properties of k- fibonacci sequence using matrix method

MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

PROPERTIES OF K- FIBONACCI SEQUENCE USING MATRIX METHOD M. B. Dhakne Dr. B. A. M. University, Aurangabad, Maharashtra, India [email protected]

A. D. Godase 1 V. P. College, Vaijapur, Maharashtra, India [email protected]

Abstract-The intention of this paper is to obtain some identities of a k-Fibonacci sequence. So in the present paper first and foremost we defined matrices Mk (n, m), Tk (n), Sk (n, m), An , E, Yn , Wn , Gn and Hn for k-Fibonacci and k- Lucas sequences and at hindmost by using matrix methods some identities are deduced for the k-Fibonacci sequence. Keywords: k-Fibonacci sequence, k-Lucas sequence, Recurrence relation, Matrices. Mathematics Subject Classification: 11B39, 11B83 1. I NTRODUCTION The Fibonacci sequence {Fn }∞ n=1 is the integer sequence and is well-known example of second order recurrence and defined recurrently by Fk,n+1 = k · Fk,n + Fk,n−1 , with F0 = 0, F1 = 1, for n ≥ 1. Some properties and generalizations

is of Fibonacci sequence have been documented in [14], [15], [16], [17], [21]. Fibonacci sequence has been generalized in several ways. Some authors have maintained the recurrence relation and changed the first two terms of the sequence. While others have maintained the first two terms of the sequence and changed the recurrence relation. But some others have changed both recurrence relation and first two terms of the sequence. One of the main and known generalization of Fibonacci numbers is the k - Fibonacci and k - Lucas numbers introduced by Falcon and Plaza in [11] and these are defined by the recurrence relation as Fk,n+1 = k · Fk,n + Fk,n−1 , with Fk,0 = 0, Fk,1 = 1, for n ≥ 1 and Lk,n+1 = k · Lk,n + Lk,n−1 , with Lk,0 = 2, Lk,1 = k, for n ≥ 1 respectively. Some identities of these sequences have been documented in [3], [6], [7], [12], [13]. 2 Characteristic equation of the √ initial recurrence√relation for k - Fibonacci sequence is, r − k · r − 1 = 0 and it‘s Characteristic roots are r1 =

k+

k2 +4 2

and r2 =

(1.1)

k−

k2 +4 . 2

These characteristic roots verify the properties p √ r1 − r2 = k2 + 4 = ∆ = δ

(1.2)

r1 + r2 = k

(1.3)

r1 .r2 = −1

The most commonly used matrix in relation to the recurrence this relation is   k 1   (1.4) M=  1 0 Matrix M is generalized using induction to  (1.5)

 Mn = 

Fk,n+1

Fk,n



Fk,n

Fk,n−1

 

where n is an integer, which for k = 1 reduces to the ordinary Q - matrix.

1Corresponding Author:[email protected]

1

MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

Now in the present paper we defined matrices Mk (n, m), Tk (n), Sk (n, m), An , E, Yn , Wn , Gn and Hn for k-Fibonacci and k- Lucas sequences and by using matrix methods many interesting identities are obtained for the k-Fibonacci sequence. 2. I DENTITIES OF k F IBONACCI AND k L UCAS NUMBERS Definition 2.1. Define 2 × 2 matrix Mk (n, m) as   Fk,n+m (−1)m+1 Fk,n   (2.6) Mk (n, m) =   Fk,n (−1)m+1 Fk,n−m where m and n are integers. Theorem 2.1. Let Mk (n, m) be a matrix as in (2.6) then  Fk,rn+m r r  (2.7) Mk (n, m) = Fk,m  Fk,rn

(−1)m+1 Fk,rn



(−1)m+1 Fk,rn−m

  

Proof. Theorem can easily proved by using Mathematical induction on n. Definition 2.2. Define 2 × 2 matrix Tk,n as (2.8)



Lk,n

Fk,n



 Tk,n = 

∆Fk,n

Lk,n

 

where n is an integer. Theorem 2.2. Let Tk,n be a matrix as in (2.8) then   m Tk,n = 2m−1 

(2.9)

Lk,nm

Fk,nm



∆Fk,nm

Lk,nm

  

Proof. Theorem can easily proved by using Mathematical induction on n. Definition 2.3. Define 2 × 2 matrix Sk (n, m) as   Sk (n, m) = 

(2.10)

