On the Periods of Biperiodic Fibonacci and Biperiodic Lucas Numbers

Hindawi Publishing Corporation Discrete Dynamics in Nature and Society Volume 2016, Article ID 7341729, 5 pages http://dx.doi.org/10.1155/2016/7341729

Research Article On the Periods of Biperiodic Fibonacci and Biperiodic Lucas Numbers Dursun TascJ and Gul Ozkan KJzJlJrmak Department of Mathematics, Gazi University, 06500 Ankara, Turkey Correspondence should be addressed to Gul Ozkan Kızılırmak; [email protected] Received 1 October 2016; Accepted 22 November 2016 Academic Editor: Allan C. Peterson Copyright © 2016 D. Tascı and G. Ozkan Kızılırmak. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. This paper is concerned with periods of Biperiodic Fibonacci and Biperiodic Lucas sequences taken as modulo prime and prime power. By using Fermat’s little theorem, quadratic reciprocity, many results are obtained.

1. Introduction Fibonacci sequence and Lucas sequence are well-known sequences among integer sequences. The Fibonacci numbers satisfy the recurrence relation 𝐹𝑛 = 𝐹𝑛−1 + 𝐹𝑛−2 with the initial conditions 𝐹0 = 0 and 𝐹1 = 1. Binet’s Formula for the Fibonacci sequence is 𝐹𝑛 =

𝛼𝑛 − 𝛽𝑛 , 𝛼−𝛽

(1)

where 𝛼 = (1 + √5)/2 and 𝛽 = (1 − √5)/2 are roots of the characteristic equation 𝑥2 − 𝑥 − 1 = 0. The positive root 𝛼 is known as “golden ratio.” Fibonacci numbers and their generalizations have many interesting properties and applications to almost every field of science and art [1–3]. In particular, Edson and Yayenie [4] introduced the Biperiodic Fibonacci sequence as follows: 𝑞0 = 0,

𝑞𝑚 = (

𝑎1−𝜉(𝑚) (𝑎𝑏)

) ⌊𝑚/2⌋

𝛼𝑚 − 𝛽𝑚 𝛼−𝛽

𝛼𝑚 − 𝛽𝑚 1 = ⌊(𝑚−1)/2⌋ ⌊(𝑚+1)/2⌋ , 𝛼−𝛽 𝑎 𝑏

if 𝑛 is even if 𝑛 is odd

(2) for 𝑛 ≥ 2,

where 𝑎 and 𝑏 are two nonzero real numbers. We take 𝑎 and 𝑏 as integers. If we take 𝑎 = 𝑏 = 1, we get the Fibonacci sequence {0, 1, 1, 2, 3, 5, 8, . . .}.

(3)

where 𝛼, 𝛽 = (𝑎𝑏 ± √𝑎2 𝑏2 + 4𝑎𝑏)/2 are the roots of the characteristic equation 𝑥2 − 𝑎𝑏𝑥 − 𝑎𝑏 = 0 and 𝜉(𝑚) = 𝑚 − 2⌊𝑚/2⌋. Moreover, (𝛼𝛽) = −𝑎𝑏, (𝛼 + 𝛽) = 𝑎𝑏.

𝑞1 = 1, {𝑎𝑞𝑛−1 + 𝑞𝑛−2 , 𝑞𝑛 = { 𝑏𝑞 + 𝑞𝑛−2 , { 𝑛−1

If we take 𝑎 = 𝑏 = 2, we get the Pell sequence {0, 1, 2, 5, 12, 29, 70, . . .}. Similarly, if we take 𝑎 = 𝑏 = 𝑘, we get the 𝑘–Fibonacci sequence [5–8]. Binet’s Formula for the Biperiodic Fibonacci sequence is given as

(4)

This sequence and its properties can be found in [1, 4]. Another well-known sequence is the Lucas sequence which satisfies the same recurrence relation as the Fibonacci sequence 𝐿 𝑛 = 𝐿 𝑛−1 + 𝐿 𝑛−2 with the initial conditions 𝐿 0 = 2 and 𝐿 1 = 1. Binet’s Formula for the Lucas sequence is 𝐿 𝑛 = 𝛼𝑛 + 𝛽𝑛 ,

(5)

2

Discrete Dynamics in Nature and Society

where 𝛼 and 𝛽 are defined in (1). Bilgici defined generalization of Lucas sequence similar to the Biperiodic Fibonacci sequence as follows:

Definition 1. The period of the Biperiodic Fibonacci sequence modulo a positive integer 𝑗 is the smallest positive integer 𝑚 such that 𝑞𝑚 ≡ 0

(mod𝑗) ,

𝑞𝑚+1 ≡ 1

(mod𝑗) .

