Open Access

Fibonacci sequences in groupoids

Advances in Difference Equations20122012:19

https://doi.org/10.1186/1687-1847-2012-19

Received: 31 January 2012

Accepted: 27 February 2012

Published: 27 February 2012

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.

Keywords

fibonacci sequencesgroupoidflexiblewrapped around

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 φn+2= φn+1 φ n for n ≥ 0, and a right--Fibonacci sequence if φn+2 = φ n φn+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(l n (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 φk+1= φ k φk-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(r n (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 (φn-1 φ n ) = (φ n φn-1) φ n for any n ≥ 1.

Proof. If < φ n > L = < φ n > R , then φ0φ1 = φ2 = φ1φ0 and hence (φ1φ0)φ1 = φ2φ1 = φ3 = φ1φ2 = φ1(φ0φ1). Similarly (φ2φ1)φ2 = φ3φ2 = φ4 = φ2φ3 = φ2 (φ1 φ2). By induction on n, we obtain φ n (φn-1 φ n ) = (φ n φn-1) φ n for any n ≥ 1.

If we assume that φ n (φn-1 φ n ) = (φ n φn-1) φ n for any n ≥ 1, then φ n φn+1= φ n (φn-1 φ n ) = (φ n φn-1) φ n = φn+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 xy := A + B(x + y) or xy := 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
A B + B ( B - 1 ) x = A C + C ( C - 1 ) 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 ( B - 1 ) x = C ( C - 1 ) 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 = 1 ± 1 + 4 B ( B - 1 ) 2 = B 1 - B ,

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
φ 2 * φ 1 = ( φ 1 * φ 0 ) * φ 1 = φ 1 * ( φ 0 * φ 1 ) = φ 1 * φ 2 = φ 3
By induction on n, we obtain
φ n * φ n - 1 = ( φ n - 1 * φ n - 2 ) * φ n - 1 = φ n - 1 * ( φ n - 2 * φ n - 1 ) = φ n - 1 * φ n

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 φn+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:
( x * y ) * x = x * ( y * x ) = 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
y = ( x * y ) * x = A + B ( x * y + x ) = A + B ( A + B ( x + y ) + x ) = A ( B + 1 ) + B ( B + 1 ) x + B 2 y
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
y = ( x * y ) * x = [ B x + ( 1 - B ) y ] * x = B [ B x + ( 1 - B ) y ] + ( 1 - B ) x = ( B 2 - B + 1 ) x + B ( 1 - B ) y

It follows that B2 - B + 1 = 0, B(1 - B) = 1 which leads to B = 1 ± 3 i 2 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, (xy) z = z (yx). If (X, ) and (X, •) are both commutative groupoids, then (xy) z = z (xy) = z (yx) 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, xy := x-y for all x, y X. Then (xy) z = (x - y) z = (x - y)2z2 = z (yx) 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., xy := (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, •)
( x y ) x = [ ( x y ) * x ] * [ x * ( x y ) ] = [ { ( x * y ) ( y * x ) } * x ] [ x * { ( x * y ) ( y * x ) } ] = [ x * { ( y * x ) ( x * y ) } ] [ { ( y * x ) ( x * y ) } * x ] = [ x * ( y x ) ] [ ( y x ) * x ] = x ( y x ) ,

proving the proposition. □

Example 3.3. Note that in the situation of Example 3.1, we have xy = (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 xy = 0 for all x, y X.

Example 3.4. In Example 3.1, if we define (X, ) := (X, •)□(X, ), i.e., xy := (xy) (yx) for all x, y X, then xy = (x - y)4 and hence (xyx = ((x - y)4 - x)4 = (x - (y - x)4)4 = x(yx). Hence (X, ) is a flexible groupoid. Note that (X, ) is not a semigroup, since 0(0z) = z16z4 = (00)z. Obviously, xy = yx 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 xy = (x - y)4, given φ0, φ1 X, we have φ2 = φ0 φ1 = φ1φ0 = (φ1 - φ0)4, and φ3 = φ2φ1 = (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 : = 1 2 ( x + y ) for any x, y R. If < φ n > is a -Fibonacci sequence on (R, ), then lim n φ n = 1 3 ( φ 0 + 2 φ 1 ) .

