IJMMS International Journal of Mathematics and Mathematical Sciences 1687-0425 0161-1712 Hindawi Publishing Corporation 10.1155/2015/823973 823973 Research Article On the Local Minima of the Order of Appearance Function http://orcid.org/0000-0003-1321-4422 Luca Florian Mosima Thato Haukkanen Pentti School of Mathematics University of the Witwatersrand Private Bag X3 Wits Johannesburg 2050 South Africa wits.ac.za 2015 1102015 2015 17 08 2015 20 09 2015 1102015 2015 Copyright © 2015 Florian Luca and Thato 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 F k , the k th member of the Fibonacci sequence. In this paper, we improve upon some results from (Marques, 2011) concerning local minima of z ( n ) .

1. Introduction

Let { F m } m 0 be the Fibonacci sequence given by F 0 = 0 , F 1 = 1 , and F n + 2 = F n + 1 + F n for all n 0 . For a positive integer n , let z ( n ) be the order of appearance of n in the Fibonacci sequence, which is the minimal positive integer k such that n F k . It is known that z ( n ) always exists and in fact z ( n ) σ ( n ) , where σ ( n ) is the sum of divisors of n . Let us say that N is a local minimum for the function z ( n ) if z ( N ) < min { z ( N - 1 ) , z ( N + 1 ) } . It is not hard to prove that if N = F n for some positive integer n 3 (so N 2 ), then N is a local minimum for z ( n ) (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 z ( n ) . That family is (1) M 1 = F m k F k : k 3 , m m k , where m k 5 is some fixed number depending on k which is not computable from the arguments in . This problem was revisited in , where a different family of local minima is given; namely, (2) M 2 = F m k + ɛ F k F k - ɛ : ɛ ± 1 , 12 k m , m > m ɛ , k , where as before m ɛ , k depends on ɛ and k and is not computable from the arguments in .

None of the above two families gives us too many examples. Indeed, let be x a large positive real number and put A ( x ) = A [ 1 , x ] . Assume that N M 1 ( x ) M 2 ( x ) . Then, using the Binet formula (3) F i = α i - β i α - β , w h e r e α , β = 1 + 5 2 , 1 - 5 2 valid for all integers i 0 , it follows that N is determined in at most 3 ways by a pair of parameters ( m , k ) with k 3 such that (4) m k - 1 c + o 1 log x a s x , where c = ( log α ) - 1 . Using the classical estimates on the summatory function of the number of divisors function we get that (5) # M 1 x M 2 x = O log x log log x . Before we formulate the main result of this paper we need one more notion. A prime factor p of F m is called primitive if z ( p ) = m . A celebrated result of Carmichael  (see  for the most general result of this type) asserts that p always exists whenever m 13 . The main result of this paper is the following.

Theorem 1.

Let n 15 and (6) N = F n a , where a is a divisor of F n subject to the following restrictions:

a F n 1 / 5 ;

there exists a primitive prime factor p of F n such that p a .

Then n is a local minimum for z ( n ) . Furthermore, each such N is representable in a unique way as N = F n / a for some integers n 15 and a satisfying (i) and (ii) above, and N is not a Fibonacci number whenever a 2 .

The inequality F a + b F a F b is valid for all positive integers a and b . To prove it, fix a , note that it trivially holds for b { 1,2 } , and then use induction on b and the recurrence formula for the Fibonacci numbers to show that it holds for all b 1 . In particular, F k F m k 1 / m . Thus, if N M 1 , then (7) N = F m k F k , where if we put n : = m k and a F k , then a F m k 1 / m F n 1 / 5 because m 5 . Additionally, n 15 because k 3 and m 5 . Further, the number N is clearly divisible by a primitive divisor of F n (in fact, by any of the primitive divisors of F n ). This argument shows that the set M 1 is contained in the set of numbers N satisfying the conditions of Theorem 1. Now Theorem 1 says that in fact the parameter m k 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 (8) # M 3 x exp exp c 1 + o 1 log log x log log log x h o l d s a s x , where c 1 = log 2 .

Theorem 2 implies that the counting function of local minima N x exceeds ( log x ) A for any positive constant A (compare with (5)). In particular, the series (9) N > 1 l o c a l m i n i m u m o f z n 1 log N A diverges for all A > 0 .

2. Proof of Theorem <xref ref-type="statement" rid="thm1">1</xref>

Suppose that n = 15 . Then F n = 610 and the only divisors a of 610 satisfying (i) of Theorem 1 is a { 1,2 } . Now one checks that z ( 609 ) , z ( 611 ) , z ( 304 ) , z ( 306 ) are all larger than z ( 305 ) = z ( 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, n 16 .

