# On the Periods of 2-Step General Fibonacci Sequences in the Generalized Quaternion Groups

Authors: Bahram Ahmadi; Hossein Doostie
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We study 2-step general Fibonacci sequences in the generalized quaternion groupsQ4n. In cases where the sequences are proved to be simply periodic, we obtain the periods of 2-step general Fibonacci sequences.


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## Body

## 1. Introduction

The study of the Fibonacci sequences in groups began with the earlier work of Wall [1] in 1960, where the ordinary Fibonacci sequences in cyclic groups were investigated. In the mid-eighties, Wilcox [2] extended the problem to the abelian groups. In 1990, Campbell et al. [3] expanded the theory to some classes of finite groups. In 1992, Knox proved that the periods of k-nacci (k-step Fibonacci) sequences in the dihedral groups are equal to 2k+2, in [4]. In the progress of this study, the article [5] of Aydin and Smith proves that the lengths of the ordinary 2-step Fibonacci sequences are equal to the lengths of the ordinary 2-step Fibonacci recurrences in finite nilpotent groups of nilpotency class 4 and a prime exponent, in 1994.Since 1994, the theory has been generalized and many authors had nice contributions in computations of recurrence sequences in groups and we may give here a brief of these attempts. In [6, 7] the definition of the Fibonacci sequence has been generalized to the ordinary 3-step Fibonacci sequences in finite nilpotent groups. Then in [8] it is proved that the period of 2-step general Fibonacci sequence is equal to the length of the fundamental period of the 2-step general recurrence constructed by two generating elements of a group of nilpotency class 2 and exponent p. In [9] Karaduman and Yavuz showed that the periods of the 2-step Fibonacci recurrences in finite nilpotent groups of nilpotency class 5 and a prime exponent are p·k(p), for 2<p≤2927, where p is a prime and k(p) is the period of the ordinary 2-step Fibonacci sequence. The main role of [10, 11] in generalizing the theory was to study the 2-step general Fibonacci sequences in finite nilpotent groups of nilpotency class 4 and exponent p and to the 2-step Fibonacci sequences in finite nilpotent groups of nilpotency class n and exponent p, respectively.One may consult [12, 13] to see the results of the Fibonacci sequences in the modular groups concerning the periodicity of 2-step Fibonacci sequences constructed by two generating elements.Going on a detailed literature in this area of research, we have to mention certain essential computation on the Fibonacci lengths of new structures like the semidirect products, the direct products, and the automorphism groups of finite groups which have been studied in [14–19]. Finally, we refer to [20] where Karaduman and Aydin studied the periodicity property of 2-step general Fibonacci sequences in dihedral groups and the goal of this paper is to calculate the periods of 2-step general Fibonacci sequences in the generalized quaternion groups.LetG=〈a0,a1,a2,…,aj-1〉 be a finite group. A k-nacci sequence in group G is a sequence {xn}n=0∞ of group elements for which each element is defined by x0=a0,x1=a1,…,xj-1=aj-1,
(1.1)xn={x0x1⋯xn-1,forj≤n<k,xn-kxn-k+1⋯xn-1,forn≥k.
This sequence of the group G is denoted by Fk(G;a0,a1,a2,…,aj-1). We also call a 2-nacci sequence of group elements a Fibonacci sequence of a finite group. A finite group G is k-nacci sequenceable if there exists a k-nacci sequence of G such that every element of the group appears in the sequence. A sequence of group elements is periodic if after a certain point, it consists only of repetitions of a fixed subsequence. The number of elements in the repeating subsequence is called the period of the sequence. For example, the sequence a,b,c,d,d,e,b,c,d,e,… is periodic after the initial element a and has period 4. We denote the period of a k-nacci sequence Fk(G;a0,a1,a2,…,aj-1) by Pk(G;a0,a1,a2,…,aj-1). A sequence of group elements is called simply periodic with period k if the first k elements in the sequence form a repeating subsequence. For example, the sequence a,b,c,d,e,f,a,b,c,d,e,f,… is simply periodic with period 6. The following theorem is well known.Theorem 1.1 (see [4]).
Everyk-nacci sequence in a finite group is simply periodic.