Proof. Since x * y = 1 2 ( x + y ) = y * x for any x, y R, < φ n > L = < φ n > R for any φ0, φ1 R.

It can be seen that φ 2 = 1 2 ( φ 0 + φ 1 ) , φ 3 = 1 2 2 ( φ 0 + 3 φ 1 ) . We let φ 3 = 1 2 2 ( A 3 φ 0 + B 3 φ 1 ) . Since
φ 4 = φ 2 * φ 3 = 1 2 φ 0 + φ 1 2 + 1 4 ( φ 0 + 3 φ 1 ) = 1 2 3 [ 3 φ 0 + 5 φ 1 ] ,
we let it by φ 4 = 1 2 3 [ A 4 φ 0 + B 4 φ 1 ] . In this fashion, if we let φ n + 2 : = 1 2 n + 1 [ A n + 2 φ 0 + B n + 2 φ 1 ] , we have
φ n + 2 = 1 2 [ φ n + φ n + 1 ] = 1 2 A n φ 0 + B n φ 1 2 n - 1 + A n + 1 φ 0 + B n + 1 φ 1 2 n = 1 2 n + 1 [ ( 2 A n + A n + 1 ) φ 0 + ( 2 B n + B n + 1 ) φ 1 ]
It follows that
A n + 2 = 2 A n + A n + 1 , A 3 = 1 , A 4 = 3 B n + 2 = 2 B n + B n + 1 , B 3 = 3 , B 4 = 5
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 = Ak+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., Ak+3= α 2 k + β(-1) k , α + β = 1 when k = 0, and 2α - β = 3 when k = 1 so that α = 4 3 , β = - 1 3 . Hence A k + 3 = 1 3 [ 2 k + 2 + ( - 1 ) k + 1 ] and B k + 3 = A k + 4 = 1 3 [ 2 k + 3 + ( - 1 ) k + 2 ] . It follows that
φ n + 3 = 1 2 k + 2 1 3 ( 2 k + 2 + ( - 1 ) k + 1 ) φ 0 + 1 3 ( 2 k + 3 + ( - 1 ) k + 2 ) φ 1 = 1 3 1 + ( - 1 ) k + 1 2 k + 2 φ 0 + 2 + ( - 1 ) k + 2 2 k + 2 φ 1

This shows that lim n φ n = 1 3 ( φ 0 + 2 φ 1 ) , 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 lim n φ n = 1 2 - A [ ( 1 - A ) φ 0 + φ 1 ] .

Proof. Given φ0, φ1 R, we consider < φ n > L . Since φ2 = φ1 φ0 = 1 + (1 - A)φ0 and φ3 = φ2φ1 = (A2-A + 1)φ1 + A(1-A)φ0, ..., if we assume that φ n := A n φ1 + B n φ0(n ≥ 2), then
φ n + 2 = φ n + 1 * φ n = ( A n + 1 φ 1 + B n + 1 φ 0 ) * ( A n φ 1 + B n φ 0 ) = A ( A n + 1 φ 1 + B n + 1 φ 0 ) + ( 1 - A ) ( A n φ 1 + B n φ 0 )
It follows that
A n + 2 = A A n + 1 + ( 1 - A ) A n , B n + 2 = A B n + 1 + ( 1 - A ) B n

Hence we obtain the characteristic equation r2 - Ar - (1 - A) = 0 with roots r = 1 or r = A - 1. Thus An+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 α = 1 A - 2 and β = 1 2 - A so that A n + 2 = ( A - 1 ) n + 2 A - 2 . For Bn+2we obtain the same characteristic equation r2 - Ar - (1 - A) = 0 with roots r = 1 or r = A - 1. Hence Bn+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 γ - 1 2 - A , δ = 1 - A 2 - A so that B n + 2 = 1 2 - A [ ( A - 1 ) n + 2 + ( 1 - A ) ] . Since 0 < A < 1, limn→∞(A - 1)n+2= 0. It follows that lim n φ n + 2 = lim n ( A n + 2 φ 1 + B n + 2 φ 0 ) = ( lim n A n + 2 ) φ 1 + ( lim n B n + 2 ) φ 0 = 1 2 - A [ φ 1 + ( 1 - A ) φ 0 ] , proving the proposition. □