Lk,n+m

(−1)m+1 Lk,n



Lk,n

(−1)m+1 Lk,n−m

 

where n, m are integers. Theorem 2.3. Let Sk (n, m) be a matrix as in (2.10) then for all integer r   Fk,2rn+m (−1)m+1 Fk,2rn  2r−1 r  (2.11) Sk (n, m)2r = Fk,m ∆  Fk,2rn (−1)m+1 Fk,2rn−m  (2.12)

Sk (n, m)2r−1

2r−2 r−1  = Fk,m ∆ 

Lk,2(r−1)n+m

(−1)m+1 Lk,2(r−1)n



Lk,2(r−1)n

(−1)m+1 Lk,2(r−1)n−m

 

Proof. Theorem can easily proved by using Mathematical induction on r. Theorem 2.4. For n, m, r ≥ 1

2



MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

i=r

(2.13)

(ri ) (−1)n(r−i)+1 ∆

∑

i−1 2

i=r i Fk,n Lk,in+m +

i=0,i−even i=2r

(2.14)

∑

2r i




2i ∆

2r−i 2

i=2r−1 2r−i Fk,in+m − Fk,n

i=0,i−even i=2r+1

(2.15)

∑

2r+1 i

i

i (ri ) (−1)n(r−i) ∆ 2 Fk,n Fk,in+m

∑

=

Fk,2nr+m

2r−i Lk,in+m Fk,n

=

2r Fk,m Lk,n

2r+1−i Lk,in+m Fk,n

=

2r+1 Fk,m ∆Lk,n

i=0,i−odd




2i ∆

2r+3−i 2

2r i

∑




2i ∆

2r−1−i 2

i=0,i−odd i=2r−1 2r+1−i Fk,in+m − Fk,n

2r+1 i

∑




2i ∆

2r+2−i 2

i=0,i−even

i=0,i−odd

Proof. Proof is obvious from equations (2.11) and (2.12).



Definition 2.4. Define 3 × 3 matrix A as  (2.16)

k2 + 1 k2 + 1 −1

  A=  

1

0

0

1

Definition 2.5. Define 3 × 3 matrix An as  i=n+1 2 ∑i=0 Fk,i   i=n 2  (2.17) An =  ∑i=0 Fk,i   i=n−1 F 2 ∑i=0 k,i

Fk,n Fk,n+2 Fk,n−1 Fk,n+1 Fk,n−2 Fk,n



 0     0

2 − ∑i=n i=0 Fk,i



 2   − ∑i=n−1 F k,i  i=0  2  − ∑i=n−2 F k,i i=0

where n, k are integers. Theorem 2.5. Let the matrices A and An have the form (2.16) and (2.17), respectively. Then for all integer n > 1 An = An

(2.18)

Proof. Using principal of Mathematical induction on n If n = 2, then  4 k + 3k2 + 1 k2 (k2 + 2) −k2 − 1   k2 + 1 k2 + 1 −1 A2 =    1 0 0 

2 ∑i=3 i=0 Fk,i

  i=2 2  =  ∑i=0 Fk,i   i=1 F 2 ∑i=0 k,i

Fk,2 Fk,4 Fk,1 Fk,3 Fk,0 Fk,2

2 − ∑i=2 i=0 Fk,i



 2  − ∑i=1 i=0 Fk,i    i=0 2  − ∑i=0 Fk,i

= A2 Hence result is true for n = 2, now suppose that the result is true for n, (n > 2) An = An

3

     

MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

Now consider An+1

n = A  A i=n+1 2 ∑i=0 Fk,i   i=n 2  =  ∑i=0 Fk,i   i=n−1 F 2 ∑i=0 k,i



Fk,n Fk,n+2

2 − ∑i=n i=0 Fk,i

Fk,n−1 Fk,n+1

2 − ∑i=n−1 i=0 Fk,i

Fk,n−2 Fk,n

2 − ∑i=n−2 i=0 Fk,i

i=n 2 2 (k+ 1) ∑i=n+1 i=0 Fk,i − ∑i=0 Fk,i

2 − ∑i=n+1 i=0 Fk,i

i=n−1 2 2 (k+ 1) ∑i=n i=0 Fk,i − ∑i=0 Fk,i

2 − ∑i=n i=0 Fk,i

i=n−2 2 2 (k+ 1) ∑i=n−1 i=0 Fk,i − ∑i=0 Fk,i

2 − ∑i=n−1 i=0 Fk,i

i=n+1 2 (k2 + 1) ∑i=0 Fk,i + Fk,n Fk,n+2

  2 i=n 2  =  (k + 1) ∑i=0 Fk,i + Fk,n−1 Fk,n+1   (k2 + 1) i=n−1 F 2 + F ∑i=0 k,n−2 Fk,n k,i

