# On the Basick-nacci Sequences in Finite Groups

Authors: Ömür Deveci; Erdal Karaduman
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

We define the basick-nacci sequences and the basic periods of these sequences in finite groups, then we obtain the basic periods of the basic k-nacci sequences and the periods of the k-nacci sequences in symmetric group S4, its subgroups, and binary polyhedral groups which related with these groups.


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## Body

## 1. Introduction

The study of Fibonacci sequences in groups began with the earlier work of Wall [1], where the ordinary Fibonacci sequences in cyclic groups were investigated. In the mid-eighties, Wilcox extended the problem to Abelian groups [2]. The theory is expanded to some finite simple groups by Campbell et al. [3]. There, they defined the Fibonacci length of the Fibonacci orbit and the basic Fibonacci length of the basic Fibonacci orbit in a 2-generator group. The concept of Fibonacci length for more than two generators has also been considered; see, for example, [4, 5]. Also, the theory has been expanded to the nilpotent groups; see, for example, [6, 7]. Other works on Fibonacci length are discussed in, for example, [8–10]. Knox proved that the periods of k-nacci (k-step Fibonacci) sequences in dihedral groups were equal to 2k+2 [11]. Deveci, Karaduman, and Campbell examined the period of the k-nacci sequences in some finite binary polyhedral groups in [12]. Recently, k-nacci sequences have been investigated; see, for example, [13, 14].This paper defines the basick-nacci sequences and the periods of these sequences in finite groups and discusses the basic periods of the basic k-nacci sequences and the periods of the k-nacci sequences in the symmetric group S4, alternating group A4, D2 four-group, and binary polyhedral groups 〈2,3,4〉 and 〈2,3,3〉 with related S4 and A4, respectively. We consider the groups S4, A4, 〈2,3,4〉, and 〈2,3,3〉 both as 2-generator and as 3-generator groups.Ak-nacci sequence in a finite group is a sequence of group elements x0,x1,x2,…,xn,… for which, given an initial (seed) set x0,x1,x2,…,xj-1, each element is defined by(1.1)xn={x0x1⋯xn-1forj≤n<k,xn-kxn-k+1⋯xn-1forn≥k.
We also require that the initial elements of the sequence x0,x1,x2,…,xj-1 generate the group, thus forcing the k-nacci sequence to reflect the structure of the group. The k-nacci sequence of a group G generated by x0,x1,x2,…,xj-1 is denoted by Fk(G;x0,x1,…,xj-1) [11].A sequence of group elements isperiodic 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,e,b,c,d,e,b,c,d,e,… is periodic after the initial element a and has period 4. A sequence of group elements is 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,a,b,c,d,e,f,… is simply periodic with period 6. In [11], Knox had denoted the period of a k-nacci sequence Fk(G;x0,x1,…,xj-1) by  Pk(G;x0,x1,…,xj-1).Definition 1.1.
For a finitely generated groupG=〈A〉, where A={a1,a2,…,an}, the sequence xi=ai+1, 0≤i≤n-1, xi+n=∏j=1nxi+j-1, i≥0 is called theFibonacci orbit of G with respect to the generating set A, denoted as FA(G) [4].Definition 1.2.
IfFA(G) is simply periodic, then the period of the sequence is called the Fibonacci lengthof G with respect to generating set A, written, LENA(G) [4].
Notice that the orbit of ak-generated group is a k-nacci sequence.
LetG be a finite j-generator group, and let X be the subset of G×G×G⋯×G︸j such that (x0,x1,…,xj-1)∈X if and only if G is generated by x0,x1,…,xj-1. We call (x0,x1,…,xj-1) a generatingj-tuplefor G.


## 2. Basic Period of  Basick-nacci Sequence

To examine the concept more fully, we study the action of automorphism group AutG of G on X and on the k-nacci sequences Fk(G:x0,x1,…xj-1), (x0,x1,…,xj-1)∈X. Now, AutG consists of all isomorphism θ:G→G and if θ∈ AutG and (x0,x1,…,xj-1)∈X, then (x0θ,x1θ,…,xj-1θ)∈X.For a subsetA⊆G and θ∈ AutG, the image of A under θ is (2.1)Aθ={aθ:a∈A}.Definition 2.1.
