# On the Periods of Biperiodic Fibonacci and Biperiodic Lucas Numbers

Authors: Dursun Tascı; Gul Ozkan Kızılırmak
Journal: Discrete Dynamics in Nature and Society
 (2016)

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## Abstract

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.


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## Body

## 1. Introduction

Fibonacci sequence and Lucas sequence are well-known sequences among integer sequences. The Fibonacci numbers satisfy the recurrence relationFn=Fn-1+Fn-2 with the initial conditions F0=0 and F1=1. Binet’s Formula for the Fibonacci sequence is(1)Fn=αn-βnα-β,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:(2)q0=0,q1=1,qn=aqn-1+qn-2,ifnis  evenbqn-1+qn-2,ifnis  oddforn≥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,….If we takea=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(3)qm=a1-ξmabm/2αm-βmα-β=1am-1/2bm+1/2αm-βmα-β,where α,β=ab±a2b2+4ab/2 are the roots of the characteristic equation x2-abx-ab=0 and ξ(m)=m-2m/2. Moreover, (4)αβ=-ab,α+β=ab.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 sequenceLn=Ln-1+Ln-2 with the initial conditions L0=2 and L1=1. Binet’s Formula for the Lucas sequence is(5)Ln=αn+βn,where α and β are defined in (1). Bilgici defined generalization of Lucas sequence similar to the Biperiodic Fibonacci sequence as follows:(6)l0=2,l1=a,ln=bln-1+ln-2,ifnis  evenaln-1+ln-2,ifnis  oddforn≥2,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].(7)ln=aξnabn+1/2αn+βn=1an/2bn+1/2αn+βn,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 mod⁡3 is(8)0,1,1,2,0,2,2,1,0,1,1,…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 mod⁡p divides p-1, while for primes p≡2,3(mod⁡5) 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 mod⁡j [16]. They investigated the Fibonacci sequence modulo p prime and then generalized to prime powers.


## 2. Period of Biperiodic Fibonacci Sequence

In this section, we investigate the Biperiodic Fibonacci sequence modulop prime and then generalize to prime powers.Definition 1.
The period of the Biperiodic Fibonacci sequence modulo a positive integerj is the smallest positive integer m such that(9)qm≡0mod⁡j,qm+1≡1mod⁡j.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(mod⁡j), then, (10)qk≡0mod⁡jqk+1≡1mod⁡j⇕m∣k,for anyk∈Z.Theorem 2.
Letp be a prime and let n be a positive integer. If (11)a≡1mod⁡pthen,(12)apn≡1mod⁡pn+1.We remark that the proof of the Theorem2 can be seen in [16].Theorem 3.
Letp 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(mod⁡p), (13)αmpk-1≡βmpk-1≡1mod⁡pk.Proof.
Fora≠0 and p is a prime integer, we have (14)qm=a1-ξmabm/2αm-βmα-β≡0mod⁡p,and then (15)αm≡βmmod⁡p.Also we obtain (16)qm≡qm+1-q1mod⁡p≡a1-ξm+1abm+1/2αm+1-βm+1α-β-α-βα-βmod⁡p≡0mod⁡p.If m is even, then (17)qm≡1abm/2ααm-1-ββm-1α-βmod⁡p.From αm≡βm(mod⁡p), we get(18)qm≡1abm/2αm-1≡0mod⁡p.Thus, αm≡βm≡1(mod⁡p). From Theorem 2,(19)αmpk-1≡βmpk-1≡1mod⁡pk.If m is odd,(20)qm≡aabm+1/2αm+1-βm+1α-β-α-βα-βmod⁡p.For αm≡βm(mod⁡p) and a/abm+1/2≠0,(21)qm≡aabm+1/2αm-1≡0mod⁡p.Therefore, αm≡βm≡1(mod⁡p). From Theorem 2,(22)αmpk-1≡βmpk-1≡1mod⁡p.Theorem 4.
Letp be an odd prime, let m denote the period of qn(mod⁡p), and let ∆=a2b2+4ab be a nonzero quadratic residue mod⁡p; then m∣p-1.Proof.
As is known from Fermat’s little theorem,(23)αp-1≡1mod⁡p,βp-1≡1mod⁡p.Thus, we have(24)qp-1≡1ap-2/2bp-1/2αp-1-βp-1α-β≡0mod⁡p,qp=1ap-1/2bp/2αp-βpα-β≡1abp-1/2α-βα-β≡1abp-1/2mod⁡p.From ab=-αβ, we have (25)abp-1/2=-αβp-11/2≡1mod⁡p,and then (26)qp≡1mod⁡p.So that (27)qp-1≡0mod⁡p,qp≡1mod⁡p.Therefore, (10) implies that m∣p-1.Lemma 5.
If∆ is a quadratic nonresidue modp, then ∆p=-∆.Lemma 6.
Letα and β be the two roots of x2-abx-ab=0 in F=Fp2. ∆ is a quadratic nonresidue modp; then βp+1=αp+1.Theorem 7.
Letp be an odd prime, let m denote the period of qn(mod⁡p), and let ∆ be a quadratic nonresidue mod⁡p; then m∣2p+2.Proof.
From the Binomial theorem, we getα-βp≡αp-βp(mod⁡p). It follows that(28)qp=1ap-1/2bp/2αp-βpα-β=1abp-1/2αp-βpα-β≡1abp-1/2α-βpα-βmod⁡p≡1abp-1/2∆p∆mod⁡p≡1abp-1/2-∆∆mod⁡p≡-1mod⁡p,qp+1=1ap/2bp+1/2αp+1-βp+1α-β=1ap-1/2bp+1/2αp+1-βp+1α-β.From Lemma 6, we obtain(29)qp+1≡0mod⁡p.Thus, m∤p+1. Also, we have(30)q2p+2=1a2p+1/2b2p+2/2α2p+2-β2p+2α-β=1abpαp+12-βp+12α-β≡0mod⁡p,q2p+3=1a2p+2/2b2p+3/2α2p+3-β2p+3α-β=1abp+1αp+12α-βp+12βα-β=1abp+1αp+12α-βα-β≡αp+12abp+1≡αp+12-αβp+1≡αp+1βp+1≡1mod⁡p.Thus, from (10), m∣2p+1.Theorem 8.
Letp be a prime, let m denote the period of qnmodp, and let m′ denote the period of qnmodpk. If mpk-1 is even then m′∣mpk-1 and if mpk-1 is odd then m′∣4mpk-1.Proof.
We have shown thatαmpk-1≡βmpk-1≡1(mod⁡pk) in Theorem 3. So that (31)qmpk-1=a1-ξmpk-1abmpk-1/2αmpk-1-βmpk-1α-β≡0mod⁡pk,qmpk-1+1=a1-ξmpk-1+1abmpk-1+1/2αmpk-1+1-βmpk-1+1α-β≡a1-ξmpk-1+1abmpk-1+1/2mod⁡pk.If mpk-1 is even, then(32)qmpk-1+1≡1abmpk-1/2mod⁡pk.From αβ=-ab and by using Theorem 3, it follows that(33)abmpk-1/2=-αβmpk-1/2=-αβmpk-11/2≡1mod⁡pk.Then,(34)qmpk-1+1≡1mod⁡pk.Thus, from (10), m′∣mpk-1.
Ifmpk-1 is odd then(35)q4mpk-1=a1-ξ4mpk-1ab4mpk-1/2α4mpk-1-β4mpk-1α-β≡0mod⁡pk,q4mpk-1+1=a1-ξ4mpk-1+1ab4mpk-1+1/2α4mpk-1+1-β4mpk-1+1α-β≡1ab2mpk-1mod⁡pk.Since αβ=-ab and αmpk-1≡βmpk-1≡1(mod⁡pk), then, (36)ab2mpk-1=-αβ2mpk-1=-αβmpk-12≡1mod⁡pk,q4mpk-1+1≡1mod⁡pk.Thus, m′∣4mpk-1.


