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Some properties of the generalized Fibonacci and Lucas sequences related to the extended Hecke groups

Abstract

In this paper, we define a sequence, which is a generalized version of the Lucas sequence, similar to the generalized Fibonacci sequence given in Koruoğlu and Şahin in Turk. J. Math. 2009, doi:10.3906/mat-0902-33. Also, we give some connections between the generalized Fibonacci sequence and the generalized Lucas sequence, and we find polynomial representations of the generalized Fibonacci and the generalized Lucas sequences, related to the extended Hecke groups given in Koruoğlu and Şahin in Turk. J. Math. 2009, doi:10.3906/mat-0902-33.

MSC:20H10, 11F06.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

In [1], Hecke introduced groups H(λ), generated by two linear fractional transformations

T(z)= 1 z andS(z)= 1 z + λ ,

where λ is a fixed positive real number. Hecke showed that H(λ) is discrete if and only if λ= λ q =2cos π q , qN, q3, or λ2. These groups have come to be known as the Hecke Groups, and we will denote them H( λ q ), H(λ) for q3, λ2, respectively. The Hecke group H( λ q ) is the Fuchsian group of the first kind when λ= λ q or λ=2, and H(λ) is the Fuchsian group of the second kind when λ>2. In this study, we focus on the case λ= λ q , q3. The Hecke group H( λ q ) is isomorphic to the free product of two finite cyclic groups of orders 2 and q, and it has a presentation

H( λ q )= T , S T 2 = S q = I C 2 C q ,[2].
(1)

The first several of these groups are H( λ 3 )=Γ=PSL(2,Z) (the modular group), H( λ 4 )=H( 2 ), H( λ 5 )=H( 1 + 5 2 ), and H( λ 6 )=H( 3 ). It is clear that H( λ q )PSL(2,Z[ λ q ]), for q4. The groups H( 2 ) and H( 3 ) are of particular interest, since they are the only Hecke groups, aside from the modular group, whose elements are completely known (see, [3]).

The extended Hecke group, denoted by H ¯ ( λ q ), has been defined in [4] and [5] by adding the reflection R(z)=1/ z ¯ to the generators of the Hecke group H( λ q ). The extended Hecke group H ¯ ( λ q ) has a presentation

T , S , R T 2 = S q = R 2 = I , R T = T R , R S = S q 1 R D 2 Z 2 D q .
(2)

The Hecke group H( λ q ) is a subgroup of index 2 in H ¯ ( λ q ). It is clear that H ¯ ( λ q )PGL(2,Z[ λ q ]) when q>3 and H ¯ ( λ 3 )=PGL(2,Z) (the extended modular group Γ ¯ ).

Throughout this paper, we identify each matrix A in GL(2,Z[ λ q ]) with −A, so that they each represent the same element of H ¯ ( λ q ). Thus, we can represent the generators of the extended Hecke group H ¯ ( λ q ) as

T= ( 0 1 1 0 ) ,S= ( 0 1 1 λ q ) andR= ( 0 1 1 0 ) .

In [6], Koruoglu and Sahin found that there is a relationship between the generalized Fibonacci numbers and the entries of matrices representations of some elements of the extended Hecke group H ¯ ( λ q ). For the elements

h=TSR= ( λ q 1 1 0 ) andf=RTS= ( 0 1 1 λ q )

in H ¯ ( λ q ), then the k th power of h and f are

h k = ( a k a k 1 a k 1 a k 2 ) and f k = ( a k 1 a k a k a k + 1 ) ,

where a 0 =0, a 1 =1, and for k2,

a k = λ q a k 1 + a k 2 .
(3)

For all k2,

a k = 1 λ q 2 + 4 [ ( λ q + λ q 2 + 4 2 ) k + 1 ( λ q λ q 2 + 4 2 ) k + 1 ] .
(4)

Notice that this real numbers sequence is a generalized version of the common Fibonacci sequence. If λ q =1, this sequence coincides with the Fibonacci sequence.