For a generating pair(x,y)∈X, the basic Fibonacci orbitF¯x,y of the basic lengthm is defined by the sequence {bi} of elements of G such that(2.2)b0=x,b1=y,bi+2=bibi+1,i≥0,
where m≥1 is the least integer with
(2.3)b0=bmθ,b1=bm+1θ,
for some θ∈ AutG. Since bm,bm+1 generate G, it follows that θ is uniquely determined.  For more information, see [3].Lemma 2.2.
Let(x0,x1,…,xj-1)∈X and let θ∈ Aut G, then (Fk(G:x0,x1,…xj-1))θ=Fk(G:x0θ,x1θ,…xj-1θ).Proof.
LetFk(G:x0,x1,…xj-1)={bi}. The result is obvious since {bi}θ={biθ} and(2.4)bi+kθ=(bibi+1⋯bi+k-1)θ=biθbi+1θ⋯bi+k-1θ.
Each generating  j-tuple(x0,x1,…,xj-1)∈X maps to |AutG| distinct elements of X under the action of elements of AutG. Hence, there are
(2.5)dj(G)=|X|/|AutG|,
(where |X| means the number of elements of X) nonisomorphic generating j-tuplesfor G. The notation dj(G) was introduced in [15].
Suppose thatω elements of AutG map Fk(G:x0,x1,…xj-1) into itself, then there are |AutG|/ω distinct k-nacci sequences Fk(G:x0θ,x1θ,…xj-1θ) for θ∈ AutG.Definition 2.3.
For aj-tuple(x0,x1,…,xj-1)∈X, the basic k-nacci sequence F¯k(G:x0,x1,…xj-1) of the basic period m is a sequence of group elements b0,b1,b2,…,bn,… for which, given an initial (seed) set b0=x0,b1=x1,b2=x2,…,bj-1=xj-1, each element is defined by(2.6)bn={b0b1⋯bn-1forj≤n<k,bn-kbn-k+1⋯bn-1forn≥k,
where m≥1 is the least integer with
(2.7)b0=bmθ,b1=bm+1θ,b2=bm+2θ,…,bk-1=bm+k-1θ,
for some θ∈ AutG. Since G is a finite j-generator group and bm, bm+1, …, bm+j-1 generate G, it follows that θ is uniquely determined. The basic k-nacci sequence F¯k(G:x0,x1,…xj-1) is finite containing m element.
In this paper, we denote the basic period of the basick-nacci sequence F¯k(G:x0,x1,…xj-1) by  BPk(G;x0,x1,…,xj-1).
From the definitions, it is clear that the periods of thek-nacci sequences and the basic k-nacci sequences in a finite group depend on the chosen generating set and the order of the generating elements.Theorem 2.4.
LetG be a finite group and (x0,x1,…,xj-1)∈X. If Pk(G;x0,x1,…,xj-1)=n and BPk(G;x0,x1,…,xj-1)=m, then m divides n, and there are n/m elements of AutG which map Fk(G:x0,x1,…xj-1) into itself.Proof.
We haven=mλ where λ is the order of automorphism θ∈ AutG since(2.8)Fk(G:x0,x1,…xj-1)=F¯k(G:x0,x1,…xj-1)∪F¯k(G:x0θ,x1θ,…xj-1θ)∪F¯k(G:x0θ2,x1θ2,…xj-1θ2)∪⋯
and  BPk(G;x0,x1,…,xj-1)=BPk(G;x0θ,x1θ,…,xj-1θ). Clearly, 1,θ,θ2,…,θλ-1 map Fk(G:x0,x1,…xj-1) into itself.


## 3. Applications

Definition 3.1.
Thepolyhedral group(l,m,n) for l,m,n>1 is defined by the presentation(3.1)〈x,y,z:xl=ym=zn=xyz=e〉,
or
(3.2)〈x,y:xl=ym=(xy)n=e〉.
