1 n <. n G F. ζ(3) c = 2ζ(2) ζ(6)log α = , Our method is general and can be applied to any Lucas sequence of general term

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1 Fibonacci Integers Florian Luca Instituto de Matemáticas, Universidad Nacional Autonoma de México, C.P , Morelia, Michoacán, México Carl Pomerance Mathematics Department, Dartmouth College, Hanover, NH 03755, USA Stephan Wagner Department of Mathematical Sciences, Stellenbosch University, Private Bag X, Matieland 7602, South Africa June 7, 200 Abstract A Fibonacci integer is an integer in the multiplicative group generated by the Fibonacci numbers. For example, 77 = 2 55/3 5) is a Fibonacci integer. Using some results about the structure of this multiplicative group, we determine a near-asymptotic formula for the counting function of the Fibonacci integers, showing that up to x the number of them is between expclog x) /2 log x) ǫ ) and exp clog x) /2 + log x) /6+ǫ), for an explicitly determined constant c. The proof is based on both combinatorial and analytic arguments.

2 Introduction Let F n ) n be the Fibonacci sequence given by F = F 2 = and F n+2 = F n+ +F n for all n and let G F be the set of integers in the multiplicative group generated by F n ) n inside Q. Hence, G F consists of all integers which can be represented as a ratio of products of Fibonacci numbers. We call the members of G F Fibonacci integers. The smallest positive integer n that is not a Fibonacci integer is 37, and the number of Fibonacci integers in [,00] is 88. One might think then that most integers are Fibonacci integers, but this is not the case. For a positive real number x let G F x) = G F [,x] be the set of Fibonacci integers in [,x]. In [5], it was shown that the estimate #G F x) A x log x) A holds for all x 2 with any constant A, where the implied constant above depends on A. Applying this with any A >, it was deduced in [5] that Here, we improve on this estimate. n G F n <. Theorem. For each fixed ǫ > 0, the estimate exp clog x) /2 log x) ǫ) #G F x) exp clog x) /2 + log x) /6+ǫ) holds for all sufficiently large x, with ζ3) c = 2ζ2) ζ6)log α = , where ζ is the Riemann zeta-function and α = + 5)/2 is the golden mean. Our method is general and can be applied to any Lucas sequence of general term u n = an b n a b or v n = a n + b n for all n, where a + b, ab are nonzero integers and a/b is not a root of. Write G u and G v for the positive integers in the multiplicative groups generated 2

3 by {u n } n and {v n } n inside Q, respectively, and for a positive real number x write G u x) and G v x) for the intersection with [,x] of G u and G v, respectively. Then, at least assuming that a and b are real, both #G u x) and #G v x) obey estimates of the same shape as what is shown in Theorem, where in the formula for the constant c we need to replace + 5)/2 by max{ a, b }. Perhaps this is still true even when a and b are complex conjugates but we have not worked out the details for this situation. To simplify the presentation, we shall deal only with the Fibonacci sequence. 2 Arithmetic considerations Let α = + 5)/2 and β = 5)/2 be the two roots of the characteristic equation X 2 X = 0 of the Fibonacci sequence. Then it is well-known that F n = αn β n holds for all n. α β For a positive integer m write Φ m X) = k m k,m)= X exp2πik/m)) Z[X] for the m-cyclotomic polynomial and let Φ m X,Y ) = X exp2πik/m)y ) Z[X,Y ] k m k,m)= be its homogenization. Further, let Φ m stand for Φ m α,β). Note that for m >, Φ m is an integer. We have F m = d m d> and by Möbius inversion, we have, for m >, Φ m = d m Φ d ) F µd) m/d, 2) where µ is the Möbius function. In particular, Φ m G F when m >. Formula ) shows that the numbers Φ m for m > generate the same group 3

