The Generalized Order-𝑘 Lucas Sequences in Finite Groups

Abstract

We study the generalized order-𝑘 Lucas sequences modulo 𝑚. Also, we define the 𝑖th generalized order-𝑘 Lucas orbit 𝑙^{𝑖,{𝛼1,𝛼2,…,𝛼𝑘−1}𝐴}(𝐺) with respect to the generating set 𝐴 and the constants 𝛼₁,𝛼₂, and 𝛼_𝑘−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.

1. Introduction

The well-known Fibonacci sequence {𝐹_𝑛} is defined as𝐹₁=𝐹₂=1,for𝑛>2,𝐹_𝑛=𝐹_𝑛−1+𝐹_𝑛−2.(1.1) We call 𝐹_𝑛 the 𝑛th Fibonacci number. The Fibonacci sequence is1,1,2,3,5,8,13,21,34,55,….(1.2)

Definition 1.1. Let 𝑓^𝑛(𝑘) denote the 𝑛th member of the 𝑘-step Fibonacci sequence defined as 𝑓^𝑛(𝑘)=^𝑘_𝑗=1𝑓^{(𝑘)𝑛−𝑗}for𝑛>𝑘(1.3) with boundary conditions 𝑓^𝑖(𝑘)=0 for 1≤𝑖<𝑘 and 𝑓^𝑘(𝑘)=1. Reducing this sequence by modulus 𝑚, we can get a repeating sequence, which we denote by 𝑓(𝑘,𝑚)=𝑓^1(𝑘,𝑚),𝑓^2(𝑘,𝑚),…,𝑓^{𝑛(𝑘,𝑚)}…,(1.4) where 𝑓^{𝑖(𝑘,𝑚)}=𝑓^𝑖(𝑘)(mod𝑚). We then have that (𝑓^1(𝑘,𝑚),𝑓^2(𝑘,𝑚),…,𝑓^{𝑘(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].
Let ℎ_𝑘(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. Let ℎ_{𝑘(𝑎1,𝑎2,…,𝑎𝑘)}(𝑚) denote the smallest period of the integer-valued recurrence relation 𝑢_𝑛=𝑢_𝑛−1+𝑢_𝑛−2+⋯+𝑢_𝑛−𝑘, 𝑢₁=𝑎₁,𝑢₂=𝑎₂,…,𝑢_𝑘=𝑎_𝑘 when each entry is reduced modulo 𝑚 [2].

Lemma 1.4. For 𝑎₁,𝑎₂,…,𝑎_𝑘,𝑥₁,𝑥₂,…,𝑥_𝑘∈ℤ with 𝑚>0, 𝑎₁,𝑎₂,…,𝑎_𝑘 not all congruent to zero modulo 𝑚 and 𝑥₁,𝑥₂,…,𝑥_𝑘 not all congruent to zero modulo 𝑚, ℎ_{𝑘(𝑎1,𝑎2,…,𝑎𝑘)}(𝑚)=ℎ_{𝑘(𝑥1,𝑥2,…,𝑥𝑘)}(𝑚),(1.5)see [2].

In [3], Taşçı and Kılıç defined the 𝑘 sequences of the generalized order-𝑘 Lucas numbers as follows:𝑙^𝑖𝑛=^𝑘_𝑗=1𝑙^{𝑖𝑛−𝑗},(1.6) for 𝑛>0 and 1≤𝑖≤𝑘, with boundary (initial) conditions𝑙^𝑖𝑛=⎧⎪⎨⎪⎩2−1if𝑖=2−𝑛,0if𝑖=1−𝑛,otherwise,(1.7) for 1−𝑘≤𝑛≤0, where 𝑙^𝑖𝑛 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, 𝑙^1𝑛=−𝐹_𝑛+1 for all 𝑛∈ℤ⁺.

In [3], it is obtained that⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣𝑙^𝑖𝑛+1𝑙^𝑖𝑛𝑙^𝑖𝑛−1𝑙⋮_{𝑖𝑛−𝑘+2}⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=𝐴⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣𝑙^𝑖𝑛𝑙^𝑖𝑛−1𝑙^𝑖𝑛−2𝑙⋮_{𝑖𝑛−𝑘+1}⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦,(1.8) where𝐴=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣⋮1100⋮1010⋮1000⋯⋯⋯⋯⋱⋮1001⋮1000⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦.(1.9) 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 𝐷_𝑛 [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 notation 𝑝 is used for a prime number.

2. Main Results and Proofs

Reducing the 𝑘 sequences of the generalized order-𝑘 Lucas numbers by modulus 𝑚, we can get a repeating sequence denoted by𝑙(𝑖,𝑚)=…,𝑙^1(𝑖,𝑚),𝑙^2(𝑖,𝑚),…,𝑙^{𝑛(𝑖,𝑚)},…for𝑛>0,1≤𝑖≤𝑘,(2.1) where 𝑙^{𝑛(𝑖,𝑚)}=𝑙^𝑖𝑛(mod𝑚). It has the same recurrence relation as that in (1.6).

Let the notation ℎ𝑙^𝑖𝑘(𝑚) denote the smallest period of 𝑙(𝑖,𝑚). It is easy to see from Lemma 1.4 that ℎ_𝑘(𝑚)=ℎ𝑙^𝑖𝑘(𝑚).