The polyhedral group(l,m,n)  is finite if and only if the number
(3.3)μ=lmn(1l+1m+1n-1)=mn+nl+lm-lmn
is positive, that is, in the cases (2,2,n), (2,3,3), (2,3,4), and (2,3,5). Its order is 2lmn/μ. A4, S4, and A5 are the groups (2,3,3), (2,3,4), and (2,3,5), respectively. Also, the groups A4, S4, and A5 being isomorphic to the groups of rotations of the regular tetrahedron, octahedron, and icosahedron. Using Tietze transformations, we may show that (l,m,n)≅(m,n,l)≅(n,l,m). For more information on these groups, see [16] and [17, pp. 67-68].Definition 3.2.
Thebinary polyhedral group〈l,m,n〉, for l,m,n>1, is defined by the presentation(3.4)〈x,y,z:xl=ym=zn=xyz〉,
or
(3.5)〈x,y:xl=ym=(xy)n〉.
The binary polyhedral group〈l,m,n〉  is finite if and only if the number k=lmn(1/l+1/m+1/n-1)=mn+nl+lm-lmn is positive. Its order is 4lmn/k.
For more information on these groups, see [17, pp. 68–71].Definition 3.3.
Letfn(k) denote the nth member of the k-step Fibonacci sequence defined as(3.6)fn(k)=∑j=1kfn-j(k)forn>k,
with boundary conditions fi(k)=0 for 1≤i<k and  fk(k)=1. Reducing this sequence by a modulo m, we can get a repeating sequence, which we denote by
(3.7)f(k,m)=(f1(k,m),f2((k,m),…,fn(k,m)…),
where fi(k,m)=fi(k)(modm). We then have that (f1(k,m),f2(k,m),…,fk(k,m))=(0,0,…0,1), and it has the same recurrence relation as in (3.6) [18].Theorem 3.4 (f(k,m) is a periodic sequence [18]).
Lethk(m) denote the smallest period of f(k,m), called the period of f(k,m) or the wall number of  the k-step Fibonacci sequence modulo m.Theorem 3.5.
The periods of the k-nacci sequences and the basic periods of the basic k-nacci sequences in the groupS4 are as follows.
if the group is defined by the presentationS4=〈x,y,z:x2=y3=z4=xyz=e〉, then
(i)
ifk=2,P2(S4;y,z,x)=18 and  BP2(S4;y,z,x)=9,(ii)
ifk>2,Pk(S4;x,y,z)=6k+6 and BPk(S4;x,y,z)=3k+3.If S4 has the presentation S4=〈x,y:x2=y3=(xy)4=e〉, then(i′)
ifk=2,P2(S4;x,y)=18 and  BP2(S4;x,y)=9,(ii′)
ifk>2,   Pk(S4;x,y)=6k+6 and BPk(S4;x,y)=3k+3.Proof.
Firstly, let us consider the 3-generator case. We first note that|x|=2, |y|=3, and |z|=4 (where |x| means the order of x).
(i)
Ifk=2, we have the sequence for the generating triple (y, z, x),(3.8)y,z,x,y2,xy2,y2xy2,z2y,z2yz3y,yxy,xyx,xy2,x,xy2x,y2x,yxy,yxz,zy,y2xy2,y,z,x,…,
which has period 18 and the basic period 9 since xθ=x, yθ=xyx, and zθ=xy2, where θ is the inner automorphism induced by conjugation by x. (ii)
Ifk=3, we have the sequence for the generating triple (x, y, z),(3.9)x,y,z,e,x,y2,xy2,xzy2,x,y,yxy2,xzy2,x,y2,yx,e,x,y,xy,z2,x,y2zy,z2,x,y,z…,
which has period 24 and the basic period 12 since xθ=x, yθ=y2, and zθ=yx where θ is an outer automorphism of order 2.
Ifk≥4, the first k elements of sequence for the generating triple (x, y, z) are(3.10)x0=x,x1=y,x2=z,x3=xyz,x4=(xyz)2…,xk-1=(xyz)2k-4.