## 2. Main Theorems

The generalized quaternion groupQ4n, (n≥2) is a group with a presentation of the form
(2.1)〈a,b∣a2n=1,an=b2,b-1ab=a-1〉.
It is easy to see that Q4n is of order 4n, a has order 2n, b has order 4, and the relation akb=ba-k holds for all k∈ℤ.First we consider the following lemma which will be used frequently without further reference.Lemma 2.1.
For an integerr, where 1≤r<2n, and for s∈{1,3}, the following relations hold in Q4n:(i)
(arbs)t=bst, where t is any even integer;(ii)
(arbs)t=arbst, where t is any odd integer.Proof.
We may use an induction method ont for both parts, simultaneously.Theorem 2.2.
P
k
(
Q
4
n
;
a
,
b
)
=
2
k
+
2.Proof.
Obviously, the order ofab is 4. If k=2, then
(2.2)x0=a,x1=b,x2=ab,x3=an-1,x4=an+2b,x5=ab,x6=a,x7=b,…,
and hence the sequence has period 6. If k=3, then
(2.3)x0=a,x1=b,x2=ab,x3=(ab)2=b2,x4=a-1,x5=a2b3,x6=ab,x7=1,x8=a,x9=b,…,
and so the period is 8. Now let k≥4. Then, the first k+1 elements of Fk(Q4n;a,b) are
(2.4)x0=a,x1=b,x2=ab,x3=(ab)2,x4=(ab)4,…,xk=(ab)2k-2.
Since ab is of order 4, this sequence reduces to
(2.5)x0=a,x1=b,x2=ab,x3=b2,xj=1,(4≤j≤k).
Thus,
(2.6)xk+1=∏i=1kxi=b(ab)b2=a-1,xk+2=∏i=2k+1‍xi=(ab)b2a-1=a2b3,xk+3=∏i=3k+2‍xi=b2a-1(a2b3)=ab,xk+4=∏i=4k+3‍xi=a-1(a2b3)(ab)=1.
It follows that xk+j=1 for j  (4≤j≤k+1). We also have
(2.7)x2k+2=∏i=k+22k+1xi=(a2b3)(ab)=a,x2k+3=∏i=k+32k+2xi=(ab)a=b.
It shows that Pk(Q4n;a,b)=2k+2.LetG=〈a,b〉 be a finite nonabelian 2-generated group. A 2-step general Fibonacci sequence in the group G is defined by x0=a,x1=b,xi=xi-2mxi-1l, for i≥2, and the integers m and l. Now we study this sequence for group Q4n.Theorem 2.3.
Letm and l be integers. If m≡0 (mod2n) or m≡0 (mod4), then 2-step general Fibonacci sequence in Q4n is not simply periodic.Proof.
First we consider the casem≡0 (mod 2n). Then the sequence is
(2.8)x0=a,x1=b,x2=ambl=bl,x3=bmbl2=bm+l2,….
Obviously, the cycle does not begin again with a and b, and hence the sequence is not simply periodic.
Now letm≡0 (mod 4). Four cases occur.
Case1 (l≡0 (mod 4)). Then
(2.9)x0=a,x1=b,x2=am,x3=aml,x4=am2+ml2,x5=a2m2l+ml3,x6=am3+3m2l2+ml4,….
Since the cycle does not begin again witha and b, the sequence is not simply periodic.
Case2 (l≡1 (mod 4)). Then
(2.10)x0=a,x1=b,x2=amb,x3=amb,x4=amb,….
Clearly, the sequence is not simply periodic.
Case 3 (l≡2 (mod 4)). Note that b2 is a central element of Q4n. Thus
(2.11)x0=a,x1=b,x2=amb2,x3=aml,x4=am2+ml2,x5=a2m2l+ml3,….
Similar to Case 1, the sequence is not simply periodic.
Case 4 (l≡3 (mod 4)). So
(2.12)x0=a,x1=b,x2=amb3,x3=amb,x4=amb3,x5=amb,….
Thus the sequence is not simply periodic.Theorem 2.4.
Letn be an even integer.(i)
Ifm≡1 (mod2n) and l≡0 (mod2n), then 2-step general Fibonacci sequence in Q4n is simply periodic with period 2.(ii)
Ifm≡-1 (mod2n) and l≡-1 (mod2n), then 2-step general Fibonacci sequence in Q4n is simply periodic with period 3.(iii)