  k2 + 1 k2 + 1 −1      1 0 0       0 1 0       

Using i=n+1

(k2 + 1)Fk,n Fk,n+1 + kFk,n−1 Fk,n+1

= k

∑

2 Fk,i

i=0

An+1

(k2 + 1)Fk,n Fk,n+1 − Fk,n−1 Fk,n =  i=n+2 2 Fk,n+1 Fk,n+3 ∑i=0 Fk,i   i=n+1 2  Fk,n Fk,n+2 =  ∑i=0 Fk,i   2 Fk,n−1 Fk,n+1 ∑i=n i=0 Fk,i

kFk,n Fk,n+2 2 − ∑i=n+1 i=0 Fk,i 2 − ∑i=n i=0 Fk,i 2 − ∑i=n−1 i=0 Fk,i

      

= An+1  Let the matrix A have the form (2.16) then  F k,n+1 Fk,n+2  k   F F k,n k,n+1  (2.19) An =  k    Fk,n−1 Fk,n  k

Fk,n Fk,n+2 Fk,n−1 Fk,n+1 Fk,n−2 Fk,n

Fk,n Fk,n+1 k Fk,n−1 Fk,n − k Fk,n−2 Fk,n−1 − k −

With det(A) = −1 The characteristic equation of the matrix A is λ 3 + (−k2 − 1)λ 2 + (−k2 − 1)λ + 1 = 0 Thus the eigenvalues of the matrix A are λ1 λ2 λ3

1 2 1 √ k + k ∆+1 2 2 11 2 1 √ = k − k ∆+1 2 2 = −1 =

Using (1.2) and (1.3), we can easily obtain λ1

= r12

λ2

= r22

4

         

MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

Definition 2.6. Define the 3 × 3 matrix B as follows 

λ12

λ22

  λ 1 B=   1

λ32



 λ3     1

λ2 1

3

detB = k(∆) 2 Let 

λ12

  λ2 C=B =  2  2 λ3

λ3

λ1n−i+3



  λ n−i+3 Di =   2  n−i+3 λ3

    





 1     1

λ2

T

and

1

λ1

Theorem 2.6. Let the matrix An = (ai, j ) have the form (2.17). Then for all i, j such that 1 ≤ i, j ≤ 3 (i)

ai, j = Definition 2.7. Define 4 × 4 matrix E as follows 

(2.20)

1

det(C j ) det(C)

0

0

0



   1 k2 + 1 k2 + 1 −1      E =  1 0 0   0     0 0 1 0

where n, k are integers. Theorem 2.7. Let the matrix E as in (2.20), then for n > 2  1 0   1 i=n Fk,n+1 Fk,n+2   k ∑i=0 Fk,i Fk,i+1 k   n Fk,n Fk,n+1 (2.21) E =  1 i=n−1  ∑i=0 Fk,i Fk,i+1  k k   1 Fk,n−1 Fk,n  ∑i=n−2 Fk,i Fk,i+1 k i=0 k

0 Fk,n Fk,n+2 Fk,n−1 Fk,n+1 Fk,n−2 Fk,n

Proof. Theorem can easily proved by using Mathematical induction on n. Theorem 2.8. For n > 0 i=n

(2.22)

∑ Fk,i Fk,i+1 =

2 Fk,n+1 + Fk,n Fk,n+2 − 1

2k2

i=0

5

0



Fk,n Fk,n+1 k Fk,n−1 Fk,n − k Fk,n−2 Fk,n−1 − k

           

−



MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

Proof. Proof is obvious from (2.21) Definition 2.8. Define 4 × 4 matrix 

 2 Fk,n

   −Fk,n Fk,n+3  Yn =  2  Fk,n+3    Fk,n Fk,n+3

(2.23)

−Fk,n Fk,n+3

2 Fk,n+3

Fk,n Fk,n+3



2 Fk,n+3

Fk,n+3 Fk,n

2 Fk,n

Fk,n Fk,n+3

2 Fk,n

−Fk,n Fk,n+3

2 Fk,n

−Fk,n Fk,n+3

2 Fk,n+3

        