The Fibonacci and the Lucas sequence have been studied extensively and generalized in many ways. For example, you can see in [712]. In this paper, firstly, we define a sequence b k , which is a generalization of the Lucas sequence. Then we give some properties of these sequences and the relationships between them. To do this, we use some results given in [1315]. In fact, in [14] and [15], Özgür found two sequences, which are the generalization of the Fibonacci sequence and the Lucas sequence, in the Hecke groups H(λ), λ2 real. But the Hecke groups H(λ) are different from the Hecke groups H( λ q ), λ q =2cos π q , qN, q3.

2 Some properties of generalized Fibonacci and generalized Lucas sequences

Firstly, we define a sequence b k by

b k = λ q b k 1 + b k 2
(5)

for k2, where b 0 =2, b 1 = λ q .

Proposition 1 For all k2,

b k = ( λ q + λ q 2 + 4 2 ) k + ( λ q λ q 2 + 4 2 ) k .
(6)

Proof To solve (6), let b k be a characteristic polynomial r k . Then we have the equation

r k = λ q r k 1 + r k 2 r 2 λ q r1=0.

The roots of this equation are

r 1 , 2 = λ q ± λ q 2 + 4 2 .

Using these roots r 1 , 2 , we can find a general formula of the general term b k . If we write b k as combinations of the roots r 1 , 2 , then we have

b k =A ( λ q + λ q 2 + 4 2 ) k +B ( λ q λ q 2 + 4 2 ) k .

To determine constants A and B, we use two boundary conditions b 0 =2 and b 1 = λ q , thus,

b 0 = 2 = A + B , b 1 = λ q = A ( λ q + λ q 2 + 4 2 ) + B ( λ q λ q 2 + 4 2 ) .

So,

2 λ q = A ( λ q + λ q 2 + 4 ) + ( 2 A ) ( λ q λ q 2 + 4 ) , A = 1 and B = 1 .

Then we obtain the formula of b k as

b k = ( λ q + λ q 2 + 4 2 ) k + ( λ q λ q 2 + 4 2 ) k .

This completes the proof. □

Notice that this formula is a generalized Lucas sequence. If λ q =1 (the modular group case q=3), we get the Lucas sequence.

Now, we have two sequences a k and b k , which are generalizations of the Fibonacci and the Lucas sequences. Let us write out the first 8 terms of a k and b k .

a k b k a 0 = 0 b 0 = 2 a 1 = 1 b 1 = λ q a 2 = λ q b 2 = λ q 2 + 2 a 3 = λ q 2 + 1 b 3 = λ q 3 + 3 λ q a 4 = λ q 3 + 2 λ q b 4 = λ q 4 + 4 λ q 2 + 2 a 5 = λ q 4 + 3 λ q 2 + 1 b 5 = λ q 5 + 5 λ q 3 + 5 λ q a 6 = λ q 5 + 4 λ q 3 + 3 λ q b 6 = λ q 6 + 6 λ q 4 + 9 λ q 2 + 2 a 7 = λ q 6 + 5 λ q 4 + 6 λ q 2 + 1 b 7 = λ q 7 + 7 λ q 5 + 14 λ q 3 + 7 λ q a 8 = λ q 7 + 6 λ q 5 + 10 λ q 3 + 4 λ q b 8 = λ q 8 + 8 λ q 6 + 20 λ q 4 + 16 λ q 2 + 2 .

Here, it is possible to extend a k and b k backward with the negative subscripts. For example, a 1 =1, a 2 = λ q , a 3 = λ q 2 +1, and so on. Therefore, we can deduce that

a k = ( 1 ) k + 1 a k
(7)

and

b k = ( 1 ) k b k .
(8)

The sequences a k and b k have some similar properties of the Fibonacci and the Lucas numbers F n and L n . Now, we investigate some properties of these sequences a k and b k .

Proposition 2

a k + a k + 4 = ( λ q 2 + 2 ) a k + 2 and b k + b k + 4 = ( λ q 2 + 2 ) b k + 2 .
(9)

Proof We will use induction on k. For k=0, we have

a 0 + a 4 =0+ λ q 3 +2 λ q = λ q ( λ q 2 + 2 ) = a 2 ( λ q 2 + 2 ) .

