Some problems related to the growth of z(n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$z(n)$\end{document}

Let (Fn)n≥0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$(F_{n})_{n\geq 0}$\end{document} be the Fibonacci sequence. The order of appearance z(n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$z(n)$\end{document} of a positive integer n is defined as z(n):=min{k≥1:n∣Fk}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$z(n):= \min \{k\geq 1: n\mid F_{k}\}$\end{document}. In 2013, Marques proved that lim infn→∞z(n)/n=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\liminf_{n\to \infty }z(n)/n=0$\end{document}. Let ϵ be a positive real number. In this paper, in particular, we generalized this Marques’ result by proving that almost all positive integers satisfy z(n)/n<ϵ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$z(n)/n<\epsilon $\end{document}.


Introduction
The Fibonacci sequence (F n ) n is one of the most famous sequences in mathematics. This sequence starts with 0 and 1 and each term afterwards is the sum of the two preceding ones. So, its first terms are 0, 1, 1, 2, 3, 5, 8, 13, and so on. A well-known non-recursive formula for the nth Fibonacci number is the Binet-formula where α := (1 + √ 5)/2 and β := (1 -√ 5)/2 (for more details see [1,4,7,20]). There are many authors interested in problems related to divisibility properties of Fibonacci numbers (see e.g. [3,8]). In this direction, we define, for any positive integer n, the order of apparition (or the rank of appearance) of n in the Fibonacci sequence, denoted by z(n), as the minimum element of the set {k ≥ 1 : n | F k }. This function is well defined by a result of Lucas [9, p. 300] (in 1878), and in fact a simple combinatorial argument yields z(n) ≤ n 2 for all positive integers n. We note that there is not a general closed formula for the z(n), and therefore Diophantine equations related to z(n) play an important role in its best comprehension (see [10,16,17,19]). A number of authors have considered, in varying degrees of generality, the problem of determining a special closed formula for z(n), when n is a number which is related to a sum or a product of terms of Fibonacci and Lucas sequences (see, for example, [5,6,11,18] and the references therein).
In 1975, Sallé [13] showed that z(n) ≤ 2n for all positive integers n and, moreover, he also found the explicit form of all integers n for which the equality is attained. In fact, it In particular, lim sup n→∞ z(n)/n = 2. However, apart from these cases this upper bound is very weak. For instance, z(2255) = 20 < 10 -2 · 2255. In fact, Marques [12] proved the existence of sharper upper bounds for the z-function which hold whenever n = 6 · 5 k , where k ≥ 0. These upper bounds become better as ω(n) increases (where, as usual, ω(n) is the number of distinct prime factors of n). By using these bounds, he deduced that However, the proof of the previous fact does not need Marques' upper bounds since, for n > 2, z(F n )/F n = n/F n tends to 0 as n → ∞ (since F n ≥ ( √ 2) n-2 ). Indeed, in order to derive this, he used the primorial number of the nth prime, denoted by p n #, which is defined as the product of the first n prime numbers. It is well known that the primorial numbers are rarer (in some sense) than Fibonacci numbers (see Remark 1 for the formal meaning).
In this paper, we are interested in two problems related to the growth of z(n). The first one is about the proportion of positive integers such that z(n)/n is sufficiently small. To be more precise, for a real number > 0, we define N = n ∈ Z >0 : z(n)/n < .
Note that N = Z >0 for all ≥ 2 (by Sallé's result). See also Fig. 1. By (2), we know that the set N is infinite for all > 0. However, by using only Fibonacci numbers in order to have z(F n )/F n < , we obtain a zero natural density set of positive integers belonging to N . Recall that if A is a set of positive integers, the natural density of A, denoted by δ(A), is the following limit (if it exists): x , Figure 2 The graphs of the functions . For example, clearly, δ(N ) = 1 for all ≥ 2. We recall that a property is said to be satisfied by almost all positive integers if the set of positive integers, for which this property holds, has natural density 1. Thus, our first result states, in particular, that almost all positive integers belong to N for all > 0. More precisely, Theorem 1 There exists a positive constant c such that, for all > 0, the estimate holds for all sufficiently large x.
As an immediate consequence, we have Corollary 1 For all > 0, one has that δ(N ) = 1.
The second problem which will be treated here is about a "Mertens-like" function related to z(n) (recall that the Mertens function M(m) := m n=1 μ(n), where μ(n) is the Möbius function). Let m be a positive integer, we define the z-summation function Z(m) := m n=1 z(n). This function can be also defined over the positive real numbers by Z( Since z(n) ≤ 2n, then we have the direct estimate for all x ≥ 1. In particular, Z(x) = O(x 2 ) (here, we use the Landau symbol O to say that f (x) =  O(g(x)) if there exists a positive constant M such that |f (x)| ≤ M|g(x)| for all sufficiently large x, i.e., the same meaning as f g). Theorem 1 implies that z(n)/n can be made arbitrarily small, with probability 1 (i.e., δ(N ) = 1), and so it suggests the existence of a substantially better upper bound for Z(x). In fact, the following theorem holds.