Where n, k are integers Theorem 2.9. For any positive integer n 4  2 2 det(Yn ) = − Fk,n + Fk,n+3 Proof. Let x = Fk,n y = Fk,n+3 Then

2 x −xy y2 xy 2 −xy y2 xy x det(Yn ) = 2 xy x2 −xy y 2 2 xy x −xy y

By interchanging C2 and C4 , C3 and C4 , R3 and R4 , changing the sign of C3 and R4 , we get 2 x xy xy y2 −xy x2 −y2 xy det(Yn ) = − 2 2 xy y −x −xy 2 −y xy xy −x2 = − x2 + y2


4

Finally theorem follows  2 4 2 det(Yn ) = − Fk,n + Fk,n+3  Definition 2.9. Define 5 × 5 matrix (2.24) 2 Fk,n Fk,n+2 + k2 Fk,n+1

  Fk,n Fk,n+2 + k2 F 2 k,n+1    0 Wn =    2  Fk,n Fk,n+2 + k2 Fk,n+1   2 2

−(Fk,n Fk,n+2 + k Fk,n+1 )

2 Fk,n

2 k2 Fk,n+1

2 Fk,n+2

Fk,n (kFk,n+1 + Fk,n+2

−kFk,n+1 Lk,n+1

(kFk,n+1 − Fk,n )Fk,n+2

2kFk,n+1 Fk,n+2

2Fk,n Fk,n+2

−2kFk,n Fk,n+1

0

−Fk,n (kFk,n+1 + Fk,n+2 )

kFk,n+1 Lk,n+1

−Fk,n+2 (kFk,n+1 − Fk,n )

2 Fk,n Fk,n+2 + k2 Fk,n+1

2 Fk,n

2 k2 Fk,n+1

2 Fk,n+2

2 Fk,n Fk,n+2 + k2 Fk,n+1

where n, k are integers.

6

2 −(Fk,n Fk,n+2 + k2 Fk,n+1 )

 

2  Fk,n Fk,n+2 + k2 Fk,n+1 

       

MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

Theorem 2.10. Let Wn be a 5 × 5 matrix as in (2.24), then  5 2 (2.25) det(Wn ) = −32 k2 Fk,n+1 + Fk,n (kFk,n+1 + Fk,n )

Proof. Put Fk,n kFk,n+1 Fk,n+2

= x = y = z

Therefore, xz + y2 x2 y2 z2 −(xz + k2 y2 ) xz + y2 x(y + z) −y(x + z) (y − x)z xz + y2 0 2yz 2xz −2xy 0 det(Wn ) = xz + y2 2 −x(y + z) y(x + z) −z(y − x) xz + y −(xz + y2 ) x2 y2 z2 xz + y2 1 x2 y2 z2 −1 1 x(y + z) −y(x + z) (y − x)z 1 2yz 2xz −2xy 0 2 2 0 = (xz + y ) 1 −x(y + z) y(x + z) −z(y − x) 1 2 2 2 −1 x y z 1 Making the row-column transformations(C5 +C1 → C1 ), (R2 + R4 → R4 ) and (−R1 + R5 → R5 ), yields 0 x2 y2 z2 −1 2 x(y + z) −y(x + z) (y − x)z 1
2 2yz 2xz −2xy 0 det(Wn ) = xz + y2 0 4 0 0 0 −2 0 0 0 0 2 x2 y2 z2 2 2 x(y + z) −y(x + z) (y − x)z = −16(xz + y ) yz xz −xy Using, z = y + x x2 y2 (x + y)2 2 2 2 2 det(Wn ) = −16(xz + y2 )2 x + 2xy −y − 2xy (y − x )z 2 y + xy x2 + xy −xy

7

MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

Making the row transformation (R1 + R3 → R3 ) x2 y2 (x + y)2 2 2 3 x2 + 2xy −y2 − 2xy (y2 − x2 )z det(Wn ) = −16(x + y + xy) 1 1 1 = −32(x2 + y2 + xy)5 Finally theorem follows  5 2 det(Wn ) = −32 k2 Fk,n+1 + Fk,n (kFk,n+1 + Fk,n )  Definition 2.10. Define n × n super- diagonal matrix Gn as  2 k + 1 k2 + 1 −1 0 0 ..........   −1 k2 + 1 k2 + 1 −1 0 ..........     0 −1 k2 + 1 k2 + 1 −1 ..........   (2.26) Gn =   .......... .......... .......... .......... .......... ..........   0 .......... .......... 0 −1 k2 + 1    0 .......... .......... 0 0 −1