For a given matrix 𝑀=[𝑏_𝑖𝑗] with 𝑏_𝑖𝑗’s being integers, 𝑀(mod𝑚) means that every entry of 𝑀 is reduced modulo 𝑚, that is, 𝑀(mod𝑚)=(𝑏_𝑖𝑗(mod𝑚)). Let ⟨𝐴⟩_𝑝𝛼={𝐴^𝑖(mod𝑝^𝛼)∣𝑖≥0} be a cyclic group, and let |⟨𝐴⟩_𝑝𝛼| denote the order of ⟨𝐴⟩_𝑝𝛼. Then, we have the following.

Theorem 2.1. ℎ_𝑘(𝑝^𝛼)=|⟨𝐴⟩_𝑝𝛼|.

Proof. It is clear that |⟨𝐴⟩_𝑝𝛼| is divisible by ℎ_𝑘(𝑝^𝛼). Then we need only to prove that ℎ_𝑘(𝑝^𝛼)is divisible by|⟨𝐴⟩_𝑝𝛼|. Let ℎ_𝑘(𝑝^𝛼)=𝜆. Then we have 𝐴^𝜆=⎡⎢⎢⎢⎢⎢⎢⎣𝑎¹¹𝑎₁₂⋯𝑎_1𝑘𝑎₂₁𝑎₂₂⋯𝑎_2𝑘𝑎⋮⋮⋱⋮_𝑘1𝑎_𝑘2⋯𝑎_𝑘𝑘⎤⎥⎥⎥⎥⎥⎥⎦.(2.2) By mathematical induction it is easy to prove that the elements of the matrix 𝐴^𝜆 are in the following forms: 𝑎𝑎₁₁₂₁=𝑓^{(𝑘)𝜆+𝑘},𝑎₁₂=𝑓^{(𝑘)𝜆+𝑘−1}+⋯+𝑓^(𝑘)𝜆+1,𝑎₁₃=𝑓^{(𝑘)𝜆+𝑘−1}+⋯+𝑓^(𝑘)𝜆+2,…,𝑎_1𝑘−1=𝑓^{(𝑘)𝜆+𝑘−1}+𝑓^{(𝑘)𝜆+𝑘−2},𝑎_1𝑘=𝑓^{(𝑘)𝜆+𝑘−1},𝑎=⋮𝑓^{(𝑘)𝜆+𝑘−1},𝑎₂₂=𝑓^{(𝑘)𝜆+𝑘−2}⋮+⋯+𝑓^𝜆(𝑘),𝑎₂₃=𝑓^{(𝑘)𝜆+𝑘−2}+⋮⋯+𝑓^(𝑘)𝜆+1,…,𝑎_1𝑘−1=𝑓⋮^{(𝑘)𝜆+𝑘−2}+𝑓^{(𝑘)𝜆+𝑘−3},𝑎_2𝑘=⋮_𝑘1𝑓^{(𝑘)𝜆+𝑘−2},=𝑓^(𝑘)𝜆+1,𝑎_𝑘2=𝑓^𝜆(𝑘)+⋯+𝑓^{(𝑘)𝜆−𝑘+2},𝑎_𝑘3=𝑓^𝜆(𝑘)+⋯+𝑓^{(𝑘)𝜆−𝑘+3},…,𝑎_𝑘𝑘−1=𝑓^𝜆(𝑘)+𝑓^{(𝑘)𝜆−1},𝑎_𝑘𝑘=𝑓^𝜆(𝑘).(2.3) We thus obtain that 𝑎𝑎_𝑖𝑖≡1(mod𝑝^𝛼)for1≤𝑖≤𝑘,_𝑖𝑗≡0(mod𝑝^𝛼)for1≤𝑖,𝑗≤𝑘suchthat𝑖≠𝑗.(2.4) So we get that 𝐴^𝜆≡𝐼(mod𝑝^𝛼), which yields that 𝑛 is divisible by |⟨𝐴⟩_𝑝𝛼|. We are done.

Definition 2.2. Let 𝐺 be a finitely generated group 𝐺=⟨𝐴⟩, where 𝐴={𝑎₁,𝑎₂,…,𝑎_𝑘} and 1≤𝑖≤𝑘. The sequence 𝑥₀=𝑎₁_𝛼1,𝑥₁=𝑎₂_𝛼2,…,𝑥_𝑘−2=𝑎_𝑘−1_𝛼𝑘−1,(2.5) where 𝑎_𝑢_𝛼𝑢=𝑎^𝑢𝑎𝑎^{𝑙𝑖𝑢−𝑘}_𝑘if𝛼_𝑢=1,_{𝑙𝑖𝑢−𝑘𝑘}𝑎_𝑢if𝛼_𝑢=2(2.6) such that 1≤𝑢≤𝑘−1 and 1≤𝛼_𝑢≤2,𝑥_𝑘−1=𝑎^𝑙𝑖0_𝑘,𝑥_𝑘+𝛽=∏^𝑘𝑗=1𝑥_{𝛽+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,…,𝛼_𝑘−1 constants, denoted by 𝑙^{𝑖,{𝛼1,𝛼2,…,𝛼𝑘−1}𝐴}(𝐺).