4 as the Fibonacci numbers. It turns out that this group is almost freely generated by the numbers Φ m ) m>. This is not exactly so because of the exceptions Φ 2 =, Φ 6 = F 6 F 2 F 3 = 2 2 = Φ 2 3, Φ 2 = F 2F 2 F 6 F 4 = 6 = Φ 3 Φ 4. 3) The numbers Φ m capture the so-called primitive prime divisors of F m, namely those primes p dividing F m that do not divide F n for any n < m. It is known that each F m, when m,2,6,2, has at least one primitive prime factor. Let Ψ m be the product of the primitive prime factors of F m with the corresponding exponents as they appear in F m. For m >, we have Ψ m Φ m. To investigate the quotient δ m = Φ m /Ψ m, note that for every positive integer k there exists some integer n such that k F n. We write zk) for the smallest such n the index of appearance of k in the Fibonacci sequence). Thus, we can rephrase the condition that p is primitive for F m as zp) = m and the definition of Ψ m as Ψ m = p ap. Further, δ m = Φ m /Ψ m = except if p ap F m zp)=m m = p k zp) for some k and prime p. If m = p k zp) and m 2, then δ m = p. All of the above properties can be found in either Section 2 of [], or in [6]. If p is prime and zp) = m, then p 0, ± mod m), and the sign in p fact equals the Legendre symbol. In particular, we see that if δ m = p 5) is an odd prime, then p = Pm), where we write Pm) for the largest prime factor of m. Thus, if m 2) is of the form p k zp), then it is uniquely of this form; that is, p and k are determined. Let M = N \ {,2,6,2}. The result that F m has a primitive prime factor for each m M shows that the group generated by F n ) n is freely generated by Φ m ) m M. This group is also freely generated by F m ) m M, but the cyclotomic numbers Φ m almost freely generate the Fibonacci integers as a mutiplicative semigroup. However, there are some Fibonacci integers not of this form, for example Φ 24 Φ 3, Φ 25 Φ 5, and Φ 37 9Φ 3 9 Φ 9. The following lemma sheds some light on the structure of G F. 4

5 Lemma. Assume that I, J are finite multisets of indices with m i,n j M, n i m j for all i I, j J and n = i I Φ ni j J Φ m j N. There exists an injection f from the multiset of prime factors of Ψ mj = p p 2...p k j J into the multiset {n i } i I where fp l ) = p k l l zp l ) for some k l N. Proof. If p is a prime factor of Ψ mj, then zp) = m j for some j J. Since m j {n i } i I, it follows that p = δ ni for some i I and n i = p k zp) for some positive integer k. In particular, if p a Ψ mj, then there are at least a values of i I with n i of the form p k zp) perhaps with different values of k), so that we may assign each factor of p to a different n i. Further, since no n i has two different representations in the form p k zp) with p prime and k > 0, we may continue this mapping for each prime factor of Ψ mj. Remark. Lemma does not tell the whole story. What is not being accounted for in the result is the contribution of the primes in j J δ m j which also need to occur in the product i I Φ n i. It turns out that these primes are negligible in our counting problem. Remark 2. We remark that a prime p is a Fibonacci integer if and only if Ψ zp) = p and δ zp) is a Fibonacci integer. Since δ zp) is either or a prime much) smaller than p, we thus have a simple algorithm for determining if a given prime is a Fibonacci integer. The first few primes which are not Fibonacci integers are 37, 43, and 53. Since Ψ n is exponentially large in n, it is easy to see that the number of prime Fibonacci integers in [,x] is Olog x). Probably there are infinitely many of them, but we do not know how to prove this. It seems to be a slightly easier assertion than the conjecture that there are infinitely many prime Fibonacci numbers, but that doesn t seem to be of much help. Yuri Bilu asked us if, in general, it is decidable whether a given natural number n is a Fibonacci integer. The arguments in the next section, and in partiqular 0), show that this is indeed the case. Remark 3. Say that a Fibonacci integer is an atom if it exceeds and it is not the product of two smaller Fibonacci integers. Let Ξ n = Ψ n if δ n is 5

