# The Generalized Order-k Lucas Sequences in Finite Groups

Authors: Ömür Deveci; Erdal Karaduman
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We study the generalized order-k Lucas sequences modulo m. Also, we define the ith generalized order-k Lucas orbit lAi,{α1,α2,…,αk-1}(G) with respect to the generating set A and the constants
α1,α2, and αk-1 for a finite group G=〈A〉. Then, we obtain the lengths of the periods of the ith generalized order-k Lucas orbits of the binary polyhedral groups 〈n,2,2〉,〈2,n,2〉,〈2,2,n〉 and the polyhedral groups (n,2,2),(2,n,2),(2,2,n) for 1≤i≤k.


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## Body

## 1. Introduction

The well-known Fibonacci sequence{Fn} is defined as(1.1)F1=F2=1,forn>2,Fn=Fn-1+Fn-2.
We call Fn the nth Fibonacci number. The Fibonacci sequence is(1.2)1,1,2,3,5,8,13,21,34,55,….Definition 1.1.
Letfn(k) denote the nth member of the k-step Fibonacci sequence defined as
(1.3)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 modulus m, we can get a repeating sequence, which we denote by
(1.4)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 (1.3) [1].Theorem 1.2.
f(k,m)  is a periodic sequence  [1].
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. For more information see [1].Definition 1.3.
Lethk(a1,a2,…,ak)(m) denote the smallest period of the integer-valued recurrence relation un=un-1+un-2+⋯+un-k, u1=a1,u2=a2,…,uk=ak when each entry is reduced modulo m [2].Lemma 1.4.
Fora1,a2,…,ak,x1,x2,…,xk∈ℤ with m>0, a1,a2,…,ak not all congruent to zero modulo m and x1,x2,…,xk not all congruent to zero modulo m,
(1.5)hk(a1,a2,…,ak)(m)=hk(x1,x2,…,xk)(m),see [2].In [3], Taşçı and Kılıç defined  the k sequences of the generalized order-k Lucas numbers as follows:(1.6)lni=∑j=1kln-ji,
for n>0 and 1≤i≤k, with boundary (initial) conditions(1.7)lni={2ifi=2-n,-1ifi=1-n,0otherwise,
for 1-k≤n≤0, where lni is the nth term of the ith sequence. When i=1 and k=2, the generalized order-k Lucas sequence reduces to the usual negative Fibonacci sequence, that is, ln1=-Fn+1 for all n∈ℤ+.In [3], it is obtained that(1.8)[ln+1ilniln-1i⋮ln-k+2i]=A[lniln-1iln-2i⋮ln-k+1i],
where(1.9)A=[111⋯11100⋯00010⋯00⋮⋮⋮⋱⋮⋮000⋯10].
The Lucas sequence, the generalized Lucas sequence, and their properties have been studied by several authors; see for example, [4–9]. The study of the Fibonacci sequences in groups began with the earlier work of Wall [10]. Knox examined the k-nacci (k-step Fibonacci) sequences in finite groups [11]. Karaduman and Aydin examined the periods of the 2-step general Fibonacci sequences in dihedral groups Dn [12]. Lü and Wang contributed to the study of the Wall number for the k-step Fibonacci sequence [1]. C. M. Campbell and P. P. Campbell examined the behaviour of the Fibonacci lengths of finite binary polyhedral groups [13]. Also, Deveci et al. obtained the periods of the k-nacci sequences in finite binary polyhedral groups [14]. Now, we extend the concept to k sequences of the generalized order-k Lucas numbers and we examine the periods of the ith generalized order-k Lucas orbits of the binary polyhedral groups 〈n,2,2〉,〈2,n,2〉,〈2,2,n〉 and the polyhedral groups (n,2,2),(2,n,2),(2,2,n) for 1≤i≤k.In this paper, the usual notationp is used for a prime number.


