Lucas Numbers and Lucas Polynomials - Science and Education

Turkish Journal of Analysis and Number Theory, 2017, Vol. 5, No. 4, 121-125 Available online at http://pubs.sciepub.com/tjant/5/4/1 ©Science and Education Publishing DOI:10.12691/tjant-5-4-1

On the 𝒌𝒌 -Lucas Numbers and Lucas Polynomials Ali Boussayoud1,*, Mohamed Kerada1, Nesrine Harrouche2

1

LMAM Laboratory and Department of Mathematics, Mohamed Seddik Ben Yahia University, Jijel, Algeria 2 Department of Mathematics, University of Jordan, Amman 11942, Jordan *Corresponding author: [email protected]

Received December 23, 2016; Revised May 11, 2017; Accepted June 07, 2017

Abstract In this paper, we introduce an operator in order to derive some new symmetric properties of 𝑘𝑘-Lucas numbers and Lucas polynomials. By making use of the operator defined in this paper, we give some new generating functions for 𝑘𝑘 -Lucas numbers and Lucas polynomials. Keywords: k-Lucas numbers, Lucas polynomials, generating functions

Cite This Article: Ali Boussayoud, Mohamed Kerada, and Nesrine Harrouche, “On the 𝑘𝑘 -Lucas Numbers and Lucas Polynomials.” Turkish Journal of Analysis and Number Theory, vol. 5, no. 4 (2017): 121-125. doi: 10.12691/tjant-5-4-1. ek = ( x1 , x2 ,…, xn )

1. Introduction Fibonacci and Lucas numbers have been studied by many researchers for a long time to get intrinsic theory and applications of these numbers in many research areas as Physics, Engineering, Architecture, Nature and Art. For example, the ratio of two consecutive numbers converges

1+ 5 which was thoroughly 2 interested in [13]. We should recall that, for k∈ ℝ+ , and k-Lucas k-Fibonacci Lk ,n Fk,n to the Golden ratio α =

{

}n∈

{

}n∈

sequences have been defined by the recursive equations [9,10]; = F k , n + 2 kFk , n +1 + Fk , n , L= k , n + 2 kLk , n +1 + Lk , n ,

with initial conditions Fk ,0 = 1, Fk ,1 = k and Lk ,0 = 2,

Lk ,1 = k , respectively. For the special case k=1, it is clear that these two sequences are simplified to the well-known Fibonacci and Lucas sequences, respectively. In this contribution, we shall define a new useful operator

∑

x11 x22 … xnin , 0 ≤ k ≤ n.

With 𝑖𝑖1 , 𝑖𝑖2 , … , 𝑖𝑖𝑛𝑛 = 0 or 1,

x11 x22 … xnin , 0 ≤ k ≤ n.

hk = ( x1 , x2 ,…, xn )

i1 +i2 +…+in = k

∑

i1 +i2 +…+in = k

i

i

i

i

With 𝑖𝑖1 , 𝑖𝑖2 , … , 𝑖𝑖𝑛𝑛 ≥ 0, First, we set 𝑒𝑒0 (𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ) = 1 and ℎ0 (𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ) = 1 (by convention). For 𝑘𝑘 > 𝑛𝑛 or k < 0 , we set

ek ( x1 , x= = 2 , …, xn ) 0 and hk ( x1 , x 2 , …, xn ) 0.

Definition 1. [1] Let 𝐵𝐵 and 𝑃𝑃 be any two alphabets, then we give 𝑆𝑆𝑛𝑛 (𝐵𝐵 − 𝑃𝑃) by the following form:

∏ p∈P(1 − pt ) ∞ = ∑ Sn ( B − P ) t n , ∏ b∈B(1 − bt ) n=0

with the condition 𝑆𝑆𝑛𝑛 (𝐵𝐵 − 𝑃𝑃) = 0 for 𝑛𝑛 < 0.