Note that if A = 1 2 in Proposition 4.2, then lim n φ n = 1 3 ( φ 0 + 2 φ 1 ) 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 φ0φ1 = φ1φ0. If <φ n > L is a left-•-Fibonacci sequence in (X, •, e) generated by φ0and φ1, then φ k + 2 = φ 1 F k + 2 φ 0 F k + 1 . In particular, if φ0 = φ1, then φ k + 2 = φ 1 F k + 3 where F k is the kthFibonacci number.

Proof. Let φ0, φ1 be elements of X such that φ0φ1 = φ1φ0. Since < φ n > L is a left-•-Fibonacci sequence in (X, •, e) generated by φ0 and φ1, we have φ 3 = φ 1 2 φ 0 , φ 4 = φ 1 3 φ 0 2 , φ 5 = φ 1 5 φ 0 5 = φ 1 F 3 φ 0 F 4 , φ 6 = φ 1 8 φ 0 5 = φ 1 F 6 φ 0 F 4 and φ 7 = φ 1 F 7 φ 0 F 6 . If we assume that φ k = φ 1 F k φ 0 F k - 1 and φ k + 1 = φ 1 F k + 1 φ 0 F k , then φ k + 2 = φ k + 1 φ k = φ 1 k + 1 φ 0 F k φ 1 F k φ 0 F k - 1 = φ 1 F k + 2 φ 0 F k + 1 . In particular, if φ0 = φ1, then φ k + 2 = φ 1 F k + 2 φ 0 F k + 1 = φ 1 F k + 3 . □

Proposition 5.3. Let (X, •, e) be a group and let φ0, φ1be elements of X such that φ0φ1 = φ1φ0. If < φ n > L is a full left-•-Fibonacci sequence in (X, •, e) generated by φ0and φ 1 , then φ - ( 2 k ) = φ 0 F k + 1 φ 1 - F 2 k and φ - ( 2 k + 1 ) = φ 0 - F 2 ( k + 1 ) φ 1 F 2 k + 1 .

Proof. Since φ1 = φ0φ-1, we have φ - 1 = φ 0 - 1 φ 1 . It follows from φ0 = φ-1φ-2 that φ - 2 = φ 0 2 φ 1 - 1 . In this fashion, since φ0φ1 = φ1φ0, we obtain φ - 3 = φ 0 - F 4 φ 1 F 3 and φ - 4 = φ 0 F 5 φ 1 - F 4 . By induction, assume that φ - ( 2 k ) = φ 0 F k + 1 φ 1 - F 2 k and φ - ( 2 k + 1 ) = φ 0 - F 2 ( k + 1 ) φ 1 F 2 k + 1 . Then we obtain
φ - ( 2 k + 2 ) = [ φ - ( 2 k + 1 ) ] - 1 φ - 2 k = φ 0 F 2 k + 2 φ 1 - F 2 k + 1 φ 0 F 2 k + 1 φ 1 - F 2 k = φ 0 F 2 k + 2 + F 2 k + 1 φ 1 - ( F 2 k + 1 + F 2 k ) = φ 0 F 2 k + 3 φ 1 - F 2 k + 2 ,
and
φ - ( 2 k + 3 ) = [ φ - ( 2 k + 2 ) ] - 1 φ - ( 2 k + 1 ) = φ 0 F 2 k + 3 φ 1 - F 2 k + 2 - 1 φ 0 - F 2 k + 2 φ 1 F 2 k + 1 = φ 0 F 2 k + 2 + F 2 k + 1 φ 1 - ( F 2 k + 2 + F 2 k + 3 ) = φ 0 F 2 k + 4 φ 1 - F 2 k + 3 ,

proving the proposition. □

Declarations

Authors’ Affiliations

(1)
Department of Applied Mathematics, Hanyang University
(2)
Department of Mathematics, Research Institute for Natural Sciences, Hanyang University
(3)
Department of Mathematics, University of Alabama

References

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Copyright

© Han et al.; licensee Springer. 2012

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