Fibonacci sequences in groupoids

Abstract

In this article, we consider several properties of Fibonacci sequences in arbitrary groupoids (i.e., binary systems). Such sequences can be defined in a left-hand way and a right-hand way. Thus, it becomes a question of interest to decide when these two ways are equivalent, i.e., when they produce the same sequence for the same inputs. The problem has a simple solution when the groupoid is flexible. The Fibonacci sequences for several groupoids and for the class of groups as special cases are also discussed.

2000 Mathematics Subject Classification: 20N02; 11B39.

1 Introduction

In this article, we consider several properties of Fibonacci sequences in arbitrary groupoids (i.e., binary systems). Such sequences can be defined in a left-hand way and a right-hand way. Thus, it becomes a question of interest to decide when these two ways are equivalent, i.e., when they produce the same sequence for the same inputs. The problem has a simple solution when the groupoid is flexible. In order to construct sufficiently large classes of flexible groupoids to make the results interesting, the notion of a groupoid (X, ∗) wrapping around a groupoid (X, •) is employed to construct flexible groupoids (X, □). Among other examples this leads to solving the problem indicated for the selective groupoids associated with certain digraphs, providing many flexible groupoids and indicating that the general groupoid problem of deciding when a groupoid (X, ∗) is wrapped around a groupoid (X, •) is of independent interest as well.

Given the usual Fibonacci-sequences [1, 2] and other sequences of this type, one is naturally interested in considering what may happen in more general circumstances. Thus, one may consider what happens if one replaces the (positive) integers by the modulo an integer n or what happens in even more general circumstances. The most general circumstance we shall deal with is the situation where (X, ∗) is actually a groupoid, i.e., the product operation ∗ is a binary operation, where we assume no restrictions a priori.

2 Fibonacci sequences in groupoids

Given a sequence < φ₀, φ₁, ..., φ _n , ... > of elements of X, it is a left-∗-Fibonacci sequence if φ_n+2= φ_n+1∗ φ _n for n ≥ 0, and a right-∗-Fibonacci sequence if φ_n+2 = φ _n ∗ φ_n+1for n ≥ 0. Unless (X, ∗) is commutative, i.e., x ∗ y = y ∗ x for all x, y ∈ X, there is no reason to assume that left-∗-Fibonacci sequences are right-∗-Fibonacci sequences and conversely. We shall begin with a collection of examples to note what if anything can be concluded about such sequences.

Example 2.1. Let (X, ∗) be a left-zero-semigroup, i.e., x ∗ y := x for any x, y ∈ X. Then φ₂ = φ₁ ∗ φ₀ = φ₁, φ₃ = φ₂ ∗ φ₁ = φ₂ = φ₁, φ₄ = φ₃ ∗ φ₂ = φ₃ = φ₁, ... for any φ₀, φ₁ ∈ X. It follows that < φ _n > _L = < φ₀, φ₁, φ₁, ... >. Similarly, φ₂ = φ₀ ∗ φ₁ = φ₀, φ₃ = φ₁ ∗ φ₂ = φ₁, φ₄ = φ₂ ∗ φ₃ = φ₂ = φ₀, ... for any φ₀, φ₁ ∈ X. It follows that < φ _n > _R = < φ₀, φ₁, φ₀, φ₁, φ₀, φ₁, ... >. In particular, if we let φ₀ := 0, φ₁ := 1, then < φ _n > _L = < 0, 1, 1, 1, 1, ... > and < φ _n > _R = < 0, 1, 0, 1, 0, 1, ... >.

A groupoid (X, ∗) is said to be a leftoid if x ∗ y = l(x), a function of x in X, for all x, y ∈ X. We denote it by (X, ∗, l).

Proposition 2.2. Let (X, ∗, l) be a leftoid and let < φ _n > be a left-∗-Fibonacci sequence on X. Then < φ _n > = < φ₀, φ₁, l(φ₁), l⁽²⁾(φ₁), ..., l⁽ⁿ⁾(φ₁), ...>, where l⁽ⁿ⁺¹⁾(x) = l(lⁿ(x)).

