K-nacci Sequences in Finite Triangle Groups

Abstract

A 𝑘-nacci sequence in a finite group is a sequence of group elements 𝑥₀,𝑥₁,𝑥₂,…,𝑥_𝑛,… for which, given an initial (seed) set 𝑥₀,𝑥₁,𝑥₂,…,𝑥_𝑗−1 , each element is defined by 𝑥_𝑛=𝑥₀𝑥₁…𝑥_𝑛−1,for𝑗≤𝑛<𝑘,and𝑥_𝑛=𝑥_𝑛−𝑘𝑥_{𝑛−𝑘+1}…𝑥_𝑛−1,for𝑛≥𝑘. We also require that the initial elements of the sequence, 𝑥₀,𝑥₁,𝑥₂,…,𝑥_𝑗−1, generate the group, thus forcing the 𝑘-nacci sequence to reflect the structure of the group. The K-nacci sequence of a group generated by 𝑥₀,𝑥₁,𝑥₂,…,𝑥_𝑗−1 is denoted by 𝐹_𝑘(𝐺;𝑥₀,𝑥₁,…,𝑥_𝑗−1) and its period is denoted by 𝑃_𝑘(𝐺;𝑥₀,𝑥₁,…,𝑥_𝑗−1) . In this paper, we obtain the period of K-nacci sequences in finite polyhedral groups and the extended triangle groups.

1. Introduction

The Fibonacci sequences and their related higher-order (tribonacci, quatranacci, 𝑘-nacci) are generally viewed as sequences of integers. In [1] the Fibonacci length of a 2-generator group is defined, thus extending the idea of forming a sequence of group elements based on a Fibonacci-like recurrence relation first introduced by Wall in [2]. There he considered the Fibonacci length of the cyclic group 𝐶_𝑛. The concept of Fibonacci length for more than two generators has also been considered, see, for example [3, 4]. Also, the theory has been expanded to the nilpotent groups, see, for example [5–7]. Other works on Fibonacci length are discussed in, for example, [8–12]. Knox proved that the periods of 𝑘-nacci (𝑘-step Fibonacci) sequences in dihedral groups are equal to 2𝑘+2 [13]. Campbell and Campbel, examined the behaviour of the Fibonacci length of the finite polyhedral, binary polyhedral groups, and related groups in [14].

This paper discusses the period of 𝑘-nacci Fibonacci sequences in the polyhedral groups (2,2,2), (𝑛,2,2), (2,𝑛,2), (2,2,𝑛) for any 𝑛 and in the extended triangle groups 𝐸(2,2,2), 𝐸(𝑛,2,2), 𝐸(2,𝑛,2), 𝐸(2,2,𝑛) for any 𝑛>2. We consider polyhedral groups both as 2-generator and as 3-generator groups. A 2-step Fibonacci sequence in the integers modulo 𝑚 can be written as 𝐹₂(𝑍_𝑚;0,1). A 2-step Fibonacci sequence of group elements is called 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 𝑥₀,𝑥₁,𝑥₂,𝑥₃,𝑥₄,𝑥₁,𝑥₂,𝑥₃,𝑥₄,𝑥₁,𝑥₂,𝑥₃,𝑥₄,… is periodic after the initial element 𝑥₀ 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 𝑥₀,𝑥₁,𝑥₂,𝑥₃,𝑥₄,𝑥₀,𝑥₁,𝑥₂,𝑥₃,𝑥₄,… is simply periodic with period 5. It is important to note that the Fibonacci length depends on the chosen generating 𝑛- tuple for a group.

Definition 1.1. For a finitely generated group 𝐺=⟨𝐴⟩ where 𝐴={𝑎₁,𝑎₂,…,𝑎_𝑛} the sequence 𝑥_𝑖=𝑎_𝑖+1, 0≤𝑖≤𝑛−1, 𝑥_𝑖+𝑛=∏^𝑛_𝑗=1𝑥_{𝑖+𝑗−1}, 𝑖≥0, is called the Fibonacci orbit of 𝐺 with respect to the generating set 𝐴, denoted 𝐹_𝐴(𝐺).
Notice that the orbit of a k-generated group is a 𝑘-nacci sequence. The orbits of (𝑛,2,2), (2,𝑛,2), (2,2,𝑛) for any 𝑛>2 and 𝐸(2,𝑞,2) for any 𝑞>2 are studied in [14].