## 2. Main Results and Proofs

Reducing thek sequences of the generalized order-k Lucas numbers by modulus m, we can get a repeating sequence denoted by(2.1)l(i,m)=(…,l1(i,m),l2(i,m),…,ln(i,m),…)forn>0,1≤i≤k,
where ln(i,m)=lni(modm). It has the same recurrence relation as that in (1.6).Let the notationhlki(m) denote the smallest period of l(i,m). It is easy to see from Lemma 1.4 that hk(m)=hlki(m).For a given matrixM=[bij] with bij’s being integers, M(modm) means that every entry of M is reduced modulo m, that is, M(modm)=(bij(modm)). Let 〈A〉pα={Ai(modpα)∣i≥0} be a cyclic group, and let |〈A〉pα| denote the order of 〈A〉pα. Then, we have the following.Theorem 2.1.
hk(pα)=|〈A〉pα|.Proof.
It is clear that|〈A〉pα| is divisible by hk(pα). Then we need only to prove that hk(pα) is divisible by |〈A〉pα|. Let hk(pα)=λ. Then we have
(2.2)Aλ=[a11a12⋯a1ka21a22⋯a2k⋮⋮⋱⋮ak1ak2⋯akk].
By mathematical induction it is easy to prove that the elements of the matrix Aλ are in the following forms:
(2.3)a11=fλ+k(k),a12=fλ+k-1(k)+⋯+fλ+1(k),a13=fλ+k-1(k)+⋯+fλ+2(k),…,a1k-1=fλ+k-1(k)+fλ+k-2(k),a1k=fλ+k-1(k),a21=fλ+k-1(k),a22=fλ+k-2(k)+⋯+fλ(k),a23=fλ+k-2(k)+⋯+fλ+1(k),…,a1k-1=fλ+k-2(k)+fλ+k-3(k),a2k=fλ+k-2(k),⋮⋮⋮⋮⋮ak1=fλ+1(k),ak2=fλ(k)+⋯+fλ-k+2(k),ak3=fλ(k)+⋯+fλ-k+3(k),…,akk-1=fλ(k)+fλ-1(k),akk=fλ(k).
We thus obtain that
(2.4)aii≡1(mod⁡pα)for1≤i≤k,aij≡0(mod⁡pα)for1≤i,j≤ksuchthati≠j.
So we get that Aλ≡I(modpα), which yields that n is divisible by |〈A〉pα|. We are done.Definition 2.2.
LetG be a finitely generated group G=〈A〉, where A={a1,a2,…,ak} and 1≤i≤k. The sequence
(2.5)x0=(a1)α1,x1=(a2)α2,…,xk-2=(ak-1)αk-1,
where
(2.6)(au)αu={auaklu-kiifαu=1,aklu-kiauifαu=2
such that 1≤u≤k-1 and 1≤αu≤2,xk-1=akl0i,xk+β=∏j=1kxβ+j-1 for β≥0, is called the ith generalized order-k Lucas orbit of G with respect to the generating set A and the α1,α2,…,αk-1 constants, denoted by lAi,{α1,α2,…,αk-1}(G).Example 2.3.
The3rd generalized order-4 Lucas orbits l{a1,a2,a3,a4}3,{1,1,1}(G), l{a1,a2,a3,a4}3,{1,2,1}(G), l{a1,a2,a3,a4}3,{1,1,2}(G), l{a1,a2,a3,a4}3,{1,2,2}(G), l{a1,a2,a3,a4}3,{2,1,1}(G), l{a1,a2,a3,a4}3,{2,2,1}(G), l{a1,a2,a3,a4}3,{2,1,2}(G), and l{a1,a2,a3,a4}3,{2,2,2}(G) of the finitely generated group G=〈A〉, where A={a1,a2,a3,a4}, respectively, are as follows:
(2.7)x0=a1a4l-33=a1,x1=a2a4l-23=a2a4-1,x2=a3a4l-13=a3a42,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0),x0=a1a4l-33=a1,x1=a4l-23a2=a4-1a2,x2=a3a4l-13=a3a42,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0),x0=a1a4l-33=a1,x1=a2a4l-23=a2a4-1,x2=a4l-13a3=a42a3,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0),x0=a1a4l-33=a1,x1=a4l-23a2=a4-1a2,x2=a4l-13a3=a42a3,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0),x0=a4l-33a1=a1,x1=a2a4l-23=a2a4-1,x2=a3a4l-13=a3a42,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0),x0=a4l-33a1=a1,x1=a4l-23a2=a4-1a2,x2=a3a4l-13=a3a42,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0),x0=a4l-33a1=a1,x1=a2a4l-23=a2a4-1,x2=a4l-13a3=a42a3,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0),x0=a4l-33a1=a1,x1=a4l-23a2=a4-1a2,x2=a4l-13a3=a42a3,x3=a4l03=e,x4+β=∏j=14xβ+j-1(β≥0).