Example 2.3. The 3rd generalized order-4 Lucas orbits 𝑙^{3,{1,1,1}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺), 𝑙^{3,{1,2,1}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺), 𝑙^{3,{1,1,2}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺), 𝑙^{3,{1,2,2}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺), 𝑙^{3,{2,1,1}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺), 𝑙^{3,{2,2,1}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺), 𝑙^{3,{2,1,2}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺), and 𝑙^{3,{2,2,2}{𝑎}_{1,𝑎2,𝑎3,𝑎4}}(𝐺) of the finitely generated group 𝐺=⟨𝐴⟩, where 𝐴={𝑎₁,𝑎₂,𝑎₃,𝑎₄}, respectively, are as follows: 𝑥₀=𝑎₁𝑎^𝑙3−3₄=𝑎₁,𝑥₁=𝑎₂𝑎^𝑙3−2₄=𝑎₂𝑎⁴⁻¹,𝑥₂=𝑎₃𝑎^𝑙3−1₄=𝑎₃𝑎²⁴,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}𝑥(𝛽≥0),₀=𝑎₁𝑎^𝑙3−3₄=𝑎₁,𝑥₁=𝑎^𝑙3−2₄𝑎₂=𝑎⁴⁻¹𝑎₂,𝑥₂=𝑎₃𝑎^𝑙3−1₄=𝑎₃𝑎²⁴,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}𝑥(𝛽≥0),₀=𝑎₁𝑎^𝑙3−3₄=𝑎₁,𝑥₁=𝑎₂𝑎^𝑙3−2₄=𝑎₂𝑎⁴⁻¹,𝑥₂=𝑎^𝑙3−1₄𝑎₃=𝑎²⁴𝑎₃,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}𝑥(𝛽≥0),₀=𝑎₁𝑎^𝑙3−3₄=𝑎₁,𝑥₁=𝑎^𝑙3−2₄𝑎₂=𝑎⁴⁻¹𝑎₂,𝑥₂=𝑎^𝑙3−1₄𝑎₃=𝑎²⁴𝑎₃,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}𝑥(𝛽≥0),₀=𝑎^𝑙3−3₄𝑎₁=𝑎₁,𝑥₁=𝑎₂𝑎^𝑙3−2₄=𝑎₂𝑎⁴⁻¹,𝑥₂=𝑎₃𝑎^𝑙3−1₄=𝑎₃𝑎²⁴,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}𝑥(𝛽≥0),₀=𝑎^𝑙3−3₄𝑎₁=𝑎₁,𝑥₁=𝑎^𝑙3−2₄𝑎₂=𝑎⁴⁻¹𝑎₂,𝑥₂=𝑎₃𝑎^𝑙3−1₄=𝑎₃𝑎²⁴,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}𝑥(𝛽≥0),₀=𝑎^𝑙3−3₄𝑎₁=𝑎₁,𝑥₁=𝑎₂𝑎^𝑙3−2₄=𝑎₂𝑎⁴⁻¹,𝑥₂=𝑎^𝑙3−1₄𝑎₃=𝑎²⁴𝑎₃,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}𝑥(𝛽≥0),₀=𝑎^𝑙3−3₄𝑎₁=𝑎₁,𝑥₁=𝑎^𝑙3−2₄𝑎₂=𝑎⁴⁻¹𝑎₂,𝑥₂=𝑎^𝑙3−1₄𝑎₃=𝑎²⁴𝑎₃,𝑥₃=𝑎^𝑙30₄=𝑒,𝑥_4+𝛽=⁴_𝑗=1𝑥_{𝛽+𝑗−1}(𝛽≥0).(2.7) 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. The 𝑖th 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 sequence 𝑙^{𝑖,{𝛼1,𝛼2,…,𝛼𝑘−1}𝐴}(𝐺) by LEN_𝐴𝑙^{𝑖,{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,…,𝛼_𝑘−1.
From the definition it is clear that the 𝑖th generalized order-𝑘 Lucas length of a group depends on the chosen generating set and the order in which the assignments of 𝑥₀,𝑥₁,…𝑥_𝑛−1 are made.
We will now address the 𝑖th 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 𝑥,𝑦,𝑧∣𝑥^𝑙=𝑦^𝑚=𝑧^𝑛=𝑥𝑦𝑧(2.8) or 𝑥,𝑦∣𝑥^𝑙=𝑦^𝑚=(𝑥𝑦)^𝑛.(2.9) 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 𝑥,𝑦,𝑧∣𝑥^𝑙=𝑦^𝑚=𝑧^𝑛=𝑥𝑦𝑧=𝑒(2.10) or 𝑥,𝑦∣𝑥^𝑙=𝑦^𝑚=(𝑥𝑦)^𝑛=𝑒.(2.11) 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. The 𝑖th 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}}𝑙^{1,{𝛼1,𝛼2}}(⟨𝑛,2,2⟩)=8for1≤𝛼₁,𝛼₂≤2,(ii)LEN_{{𝑥,𝑦,𝑧}}𝑙^{2,{𝛼1,𝛼2}}(⟨𝑛,2,2⟩)=ℎ₃(2𝑛)for1≤𝛼₁,𝛼₂≤2,(iii)(1)LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,1}(⟨𝑛,2,2⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,2}(⟨𝑛,2,2⟩)=8,(2)LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,1}(⟨𝑛,2,2⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{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∣𝑥^𝑛=𝑦²=𝑧²=𝑥𝑦𝑧⟩, |𝑥|=2𝑛 (where |𝑥| means the order of𝑥), |𝑦|=4, |𝑧|=4, 𝑥=𝑧𝑦³, 𝑦=𝑥⁻¹𝑧, and 𝑧=𝑥𝑦.
(i) The 1st 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: 𝑥₀=𝑥,𝑥₁=𝑥𝑦₇,𝑥₂=𝑧⁻¹,𝑥₃=𝑒,𝑥₄=𝑦𝑧⁻¹,𝑥₅=𝑥𝑧⁻¹,𝑥₆=𝑧⁻¹,=𝑥^𝑛,𝑥₈=𝑥,𝑥₉=𝑦,𝑥₁₀=𝑧⁻¹,….(2.12) 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}}𝑙^{1,{𝛼1,𝛼2}}(⟨𝑛,2,2⟩)=8 for 1≤𝛼₁,𝛼₂≤2.
