See discussions, stats, author profiles for this publication at: http://wwwresearchgatenet/publication/7839139 Some New Properties for k-fibonacci k- Lucas Numbers using Matrix Methods RESEARCH JUNE 015 DOI: 101310/RG1173677 READS 15 1 AUTHOR: A D Godase V P College, Vaijapur 3 PUBLICATIONS 95 CITATIONS SEE PROFILE All in-text references underlined in blue are linked to publications on ResearchGate, letting you access read them immediately Available from: A D Godase Retrieved on: 09 November 015
Some New Properties for k Fibonacci k Lucas Numbers using Matrix Methods A D Godase 1 V P College, Vaijapur, Maharashtra, India mathematicsdept@vpcollegenet M B Dhakne Dr B A M University, Aurangabad, Maharashtra, India ijntindia@gmailcom Abstract In this paper, some new sums for k Fibonacci k Lucas numbers are derived proved k k + by using matrices S 0 k 1 k R + The sums we derived are not 1 0 encountered in the k Fibonacci k Lucas numbers literature Keywords: k-fibonacci sequence; k-lucas sequence;recurrence relation; Matrix algebra Mathematics Subject Classification (010): 11B39, 11B37,11B99 1 Introduction Many authors defined generalized Fibonacci sequences by varying initial conditions recurrence relation 1 1This paper represents an interesting investigation about some special relations between matrices k Fibonacci numbers,k Lucas numbersthis investigation is valuable to obtain new k Fibonacci,k Lucas identities by different methodsthis paper contributes to k Fibonacci,k Lucas numbers literature, encourage many researchers to investigate the properties of such number sequences Some Preliminary Results In this section, some definitions preliminary results are given which will be used in this paper Definition 1 The k Fibonacci sequence {F k,n } n N is defined as, F k,n+1 kf k,n + F k,n 1, with F k,0 0, F k,1 1,for n 1 1 Corresponding Author:mathematicsdept@vpcollegenet 1
Definition The k Lucas sequence {L k,n } n N is defined as, L k,n+1 kl k,n + L k,n 1, with L k,0, L k,1 k,for n 1 Characteristic equation of the initial recurrence relation is, characteristic roots are r kr 1 0, (1) r 1 k + k + r k k +, Characteristic roots verify the properties, r 1 r k +, () r 1 + r k, (3) Binet forms for F k,n L k,n are r 1 r 1 () F k,n rn 1 rn r 1 r (5) L k,n r n 1 + r n (6) From the definition of the k Fibonacci numbers,one may deduce the value of any k-fibonacci number by simple substitution on the corresponding F k,n For example,the seventh element of the Fibonacci sequence, F,7 is 573 By doing k 1; ; 3; the respective k-fibonacci sequences are obtained Sequence {F 1,n } is the classical Fibonacci sequence {F,n } is the Pell sequence It is worthy to be noted that only the first 10 k Fibonacci sequences are