6 a Fibonacci integer, and otherwise, let Ξ n = Φ n cf. Remark 2). Using the thoughts behind Lemma it is possible to characterize the atoms as the union of {Ξ n : n M} and { l i= Φ } p k i i m : m M, Ξ m = p... p l, l 2, and p k Ξ m,...,pk l l m M, m where p,...,p l denote primes. The Fibonacci integers do not enjoy unique factorization into atoms. Here are three examples based on the fact that Φ 9 = 37 3: Let nj,l) denote the atom Φ 37j 9Φ 3l 9/Φ 9, where j,l > 0, and note that Φ 37 9 Φ 3 9 = Φ 9 n,), Φ n,) = Φ 37 9 n2,), n,) n2,2) = n,2) n2,). Such redundancies complicate the possible attainment of an asymptotic formula for the distribution of Fibonacci integers. Lemma 2. The inequality holds for all integers m >. α φm) Φ m < α φm)+ 4) Proof. The lemma holds with equality at the lower bound when m = 2, so assume m 3. From 2), we get Φ m = d m F µd) m/d = α m/d β m/d) µd) = α φm) d m d m so that since β/α = α 2, L m := log Φ m /α φm)) = d m µd)log α 2 ) m/d). ) ) β m/d µd), α It suffices to show that L m < log α. Note that if k N, log α 2 ) j) < log α 2j ) j>k j>k = i i α 2ik α 2i < log α 2k). 6

7 If m is not squarefree, we have L m < log α 2 ) j) < log α 2 ) = log α. j> Suppose m is squarefree and p is the smallest odd prime factor of m. Then p + m. If m is odd, L m < log + α 2 ) log + α 2p) + log α 2 ) j) If m is even, j>p+ < log + α 2 ) log + α 2p) + log α 2p 2 ) = log + α 2) log + α 2p) log α 2p 2) < log + α 2) < log α. L m < log + α 2 ) log α 4) log + α 2p) + log α 2 ) j) j>p+ < log α 2 ) log + α 2p) + log α 2p 2 ) = log α 2) log + α 2p) log α 2p 2) < log α 2) = log α. This completes the proof. 3 Combinatorial arguments Let G be the multiplicative semigroup freely generated by {Φ m } m M inside the set N of natural numbers and for a positive real number x let G x) = G [,x]. In the next section, with tools specific to complex analysis, we will prove the following theorem. Theorem 2. For each fixed ǫ > 0, the estimate exp clog x) /2 log x) ǫ) #G x) exp c log x + log x) ǫ) 5) holds for all sufficiently large x. In this section, we shall show how to use this theorem and the results of the previous section to complete the proof of our main result, Theorem. 7

8 It suffices to deal with the upper bound on #G F x), since the lower bound is an immediate consequence of Theorem 2 and the fact that each Φ m for m M is a Fibonacci integer. Let x be large and assume that N = Φ ni Φ m j x i I j J is an integer. We need to bound from above the number of such possible N s. For each m M, let lm) = ΩΨ m ) be the number of primitive prime factors of Φ m counted with multiplicity. Let f be an injection from the multiset of prime factors of j J Ψ mj into the multiset {n i } i I as guaranteed by Lemma. For j J, let p j, p j,2 p j,lmj ) be the multiset of prime factors of Ψ mj, and for l lm j ), let n j,l = fp j,l ), so that n j,l = p k j,l j,l m j for some positive integer k j,l. Thus, by a change in notation, we wish to count the number of numbers of the form N = j J Φ mj lm j ) l= Φ nj,l i I Φ ni x. 6) Note that numbers N in 6) are not necessarily integers since if some Φ mj has a non-primitive prime factor, we have not necessarily arranged for it to be cancelled with some corresponding prime among the Φ n s, see Remark.) We thus have #G F x) #G x/w), 7) w where w ranges over rationals of the form of the parenthetical double product in 6). Fix some j,l and look at m := m j and n := n j,l = p k m, where p = p j,l is a primitive prime factor of Φ m and k. By Lemma 2, we have We claim that Indeed, if p = 2, then m = 3 and k 3, so that Φ n Φ m α φn) φm) 2. 8) φn) φm) 2 φn). 9) 2 φn) φm) 2 = 3 n 4 6 n = 2 φn). 8