(ii) Firstly, let us consider the orbits 𝑙^{2,{1,1}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) and 𝑙^{2,{2,1}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩). The orbits 𝑙^{2,{1,1}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) and 𝑙^{2,{2,1}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) are the same and are as follows: 𝑥₀𝑥₇=𝑥,𝑥₁=𝑥⁻¹,𝑥₂=𝑧²,𝑥₃=𝑧²,𝑥₄=𝑥⁻¹,𝑥₅=𝑥⁻¹,𝑥₆=𝑥⁻²𝑧²,=𝑥⁻⁴𝑧²,𝑥₈=𝑥⁻⁷,𝑥₉=𝑥⁻¹³,𝑥₁₀=𝑥⁻²⁴𝑧²,𝑥₁₁=𝑥⁻⁴⁴𝑧²,….(2.13) For 𝑚>3 we can see that the sequence will separate into some natural layers and each layer will be of the form 𝑥_𝑚=⎧⎪⎨⎪⎩𝑥^𝑢𝑚𝑥if𝑚≡0(mod4),_𝑢𝑚𝑥if𝑚≡1(mod4),_𝑢𝑚𝑥𝑧²if𝑚≡2(mod4),_𝑢𝑚𝑧²if𝑚≡3(mod4),(2.14) where 𝑢_𝑚=𝑢_𝑚−3+𝑢_𝑚−2+𝑢_𝑚−1,𝑢₀=1,𝑢₁=−1,𝑢₂=0.(2.15) Now the proof is finished when we note that the sequence will repeat when 𝑥_ℎ3(2𝑛)=𝑥,𝑥_{ℎ3(2𝑛)+1}=𝑥⁻¹, and 𝑥_{ℎ3(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}}𝑙^2,{1,1}(⟨𝑛,2,2⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{2,1}(⟨𝑛,2,2⟩), we have 𝑥_𝐿=𝑥^𝑢𝐿,𝑥_𝐿+1=𝑥^𝑢𝐿+1,𝑥_𝐿+2=𝑥^𝑢𝐿+2𝑧².(2.16) Using Lemma 1.4, we obtain 𝑢_𝐿=𝑢₀=1,𝑢_𝐿+1=𝑢₁=−1, and 𝑢_𝐿+2=𝑢₂=0. In this case the above equalities give 𝑥_𝐿=𝑥^𝑢𝐿=𝑥,𝑥_𝐿+1=𝑥^𝑢𝐿+1=𝑥⁻¹,𝑥_𝐿+2=𝑥^𝑢𝐿+2𝑧²=𝑥⁰𝑧²=𝑧².(2.17) The smallest nontrivial integer satisfying the above conditions occurs when the period is ℎ₃(2𝑛).
Secondly, let us consider the orbits 𝑙^{2,{1,2}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) and 𝑙^{2,{2,2}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩). The orbits 𝑙^{2,{1,2}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) and 𝑙^{2,{2,2}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) are the same and are as follows: 𝑥𝑥₀₇=𝑥,𝑥₁=𝑥,𝑥₂=𝑧²,𝑥₃=𝑥²𝑧²,𝑥₄=𝑥³,𝑥₅=𝑥⁵,𝑥₆=𝑥¹⁰𝑧²,=𝑥¹⁸𝑧²,𝑥₈=𝑥³³,𝑥₉=𝑥⁶¹,𝑥₁₀=𝑥¹¹²𝑧²,𝑥₁₁=𝑥²⁰⁶𝑧²,….(2.18) For 𝑚>3 we can see that the sequence will separate into some natural layers and each layer will be of the form 𝑥_𝑚=⎧⎪⎨⎪⎩𝑥^𝑣𝑚𝑥if𝑚≡0(mod4),_𝑣𝑚𝑥if𝑚≡1(mod4),_𝑣𝑚𝑥𝑧²if𝑚≡2(mod4),_𝑣𝑚𝑧²if𝑚≡3(mod4),(2.19) where 𝑣_𝑚=𝑣_𝑚−3+𝑣_𝑚−2+𝑣_𝑚−1,𝑣₀=1,𝑣₁=1,𝑣₂=0.(2.20) Now the proof is finished when we note that the sequence will repeat when 𝑥_ℎ3(2𝑛)=𝑥,𝑥_{ℎ3(2𝑛)+1}=𝑥 and 𝑥_{ℎ3(2𝑛)+2}=𝑧². Letting 𝐿=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,2}(⟨𝑛,2,2⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{2,2}(⟨𝑛,2,2⟩), we have 𝑥_𝐿=𝑥^𝑣𝐿,𝑥_𝐿+1=𝑥^𝑣𝐿+1,𝑥_𝐿+2=𝑥^𝑣𝐿+2𝑧².(2.21) Using Lemma 1.4, we obtain 𝑣_𝐿=𝑣₀=1,𝑣_𝐿+1=𝑣₁=1, and 𝑣_𝐿+2=𝑣₂=0. In this case the above equalities give 𝑥_𝐿=𝑥^𝑣𝐿=𝑥,𝑥_𝐿+1=𝑥^𝑣𝐿+1=𝑥,𝑥_𝐿+2=𝑥^𝑣𝐿+2𝑧²=𝑥⁰𝑧²=𝑧².(2.22) The smallest nontrivial integer satisfying the above conditions occurs when the period is ℎ₃(2𝑛).
(iii) (1)The orbits 𝑙^{3,{1,1}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) and 𝑙^{3,{1,2}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) are the same and are as follows: 𝑥₅𝑥₀=𝑥𝑧⁻¹,𝑥₁=𝑦³,𝑥₂=𝑒,𝑥₃=𝑥^𝑛+2,𝑥₄=𝑦𝑥²,=𝑦³,𝑥₆=𝑥^𝑛,𝑥₇=𝑥^𝑛−2,𝑥₈=𝑥𝑧⁻¹,𝑥₉=𝑦³,𝑥₁₀=𝑒,….(2.23) So, we get LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,1}(⟨𝑛,2,2⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,2}(⟨𝑛,2,2⟩)=8. (2)The orbits 𝑙^{3,{2,1}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) and 𝑙^{3,{2,2}{𝑥,𝑦,𝑧}}(⟨𝑛,2,2⟩) are the same and are as follows: 𝑥₀𝑥=₇𝑦³,𝑥₁=𝑥^𝑛+1,𝑥₂=𝑒,𝑥₃=𝑦𝑥,𝑥₄=𝑥⁻¹,𝑥₅=𝑥⁻²,𝑥₆=𝑦³𝑥⁻³,𝑥₁₃=𝑥⁻³𝑦𝑥⁻³,𝑥₈=𝑦³,𝑥₉=𝑥^𝑛+1,𝑥₁₀=𝑥⁴,𝑥₁₁=𝑦𝑥⁵,𝑥₁₂=𝑦³,=𝑥⁻¹,𝑥₁₄=𝑥⁻⁶,𝑥₁₅=𝑦³𝑥⁻⁷,𝑥₁₆=𝑦³,𝑥₁₇=𝑥^𝑛+1,𝑥₁₈=𝑥⁸,….(2.24) The sequence can be said to form layers of length eight. Using the above, the sequence becomes 𝑥𝑥₀=𝑦³,𝑥₁=𝑥^𝑛+1,𝑥₂=𝑒,…,₈𝑥=𝑦³,𝑥₉=𝑥^𝑛+1,𝑥₁₀=𝑥⁴,…,₁₆𝑥=𝑦³,𝑥₁₇=𝑥^𝑛+1,𝑥₁₈=𝑥⁸,…,_8𝑖=𝑦³,𝑥_8𝑖+1=𝑥^𝑛+1,𝑥_8𝑖+2=𝑥^4𝑖,….(2.25) So we need the smallest 𝑖∈ℕ such that 4𝑖=2𝑛𝑘 for 𝑘∈ℕ.
If 𝑛 is even, then 𝑖=𝑛/2. Thus, 8𝑖=4𝑛 and LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,1}(⟨𝑛,2,2⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,2}(⟨𝑛,2,2⟩)=4𝑛.
If 𝑛 is odd, then 𝑖=𝑛. Thus, 8𝑖=8𝑛 and LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,1}(⟨𝑛,2,2⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,2}(⟨𝑛,2,2⟩)=8𝑛.