referenced in The On- Line Encyclopedia of Integer Sequences 11 with the numbers given in Table 1 For k even with 1 k 6 sequences {F k,n } are referenced without the first term F k,0 0 in 11 The first 11k-Fibonacci sequences as numbered in The On-Line Encyclopedia of Integer Sequences 10: F 1,n F,n F 3,n F,n F 5,n F 6,n F 7,n F 8,n F 9,n F 10,n F 11,n A00005 A00019 A006190 A001076 A05918 A005668 A0513 A0105 A099371 A0101 A09666
The most commonly used matrix in relation to the recurrence relation(1) is k 1 M 1 0 (7) which for k 1 reduces to the ordinary Q-matrix,8 In this paper we define more general matrices M k (n, m), T k (n), S k (n, m) We use these matrices to develop various summation identities involving terms from the sequences F k,n L k,n Several identities for F k,n L k,n are proved using Binet forms in 1,3 5Some of these are listed below F k,n+1 + F k,n 1 L k,n (8) F k,n+1 + L k,n 1 F k,n (9) F k,n ( 1) n F k,n (10) F k,m+n ( 1) m L k,n m F k,m L k,n (11) L k,m+n ( 1) m L k,n m F k,m F k,n (1) F k,m+n F k,n m F k,n ( 1)n m+1 F k,m (13) L k,m+n L k,n m L k,n ( 1)n m F k,m (1) F k,m+n F k,r+m ( 1) m F k,n F k,r F k,m F k,n+r+m (15) L k,mn L k,n + F k,mn F k,n L k,(m+1)n (16) F k,mn L k,n + L k,mn F k,n F k,(m+1)n (17) L k,m+n + ( 1)m 1 L k,n F k,n+mf k,m (18) L k,m+n L k,n + ( 1) m+1 L k,n m L k,n F k,n F k,m (19) F k,m+rn F k,n+m + ( 1) m+1 F k,rn F k,n F k,(r+1)n+m F k,m (0) In 1matrix M is generalized using PMI M n Fk,n+1 F k,n where n is an integer F k,n F k,n 1 (1) 3
Lemma 3 If X is a square matrix with X kx +I, then X n F k,n X +F k,n 1 I, for all n Z Corollary Let,M k 1 1 0, then M n Fk,n+1 F k,n F k,n F k,n 1 Proof Since, M km + I F k,n M + F k,n 1 I(UsingLemma1) kfk,n F k,n Fk,n 1 0 + F k,n 0 0 F k,n 1 Fk,n+1 F k,n, F k,n F k,n 1 for all n Z Corollary 5 Let,S k k k Lk,n, then S n F k,n ( )F k,n L k,n,for every n Z Lemma 6 L k,n F k,n ( 1)n for all,n Z Proof Since, Moreover since, We get, Thus it follows that, det(s) 1 det(s n ) det(s) n ( 1) n S n Lk,n F k,n det(s n ) L k,n F k,n L k,n F k,n L k,n F k,n ( 1)n, for all n Z
Lemma 7 L k,n+m L k,n L k,m + F k,n F k,m F k,n+m F k,n L k,m + L k,n F k,m for all n, m Z Proof Since, S n+m S n S m Lk,n F k,n Lk,m F k,m F k,n L k,n F k,m L k,m Lk,n L k,m + F k,n F k,m L k,n F k,m +F k,n L k,m L k,n F k,m +F k,n L k,m L k,n L k,m + F k,n F k,m But, Gives, Lemma 8 S n+m Lk,n+m F k,n+m F k,n+m L k,n+m L k,n+m L k,n L k,m + F k,n F k,m F k,n+m F k,n L k,m + L k,n F k,m ( 1) m L k,n m L k,n L k,m F k,n F k,m ( 1) m F k,n m F k,n L k,m L k,n F k,m for all n, m Z for all n, m Z Proof Since, S n m S n S m S n S m 1 Lk,m S n ( 1) m Lk,n ( 1) m F k,m F k,m L k,m F k,n Lk,m F k,m F k,n L k,n F k,m L k,m Lk,n L k,m F k,n F k,m L k,n F k,m F k,n L k,m L k,n F k,m F k,n L k,m L k,n L k,m F k,n F k,m 5