Proof. If < φ _n > is a left-∗-Fibonacci sequence on X, then φ₂ = φ₁ ∗ φ₀ = l(φ₁) and φ₃ = φ₂ ∗ φ₁ = l(φ₂) = l⁽²⁾(φ₁). It follows that φ_k+1= φ _k ∗ φ_k-1= 1(φ _k ) = l^(k)(φ₁). □

A groupoid (X, ∗) is said to be a rightoid if x ∗ y = r(y), a function of y in X, for all x, y ∈ X. We denote it by (X, ∗, r).

Proposition 2.2'. Let (X, ∗, r) be a rightoid and let < φ _n > be a right-∗-Fibonacci sequence on X. Then < φ _n > = < φ₀, φ₁, r (φ₀), r(φ₀), r⁽²⁾(φ₀), r⁽²⁾ (φ₁), r⁽³⁾ (φ₀), r(3)(φ), ..., r⁽ⁿ⁾(φ₀), r⁽ⁿ⁾(φ₁), ... where r⁽ⁿ⁺¹⁾(x) = r(rⁿ(x)).

In particular, if l (r, resp.) is a constant map in Proposition 2.2 (or Proposition 2.2', resp.), say l(φ₀) = l(φ₁) = a for some x ∈ X, then < φ _n > = < φ₀, φ₁, a, a, ... >.

Theorem 2.3. Let < φ _n > _L and < φ _n > _R be the left-∗-Fibonacci and the right-∗-Fibonacci sequences generated by φ₀and φ₁. Then < φ _n > _L = < φ _n > _R if and only if φ _n ∗ (φ_n-1∗ φ _n ) = (φ _n ∗φ_n-1) ∗ φ _n for any n ≥ 1.

Proof. If < φ _n > _L = < φ _n > _R , then φ₀∗φ₁ = φ₂ = φ₁∗φ₀ and hence (φ₁∗φ₀)∗φ₁ = φ₂∗φ₁ = φ₃ = φ₁∗φ₂ = φ₁∗(φ₀∗φ₁). Similarly (φ₂∗φ₁)∗φ₂ = φ₃∗φ₂ = φ₄ = φ₂∗φ₃ = φ₂∗ (φ₁∗ φ₂). By induction on n, we obtain φ _n ∗ (φ_n-1∗ φ _n ) = (φ _n ∗ φ_n-1) ∗ φ _n for any n ≥ 1.

If we assume that φ _n ∗ (φ_n-1∗ φ _n ) = (φ _n ∗ φ_n-1) ∗ φ _n for any n ≥ 1, then φ _n ∗ φ_n+1= φ _n ∗ (φ_n-1∗ φ _n ) = (φ _n ∗ φ_n-1) ∗ φ _n = φ_n+1∗ φ _n for any n ≥ 1. □

A groupoid (X, ∗) is said to be flexible if (x ∗ y) ∗ x = x ∗ (y ∗ x) for any x, y ∈ X.

Proposition 2.4. Let X := R be the set of all real numbers and let A, B ∈ R. Then any groupoid (X, ∗) of the types x∗y := A + B(x + y) or x∗y := Bx + (1 - B)y for any x, y ∈ X is flexible.