9 If p = 3, then m = 4 and k 2, so that φn) φm) 2 = 3 n n = 2 3 φn). In the remaining cases, we have p 5 and φm) 4, so that φn) φm) 2 φp k )φm) 3 2 φm) 5 8 φpk )φm) = 5 8 φn). In each case we have 9). It follows from 6), 8), and 9) that j J lm j ) l= φn j,l ) 2 log x, log α and so using the minimal order of φ, we have j J lm j ) l= n j,l K, 0) where K = κ log x log log log x and κ is an absolute computable constant. Since n j,l = p k j,l j,l m j, we also have each p j, m j K. So, the inequality Φ mj Ψ mj p lm j) j, m j and Lemma 2 imply that lm j ) m j /log log xlog log log x). Thus, lm j ) m j log log x) 2 ) for sufficiently large x, each j J, and all J as above. Let L = log x) /6. We say a multiset {n j,l } is good if for each j there are at most L distinct primes p j,l with exponents k j,l at least 2. We say {n j,l } is bad if for each j, there are more than L distinct primes p j,l with exponents k j,l at least 2. Each multiset in our problem can be partitioned into a good multiset and a bad multiset. With T,U,M positive integers, let N T be the set of good multisets with T distinct m j s and let N U,M be the set of bad multisets with U distinct m j s, where the sum of the distinct m j s is M. Let W T denote the set of rationals w in [,x] of the form w = j J Φ mj 9 lm j ) l= Φ nj,l,

10 where the multiset {n j,l } is in N T and let w T be the least member of W T, with w T = if W T is empty. Similarly define W U,M and w U,M for multisets in N U,M. We have from 7) that #G F x) T,U,M w W T w W U,M #G x ww ) T,U,M #G x w T w U,M ) N T N U,M, 2) where T,U,M run up to K, and N T = + #N T, N U,M = + #N U,M. We wish to count the number of multisets {n j,l } arising with certain constraints. Such a multiset uniquely determines the corresponding multiset {m j }, and so we count by first choosing this simpler multiset and then extending to {n j,l }. The number of ways of choosing a multiset {m j } with mj K and with T distinct m j s is at most K 2T exp3t log log x) for x large. Given some m j, the number of corresponding multisets {n j,l } is at most 2log K) lmj) exp2m j log log x) for large x, since lm j ) < m j and the number of choices for an exponent k j,l for the prime p j,l is at most log K/log p j,l < 2log K. In the case where we know that the number of distinct primes p j,l which have exponents at least 2 is at most L, the number of ways of choosing these distinct primes is at most K L. Fixing one such prime p, the number of ways of choosing exponents for all of its copies is at most K 2log K for large x. Indeed, with Z = 2log K, we are counting integer vectors v,...,v Z ) where v k is the number of copies of p with exponent k. If there are s copies of p in all, then each v k s. So, there are at most s + ) Z choices, and it remains to note that since s K/2, the above quantity is smaller than K 2log K for large values of x. So, the total number of ways to choose these L distinct primes and exponents for them and their copies is at most K L+2Llog K exp3llog log x) 2 ). We conclude that N T N U,M exp 3T + U + M + L)log log x) 2). 3) Thus, to use 2), we wish to have upper bounds for T,U,M and lower bounds for the numbers w T and w U,M. We will see that these tasks are related. Suppose w T arises from the multiset pair {m j }, {n j,l }. Without loss of generality, we may assume that m,...,m T are distinct. Since each 0

11 p j,l m j and m j 3, we have by ) j J lm j ) l= n j,l T j= lm j ) l= p j,l m j T+2 log log x) 2 j=3 T lm j )m j )m j j= j 2 j ) /4 log log x) 2T 4. We conclude from Lemma 2, 9), 0) and the minimal order of φ that T 4 ) w T exp log log x) 3 4) for all large x. We achieve two lower bounds for w U,M as follows. For {n j,l } in N U,M, with corresponding multiset {m j }, we assume that m,...,m U are distinct. For each j U there are more than L distinct primes p j,l with exponent at least 2 in n j,l. Each of these primes satisfies p j,l 0, ± mod m j ), so that l p2 j,l L3 m 2 j and n j,l L 3 m 3 j. As above, we deduce that L 3 ) m 3 j w U,M exp 5) log log x for all large x. In addition, the primes associated with m j are all different from the primes associated with m j when m j m j, so there are more than UL distinct primes p j,l with exponents at least 2 among the various n j,l s. The sum of their squares is at least of order U 3 L 3, and since each m j > L using lm j ) > L), we have n j,l U 3 L 4. We thus deduce that U 3 L 4 ) w U,M exp 6) log log x for all large x. By Hölder s inequality, we have M = U U m j j= j= so that from 6) and then 5), we obtain m 3 j /3 U 2/3, U log w U,M) /3 L 4/3 log log x) /3, M log w U,M) 5/9 L 7/9 log log x) 5/9 7)