Definition 2. [2] Let 𝑔𝑔 be any function on R n , then we consider the divided difference operator as the following form

 g ( x1 , …, xi , xi +1 , …, xn )  denoted by δ pk p for which we can formulate, extend and 1 2    − g ( x1 , … xi −1 , xi +1 , xi , …, xn )   prove new results based on our previous ones [4,5,6]. In ∂ xi , xi +1 ( g ) = . order to determine generating functions for k-Fibonacci xi − xi +1 numbers, 𝑘𝑘 -Lucas numbers and Lucas polynomials, we Definition 3. [7] The symmetrizing operator 𝛿𝛿𝑝𝑝𝑘𝑘1 𝑝𝑝 2 is combine between our indicated past techniques and these defined by presented polishing approaches. Let 𝑘𝑘 and 𝑛𝑛 be two positive integers and {𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 } p1k g ( p1 ) − p2k g ( p2 ) are set of given variables, recall [8] that = the 𝑘𝑘 -th δ pk p g ( p1 ) for all k ∈ . 1 2 p1 − p2 elementary symmetric function 𝑒𝑒𝑘𝑘 (𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ) and the 𝑘𝑘 -th complete homogeneous symmetric function Remark 1. Let 𝑃𝑃 = {𝑝𝑝1 , 𝑝𝑝2 } an alphabet, we have ℎ𝑘𝑘 (𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ) are defined respectively by

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Turkish Journal of Analysis and Number Theory

hk ( p1 , p2 )= Sk ( p1 + p2 )= δ pk p ( p1 ) . 1 2

lim

= σ r.

n →∞ Lk , n

2. The 𝒌𝒌-Lucas Numbers and Properties

The 𝑘𝑘 -Lucas numbers have been defined in [11] for any number 𝑘𝑘 as follows. Definition 4. [11] For any positive real number, the 𝑘𝑘 -Lucas numbers, say �𝐿𝐿𝑘𝑘,𝑛𝑛 �𝑛𝑛∈ℕ is defined recurrently by

Lk ,n +1 =+ kLk ,n Lk ,n −1 for n ≥ 1,

Lk ,n + r

(2.1)

with initial conditions 𝐿𝐿𝑘𝑘 ,0 = 2, 𝐿𝐿𝑘𝑘,1 = 𝑘𝑘. Note that if 𝑘𝑘 is a real variable 𝑥𝑥 then 𝐿𝐿𝑘𝑘 ,𝑛𝑛 = 𝐿𝐿𝑥𝑥 ,𝑛𝑛 and they correspond to the Lucas polynomials defined by

2 if n = 0  if n 1 . Ln +1 ( x ) = x =  xL x + L  n ( ) n −1 ( x ) if n > 1

In addition, the general term of the 𝑘𝑘 -Lucas numbers may be obtained by the formula [10]:

= Lk ,n kFk ,n −1 + Fk ,n +1.

3. On the Symmetric Functions of Some Numbers and Polynomails Theorem 1. [4] Let 𝑃𝑃 and 𝐵𝐵 be two alphabets, respectively, {𝑝𝑝1 , 𝑝𝑝2 } and {𝑏𝑏1 , 𝑏𝑏2 , 𝑏𝑏3 } , then we have ∞

∑ hn (b1, b2 , b3 )hn ( p1, p2 ) t n

n =0

 1 − p1 p2 e2 ( b1 , b2 , b3 ) t 2     − p p e (b , b , b ) h ( p , p ) t3   1 2 2 1 2 3 1 1 2 

(3.1)

= . Particular cases of the 𝑘𝑘 -Lucas numbers are ∞ ∞ n n n n e ( b , b , b ) p t e ( b , b , b ) p t ∑ n 0= ∑n 0 n 1 2 3 2 n 1 2 3 1 • If 𝑘𝑘 = 1, the classical Lucas numbers is obtained: =

)(

(

L0 2,= L1 1, and =

In the case 𝐵𝐵 = {1} based on Theorem 1, we deduce the following Lemmas. Lemma 1. Given an alphabet 𝑃𝑃 = {𝑝𝑝1 , −𝑝𝑝2 }, we have

Ln +1 = Ln + Ln −1 for n ≥ 1:

= {Ln }n∈

{2,1,3, 4, 7,…} .

