International Journal of Mathematics and Mathematical Sciences
Volume 2011, Issue 1 407643
Research Article
Open Access

On Integer Numbers with Locally Smallest Order of Appearance in the Fibonacci Sequence

Diego Marques, 

Corresponding Author

Diego Marques

Departament of Mathematics, University of Brasilia, Brasilia-DF 70910-900, Brazil unb.br

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First published: 20 April 2011
https://doi.org/10.1155/2011/407643
Citations: 5
Academic Editor: Ilya M. Spitkovsky

Abstract

Let Fn be the nth Fibonacci number. The order of appearance z(n) of a natural number n is defined as the smallest natural number k such that n divides Fk. For instance, for all n = Fm ≥ 5, we have z(n − 1) > z(n) < z(n + 1). In this paper, we will construct infinitely many natural numbers satisfying the previous inequalities and which do not belong to the Fibonacci sequence.

1. Introduction

Let (Fn) n≥0 be the Fibonacci sequence given by Fn+2 = Fn+1 + Fn, for n ≥ 0, where F0 = 0 and F1 = 1. A few terms of this sequence are
(1.1)

The Fibonacci numbers are well known for possessing wonderful and amazing properties (consult [1] together with its very extensive annotated bibliography for additional references and history). In 1963, the Fibonacci Association was created to provide enthusiasts an opportunity to share ideas about these intriguing numbers and their applications. Also, in the issues of The Fibonacci Quarterly, we can find many new facts, applications, and relationships about Fibonacci numbers.

Let n be a positive integer number, the order (or rank) of appearance of n in the Fibonacci sequence, denoted by z(n), is defined as the smallest positive integer k, such that n∣Fk (some authors also call it order of apparition, or Fibonacci entry point). There are several results about z(n) in the literature. For instance, every positive integer divides some Fibonacci number, that is, z(n) < ∞ for all n ≥ 1. The proof of this fact is an immediate consequence of the Théorème Fondamental of Section XXVI in [2, page 300]. Also, it is a simple matter to prove that z(Fn − 1) > z(Fn) < z(Fn + 1), for n ≥ 5. In fact, if z(Fm + ϵ) = jϵ with ϵ ∈ {±1}, then Fm + ϵ divides , for some j ≥ 5 and thus with u ≥ 2. Therefore, the inequality gives z(Fm + ϵ) = jϵ > m = z(Fm). So the order of appearance of a Fibonacci number is locally smallest in this sense. On the other hand, there are integers n for which z(n) is locally smallest but which are not Fibonacci numbers, for example, n = 11,17,24,26,29,36,38,41,44,48, …. So, a natural question arises: are there infinitely many natural numbers n that do not belong to the Fibonacci sequence and such that z(n − 1) > z(n) < z(n + 1)?

In this note, we give an affirmative answer to this question by proving the following.

Theorem 1.1. Given an integer k ≥ 3, the number Nm : = Fmk/Fk has order of appearance mk, for all m ≥ 5. In particular, it is not a Fibonacci number. Moreover, one has

(1.2)
for all sufficiently large m.

2. Proof of Theorem 1.1

We recall that the problem of the existence of infinitely many prime numbers in the Fibonacci sequence remains open; however, several results on the prime factors of a Fibonacci number are known. For instance, a primitive divisor p of Fn is a prime factor of Fn that does not divide . In particular, z(p) = n. It is known that a primitive divisor p of Fn exists whenever n ≥ 13. The above statement is usually referred to the Primitive Divisor Theorem (see [3] for the most general version).

Now, we are ready to deal with the proof of the theorem.

Since Nm divides Fmk, then z(Nm) ≤ mk. On the other hand, if Nm divides Fj, then we get the relation
(2.1)
where t is a positive integer number. Since mk ≥ 15, the Primitive Divisor Theorem implies that j ≥ mk. Therefore, z(Nm) ≥ mk yielding z(Nm) = mk. Now, if Nm is a Fibonacci number, say Ft, we get t = z(Nm) = mk which leads to an absurdity as Fk = 1 (keep in mind that k ≥ 3). Therefore, Nm is not a Fibonacci number, for all m ≥ 5.

Now, it suffices to prove that z(Nm + ϵ) > mk = z(Nm), or equivalently, if Nm ± 1 divides Fj, then j > mk, for all sufficiently large m, where ϵ ∈ {±1}.

Let u be a positive integer number such that Fj = u(Nm + ϵ). If u ≥ Fk + 1, we have
(2.2)
where in the last inequality above, we used the fact that Fmk > FmFk > ϵ(Fk + 1)Fk, for m > k ≥ 3. Thus, j > mk as desired. For finishing the proof, it suffices to show that there exist only finitely many pairs (k, j) of positive integers, such that
(2.3)
or equivalently,
(2.4)
Towards a contradiction, suppose that (2.4) have infinitely many solutions (un, mn, jn) with un ∈ {1, …, Fk} and n ≥ 1. Hence, (mn) n and (jn) n are unbounded sequences. Since (un) n is bounded, we can assume, without loss of generality, that un is a constant, say u, for all sufficiently large n (by the reordering of indexes if necessary). Now, by (2.4), we get
(2.5)
On the other hand, the well-known Binet′s formula
(2.6)
leads to
(2.7)
Thus,
(2.8)
Combining (2.5) and (2.8), we get
(2.9)
Since mnk − jn is an integer and |α | > 1, we have that mnk − jn must be constant with respect to n, say t, for all n sufficiently large. Therefore, (2.9) yields the relation αt = Fk/u ∈ ℚ and so t = 0 (because αt is irrational for all nonzero rational number). But, this leads to (by (2.4))
(2.10)
which is absurd. This completes the proof of the theorem.

Acknowledgments

The author would like to express his gratitude to the anonymous referees for carefully examining this paper and providing a number of important comments, critics, and suggestions. One of their suggestions leads us to Theorem 1.1. The author also thanks FEMAT and CNPq for the financial support.

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