On the Parity of the Order of Appearance in the Fibonacci Sequence

: Let ( F n ) n ≥ 0 be the Fibonacci sequence. The order of appearance function (in the Fibonacci sequence) z : Z ≥ 1 → Z ≥ 1 is deﬁned as z ( n ) : = min { k ≥ 1 : F k ≡ 0 ( mod n ) } . In this paper, among other things, we prove that z ( n ) is an even number for almost all positive integers n (i.e., the set of such n has natural density equal to 1).


Remark 1.
We remark that the arithmetic progressions of Theorem 1 are the largest ones (in the sense of natural density) belonging to E z . In fact, for other arithmetic progressions of density of at least 1/4, we have infinitely many terms which are outside of E z . Some of these terms can be made explicit. For example: We point out that Theorem 1 implies, in particular, that the natural density of E z is at least 1/2 (since (4n − 1) n and (4n) n are disjoint arithmetic progressions with natural density equal to 1/4). In fact, it is possible to obtain a better lower bound for δ(E z ), by noting that E z also contains (3n) n which has density 1/3. Since the last sequence is not disjoint from the previous ones, we may use the inclusion-exclusion principle to deduce that the natural density of E z is at least Table 3 suggests that for almost all positive integers n, one has that z(n) is even (i.e., one expects that δ(E z ) = 1). Note that a sieve method of searching by arithmetic progressions lying in E z does not seem to reach the value 1.

Remark 2.
Note that all previously provided infinite sequences belonging to Z ≥1 \E z have exponential growth (e.g., (5 n ) n ). In fact, the previous theorem ensures, in particular, that no infinite arithmetic progression can be within Z ≥1 \E z (since its natural density is 0).
The proof of the theorems combines Diophantine properties of z(n) with analytical tools concerning primes in arithmetic progressions.

Auxiliary Results
In this section, we shall present some results which will be very important tools in the proof. The first ingredient is related to the value of z(p k ), for a prime number p and k ≥ 1: Lemma 1 (Theorem 2.4 of [12]). We have that z(2 k ) = 3 · 2 k−1 , for all k ≥ 2, and z(3 k ) = 4 · 3 k−1 , for all k ≥ 1. In general, it holds that The next result provides some divisibility properties of the Pisano period for prime numbers.
The next tool is a kind of "formula" for z(n) depending on z(p a ) for all primes p dividing n. The proof of this fact can be found in [14].
In order to prove Theorem 2, we need an analytic tool related to the profusion of integers having factorization allowing only some classes of primes. The following notation will be used throughout this work: let P be the set of prime numbers and for an integer q ≥ 2, set P(a, q) as the set of all prime numbers of the form a + kq, for some integer k ≥ 0 (Dirichlet's theorem on arithmetic progressions) that ensures that P(a, q) is an infinite set whenever gcd(a, q) = 1). Let N a,q = {n ≥ 1 : p | n ⇒ p ∈ P(a, q)} be the set of all positive integers whose prime factors lie exclusively in P(a, q). Additionally, denote β := 1/φ(q) and which has an analytic continuation to a neighborhood of s = 1. Here, as usual, ζ(s) denotes the Riemann zeta function.
Our last auxiliary lemma is a particular case of a theorem due to Chang and Martin [15]. We state this more precisely in the next lemma. [15]). For any integer q ≥ 3, there exists a positive absolute constant C such that uniformly for q ≤ (log x) 1/3 , we have
Now, we are ready to deal with the proof of the theorems.

The Proof of Theorem 1
The Case z(3n) Proof. First, in order to use Lemma 4, one may write n = 3 k m, where k ≥ 0 and gcd(m, 3) = 1. So, Lemma 4 yields that where we used Lemma 1. Hence we deduce that 4 divides z(3n).
The Case z(4n + 3) This case is more delicate, since 4n + 3 is a prime number for infinitely many values of n. To overcome this difficulty, we deal first with these prime values. Claim 1. If p = 4k + 3 is a prime number, then z(p) is even. The proof splits into two cases concerning the residues of p modulo 5. In fact, if p ≡ ±1 (mod 5), then Lemma 2 (i) implies that π(p) divides p − 1. Striving for a contradiction, suppose that z(p) is odd. Then, by Lemma 3, we have that π(p) = 4z(p) leading to the absurdity that 4z(p) divides p − 1 = 4n + 2. Thus, z(p) is an even number. Now, for the case in which p ≡ ±2 (mod 5), one has, by Lemma 2 (ii), that π(p) = 2(p + 1)/t, for some odd integer t. Thus, by Lemma 3, we obtain for some i ∈ {0, 1, 2}. Thus, z(p) = 2 3−i (n + 1)/t is an even number, since 3 − i ≥ 1 and t ≡ 1 (mod 2). This finishes the proof of the claim.
Note that since z(n) is a multiple of z(p), for all prime p in the factorization of n (by Lemma 4), then a sufficient condition for z(n) to be even is for n to have some prime factor of the form 4k + 3 (since z(4k + 3) is even, by Theorem 1). Therefore, Since we can apply Lemma 5 in order to get an upper bound for the cardinality of the previous set. Thus, Lemma 5 (for β = 1/φ(4) = 1/2) implies the existence of a positive absolute constant C such that for all x ≥ e 64 ≈ 6.2 × 10 27 (where we used that Γ(1/2) = √ π). Moreover, we have that (1,4) (1 − p −s ) −1 (6) and Since both sets P(1, 4) and P(3, 4) have the same density inside the set of all primes (namely, 1/2), then G 1,4 (s) G 3,4 (s) (i.e., G 1,4 (s) G 3,4 (s) and G 3,4 (s) G 1,4 (s)). Now, we multiply (6) and (7) to arrive at yielding that the natural density of E (z(p)) z is at least 1/p. In the case in which p = 3, we infer that δ(E (4) z ) ≥ 1/3 (observe that z(3) = 4). Moreover, we believe that Conjecture 1. The natural density of E (4) z is equal to 1.
The main difficulty in proving the last statement is to find some infinite arithmetic progression (containing infinitely many prime numbers) lying completely in E (4) z (in the spirit of Theorem 1). In general, we conclude by posing the following questions:

Conclusions
In this paper, we study a Diophantine problem related to the arithmetic function z : Z ≥1 → Z ≥1 , defined as z(n) := min{k ≥ 1 : n | F k } (the so-called order of appearance in the Fibonacci sequence). The problem is to establish some properties of E z := {n ≥ 1 : 2 | z(n)}. Indeed, we provide three infinite arithmetic progressions lying entirely within E z (which implies, in particular, that the natural density of E z is at least 2/3). Furthermore, we provide a lower bound for #(E z ∩ [1, x]) (for all sufficiently large x) which, in particular, confirms the expectation: the natural density of E z is 1. We close the work with some comments, questions, and a conjecture for further research. The proof combines arithmetical and analytical tools in number theory.