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

ISSN: 2347-2529

Available online atwww.ijaamm.com

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. Dhakne}

2

1_{Department of Mathematics, V. P. College,Vaijapur, India}

2_{Department 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 ofk−Fibonacci andk−Lucas numbers are derived and proved by using matrices S=

_k

2 1 2

k2₊₄

2

k

2

andM =

 0 1

k2₊₄

0 ‹

. The identities we proved are not encountered in thek−Fibonacci andk−Lucas numbers literature.

MSC:

11B39• 11B83

Keywords:

k−Fibonacci numbers•k−Lucas numbers• Fibonacci Matrix c

2014 IJAAMM all rights reserved.

1.

Introduction

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

Definition 1.1.

Thek−Fibonacci sequence{Fk,n}n∈Nis defined as,F_k,0=0,F_k,1=1 andF_k,n+1=k Fk,n+Fk,n−1forn≥1

Definition 1.2.

Thek−Lucas sequence{Lk,n}n∈Nis defined as,L_k,0=2,L_k,1=kandL_k,n+1=k Lk,n+Lk,n−1forn≥1

2.

Main theorems

Lemma 2.1.

If X is a square matrix with X2_{=k X+I , then X}n_=F

k,nX+Fk,n−1I , for all n∈Z

Proof. Ifn=0 then result is obivious, Ifn=1 then

(X)1=Fk,1X+Fk,0I

=1X+I =X

∗ _{Corresponding author.}

E-mail address:ashokgodse2012@gmail.com

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

Xn=Fk,nX+Fk,n−1I,for alln∈Z

Assume that,Xn_=F

k,nX+Fk,n−1I, and prove that,Xn+1=Fk,n+1X+Fk,nI,

Consider,

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

=X2Fk,n+X Fk,n−1=X(X Fk,n+Fk,n−1)

=X(Xn)

=Xn+1

Hence,Xn+1_=F

k,n+1X+Fk,nI,

By Induction,Xn_=F

k,nX+Fk,n−1I, for alln∈Z

We now show that,X−(n)₌

Fk,−nX+Fk,−n−1I, for alln∈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) +Fk,n−1I(−1)nX−n =−F_k,nX+Fk,n+1I

X−n = (−1)n+1Fk,nX+ (−1)nFk,n+1I

Since,Fk,−n= (−1)n+1Fk,n,Fk,−n−1= (−1)nFk,n+1, thereforeX−n=Fk,−nX+Fk,−n−1I,givesX−(n)=Fk,−nX+Fk,−n+1I,

for alln∈Z+. Hence proof.

Corollary 2.1.

Let, M = 

k

1 1 0

‹

,then Mn₌ 

Fk,n+1Fk,n

Fk,nFk,n−1 ‹

Proof. Since,

M2=k M+I=Fk,nM+Fk,n−1I Using Lemma2.1

= 

k Fk,n

Fk,n

Fk,n

0

‹ +



Fk,n−1

0

0

Fk,n−1 ‹

= 

Fk,n+1 Fk,n

Fk,n

Fk,n−1 ‹

,

for alln∈Z

Hence proof.

Corollary 2.2.

Let, S= _k

2 k2₊₄

2 k 2

k 2

,then Sn_{= L}

k,n

2

(k2_+4)F

k,n

2 Fk,n

2 Lk,n

2

,for every n∈Z

Lemma 2.2.

L2_k,n−(k2+4)F_k2_,n=4(−1)n_{,for all n∈Z}

Proof. Since,d e t(S) =−1, d e t(Sn_{) = [d e t(S)]}n_{= (−1)}n_,

Moreover since,

Sn= _Lk,n

2 Fk,n

2

(k2_+4)F

k,n

2 Lk,n

2

,

We get

d e t(Sn) =L 2 k,n

4 −

(k2_+4)F2 k,n

4 , Thus it follows thatL2

k,n−(k2+4)Fk2,n=4(−1)n,for alln∈Z

Hence proof.

Lemma 2.3.

2Lk,n+m=Lk,nLk,m+ (k2+4)Fk,nFk,m and 2Fk,n+m=Fk,nLk,m+Lk,nFk,m

for all n,m∈Z

Proof. Since,

Sn+m=Sn·Sm

= Lk,n

2 Fk,n

2

(k2_+4)F

k,n

2 Lk,n

2

.