Thus, using the above information, sequence reduces to
(3.11)x0=x,x1=y,x2=z,x3=e,…,e,xk-1=e,
where xj=e for 3≤j≤k-1.  Thus,
(3.12)xk=e,xk+1=x,xk+2=y2,xk+3=xy2,xk+4=xzy2,xk+5=e,…,e,x2k+1=e,x2k+2=x,x2k+3=y,x2k+4=yxy2,x2k+5=xzy2,x2k+6=e,…,e,x3k+2=e,x3k+3=x,x3k+4=y2,x3k+5=yx,x3k+6=e,…,e,x4k+3=e,x4k+4=x,x4k+5=y,x4k+6=xy,x4k+7=z2,x4k+8=e,…,e,x5k+4=e,x5k+5=x,x5k+6=y2,x5k+7=zy,x5k+8=z2,x5k+9=e,…,e,x6k+5=e,
where xj=e for  k+5≤j≤2k+1, 2k+6≤j≤3k+2, 3k+6≤j≤4k+3, 4k+8≤j≤5k+4, and 5k+9≤j≤6k+5.
We also have(3.13)x6k+6=∏i=5k+66k+5xi=x,x6k+7=∏i=5k+76k+6xi=y,x6k+8=∏i=5k+86k+7xi=z.
Since the elements succeeding x6k+6, x6k+7, and x6k+8 depend on x,  y, and z for their values, the cycle begins again with the 6k+6th element, that is, x0=x6k+6,x1=x6k+7,x2=x6k+8,…. Thus, Pk(S4;x,y,z)=6k+6.
It is easy to see from the  above sequence that(3.14)x3k+3=x,x3k+4=y2,x3k+5=yx,x3k+6=e,…,e,x4k+2=e.BPk(S4;x,y,z)=3k+3 since xθ=x, yθ=y2, and zθ=yx where θ is an outer automorphism of order 2.
Secondly, let us consider the 2-generator case. We first note that|x|=2, |y|=3, and |xy|=4.
(i′)
Ifk=2,P2(S4;x,y)=18 and  BP2(S4;x,y)=9 since xθ=x and yθ=xyx where θ is the inner automorphism induced by conjugation by x.(ii′)
Ifk>2,  Pk(S4;x,y)=6k+6 and BPk(S4;x,y)=3k+3 since xθ=x and yθ=y2 where θ is an outer automorphism of order 2.The proofs are similar to above and are omitted.Theorem 3.6.
The periods of the k-nacci sequences and the basic periods of the basic k-nacci sequences in the binary polyhedral group〈2,3,4〉 are as follows.
If the group is defined by the presentation〈2,3,4〉=〈x,y,z:x2=y3=z4=xyz〉, then
(i)
ifk=2,P2(〈2,3,4〉;y,z,x)=18 and  BP2(〈2,3,4〉;y,z,x)=9,(ii)
ifk>2,Pk(〈2,3,4〉;x,y,z)=6k+6 and BPk(〈2,3,4〉;x,y,z)=6k+6.If the group is defined by the presentation 〈2,3,4〉=〈x,y:x2=y3=(xy)4〉, then(i′)
ifk=2,P2(〈2,3,4〉;x,y)=18 and  BP2(〈2,3,4〉;x,y)=9,(ii′)
ifk>2,  Pk(〈2,3,4〉;x,y)=6k+6 and BPk(〈2,3,4〉;x,y)=6k+6.Proof.
Firstly, let us consider the 2-generator case. We first note that|x|=4, |y|=6, and |xy|=8.
(i′)
Ifk=2, we have the sequence for the generating pair (x, y),(3.15)x,y,xy,yxy,xy2xy,xyxy2x,y2xy2,xy5x,xy,x3,xyx3,yx3,y2xy2,y2xyx,yxy2,yxy,y2,y4x,x,y,…,
which has period 18 and the basic period 9 since xθ=x3 and yθ=x3yx where θ is a outer automorphism of order 2.  (ii′)
Ifk=3, we have the sequence for the generating pair (x, y),(3.16)x,y,xy,(xy)2,x,y2,y5xy,(xy)2,x,y,(xy)3,(xy)4,x3,y2,xy2,(yx)2,x3,y,yxy2,(yx)2,x3,y2,y4x,e,x,y,xy,…,
which has period 24 and the basic period 24 since xθ=x and yθ=y where θ is an inner automorphism induced by conjugation by x2.
