On Integer Numbers with Locally Smallest Order of ... › ... › Volume2011 › 407643.pdf · In this paper, we will construct inﬁnitely many natural numbers satisfying the previous

Hindawi Publishing CorporationInternational Journal of Mathematics and Mathematical SciencesVolume 2011, Article ID 407643, 4 pagesdoi:10.1155/2011/407643

Research ArticleOn Integer Numbers with Locally SmallestOrder of Appearance in the Fibonacci Sequence

Diego Marques

Departament of Mathematics, University of Brasilia, Brasilia-DF 70910-900, Brazil

Correspondence should be addressed to Diego Marques, [email protected]

Received 13 December 2010; Revised 7 February 2011; Accepted 27 February 2011

Academic Editor: Ilya M. Spitkovsky

Copyright q 2011 Diego Marques. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

Let Fn be the nth Fibonacci number. The order of appearance z(n) of a natural number n is definedas the smallest natural number k such that n divides Fk. For instance, for all n = Fm ≥ 5, wehave z(n − 1) > z(n) < z(n + 1). In this paper, we will construct infinitely many natural numberssatisfying 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 andF1 = 1. A few terms of this sequence are

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, . . . . (1.1)

The Fibonacci numbers are well known for possessing wonderful and amazingproperties (consult [1] together with its very extensive annotated bibliography for additionalreferences and history). In 1963, the Fibonacci Association was created to provide enthusiastsan 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, andrelationships about Fibonacci numbers.

Let n be a positive integer number, the order (or rank) of appearance of n in the Fibonaccisequence, denoted by z(n), is defined as the smallest positive integer k, such that n | Fk (someauthors also call it order of apparition, or Fibonacci entry point). There are several results aboutz(n) in the literature. For instance, every positive integer divides some Fibonacci number, thatis, z(n) < ∞ for all n ≥ 1. The proof of this fact is an immediate consequence of the TheoremeFondamental 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 ,

2 International Journal of Mathematics and Mathematical Sciences

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 locallysmallest in this sense. On the other hand, there are integers n for which z(n) is locally smallestbut 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 belongto 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, forall m ≥ 5. In particular, it is not a Fibonacci number. Moreover, one has

z(Nm − 1) > z(Nm) < z(Nm + 1), (1.2)

for all sufficiently largem.

2. Proof of Theorem 1.1

We recall that the problem of the existence of infinitely many prime numbers in the Fibonaccisequence remains open; however, several results on the prime factors of a Fibonacci numberare known. For instance, a primitive divisor p of Fn is a prime factor of Fn that does not divide∏n−1

j=1Fj . In particular, z(p) = n. It is known that a primitive divisor p of Fn exists whenevern ≥ 13. The above statement is usually referred to the Primitive Divisor Theorem (see [3] forthe 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, ifNm divides Fj , then we

get the relation

FkFj = tFmk, (2.1)

where t is a positive integer number. Since mk ≥ 15, the Primitive Divisor Theorem impliesthat j ≥ mk. Therefore, z(Nm) ≥ mk yielding z(Nm) = mk. Now, ifNm is a Fibonacci number,say Ft, we get t = z(Nm) = mk which leads to an absurdity as Fk = 1 (keep in mind thatk ≥ 3). Therefore,Nm is not a Fibonacci number, for allm ≥ 5.

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

Let u be a positive integer number such that Fj = u(Nm + ε). If u ≥ Fk + 1, we have

Fj ≥(

1 +1Fk

)

Fmk + ε(Fk + 1) > Fmk, (2.2)

where in the last inequality above, we used the fact that Fmk > FmFk > ε(Fk + 1)Fk, form > k ≥ 3. Thus, j > mk as desired. For finishing the proof, it suffices to show that there existonly finitely many pairs (k, j) of positive integers, such that

Fj

Nm + ε= u ∈ {1, . . . , Fk}, (2.3)

International Journal of Mathematics and Mathematical Sciences 3

or equivalently,

uFmk − FkFj = −εuFk. (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, forall sufficiently large n (by the reordering of indexes if necessary). Now, by (2.4), we get

limn→∞

Fmnk

Fjn

=Fk

u. (2.5)

On the other hand, the well-known Binet’s formula

Fn =αn−(−1)nα−n

√5

, where α =1 +

√5

2, (2.6)

leads to

Fmnk

Fjn

=αmnk−jn − (−1)mnkα−mnk−jn

1 − (−1)jnα−2jn. (2.7)

Thus,

limn→∞

Fmnk

Fjn

= limn→∞

αmnk−jn . (2.8)

Combining (2.5) and (2.8), we get

limn→∞

αmnk−jn =Fk

u. (2.9)

Since mnk − jn is an integer and |α| > 1, we have that mnk − jn must be constant withrespect to n, say t, for all n sufficiently large. Therefore, (2.9) yields the relation αt = Fk/u ∈ �and so t = 0 (because αt is irrational for all nonzero rational number). But, this leads to(by (2.4))

εF2k = εuFk = FkFmnk − FkFmnk = 0, (2.10)

which is absurd. This completes the proof of the theorem.

4 International Journal of Mathematics and Mathematical Sciences

Acknowledgments

The author would like to express his gratitude to the anonymous referees for carefully exam-ining 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 CNPqfor the financial support.

References

[1] T. Koshy, Fibonacci and Lucas Numbers with Applications, Pure and Applied Mathematics (New York),Wiley-Interscience, New York, NY, USA, 2001.

[2] E. Lucas, “Theorie des fonctions numeriques simplement periodiques,” American Journal of Mathemat-ics, vol. 1, no. 4, pp. 289–321, 1878.

[3] Yu. Bilu, G. Hanrot, and P. M. Voutier, “Existence of primitive divisors of Lucas and Lehmer numbers,”Journal fur die Reine und Angewandte Mathematik, vol. 539, pp. 75–122, 2001.

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On Integer Numbers with Locally Smallest Order of ... › ... › Volume2011 › 407643.pdf · In this paper, we will construct inﬁnitely many natural numbers satisfying the previous

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