0



      0   ..........     k2 + 1    2 k +1 0

n×n

where n, k are integers. Theorem 2.11. For n > 1 i=n+1

det(Gn ) =

(2.27)

∑

2 Fk,i

i=0

where i=2 2 det(G1 ) = ∑ Fk,i = k2 + 1 i=0

Proof. Using the Laplace expansion of a determinant gives (2.28)

det(Gn ) = (k2 + 1)det(Gn − 1) + (k2 + 1)det(Gn − 2) − det(Gn − 3)

(2.29)

2 det(G1 ) = ∑ Fk,i

i=2 i=0

i=3 2 det(G2 ) = ∑ Fk,i

(2.30)

i=0

i=4 2 det(G3 ) = ∑ Fk,i

(2.31)

i=0

By (2.28) - (2.31) and using i=n+1

(k2 + 1)

∑

i=0

i=n

i=n−1

2 2 − Fk,i + (k2 + 1) ∑ Fk,i i=0

∑

i=0

i=n+1

det(Gn ) =

∑

i=0

8

2 Fk,i

i=n+2 2 Fk,i =

∑

i=0

2 Fk,i

MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

 Definition 2.11. Define n × n matrix Hn as  2 k + 1 k2 + 1 −1 0   −1 k2 + 1 k2 + 1 −1     0 −1 k2 + 1 k2 + 1  (2.32) Hn =   .......... .......... .......... ..........    0 .......... .......... 0    0 .......... .......... 0

0

..........

0



      −1 .......... 0   .......... .......... ..........     2 2 −1 k +1 k +1    0 −1 0 0

..........

0

n×n

where n, k are integers. The matrix Hn is obtained from matrix Gn by deleting (n, n)th entry. Theorem 2.12. For n > 1 det(Hn ) = Fk,n−1 Fk,n+1

(2.33) Where

det(Hn ) = (k2 + 1) Proof. Similar to equation (2.31), we obtain (2.34)

det(Hn ) = (k2 + 1)det(Hn − 1) + (k2 + 1)det(Hn − 2) − det(Hn − 3)

We have (2.35)

det(H2 ) = Fk,1 Fk,3

(2.36)

det(H3 ) = Fk,2 Fk,4

(2.37)

det(H4 ) = Fk,3 Fk,5

By (2.34) - (2.37) and using (k2 + 1)Fk,n+1 Fk,n+3 + (k2 + 1)Fk,n Fk,n+2 − Fk,n−1 Fk,n+1 = Fk,n+2 Fk,n+4 det(Hn ) = Fk,n−1 Fk,n+1  3. C ONCLUSION In this paper several properties of k Fibonacci and k Lucas sequences have been obtained by matrix methods. R EFERENCES [1] K. Atanassov, L. Atanassov, D. Sasselov, A New Perspective to the Generalization of the Fibonacci Sequence, The Fibonacci Quarterly 23 (1)(1985), 21-28. [2] K. T. Atanassov, Remark on a New Direction of Generalized Fibonacci Sequence, The Fibonacci Quarterly 33 (3)(1995) 249-250. [3] A. D. Godase, M. B. Dhakne, On the properties of k Fibonacci and k Lucas numbers, International Journal of Advances in Applied Mathematics and Mechanics, 02 (2014), 100 - 106. [4] A. D. Godase, M. B. Dhakne, Fundamental Properties of Multiplicative Coupled Fibonacci Sequences of Fourth Order Under Two Specific Schemes, International Journal of Mathematical Archive , 04, No. 6 (2013), 74 - 81. [5] A. D. Godase, M. B. Dhakne, Recurrent Formulas of The Generalized Fibonacci Sequences of Fifth Order, ”International Journal of Mathematical Archive” , 04, No. 6 (2013), 61 - 67. [6] A. D. Godase, M. B. Dhakne, Summation Identities for k Fibonacci and k Lucas Numbers using Matrix Methods, International Journal of Mathematics and Scientific Computing , 05, No. 2 (2015), 75 - 78. [7] A. D. Godase, M. B. Dhakne, Determinantal Identities for k Lucas Sequence, Journal of New Theory , 12 (2015), 01 - 07.

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MAYFEB Journal of Mathematics Vol 1 (2017) - Pages 11-20

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