Ifk=4, we have the sequence for the generating pair (x, y),(3.17)x,y,xy,(xy)2,(xy)4,x3,y2,y5xy,(xy)2,e,x,y,(xy)3,(xy)4,e,x3,y2,xy2,(yx)2,x2,x,y,yxy2,(yx)2,e,x3,y2,y4x,e,e,x,y,xy,(xy)2,…,
which has period 30 and the basic period 30 since xθ=x and yθ=y where θ is an inner automorphism induced by conjugation by x2.
Ifk≥5, the first k elements of sequence for the generating pair (x, y) are(3.18)x0=x,x1=y,x2=xy,x3=(xy)2,x4=(xy)4,x5=(xy)8…,xk-1=(xy)2k-3.
Thus, using the above information, sequence reduces to
(3.19)x0=x,x1=y,x2=xy,x3=(xy)2,x4=(xy)4,x5=e,…,e,xk-1=e,
where xj=efor5≤j≤k-1. Thus,
(3.20)xk=e,xk+1=x3,xk+2=y2,xk+3=y5xy,xk+4=(xy)2,xk+5=e,…,e,x2k+1=e,x2k+2=x,x2k+3=y,x2k+4=(xy)3,x2k+5=(xy)4,x2k+6=e,…,e,(3.21)x3k+2=e,x3k+3=x3,x3k+4=y2,x3k+5=xy2,x3k+6=(yx)2,x3k+7=x2,x3k+8=e…,e,x4k+3=e,x4k+4=x,x4k+5=y,x4k+6=yxy2x4k+7=(yx)2,x4k+8=e,…,e,x5k+4=e,x5k+5=x3,x5k+6=y2,x5k+7=y4x,x5k+8=e,…,e,x6k+5=e,
where xj=e for  k+5≤j≤2k+1, 2k+6≤j≤3k+2, 3k+8≤j≤4k+3, 4k+8≤j≤5k+4, and 5k+8≤j≤6k+5.
We also have(3.22)x6k+6=∏i=5k+66k+5xi=x,x6k+7=∏i=5k+76k+6xi=y.
Since the elements succeeding x6k+6, x6k+7 depend on x and y for their values, the cycle begins again with the 6k+6th element, that is,x0=x6k+6,x1=x6k+7,…. Thus, Pk(〈2,3,4〉;x,y)=6k+6 and BPk(〈2,3,4〉;x,y)=6k+6 since xθ=x and yθ=y where θ is an inner automorphism induced by conjugation by x2.
Secondly, let us consider the 3-generator case. We first note that|x|=4, |y|=6, and |z|=8.
(i)
Ifk=2,P2(〈2,3,4〉;y,z,x)=18 and  BP2(〈2,3,4〉;y,z,x)=9 since xθ=x3, yθ=x3yx, and zθ=xy2 where θ is an outer automorphism of order 2.(ii)
Ifk>2,Pk(〈2,3,4〉;x,y,z)=6k+6 and BPk(〈2,3,4〉;x,y,z)=6k+6 since xθ=x and yθ=y where θ is an inner automorphism induced by conjugation by x2.The proofs are similar to the proofs of Theorems 3.5.(i) and 3.5.(ii) and are omitted.Theorem 3.7.
The periods of the k-nacci sequences and the basic periods of the basic k-nacci sequences in the groupA4 are as follows.
If the group is defined by the presentationA4=〈x,y,z:x2=y3=z3=xyz=e〉, then
(i)
ifk=2, P2(A4;y,z,x)=16 and  BP2(A4;y,z,x)=4,(ii)
ifk>2,
(3.23)Pk(A4;x,y,z)={3BPk(A4;x,y,z),k≡0mod⁡4,2BPk(A4;x,y,z),k≡2mod⁡4,2BPk(A4;x,y,z),otherwise,BPk(A4;x,y,z)={u1hk(3),k≡0mod⁡4,u2hk(3),k≡2mod⁡4,u3hk(3),otherwise,where u1,u2,u3∈N, and hk(3) denote the wall number of the k-step Fibonacci sequence modulo 3.