Assume that N = F n / a satisfies the conditions of Theorem 1. Then z ( N ) = n . Indeed, N F n , so z ( N ) n . On the other hand, if N F m for some positive integer m , then (ii) of Theorem 1 shows that p N F m for some prime p with z ( p ) = n ; therefore m n . Thus, z ( N ) = n . Assume now that z ( N + ɛ ) n for some ɛ { ± 1 } . We then get an equation of the form (10) F n a + ɛ = F m b for some positive integers m n and b F m . Since (11) F n a F n 4 / 5 F 16 4 / 5 > 200 , we get (12) F m b = F n a + ɛ F n a - 1 > 2 F n 3 a , so (13) b < 3 a 2 F m F n 3 a 2 . Let us see that in fact m < n . Indeed, if m = n , then multiplying both sides of (10) by a b , we get F n ( a - b ) = ɛ a b . This implies first that a b and secondly that (14) F n a b . But this conclusion is impossible because it leads, by (13), to (15) F n a b < 3 a 2 2 1.5 F n 2 / 5 , which is false for n 16 . Hence, m n - 1 . Since (16) F m F n F n - 1 F n 2 3 , for n 16 , together with inequality (13), we get b < a . We will use the inequality (17) α k - 2 F k α k - 1 valid for all integers k 2 . We then have, using inequality (17) with k = m and n , respectively, that (18) α n - 2 α m - 1 F m b F n a - 1 F n 4 / 5 - 1 > F n 4 / 5 α 2 / 5 α 4 n / 5 - 2 , so m = n - k for some integer k [ 1 , n / 5 + 1 ) . Using Binet formula (3) with i = m and i = n , respectively, (10) is equivalent to (19) b α n - β n + ɛ 5 a b = a α n - k - β n - k , which can be regrouped as (20) α n b - a α - k = b β n - ɛ 5 a b - a β n - k . The number η b - a α - k is an algebraic integer in K Q ( 5 ) which is not zero; otherwise α k = b / a Q , which is impossible for positive integers k . Thus, the norm of η over K is an integer which is at least 1 in absolute value. Hence, (21) b - a α - k b - a β - k 1 , giving (22) b - a α - k 1 b + a α k > 1 F n 1 / 5 1 + α n / 5 + 1 > 1 α n - 1 / 5 1 + α n + 5 / 5 . Inserting (22) into (20) and using also (17), we get (23) α 4 n + 1 / 5 1 + α n + 5 / 5 < α n b - a α - k = b β n - ɛ 5 a b - a β n - k b α n + 5 a b + a α n - k < F n 1 / 5 α n + 5 F n 2 / 5 + F n 1 / 5 α 4 n / 5 - 1 < 5 + 0.1 F n 2 / 5 < 5 + 0.1 α 2 n - 1 / 5 , where we used the fact that α 4 n / 5 - 1 > 20 for n 16 . The above inequality leads to the conclusion that x : = α n / 5 satisfies the inequality (24) α 3 / 5 x 2 - 5 + 0.1 α x - 5 + 0.1 < 0 . However, the largest root of the quadratic polynomial from the left-hand side above is 4.607 < 4.664 = α 3.2 α n / 5 = x for n 16 , so quadratic (24) in x cannot be negative for n 16 , which is a contradiction.

The remaining assertions of the theorem are easy. To see unicity, assume that N = F n / a = F n / a are two representations of the same N satisfying conditions (i) and (ii) of the theorem. If n = n , then a = a and we are through. If n n , suppose without loss of generality that n > n . Then, by (ii), there is some primitive prime factor p of F n which divides N = F n / a . Since p is primitive for F n it cannot divide F n , which is a multiple of N , a contradiction. In particular, if n 15 and a 2 , then N = F n / a cannot have another representation of the form F n / a with a = 1 (so n > n 15 ), so it cannot be a Fibonacci number.

The theorem is therefore proved.

3. Proof of Theorem <xref ref-type="statement" rid="thm2">2</xref>

Let x be large and let n be such that (25) 15 n c log x . Since F n < α n x by (17), it follows that any number N = F n / a satisfying the conditions of Theorem 1 with n satisfying (25) is in M 3 ( x ) . We now choose n maximal satisfying inequality (25) of the form (26) n = i = 1 m p i , where 2 = p 1 < p 2 < denotes the sequence of all primes. By the Prime Number Theorem, we have (27) n = e 1 + o 1 p m , as m , showing that p m = log log x as x . By the Prime Number Theorem again, we get that (28) m = 1 + o 1 log log x log log log x as x . Now let a be a divisor of F n / p m . For large x , we have p m 5 ; therefore a F n / p m F n 1 / p m F n 1 / 5 , so condition (i) from Theorem 1 is satisfied. Condition (ii) is also satisfied and in fact any primitive prime factor of F n will divide F n / F n / p m , which is a divisor of F n / a . Now by the Primitive Divisor Theorem, for every divisor d { 1,2 , 6 } of n / p m , F d has a primitive prime factor p d which of course divides F n / p m . This shows that F n / p m has at least (29) τ n p m - 3 = τ n 2 - 3 = 2 m - 1 - 3 distinct prime factors, where τ ( k ) denotes the number of divisors of the positive integer k . Hence, the number of such convenient a ’s is at least as large as the number of square-free integers built up with prime factors from a set of 2 m - 1 - 3 distinct primes, and this number is at least as large as (30) 2 2 m - 1 - 3 . Thus, (31) # M 3 x 2 2 m - 1 - 3 , where m 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.

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