Proof. Define a binary operation "∗" on X by x ∗ y := A + Bx + Cy for any x, y ∈ X, where A, B, C ∈ X. Assume that (X, ∗) is flexible. Then (x ∗ y) ∗ x = x ∗ (y ∗ x) for any x, y ∈ X. It follows that A + B(x + y) + Cx = A + Bx + C(y ∗ x). It follows that

$$AB+B\left(B-1\right)x=AC+C\left(C-1\right)x$$

(1)

for any x ∈ X. If we let x := 0 in (1), then we obtain AB = AC. If A ≠ 0, then B = C and x ∗ y = A + B(x + y). If A = 0, then it follows from (1) that

$$B\left(B-1\right)x=C\left(C-1\right)x$$

(2)

for any x ∈ X. If we let x := 1 in (2), then we obtain B(B - 1) = C(C - 1) and hence

$$C=\frac{1\pm \sqrt{1+4B\left(B-1\right)}}{2}=\left\{\begin{array}{c}B\hfill \\ 1-B,\hfill \end{array}\right.$$

i.e., x ∗ y = Bx + By or x ∗ y = Bx + (1 - B)y. This proves the proposition. □

Proposition 2.5. Let (X, ∗) be a flexible groupoid. Then (X, ∗) is commutative if and only if < φ₀, φ₁, ... > _L = < φ₀, φ₁, ... > _R for any φ₀, φ₁ ∈ X.

Proof. Given φ₀, φ₁ ∈ X, φ₀ ∗ φ₁ = φ₁ ∗ φ₀ = φ₂ since (X, ∗) is commutative. Since (X, ∗) is flexible, we obtain

$$\begin{array}{ll}\hfill {\phi }_2*{\phi }_1& =\left({\phi }_1*{\phi }_0\right)*{\phi }_1\\ ={\phi }_1*\left({\phi }_0*{\phi }_1\right)\\ ={\phi }_1*{\phi }_2={\phi }_3\end{array}$$

By induction on n, we obtain

$$\begin{array}{ll}\hfill {\phi }_n*{\phi }_{n-1}& =\left({\phi }_{n-1}*{\phi }_{n-2}\right)*{\phi }_{n-1}\\ ={\phi }_{n-1}*\left({\phi }_{n-2}*{\phi }_{n-1}\right)\\ ={\phi }_{n-1}*{\phi }_n\end{array}$$

Hence < φ₀, φ₁, ... > _L = < φ₀, φ₁, ... > _R . The converse is trivial, we omit the proof. □

Example 2.6. Let X := R be the set of all real numbers and let x ∗ y := -(x + y) for any x, y ∈ X. Consider aright-∗-Fibonacci sequence < φ _n > _R , where φ₀, φ₁ ∈ X. Since φ₂ = φ₀∗ φ₁ = -(φ₀ + φ₁), φ₃ = φ₁ ∗ φ₂ = -[φ₁ + φ₂] = -[φ₁-(φ₀ + φ₁)] = φ₀, φ₄ = φ₂ ∗ φ₃ = φ₁, ..., we obtain < φ _n > _R = < φ₀, φ₁, -(φ₀ + φ₁), φ₀, φ₁, -(φ₀ + φ₁), φ₀, φ₁, -(φ₀ + φ₁), φ₀, φ₁, ... > and φ_n+3= φ _n (n = 0, 1, 2, ...). Since (X, ∗) is commutative and flexible, by Proposition 2.5, < φ _n > _L = < φ _n > _R .

Proposition 2.7. Let (X, ∗) be a groupoid satisfying the following condition:

$$\left(x*y\right)*x=x*\left(y*x\right)=y$$

(3)

for any x, y ∈ X. Then < φ _n > _L = < φ _n > _R if φ₀ ∗ φ₁ = φ₁ ∗ φ₀.

Proof. Straightforward. □

Proposition 2.8. The linear groupoid (R, ∗), with x ∗ y := A - (x + y), ∀x, y ∈ R, where A ∈ R, is the only linear groupoid satisfying the condition (3).