12 for all large x. Note that for w x, log x ) /2 = log x) /2 log w w log x = log x) /2 log w 2log x) /2. ) /2 log x) /2 log w ) 2log x Thus, from Theorem 2, we have ) x #G exp clog x) /2 clogw ) Tw U,M ) w T w U,M 2log x) /2 + log x) ǫ for all large x. With 2) we thus get, #G F x) exp clog x) /2 + log x) ǫ) T,U,M exp clogw ) Tw U,M ) 2log x) /2 N T N U,M. 8) Since the sum in 8) has at most K + ) 3 terms, it suffices to show that each term is at most explog x) /6+ǫ/2 ). Suppose 2/3 < a and we have shown that all of the terms in 8) with w T w U,M > explog x) a log log x) 5 ) are negligible. We definitely have this for a = since no term satisfies this inequality.) Then 4) gives T log x) a/4 log log x) 2 and 7) gives both U log x) a/3 2/9 log log x) 2 and M < log x) 5a/9 7/54 log log x) 4. Note that a/3 2/9 < 5a/9 7/54 < a/4, so from 3), N T N U,M exp 4log x) a/4 log log x) 4). Since a/4+/2 > 2/3, we thus may replace a with a/4+/2 in the argument, since those terms with w T w U,M > explog x) a/4+/2 log log x) 5 ) are now seen to be negligible in 8) because a/4 < a /2. In a finite number of steps, we reach a value of a with 2/3 < a < 2/3 + ǫ, and then all remaining terms in 8) are smaller than explog x) /6+ǫ/2 ). This completes the proof of Theorem. Remark 4. To improve on the exponent /6 in Theorem it would seem that a finer knowledge of the prime factors of the numbers Φ m would be needed. It seems reasonable that for all large m there is a prime factor of Φ m that is larger than any fixed power of m, and if this is the case, the exponent /6 can be replaced with 0. Assuming a strong form of the abc conjecture, it follows that for each fixed ǫ > 0 the number Φ m has a squarefree divisor larger than Φ ǫ m for all sufficiently large m see [7]). Using this, our argument would give exponent /8 in place of /6 in Theorem. 2

13 4 Analytic arguments In the previous section, we reduced the problem to counting the number of positive integers in [,x] that belong to the semigroup G freely generated by {Φ m } m M. We need to prove that this number satisfies the estimate given by Theorem 2. Note first that the Dirichlet series associated with G is given by Dz) := n z = Φ z n G m M m ). In view of the exponential growth of Φ m, the series Dz) converges for Rz) > 0. Now we apply the following well-known variant of Perron s formula see, e.g., [4]): Gx) := n x n G n ) = r+i x 2πi r i Dz)x z dz 9) zz + ) for any r > 0. It will later turn out to be advantageous to work with this variant rather than Perron s formula itself. The integral in 9) is estimated by means of the saddle-point method. In order to choose r appropriately, we have to study the behavior of Dz) as z 0. We have log Dz) = log Φ z ) m = log exp log Φ m )z)). m M m M 20) Now we apply the Mellin transform to this harmonic sum; the Mellin transform of log e z ) is given by Γs)ζs+), which implies that the Mellin transform of log Dz) is Γs)ζs + )Cs), where Cs) = m Mlog Φ m ) s. Next we need information on the analytic behavior of the Dirichlet series Cs), which is provided in the following lemma: Lemma 3. The Dirichlet series Cs) satisfies Cs) = log α) s m φm) s + As), where As) is analytic on Rs) > 0 and satisfies As) = O s ) for ǫ Rs) ǫ, where the implied constant only depends on ǫ. 3