But, Gives, S n m Lk,n m F k,n m F k,n m L k,n m ( 1) m L k,n m L k,n L k,m F k,n F k,m ( 1) m F k,n m F k,n L k,m L k,n F k,m for all n, m Z Lemma 9 ( 1) m L k,n m + L k,n+m L k,n L k,m ( 1) m F k,n m + F k,n+m F k,n L k,m for all n, m Z Proof By definition of the matrix S n,it can be seen that S n+m + ( 1) m S n m Lk,n+m +( 1) m L k,n m F k,n+m +( 1) m F k,n m F k,n+m +( 1) m F k,n m L k,n+m +( 1) m L k,n 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 Lk,n F k,n Lk,m F k,n L k,n F k,m Lk,n F k,n F k,n L k,n Lk,m L k,n F k,n L k,m F k,n L k,m L k,m L k,n Lk,m 0 F k,m L k,m 0 L k,m Lk,m + ( 1) m F k,m F k,m L k,m Gives, ( 1) m L k,n m + L k,n+m L k,n L k,m ( 1) m F k,n m + F k,n+m F k,n L k,m for all n, m Z 6
Lemma 10 8F k,x+y+z L k,x L k,y F k,z + F k,x L k,y L k,z + L k,x F k,y L k,z + F k,x F k,y F k,z 8L k,x+y+z L k,x L k,y L k,z + L k,x F k,y F k,z + F k,x L k,y F k,z + F k,x F k,y L k,z for all x, y, z Z Proof By definition of the matrix S n,it can be seen that S x+y+z Lk,x+y+z F k,x+y+z F k,x+y+z L k,x+y+z On the other h, S x+y+z S x+y S z Lk,x+y F k,x+y Lk,z F k,z F k,x+y L k,x+y F k,z L k,z Lk,x+y L k,z + F k,x+y F k,z L k,x+y F k,z +F k,x+y L k,z L k,z F k,x+y +F k,z L k,x+y L k,x+y L k,z + F k,x+y F k,z Using, L k,x+y L k,x L k,y + F k,x F k,y F k,x+y L k,y F k,x + F k,y L k,x Gives, 8F k,x+y+z L k,x L k,y F k,z + F k,x L k,y L k,z + L k,x F k,y L k,z + F k,x F k,y F k,z 8L k,x+y+z L k,x L k,y L k,z + L k,x F k,y F k,z + F k,x L k,y F k,z + F k,x F k,y L k,z for all x, y, z Z Theorem 11 L k,x+y ( 1)x+y+1 F k,z x L k,x+y F k,y+z ( 1) x+z F k,y+z ( 1)y+z L k,z x for all x, y, z Z Proof Consider matrix multiplication given below Lk,x F k,z F k,x L k,z Lk,y F k,y Lk,x+y F k,y+z 7
Now, det Lk,x F k,z F k,x L k,z L k,xl k,z F k,x F k,z ( 1)x L k,z x Q 0 Therefore we can write Lk,y F k,y 1 Q Lk,x F k,x F k,z L k,z Lk,z F k,x F k,z L k,x 1 Lk,x+y F k,y+z Lk,x+y F k,y+z Gives, Since, We get, L k,y ( 1)x L k,z L k,x+y F k,x F k,y+z L k,z x F k,y ( 1)x L k,x F k,z+y F k,z L k,y+x L k,z x L k,y F k,y ( 1)y L k,z L k,x+y F k,x F k,y+z ( ) L k,x F k,z+y F k,z L k,y+x ( 1) y L k,z x Using Lemma Lemma 6, (L k,z L k,x+y L k,zf k,x+y F k,y+z + Fk,x F k,y+z ) (L k,x F k,y+z L k,xf k,z F k,y+z L k,x+y + Fk,z L k,x+y ) ( 1) y L k,z x Gives, L k,x+y ( 1)x+y+1 F k,z x L k,x+y F k,y+z ( 1) x+z F k,y+z ( 1)y+z L k,z x for all x, y, z Z 8