Theorem 2.6. The 𝑖th 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 𝑥,𝑦∣𝑥^𝑛=𝑦²=(𝑥𝑦)²,|𝑥|=2𝑛,||𝑦||=4,𝑥𝑦=𝑦𝑥⁻¹,𝑦𝑥=𝑥⁻¹𝑦.(2.26) Firstly, let us consider the orbits 𝑙^{1,{1}{𝑥,𝑦}}(⟨𝑛,2,2⟩) and 𝑙^{1,{2}{𝑥,𝑦}}(⟨𝑛,2,2⟩). The orbits 𝑙^{1,{1}{𝑥,𝑦}}(⟨𝑛,2,2⟩) and 𝑙^{1,{2}{𝑥,𝑦}}(⟨𝑛,2,2⟩) are the same and are as follows: 𝑥₀=𝑥,𝑥₁=𝑦⁻¹,𝑥₂=𝑥𝑦⁻¹,𝑥₃=𝑥^𝑛−1,𝑥₄=𝑥²𝑦,𝑥₅=𝑦⁻¹𝑥⁻¹,𝑥₆=𝑥,𝑥₇=𝑦⁻¹,….(2.27) So, we get LEN_{𝑥,𝑦}𝑙^1,{1}(⟨𝑛,2,2⟩)=LEN_{𝑥,𝑦}𝑙^1,{2}(⟨𝑛,2,2⟩)=6.
Secondly, let us consider the orbit 𝑙^{2,{1}{𝑥,𝑦}}(⟨𝑛,2,2⟩). The orbit 𝑙^{2,{1}{𝑥,𝑦}}(⟨𝑛,2,2⟩) is as follows: 𝑥₀=𝑥𝑦⁻¹,𝑥₁=𝑥^𝑛,𝑥₂=𝑥𝑦,𝑥₃=𝑥𝑦⁻¹,𝑥₄=𝑒,𝑥₅=𝑥𝑦⁻¹,𝑥₆=𝑥𝑦⁻¹,𝑥₇=𝑥^𝑛,….(2.28) So, we get LEN_{𝑥,𝑦}𝑙^2,{1}(⟨𝑛,2,2⟩)=6.
Thirdly, let us consider the orbit 𝑙^{2,{2}{𝑥,𝑦}}(⟨𝑛,2,2⟩). The orbit 𝑙^{2,{2}{𝑥,𝑦}}(⟨𝑛,2,2⟩) is as follows: 𝑥₀=𝑦⁻¹𝑥,𝑥₁=𝑥^𝑛,𝑥₂=𝑦𝑥,𝑥₃=𝑦⁻¹𝑥,𝑥₄=𝑒,𝑥₅=𝑦⁻¹𝑥,𝑥₆=𝑦⁻¹𝑥,𝑥₇=𝑥^𝑛,….(2.29) So, we get LEN_{𝑥,𝑦}𝑙^2,{2}(⟨𝑛,2,2⟩)=6.