2. The Groups (2,2,2), (𝑛,2,2), (2,𝑛,2), and (2,2,𝑛)

Definition 2.1. 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⟩ (2.1) or ⟨𝑥,𝑦∶𝑥^𝑙=𝑦^𝑚=(𝑥𝑦)^𝑛=𝑒⟩. (2.2) The polyhedral group (𝑙,𝑚,𝑛) is finite if and only if the number 1𝜇=𝑙𝑚𝑛(𝑙+1𝑚+1𝑛−1)=𝑚𝑛+𝑛𝑙+𝑙𝑚−𝑙𝑚𝑛 (2.3) is positive, that is, in the case (2,2,𝑛), (2,3,3), (2,3,3), (2,3,4), (2,3,5). Its order is 2𝑙𝑚𝑛/𝜇. Using Tietze transformations, we may show that (𝑙,𝑚,𝑛)≅(𝑚,𝑛,𝑙)≅(𝑛,𝑙,𝑚). For more information on these groups see [15] and [16, pages 67–68]. The groups considered in Theorems 2.3 and 2.4 are the same group, namely, 𝐷_𝑛, the dihedral group of 2𝑛 elements, except the generators 𝑥,𝑦, and 𝑧 are different from one theorem to the other.

Theorem 2.2. Let 𝐺₂ be the group defined by the presentation 𝐺₂=⟨𝑥,𝑦,𝑧∶𝑥²=𝑦²=𝑧²=𝑥𝑦𝑧=𝑒⟩. Then 𝑃_𝑘(𝐺₂,𝑥,𝑦,𝑧)=𝑘+1.

Proof. Firstly, let us consider the 2-generator case. Notice that 𝐺₂ is 𝑍₂⊕𝑍₂ and 𝑃_𝑘(𝑍₂;0,1)=𝑘+1. Under these identifications, since the period of a Fibonacci sequence in a direct product of groups is the least common multiple of the periods in each the factors we get 𝑃_𝑘(𝐺₂;𝑥,𝑦)=𝑘+1. On the other hand, since 𝑧=𝑥𝑦 the formulas in the “three generator case” with recurrences of period 𝑘+1 are the same as the formulas the two generator case as long as 𝑘≥4.

Theorem 2.3. Let 𝐺_𝑛,𝑛>2, be the group defined by the presentation ⟨𝑥,𝑦,𝑧∶𝑥^𝑛=𝑦²=𝑧²=𝑥𝑦𝑧=𝑒⟩. Then 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦,𝑧)=2𝑘+2.

