On the properties of k-fibonacci and k-lucas numbers

Transcription

1 Int J Adv Appl Math Mech (1) (01) ISSN: Available online at wwwijaammcom International Journal of Advances in Applied Mathematics Mechanics On the properties of k-fibonacci k-lucas numbers Research Article A D Godase 1 M B Dhakne 1 Department of Mathematics V P CollegeVaijapur India Department of Mathematics Dr B A M UniversityAurangabad India Received 07 July 01; accepted (in revised version) 6 August 01 Abstract: MSC: In this paper some properties of k Fibonacci k Lucas numbers are derived proved by using matrices k k + S 0 k + 1 k M The identities we proved are not encountered in the k Fibonacci 1 0 k Lucas numbers literature 11B39 11B83 Keywords: k Fibonacci numbers k Lucas numbers Fibonacci Matrix c 01 IJAAMM all rights reserved 1 Introduction This paper represents an interesting investigation about some special relations between matrices k Fibonacci numbersk Lucas numbersthis investigation is valuable to obtain new k Fibonacci k Lucas identities by different methodsthis paper contributes to k Fibonaccik Lucas numbers literature encourage many researchers to investigate the properties of such number sequences Definition 11 The k Fibonacci sequence { } n N is defined as F k 0 0 F k k + 1 for n 1 Definition 1 The k Lucas sequence { } n N is defined as L k 0 L k 1 k +1 k + 1 for n 1 Main theorems Lemma 1 If X is a square matrix with X k X + I then X n X + 1 I for all n Z If n 0 then result is obivious If n 1 then (X ) 1 F k 1 X + F k 0 I 1X + I X Corresponding author address: ashokgodse01@gmailcom 100

2 Hence result is true for n 1 It can be shown by induction that A D Godase M B Dhakne / Int J Adv Appl Math Mech (1) (01) X n X + 1 I for all n Z Assume that X n X + 1 I prove thatx n+1 +1 X + I Consider +1 X + I ( X + 1 I )X + I (k X + I ) + X 1 X + X 1 X (X + 1 ) X (X n ) X n+1 Hence X n+1 +1 X + I By Induction X n X + 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 Y (k I X ) k I k X + X k I k X + k X + I k I k X + I k(k I X ) + X + I k Y + I Therefore Y k Y + I This shows that Y n Y + 1 I i e ( X 1 ) n (k I X ) + 1 I ( 1) n X n X + +1 I X n ( 1) n+1 X + ( 1) n +1 I Since F k n ( 1) n+1 F k n 1 ( 1) 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 + Corollary 1 k Let M 1 Since 1 0 then M n Fk n+1 1 M km + I M + 1 I Using Lemma 1 k Fk n Fk + n Fk n+1 1 for all n Z Corollary Let S k k k + k then S n n (k +) for every n Z Lemma L k n (k + )F k n ( 1)n for all n Z 101

3 On the properties of k-fibonacci k-lucas numbers Since d e t (S) 1 d e t (S n ) [d e t (S)] n ( 1) n Moreover since We get S n n (k +) d e t (S n ) L k n (k + )Fk n Thus it follows that L k n (k + )F k n ( 1)n for all n Z Lemma 3 +m + (k + ) F k m +m + F k m for all nm Z But Since S n+m S n S m S n+m n (k +) n +(k +) F k m F k m + n+m +m (k +)+m +m m F k m (k +)F k m (k +)[ F k m + +(k +) F k m +m + (k + ) F k m +m + F k m for all nm Z Lemma ( 1) m m (k + ) F k m ( 1) m m F k m for all nm Z Since S n m S n S m S n [S m ] 1 S n ( 1) m m F k m ( 1) m n (k +) n (k +) F k m F k m (k +)F k m m F k m (k +)F k m (k +)[ F k m (k +) F k m 10

4 A D Godase M B Dhakne / Int J Adv Appl Math Mech (1) (01) But S n m n m m (k +) m m ( 1) m m (k + ) F k m ( 1) m m F k m for all nm Z Lemma 5 ( 1) m m + +m ( 1) m m + +m for all nm Z By definition of the matrix S n it can be seen that S n+m + ( 1) m S n m n+m +( 1) m m +m +( 1) m m (k +)[+m +( 1) m m +m +( 1) m m On the other h S n+m + ( 1) m S n m S n S m + ( 1) m S n S m S n [S m + ( 1) m S m ] n n m (k +) (k +) m F k m m 0 (k +) 0 (k +)F k m + ( 1) m m F k m (k +)F k m ( 1) m m + +m ( 1) m m + +m for all nm Z Lemma 6 8F k x +y +z L k x L k y + F k x L k y + L k x F k y + (k + )F k x F k y 8L k x +y +z L k x L k y + (k + )[L k x F k y + F k x L k y + F k x F k y for all x y z Z 103