𝑙0 = 2, 𝑙1 = 𝑎, {𝑏𝑙𝑛−1 + 𝑙𝑛−2 , 𝑙𝑛 = { 𝑎𝑙 + 𝑙 , { 𝑛−1 𝑛−2

if 𝑛 is even if 𝑛 is odd

(6) for 𝑛 ≥ 2,

By the definition above, the only members that can possibly come back to the starting point are multiples of 𝑚. This can be summed up in the statement that if 𝑚 is the period of 𝑞𝑛 (mod𝑗), then,

where 𝑎 and 𝑏 are two nonzero real numbers. We take 𝑎 and 𝑏 as integers. This sequence and other generalizations of Lucas sequence with their properties can be found in [9, 10]. If we take 𝑎 = 𝑏 = 1, we get the Lucas sequence {2, 1, 3, 4, 7, 11, 18, . . .}. Also, if we take 𝑎 = 𝑏 = 𝑘, we get the 𝑘–Lucas sequence [11]. 𝑙𝑛 = (

𝑎𝜉(𝑛) (𝑎𝑏)

⌊(𝑛+1)/2⌋

𝑞𝑘 ≡ 0

(mod𝑗)

𝑞𝑘+1 ≡ 1

(mod𝑗) ⇕

(10)

𝑚 | 𝑘, for any 𝑘 ∈ 𝑍.

) 𝛼𝑛 + 𝛽𝑛

1 = ⌊𝑛/2⌋ ⌊(𝑛+1)/2⌋ (𝛼𝑛 + 𝛽𝑛 ) , 𝑎 𝑏

(7)

Theorem 2. Let 𝑝 be a prime and let 𝑛 be a positive integer. If 𝑎≡1

where 𝛼, 𝛽, and 𝜉 are defined in (3). On the other hand several researchers have made significant studies about the period of the recurrence sequences [2]. Wall [12] defined the period-length of the recurring series obtained by reducing a Fibonacci series by a modulus 𝑚. As an example, the Fibonacci sequence mod3 is 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, . . .

(9)

(8)

and has period 8. Vinson [3] and Robinson [13] both extended Wall’s study. Moreover, they studied the Fibonacci sequence for prime moduli and showed that for primes 𝑝 ≡ 1, 4 (mod5) the period-length of the Fibonacci sequence mod𝑝 divides 𝑝 − 1, while for primes 𝑝 ≡ 2, 3 (mod5) the period-length of the Fibonacci sequence mod 𝑝 divides 2(𝑝+ 1). Gupta et al. [14] give alternative proofs of this results that also use the Fibonacci matrix. They place the roots of its characteristic polynomial in an appropriate splitting fields. Renault [15] examined the behaviour of the (𝑎, 𝑏)-Fibonacci sequence under a modulus. Lucas studied the (𝑎, 𝑏)-Fibonacci sequence extensively. He assigned Δ = 𝑎2 + 4𝑏 and deduced that if Δ is quadratic residue (that is, a nonzero perfect square) mod 𝑝, then 𝛼(𝑝) | 𝑝 − 1. If Δ is quadratic nonresidue then 𝛼(𝑝) | 𝑝 + 1. Also, Rogers and Campbell studied the period of the Fibonacci sequence mod𝑗 [16]. They investigated the Fibonacci sequence modulo 𝑝 prime and then generalized to prime powers.

(mod𝑝)

(11)

then, 𝑛

𝑎𝑝 ≡ 1

(mod𝑝𝑛+1 ) .