∞

∑ hn ( p1,[− p2 ]) t n = 1− p − p

• If 𝑘𝑘 = 2, the Pell-Lucas numbers appears:

Qn +1 = 2Qn + Qn −1 for n ≥ 1

such that

The well-known Binet's formula in the Lucas numbers theory allows us to express the 𝑘𝑘 -Lucas number in function of the roots 𝑟𝑟1 and 𝑟𝑟2 of the characteristic equation, associated to the recurrence relation (2.1): (2.2)

Proposition 1. (Binet's formula) The nth 𝑘𝑘 -Lucas number is given by Lk ,n

r n − r2n , = 1 r1 − r2

where 𝑟𝑟1 , 𝑟𝑟2 are the roots of the characteristic equation (2.2) and 𝑟𝑟1 > 𝑟𝑟2 . Proof . The roots of the characteristic equation (2.2) are

k − k2 + 4 k + k2 + 4 and r2 = . 2 2 Note that, since 𝑘𝑘 > 0, the 𝑟𝑟2 < 0 < 𝑟𝑟1 and |𝑟𝑟2 | < |𝑟𝑟1 |, 𝑟𝑟1 + 𝑟𝑟2 = 𝑘𝑘 and 𝑟𝑟1 . 𝑟𝑟2 = −1, 𝑟𝑟1 − 𝑟𝑟2 = √𝑘𝑘 2 + 4. If 𝜎𝜎 denotes the positive root of the characteristic equation, the general term may be written in the form [10] r1 =

Lk ,n =

σ n − σ −n σ +σ

2

) t − p1 p2t 2

. (3.2)

∞

formula (3.2) can be written as: ∞

n n p ∑ ∞ 0= p1n t n − p2 ∑ ∞ n 0[ − p2 ] t p1 − [− p2 ]

n n 1 n δ p1[ − p2 ] ∑ p= 1t = n =0

= =

∞

∑

n =0 ∞

p1n +1 − [− p2 ]n +1 n t p1 − [− p2 ]

∑ hn ( p1,[− p2 ])t n .

n =0

white the right -hand side can be expressed as

p1 1−1p t − [− p2 ] 1−[ −1p ]t 1 1 2 δ p1[ − p2 ] = 1 − p1t p1 − [− p2 ] = = =

−1

and the limit of the quotient of two terms is

1

1

∑ n≥0 p1t n and (1 − p1t ) be two sequences ∞ ∑ n≥0 p1nt n = 1 −1p1t , the left-hand side of the

Proof. Let

{2, 2, 6,14,34,…} .

r 2= kr + 1.

(

n =0

= Q0 2,= Q1 2, and

= {Qn }n∈

)

=

p1 1−1p t − [− p2 ] 1+ 1p

2t

1

p1 − [− p2 ] p1 (1 + p2t ) + p2 (1 − p1t ) ( p1 + p2 )(1 − p1t )(1 + p2t ) p1 + p2 ( p1 + p2 )(1 − ( p1 − p2 )t − p1 p2t 2 ) 1 1 − ( p1 − p2 )t − p1 p2t 2

Turkish Journal of Analysis and Number Theory

This completes the proof. Lemma 2. Given an alphabet 𝑃𝑃 = {𝑝𝑝1 , −𝑝𝑝2 } , we have

∞

∑ hn ( p1,[− p2 ]) t n=

∞

1

∑ Fk ,nt n=

1 − kt − t 2

= n 0= n 0

∞

p − p + p1 p2t . (3.3) ( 1 2 ) t − p1 p2t 2

1 2 ∑ hn+1 ( p1,[− p2 ]) t n = 1− p − p

n =0

Proof. Let

123

∞

∑ n≥0 p1t n

and (1 − p1t ) be two sequences

∞ pnt n n≥0 1

such that ∑ = 1 , the left-hand side of the 1 − p1t formula (3.3) can be written as: ∞

δ 2 p1[ − p2 ] ∑ p1n t n = n =0

n n p12 ∑ ∞ n =0 p1 t p1 − [− p2 ]

− = =

[− p2 ]

2

∑

n =0 ∞

n n ∑∞ n =0[ − p2 ] t

p1n + 2 − [− p2 ]n + 2 n t p1 − [− p2 ]

∑ hn+1 ( p1,[− p2 ])t n .

