F_{k ,x−z}

Now consider,

[Lk ,zF_{k ,x+y}− Lk ,xF_{k ,y+z}]²− (k²+ 4)[Fk ,xF_{k ,z+y}− Fk ,zF_{k ,y+x}]²= 4(−1)^yF_{k ,x−z}² Using Lemma2.4and Lemma2.6, We get

F_{k ,x+y}² − Lk ,x−zF_{k ,x+y}F_{k ,y+z}+ (−1)^x+zF_{k ,y}² _+z= (−1)^y+zF_{k ,z−x}² for all x , y , z ∈ Z , x 6= z .

3. Conclusions

The conclusions arising from the work are as follows:

Some new identities have been obtained for the k−Fibonacci and k −Lucas sequences.

ACKNOWLEDGEMENTS.

The authors wish to thank two anonymous referees for their suggestions,which led to substantial improvement of this paper.

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