It is well known that 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 the period of the sequence.Theorem 2.4.
Theith generalized order-k Lucas orbits in a finite group are periodic.Proof.
The proof is similar to the proof of Theorem  1 in [10] and is omitted.
We denote the length of the period of the sequencelAi,{α1,α2,…,αk-1}(G) by LENAli,{α1,α2,…,αk-1}(G) and call it the ith generalized order-k Lucas length of G with respect to the generating set A and the constants α1,α2,…,αk-1.
From the definition it is clear that theith generalized order-k Lucas length of a group depends on the chosen generating set and the order in which the assignments of x0,x1,…xn-1 are made.
We will now address theith generalized order-k Lucas lengths in specific classes of groups.
The binary polyhedral group〈l,m,n〉,  for  l,m,n>1,  is defined by the presentation
(2.8)〈x,y,z∣xl=ym=zn=xyz〉
or
(2.9)〈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 [15, pages 68–71].
The polyhedral group(l,m,n),  for  l,m,n>1,  is defined by the presentation
(2.10)〈x,y,z∣xl=ym=zn=xyz=e〉
or
(2.11)〈x,y∣xl=ym=(xy)n=e〉.
The 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 2lmn/k.
For more information on these groups, see [15, pages 67-68].Theorem 2.5.
Theith generalized order-3 Lucas lengths of the binary polyhedral group 〈n,2,2〉 for every i integer such that 1≤i≤3 and the generating triple {x,y,z}  are as follows:(i)
LEN{x,y,z}l1,{α1,α2}(〈n,2,2〉)=8for1≤α1,α2≤2,(ii)
LEN{x,y,z}l2,{α1,α2}(〈n,2,2〉)=h3(2n)for1≤α1,α2≤2,(iii)
(1)
LEN{x,y,z}l3,{1,1}(〈n,2,2〉)=LEN{x,y,z}l3,{1,2}(〈n,2,2〉)=8,(2)
LEN{x,y,z}l3,{2,1}(〈n,2,2〉)=LEN{x,y,z}l3,{2,2}(〈n,2,2〉)=4n if n is even, 8n if n is odd.Proof.
We prove the result by direct calculation. We first note that in the group defined by〈x,y,z∣xn=y2=z2=xyz〉, |x|=2n (where |x| means the order ofx), |y|=4, |z|=4, x=zy3, y=x-1z, and z=xy.
(i) The1st generalized order-3 Lucas orbits of the group 〈n,2,2〉 for generating triple {x,y,z} and every constant α1,α2 such that 1≤α1,α2≤2 are the same and are as follows:
(2.12)x0=x,x1=y,x2=z-1,x3=e,x4=yz-1,x5=xz-1,x6=z-1,x7=xn,x8=x,x9=y,x10=z-1,….
Since the elements succeeding x8,x9,andx10 depend on x, y, and z-1 for their values, the cycle is again the 8th element; that is, x0=x8,x1=x9,x2=x10,…. Thus, LEN{x,y,z}l1,{α1,α2}(〈n,2,2〉)=8 for 1≤α1,α2≤2.