Proof. By Proposition 2.4, we consider two cases: x ∗ y := A + B(x + y) or x ∗ y := Bx + (1- B)y where A, B ∈ R. Assume that x ∗ y := A + B(x + y). Since y = (x ∗ y) ∗ x, we have

$$\begin{array}{ll}\hfill y& =\left(x*y\right)*x\\ =A+B\left(x*y+x\right)\\ =A+B\left(A+B\left(x+y\right)+x\right)\\ =A\left(B+1\right)+B\left(B+1\right)x+B^{2}y\end{array}$$

It follows that B² = 1, B(B + 1) = 0, A(B + 1) = 0. If B = 1, then 0 = B(B + 1) = 2, a contradiction. If B = -1, then A is arbitrary. Hence x ∗ y = A - (x + y). Assume that x ∗ y := Bx + (1 - B)y. Since y = (x ∗ y) ∗ x, we have

$$\begin{array}{ll}\hfill y& =\left(x*y\right)*x\\ =\left[Bx+\left(1-B\right)y\right]*x\\ =B\left[Bx+\left(1-B\right)y\right]+\left(1-B\right)x\\ =\left(B^2-B+1\right)x+B\left(1-B\right)y\end{array}$$

It follows that B² - B + 1 = 0, B(1 - B) = 1 which leads to $B=\frac{1\pm \sqrt{3i}}{2}\notin R$, a contradiction.

This proves the proposition. □

3 Flexibility in Bin(X)

Given groupoids (X, ∗) and (X, •), we consider (X, ∗) to be wrapped around (X, •) if for all x, y, z ∈ X, (x • y) ∗ z = z ∗ (y • x). If (X, ∗) and (X, •) are both commutative groupoids, then (x • y) ∗ z = z ∗ (x • y) = z ∗ (y • x) and (x ∗ y) • z = z • (x ∗ y) for all x, y, z ∈ X, i.e., (X, ∗) and (X, •) are wrapped around each other.

Example 3.1. Let X := R be the set of all real numbers and let x ∗ y := x²y², x • y := x-y for all x, y ∈ X. Then (x • y) ∗ z = (x - y) ∗ z = (x - y)²z² = z ∗ (y • x) for all x, y, z ∈ X, i.e., (X, ∗) is wrapped around (X, •). On the other hand, (x ∗ y) • z = x²y² - z and z • (y ∗ x) = z- y²x² so that (X, •) is not wrapped around (X, ∗).

The notion of the semigroup (Bin(X), □) was introduced by Kim and Neggers [3]. They showed that (Bin(X), □) is a semigroup, i.e., the operation □ as defined in general is associative. Furthermore, the left-zero semigroup is an identity for this operation.

Proposition 3.2. Let (X, ∗) be wrapped around (X, •). If we define (X, □) := (X, ∗)□(X, •), i.e., x□y := (x ∗ y) • (y ∗ x) for all x, y ∈ X, then (X, □) is flexible.

Proof. Given x, y, z ∈ X, since (X, ∗) is wrapped around (X, •), we obtain (X, ∗) be wrapped around (X, •)

$$\begin{array}{ll}\hfill \left(x\square y\right)\square x& =\left[\left(x\square y\right)*x\right]*\left[x*\left(x\square y\right)\right]\\ =\left[\left\{\left(x*y\right)\bullet \left(y*x\right)\right\}*x\right]\bullet \left[x*\left\{\left(x*y\right)\bullet \left(y*x\right)\right\}\right]\\ =\left[x*\left\{\left(y*x\right)\bullet \left(x*y\right)\right\}\right]\bullet \left[\left\{\left(y*x\right)\bullet \left(x*y\right)\right\}*x\right]\\ =\left[x*\left(y\square x\right)\right]\bullet \left[\left(y\square x\right)*x\right]\\ =x\square \left(y\square x\right),\end{array}$$

proving the proposition. □

Example 3.3. Note that in the situation of Example 3.1, we have x□y = (x ∗ y) • (y ∗ x) = x²y² - y²x² = 0 for all x, y ∈ X, i.e., (X, □) = (X, ∗)□(X, •) is a trivial groupoid (X, □, t) where t = 0 and x□y = 0 for all x, y ∈ X.