Lk,m

2 Fk,m

2

(k2_+4)F

k,m

2 Lk,m

2

= ‚

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

4 Lk,nFk,m+Fk,nLk,m

4

(k2_+4)[L

k,nFk,m+Fk,nLk,m

4

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

4

Œ

But,

Sn+m= _Lk,n+m

2 Fk,n+m

2

(k2_+4)F

k,n+m

2 Lk,n+m

2

,

Gives,

2Lk,n+m=Lk,nLk,m+ (k2+4)Fk,nFk,m

and

2Fk,n+m=Fk,nLk,m+Lk,nFk,m

for alln,m∈Z

Hence proof.

Lemma 2.4.

2(−1)mLk,n−m=Lk,nLk,m−(k2+4)Fk,nFk,m

and

2(−1)mFk,n−m=Fk,nLk,m−Lk,nFk,m

for all n,m∈Z .

Proof. Since,

Sn−m=Sn·S−m

=Sn·[Sm]−1

=Sn·(−1)m Lk,m

2 −Fk,m

2

−(k2_+4)F

k,m

2 Lk,m

2

= (−1)m _Lk,n

2 Fk,n

2

(k2_+4)F

k,n

2 Lk,n

2

.

_Lk,m 2 −Fk,m

2

−(k2_+4)F

k,m

2 Lk,m

2

= ‚

Lk,nLk,m−(k2+4)Fk,nFk,m

4 Lk,nFk,m−Fk,nLk,m

4

(k2_+4)[L

k,nFk,m−Fk,nLk,m

4

Lk,nLk,m−(k2+4)Fk,nFk,m

4

Œ

But,

Sn−m= _Lk,n−m

2 Fk,n−m

2

(k2_+4)F

k,n−m

2 Lk,n−m

2

,

Gives,

2(−1)mLk,n−m=Lk,nLk,m−(k2+4)Fk,nFk,m

and

2(−1)mFk,n−m=Fk,nLk,m−Lk,nFk,m

for alln,m∈Z.

Lemma 2.5.

(−1)mLk,n−m+Lk,n+m=Lk,nLk,m

and

(−1)mFk,n−m+Fk,n+m=Fk,nLk,m

for all n,m∈Z .

Proof. By definition of the matrixSn_{, it can be seen that}

Sn+m+ (−1)mSn−m= ‚ _L

k,n+m+(−1)mLk,n−m

2 Fk,n+m+(−1)mFk,n−m

2

(k2_+4)[F

k,n+m+(−1)mFk,n−m

2 Lk,n+m+(−1)mLk,n−m

2

Œ

On the other hand,

Sn+m+ (−1)mSn−m=SnSm+ (−1)mSnS−m =Sn[Sm+ (−1)mS−m]

= _Lk,n

2 Fk,n

2

(k2_+4)F

k,n

2 Lk,n

2

.

_Lk,m 2 Fk,m

2

(k2_+4)F

k,m

2 Lk,m

2

+ (−1)m _Lk,m

2 −Fk,m

2

−(k2_+4)F

k,m

2 Lk,m

2

= _Lk,n

2 Fk,n

2

(k2_+4)F

k,n

2 Lk,n

2

.



Lk,m

0 0

Lk,m ‹

=

_Lk,mLk,n 2 Fk,nLk,m

2

(k2_+4)F

k,nLk,m

2 Lk,mLk,n

2

Gives,

(−1)mLk,n−m+Lk,n+m=Lk,nLk,m

and

(−1)mFk,n−m+Fk,n+m=Fk,nLk,m

for alln,m∈Z.

Lemma 2.6.

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

Proof. By definition of the matrixSn_{,it can be seen that}

Sx+y+z= ‚ _L

k,x+y+z

2 Fk,x+y+z

2

(k2_+4)F

k,x+y+z

2 Lk,x+y+z

2

Œ

On the other hand,

Sx+y+z=Sx+ySz

= ‚ _L

k,x+y

2 Fk,x+y

2

(k2_+4)F

k,x+y

2 Lk,x+y

2 Œ

.