Theorem 2.7. The 𝑖th 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}}𝑙^{1,{𝛼1,𝛼2}}⟨2,𝑛,2⟩=8for1≤𝛼₁,𝛼₂≤2,(ii) LEN_{{𝑥,𝑦,𝑧}}𝑙^{2,{𝛼1,𝛼2}}⟨2,𝑛,2⟩=8for1≤𝛼₁,𝛼₂≤2,(iii) LEN_{{𝑥,𝑦,𝑧}}𝑙^{3,{𝛼1,𝛼2}}⟨2,𝑛,2⟩=ℎ₃(2𝑛)for1≤𝛼₁,𝛼₂≤2.

Proof. The proof is similar to the proof of Theorem 2.5 and is omitted.

Theorem 2.8. The 𝑖th 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 Theorem 2.6 and is omitted.

Theorem 2.9. The 𝑖th 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)LEN_{{𝑥,𝑦,𝑧}}𝑙^{1,{𝛼1,𝛼2}}(⟨2,2,𝑛⟩)=4𝑛8𝑛if𝑛iseven,if𝑛isoddfor1≤𝛼₁,𝛼₂≤2,(2.30)(ii)LEN_{{𝑥,𝑦,𝑧}}𝑙^{2,{𝛼1,𝛼2}}(⟨2,2,𝑛⟩)=⎧⎪⎨⎪⎩2𝑛4𝑛if𝑛≡0mod4,8𝑛if𝑛≡2mod4,otherwisefor1≤𝛼₁,𝛼₂≤2,(2.31)(iii)LEN_{{𝑥,𝑦,𝑧}}𝑙^{3,{𝛼1,𝛼2}}(⟨2,2,𝑛⟩)=8for1≤𝛼₁,𝛼₂≤2(2.32)