Example 3.4. In Example 3.1, if we define (X, ∇) := (X, •)□(X, ∗), i.e., x∇y := (x • y) ∗ (y • x) for all x, y ∈ X, then x∇y = (x - y)⁴ and hence (x∇y)Δx = ((x - y)⁴ - x)⁴ = (x - (y - x)⁴)⁴ = x∇(y∇x). Hence (X, ∇) is a flexible groupoid. Note that (X, ∇) is not a semigroup, since 0∇(0∇z) = z¹⁶ ≠ z⁴ = (0∇0)∇z. Obviously, x∇y = y∇x for all x, y ∈ X. By applying Proposition 2.5, we obtain < φ₀, φ₁, ... > _L = < φ₀, φ₁, ... > _R for any φ₀, φ₁ ∈ (X, ∇).

We obtain a Fibonacci-∇-sequence in the groupoid (X, ∇) discussed in Example 3.4 as follows:

Example 3.5. Consider a groupoid (X, ∇) in Example 3.4. Since x∇y = (x - y)⁴, given φ₀, φ₁ ∈ X, we have φ₂ = φ₀ ∇ φ₁ = φ₁∇φ₀ = (φ₁ - φ₀)⁴, and φ₃ = φ₂∇φ₁ = (φ₂ - φ₁)⁴ = [(φ₁ - φ₀)⁴ - φ₁]⁴. In this fashion, we have φ₄ = [[(φ₁ - φ₀)⁴ - φ₁]⁴ - (φ₁ - φ₀)⁴]⁴. In particular, if we let φ₀ = φ₁ = 1, then φ₂ = 0, φ₃ = φ₄ = 1, φ₅ = 0, φ₆ = φ₇ = 1, φ₈ = 0, .... Hence < 1, 1, 0, 1, 1, 0, 1, 1, 0, ... > is a Fibonacci-∇-sequence in (X, ∇).

4 Limits of ∗-Fibonacci sequences

In this section, we discuss the limit of left(right)-∗-Fibonacci sequences in a real groupoid (R, ∗).

Proposition 4.1. Define a binary operation ∗ on R by$x*y:=\frac{1}{2}\left(x+y\right)$for any x, y ∈ R. If < φ _n > is a ∗-Fibonacci sequence on (R, ∗), then$\underset{n\to \infty }{\text{lim}}{\phi }_n=\frac{1}{3}\left({\phi }_0+2{\phi }_1\right)$.

Proof. Since $x*y=\frac{1}{2}\left(x+y\right)=y*x$ for any x, y ∈ R, < φ _n > _L = < φ _n > _R for any φ₀, φ₁ ∈ R.

It can be seen that ${\phi }_2=\frac{1}{2}\left({\phi }_0+{\phi }_1\right),{\phi }_3=\frac{1}{2^2}\left({\phi }_0+3{\phi }_1\right)$. We let ${\phi }_3=\frac{1}{2^2}\left(A_3{\phi }_0+B_3{\phi }_1\right)$. Since

$$\begin{array}{ll}\hfill {\phi }_4& ={\phi }_2*{\phi }_3\\ =\frac{1}{2}\left[\frac{{\phi }_0+{\phi }_1}{2}+\frac{1}{4}\left({\phi }_0+3{\phi }_1\right)\right]\\ =\frac{1}{2^3}\left[3{\phi }_0+5{\phi }_1\right],\end{array}$$

we let it by ${\phi }_4=\frac{1}{2^3}\left[A_4{\phi }_0+B_4{\phi }_1\right]$. In this fashion, if we let ${\phi }_{n+2}:=\frac{1}{2^{n+1}}\left[A_{n+2}{\phi }_0+B_{n+2}{\phi }_1\right]$, we have

$$\begin{array}{ll}\hfill {\phi }_{n+2}& =\frac{1}{2}\left[{\phi }_n+{\phi }_{n+1}\right]\\ =\frac{1}{2}\left[\frac{A_n{\phi }_0+B_n{\phi }_1}{2^{n-1}}+\frac{A_{n+1}{\phi }_0+B_{n+1}{\phi }_1}{2^n}\right]\\ =\frac{1}{2^{n+1}}\left[\left(2A_n+A_{n+1}\right){\phi }_0+\left(2B_n+B_{n+1}\right){\phi }_1\right]\end{array}$$