(ii) Firstly, let us consider the orbitsl{x,y,z}2,{1,1}(〈n,2,2〉) and l{x,y,z}2,{2,1}(〈n,2,2〉). The orbits l{x,y,z}2,{1,1}(〈n,2,2〉) and l{x,y,z}2,{2,1}(〈n,2,2〉) are the same and are as follows:
(2.13)x0=x,x1=x-1,x2=z2,x3=z2,x4=x-1,x5=x-1,x6=x-2z2,x7=x-4z2,x8=x-7,x9=x-13,x10=x-24z2,x11=x-44z2,….
For m>3 we can see that the sequence will separate into some natural layers and each layer will be of the form
(2.14)xm={xumifm≡0(mod⁡4),xumifm≡1(mod⁡4),xumz2ifm≡2(mod⁡4),xumz2ifm≡3(mod⁡4),
where
(2.15)um=um-3+um-2+um-1,u0=1,u1=-1,u2=0.
Now the proof is finished when we note that the sequence will repeat when xh3(2n)=x,xh3(2n)+1=x-1, and xh3(2n)+2=z2, where h3(2n) is the 3-step Wall number of the positive integer 2n and h3(2n)=4μ(μ∈N). Letting L=LEN{x,y,z}l2,{1,1}(〈n,2,2〉)=LEN{x,y,z}l2,{2,1}(〈n,2,2〉), we have
(2.16)xL=xuL,xL+1=xuL+1,xL+2=xuL+2z2.
Using Lemma 1.4, we obtain uL=u0=1,uL+1=u1=-1, and uL+2=u2=0. In this case the above equalities give
(2.17)xL=xuL=x,xL+1=xuL+1=x-1,xL+2=xuL+2z2=x0z2=z2.
The smallest nontrivial integer satisfying the above conditions occurs when the period is h3(2n).
Secondly, let us consider the orbitsl{x,y,z}2,{1,2}(〈n,2,2〉) and l{x,y,z}2,{2,2}(〈n,2,2〉). The orbits l{x,y,z}2,{1,2}(〈n,2,2〉) and l{x,y,z}2,{2,2}(〈n,2,2〉) are the same and are as follows:
(2.18)x0=x,x1=x,x2=z2,x3=x2z2,x4=x3,x5=x5,x6=x10z2,x7=x18z2,x8=x33,x9=x61,x10=x112z2,x11=x206z2,….
For m>3 we can see that the sequence will separate into some natural layers and each layer will be of the form
(2.19)xm={xvmifm≡0(mod⁡4),xvmifm≡1(mod⁡4),xvmz2ifm≡2(mod⁡4),xvmz2ifm≡3(mod⁡4),
where
(2.20)vm=vm-3+vm-2+vm-1,v0=1,v1=1,v2=0.
Now the proof is finished when we note that the sequence will repeat when xh3(2n)=x,xh3(2n)+1=x and xh3(2n)+2=z2. Letting L=LEN{x,y,z}l2,{1,2}(〈n,2,2〉)=LEN{x,y,z}l2,{2,2}(〈n,2,2〉), we have
(2.21)xL=xvL,xL+1=xvL+1,xL+2=xvL+2z2.
Using Lemma 1.4, we obtain vL=v0=1,vL+1=v1=1, and vL+2=v2=0. In this case the above equalities give
(2.22)xL=xvL=x,xL+1=xvL+1=x,xL+2=xvL+2z2=x0z2=z2.
The smallest nontrivial integer satisfying the above conditions occurs when the period is h3(2n).
(iii)
(1)
The orbitsl{x,y,z}3,{1,1}(〈n,2,2〉) and l{x,y,z}3,{1,2}(〈n,2,2〉) are the same and are as follows:
(2.23)x0=xz-1,x1=y3,x2=e,x3=xn+2,x4=yx2,x5=y3,x6=xn,x7=xn-2,x8=xz-1,x9=y3,x10=e,….