(12)

We remark that the proof of the Theorem 2 can be seen in [16]. Theorem 3. Let 𝑝 be a prime, let 𝑘 be a positive integer, and let 𝛼 and 𝛽 be the fundamental roots of the Biperiodic Fibonacci sequence. If 𝑚 is the period of 𝑞𝑛 (mod𝑝), 𝛼𝑚𝑝

𝑘−1

≡ 𝛽𝑚𝑝

𝑘−1

≡1

(mod𝑝𝑘 ) .

(13)

Proof. For 𝑎 ≠ 0 and 𝑝 is a prime integer, we have 𝑞𝑚 = (

(𝑎1−𝜉(𝑚) ) (𝑎𝑏)

) ⌊𝑚/2⌋

𝛼𝑚 − 𝛽𝑚 ≡0 𝛼−𝛽

(mod𝑝) ,

(14)

and then 𝛼𝑚 ≡ 𝛽𝑚

(mod𝑝) .

(15)

Also we obtain 𝑞𝑚 ≡ 𝑞𝑚+1 − 𝑞1 ≡(

(mod𝑝)

𝑎1−𝜉(𝑚+1) (𝑎𝑏)

⌊(𝑚+1)/2⌋

)

𝛼𝑚+1 − 𝛽𝑚+1 𝛼 − 𝛽 − 𝛼−𝛽 𝛼−𝛽

(16) (mod𝑝)

≡0

(mod𝑝) .

If 𝑚 is even, then

2. Period of Biperiodic Fibonacci Sequence In this section, we investigate the Biperiodic Fibonacci sequence modulo 𝑝 prime and then generalize to prime powers.

𝑞𝑚 ≡ (

1 (𝑎𝑏)

) 𝑚/2

𝛼 (𝛼𝑚 − 1) − 𝛽 (𝛽𝑚 − 1) 𝛼−𝛽

(17) (mod𝑝) .


3

From 𝛼𝑚 ≡ 𝛽𝑚 (mod𝑝), we get 𝑞𝑚 ≡ (

1 (𝑎𝑏)𝑚/2

) (𝛼𝑚 − 1) ≡ 0

(mod𝑝) .

Lemma 5. If 󳵻 is a quadratic nonresidue mod𝑝, then (√󳵻)𝑝 = −√󳵻. (18)

Thus, 𝛼𝑚 ≡ 𝛽𝑚 ≡ 1 (mod𝑝). From Theorem 2, 𝛼𝑚𝑝

𝑘−1

≡ 𝛽𝑚𝑝

𝑘−1

≡1

(mod𝑝𝑘 ) .

(19)

If 𝑚 is odd, 𝑞𝑚 ≡ (

𝑎

(𝑎𝑏)

)( (𝑚+1)/2

𝛼𝑚+1 − 𝛽𝑚+1 𝛼 − 𝛽 − ) 𝛼−𝛽 𝛼−𝛽

(20)

Lemma 6. Let 𝛼 and 𝛽 be the two roots of 𝑥2 − 𝑎𝑏𝑥 − 𝑎𝑏 = 0 in 𝐹 = 𝐹𝑝2 . 󳵻 is a quadratic nonresidue mod 𝑝; then 𝛽𝑝+1 = 𝛼𝑝+1 . Theorem 7. Let 𝑝 be an odd prime, let 𝑚 denote the period of 𝑞𝑛 (mod𝑝), and let 󳵻 be a quadratic nonresidue mod𝑝; then 𝑚 | 2𝑝 + 2. Proof. From the Binomial theorem, we get (𝛼 − 𝛽)𝑝 ≡ 𝛼𝑝 − 𝛽𝑝 (mod𝑝). It follows that

(mod𝑝) .

For 𝛼𝑚 ≡ 𝛽𝑚 (mod𝑝) and 𝑎/(𝑎𝑏)(𝑚+1)/2 ≠ 0, 𝑞𝑚 ≡ (

𝑎 (𝑎𝑏)(𝑚+1)/2

) (𝛼𝑚 − 1) ≡ 0

𝑞𝑝 = (

(mod𝑝) .

(21)

=(

Therefore, 𝛼𝑚 ≡ 𝛽𝑚 ≡ 1 (mod𝑝). From Theorem 2, 𝛼𝑚𝑝

𝑘−1

≡ 𝛽𝑚𝑝

𝑘−1

≡1

(mod𝑝) .