Proof. Let us consider the 3-generator case. We first note that the orders of 𝑥,𝑦, and 𝑧 are 𝑛,2,2, respectively. If 𝑘=2, we have the sequence 𝑥,𝑦,𝑧,𝑦𝑧,𝑧𝑦𝑧,𝑧,𝑥,𝑦,…, (2.4) which has period 6. If 𝑘=3, we have the sequence 𝑥,𝑦,𝑧,𝑥𝑦𝑧=𝑒,𝑦𝑧,𝑧𝑦𝑧,𝑧,𝑒,𝑥,𝑦,𝑧,…, (2.5) which has period 8. If 𝑘≥4, the first 𝑘 elements of sequence are 𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧,𝑥₃=𝑥𝑦𝑧,𝑥₄=(𝑥𝑦𝑧)²,…,𝑥_𝑘−1=(𝑥𝑦𝑧)^2𝑘−3. (2.6) Thus, using the above information the sequence reduces to 𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧,𝑥₃=𝑒,𝑥₄=𝑒,…,𝑒, (2.7) where 𝑥_𝑗=𝑒for3≤𝑗≤𝑘−1. Thus, 𝑥_𝑘=^𝑘−1_𝑖=0𝑥_𝑖=(𝑥𝑦)^2𝑘−2=𝑒,𝑥_𝑘+1=^𝑘_𝑖=1𝑥_𝑖=𝑦𝑧=𝑥^𝑛−1,𝑥_𝑘+2=^𝑘+1_𝑖=2𝑥_𝑖𝑥=𝑧𝑦𝑧=𝑥𝑧,_𝑘+3=^𝑘+2_𝑖=3𝑥_𝑖=𝑧,𝑥_𝑘+4=^𝑘+3_𝑖=4𝑥_𝑖=𝑒,…,𝑒. (2.8) It follows that 𝑥_𝑘+𝑗=𝑒for4≤𝑗≤𝑘. We also have, 𝑥_𝑘+𝑘+1=^𝑘+𝑘_𝑖=𝑘+1𝑥_𝑖=𝑒,𝑥_𝑘+𝑘+2=^𝑘+𝑘+1_𝑖=𝑘+2𝑥_𝑖𝑥=𝑥,_𝑘+𝑘+3=^𝑘+𝑘+2_𝑖=𝑘+3𝑥_𝑖=𝑦,𝑥_𝑘+𝑘+4=^𝑘+𝑘+3_𝑖=𝑘+4𝑥_𝑖=𝑧. (2.9) Since the elements succeeding 𝑥_2𝑘+2,𝑥_2𝑘+3,𝑥_2𝑘+4, depend on 𝑥,𝑦, and 𝑧 for their values, the cycle begins again with the 2𝑘+2nd element; that is, 𝑥₀=𝑥_2𝑘+2,𝑥₁=𝑥_2𝑘+3,𝑥₂=𝑥_2𝑘+4,…. Thus, 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦,𝑧)=2𝑘+2.
Similarly, it is easy to show that for 2-generator, 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦,𝑧)=2𝑘+2 in (𝑛,2,2), and it can be shown that 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦,𝑧)=2𝑘+2 for (2,𝑛,2).
Because of (𝑛,2,2)≅(2,𝑛,2)≅(2,2,𝑛)≅𝐷_𝑛 for any 𝑛>2 and using Tietze transformations we can obtain the same presentation for this groups, it is easy to show that for 2-generator 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦)=2𝑘+2 in the groups (𝑛,2,2),(2,𝑛,2), and (2,2,𝑛).

Theorem 2.4. Let 𝐺_𝑛,𝑛>2, be the group defined by the presentation ⟨𝑥,𝑦,𝑧∶𝑥²=𝑦²=𝑧^𝑛=𝑥𝑦𝑧=𝑒⟩
(i) 𝑃₂(𝐺_𝑛;𝑦,𝑥,𝑧)=6:
(ii)
𝑃₄(𝐺_𝑛5;𝑥,𝑦,𝑧)=𝑛(2),𝑛≡0mod4,5𝑛,𝑛≡2mod4,10𝑛,otherwise, (2.10)
(iii) 𝑘≥5.
(1)If there is no 𝑡∈[3,𝑘−2] such that 𝑡 is a odd factor of 𝑛, then 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦,𝑧)=𝑛(𝑘+12),𝑛≡0mod4,𝑛(k+1),𝑛≡2mod4,2𝑛(𝑘+1),otherwise. (2.11)(2)Let 𝛼 be the biggest odd factor of 𝑛 in [3,𝑘−2]. Then two cases occur:(i^′)if 𝛼3^𝑗∉[3,𝑘−2] for 𝑗∈𝑁, then 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛼(𝑛(𝑘+12)),𝑛≡0mod4,𝛼(𝑛(𝑘+1)),𝑛≡2mod4,𝛼(2𝑛(𝑘+1)),otherwise; (2.12)(ii^′)if 𝛽 is the biggest odd number which is in [3,𝑘−2] and 𝛽=𝛼3^𝑗 for 𝑗∈𝑁, then 𝑃_𝑘(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛽(𝑛(𝑘+12)),𝑛≡0mod4,𝛽(𝑛(𝑘+1)),𝑛≡2mod4,𝛽(2𝑛(𝑘+1)),otherwise. (2.13)