So, we get LEN{x,y,z}l3,{1,1}(〈n,2,2〉)=LEN{x,y,z}l3,{1,2}(〈n,2,2〉)=8.(2)
The orbitsl{x,y,z}3,{2,1}(〈n,2,2〉) and l{x,y,z}3,{2,2}(〈n,2,2〉) are the same and are as follows:
(2.24)x0=y3,x1=xn+1,x2=e,x3=yx,x4=x-1,x5=x-2,x6=y3x-3,x7=x-3yx-3,x8=y3,x9=xn+1,x10=x4,x11=yx5,x12=y3,x13=x-1,x14=x-6,x15=y3x-7,x16=y3,x17=xn+1,x18=x8,….
The sequence can be said to form layers of length eight. Using the above, the sequence becomes
(2.25)x0=y3,x1=xn+1,x2=e,…,x8=y3,x9=xn+1,x10=x4,…,x16=y3,x17=xn+1,x18=x8,…,x8i=y3,x8i+1=xn+1,x8i+2=x4i,….
So we need the smallest i∈ℕ such that 4i=2nk for k∈ℕ.
Ifn is even, then i=n/2. Thus, 8i=4n and LEN{x,y,z}l3,{2,1}(〈n,2,2〉)=LEN{x,y,z}l3,{2,2}(〈n,2,2〉)=4n.
Ifn is odd, then i=n. Thus, 8i=8n and LEN{x,y,z}l3,{2,1}(〈n,2,2〉)=LEN{x,y,z}l3,{2,2}(〈n,2,2〉)=8n.Theorem 2.6.
Theith generalized order-2 Lucas lengths of the binary polyhedral group 〈n,2,2〉 for every i such that 1≤i≤2 and the generating pair {x,y} are 6.Proof.
We prove the result by direct calculation. We first note that in the group defined by(2.26)〈x,y∣xn=y2=(xy)2〉,|x|=2n,|y|=4,xy=yx-1,yx=x-1y.
Firstly, let us consider the orbits l{x,y}1,{1}(〈n,2,2〉) and l{x,y}1,{2}(〈n,2,2〉). The orbits l{x,y}1,{1}(〈n,2,2〉) and l{x,y}1,{2}(〈n,2,2〉) are the same and are as follows:
(2.27)x0=x,x1=y-1,x2=xy-1,x3=xn-1,x4=x2y,x5=y-1x-1,x6=x,x7=y-1,….
So, we get LEN{x,y}l1,{1}(〈n,2,2〉)=LEN{x,y}l1,{2}(〈n,2,2〉)=6.
Secondly, let us consider the orbitl{x,y}2,{1}(〈n,2,2〉). The orbit l{x,y}2,{1}(〈n,2,2〉) is as follows:
(2.28)x0=xy-1,x1=xn,x2=xy,x3=xy-1,x4=e,x5=xy-1,x6=xy-1,x7=xn,….
So, we get LEN{x,y}l2,{1}(〈n,2,2〉)=6.
Thirdly, let us consider the orbitl{x,y}2,{2}(〈n,2,2〉). The orbit l{x,y}2,{2}(〈n,2,2〉) is as follows:
(2.29)x0=y-1x,x1=xn,x2=yx,x3=y-1x,x4=e,x5=y-1x,x6=y-1x,x7=xn,….
So, we get LEN{x,y}l2,{2}(〈n,2,2〉)=6.Theorem 2.7.
Theith generalized order-3 Lucas lengths of the binary polyhedral group 〈2,n,2〉 for every i integer such that 1≤i≤3 and the generating triple {x,y,z} are as follows:(i)
LEN{x,y,z}l1,{α1,α2}〈2,n,2〉=8for1≤α1,α2≤2,(ii)
LEN{x,y,z}l2,{α1,α2}〈2,n,2〉=8for1≤α1,α2≤2,(iii)
LEN{x,y,z}l3,{α1,α2}〈2,n,2〉=h3(2n)for1≤α1,α2≤2.Proof.
The proof is similar to the proof of Theorem2.5 and is omitted.Theorem 2.8.