Theorem 1 L k,x+y ( 1)x+z L k,z x L k,x+y L k,y+z + ( 1) x+z L k,y+z ( 1)y+z+1 F k,z x for all x, y, z Z, x z Proof Consider matrix multiplication, Lk,x L k,z F k,x F k,z Lk,y F k,y Lk,x+y L k,y+z Now, det Lk,x L k,z F k,x F k,z ( 1)x F k,z x P 0, (ifx z) Therefore for x z, we can write Lk,y F k,y 1 P Lk,x F k,x L k,z F k,z (k +)F k,z L k,z 1 Lk,x+y L k,y+z F k,x Lk,x+y L k,x L k,y+z Gives, Since, We get, L k,y ( 1)x F k,z L k,x+y F k,x L k,y+z F k,z x F k,y ( 1)x L k,x L k,z+y L k,z L k,y+x F k,z x L k,y F k,y ( 1)y F k,z L k,x+y F k,x L k,y+z L k,x L k,z+y L k,z L k,y+x ( 1) y F k,z x Using Lemma Lemma 6,We obtain L k,x+y ( 1)x+z L k,z x L k,x+y L k,y+z + ( 1) x+z L k,y+z ( 1)y+z+1 F k,z x for all x, y, z Z, x z 9
Theorem 13 F k,x+y L k,x zf 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 Proof Consider matrix multiplication, Fk,x F k,z F k,x L k,z Lk,y F k,y Fk,x+y F k,y+z Now, det Fk,x F k,z F k,x L k,z ( 1)z F k,x z T 0, (ifx z) Therefore for x z, we get, Lk,y F k,y 1 T Fk,x F k,x F k,z L k,z Lk,z L k,x F k,z F k,x 1 Fk,x+y F k,y+z Fk,x+y F k,y+z Gives, Now consider,, L k,y ( 1)z L k,z F k,x+y L k,x F k,y+z F k,x z F k,y ( 1)z F k,x F k,z+y F k,z F k,y+x F k,x z L k,z F k,x+y L k,x F k,y+z F k,x F k,z+y F k,z F k,y+x ( 1) y F k,x z Using Lemma Lemma 6,We get F k,x+y L k,x zf 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 10
3 Main Results Theorem 31 Let n N m, t Z with m 0,Then L k,mj+t L k,t L k,mn+m+t + ( 1) m (L k,mn+t L k,t m ) 1 + ( 1) m L k,m (1) F k,mj+t F k,t F k,mn+m+t + ( 1) m (F k,mn+t F k,t m ) 1 + ( 1) m L k,m () Proof It is known that, If det(i S m ) 0, then we can write I (S m ) n+1 (I S m ) (S m ) j 1 I (S m ) 1 (I (S m ) n+1 )S t (S mj+t ) From lemma () det(i S m ) (1 L k,m ) (F k,m) (L k,mj+t) (F k,mj+t) 1 (F 1 k,mj+t) (L k,mj+t) 1 + ( 1) m L k,m 0 For m 0,If we take D 1 + ( 1) m L k,m,then we get (I S m ) 1 1 1 (L k,m) (F k,m) (F D k,m ) 1 (L k,m) 1 (1 (L k,m) )I + (F k,m) R D Thus it is seen that, (I S m ) 1 (S t S mn+m+t ) 1 D (1 (L k,m) )I + (F k,m) R (S t S mn+m+t ) (3) (I S m ) 1 (S t S mn+m+t ) 1 D (1 (L k,m) )(S t S mn+m+t ) + (F k,m) R(S t S mn+m+t ) () By Corollary () (3) we get R(S t S mn+m+t ) (F k,t F k,mn+m+t ) (L k,t L k,mn+m+t ) (L k,t L k,mn+m+t ) (F k,t F k,mn+m+t ) (5) 11
Using in 3, we obtain (I S m ) 1 (S t S mn+m+t ) 1 D (1 L k,m ) + F k,m D (L k,t L k,mn+m+t ) (F k,t F k,mn+m+t ) (F k,t F k,mn+m+t ) (L k,t L k,mn+m+t ) (F k,t F k,mn+m+t ) (L k,t L k,mn+m+t ) (L k,t L k,mn+m+t ) (F k,t F k,mn+m+t ) Thus from () it follows that L k,mj+t 1 D 1 D Using Lemma(6) (1 L k,m )(L k,t L k,mn+m+t ) + F k,m (F k,t F k,mn+m+t L k,t L k,mn+m+t L k,ml k,t F k,m F k,t + L k,ml k,mn+m+t F k,m F k,mn+m+t On the other h, by (3) () we get F k,mj+t 1 D 1 D Using Lemma(6) it follows that L k,mj+t L k,t L k,mn+m+t + ( 1) m (L k,mn+t L k,t m ) 1 + ( 1) m L k,m (1 L k,m )(F k,t F k,mn+m+t ) + F k,m (L k,t L k,mn+m+t F k,t F k,mn+m+t L k,mf k,t F k,m L k,t + L k,mf k,mn+m+t F k,m L k,mn+m+t F k,mj+t F k,t F k,mn+m+t + ( 1) m (F k,mn+t F k,t m ) 1 + ( 1) m L k,m Theorem 3 Let n N m, t Z,Then ( 1) j L k,mj+t L k,t L k,mn+m+t + ( 1) m (L k,mn+t L k,t m ) 1 + ( 1) m (6) L k,m ( 1) j F k,mj+t F k,t F k,mn+m+t + ( 1) m (F k,mn+t F k,t m ) 1 + ( 1) m (7) L k,m 1