Lk,z

2 Fk,z

2

(k2_+4)F

k,z

2 Lk,z

2

= ‚ _L

k,x+yLk,z+(k2+4)Fk,x+yFk,z

4 Lk,zFk,x+y+Fk,zLk,x+y

4

(k2_+4)[L

k,x+yFk,z+Fk,x+yLk,z

4

Lk,x+yLk,z+(k2+4)Fk,x+yFk,z

4

Œ

Using,

2Lk,x+y=Lk,xLk,y+ (k2+4)Fk,xFk,y

2Fk,x+y=Lk,yFk,x+ (k2+4)Fk,yLk,x

Gives,

8Fk,x+y+z=Lk,xLk,yFk,z+Fk,xLk,yLk,z+Lk,xFk,yLk,z+ (k2+4)Fk,xFk,yFk,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 allx,y,z∈Z.

Theorem 2.1.

L_k2_,x+y−(k2+4)(−1)x+y+1Fk,z−xLk,x+yFk,y+z−(k2+4)(−1)x

+z

F_k2_,y+z= (−1)y+zL_k2_,z−x

for all x,y,z∈Z .

Proof. Consider matrix multiplication given below. That is,

Lk,x

2 Fk,z

2

(k2_+4)F

k,x

2 Lk,z

2

.



Lk,y

Fk,y ‹

= 

Lk,x+y

Fk,y+z ‹

Now,

d e t

_Lk,x 2 Fk,z

2

(k2_+4)F

k,x

2 Lk,z

2

=Lk,xLk,z−(k 2_+4)F

k,xFk,z

4

=(−1) x_L

k,z−x

2

=Q6=0

Therefore we can write



Lk,y

Fk,y ‹

= _Lk,x

2 Fk,z

2

(k2_+4)F

k,x

2 Lk,z

2 −1

·



Lk,x+y

Fk,y+z ‹

= 1 Q

_Lk,z 2 −Fk,z

2

−(k2_+4)F

k,x

2 Lk,x

2

.



Lk,x+y

Fk,y+z ‹

Gives,

Lk,y= (−1)x_[L

k,zLk,x+y−(k2+4)Fk,xFk,y+z]

Lk,z−x

and

Fk,y= (−1)x_[L

k,xFk,z+y−Fk,zLk,y+x]

Lk,z−x

Since,

L_k2_,y−(k2+4)F_k2_,y=4(−1)y

We get,

[Lk,zLk,x+y−(k2+4)Fk,xFk,y+z]2−(k2+4)2[Lk,xFk,z+y−Fk,zLk,y+x]2=4(−1)yL2k,z−x

Using Lemma2.4and Lemma2.6,

€

L_k2_,zL2_k,x+y−2(k2+4)Lk,zFk,x+yFk,y+z+ (k2+4) 2

F_k2_,xF_k2_,y+zŠ

−(k2+4)(L_k2_,xF_k2_,y+z−2Lk,xFk,zFk,y+zLk,x+y+F 2 k,zL

2

k,x+y) =4(−1) y

L_k2_,z−x

Gives,

L_k2_,x+y−(k2+4)(−1)x+y+1Fk,z−xLk,x+yFk,y+z−(k2+4)(−1)x+zFk2,y+z= (−1) y+z_L2

k,z−x

for allx,y,z∈Z.

Theorem 2.2.

L_k2_,x+y−(−1)x+zLk,z−xLk,x+yLk,y+z+ (−1)x+zL2k,y+z= (−1)

y+z+1_(k2_+4)F2 k,z−x

for all x,y,z∈Z,x6=z .

Proof. Consider matrix multiplication,

‚ _L k,x

2 Lk,z

2

(k2_+4)F

k,x

2 (k2_+4)F

k,z

2 Œ

.



Lk,y

Fk,y ‹

= 

Lk,x+y

Lk,y+z ‹

Now,

d e t

‚ _L k,x

2 Lk,z

2

(k2_+4)F

k,x

2 (k2_+4)F

k,z

2 Œ

=(k

2_+4)(−1)x

Fk,z−x

2

=P6=0, ( ifx6=z)

Therefore forx6=z, we can write



Lk,y

Fk,y ‹

= ‚ _L

k,x

2 Lk,z

2

(k2_+4)F

k,x

2 (k2_+4)F

k,z

2 Œ−1

.