Theith generalized order-2 Lucas lengths of the binary polyhedral group 〈2,n,2〉 for every i such that 1≤i≤2 and the generating pair {x,y} are 6.Proof.
The proof is similar to the proof of Theorem2.6 and is omitted.Theorem 2.9.
Theith generalized order-3 Lucas lengths of the binary polyhedral group 〈2,2,n〉 for every i integer such that 1≤i≤3 and the generating triple {x,y,z} are as follows:(i)
(2.30)LEN{x,y,z}l1,{α1,α2}(〈2,2,n〉)={4nifniseven,8nifnisoddfor1≤α1,α2≤2,(ii)
(2.31)LEN{x,y,z}l2,{α1,α2}(〈2,2,n〉)={2nifn≡0mod⁡4,4nifn≡2mod⁡4,8notherwisefor1≤α1,α2≤2,(iii)
(2.32)LEN{x,y,z}l3,{α1,α2}(〈2,2,n〉)=8for1≤α1,α2≤2Proof.
We prove the result by direct calculation. We first note that in the group defined by〈x,y,z∣x2=y2=zn=xyz〉,|x|=4,|y|=4,|z|=2n,x=yz,y=xz-1 and z=yx-1.
(i) the1st generalized order-3 Lucas orbits of the group 〈2,2,n〉 for generating triple {x,y,z} and every constant α1,α2 such that 1≤α1,α2≤2 are the same and are as follows:
(2.33)x0=x,x1=y,x2=z-1,x3=yz2y,x4=y2z3y,x5=y,x6=y2z,x7=y2z4,x8=xz4,x9=y,x10=z-1,x11=yz6y,x12=y2z7y,x13=y,x14=y2z,x15=y2z8,x16=xz8,x17=y,x18=z-1,….
The sequence can be said to form layers of length eight. Using the above, the sequence becomes
(2.34)x0=x,x1=y,x2=z-1,…,x8=xz4,x9=y,x10=z-1,…,x16=xz8,x17=y,x18=z-1,…,x8i=xz4i,x8i+1=y,x8i+2=z-1,….
So, we need the smallest i∈ℕ such that 4i=2nk for k∈ℕ.
Ifn is even, then i=n/2. Thus, 8i=4n and LEN{x,y,z}l1,{α1,α2}(〈2,2,n〉)=4n for 1≤α1,α2≤2.
Ifn is odd, then i=n. Thus, 8i=8n and LEN{x,y,z}l1,{α1,α2}(〈2,2,n〉)=8n for 1≤α1,α2≤2.
(ii) The orbitsl{x,y,z}2,{1,1}(〈2,2,n〉) and l{x,y,z}2,{2,1}(〈2,2,n〉) are the same and are as follows:
(2.35)x0=x,x1=yz-1,x2=z2,x3=zn,x4=xzn,x5=z2x,x6=xz2x,x7=xz4x,x8=z8x,x9=yz-1,x10=z2,x11=zn+8,x12=xzn+8,x13=z2x,x14=xz2x,x15=xz12x,x16=z16x,x17=yz-1,x18=z2,….
The sequence can be said to form layers of length eight. Using the above, the sequence becomes
(2.36)x0=x,x1=yz-1,x2=z2,…,x8=z8x,x9=yz-1,x10=z2,…,x16=z16x,x17=yz-1,x18=z2,…,x8i=z8ix,x8i+1=yz-1,x8i+2=z2,….
So, we need the smallest i∈ℕ such that 4i=2nk for k∈ℕ.
Ifn≡0mod4, then i=n/4. Thus, 8i=2n and LEN{x,y,z}l2,{1,1}(〈2,2,n〉)=LEN{x,y,z}l2,{1,1}(〈2,2,n〉)=2n.
Ifn≡2mod4, then i=n/2. Thus, 8i=4n and LEN{x,y,z}l2,{1,1}(〈2,2,n〉)=LEN{x,y,z}l2,{1,1}(〈2,2,n〉)=4n.