Proof. We consider 𝐺_𝑛 as 𝐷_𝑛, the dihedral group of 2𝑛 elements. Now 𝐷_𝑛 being the group of symmetries of the regular polygon with 𝑛 elements admits a presentation as the group generated by the two matrices:𝑎∶=cos(2𝜋𝑛)−sin(2𝜋𝑛)sin(2𝜋𝑛)cos(2𝜋𝑛),𝑏∶=100−1. (2.14) Under these identifications, we can take 𝑧=(^{cos(2𝜋/𝑛)−sin(2𝜋/𝑛)}_{sin(2𝜋/𝑛)cos(2𝜋/𝑛)}),𝑦=(¹⁰₀₋₁), and 𝑥=(^{cos(2𝜋/𝑛)−sin(2𝜋/𝑛)}_{−sin(2𝜋/𝑛)−cos(2𝜋/𝑛)}).(i)If 𝑘=2, we have the sequence𝑥₀=𝑦,𝑥₁=𝑥,𝑥₂=𝑧,𝑥₃=cos(4𝜋𝑛)−sin(4𝜋𝑛)−sin(4𝜋𝑛)−cos(4𝜋𝑛)=𝑥𝑧,𝑥₄𝑥=𝑥,₅=cos(2𝜋𝑛)sin(2𝜋𝑛)−sin(2𝜋𝑛)cos(2𝜋𝑛)=𝑥𝑦,𝑥₆=𝑦,𝑥₇=𝑥,𝑥₈=𝑧,…. (2.15) Thus we get 𝑃₂(𝐺_𝑛;𝑦,𝑥,𝑧)=6.(ii)If 𝑘=4, we have the sequence𝑥,𝑦,𝑧,𝑥𝑦𝑧=1001=𝑒,(𝑥𝑦𝑧)²=1001=𝑒,cos(2𝜋𝑛)−sin(2𝜋𝑛)−sin(2𝜋𝑛)−cos(2𝜋𝑛)=𝑥,100−1=𝑦,cos(2𝜋𝑛)sin(2𝜋𝑛)−sin(2𝜋𝑛)cos(2𝜋𝑛)=𝑥𝑦,cos(4𝜋𝑛)sin(4𝜋𝑛)−sin(4𝜋𝑛)cos(4𝜋𝑛)=𝑧⁻²,cos(6𝜋𝑛)sin(6𝜋𝑛)sin(6𝜋𝑛)−cos(6𝜋𝑛)=𝑧⁴𝑥,cos(14𝜋𝑛)sin(14𝜋𝑛)sin(14𝜋𝑛)−cos(14𝜋𝑛)=𝑧⁸𝑥,100−1=𝑦,cos(2𝜋𝑛)−sin(2𝜋𝑛)sin(2𝜋𝑛)cos(2𝜋𝑛)=𝑧,cos(8𝜋𝑛)−sin(8𝜋𝑛)sin(8𝜋𝑛)cos(8𝜋𝑛)=𝑧⁴,cos(24𝜋𝑛)−sin(24𝜋𝑛)sin(24𝜋𝑛)cos(24𝜋𝑛)=𝑧¹²,cos(34𝜋𝑛)−sin(34𝜋𝑛)−sin(34𝜋𝑛)−cos(34𝜋𝑛)=𝑥𝑧¹⁶,…. (2.16) Now we consider what happens to the 4-nacci sequence when we have a section of the form …,𝑧^𝜏𝑥,𝑧𝑥,𝑧,…: 𝑧^𝜏𝑥,𝑧𝑥=𝑦,𝑧,𝑧^𝜀,𝑧^𝜏+𝜀,𝑥𝑧^2𝜀+𝜏,𝑦,𝑥𝑧𝑥=𝑥𝑦,𝑥𝑧^𝜀+2𝑥=𝑧^−(𝜀+2),𝑥𝑧^3𝜀+𝜏+4𝑥=𝑧^{−(3𝜀+𝜏+4)},𝑧^4𝜀+𝜏+8𝑥,𝑧𝑥=𝑦,𝑧,…. (2.17) The 4-nacci sequence can be said to form layers of length 10. Using the above, the 4-nacci sequence becomes 𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧,𝑥₃𝑥=𝑒,…,₁₀=𝑧⁸𝑥,𝑥₁₁=𝑧𝑥=𝑦,𝑥₁₂=𝑧,𝑥₁₃=𝑧⁴𝑥,…,₂₀=𝑧³²𝑥,𝑥₂₁=𝑧𝑥=𝑦,𝑥₂₂=𝑧,𝑥₂₃=𝑧⁸𝑥,…,_10𝑖=𝑧^8𝑖2𝑥,𝑥_10𝑖+1=𝑧𝑥=𝑦,𝑥_10𝑖+2=𝑧,𝑥_10𝑖+3=𝑧^4𝑖,…, (2.18) where 𝑧^8𝑖2=(^{cos8𝑖2(2𝜋/𝑛)−sin8𝑖2(2𝜋/𝑛)}_{sin8𝑖2(2𝜋/𝑛)cos8𝑖2(2𝜋/𝑛)}) and 𝑧^4𝑖=(^{cos4𝑖(2𝜋/𝑛)−sin4𝑖(2𝜋/𝑛)}_{sin4𝑖(2𝜋/𝑛)cos4𝑖(2𝜋/𝑛)}).
So, we need the smallest 𝑖∈𝑁 such that 8𝑖²=𝑛𝑣₁ and 4𝑖=𝑛𝑣₂ for 𝑣₁,𝑣₂∈𝑁.
If 𝑛≡0(mod4), 𝑧^8𝑖2=(¹⁰₀₁) and 𝑧^4𝑖=(¹⁰₀₁) for 𝑖=𝑛/4.
Thus, 10𝑖=(5/2)𝑛 and 𝑃₄=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝑛((𝑘+1)/2)=(5/2)𝑛.
If 𝑛≡2(mod4), 𝑧^8𝑖2=(¹⁰₀₁) and 𝑧^4𝑖=(¹⁰₀₁) for 𝑖=𝑛/2.
Thus, 10𝑖=5𝑛 and 𝑃₄=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝑛(𝑘+1)=5𝑛.
If 𝑛≡1(mod4) or 𝑛≡3(mod4), 𝑧^8𝑖2=(¹⁰₀₁) and 𝑧^4𝑖=(¹⁰₀₁) for 𝑖=𝑛.
Thus, 10𝑖=10𝑛 and 𝑃₄=(𝐺_𝑛;𝑥,𝑦,𝑧)=2𝑛(𝑘+1)=10𝑛.
(iii)If 𝑘≥5, the first 𝑘+1 elements of the sequence are𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧,𝑥₃=𝑧^𝑛,𝑥₄=𝑧^2𝑛,…,𝑥_𝑘=𝑧^{2𝑘−3𝑛}. (2.19) Thus, using the above information, the sequence reduces 𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧,𝑥₃=𝑒,𝑥₄=𝑒,…,𝑒, (2.20) where 𝑥_𝑗=𝑒 for 3≤𝑗≤𝑘.
Now we consider what happens to the 𝑘-nacci sequence when we have a section of the form …,𝑧^𝜏𝑥,𝑧𝑥,𝑧,…∶