Proof. We prove the result by direct calculation. We first note that in the group defined by ⟨𝑥,𝑦,𝑧∣𝑥²=𝑦²=𝑧^𝑛=𝑥𝑦𝑧⟩,|𝑥|=4,|𝑦|=4,|𝑧|=2𝑛,𝑥=𝑦𝑧,𝑦=𝑥𝑧⁻¹ and 𝑧=𝑦𝑥⁻¹.
(i) the 1st 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: 𝑥𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧⁻¹,𝑥₃=𝑦𝑧²𝑦,𝑥₄=𝑦²𝑧³𝑦,𝑥₅=𝑦,𝑥₆=𝑦²𝑧,₇𝑥=𝑦²𝑧⁴,𝑥₈=𝑥𝑧⁴,𝑥₉=𝑦,𝑥₁₀=𝑧⁻¹,𝑥₁₁=𝑦𝑧⁶𝑦,𝑥₁₂=𝑦²𝑧⁷𝑦,₁₃=𝑦,𝑥₁₄=𝑦²𝑧,𝑥₁₅=𝑦²𝑧⁸,𝑥₁₆=𝑥𝑧⁸,𝑥₁₇=𝑦,𝑥₁₈=𝑧⁻¹,….(2.33) The sequence can be said to form layers of length eight. Using the above, the sequence becomes 𝑥𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧⁻¹,…,₈𝑥=𝑥𝑧⁴,𝑥₉=𝑦,𝑥₁₀=𝑧⁻¹,…,₁₆𝑥=𝑥𝑧⁸,𝑥₁₇=𝑦,𝑥₁₈=𝑧⁻¹,…,_8𝑖=𝑥𝑧^4𝑖,𝑥_8𝑖+1=𝑦,𝑥_8𝑖+2=𝑧⁻¹,….(2.34) So, we need the smallest 𝑖∈ℕ such that 4𝑖=2𝑛𝑘 for 𝑘∈ℕ.
If 𝑛 is even, then 𝑖=𝑛/2. Thus, 8𝑖=4𝑛 and LEN_{{𝑥,𝑦,𝑧}}𝑙^{1,{𝛼1,𝛼2}}(⟨2,2,𝑛⟩)=4𝑛 for 1≤𝛼₁,𝛼₂≤2.
If 𝑛 is odd, then 𝑖=𝑛. Thus, 8𝑖=8𝑛 and LEN_{{𝑥,𝑦,𝑧}}𝑙^{1,{𝛼1,𝛼2}}(⟨2,2,𝑛⟩)=8𝑛 for 1≤𝛼₁,𝛼₂≤2.
(ii) The orbits 𝑙^{2,{1,1}{𝑥,𝑦,𝑧}}(⟨2,2,𝑛⟩) and 𝑙^{2,{2,1}{𝑥,𝑦,𝑧}}(⟨2,2,𝑛⟩) are the same and are as follows: 𝑥𝑥₀=𝑥,𝑥₁=𝑦𝑧⁻¹,𝑥₂=𝑧²,𝑥₃=𝑧^𝑛,𝑥₄=𝑥𝑧^𝑛,𝑥₅=𝑧²𝑥,𝑥₆=𝑥𝑧²𝑥,₇𝑥₁₃=𝑥𝑧⁴𝑥,𝑥₈=𝑧⁸𝑥,𝑥₉=𝑦𝑧⁻¹,𝑥₁₀=𝑧²,𝑥₁₁=𝑧^𝑛+8,𝑥₁₂=𝑥𝑧^𝑛+8,=𝑧²𝑥,𝑥₁₄=𝑥𝑧²𝑥,𝑥₁₅=𝑥𝑧¹²𝑥,𝑥₁₆=𝑧¹⁶𝑥,𝑥₁₇=𝑦𝑧⁻¹,𝑥₁₈=𝑧²,….(2.35) The sequence can be said to form layers of length eight. Using the above, the sequence becomes 𝑥𝑥₀=𝑥,𝑥₁=𝑦𝑧⁻¹,𝑥₂=𝑧²,…,₈𝑥=𝑧⁸𝑥,𝑥₉=𝑦𝑧⁻¹,𝑥₁₀=𝑧²,…,₁₆𝑥=𝑧¹⁶𝑥,𝑥₁₇=𝑦𝑧⁻¹,𝑥₁₈=𝑧²,…,_8𝑖=𝑧^8𝑖𝑥,𝑥_8𝑖+1=𝑦𝑧⁻¹,𝑥_8𝑖+2=𝑧²,….(2.36) So, we need the smallest 𝑖∈ℕ such that 4𝑖=2𝑛𝑘 for 𝑘∈ℕ.
If 𝑛≡0mod4, then 𝑖=𝑛/4. Thus, 8𝑖=2𝑛 and LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,1}(⟨2,2,𝑛⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,1}(⟨2,2,𝑛⟩)=2𝑛.
If 𝑛≡2mod4, then 𝑖=𝑛/2. Thus, 8𝑖=4𝑛 and LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,1}(⟨2,2,𝑛⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,1}(⟨2,2,𝑛⟩)=4𝑛.
If 𝑛≡1mod4 or 𝑛≡3mod4, then 𝑖=𝑛. Thus, 8𝑖=8𝑛 and LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,1}(⟨2,2,𝑛⟩)=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,1}(⟨2,2,𝑛⟩)=8𝑛.
The orbits 𝑙^{2,{1,1}{𝑥,𝑦,𝑧}}(⟨2,2,𝑛⟩) and 𝑙^{2,{2,1}{𝑥,𝑦,𝑧}}(⟨2,2,𝑛⟩) are the same. The proofs for these orbits are similar to the above and are omitted.
(iii) The orbits 𝑙^3,{1,1}(⟨2,2,𝑛⟩),𝑙^3,{1,2}(⟨2,2,𝑛⟩),𝑙^3,{2,1}(⟨2,2,𝑛⟩), and 𝑙^3,{2,2}(⟨2,2,𝑛⟩), respectively, are as follows: 𝑥₀=𝑥𝑦,𝑥₁=𝑥𝑧,𝑥₂=𝑒,𝑥₃=𝑧^𝑛+2,𝑥₄=𝑥𝑧^𝑛+3,𝑥₅=𝑥𝑧,₆𝑥=𝑧^𝑛,𝑥₇=𝑥𝑧²𝑥,𝑥₈=𝑦,𝑥₉=𝑥𝑧,𝑥₁₀=𝑒,…,₀=𝑥𝑦,𝑥₁=𝑧²𝑦,𝑥₂=𝑒,𝑥₃=𝑥𝑧𝑦,𝑥₄=𝑧⁴𝑦³,𝑥₅=𝑧²𝑦,₆𝑥=𝑧^𝑛,𝑥₇=𝑧^𝑛+2,𝑥₈=𝑦,𝑥₉=𝑧²𝑦,𝑥₁₀=𝑒,…,₀=𝑥𝑥𝑧,𝑥₁=𝑥𝑧,𝑥₂=𝑒,𝑥₃=𝑧^𝑛,𝑥₄=𝑥𝑧^𝑛+1,𝑥₅=𝑥𝑧,₆𝑥=𝑧^𝑛,𝑥₇=𝑧^𝑛,𝑥₈=𝑥𝑧,𝑥₉=𝑥𝑧,𝑥₁₀=𝑒,…,₀=𝑥𝑥𝑧,𝑥₁=𝑧²𝑦,𝑥₂=𝑒,𝑥₃=𝑦𝑧⁴𝑦,𝑥₄=𝑧^𝑛+6𝑦,𝑥₅=𝑧²𝑦,₆=𝑧^𝑛,𝑥₇=𝑧^𝑛+4,𝑥₈=𝑥𝑧,𝑥₉=𝑧²𝑦,𝑥₁₀=𝑒,…,(2.37) which have period 8.

Theorem 2.10. The 𝑖th 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}𝑙^1,{1}(⟨2,2,𝑛⟩)=LEN_{𝑥,𝑦}𝑙^1{2}(⟨2,2,𝑛⟩)=6,(ii)LEN_{𝑥,𝑦}𝑙^2,{1}(⟨2,2,𝑛⟩)=LEN_{𝑥,𝑦}𝑙^2,{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 orbits 𝑙^{1,{1}{𝑥,𝑦}}(⟨2,2,𝑛⟩) and 𝑙^{1,{2}{𝑥,𝑦}}(⟨2,2,𝑛⟩) are the same and are as follows: 𝑥₀=𝑥,𝑥₁=𝑦³,𝑥₂=𝑥𝑦³,𝑥₃=𝑦𝑥𝑦,𝑥₄=𝑦³,𝑥₅=𝑦𝑥,𝑥₆=𝑥,𝑥₇=𝑦³,…,(2.38) which have period 6.(ii)The orbits 𝑙^{2,{1}{𝑥,𝑦}}(⟨2,2,𝑛⟩) and 𝑙^{2,{2}{𝑥,𝑦}}(⟨2,2,𝑛⟩) are the same and are as follows: 𝑥₀=(𝑥𝑦)^𝑛−1,𝑥₁=(𝑥𝑦)^𝑛,….(2.39)We consider the recurrence relation defined by the following: 𝑢_𝑚=𝑢_𝑚−2+𝑢_𝑚−1,𝑢₀=𝑛−1,𝑢₁=𝑛.(2.40) Then a routine induction shows that 𝑥_𝑚=(𝑥𝑦)^𝑢𝑚. Using Lemma 1.4, we obtain 𝑢_𝐿=𝑢₀=𝑛−1 and  𝑢_𝐿+1=𝑢₁=𝑛. In this case the equalities 𝑥_𝑚=(𝑥𝑦)^𝑢𝑚 give 𝑥_𝐿=(𝑥𝑦)^𝑢𝐿=(𝑥𝑦)^𝑛−1,𝑥_𝐿+1=(𝑥𝑦)^𝑢𝐿+1=(𝑥𝑦)^𝑛.(2.41) The smallest nontrivial integer satisfying the above conditions occurs when the period is ℎ₂(2𝑛).