Ifn≡1mod4 or n≡3mod4, then i=n. Thus, 8i=8n and LEN{x,y,z}l2,{1,1}(〈2,2,n〉)=LEN{x,y,z}l2,{1,1}(〈2,2,n〉)=8n.
The orbitsl{x,y,z}2,{1,1}(〈2,2,n〉) and l{x,y,z}2,{2,1}(〈2,2,n〉) are the same. The proofs for these orbits are similar to the above and are omitted.
(iii) The orbitsl3,{1,1}(〈2,2,n〉),l3,{1,2}(〈2,2,n〉),l3,{2,1}(〈2,2,n〉), and l3,{2,2}(〈2,2,n〉), respectively, are as follows:
(2.37)x0=y,x1=xz,x2=e,x3=zn+2,x4=xzn+3,x5=xz,x6=zn,x7=xz2x,x8=y,x9=xz,x10=e,…,x0=y,x1=z2y,x2=e,x3=xzy,x4=z4y3,x5=z2y,x6=zn,x7=zn+2,x8=y,x9=z2y,x10=e,…,x0=xz,x1=xz,x2=e,x3=zn,x4=xzn+1,x5=xz,x6=zn,x7=zn,x8=xz,x9=xz,x10=e,…,x0=xz,x1=z2y,x2=e,x3=yz4y,x4=zn+6y,x5=z2y,x6=zn,x7=zn+4,x8=xz,x9=z2y,x10=e,…,
which have period 8.Theorem 2.10.
Theith generalized order-2 Lucas lengths of the binary polyhedral group 〈2,2,n〉 for every i integer such that 1≤i≤2 and the generating triple {x,y} are as follows:(i)
LEN{x,y}l1,{1}(〈2,2,n〉)=LEN{x,y}l1{2}(〈2,2,n〉)=6,(ii)
LEN{x,y}l2,{1}(〈2,2,n〉)=LEN{x,y}l2,{2}(〈2,2,n〉)=h2(2n).Proof.
We prove the result by direct calculation. We first note that in the group defined by〈x,y∣x2=y2=(xy)n〉,|x|=4,|y|=4, and |xy|=2n.(i)
The orbitsl{x,y}1,{1}(〈2,2,n〉) and l{x,y}1,{2}(〈2,2,n〉) are the same and are as follows:
(2.38)x0=x,x1=y3,x2=xy3,x3=yxy,x4=y3,x5=yx,x6=x,x7=y3,…,
which have period 6.(ii)
The orbitsl{x,y}2,{1}(〈2,2,n〉) and l{x,y}2,{2}(〈2,2,n〉) are the same and are as follows:
(2.39)x0=(xy)n-1,x1=(xy)n,….We consider the recurrence relation defined by the following:
(2.40)um=um-2+um-1,u0=n-1,u1=n.
Then a routine induction shows that xm=(xy)um. Using Lemma 1.4, we obtain uL=u0=n-1 and  uL+1=u1=n. In this case the equalities xm=(xy)um give
(2.41)xL=(xy)uL=(xy)n-1,xL+1=(xy)uL+1=(xy)n.
The smallest nontrivial integer satisfying the above conditions occurs when the period is h2(2n).Theorem 2.11.
Theith generalized order-3 Lucas lengths of the polyhedral group (n,2,2) for every i integer such that 1≤i≤3 and the generating triple {x,y,z} are as follows:(i)
LEN{x,y,z}l1,{α1,α2}((n,2,2))=6 for 1≤α1,α2≤2,(ii)
LEN{x,y,z}l2,{α1,α2}((n,2,2))=h3(n) for 1≤α1,α2≤2,(iii)
(1)
LEN{x,y,z}l3,{1,1}((n,2,2))=LEN{x,y,z}l3,{1,2}((n,2,2))=8,(2)
LEN{x,y,z}l3,{2,1}((n,2,2))=LEN{x,y,z}l3,{2,2}((n,2,2))=4.Proof.
(i) We follow the proof given in [13].
