On the Local Minima of the Order of Appearance Function

Copyright © 2015 F. Luca and T. Mosima. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The order of appearance z(n) of the positive integer n is the smallest positive integer k such that n divides Fk, the kth member of the Fibonacci sequence. In this paper, we improve upon some results from (Marques, 2011) concerning local minima of z(n).


Introduction
Let {  } ≥0 be the Fibonacci sequence given by  0 = 0,  1 = 1, and  +2 =  +1 +  for all  ≥ 0. 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 () < min{(−1), (+ 1)}.It is not hard to prove that if  =   for some positive integer  ̸ = 3 (so  ̸ = 2), then  is a local minimum for () (see Page 1 in [1]).
In Theorem 1.1 in [1], Marques exhibited a family of positive integers which are not members of the Fibonacci sequence but are local minima for ().That family is where   ≥ 5 is some fixed number depending on  which is not computable from the arguments in [1].This problem was revisited in [2], where a different family of local minima is given; namely, where as before  , depends on  and  and is not computable from the arguments in [2].None of the above two families gives us too many examples.Indeed, let be  a large positive real number and Then, using the Binet formula valid for all integers  ≥ 0, it follows that  is determined in at most 3 ways by a pair of parameters (, ) with  ≥ 3 such that where  = (log ) −1 .Using the classical estimates on the summatory function of the number of divisors function we get that Before 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 [3] (see [4] for the most general result of this type) asserts that  always exists whenever  ≥ 13.The main result of this paper is the following.
Theorem 1.Let  ≥ 15 and where  is a divisor of   subject to the following restrictions: (i)  ≤  1/5  ; (ii) there exists a primitive prime factor  of   such that  ∤ .

International Journal of Mathematics and Mathematical Sciences
Then  is a local minimum for ().Furthermore, each such  is representable in a unique way as  =   / for some integers  ≥ 15 and  satisfying (i) and (ii) above, and  is not a Fibonacci number whenever  ≥ 2.
The inequality  + ≥     is valid for all positive integers  and .To prove it, fix , note that it trivially holds for  ∈ {1, 2}, and then use induction on  and the recurrence formula for the Fibonacci numbers to show that it holds for all  ≥ 1.In particular,   ≤  1/  .Thus, if  ∈ M 1 , then where if we put  :=  and  :=   , then  ≤  1/  ≤  This argument shows that the set M 1 is contained in the set of numbers  satisfying the conditions of Theorem 1.Now Theorem 1 says that in fact the parameter   from M 1 can always be taken to be 5. Putting M 3 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 (log )  for any positive constant  (compare with (5)).In particular, the series diverges for all  > 0.

Proof of Theorem 1
Suppose that  = 15.Then   = 610 and the only divisors  of 610 satisfying (i) of Theorem 1 is  ∈ {1, 2}.Now one checks that (609), (611), (304), (306) are all larger than (305) = (610) = 15.One does not even have to compute the above orders of appearance; one only has to factor the first 15 members of the Fibonacci sequence in order to convince oneself that none of them is a multiple of 609 or of 611 or of 304 or of 306.From now on,  ≥ 16.
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  ∈ {±1}.We then get an equation of the form for some positive integers  ≤  and  |   .Since we get so Let 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  ≥ 16.Hence,  ≤  − 1.Since for  ≥ 16, together with inequality (13), we get  < .We will use the inequality valid for all integers  ≥ 2. We then have, using inequality (17) with  =  and , respectively, that so  =  −  for some integer  ∈ [1,/5 + 1).Using Binet formula (3) with  =  and  = , respectively, (10) is equivalent to which can be regrouped as The number  :=  −  − is an algebraic integer in K := Q( √ 5) which is not zero; otherwise   = / ∈ Q, which is International Journal of Mathematics and Mathematical Sciences 3 impossible for positive integers .Thus, the norm of  over K is an integer which is at least 1 in absolute value.Hence, giving Inserting ( 22) into (20) and using also (17), we get where we used the fact that  4/5−1 > 20 for  ≥ 16.The above inequality leads to the conclusion that  :=  /5 satisfies the inequality  3/5  2 − ( √ 5 + 0.1)  − ( √ 5 + 0.1) < 0.
However, the largest root of the quadratic polynomial from the left-hand side above is 4.607 ⋅ ⋅ ⋅ < 4.664 ⋅ ⋅ ⋅ =  3.2 ≤  /5 =  for  ≥ 16, so quadratic (24) in  cannot be negative for  ≥ 16, 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  ≥ 15 and  ≥ 2, then  =   / cannot have another representation of the form    /  with   = 1 (so   >  ≥ 15), so it cannot be a Fibonacci number.
The theorem is therefore proved.

Proof of Theorem 2
Let  be large and let  be such that where  satisfies estimate (28), which leads to the desired conclusion of the theorem.

3 (
, 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 2 −1 − 3 distinct primes, and this number is at least as large as 2 2 −1 −3 .) ≥ 2 2 −1 −3 , satisfying (25) is in M 3 ().We now choose  maximal satisfying inequality (25) of the form Now let  be a divisor of  /  .For large , we have   ≥ 5; therefore  ≤  /  ≤  1/  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  ∉ {1, 2, 6} of /  ,   has a primitive prime factor   which of course divides  /  .This shows that  /  has at least 15 ≤  ≤  log .(25)Since  <   ≤  by (17), it follows that any number  =   / satisfying the conditions of Theorem 1 with