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

Authors: Diego Marques
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

LetFn 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.


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## Body

## 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)0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,….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.Letn 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 Fjϵ, for some j≥5 and thus Fjϵ=u(Fm+ϵ) with u≥2. Therefore, the inequality Fjϵ≥2Fm+2ϵ>Fm 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 integerk≥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)z(Nm-1)>z(Nm)<z(Nm+1),
for all sufficiently large m.


## 2. Proof of Theorem1.1

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, aprimitive divisorp of Fn is a prime factor of Fn that does not divide ∏j=1n-1Fj. 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.SinceNm divides Fmk, then z(Nm)≤mk. On the other hand, if Nm divides Fj, then we get the relation
(2.1)FkFj=tFmk,
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 thatz(Nm+ϵ)>mk=z(Nm), or equivalently, if Nm±1 divides Fj, then j>mk, for all sufficiently large m, where ϵ∈{±1}.Letu be a positive integer number such that Fj=u(Nm+ϵ). If u≥Fk+1, we have
(2.2)Fj≥(1+1Fk)Fmk+ϵ(Fk+1)>Fmk,
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)FjNm+ϵ=u∈{1,…,Fk},
or equivalently,
(2.4)uFmk-FkFj=-ϵuFk.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)lim⁡n→∞FmnkFjn=Fku.
On the other hand, the well-known Binet's formula
(2.6)Fn=αn-(-1)nα-n5,whereα=1+52,
leads to
(2.7)FmnkFjn=αmnk-jn-(-1)mnkα-mnk-jn1-(-1)jnα-2jn.
Thus,
(2.8)lim⁡n→∞FmnkFjn=lim⁡n→∞αmnk-jn.Combining (2.5) and (2.8), we get
(2.9)lim⁡n→∞αmnk-jn=Fku.Sincemnk-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)ϵFk2=ϵuFk=FkFmnk-FkFmnk=0,
which is absurd. This completes the proof of the theorem.



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*Source: 407643-2011-04-20.xml*