For k=1, we get

a 1 + a 5 = 1 + λ q 4 + 3 λ q 2 + 1 = ( λ q 2 + 2 ) ( λ q 2 + 1 ) = ( λ q 2 + 2 ) a 3 .

Now let us assume that the proposition holds for k=2,,n. We show that it holds for k=n+1. By assumption, we have

a n 1 + a n + 3 = ( λ q 2 + 2 ) a n + 1 and a n + a n + 4 = ( λ q 2 + 2 ) a n + 2 .

From (3), we obtain

a n + 1 + a n + 5 = ( λ q a n + a n 1 ) + ( λ q a n + 4 + a n + 3 ) = λ q ( a n + a n + 4 ) + a n 1 + a n + 3 = λ q ( λ q 2 + 2 ) a n + 2 + ( λ q 2 + 2 ) a n + 1 = ( λ q 2 + 2 ) ( λ q a n + 2 + a n + 1 ) = ( λ q 2 + 2 ) a n + 3 .

Then we get

a k + a k + 4 = ( λ q 2 + 2 ) a k + 2 .

Similarly, it can be shown that

b k + b k + 4 = ( λ q 2 + 2 ) b k + 2 .

 □

Proposition 3

b k = a k + 1 + a k 1 .
(10)

Proof We will use the induction method on k. If k=1, then

b 1 = a 2 + a 0 .

We suppose that the equation holds for k=2,3,,n1, i.e.,

b n 1 = a n + 1 + a n 1 .

Now, we show that the equation holds for k=n. Then we have

b n = ( λ q 2 + 2 ) b n 2 b n 4 = ( λ q 2 + 2 ) ( a n 1 + a n 3 ) ( a n 5 + a n 3 ) = ( λ q 2 + 2 ) a n 1 a n 3 + ( λ q 2 + 2 ) a n 3 a n 5 = a n + 1 + a n 1 .

 □

Proposition 4

b k + b k + 2 = ( λ q 2 + 4 ) a k + 1 .
(11)

Proof For k=0, we have

b 0 + b 2 = 2 + λ q 2 + 2 = λ q 2 + 4 = ( λ q 2 + 4 ) a 1 .

For k=1, we have

b 1 + b 3 = λ q + λ q 3 + 3 λ q = λ q 3 + 4 λ q = λ q ( λ q 2 + 4 ) .

Now, we assume that the proposition holds for k=2,,n. We show that it holds for k=n+1. By assumption, we have

b n + b n + 2 = ( λ q 2 + 4 ) a n + 1 and b n 1 + b n + 1 = ( λ q 2 + 4 ) a n .

Then we find

b n + 1 + b n + 3 = ( λ q b n + b n 1 ) + ( λ q b n + 2 + b n + 1 ) = λ q ( b n + b n + 2 ) + ( b n 1 + b n + 1 ) = λ q ( λ q 2 + 4 ) a n + 1 + ( λ q 2 + 4 ) a n = ( λ q 2 + 4 ) ( λ q a n + 1 + a n ) = ( λ q 2 + 4 ) a n + 2 .

 □

Proposition 5

a k 3 + a k + 3 = ( λ q 2 + 1 ) b k .
(12)

Proof We will use induction on k. For k=0, we find

a 3 + a 3 = ( 1 ) 4 a 3 + a 3 = 2 a 3 = 2 ( λ q 2 + 1 ) b 0 .

For k=1, we get

a 2 + a 4 = ( 1 ) 3 a 2 + a 4 = a 2 + a 4 = λ q + λ q 3 + 2 λ q = λ q 3 + λ q = λ q ( λ q 2 + 1 ) = b 1 ( λ q 2 + 1 ) .