≡(

(22)

Theorem 4. Let 𝑝 be an odd prime, let 𝑚 denote the period of 𝑞𝑛 (mod𝑝), and let 󳵻 = 𝑎2 𝑏2 + 4𝑎𝑏 be a nonzero quadratic residue mod𝑝; then 𝑚 | 𝑝 − 1.

≡( ≡(

Proof. As is known from Fermat’s little theorem, 𝛼𝑝−1 ≡ 1

(mod𝑝) ,

𝛽𝑝−1 ≡ 1

(mod𝑝) .

(𝑎𝑏) (𝑎𝑏)

𝑞𝑝+1 = (

𝛼𝑝−1 − 𝛽𝑝−1 1 ) ≡0 𝛼−𝛽 𝑎⌊(𝑝−2)/2⌋ 𝑏⌊(𝑝−1)/2⌋

=(

𝛼𝑝 − 𝛽𝑝 𝛼−𝛽

𝛼𝑝 − 𝛽𝑝 𝛼−𝛽

1

) (𝑝−1)/2

(𝛼 − 𝛽) 𝛼−𝛽

1

(√󳵻)

) (𝑝−1)/2

1 (𝑎𝑏)

)

1

(𝑎𝑏)(𝑝−1)/2

≡ −1 (23)

Thus, we have 𝑞𝑝−1 ≡ (

1 𝑎⌊(𝑝−1)/2⌋ 𝑏⌊𝑝/2⌋

(𝑝−1)/2

) )

𝑝

(mod𝑝)

𝑝

(mod𝑝)

√󳵻 −√󳵻 √󳵻

(28)

(mod𝑝)

(mod𝑝) ,

𝛼𝑝+1 − 𝛽𝑝+1 1 ) 𝛼−𝛽 𝑎⌊𝑝/2⌋ 𝑏⌊(𝑝+1)/2⌋ 𝛼𝑝+1 − 𝛽𝑝+1 1 ) . 𝛼−𝛽 𝑎(𝑝−1)/2 𝑏(𝑝+1)/2

(mod𝑝) , 𝑞𝑝 = (

From Lemma 6, we obtain

𝛼𝑝 − 𝛽𝑝 ) ⌊(𝑝−1)/2⌋ ⌊𝑝/2⌋ 𝛼−𝛽 𝑎 𝑏 1

(24) 𝑞𝑝+1 ≡ 0

𝛼−𝛽 1 ) ≡( ≡ (𝑝−1)/2 (𝑝−1)/2 𝛼 − 𝛽 (𝑎𝑏) (𝑎𝑏) 1

From 𝑎𝑏 = −𝛼𝛽, we have 𝑝−1 1/2

)

≡1

(mod𝑝) ,

(25)

and then 𝑞𝑝 ≡ 1

(mod𝑝) .

(26)

So that 𝑞𝑝−1 ≡ 0

(mod𝑝) ,

𝑞𝑝 ≡ 1

(mod𝑝) .

Therefore, (10) implies that 𝑚 | 𝑝 − 1.

(29)

Thus, 𝑚 ∤ 𝑝 + 1. Also, we have

(mod𝑝) .

(𝑎𝑏)(𝑝−1)/2 = ((−𝛼𝛽)

(mod𝑝) .

𝑞2𝑝+2 = (

(𝛼 1 =( 𝑝) (𝑎𝑏) 𝑞2𝑝+3 = (

(27)

𝛼2𝑝+2 − 𝛽2𝑝+2 1 ) 𝛼−𝛽 𝑎⌊(2𝑝+1)/2⌋ 𝑏⌊(2𝑝+2)/2⌋ 𝑝+1 2

2

) − (𝛽𝑝+1 ) 𝛼−𝛽

1 𝑎⌊(2𝑝+2)/2⌋ 𝑏⌊(2𝑝+3)/2⌋

(𝛼 1 ) =( 𝑝+1 (𝑎𝑏)

)

