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

Abstract

Let 𝐹_𝑛 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 𝐹_𝑘. For instance, for all 𝑛=𝐹_𝑚≥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 (𝐹_𝑛)_𝑛≥0 be the Fibonacci sequence given by 𝐹_𝑛+2=𝐹_𝑛+1+𝐹_𝑛, for 𝑛≥0, where 𝐹₀=0 and 𝐹₁=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 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∣𝐹_𝑘 (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(𝐹_𝑛−1)>𝑧(𝐹_𝑛)<𝑧(𝐹_𝑛+1), for 𝑛≥5. In fact, if 𝑧(𝐹_𝑚+𝜖)=𝑗_𝜖 with 𝜖∈{±1}, then 𝐹_𝑚+𝜖 divides 𝐹_𝑗𝜖, for some 𝑗≥5 and thus 𝐹_𝑗𝜖=𝑢(𝐹_𝑚+𝜖) with 𝑢≥2. Therefore, the inequality 𝐹_𝑗𝜖≥2𝐹_𝑚+2𝜖>𝐹_𝑚 gives 𝑧(𝐹_𝑚+𝜖)=𝑗_𝜖>𝑚=𝑧(𝐹_𝑚). 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 𝑁_𝑚∶=𝐹_𝑚𝑘/𝐹_𝑘 has order of appearance 𝑚𝑘, for all 𝑚≥5. In particular, it is not a Fibonacci number. Moreover, one has 𝑧𝑁_𝑚−1>𝑧𝑁_𝑚<𝑧𝑁_𝑚+1, (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 𝐹_𝑛 is a prime factor of 𝐹_𝑛 that does not divide ∏^𝑛−1_𝑗=1𝐹_𝑗. In particular, 𝑧(𝑝)=𝑛. It is known that a primitive divisor 𝑝 of 𝐹_𝑛 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 𝑁_𝑚 divides 𝐹_𝑚𝑘, then 𝑧(𝑁_𝑚)≤𝑚𝑘. On the other hand, if 𝑁_𝑚 divides 𝐹_𝑗, then we get the relation 𝐹_𝑘𝐹_𝑗=𝑡𝐹_𝑚𝑘, (2.1) where 𝑡 is a positive integer number. Since 𝑚𝑘≥15, the Primitive Divisor Theorem implies that 𝑗≥𝑚𝑘. Therefore, 𝑧(𝑁_𝑚)≥𝑚𝑘 yielding 𝑧(𝑁_𝑚)=𝑚𝑘. Now, if 𝑁_𝑚 is a Fibonacci number, say 𝐹_𝑡, we get 𝑡=𝑧(𝑁_𝑚)=𝑚𝑘 which leads to an absurdity as 𝐹_𝑘=1 (keep in mind that 𝑘≥3). Therefore, 𝑁_𝑚 is not a Fibonacci number, for all 𝑚≥5.

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

Let 𝑢 be a positive integer number such that 𝐹_𝑗=𝑢(𝑁_𝑚+𝜖). If 𝑢≥𝐹_𝑘+1, we have 𝐹_𝑗≥11+𝐹_𝑘𝐹_𝑚𝑘+𝜖𝐹_𝑘+1>𝐹_𝑚𝑘, (2.2) where in the last inequality above, we used the fact that 𝐹_𝑚𝑘>𝐹_𝑚𝐹_𝑘>𝜖(𝐹_𝑘+1)𝐹_𝑘, 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 𝐹_𝑗𝑁_𝑚+𝜖=𝑢∈1,…,𝐹_𝑘, (2.3) or equivalently, 𝑢𝐹_𝑚𝑘−𝐹_𝑘𝐹_𝑗=−𝜖𝑢𝐹_𝑘. (2.4)

Towards a contradiction, suppose that (2.4) have infinitely many solutions (𝑢_𝑛,𝑚_𝑛,𝑗_𝑛) with 𝑢_𝑛∈{1,…,𝐹_𝑘} and 𝑛≥1. Hence, (𝑚_𝑛)_𝑛 and (𝑗_𝑛)_𝑛 are unbounded sequences. Since (𝑢_𝑛)_𝑛 is bounded, we can assume, without loss of generality, that 𝑢_𝑛 is a constant, say 𝑢, for all sufficiently large 𝑛 (by the reordering of indexes if necessary). Now, by (2.4), we get lim_𝑛→∞𝐹_𝑚𝑛𝑘𝐹_𝑗𝑛=𝐹_𝑘𝑢. (2.5) On the other hand, the well-known Binet's formula 𝐹_𝑛=𝛼^𝑛−(−1)^𝑛𝛼^−𝑛√5√,where𝛼=1+52, (2.6) leads to 𝐹_𝑚𝑛𝑘𝐹_𝑗𝑛=𝛼^{𝑚𝑛𝑘−𝑗𝑛}−(−1)^𝑚𝑛𝑘𝛼^{−𝑚𝑛𝑘−𝑗𝑛}1−(−1)^𝑗𝑛𝛼^−2𝑗𝑛. (2.7) Thus, lim_𝑛→∞𝐹_𝑚𝑛𝑘𝐹_𝑗𝑛=lim_𝑛→∞𝛼^{𝑚𝑛𝑘−𝑗𝑛}. (2.8)

Combining (2.5) and (2.8), we get lim_𝑛→∞𝛼^{𝑚𝑛𝑘−𝑗𝑛}=𝐹_𝑘𝑢. (2.9)

Since 𝑚_𝑛𝑘−𝑗_𝑛 is an integer and |𝛼|>1, we have that 𝑚_𝑛𝑘−𝑗_𝑛 must be constant with respect to 𝑛, say 𝑡, for all 𝑛 sufficiently large. Therefore, (2.9) yields the relation 𝛼^𝑡=𝐹_𝑘/𝑢∈ℚ and so 𝑡=0 (because 𝛼^𝑡 is irrational for all nonzero rational number). But, this leads to (by (2.4)) 𝜖𝐹²_𝑘=𝜖𝑢𝐹_𝑘=𝐹_𝑘𝐹_𝑚𝑛𝑘−𝐹_𝑘𝐹_𝑚𝑛𝑘=0, (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.