Theorem 2 We have that
The proof of these results combines some new (sharper upper bounds for z(n) due to Marques) and classical results (such as a result due to Sathé and Selberg) in number theory.
The graphs of the functions in Fig. 2 show that our upper bound (4) is much sharper than upper bound (3).

Auxiliary results
A consequence of Marques' theorems (in [12]) will be an essential ingredient in our proof. Therefore, we shall present his results as lemmas (in what follows ν 2 (n) denotes the 2-adic valuation of n, i.e., the largest exponent of 2 in the prime factorization of n).
where, as usual, ( a q ) denotes the Legendre symbol of a with respect to a prime q > 2.
(ii) If ν 2 (n) = 1, then if ω(n) = 2 and 5 | n; 2n if ω(n) = 2 and 5 n; if ω(n) = 2 and 5 | n; n if ω(n) = 2 and 5 n; The next two lemmas are powerful results in analytic number theory which are related to positive integers with fixed number of distinct prime factors.

Lemma 4 (Sathé-Selberg formula) For any positive constant A, we have
In the previous statement Γ (z) = ∞ 0 x z-1 e -x dx (for x > 0) is the well-known gamma function and, as usual, for two positive functions f (x) and g(x), we say that f ∼ g (i.e., f is asymptotically equivalent to g) if f (x)/g(x) tends to 1 as x → ∞.
Remark 1 In the text right after (2), we said that the primorial numbers are rarer than Fibonacci numbers. Indeed, this is true because, in light of the previous definition, we have F n ∼ α n / √ 5 while p n # ∼ n (1+o(1))n (where, in this case, o(1) means a function which tends to 0 as n → ∞).
As usual, from now on we utilize the notation [a, b] = {a, a + 1, . . . , b} for any integers a < b. Now we are ready to deal with the proof of our results.

The proofs 3.1 The proof of Theorem 1
We can deduce by Lemmas 1, 2, and 3 that n for all n > 1. Thus, to obtain the inequality z(n)/n < , it suffices that 7 · (2/3) ω(n) < . This inequality holds when ω(n) > log( /7)/ log(2/3). Therefore, for k := log( /7)/ log(2/3) , we have Thus, to finish the proof, it is enough to show that for all t. Now, the proof conveniently splits into two cases: Note that since G(z) converges uniformly and absolutely in any bounded set, we have max z∈ [0,2] {|G(z)|} ≤ C, for some positive constant C. Now, by Lemma 4 for A = 2, we get |G(z t )| ≤ C (for z t := (t -1)/ log log x < 2) and By the Stirling's formula (i.e., n! ≥ √ 2π n n+1/2 e -n ), the previous inequality can be written as However (by a straightforward calculation), for a given A > 1, the function f (y) := (A/y) y has its maximum value e A/e which is attained at y = A/e. Thus, if A = e log log x, then f (t -1) ≤ e log log x = log x and so Case 2 If t > 2 log log x. In this case, for ω(n) = t > 2 log log x, then ω(n) -log log x > log log x and therefore Since n≤x (ω(n) -log log x) 2 = O(x log log x) (by Lemma 5), we deduce that In conclusion, inequalities (6) and (7) imply (5), as desired.

Proof of Corollary 1
By definition of natural density and by Theorem 1, we have which implies the desired result.

The proof of Theorem 2
By using that z(n) ≤ 7 · (2/3) ω(n) n, we have that where h(x) = max{ω(t) : t ≤ x}. This inequality can be written as In order to obtain an upper bound for the first sum in the right hand side above, we apply Lemma 4 for A = 1 and z k := (k -1)/ log log x < 1. Thus Here, we used that G(z k ) = O(1) (as in the proof of Theorem 1). Therefore, For the second sum in the right hand side of (8), we have where we used that log log x + 1 > log log x. Since 2/3 < 1/ 3 √ e, then By combining (8), (9) and (10), we arrive at Z(x) x x (log x) 1/3 + x (log x) 1/3 x 2 (log x) 1/3 as desired.