Proof We prove the theorem in two cases by taking n as an even odd natural number Case:1 If n is an even natural number I + (S m ) n+1 (I + S m ) (S m ) j Since, det(i + S m ) 1 + ( 1) m + L k,m 0, then using Lemma6 we can write 1 I+(S m ) 1 (I+(S m ) n+1 )S t ( 1) j (S mj+t ) 1 If we take d 1 + ( 1) m + L k,m,then we get (I + S m ) 1 1 1 + (L k,m) d 1 (1 + (L k,m) d Thus it is seen that, (I + S m ) 1 (S t + S mn+m+t ) 1 d By Corollary () R S + S 1 we get R(S t + S mn+m+t ) Using (8) in (7), we obtain ( 1)j (L k,mj+t ) ( 1)j (F k,mj+t ) 1 (F k,m) ( F k,m ) 1 + (L k,m) )I (F k,m) R (1 + (L k,m) (I + S m ) 1 (S t + S mn+m+t ) 1 d (1 + L k,m ) Thus from (7) it follows that ( 1) j L k,mj+t 1 d 1 d ( 1)j (F k,mj+t ) ( 1)j (L k,mj+t ) )I (F k,m) R (S t + S mn+m+t ) (F k,t+f k,mn+m+t ) (L k,t+l k,mn+m+t ) (L k,t +L k,mn+m+t ) (F k,t+f k,mn+m+t ) F k,m d (L k,t+l k,mn+m+t ) (F k,t+f k,mn+m+t ) (F k,t +F k,mn+m+t ) (L k,t +L k,mn+m+t ) (F k,t+f k,mn+m+t ) (L k,t+l k,mn+m+t ) (L k,t +L k,mn+m+t ) (F k,t+f k,mn+m+t ) (1 + L k,m )(L k,t + L k,mn+m+t ) F k,m (F k,t + F k,mn+m+t L k,t + L k,mn+m+t + L k,ml k,t F k,m F k,t + (L k,m L k,mn+m+t F k,m F k,mn+m+t ) Using Lemma(6) fact that d 1 + ( 1) m + L k,m we obtain ( 1) j L k,mj+t L k,t + L k,mn+m+t + ( 1) m (L k,mn+t + L k,t m ) 1 + ( 1) m + L k,m (8) (9) 13
Similarly it can be easily seen that by using Lemma(6) ( 1) j F k,mj+t F k,t + F k,mn+m+t + ( 1) m (F k,mn+t + F k,t m ) 1 + ( 1) m + L k,m Case: If n is an odd natural number Since n is an odd natural number, we get ( 1) j L k,mj+t 1 ( 1) j L k,mj+t L k,mn+t (30) Since n is an odd natural number then (n 1) is an even Then taking (n 1) in (5) using it in (9), it follows that, ( 1) j L k,mj+t L k,t + L k,mn+t + ( 1) m (L k,mn m+t + L k,t m ) 1 + ( 1) m L k,mn+t + L k,m Using Lemma(7) we obtain L k,t + ( 1) m (L k,mn m+t + L k,t m ) ( 1) m L k,mn+t L k,m L k,mn+t 1 + ( 1) m + L k,m ( 1) j L k,mj+t L k,t + L k,mn m+t + ( 1) m (L k,t m L k,mn+t ) 1 + ( 1) m (31) + L k,m In similar way it can be seen that, ( 1) j F k,mj+t Using (5) in (3), it follows that, 1 ( 1) j F k,mj+t F k,mn+t (3) ( 1) j F k,mj+t F k,t + F k,mn+t + ( 1) m (F k,mn m+t + L k,t m ) 1 + ( 1) m F k,mn+t + L k,m Using Lemma(7) we get expected formula F k,t + ( 1) m (F k,mn m+t + F k,t m ) ( 1) m F k,mn+t L k,m L k,mn+t 1 + ( 1) m + L k,m ( 1) j F k,mj+t F k,t L k,mn+m+t + ( 1) m (F k,t m F k,mn+t ) 1 + ( 1) m (33) + L k,m Conclusion: Some new summation identities have been obtained for the k Fibonacci k Lucas sequences 1
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