𝑥_2𝑘+2=^2𝑘+1_𝑖=𝑘+2𝑥_𝑖=𝑧^𝜏𝑥,𝑥_2𝑘+2+1=^2𝑘+2_𝑖=𝑘+3𝑥_𝑖=𝑧𝑥=𝑦,𝑥_2𝑘+2+2=^2𝑘+3_𝑖=𝑘+4𝑥_𝑖𝑥=𝑧,_2𝑘+2+3=^2𝑘+4_𝑖=𝑘+5𝑥_𝑖=𝑧^𝜀,𝑥_2𝑘+2+4=^2𝑘+5_𝑖=𝑘+6𝑥_𝑖=𝑧^𝑐,𝑥_2𝑘+2+5=^2𝑘+6_𝑖=𝑘+7𝑥_𝑖=𝑧^𝑢1𝑥,…,_2𝑘+2+𝑘=^3𝑘+1_𝑖=2𝑘+2𝑥_𝑖=𝑧^𝑢𝑘−4,…. (2.21) The 𝑘-nacci sequence can be said to form layers of length (2𝑘+2). Using the above, the 𝑘-nacci sequence becomes 𝑥₀=𝑥,𝑥₁=𝑦,𝑥₂=𝑧,𝑥₃=𝑒,…,𝑥_𝑘=𝑧^{2𝑘−3𝑛}𝑥=𝑒,…,_{𝑖(2𝑘+2)}=𝑧^𝜏𝑥,𝑥_{𝑖(2𝑘+2)+1}=𝑧𝑥,𝑥_{𝑖(2𝑘+2)+2}=𝑧,𝑥_{𝑖(2𝑘+2)+3}=𝑧^4𝑖,𝑥_{𝑖(2𝑘+2)+4}=𝑧^8𝑖2+4𝑖,𝑥_{𝑖(2𝑘+2)+5}=𝑧^𝑢1,…,𝑥_{𝑖(2𝑘+2)+𝑘}=𝑧^𝑢𝑘−4,…. (2.22) So, we need the smallest 𝑖∈𝑁 such that 4𝑖=𝑛𝑣₁ and 8𝑖²+4𝑖=𝑛𝑣₂ for 𝑣₁,𝑣₂∈𝑁. (1)If there is no 𝑡∈[3,𝑘−2] such that 𝑡 is an odd factor of 𝑛, there are 3 subcases.Case 1. If 𝑛≡0(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4 , 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝑛/4. So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝑛((𝑘+1)/2) since 𝑖(2𝑘+2)=𝑛((𝑘+1)/2) (where by 𝑛∣𝜏 we mean that 𝑛 divides 𝜏).Case 2. If 𝑛≡2(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝑛/2. So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝑛(𝑘+1) since 𝑖(2𝑘+2)=𝑛(𝑘+1).Case 3. If 𝑛≡1(mod4) or 𝑛≡3(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝑛. So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=2𝑛(𝑘+1) since 𝑖(2𝑘+2)=2𝑛(𝑘+1).(2)Let 𝛼 odd be the biggest factor of 𝑛 in [3,𝑘−2]. Then two cases occur:(i^′)If 𝛼3^𝑗∉[3,𝑘−2] for 𝑗∈𝑁, then there are 3 subcases.Case 1. If 𝑛≡0(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝛼(𝑛/4). So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛼(𝑛((𝑘+1)/2)) since 𝑖(2𝑘+2)=𝛼(𝑛((𝑘+1)/2)).Case 2. If 𝑛≡2(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝛼(𝑛/2). So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛼(𝑛(𝑘+1)) since 𝑖(2𝑘+2)=𝛼(𝑛(𝑘+1)).Case 3. If 𝑛≡1(mod4) or 𝑛≡3(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁) and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝛼𝑛. So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛼(2𝑛(𝑘+1)) since 𝑖(2𝑘+2)=𝛼(2𝑛(𝑘+1)).(ii’)If 𝛽 is the biggest odd number which is in [3,𝑘−2] and 𝛽=𝛼3^𝑗 for 𝑗∈𝑁, then there are 3 subcases.Case 1. If 𝑛≡0(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝛽(𝑛/4). So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛽(𝑛((𝑘+1)/2)) since 𝑖(2𝑘+2)=𝛽(𝑛((𝑘+1)/2)).Case 2. If 𝑛≡2mod4 and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝛽(𝑛/2). So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛽(𝑛(𝑘+1)) since 𝑖(2𝑘+2)=𝛽(𝑛(𝑘+1)).Case 3. If 𝑛≡1(mod4) or 𝑛≡3(mod4) and 𝑛∣𝜏,𝑛∣𝑢₁,…,𝑛∣𝑢_𝑘−4, 𝑧^4𝑖=(¹⁰₀₁), and 𝑧^8𝑖2+4𝑖=(¹⁰₀₁) for 𝑖=𝛽𝑛. So, we get 𝑃_𝑘=(𝐺_𝑛;𝑥,𝑦,𝑧)=𝛽(2𝑛(𝑘+1)) since 𝑖(2𝑘+2)=𝛽(2𝑛(𝑘+1)).This completes the proof.