≡0

(mod𝑝) ,

𝛼2𝑝+3 − 𝛽2𝑝+3 𝛼−𝛽

𝑝+1 2

2

) 𝛼 − (𝛽𝑝+1 ) 𝛽 𝛼−𝛽

4

Discrete Dynamics in Nature and Society 2

Since 𝛼𝛽 = −𝑎𝑏 and 𝛼𝑚𝑝

(𝛼𝑝+1 ) 1 𝑝+1 2 𝛼 − 𝛽 =( (𝛼 ) ) ≡ 𝛼 − 𝛽 (𝑎𝑏)𝑝+1 (𝑎𝑏)𝑝+1 ≡

(𝛼

𝑝+1 2

)

≡

𝑝+1

(−𝛼𝛽)

(𝑎𝑏)2𝑚𝑝

𝑝+1

𝛼 ≡1 𝛽𝑝+1

Theorem 8. Let 𝑝 be a prime, let 𝑚 denote the period of 𝑞𝑛 (mod𝑝), and let 𝑚󸀠 denote the period of 𝑞𝑛 (mod𝑝𝑘 ). If 𝑚𝑝𝑘−1 is even then 𝑚󸀠 | 𝑚𝑝𝑘−1 and if 𝑚𝑝𝑘−1 is odd then 𝑚󸀠 | 4𝑚𝑝𝑘−1 .

𝑞𝑚𝑝𝑘−1 = (

(𝑎𝑏)⌊𝑚𝑝

𝑘−1

)

𝑘−1 /2⌋

)

𝑚𝑝𝑘−1

𝛼𝑚𝑝

𝑞4𝑚𝑝𝑘−1 +1 ≡ 1

𝑞𝑚𝑝𝑘−1 +1 = (

If 𝑚𝑝

(𝑎𝑏)

⌊(𝑚𝑝𝑘−1 +1)/2⌋

)

1−𝜉(𝑚𝑝𝑘−1 +1)

𝑎

≡ 𝑘−1

𝑎

𝑚𝑝𝑘−1

(𝑎𝑏)

𝑘−1

− 𝛽𝑚𝑝 𝛼−𝛽

≡0

Definition 9. The period of the Biperiodic Lucas sequence modulo a positive integer 𝑗 is the smallest positive integer 𝑡 such that

𝑘−1

𝑚𝑝𝑘−1 +1

(31)

𝑚𝑝𝑘−1 +1

−𝛽 𝛼−𝛽

𝑘−1

/2

= (−𝛼𝛽)

1

𝑚𝑝𝑘−1 /2

(mod𝑝 ) .

𝑘−1 /2

= ((−𝛼𝛽)

𝑞𝑚𝑝𝑘−1 +1 ≡ 1

𝑚𝑝𝑘−1

)

1/2

(33)

(mod𝑝𝑘 ) .

(𝑎𝑏)⌊4𝑚𝑝

𝑘−1 /2⌋

)

𝛼4𝑚𝑝

𝑘−1

− 𝛽4𝑚𝑝 𝛼−𝛽

(34)

≡0

≡

𝑙𝑝−1 = (

(35)

(𝑎𝑏)⌊(4𝑚𝑝 1

(𝑎𝑏)

2𝑚𝑝𝑘−1

+1)

𝑘−1 +1)/2⌋

)

𝛼4𝑚𝑝

(mod𝑝𝑘 ) .

𝑘−1

+1

(38)

𝛼𝑝−1 ≡ 1

(mod𝑝) ,

𝛽𝑝−1 ≡ 1

(mod𝑝) ,

(39)

so

𝑞4𝑚𝑝𝑘−1 +1 𝑘−1

(mod𝑗)

Proof. We use Fermat’s little theorem to get

𝑘−1

(mod𝑝 ) ,

𝑎1−𝜉(4𝑚𝑝

(mod𝑗) ,

Theorem 10. Let 𝑝 be an odd prime, let 𝑡 denote the period of 𝑙𝑛 (mod𝑝), and let 󳵻 be a nonzero quadratic residue mod 𝑝; then 𝑡 | 𝑝 − 1.

𝑘

=(

(37)

for any 𝑘 ∈ 𝑍,

≡1

Thus, from (10), 𝑚󸀠 | 𝑚𝑝𝑘−1 . If 𝑚𝑝𝑘−1 is odd then 𝑞4𝑚𝑝𝑘−1 = (

(mod𝑗) .

𝑡 | 𝑘.

