A CLOSED FORM FOR A THIRD-ORDER FIBONACCI-LIKE SEQUENCE.

Transcription

1 A CLOSED FORM FOR A THIRD-ORDER FIBONACCI-LIKE SEQUENCE. Cristina M. Marques Universidade Federal de Minas Gerais, Caixa Postal 702. Departamento de Matemática, Belo Horizonte , MG, Brazil marques@mat.ufmg.br B.J.O. Franco Universidade Federal de Minas Gerais, Caixa Postal 702. Departamento de Matemática, Belo Horizonte , MG, Brazil benbj@terra.com.br 1 ABSTRACT A closed form for the third-order Fibonacci-like sequence given by S n = S n 1 + 2S n 2 S n 3 with S 0 = 0, S 1 = 1, S 2 = 2 e S 3 = 4, is proposed. 2 INTRODUCTION In the last decades, second-order Fibonacci sequences have been intensively used in the study of quasiperiodic systems. One of the quasiperiodic lattices is the second-order Fibonacci chain (one-dimensional Fibonacci crystal) [3]. A third-order Fibonacci-like chain was considered by Terauchy et al [6] on the study of the diffraction pattern of this chain grown by molecular beam epitaxy (MBE). 1

2 The second-order Fibonacci sequence is defined by (see [7]): F n = F n 1 + F n 2 (1) where F 0 = 0 and F 1 = 1. (2) Numerical valores for F n are 0, 1, 1, 2, 3, 5, 8, 13, 21,... and a closed form for F n is given by [( F n = ) n ( 5 1 ) n ] (The Binet theorem) (3) In this short note we deal with a closed form for the sequence given by (see [1, 2]): S n = S n 1 + 2S n 2 S n 3 (4) where S 0 = 0, S 1 = 1, S 2 = 2 and S 3 = 4. (5) The corresponding transformation matrix is T = (6) and the secular equation is given by x 3 x 2 2x + 1 = 0. (7) Its positive solutions are λ 1 and 1/λ 2, where λ 1 and λ 2 are the diagonals of a heptagonal polygon with unitary sides. Approximate values for λ 1 and λ 2 are and 2.247, respectively. (We have λ 1 = 2 cos φ and λ 2 = 2 cos 2 φ + cos 2φ, where φ = π 7.) 3 Fibonacci-like sequences of third-order A Fibonacci-like sequence of third-order is defined by S n = ps n 1 + qs n 2 + rs n 3 (8) 2

3 where p, q and r are non-zero integers. If p = q = r = 1, we have a third-order Fibonacci sequence (the so-called Tribonacci sequence). Here we consider the Fibonacci-like sequences of third-order, that is, p = 1, q = 2 and r = 1 in (8): S n = S n 1 + 2S n 2 S n 3. (9) We define S = { Fibonacci-like sequences of third-order } and denote v, w S by v = (v 0, v 1, v 2, v 3,...) and w = (w 1, w 2, w 3,...). In S, we define the sum of v and w as v + w = (v 0, v 1, v 2, v 3,...) + (w 1, w 2, w 3,...) = (v 1 + w 1, v 2 + w 2, v 3 + w 3,...) and, for λ R, the scalar multiplication as λv = λ(v 0, v 1, v 2, v 3,...) = (λv 1, λv 2, λv 3,...). With this operations, (S, +,.) is a real vector space. A non-zero geometric progression v = (x, x 2, x 3, x 4,...) (10) is a third-order sequence if x 4 = x 3 + 2x 2 x. (11) that is, if x is a root of the irredutible polynomial over Q: p(x) = x 3 x 2 2x + 1 = 0. (12) If α is a root of p(x), in the extension field Q(α), p(x) can be factored as p(x) = (x α)[x 2 (α 1)x + (α 2 α 2)] (13) The roots of the polynomial x 2 (α 1)x + (α 2 α 2) = 0 (14) 3

4 are given by β = α 2 α 1, (15) γ = 2 α 2, (16) where α = 1/λ 2 and β = λ 1. Hence, a geometric progression v = (x, x 2, x 3, x 4,...) is a third-order sequence of S if x = α or x = β or x = γ. Note that the roots β and γ of p(x) can be expressed in terms of α because the Galois group Gal(Q(α) Q ) of p(x) is cyclic of order 3.[4] We will prove now that the three geometric progressions v 1 = (1, α, α 2, α 3,...) v 2 = (1, β, β 2, β 3,...) v 3 = (1, γ, γ 2, γ 3,...) (17) form a basis for S and thus S has dimension 3. To prove that the set B = {v 1, v 2, v 3 } is a basis, let us consider av 1 + bv 2 + cv 3 = 0. (18) for real numbers a, b and c. It is clear that (a, b, c) is a solution of the hommogeneous system X + Y + Z = 0 αx + βy + γz = 0. α 2 X + β 2 Y + γ 2 Z = 0 If there exists a nontrivial solution, the matrix M = α β γ α 2 β 2 γ 2 would be singular, that is det(m) = 0. But det(m) = (γ α)(γ β)(β α) is not equal zero. This shows that a = b = c = 0 is the unique solution of (18). To prove that v = (S 0, S 1, S 2, S 3,... ) S is a linear combination of elements of B, we have to prove that the linear system that results from the equation v = Xv 1 + Y v 2 + Zv 3 4

5 has a solution. But this linear system is given by X + Y + Z = S 0 αx + βy + γz = S 1 α 2 X + β 2 Y + γ 2 Z = S 2 and has coefficient matrix given by M = α β γ. α 2 β 2 γ 2 Since M in non-singular (that is, det(m) 0) the unique solution of the sistem is given by Cramer s rule: S det S 1 β γ X = S 2 β 2 γ 2 det(m), Y = Z = det det 1 S 0 1 α S 1 γ α 2 S 2 γ 2 det(m) 1 1 S 0 α β S 1 α 2 β 2 S 2 det(m). So, if then we have S 0 = 0, S 1 = 1, S 2 = 2 e S 3 = 4 (19) S n = (3α2 5) 7 α n + ( 3α2 + 3α + 4) (2 α 2 ) n + 7 ( 3α + 1) (α 2 α 1) n. 7 This form is different of others recent results, [5] because it depends of only one root of equation. 5

6 References [1] B.J.O.Franco, Physics Letters A 178(1993) [2] B.J.O.Franco et al, Phys. sta. sol.(b) 182(1994)K57-K62. [3] Marjorie Senechal, Quasicrystals and Geometry, Cambridge University Press, 1995, page [4] Stewart, Ian, Galois Theory, Chapman and Hall, [5] Sloane N. J.A, The on-line Encyclopedia of Intergers Sequences (A006053) [6] Terauchy et al, J. Phys. Soc. Japan 59(1990)405. [7] N.N.Voroboyov, Fibonacci Numbers, Boston, D.C.Heath and Co.,