## 3. Period of Biperiodic Lucas Sequence

In this section, we investigate the Biperiodic Lucas sequence modulop a prime similar to Biperiodic Fibonacci sequence.Definition 9.
The period of the Biperiodic Lucas sequence modulo a positive integerj is the smallest positive integer t such that(37)lt≡2mod⁡j,lt+1≡amod⁡j.For the same reasons as the Biperiodic Fibonacci sequence we have that ift is the period of ln(mod⁡j), then (38)lk≡0mod⁡j,lk+1≡1mod⁡j⇕t∣k.for  anyk∈Z,Theorem 10.
Letp be an odd prime, let t denote the period of ln(mod⁡p), and let ∆ be a nonzero quadratic residue modp; then t∣p-1.Proof.
We use Fermat’s little theorem to get(39)αp-1≡1mod⁡p,βp-1≡1mod⁡p,so(40)lp-1=1ap-1/2bp/2αp-1+βp-1≡2abp-1/2mod⁡p.We know abp-1/2≡1(mod⁡p); thus(41)lp-1≡2mod⁡p.Also, we have(42)lp=1ap/2bp+1/2αp+βp≡1ap-1/2bp+1/2α+βmod⁡p≡ababp-1/2≡abbmod⁡p≡amod⁡p.By using (10), we get t∣p-1.Theorem 11.
Letp be an odd prime, let t denote the period of ln(mod⁡p), and let ∆ be a quadratic nonresidue mod⁡p; then t∣2p+2.Proof.
From Lemma6 and (3), we get (43)l2p+2=1a2p+2/2b2p+3/2α2p+2+β2p+2≡1abp+1αp+12+βp+12mod⁡p≡2βp+12abp+1mod⁡p≡2βp+12αp+1βp+1mod⁡p≡2βp+1αp+1mod⁡p≡2mod⁡p,l2p+3=1a2p+3/2b2p+4/2α2p+3+β2p+3=1ap+1bp+2α2p+3+β2p+3=1ap+1bp+2αp+12α+βp+12β=1α+βαp+12abp+1b=1abαp+12abp+1b=1aαp+12αp+1βp+1=1aαp+1βp+1=1amod⁡p.
Thus, from (10), m∣2p+2.



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