In the case of 2-generator the group has the presentation ⟨𝑥,𝑦∶𝑥²=𝑦²=(𝑥𝑦)^𝑛=𝑒⟩ and the period is the same as in the 3-generator case and proof is similar.

3. The Groups 𝐸(2,2,2), 𝐸(𝑛,2,2), 𝐸(2,𝑛,2), and 𝐸(2,2,𝑛)

Definition 3.1. The extended triangle group 𝐸(𝑝,𝑞,𝑟), for 𝑝,𝑞,𝑟>1, is defined by the presentation ⟨𝑥,𝑦,𝑧∶𝑥²=𝑦²=𝑧²=(𝑥𝑦)^𝑝=(𝑦𝑧)^𝑞=(𝑧𝑥)^𝑟=𝑒⟩. (3.1) The extended triangle groups are a very important class of groups closely linked to automorphism groups of regular maps, see [17]. The triangle groups (polyhedral groups), (𝑝,𝑞,𝑟) are index two subgroups of extended triangle groups. To see this, let 𝑋=𝑥𝑦,𝑌=𝑦𝑧 and 𝑍=𝑧𝑥 in 𝐸(𝑝,𝑞,𝑟) and then use the obvious epimorphism. We get the following three cases for 𝐸(𝑝,𝑞,𝑟):(1)the Euclidean case if 1/𝑝+1/𝑞+1/𝑟=1,(2)the elliptic case if 1/𝑝+1/𝑞+1/𝑟>1,(3)the hyperbolic case if 1/𝑝+1/𝑞+1/𝑟<1.The group 𝐸(𝑝,𝑞,𝑟) is finite if and only if 1/𝑝+1/𝑞+1/𝑟>1.
For more information on these groups, see [14, 18].