n =0

which represents a generatings functions for 𝑘𝑘 -Fibonacci numbers (with 𝑝𝑝1 = 𝑟𝑟1 𝑎𝑎𝑎𝑎𝑎𝑎 [−𝑝𝑝2 ] = 𝑟𝑟2 ). ∞

k +t

, ∑ hn+1 ( p1,[− p2 ]) t n = 1 − kt − kt 2

p12 1−1p t − [− p2 ]2 1−[ −1p ]t 1 1 2 δ p1[ − p2 ] = 1 − p1t p1 − [− p2 ] 2

p2t ) − = ( p1 + p2 )(1 − p1t )(1 + p2t )

p22 (1 − p1t )

p12 + p12 p2t − p22 + p22 p1t = ( p1 + p2 )(1 − ( p1 − p2 )t − p1 p2t 2 ) =

p12 − p22 + p1 p2 ( p1 + p2 )t ( p1 + p2 )(1 − ( p1 − p2 )t − p1 p2t 2 )

=

p1 − p2 + p1 p2t . 1 − ( p1 − p2 )t − p1 p2t 2

This completes the proof. Taking 𝑝𝑝1 − 𝑝𝑝2 = 1 and 𝑝𝑝1 𝑝𝑝2 = 1 in (3.2) and (3.3), we obtain the generating functions given by Boussayoud et al [5] which arises 1) The generating function of the Fibonacci numbers ∞

1

∑ Fnt n = 1 − t − t 2 .

which represents a new generatings functions. • Multiplying the equation (3.4) by (2 + 𝑘𝑘 2 ) and subtract it from (3.5) by (𝑘𝑘) , we obtain

∑ [( 2 + k 2 ) hn ( p1,[− p2 ]) − khn+1 ( p1,[− p2 ])]t n

n =0

=

2 − kt 1 − kt − t 2

,

from which we have the following theorem. Theorem 3. For n ∈ ℕ , the generating function of the 𝑘𝑘 -Lucas numbers is given by ∞

∑ Lnt n = 1 − t − t 2 .

n =0

( −1)n+1 Fk ,n ,

2) Lk ,− n =

( −1)n Lk ,n

hold for all 𝑛𝑛 ≥ 0.

𝑝𝑝1 𝑝𝑝2 = 1 and 𝑝𝑝1 − 𝑝𝑝2 = 𝑘𝑘 substituting in (3.2) and (3.3) we end up with Choosing 𝑝𝑝1 and 𝑝𝑝2 such that �

• Put 𝑘𝑘 = 2 in the relationship (3.6) we have ∞

2 − 2t

∑Qnt n = 1 − 2t − t 2 ,

n =0

which represents a generating function for Pell-Lucas numbers [5]. Replacing 𝑡𝑡 by (−𝑡𝑡) in (3.4) and (3.6), we have the following theorems. Theorem 4. We have the following a new generating function of the 𝑘𝑘 -Fibonacci numbers at negative indices as ∞

1

∑ Fk ,−nt n = t 2 − kt − 1 .

n =0

Proof. The ordinary generating function associated is defined by

G ( Fk , n , t ) =

∞

∑ Fk , nt n .

n =0

∑ Fk , nt n =

Fk , 0t 0 + Fk ,1t +

∞

∑ Fk , nt n

= n 0= n 2 ∞ ∞ =1 + kt + kFk , n −1t n + Fk , n − 2t n . = n 2= n 2

Proposition 2. [10,12] The relations 1) Fk ,− n =

(3.6)

n =0

∞

2) The generating function of the Lucas numbers

2−t

2 − kt

∑ Lk ,nt n = 1 − kt − t 2 .

Using the initial conditions, we get

n =0

∞

(3.5)

n =0

white the right-hand side can be expressed as:

p12 (1 +

(3.4)

∞

p1 − [− p2 ] ∞

,

∑

Consider that written by

∑

j= n − 2 and p= n − 1 . Then can be ∞

∞

=1 + kt + t ∑kFk , n t n + t 2 ∑ Fk , n t n

= n 1 =j 0 ∞

∞

=1 + kt + kt ∑ Fk , n t n − kt + t 2 ∑ Fk , n t n ,

= p 0=j 0

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Turkish Journal of Analysis and Number Theory

which is equivalent to

(

1 − kt − t 2

Proof. The ordinary generating function associated is defined by

∞

) n∑=0Fk , nt n =1

⇒

G ( Ln ( x), t ) =

∞

1

. ∑ Fk , nt n = 1 − kt − t 2

∞

Replacing t by (−t ) , we have

∑ Ln ( x)t n

= n 0= n 2 ∞

=2 + xt +

∞

=2 + xt + x ∑ Ln −1 ( x)t n +

n

1

∞

∞

This completes the proof. Theorem 5. We have the following a new generating function of the k -Lucas numbers at negative indices as

2 + kt

∑ Lk ,−nt n = 1 + kt − t 2 .