It follows that

$$\begin{array}{ll}\hfill A_{n+2}& =2A_n+A_{n+1},A_3=1,A_4=3\\ \hfill B_{n+2}& =2B_n+B_{n+1},B_3=3,B_4=5\end{array}$$

so that the evolution for A _k is < 1, 3, 5, 11, 21, 43, 85, ... > = < A₃, A₄, A₅, ... > and the evolution for B _k is < 3, 5, 11, 21, 43, 85, ... > so that B _k = A_k+1. If we wish to solve explicitly for A _k , we note that the corresponding characteristic equation is r² - r - 2 = 0 with roots r = 2 or r = -1, i.e., A_k+3= α 2^k+ β(-1)^k, α + β = 1 when k = 0, and 2α - β = 3 when k = 1 so that $\alpha =\frac{4}{3},\beta =-\frac{1}{3}$. Hence $A_{k+3}=\frac{1}{3}\left[2^{k+2}+{\left(-1\right)}^{k+1}\right]$ and $B_{k+3}=A_{k+4}=\frac{1}{3}\left[2^{k+3}+{\left(-1\right)}^{k+2}\right]$. It follows that

$$\begin{array}{ll}\hfill {\phi }_{n+3}& =\frac{1}{2^{k+2}}\left[\frac{1}{3}\left(2^{k+2}+{\left(-1\right)}^{k+1}\right){\phi }_0+\frac{1}{3}\left(2^{k+3}+{\left(-1\right)}^{k+2}\right){\phi }_1\right]\\ =\frac{1}{3}\left[\left\{1+\frac{{\left(-1\right)}^{k+1}}{2^{k+2}}\right\}{\phi }_0+\left\{2+\frac{{\left(-1\right)}^{k+2}}{2^{k+2}}\right\}{\phi }_1\right]\end{array}$$

This shows that $\underset{n\to \infty }{\text{lim}}{\phi }_n=\frac{1}{3}\left({\phi }_0+2{\phi }_1\right)$, proving the proposition.

Proposition 4.2. Define a binary operation ∗ on R by x ∗ y := Ax + (1 - A)y, 0 < A < 1, for any x, y ∈ R. If < φ _n > _L is a left-∗-Fibonacci sequence on (R, ∗), then$\underset{n\to \infty }{\text{lim}}{\phi }_n=\frac{1}{2-A}\left[\left(1-A\right){\phi }_0+{\phi }_1\right]$.

Proof. Given φ₀, φ₁ ∈ R, we consider < φ _n > _L . Since φ₂ = φ₁ ∗ φ₀ = Aφ₁ + (1 - A)φ₀ and φ₃ = φ₂∗φ₁ = (A²-A + 1)φ₁ + A(1-A)φ₀, ..., if we assume that φ _n := A _n φ₁ + B _n φ₀(n ≥ 2), then

$$\begin{array}{ll}\hfill {\phi }_{n+2}& ={\phi }_{n+1}*{\phi }_n\\ =\left(A_{n+1}{\phi }_1+B_{n+1}{\phi }_0\right)*\left(A_n{\phi }_1+B_n{\phi }_0\right)\\ =A\left(A_{n+1}{\phi }_1+B_{n+1}{\phi }_0\right)+\left(1-A\right)\left(A_n{\phi }_1+B_n{\phi }_0\right)\end{array}$$

It follows that

$$\begin{array}{ll}\hfill A_{n+2}& =A{A}_{n+1}+\left(1-A\right)A_n,\\ \hfill B_{n+2}& =A{B}_{n+1}+\left(1-A\right)B_n\end{array}$$