Then,

)

𝑙𝑡+1 ≡ 𝑎

⇕

(32)

(mod𝑝 ) .

𝑘−1

(mod𝑗) ,

𝑙𝑘+1 ≡ 1

𝑘

𝑘

𝑎1−𝜉(4𝑚𝑝

𝑙𝑡 ≡ 2

For the same reasons as the Biperiodic Fibonacci sequence we have that if 𝑡 is the period of 𝑙𝑛 (mod𝑗), then

From 𝛼𝛽 = −𝑎𝑏 and by using Theorem 3, it follows that (𝑎𝑏)𝑚𝑝

(mod𝑝𝑘 ) .

𝑙𝑘 ≡ 0

(𝑎𝑏)𝑚𝑝

2

) ≡1

3. Period of Biperiodic Lucas Sequence

is even, then 𝑞𝑚𝑝𝑘−1 +1 ≡

𝑚𝑝𝑘−1

≡ 1 (mod𝑝 ) in

(mod𝑝𝑘 ) .

⌊(𝑚𝑝𝑘−1 +1)/2⌋

= ((−𝛼𝛽)

In this section, we investigate the Biperiodic Lucas sequence modulo 𝑝 a prime similar to Biperiodic Fibonacci sequence.

𝑘

≡𝛽

𝛼

≡ 1 (mod𝑝𝑘 ), then,

Thus, 𝑚󸀠 | 4𝑚𝑝𝑘−1 .

(mod𝑝𝑘 ) , 1−𝜉(𝑚𝑝𝑘−1 +1)

𝑘−1

(mod𝑝𝑘 ) , (36)

Thus, from (10), 𝑚 | 2𝑝 + 1.

𝑎1−𝜉(𝑚𝑝

2𝑚𝑝𝑘−1

= (−𝛼𝛽)

≡ 𝛽𝑚𝑝

(mod𝑝) . (30)

Proof. We have shown that 𝛼 Theorem 3. So that

𝑘−1

𝑘−1

− 𝛽4𝑚𝑝 𝛼−𝛽

𝑘−1

1 2 ) (𝛼𝑝−1 + 𝛽𝑝−1 ) ≡ (𝑝−1)/2 𝑎⌊(𝑝−1)/2⌋ 𝑏⌊𝑝/2⌋ (𝑎𝑏)

(40)

(mod𝑝) .

+1

We know (𝑎𝑏)(𝑝−1)/2 ≡ 1 (mod𝑝); thus 𝑙𝑝−1 ≡ 2

(mod𝑝) .

(41)


5

References

Also, we have 𝑙𝑝 = ( ≡( ≡

1 ) (𝛼𝑝 + 𝛽𝑝 ) ⌊𝑝/2⌋ ⌊(𝑝+1)/2⌋ 𝑎 𝑏 1 𝑎(𝑝−1)/2 𝑏(𝑝+1)/2 𝑎𝑏

(𝑎𝑏)(𝑝−1)/2

≡𝑎

) (𝛼 + 𝛽)

(mod𝑝) (42)

𝑎𝑏 ≡ 𝑏

(mod𝑝)

(mod𝑝) .

By using (10), we get 𝑡 | 𝑝 − 1. Theorem 11. Let 𝑝 be an odd prime, let 𝑡 denote the period of 𝑙𝑛 (mod𝑝), and let 󳵻 be a quadratic nonresidue mod𝑝; then 𝑡 | 2𝑝 + 2. Proof. From Lemma 6 and (3), we get 𝑙2𝑝+2 = ( ≡(

1 𝑎⌊(2𝑝+2)/2⌋ 𝑏⌊(2𝑝+3)/2⌋

) (𝛼2𝑝+2 + 𝛽2𝑝+2 )

2 2 1 ) ((𝛼𝑝+1 ) + (𝛽𝑝+1 ) ) 𝑝+1 (𝑎𝑏)

(mod𝑝)

2

≡

2 (𝛽𝑝+1 )

(mod𝑝)

(𝑎𝑏)𝑝+1 2

≡ ≡

2 (𝛽𝑝+1 ) 2𝛽𝑝+1 𝛼𝑝+1

≡2 𝑙2𝑝+3

(mod𝑝)