The proofs of (ii) and (iii) are similar to the proofs of Theorem2.5(ii) and 2.5(iii) and are omitted.Theorem 2.12.
Theith generalized order-2 Lucas lengths of the polyhedral group (n,2,2) for every i integer such that 1≤i≤2 and the generating triple {x,y} are as follows:(i)
LEN{x,y}l1,{1}((n,2,2))=LEN{x,y}l1,{2}((n,2,2))=6,(ii)
LEN{x,y}l2,{1}((n,2,2))=LEN{x,y}l2,{2}((n,2,2))=3.Proof.
(i) The orbitsl1,{1}((n,2,2)) and l1,{2}((n,2,2)) are the natural extension of the result of dihedral groups given in [16].
(ii) The orbitsl{x,y}2,{1}((n,2,2)) and l{x,y}2,{2}((n,2,2)), respectively, are as follows:
(2.42)x0=xy,x1=e,x2=xy,x3=xy,x4=e,…,x0=yx,x1=e,x2=yx,x3=yx,x4=e,…,
which have period 3.Theorem 2.13.
Theith generalized order-3 Lucas lengths of the polyhedral group (2,n,2) for every i integer such that 1≤i≤3 and the generating triple {x,y,z} are as follows:(i)
LEN{x,y,z}l1,{α1,α2}((2,n,2))=6 for 1≤α1,α2≤2,(ii)
(1)
LEN{x,y,z}l2,{1,1}((2,n,2))=LEN{x,y,z}l2,{2,1}((2,n,2))=4,(2)
LEN{x,y,z}l2,{1,2}((2,n,2))=LEN{x,y,z}l2,{2,2}((2,n,2))=8,(iii)
LEN{x,y,z}l3,{α1,α2}((2,n,2))=h3(n)for1≤α1,α2≤2.Proof.
(i) We follow the proof given in [13].
The proofs of (ii) and (iii) are similar to the proofs of Theorem2.5(ii) and 2.5(iii) and are omitted.Theorem 2.14.
Theith generalized order-2 Lucas lengths of the polyhedral group (2,n,2) for every i such that 1≤i≤2 and the generating pair {x,y} are 6.Proof .
The proof is similar to the proof of Theorem2.6 and is omitted.Theorem 2.15.
Theith generalized order-3 Lucas lengths of the polyhedral group (2,2,n) for every i integer such that 1≤i≤3 and the generating triple {x,y,z} are as follows:(i)
(2.43)LEN{x,y,z}l1,{α1,α2}(2,2,n)={2nifn≡0mod⁡4,4nifn≡2mod⁡4,8notherwise,for1≤α1,α2≤2(ii)
(2.44)LEN{x,y,z}l2,{α1,α2}((2,2,n))={nifn≡0mod⁡8,2nifn≡4mod⁡8,4nifn≡2mod⁡8,8notherwise,for1≤α1,α2≤2(iii)
(1)
(2.45)LEN{x,y,z}l3,{1,1}((2,2,n))=LEN{x,y,z}l3,{1,2}((2,2,n))=LEN{x,y,z}l3,{2,2}((2,2,n))=8,(2)
(2.46)LEN{x,y,z}l3,{2,1}((2,2,n))=4.Proof.
The proof is similar to the proof of Theorem2.9 and is omitted.Theorem 2.16.
Theith generalized order-2 Lucas lengths of the polyhedral group (2,2,n) for every i integer such that 1≤i≤2 and the generating triple {x,y} are as follows:(i)
LEN{x,y}l1,{1}((2,2,n))=LEN{x,y}l1,{2}((2,2,n))=6,(ii)
LEN{x,y}l2,{1}((2,2,n))=LEN{x,y}l2,{2}((2,2,n))=h2(n).Proof.
(i) The orbitsl1,{1}((2,2,n)) and l1,{2}((2,2,n)) are the natural extension of the result of dihedral groups given in [16].
(ii) The proof is similar to the proof of Theorem2.10(ii) and is omitted.



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*Source: 464580-2012-02-26.xml*