Theorem 2.11. The 𝑖th 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}}𝑙^{1,{𝛼1,𝛼2}}((𝑛,2,2))=6 for 1≤𝛼₁,𝛼₂≤2,(ii)LEN_{{𝑥,𝑦,𝑧}}𝑙^{2,{𝛼1,𝛼2}}((𝑛,2,2))=ℎ₃(𝑛) for 1≤𝛼₁,𝛼₂≤2,(iii)(1)LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,1}((𝑛,2,2))=LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,2}((𝑛,2,2))=8,(2)LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,1}((𝑛,2,2))=LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{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 Theorem 2.5(ii) and 2.5(iii) and are omitted.

Theorem 2.12. The 𝑖th 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}𝑙^1,{1}((𝑛,2,2))=LEN_{𝑥,𝑦}𝑙^1,{2}((𝑛,2,2))=6,(ii)LEN_{𝑥,𝑦}𝑙^2,{1}((𝑛,2,2))=LEN_{𝑥,𝑦}𝑙^2,{2}((𝑛,2,2))=3.

Proof. (i) The orbits 𝑙^1,{1}((𝑛,2,2)) and 𝑙^1,{2}((𝑛,2,2)) are the natural extension of the result of dihedral groups given in [16].
(ii) The orbits 𝑙^{2,{1}{𝑥,𝑦}}((𝑛,2,2)) and 𝑙^{2,{2}{𝑥,𝑦}}((𝑛,2,2)), respectively, are as follows: 𝑥𝑥₀=𝑥𝑦,𝑥₁=𝑒,𝑥₂=𝑥𝑦,𝑥₃=𝑥𝑦,𝑥₄=𝑒,…,₀=𝑦𝑥,𝑥₁=𝑒,𝑥₂=𝑦𝑥,𝑥₃=𝑦𝑥,𝑥₄=𝑒,…,(2.42) which have period 3.

Theorem 2.13. The 𝑖th 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}}𝑙^{1,{𝛼1,𝛼2}}((2,𝑛,2))=6 for 1≤𝛼₁,𝛼₂≤2,(ii)(1)LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,1}((2,𝑛,2))=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{2,1}((2,𝑛,2))=4,(2)LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{1,2}((2,𝑛,2))=LEN_{{𝑥,𝑦,𝑧}}𝑙^2,{2,2}((2,𝑛,2))=8,(iii)LEN_{{𝑥,𝑦,𝑧}}𝑙^{3,{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 Theorem 2.5(ii) and 2.5(iii) and are omitted.

Theorem 2.14. The 𝑖th 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 Theorem 2.6 and is omitted.

Theorem 2.15. The 𝑖th 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)LEN_{{𝑥,𝑦,𝑧}}𝑙^{1,{𝛼1,𝛼2}}(2,2,𝑛)=⎧⎪⎨⎪⎩2𝑛4𝑛if𝑛≡0mod4,8𝑛if𝑛≡2mod4,otherwise,𝑓𝑜𝑟1≤𝛼₁,𝛼₂≤2(2.43)(ii)LEN_{{𝑥,𝑦,𝑧}}𝑙^{2,{𝛼1,𝛼2}}((2,2,𝑛))=⎧⎪⎨⎪⎩𝑛2𝑛if𝑛≡0mod8,4𝑛if𝑛≡4mod8,8𝑛if𝑛≡2mod8,otherwise,𝑓𝑜𝑟1≤𝛼₁,𝛼₂≤2(2.44)(iii)(1)LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,1}((2,2,𝑛))==LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{1,2}((2,2,𝑛))LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,2}((2,2,𝑛))=8,(2.45)(2)LEN_{{𝑥,𝑦,𝑧}}𝑙^3,{2,1}((2,2,𝑛))=4.(2.46)

Proof. The proof is similar to the proof of Theorem 2.9 and is omitted.

Theorem 2.16. The 𝑖th 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}𝑙^1,{1}((2,2,𝑛))=LEN_{𝑥,𝑦}𝑙^1,{2}((2,2,𝑛))=6,(ii)LEN_{𝑥,𝑦}𝑙^2,{1}((2,2,𝑛))=LEN_{𝑥,𝑦}𝑙^2,{2}((2,2,𝑛))=ℎ₂(𝑛).

Proof. (i) The orbits 𝑙^1,{1}((2,2,𝑛)) and 𝑙^1,{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 Theorem 2.10(ii) and is omitted.

Acknowledgment

This Project was supported by the Commission for the Scientific Research Projects of Kafkas University. The Project number is 2010-FEF-61.