Theorem 3.2. Let 𝐸₂ be the group defined by the presentation ⟨𝑥,𝑦,𝑧∶𝑥²=𝑦²=𝑧²=(𝑥𝑦)²=(𝑦𝑧)²=(𝑧𝑥)²=𝑒⟩. Then 𝑃_𝑘(𝐸₂;𝑥,𝑦,𝑧)=𝑘+1 for 𝑘>2.

Proof. Since 𝐸₂ can be identified with 𝑍₂⊕𝑍₂⊕𝑍₂ and 𝑥,𝑦,𝑧 with (1,0,0),(0,1,0),and(0,0,1), respectively, from a similar argument applied to Theorem 2.2, we get 𝑃_𝑘(𝐸₂;𝑥,𝑦,𝑧)=𝑘+1.

Theorem 3.3. Let 𝐸_𝑛,𝑛>2, be the group defined by the presentation ⟨𝑥,𝑦,𝑧∶𝑥²=𝑦²=𝑧²=(𝑥𝑦)²=(𝑦𝑧)²=(𝑧𝑥)^𝑛=𝑒⟩(i)𝑃_4,5(𝐸_𝑛;𝑥,𝑦,𝑧)=𝑛(𝑘+12),𝑛≡0mod4,𝑛(k+1),𝑛≡2mod4,2𝑛(𝑘+1),otherwise; (3.2)(ii)let 𝑘≥6.(1)If there is no 𝑡∈[3,𝑘−3] such that 𝑡 is an odd factor of 𝑛, then 𝑃_𝑘(𝐸_𝑛;𝑥,𝑦,𝑧)=𝑛(𝑘+12),𝑛≡0mod4,𝑛(k+1),𝑛≡2mod4,2𝑛(𝑘+1),otherwise. (3.3)(2)Let 𝛼 be the biggest odd factor of 𝑛 in [3,𝑘−3]. Then two cases occur:(i^′)if 𝛼3^𝑗∉[3,𝑘−3] for 𝑗∈𝑁, then 𝑃_𝑘(𝐸_𝑛;𝑥,𝑦,𝑧)=𝛼(𝑛(𝑘+12)),𝑛≡0mod4,𝛼(𝑛(𝑘+1)),𝑛≡2mod4,𝛼(2𝑛(𝑘+1)),otherwise; (3.4)(ii^′)if 𝛽 is be the biggest odd number which is in [3,𝑘−3] and 𝛽=𝛼3^𝑗 for 𝑗∈𝑁, then𝑃_𝑘(𝐸_𝑟;𝑥,𝑦,𝑧)=𝛽(𝑛(𝑘+12)),𝑛≡0mod4,𝛽(𝑛(𝑘+1)),𝑛≡2mod4,𝛽(2𝑛(𝑘+1)),otherwise. (3.5)

Proof. Since 𝑦 has order 2 and commutes with 𝑥 and 𝑧 it follows that 𝐸_𝑛=𝑍₂⊕𝐷_𝑛. As a group of matrices, the can be identified with a group of 3×3 matrices of form ±100a, (3.6) where a is a 2×2 matrix in dihedral group generated by 𝑎 and 𝑏 shown at (2.14). Here, 𝑥=1000cos(2𝜋𝑛)−sin(2𝜋𝑛)0−sin(2𝜋𝑛)−cos(2𝜋𝑛),𝑦=−100010001,𝑧=10001000−1. (3.7) Now, since the period of a Fibonacci sequence in a direct product of groups is the least common multiple of the periods in each the factors and from a similar argument applied to Theorem 2.4 the proof is done.

Acknowledgment

The authors thank the referees for their valuable suggestions which improved the presentation of the paper.