𝛼𝑝+1 𝛽𝑝+1

(mod𝑝)

(mod𝑝) ,

(43)

1 = ( ⌊(2𝑝+3)/2⌋ ⌊(2𝑝+4)/2⌋ ) (𝛼2𝑝+3 + 𝛽2𝑝+3 ) 𝑎 𝑏 =( =( = = =

1 𝑎𝑝+1 𝑏𝑝+2 1

) (𝛼2𝑝+3 + 𝛽2𝑝+3 ) 2

𝑎𝑝+1 𝑏𝑝+2 1

(𝛼 +

2

) ((𝛼𝑝+1 ) 𝛼 + (𝛽𝑝+1 ) 𝛽)

2 𝛽) (𝛼𝑝+1 )

1 2 (𝑎𝑏) (𝛼𝑝+1 )

(𝑎𝑏)𝑝+1 𝑏

(𝑎𝑏)𝑝+1 𝑏 =

1 1 𝛽𝑝+1 = 𝑎𝛼𝑝+1 𝑎

1 𝑎 (𝛼𝑝+1 )

2

𝛼𝑝+1 𝛽𝑝+1

(mod𝑝) .

Thus, from (10), 𝑚 | 2𝑝 + 2.

Competing Interests The authors declare that they have no competing interests.

[1] M. Edson, S. Lewis, and O. Yayenie, “The k-periodic Fibonacci sequence and extended Binet’s formula,” Integers, vol. 11, pp. 739–751, 2011. [2] J. Kramer and V. E. Hoggatt, Special Cases of Fibonacci Periodicity, 1972. [3] J. Vinson, “The relation of the period modulo m to the rank of apparition of m in the fibonacci sequence,” The Fibonacci Quarterly, pp. 37–45, 1963. [4] M. Edson and O. Yayenie, “A new generalization of Fibonacci sequence and extended Binet’s formula,” Integers. Electronic Journal of Combinatorial Number Theory, vol. 9, pp. 639–654, 2009. ´ Plaza, “On the Fibonacci k-numbers,” Chaos, [5] S. Falc´on and A. Solitons & Fractals, vol. 32, no. 5, pp. 1615–1624, 2007. ´ Plaza, “k-Fibonacci sequences modulo m,” [6] S. Falcon and A. Chaos, Solitons and Fractals, vol. 41, no. 1, pp. 497–504, 2009. ´ Plaza, “The k-Fibonacci sequence and the [7] S. Falc´on and A. Pascal 2-triangle,” Chaos, Solitons & Fractals, vol. 33, no. 1, pp. 38–49, 2007. [8] S. Falcon and A. Plaza, “On k-Fibonacci sequences and polynomials and their derivatives,” Chaos, Solutions and Fractals, vol. 39, no. 3, pp. 1005–1019, 2009. [9] G. Bilgici, “Two generalizations of Lucas sequence,” Applied Mathematics and Computation, vol. 245, pp. 526–538, 2014. [10] G. Bilgici, “New generalizations of Fibonacci and Lucas sequences,” Applied Mathematical Sciences, vol. 8, no. 29, pp. 1429–1437, 2014. [11] S. Falcon, “On the k-Lucas numbers,” International Journal of Contemporary Mathematical Sciences, vol. 21, pp. 1039–1050, 2011. [12] D. D. Wall, “Fibonacci series modulo m,” The American Mathematical Monthly, vol. 67, no. 6, pp. 525–532, 1960. [13] D. W. Robinson, “The fibonacci matrix modulo m,” The Fibonacci Quarterly, vol. 1, pp. 29–36, 1963. [14] S. Gupta, P. Rockstroh, and F. E. Su, “Splitting fields and periods of Fibonacci sequences modulo primes,” Mathematics Magazine, vol. 85, no. 2, pp. 130–135, 2012. [15] M. Renault, “The period, rank, and order of the (a, b)−fibonacci sequence Mod m,” Mathematics Magazine, vol. 86, no. 5, pp. 372–380, 2013. [16] N. Rogers and C. W. Campbell, The period of the fibonacci sequence modulo j [Ph.D. thesis], The University Of Arizona, Tucson, Ariz, USA, 2007.

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