Hence we obtain the characteristic equation r² - Ar - (1 - A) = 0 with roots r = 1 or r = A - 1. Thus A_n+2= α(A - 1)ⁿ⁺²+ β for some α, β ∈ R. Since A = A₂ = α(A - 1)² + β and A² - A + 1 = A₃ = α(A - 1)³ + β, we obtain $\alpha =\frac{1}{A-2}$ and $\beta =\frac{1}{2-A}$ so that $A_{n+2}=\frac{{\left(A-1\right)}^{n+2}}{A-2}$. For B_n+2we obtain the same characteristic equation r² - Ar - (1 - A) = 0 with roots r = 1 or r = A - 1. Hence B_n+2= γ(A - 1)ⁿ⁺²+ δ for some γ, δ ∈ R. Since 1 - A = B₂ = γ(A - 1)² + β, A(1 - A) = B₃ = γ(A - 1)³ + δ, we obtain $\gamma -\frac{1}{2-A},\delta =\frac{1-A}{2-A}$ so that $B_{n+2}=\frac{1}{2-A}\left[{\left(A-1\right)}^{n+2}+\left(1-A\right)\right]$. Since 0 < A < 1, lim_n→∞(A - 1)ⁿ⁺²= 0. It follows that $\underset{n\to \infty }{\text{lim}}{\phi }_{n+2}=\underset{n\to \infty }{\text{lim}}\left(A_{n+2}{\phi }_1+B_{n+2}{\phi }_0\right)=\left(\underset{n\to \infty }{\text{lim}}{A}_{n+2}\right){\phi }_1+\left(\underset{n\to \infty }{\text{lim}}{B}_{n+2}\right){\phi }_0=\frac{1}{2-A}\left[{\phi }_1+\left(1-A\right){\phi }_0\right]$, proving the proposition. □

Note that if $A=\frac{1}{2}$ in Proposition 4.2, then $\underset{n\to \infty }{\text{lim}}{\phi }_n=\frac{1}{3}\left({\phi }_0+2{\phi }_1\right)$ as in Proposition 4.1. Note that (R, ∗) in Proposition 4.2 is neither a semigroup nor commutative and we may consider a right-∗-Fibonacci sequence < φ _n > _R on (R, ∗).

5 Fibonacci sequences in a group

In this section, we discuss ∗-Fibonacci sequence in groups.

Example 5.1. Suppose that X = S₄ is a symmetric group of order 4 and suppose that φ₀ = (13), φ₁ = (12). We wish to determine < φ _n >_L. Since φ₂ = φ₁φ₀ = (12)(13) = (123), φ₃ = φ₂φ₁ = (123)(13) = (12), φ₄ = φ₃φ₂ = (12)(123), ..., we obtain < φ _n > _L = < (13), (12), (123), (12), (13), (132), (23), (13), (123), (12), (13), ... >, i.e., it is periodic of period 6.

Proposition 5.2. Let (X, •, e) be a group and let φ₀, φ₁be elements of X such that φ₀•φ₁ = φ₁ • φ₀. If <φ _n > _L is a left-•-Fibonacci sequence in (X, •, e) generated by φ₀and φ₁, then${\phi }_{k+2}={\phi }_1^{F_{k+2}}{\phi }_0^{F_{k+1}}$. In particular, if φ₀ = φ₁, then${\phi }_{k+2}={\phi }_1^{F_{k+3}}$where F _k is the k^thFibonacci number.