(3.7)

∑Q−nt

n =0

=

1 + 2t − t 2

,

∞

∞

=2 + xt + xt ∑ L p ( x)t p − 2 xt + t 2 ∑ L j ( x)t j ,

∞

1

(3.8)

n =0

which represents a generating function of the Fibonacci polynomials

n =0

∞

⇒

x+t , ∑ hn+1 ( p1,[− p2 ]) t n = 1 − xt − kt 2 n =0

(3.9)

which represents a new generatings functions. • Multiplying the equation (3.8) by (2 + 𝑥𝑥 2 ) and subtract it from (3.9) by (𝑥𝑥) , we obtain

)

∑ [ x2 + 2 hn ( p1,[− p2 ]) − xhn+1 ( p1,[− p2 ])]t n

2 − xt

. ∑ Ln ( x)t n = 1 − xt − t 2

n =0

This completes the proof. Replacing 𝑡𝑡 by (−𝑡𝑡) in (3.8) and (3.10), we have the following theorems. Theorem 7. We have the following a new generating function of the Fibonacci polynomials at negative coefficient as ∞

∞

1 − xt − t 2

∞

(1 − xt − t 2 ) ∑ Ln ( x)t n =2 − xt

∑ Fn ( x)t n = 1 − xt − t 2 ,

2 − xt

∞

=2 + xt + xt ( ∑ L p ( x)t p − L0 ( x)) + t 2 ∑ L j ( x)t j

which is equivalent to

which represents a generating function for Pell-Lucas numbers at negative indices [3]. 𝑝𝑝 𝑝𝑝 = 1 and Choosing 𝑝𝑝1 and 𝑝𝑝2 such that � 1 2 𝑝𝑝1 − 𝑝𝑝2 = 𝑥𝑥 substituting in (3.2) and (3.3) we end up with ∞

= p 1 =j 0

= p 0=j 0

• Put k = 2 in the relationship (3.7) we have

2 + 2t

∞

=2 + xt + x( ∑ L p ( x)t p +1 ) + t 2 ∑ L j ( x)t j

= p 0=j 0

n =0

n

j= n − 2 and p= n − 1 . Then can be

Consider that written by

1 ⇒ ∑ Fk ,− n t = . 2 t − kt − 1 n =0

∞

∞

∑ Ln−2 ( x)t n .

= n 2= n 2

n

∞

∑ ( xLn−1 ( x) + Ln−2 ( x))t n

n=2 ∞

n =0

=

∞

1 , ∑ (−1) Fk , nt = 1 + kt − t 2 n =0 n

∑ (−1)n+1 Fk , nt n = t 2 − kt − 1

( n =0

n =0

∑ Ln ( x)t n = L0 ( x)t 0 + L1 ( x)t +

therefore

∞

∑ Ln ( x)t n .

Using the initial conditions, we get

n =0

∞

∞

1

. ∑ Fn (− x)t n = 1 + xt − t 2

n =0

Theorem 8. We have the following a new generating function of the Lucas polynomials at negative coefficient as ∞

2 + xt

. ∑ Ln (− x)t n = 1 + xt − t 2

n =0

.

4. Conclusion

Thus we get the following theorem. Theorem 6. We have the following a generating function of the Lucas polynomials as ∞

2 − xt

∑ n=0 Ln ( x)t n = 1 − xt − t 2 .

(3.10)

In this paper, a new theorem has been proposed in order to determine the generating functions. The proposed theorem is based on the symmetric functions. The obtained results agree with the results obtained in some previous works.

Turkish Journal of Analysis and Number Theory

Acknowledgments

[6] [7]

The authors would like to thank the anonymous referees for their valuable comments and suggestions.

[8]

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[9]

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