Now, let us suppose that the proposition holds for k=2,,n. We show that it holds for k=n+1. By assumption, a n 3 + a n + 3 =( λ q 2 +1) b n and a n 4 + a n + 2 =( λ q 2 +1) b n 1 . Hence we get

a n 2 + a n + 4 = λ q a n 3 + a n 4 + λ q a n + 3 + a n + 2 = λ q ( a n 3 + a n + 3 ) + a n 4 + a n + 2 = λ q ( λ q 2 + 1 ) b n + ( λ q 2 + 1 ) b n 1 = ( λ q 2 + 1 ) ( λ q b n + b k 1 ) = ( λ q 2 + 1 ) b n + 1 .

 □

Proposition 6

a 2 k = a k b k .
(13)

Proof We will use the induction method on k. For k=0, we have

a 0 b 0 =0= a 0 .

For k=1, we have

a 1 b 1 = λ q = a 2 .

We suppose that the equation holds for k=2,,n1, i.e.,

a 2 ( n 1 ) = a n 1 b n 1 .

Now, we show that the equation holds for k=n. By equalities (3), (9) and (10),

a n b n = a n ( a n + 1 + a n 1 ) = a n ( ( λ q 2 + 2 ) a n 1 a n 3 ) + a n 1 ( ( λ q 2 + 2 ) a n 2 a n 4 ) = ( λ q 2 + 2 ) a n a n 1 + ( λ q 2 + 2 ) a n 1 a n 2 a n a n 3 a n 1 a n 4 = ( λ q 2 + 2 ) a n 1 ( a n + a n 2 ) a n a n 3 a n 1 a n 4 = ( λ q 2 + 2 ) a n 1 b n 1 a n a n 3 a n 1 a n 4 = ( λ q 2 + 2 ) a n 1 b n 1 a n 3 ( λ q a n 1 + a n 2 ) a n 1 ( a n 2 λ q a n 3 ) = ( λ q 2 + 2 ) a n 1 b n 1 a n 3 a n 2 a n 1 a n 2 = ( λ q 2 + 2 ) a n 1 b n 1 a n 2 ( a n 3 + a n 1 ) = ( λ q 2 + 2 ) a n 1 b n 1 a n 2 b n 2 = ( λ q 2 + 2 ) a 2 n 2 a 2 n 4 (by assumption) = a 2 n .

 □

Proposition 7

b k 2 ( λ q 2 + 4 ) a k 2 =4 ( 1 ) k .
(14)

Proof Using (10) and the definitions of a k and b k , we have

b k 2 ( λ q 2 + 4 ) a k 2 = ( a k 1 + a k + 1 ) 2 ( λ q 2 + 4 ) a k 2 = a k 1 2 + 2 a k 1 a k + 1 + a k + 1 2 λ q 2 a k 2 4 a k 2 = a k 1 2 + 2 a k 1 ( λ q a k + a k 1 ) + ( λ q a k + a k 1 ) 2 λ q 2 a k 2 4 a k 2 = a k 1 2 + 2 λ q a k 1 a k + 2 a k 1 2 + λ q 2 a k 2 + 2 λ q a k a k 1 + a k 1 2 λ q 2 a k 2 4 a k 2 = 4 a k 1 2 + 4 λ q a k 1 a k 4 a k 2 = 4 a k 1 ( a k 1 + λ q a k ) 4 a k 2 = 4 a k 1 a k + 1 4 a k 2 = 4 ( a k 1 a k + 1 a k 2 ) .

In [10], Yayenie and Edson obtained a generalization of Cassini’s identity for the positive real numbers a and b. If we take a= λ q and b= λ q in generalized Cassini’s identity, we get

a k 1 a k + 1 a k 2 = ( 1 ) n ,

and so,

b k 2 ( λ q 2 + 4 ) a k 2 =4 ( 1 ) n .

 □

Proposition 8

a k a k + 3 a k + 1 a k + 2 = ( 1 ) k + 1 λ q .
(15)

Proof We will use the induction method on k. For k=0, we have

a 0 a 3 a 1 a 2 = λ q = ( 1 ) 1 λ q .