Ifm≡1 (mod2n) and l≡1 (mod2n), then 2-step general Fibonacci sequence in Q4n is simply periodic with period 6.Proof.
(i) Sincen is even, m≡1 (mod 4) and l≡0 (mod 4). Thus
(2.13)x0=a,x1=b,x2=a,x3=b,…,
and the period is 2.
(ii) Sincen is even, m≡3 (mod 4) and l≡3 (mod 4). Note that m and l are odd integers. So the sequence reduces to
(2.14)x0=a,x1=b,x2=a-1b3,x3=a,x4=b,…,
and the period is 3.
(iii) Becausen is even, m≡1 (mod 4) and l≡1 (mod 4). Furthermore, m and l are odd integers. Then the sequence reduces to
(2.15)x0=a,x1=b,x2=ab,x3=an-1,x4=a2b3,x5=ab,x6=a,x7=b,…,
and the period is 6.Theorem 2.5.
Letn be an odd integer.(i)
Ifm≡1 (mod2n), m≡1 (mod4), l≡0 (mod4) and (l,n)=1, then 2-step general Fibonacci sequence in Q4n is simply periodic with period 2n.(ii)
Ifm≡1 (mod2n), m≡3 (mod4), l≡0 (mod4) and (l,n)=1, then 2-step general Fibonacci sequence in Q4n is simply periodic with period 4n.Proof.
(i) By induction onk≥0, we can show that x2k=a and x2k+1=a-klb. In particular, the period must be even. Now we have
(2.16)x2k+1=b⟺2n∣kl⟺n∣k(because(l,n)=1),
and so the period is 2n.
(ii) By induction onk≥0, it may be shown that
(2.17)x2k=a,x2k+1={a-klbifkiseven,a-klb3ifkisodd.
Therefore, the period must be even. Now we have
(2.18)x2k+1=b⟺2n∣kl⟺2n∣k(because(l,n)=1andnisodd),
and thus the period is 4n.Theorem 2.6.
Letn and l be odd integers, m≡-1 (mod2n) and (l+1,n)=1.(i)
Ifm≡1 (mod4), then 2-step general Fibonacci sequence in Q4n is simply periodic with period 6n.(ii)
Ifm≡3 (mod4) and l≡1 (mod4), then 2-step general Fibonacci sequence in Q4n is simply periodic with period 6n.(iii)
Ifm≡3 (mod4) and l≡3 (mod4), then 2-step general Fibonacci sequence in Q4n is simply periodic with period 3n.Proof.
(i) Two cases occur.
Case
1 (l≡1 (mod 4)). By induction on k≥0, we can prove that the following relations hold:
(2.19)x6k=a,x6k+1=a-2k(l+1)b,x6k+2=a-(2k(l+1)+1)b,x6k+3=an+1,x6k+4=a-(n+(2k+1)(l+1))b,x6k+5=a-((2k+1)(l+1)+1)b.
Consequently, the period must be a multiple of 6. Now we have
(2.20)x6k+1=b⟺2n∣2k(l+1)⟺n∣k(because(l+1,n)=1),
and hence the period is 6n.
Case
2 (l≡3 (mod 4)). The proof is similar to Case 1 except that
(2.21)x6k+2=a-(n+2k(l+1)+1)b,x6k+4=a-(2k+1)(l+1)b.
(ii) It is easily shown that the following relations hold for everyk≥0:
(2.22)x3k=a,x6k=a,x6k+1=a-2k(l+1)b,x6k+2=a-(2k(l+1)+1)b,x6k+3=a,x6k+4=a-(n+(2k+1)(l+1))b,x6k+5=a-(n+1+(2k+1)(l+1))b.
Particularly, the period must be a multiple of 3. Since n+(2k+1)(l+1) is odd, then it is not a multiple of 2n. Thus x6k+4≠b. Further,
(2.23)x6k+1=b⟺2n∣2k(l+1)⟺n∣k(because(l+1,n)=1),
and hence the period is 6n.
(iii) By induction onk≥0, we can prove the following relations:
(2.24)x3k=a,x3k+1=a-k(l+1)b,x3k+2=a-(n+1+k(l+1))b.
Therefore, the period must be a multiple of 3. Now we have
(2.25)x3k+1=b⟺2n∣k(l+1)⟺n∣k(because(l+1,n)=1),
and so the period is 3n.



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