Proof. Let φ₀, φ₁ be elements of X such that φ₀ • φ₁ = φ₁• φ₀. Since < φ _n > _L is a left-•-Fibonacci sequence in (X, •, e) generated by φ₀ and φ₁, we have ${\phi }_3={\phi }_1^2{\phi }_0,{\phi }_4={\phi }_1^3{\phi }_0^2,{\phi }_5={\phi }_1^5{\phi }_0^5={\phi }_1^{F_3}{\phi }_0^{F_4},{\phi }_6={\phi }_1^8{\phi }_0^5={\phi }_1^{F_6}{\phi }_0^{F_4}$ and ${\phi }_7={\phi }_1^{F_7}{\phi }_0^{F_6}$. If we assume that ${\phi }_k={\phi }_1^{F_k}{\phi }_0^{F_{k-1}}$ and ${\phi }_{k+1}={\phi }_1^{F_{k+1}}{\phi }_0^{F_k}$, then ${\phi }_{k+2}={\phi }_{k+1}{\phi }_k={\phi }_1^{k+1}{\phi }_0^{F_k}{\phi }_1^{F_k}{\phi }_0^{F_{k-1}}={\phi }_1^{F_{k+2}}{\phi }_0^{F_{k+1}}$. In particular, if φ₀ = φ₁, then ${\phi }_{k+2}={\phi }_1^{F_{k+2}}{\phi }_0^{F_{k+1}}={\phi }_1^{F_{k+3}}$. □

Proposition 5.3. Let (X, •, e) be a group and let φ₀, φ₁be elements of X such that φ₀•φ₁ = φ₁ • φ₀. If < φ _n > _L is a full left-•-Fibonacci sequence in (X, •, e) generated by φ₀and φ ₁ , then${\phi }_{-\left(2k\right)}={\phi }_0^{F_{k+1}}{\phi }_1^{-F_{2k}}$and${\phi }_{-\left(2k+1\right)}={\phi }_0^{-F_{2\left(k+1\right)}}{\phi }_1^{F_{2k+1}}$.

Proof. Since φ₁ = φ₀φ_-1, we have ${\phi }_{-1}={\phi }_0^{-1}{\phi }_1$. It follows from φ₀ = φ_-1φ_-2 that ${\phi }_{-2}={\phi }_0^2{\phi }_1^{-1}$. In this fashion, since φ₀•φ₁ = φ₁•φ₀, we obtain ${\phi }_{-3}={\phi }_0^{-F_4}{\phi }_1^{F_3}$ and ${\phi }_{-4}={\phi }_0^{F_5}{\phi }_1^{-F_4}$. By induction, assume that ${\phi }_{-\left(2k\right)}={\phi }_0^{F_{k+1}}{\phi }_1^{-F_{2k}}$ and ${\phi }_{-\left(2k+1\right)}={\phi }_0^{-F_{2\left(k+1\right)}}{\phi }_1^{F_{2k+1}}$. Then we obtain

$$\begin{array}{ll}\hfill {\phi }_{-\left(2k+2\right)}& ={\left[{\phi }_{-\left(2k+1\right)}\right]}^{-1}{\phi }_{-2k}\\ ={\phi }_0^{F_{2k+2}}{\phi }_1^{-F_{2k+1}}{\phi }_0^{F_{2k+1}}{\phi }_1^{-F_{2k}}\\ ={\phi }_0^{F_{2k+2}+F_{2k+1}}{\phi }_1^{-\left(F_{2k+1}+F_{2k}\right)}\\ ={\phi }_0^{F_{2k+3}}{\phi }_1^{-F_{2k+2}},\end{array}$$

and

$$\begin{array}{ll}\hfill {\phi }_{-\left(2k+3\right)}& ={\left[{\phi }_{-\left(2k+2\right)}\right]}^{-1}{\phi }_{-\left(2k+1\right)}\\ ={\left[{\phi }_0^{F_{2k+3}}{\phi }_1^{-F_{2k+2}}\right]}^{-1}{\phi }_0^{-F_{2k+2}}{\phi }_1^{F_{2k+1}}\\ ={\phi }_0^{F_{2k+2}+F_{2k+1}}{\phi }_1^{-\left(F_{2k+2}+F_{2k+3}\right)}\\ ={\phi }_0^{F_{2k+4}}{\phi }_1^{-F_{2k+3}},\end{array}$$

proving the proposition. □