If the group is defined by the presentationA4=〈x,y:x2=y3=(xy)3=e〉, then
(i′)
ifk=2, P2(A4;x,y)=16 and  BP2(A4;x,y)=4,(ii′)
ifk>2,
(3.24)Pk(A4;x,y)={3BPk(A4;x,y),k≡0mod⁡4,2BPk(A4;x,y),k≡2mod⁡4,2BPk(A4;x,y),otherwise,BPk(A4;x,y)={u1hk(3),k≡0mod⁡4,u2hk(3),k≡2mod⁡4,u3hk(3),otherwise,where u1,u2,u3∈N.Proof.
Firstly, let us consider the 2-generator case. We process as similar to the proof of Theorem3.6 We first note that |x|=2, |y|=3, and |xy|=3.
(i′)
Ifk=2, we have the sequence for the generating pair (x, y),(3.25)x,y,xy,yxy,yxy2,(xy)2,xy2,y,x,yx,xyx,y2x,yxy2,yxy,y2,yx,x,y,…,
which has period 16 and the basic period 4 since xθ=yxy2 and yθ=yxy where θ is an outer automorphism of order 4. (ii′)
Ifk>2,let k be even, then the first k elements of sequence for the generating pair (x, y) are
(3.26)x0=x,x1=y,x2=xy,x3=(xy)2,x4=xy,x5=(xy)2…,xk-2=xy,xk-1=(xy)2.
If  k≡0mod4,
(3.27)xu1hk(3)-(k-2)=e,xu1hk(3)-(k-1)=e,…,e,xu1hk(3)-1=e,xu1hk(3)=y2xy,xu1hk(3)+1=yx,….Pk(A4;x,y)=3BPk(A4;x,y) and BPk(A4;x,y)=u1hk(3) since xθ=yxy2 and yθ=xyx where θ is an outer automorphism of order 3.
Ifk≡2mod4,(3.28)xu2hk(3)-(k-2)=e,xu2hk(3)-(k-1)=e,…,e,xu2hk(3)-1=e,xu2hk(3)=x,xu2hk(3)+1=xy,….Pk(A4;x,y)=2BPk(A4;x,y) and BPk(A4;x,y)=u2hk(3) since xθ=x and yθ=xy where θ is a outer automorphism of order 2.
Letk be odd, then the first k elements of sequence are for the generating pair (x, y),(3.29)x0=x,x1=y,x2=xy,x3=(xy)2,x4=xy,x5=(xy)2…,xk-2=(xy)2,xk-1=xy.
Also,
(3.30)xu3hk(3)-(k-2)=e,xu3hk(3)-(k-1)=e,…,e,xu3hk(3)-1=e,xu3hk(3)=x,xu3hk(3)+1=yx,….Pk(A4;x,y)=2BPk(A4;x,y) and BPk(A4;x,y)=u3hk(3) since xθ=x and yθ=yx where θ is an outer automorphism of order 2.
Secondly, let us consider the 3-generator case. We first note that|x|=2, |y|=3, and |z|=3.
(i)
Ifk=2,P2(A4;y,z,x)=16 and  BP2(A4;y,z,x)=4 since xθ=y2xy, yθ=yxy, and zθ=yx where θ is an outer automorphism of order 4.(ii)
Ifk>2, let k≡0mod4, then Pk(A4;x,y,z)=3BPk(A4;x,y,z) and BPk(A4;x,y,z)=u1hk(3) since xθ=y2xy, yθ=xyx, and zθ=zx where θ is an outer automorphism of order 3; let  k≡2mod4, then Pk(A4;x,y,z)=2BPk(A4;x,y,z) and BPk(A4;x,y,z)=u2hk(3) since xθ=x, yθ=yx, and zθ=yz2 where θ is an outer automorphism of order 2; let k be odd; then Pk(A4;x,y,z)=2BPk(A4;x,y,z) and BPk(A4;x,y,z)=u3hk(3) since xθ=x, yθ=xy, and zθ=zx where θ is an outer automorphism of order 2.
The proofs are similar to the proofs of Theorems3.5.(i) and 3.5.(i.i) and are omitted.Theorem 3.8.
The periods of the k-nacci sequences and the basic periods of the basic k-nacci sequences in the binary polyhedral group〈2,3,3〉 are as follows.