For k=1, we have

a 1 a 4 a 2 a 3 = λ q 3 + 2 λ q λ q ( λ q 2 + 1 ) = ( 1 ) 2 λ q .

Now, we assume that the proposition holds for k=2,,n. We show that it holds for k=n+1. From assumption a n a n + 3 a n + 1 a n + 2 = ( 1 ) n + 1 λ q , and, thus,

a n + 1 a n + 4 a n + 2 a n + 3 = a n + 1 ( λ q a n + 3 + a n + 2 ) a n + 3 ( λ q a n + 1 + a n ) = λ q a n + 1 a n + 3 + a n + 1 a n + 2 λ q a n + 3 a n + 1 a n + 3 a n = a n + 1 a n + 2 a n + 3 a n = ( 1 ) n + 1 λ q = ( 1 ) n + 2 λ q .

 □

Proposition 9

a 2 m + 2 a k a 2 m a k 2 = a 2 m + k λ q .
(16)

Let m be fixed. We will use the induction method on k. For k=0, we have

a 2 m + 2 a 0 a 2 m a 2 = λ q a 2 m ,

since a 0 =0 and a 2 = ( 1 ) 3 a 2 = λ q . For k=1, we find

a 2 m + 2 a 1 a 2 m a 1 = a 2 m + 2 a 2 m = λ q a 2 m + 1 + a 2 m a 2 m = λ q a 2 m + 1 ,

since a 1 =1 and a 1 =1. Now, we assume that the proposition holds for k=2,,n. We show that it holds for k=n+1. By assumption,

a 2 m + 2 a n a 2 m a n 2 = λ q a 2 m + n

and

a 2 m + 2 a n 1 a 2 m a n 3 = λ q a 2 m + n 1 .

Thus, we have

a 2 m + 2 a n + 1 a 2 m a n 1 = a 2 m + 2 ( λ q a n + a n 1 ) a 2 m ( λ q a n 2 + a n 3 ) = λ q ( a 2 m + 2 a n a 2 m a n 2 ) + ( a 2 m + 2 a n 1 a 2 m a n 3 ) = λ q λ q a 2 m + n + λ q a 2 m + n 1 = λ q ( λ q a 2 m + n + a 2 m + n 1 ) = λ q a 2 m + n + 1 .

Now, we give a formula for a k and b k .

Proposition 10 For all k1,

a k = { 1 2 k 1 i = 0 k 2 2 ( k 2 i + 1 ) λ q k ( 2 i + 1 ) ( λ q 2 + 4 ) i if  k  is even, 1 2 k 1 i = 0 k 1 2 ( k 2 i + 1 ) λ q k ( 2 i + 1 ) ( λ q 2 + 4 ) i if  k  is odd
(17)

and

b k = { 1 2 k 1 i = 0 k 2 ( k 2 i ) λ q k 2 i ( λ q 2 + 4 ) 2 i if  k  is even, 1 2 k 1 i = 0 k 1 2 ( k 2 i 1 ) λ q k ( 2 i 1 ) ( λ q 2 + 4 ) 2 i if  k  is odd.
(18)

Proof Let k be even. By (4),

a k = 1 λ q 2 + 4 [ ( λ q + λ q 2 + 4 2 ) k ( λ q λ q 2 + 4 2 ) k ] = 1 2 k 1 λ q 2 + 4 [ ( k 1 ) λ q k 1 λ q 2 + 4 + ( k 3 ) λ q k 3 ( λ q 2 + 4 ) 3 + + ( k k 1 ) λ q ( λ q 2 + 4 ) k 1 ] = 1 2 k 1 [ ( k 1 ) λ q k 1 + ( k 3 ) λ q k 3 ( λ q 2 + 4 ) 2 + + ( k k 1 ) λ q ( λ q 2 + 4 ) k 2 ] = 1 2 k 1 [ i = 0 k 2 2 ( k 2 i + 1 ) λ q k ( 2 i + 1 ) ( λ q 2 + 4 ) 2 i ] .

