#### Abstract

The order of appearance of the positive integer is the smallest positive integer such that divides , the th member of the Fibonacci sequence. In this paper, we improve upon some results from (Marques, 2011) concerning local minima of .

#### 1. Introduction

Let be the Fibonacci sequence given by , , and for all . For a positive integer , let be the order of appearance of in the Fibonacci sequence, which is the minimal positive integer such that . It is known that always exists and in fact , where is the sum of divisors of . Let us say that is a local minimum for the function if . It is not hard to prove that if for some positive integer (so ), then is a local minimum for (see Page 1 in ).

In Theorem 1.1 in , Marques exhibited a family of positive integers which are not members of the Fibonacci sequence but are local minima for . That family is where is some fixed number depending on which is not computable from the arguments in . This problem was revisited in , where a different family of local minima is given; namely, where as before depends on and and is not computable from the arguments in .

None of the above two families gives us too many examples. Indeed, let be a large positive real number and put . Assume that . Then, using the Binet formulavalid for all integers , it follows that is determined in at most ways by a pair of parameters with such that where . Using the classical estimates on the summatory function of the number of divisors function we get thatBefore we formulate the main result of this paper we need one more notion. A prime factor of is called primitive if . A celebrated result of Carmichael  (see  for the most general result of this type) asserts that always exists whenever . The main result of this paper is the following.

Theorem 1. Let andwhere is a divisor of subject to the following restrictions: (i);(ii)there exists a primitive prime factor of such that .Then is a local minimum for . Furthermore, each such is representable in a unique way as for some integers and satisfying (i) and (ii) above, and is not a Fibonacci number whenever .

The inequality is valid for all positive integers and . To prove it, fix , note that it trivially holds for , and then use induction on and the recurrence formula for the Fibonacci numbers to show that it holds for all . In particular, . Thus, if , then where if we put and , then because . Additionally, because and . Further, the number is clearly divisible by a primitive divisor of (in fact, by any of the primitive divisors of ). This argument shows that the set is contained in the set of numbers satisfying the conditions of Theorem 1. Now Theorem 1 says that in fact the parameter from can always be taken to be . Putting for the set of numbers satisfying the conditions of Theorem 1, we have the following estimate.

Theorem 2. The estimate where .

Theorem 2 implies that the counting function of local minima exceeds for any positive constant (compare with (5)). In particular, the series diverges for all .

#### 2. Proof of Theorem 1

Suppose that . Then and the only divisors of satisfying (i) of Theorem 1 is . Now one checks that , , , are all larger than . One does not even have to compute the above orders of appearance; one only has to factor the first members of the Fibonacci sequence in order to convince oneself that none of them is a multiple of or of or of or of . From now on, .

Assume that satisfies the conditions of Theorem 1. Then . Indeed, , so . On the other hand, if for some positive integer , then (ii) of Theorem 1 shows that for some prime with ; therefore . Thus, . Assume now that for some . We then get an equation of the formfor some positive integers and . Since we get soLet us see that in fact . Indeed, if , then multiplying both sides of (10) by , we get . This implies first that and secondly that But this conclusion is impossible because it leads, by (13), to which is false for . Hence, . Since for , together with inequality (13), we get . We will use the inequalityvalid for all integers . We then have, using inequality (17) with and , respectively, that so for some integer . Using Binet formula (3) with and , respectively, (10) is equivalent to which can be regrouped asThe number is an algebraic integer in which is not zero; otherwise , which is impossible for positive integers . Thus, the norm of over is an integer which is at least in absolute value. Hence, givingInserting (22) into (20) and using also (17), we get where we used the fact that for . The above inequality leads to the conclusion that satisfies the inequalityHowever, the largest root of the quadratic polynomial from the left-hand side above is for , so quadratic (24) in cannot be negative for , which is a contradiction.

The remaining assertions of the theorem are easy. To see unicity, assume that are two representations of the same satisfying conditions (i) and (ii) of the theorem. If , then and we are through. If , suppose without loss of generality that . Then, by (ii), there is some primitive prime factor of which divides . Since is primitive for it cannot divide , which is a multiple of , a contradiction. In particular, if and , then cannot have another representation of the form with (so ), so it cannot be a Fibonacci number.

The theorem is therefore proved.

#### 3. Proof of Theorem 2

Let be large and let be such thatSince by (17), it follows that any number satisfying the conditions of Theorem 1 with satisfying (25) is in . We now choose maximal satisfying inequality (25) of the form where denotes the sequence of all primes. By the Prime Number Theorem, we have as , showing that as . By the Prime Number Theorem again, we get thatas . Now let be a divisor of . For large , we have ; therefore , so condition (i) from Theorem 1 is satisfied. Condition (ii) is also satisfied and in fact any primitive prime factor of will divide , which is a divisor of . Now by the Primitive Divisor Theorem, for every divisor of , has a primitive prime factor which of course divides . This shows that has at least distinct prime factors, where denotes the number of divisors of the positive integer . Hence, the number of such convenient ’s is at least as large as the number of square-free integers built up with prime factors from a set of distinct primes, and this number is at least as large as Thus, where satisfies estimate (28), which leads to the desired conclusion of the theorem.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.