If the group is defined by the presentation〈2,3,3〉=〈x,y,z:x2=y3=z3=xyz〉, then
(i)
ifk=2, P2(〈2,3,3〉;y,z,x)=48 and  BP2(〈2,3,3〉;y,z,x)=12,(ii)
ifk>2,(3.31)Pk(〈2,3,3〉;x,y,z)={3BPk(〈2,3,3〉;x,y,z),k≡0mod⁡4,BPk(〈2,3,3〉;x,y,z),k≢0mod⁡4,(3.32)BPk(〈2,3,3〉;x,y,z)={v1hk(6),k≡0mod⁡4,v2hk(6),k≢0mod⁡4,
where v1,v2∈N, and hk(6) denote the wall number of the k-step Fibonacci sequence modulo 6.
If the group is defined by the presentation〈2,3,3〉=〈x,y:x2=y3=(xy)3〉, then
(i′)
ifk=2, P2(〈2,3,3〉;x,y)=48 and  BP2(〈2,3,3〉;x,y)=12,(ii′)
ifk>2,
(3.33)Pk(〈2,3,3〉;x,y)={3BPk(〈2,3,3〉;x,y),k≡0mod⁡4,BPk(〈2,3,3〉;x,y),k≢0mod⁡4,(3.34)BPk(〈2,3,3〉;x,y)={v1hk(6),k≡0mod⁡4,v2hk(6),k≢0mod⁡4,where v1,v2∈N.Proof.
Firstly, let us consider the 3-generator case.  We first note that|x|=4, |y|=6, and |z|=6.(i)
Ifk=2,P2(〈2,3,3〉;y,z,x)=48 and  BP2(〈2,3,3〉;y,z,x)=12 since xθ=y2xy, yθ=xz4x, and zθ=y2xy2 where θ is an outer automorphism of order 4.(ii)
Ifk>2,let k≡0mod4, then Pk(〈2,3,3〉;x,y,z)=3BPk(〈2,3,3〉;x,y,z) and BPk(〈2,3,3〉;x,y,z)=v1hk(6) since xθ=yxy5, yθ=z3xy, and zθ=xy2x where θ is an inner automorphism induced by conjugation by z3yx;
letk≢0mod4, then Pk(〈2,3,3〉;x,y,z)=BPk(〈2,3,3〉;x,y,z) and BPk(〈2,3,3〉;x,y,z)=v2hk(6) since xθ=x, yθ=y, and zθ=z where θ is an inner automorphism induced by conjugation by x2.
The proofs are similar to the proofs of Theorems3.5.(i) and 3.5.(ii) and are omitted.
Secondly, let us consider the 2-generator case. We first note that|x|=4, |y|=6, and |xy|=6.
(i′)
Ifk=2,P2(〈2,3,3〉;x,y)=48 and  BP2(〈2,3,3〉;x,y)=12 since xθ=yxy2 and yθ=y2x where θ is an outer automorphism of order 4.(ii′)
Ifk>2, let k≡0mod4, then Pk(〈2,3,3〉;x,y)=3BPk(〈2,3,3〉;x,y) and BPk(〈2,3,3〉;x,y)=v1hk(6) since xθ=y5xy, yθ=yx, and zθ=xy2x where θ is an inner automorphism induced by conjugation by y5x,
letk≢0mod4, then Pk(〈2,3,3〉;x,y)=BPk(〈2,3,3〉;x,y) and BPk(〈2,3,3〉;x,y)=v2hk(6) since xθ=x and yθ=y where θ is an inner automorphism induced by conjugation by x2.
The proofs are similar to the proofs of Theorem3.6.(i′) and Theorem 3.6.(ii′) and are omitted.Theorem 3.9.
The periods of the k-nacci sequences arek+1, and the basic period of the basic k-nacci sequences is k+1 in D2  four-group.Proof.
We have the presentationD2=〈x,y:x2=y2=e,xy=yx〉. Pk(D2;x,y)=k+1;  see [14] for a proof  and BPk(D2;x,y)=k+1 since xθ=x and yθ=y where θ is an inner automorphism induced by conjugation by x.



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*Source: 639476-2011-05-23.xml*