Similarly, if k is odd, then we get

a k = 1 2 k 1 i = 0 k 1 2 ( k 2 i + 1 ) λ q k ( 2 i + 1 ) ( λ q 2 + 4 ) i .

 □

Proposition 11

i = 1 k + 1 a i = a k + 2 + a k + 1 1 λ q
(19)

and

i = 1 k + 1 b i = b k + 2 + b k + 1 ( λ q + 2 ) λ q .
(20)

Proof From (3), we have

a k + 2 a k + 1 = λ q a k + 1 + a k a k + 1 = ( λ q 1 ) a k + 1 + a k ,

and so,

n = 0 a 2 a 1 = ( λ q 1 ) a 1 + a 0 , n = 1 a 3 a 2 = ( λ q 1 ) a 2 + a 1 , n = k 1 a k + 1 a k = ( λ q 1 ) a k + a k 1 , n = k a k + 2 a k + 1 = ( λ q 1 ) a k + 1 + a k .

If we sum both sides, then we obtain

a k + 2 a 1 = ( λ q 1 ) ( a 1 + a 2 + + a k + 1 ) + ( a 0 + a 1 + + a k ) = λ q ( a 1 + a 2 + + a k + 1 ) + a 0 a k + 1 .

Since a 0 =0 and a 1 =1, we have

a k + 2 1 = λ q ( a 1 + a 2 + + a k + 1 ) a k + 1 , a k + 2 + a k + 1 1 = λ q ( a 1 + a 2 + + a k + 1 ) , a k + 2 + a k + 1 1 λ q = a 1 + a 2 + + a k + 1 , i = 1 k + 1 a i = a k + 2 + a k + 1 1 λ q .

Similarly, it is easily seen that

i = 1 k + 1 b i = b k + 2 + b k + 1 ( λ q + 2 ) λ q .

 □

3 Polynomial representations of a k and b k

Before we find the polynomial representations of a k and b k , note the following identities

( k p ) +2 ( k + 1 p 1 ) ( k p 2 ) = ( k + 2 p )
(21)

and

( k p ) + ( k 1 p 1 ) = ( k 1 p 1 ) p + k p
(22)

Theorem 1 Let { a k } denote the generalized Fibonacci sequence. Then, the polynomial representations of a 2 k and a 2 k + 1 are

a 2 k = ( λ q ) 2 k 1 + ( 2 k 2 1 ) ( λ q ) 2 k 3 + ( 2 k 3 2 ) ( λ q ) 2 k 5 + + ( k + 2 k 3 ) ( λ q ) 3 + ( k + 1 k 2 ) ( λ q )

and

a 2 k + 1 = ( λ q ) 2 k + ( 2 k 1 ) ( λ q ) 2 k 2 + ( 2 k 2 1 ) 2 k 3 2 ( λ q ) 2 k 4 + ( 2 k 3 2 ) 2 k 5 3 ( λ q ) 2 k 6 + + ( k + 1 k 2 ) 3 k 1 ( λ q ) 2 + 1 .

Proof We will use the induction method on k. For k=1, we have a 2 = λ q , and for k=2, we have a 4 = ( λ q ) 3 +2 λ q . Now, suppose that the equality is true for k=1,2,,n. We will show that it holds for k=n+1. By assumption,

a 2 n 2 = ( λ q ) 2 n 3 + ( 2 n 4 1 ) ( λ q ) 2 n 5 + ( 2 n 5 2 ) ( λ q ) 2 n 7 + + ( n + 1 n 4 ) ( λ q ) 3 + ( n n 3 ) ( λ q )

and

a 2 n = ( λ q ) 2 n 1 + ( 2 n 2 1 ) ( λ q ) 2 n 3 + ( 2 n 3 2 ) ( λ q ) 2 n 5 + + ( n + 2 n 3 ) ( λ q ) 3 + ( n + 1 n 2 ) ( λ q ) .