5 On the properties of k-fibonacci k-lucas numbers By definition of the matrix S n it can be seen that S x +y +z On the other h x +y +z F k x +y +z (k +)F k x +y +z L k x +y +z S x +y +z S x +y S z x +y F k x +y (k +)F k x +y L k x +y x +y +(k +)F k x +y F k x +y + L k x +y z (k +) (k +)[L k x +y +F k x +y L k x +y +(k +)F k x +y Using L k x +y L k x L k y + (k + )F k x F k y F k x +y L k y F k x + (k + )F k y L k x 8F k x +y +z L k x L k y + F k x L k y + L k x F k y + (k + )F k x F k y 8L k x +y +z L k x L k y + (k + )[L k x F k y + F k x L k y + F k x F k y for all x y z Z Theorem 1 L k x +y (k + )( 1) x +y +1 x L k x +y F k y +z (k + )( 1) x +z F k y +z ( 1)y +z L k z x for all x y z Z Consider matrix multiplication given below That is x y x +y F k y F k y +z Now (k +)F k x d e t x (k +)F k x L k x (k + )F k x ( 1)x x Q 0 Therefore we can write y F k y x 1 Q z (k +)F k x (k +)F k x L k x 1 x +y F k y +z x +y F k y +z L k y ( 1)x [ L k x +y (k + )F k x F k y +z ] x 10

6 Since We get A D Godase M B Dhakne / Int J Adv Appl Math Mech (1) (01) F k y ( 1)x [L k x +y L k y +x ] x L k y (k + )F k y ( 1)y [ L k x +y (k + )F k x F k y +z ] (k + ) [L k x +y L k y +x ] ( 1) y L k z x Using Lemma Lemma 6 L k z L k x +y (k + ) F k x +y F k y +z + (k + ) F k x F k y +z (k + )(L k x F k y +z L k x F k y +z L k x +y + F k z L k x +y ) ( 1)y L k z x L k x +y (k + )( 1) x +y +1 x L k x +y F k y +z (k + )( 1) x +z F k y +z ( 1)y +z L k z x for all x y z Z Theorem L k x +y ( 1)x +z x L k x +y L k y +z + ( 1) x +z L k y +z ( 1)y +z +1 (k + )F k z x for all x y z Z x z Now Consider matrix multiplication y F k y x d e t x (k +)F k x (k +) (k +)F k x (k +) Therefore for x z we can write Since We get y F k y x 1 P (k +) (k +)F k x (k +) x +y L k y +z (k + )( 1) x x P 0 ( if x z ) 1 x +y L k y +z (k +)F k x L k x L k y ( 1)x [ L k x +y F k x L k y +z ] x F k y ( 1)x [L k x +y L k y +x ] (k + ) x L k y (k + )F k y ( 1)y x +y L k y +z (k + )[ L k x +y F k x L k y +z ] [L k x +y L k y +x ] (k + )( 1) y F k z x Using Lemma Lemma 6 We obtain L k x +y ( 1)x +z x L k x +y L k y +z + ( 1) x +z L k y +z ( 1)y +z +1 (k + )F k z x for all x y z Z x z 105

7 On the properties of k-fibonacci k-lucas numbers Theorem 3 F k x +y L k x z F k x +y F k y +z + ( 1) x +z F k y +z ( 1)y +z F k z x for all x y z Z x z Now Consider matrix multiplication y Fk x d e t Fk x F k x F k x F k y Therefore for x z we get y F k y Fk x 1 R z Fk x +y F k y +z ( 1)z F k x z R 0 ( if x z ) F k x 1 Fk x +y F k y +z L k x F k x L k y ( 1)z [ F k x +y L k x F k y +z ] F k x z F k y ( 1)z [F k x +y F k y +x ] F k x z Now consider Fk x +y F k y +z [ F k x +y L k x F k y +z ] (k + )[F k x +y F k y +x ] ( 1) y F k x z Using Lemma Lemma 6 We get F k x +y L k x z F k x +y F k y +z + ( 1) x +z F k y +z ( 1)y +z F k z x for all x y z Z x z 3 Conclusions The conclusions arising from the work are as follows: Some new identities have been obtained for the k Fibonacci k Lucas sequences ACKNOWLEDGEMENTS The authors wish to thank two anonymous referees for their suggestionswhich led to substantial improvement of this paper References [1] ME Waddill Matrices Generalized Fibonacci Sequences The Fibonacci Quarterly 55(1) (197) [] T Koshy Fibonacci Lucas numbers with applications Wiley-Intersection Pub 001 [3] S Falcon A Plaza On the k Fibonacci numbers Chaos Solitons Fractals 5(3) (007) [] AF Horadam Basic Properties of a Certain Generalized Sequence of Numbers The Fibonacci Quarterly 3(3) (1965) [5] P Filipponi A F Horadam A Matrix Approach to Certain Identities The Fibonacci Quarterly 6() (1988) [6] H W Gould A History of the Fibonacci g-matrix a Higher-Dimensional Problem The Fibonacci Quarterly 19(3) (1981) [7] S Vajda Fibonacci Lucas numbers the Golden Section:Theory applications Chichester:Ellis Horwood 1989 [8] A F Horadam Jacobstal Representation of Polynomials The Fibonacci Quarterly 35() (1997) [9] G StrangIntroduction to Linear Algebra Wellesley-Cambridge:Wellesley MA Pub