Lk,x+y

Lk,y+z ‹

= 1 P

_(k2_+4)F

k,z

2 −Lk,z

2

−(k2_+4)F

k,x

2 Lk,x

2

.



Lk,x+y

Lk,y+z ‹

Gives,

Lk,y= (−1)x_[F

k,zLk,x+y−Fk,xLk,y+z]

Fk,z−x

and

Fk,y= (−1)x_[L

k,xLk,z+y−Lk,zLk,y+x] (k2_+4)F

k,z−x

Since,

L_k2_,y−(k2+4)F_k2_,y=4(−1)y

We get,

(k2+4)[Fk,zLk,x+y−Fk,xLk,y+z]2−[Lk,xLk,z+y−Lk,zLk,y+x]2=4(k2+4)(−1)yFk2,z−x

Using Lemma2.4and Lemma2.6, We obtain

L_k2_,x+y−(−1)x+zLk,z−xLk,x+yLk,y+z+ (−1)x+zL2k,y+z= (−1)

y+z+1_(k2_+4)F2 k,z−x

for allx,y,z∈Z,x6=z.

Theorem 2.3.

F_k2_,x+y−Lk,x−zFk,x+yFk,y+z+ (−1)x+zFk2,y+z= (−1) y+z_F2

k,z−x

for all x,y,z∈Z,x6=z .

Proof. Consider matrix multiplication,

Fk,x

2 Fk,z

2 Fk,x

2 Lk,z

2

.



Lk,y

Fk,y ‹

= 

Fk,x+y

Fk,y+z ‹

Now,

d e t

Fk,x

2 Fk,z

2 Fk,x

2 Lk,z

2

=(−1) z

Fk,x−z

2

=R6=0, (ifx6=z)

Therefore forx6=z, we get,



Lk,y

Fk,y ‹

= _Fk,x

2 Fk,z

2 Fk,x

2 Lk,z

2 −1

.



Fk,x+y

Fk,y+z ‹

= 1 R

_Lk,z 2 −Fk,z

2

−Lk,x

2 Fk,x

2

.



Fk,x+y

Fk,y+z ‹

Gives,

Lk,y= (−1)z_[L

k,zFk,x+y−Lk,xFk,y+z]

Fk,x−z

and

Fk,y= (−1)z_[F

k,xFk,z+y−Fk,zFk,y+x]

Fk,x−z

Now consider,

[Lk,zFk,x+y−Lk,xFk,y+z]2−(k2+4)[Fk,xFk,z+y−Fk,zFk,y+x]2=4(−1)yFk2,x−z

Using Lemma2.4and Lemma2.6, We get

F_k2_,x+y−Lk,x−zFk,x+yFk,y+z+ (−1)x

+z_F2

k,y+z= (−1) y+z_F2

k,z−x

for allx,y,z∈Z,x6=z.

3.

Conclusions

The conclusions arising from the work are as follows:

Some new identities have been obtained for thek−Fibonacci andk−Lucas sequences.

ACKNOWLEDGEMENTS.

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

References

[1] M.E. Waddill, Matrices and Generalized Fibonacci Sequences, The Fibonacci Quarterly 55(12.4) (1974) 381-86.

[2] T. Koshy, Fibonacci and Lucas numbers with applications, Wiley-Intersection Pub. 2001.

[3] S. Falcon, A. Plaza, On thek−Fibonacci numbers Chaos, Solitons Fractals 5(32) (2007) 1615-1624.

[4] A.F. Horadam, Basic Properties of a Certain Generalized Sequence of Numbers, The Fibonacci Quarterly 3(3) (1965) 161-176.

[5] P. Filipponi, A. F. Horadam, A Matrix Approach to Certain Identities, The Fibonacci Quarterly 26(2) (1988) 115-126.

[6] H. W. Gould, A History of the Fibonacci g-Matrix and a Higher-Dimensional Problem, The Fibonacci Quarterly 19(3) (1981) 50-57.

[7] S. Vajda, Fibonacci and Lucas numbers, and the Golden Section:Theory and applications, Chichester:Ellis Hor-wood, 1989.

[8] A. F. Horadam, Jacobstal Representation of Polynomials, The Fibonacci Quarterly 35(2) (1997) 137-148.

[9] G. Strang,Introduction to Linear Algebra, Wellesley-Cambridge:Wellesley, MA Pub., 2003.