From (9), we have a 2 k + 2 =( λ q 2 +2) a 2 k a 2 k 2 , and by definition of a k , we get

a 2 n + 2 = ( λ q 2 + 2 ) [ ( λ q ) 2 n 1 + ( 2 n 2 1 ) ( λ q ) 2 n 3 + ( 2 n 3 2 ) ( λ q ) 2 n 5 + + ( n + 2 n 3 ) ( λ q ) 3 + ( n + 1 n 2 ) ( λ q ) ] [ ( λ q ) 2 n 3 + ( 2 n 4 1 ) ( λ q ) 2 n 5 + ( 2 n 5 2 ) ( λ q ) 2 n 7 + + ( n + 1 n 4 ) ( λ q ) 3 + ( n n 3 ) ( λ q ) ] = ( λ q ) 2 n + 1 + ( ( 2 n 2 1 ) + 2 ) ( λ q ) 2 n 1 + [ ( 2 n 3 2 ) + 2 ( 2 n 2 1 ) ] ( λ q ) 2 n 3 + + [ ( n + 1 n 2 ) + 2 ( n + 2 n 3 ) ] ( λ q ) 3 + 2 ( n + 1 n 2 ) λ q .

From (21), we get

a 2 n + 2 = ( λ q ) 2 n + 1 + ( 2 n 1 ) ( λ q ) 2 n 1 + ( 2 n 1 2 ) ( λ q ) 2 n 3 + + ( n + 3 n 2 ) ( λ q ) 3 + ( n + 2 n 1 ) ( λ q ) .

Now, we compute a 2 k + 1 . By definition of a k , we get

a 2 k + 1 = 1 λ q ( a 2 k + 2 a 2 k ) = 1 λ q [ ( ( λ q ) 2 k + 1 + ( 2 k 1 ) ( λ q ) 2 k 1 + ( 2 k 1 2 ) ( λ q ) 2 k 3 + + ( k + 3 k 2 ) ( λ q ) 3 + ( k + 2 k 1 ) ( λ q ) ) ( ( λ q ) 2 k 1 + ( 2 k 2 1 ) ( λ q ) 2 k 3 + ( 2 k 3 2 ) ( λ q ) 2 k 5 + + ( k + 2 k 3 ) ( λ q ) 3 + ( k + 1 k 2 ) ( λ q ) ) ] .

From (22), we get

a 2 k + 1 = ( λ q ) 2 k + ( 2 k 1 ) ( λ q ) 2 k 2 + ( 2 k 2 1 ) 2 k 3 2 ( λ q ) 2 k 4 + ( 2 k 3 2 ) 2 k 5 3 ( λ q ) 2 k 6 + + ( k + 1 k 2 ) 3 k 1 ( λ q ) 2 + 1 .

 □

Theorem 2 Let { b k } denote the generalized Lucas sequence. Then, the polynomial representations of b 2 k and b 2 k + 1 are

b 2 k = ( λ q ) 2 k + ( 2 k ) ( λ q ) 2 k 2 + ( 2 k 3 1 ) 2 k 2 ( λ q ) 2 k 4 + ( 2 k 4 2 ) 2 k 3 ( λ q ) 2 k 6 + + ( k k 2 ) 2 k k 1 ( λ q ) 2 + 2

and

b 2 k + 1 = ( λ q ) 2 k + 1 + ( 2 k + 1 ) ( λ q ) 2 k 1 + ( 2 k 2 1 ) 2 k + 1 2 ( λ q ) 2 k 3 + ( 2 k 3 2 ) 2 k + 1 3 ( λ q ) 2 k 5 + + ( k + 1 k 2 ) 2 k + 1 k 1 ( λ q ) .

Proof From (10), it is easy to find the polynomial representations of b 2 k and b 2 k + 1 . □

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İkikardes, S., Sarıgedik, Z. Some properties of the generalized Fibonacci and Lucas sequences related to the extended Hecke groups. J Inequal Appl 2013, 398 (2013). https://doi.org/10.1186/1029-242X-2013-398

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  • DOI: https://doi.org/10.1186/1029-242X-2013-398

Keywords

  • extended Hecke group
  • generalized Fibonacci sequence
  • generalized Lucas sequence