14 Proof. Write L m = log Φ m φm)log α as in the proof of Lemma 2, and recall that L m log α for arbitrary m >. Thus, we have As) = Cs) log α) s m φm) s = log α) s s + 4 s ) + m M log Φm ) s log α) s φm) s). The first part is analytic and bounded in the indicated region, so it remains to consider the sum over M, which we denote A 0 s). We have A 0 s) = log α) ) s φm) s L s m + ), φm)log α m M so that A 0 s) log α) s log α) s s. m M φm)< s m M φm)< s φm) s + φm) ǫ + s m M φm) s m M φm) s φm) s sl m φm)log α φm) ǫ This shows that the sum converges absolutely and uniformly on compact subsets of the half-plane Rs) > 0, and so we have that A 0 s), and hence As), is analytic on this half-plane. This completes the proof of the lemma. The Dirichlet series m φm) s was studied, for instance, in [2]. Since φm) is multiplicative, we have the Euler product m φm) s = φp α ) s = ) + p ) s p s ) p α 0 p = p s ) + p ) s p s) p p = ζs) + p ) s p s). p 4

15 The second factor converges for Rs) = σ ǫ > 0, and it is easy to show that it grows subexponentially as Is) in this region. Indeed, noting that /p) s = + O s p ) if p > s, we find log p + p ) s p s) = p log + p ) s p s) = log + p ) s p s) + log p s /p) s)) p s p> s = log + Op σ ) ) + log + O s p σ ) ) p s p> s p s p σ + s p> s p σ s σ s ǫ. Since Γs) decreases exponentially as Is), this is therefore also the case for Γs)ζs + )Cs), which allows us to apply the Mellin inversion formula: we have log Dr) = 2πi 2+i 2 i Γs)ζs + )Cs)r s ds. If we shift the path of integration to Rs) = ǫ and pick up the residue at s = see [3] for details), we obtain where the constant A is given by Γ)ζ2) log α p log Dr) = A r + Or ǫ ), + p ) p ) = ζ2) p 2 p + log α pp ) p = ζ2) log α By similar arguments, we find that p p 6 p 2 ) p 3 ) = ζ2)2 ζ3) ζ6)log α. d dz log Dz) = log Φ m Φ z m M m has Mellin transform Γs)ζs)Cs ), which yields d dz log Dz) z=r = A r 2 + Or ǫ ), 5

16 and in the same manner Finally, we have, if z = r + it, d 3 log Dz) dz3 = d 2 dz 2 log Dz) z=r = 2A r 3 + Or 2 ǫ ). m M m M log Φ m ) 3 Φ z m Φz m + ) Φ z m )3 log Φ m ) 3 Φ r mφ r m + ) Φ r m ) 3 = Or 4 ), uniformly in t. Now we can expand log Dz) into the series: ) ) ) A log Dr) it r 2 + Or ǫ ) t2 2A t 3 2 r 3 + Or 2 ǫ ) + O r 4. If we restrict ourselves to the central part t T = r 7/5, then this gives us when 0 < r <. Using z = t ) ) + O, r r log Dz) = log Dr) iat r 2 At2 r 3 + Or/5 ), z + = + O t )) = + Or), r + we consequently have Dz)x z zz + ) = Dr)xr exp iat ) r r 2 At2 r 3 + it log x + Or/5 ). Now choose r in such a way that the linear terms in the exponent cancel, i.e., r = A/log x. Then the central part in the integral in 9) is r+it Dz)x z Dr)xr T ) dz = exp At2 2πi zz + ) r 2π r 3 + Or )) /5 dt. r it Completing the integral on the right to the entire interval, ) only yields an exponentially small error term, hence we have 2πi r+it r it T Dz)x z Dr)xr dz = zz + ) r = Dr)xr r 2 πa ) 2 πa r3/2 + Or /5 ) ) + Or /5 ). 6

17 It remains to estimate the part with t r 7/5 in the integral in 9). The part t r can be trivially estimated, since Dz) Dr) : 2πi Dz)x z r+ir zz + ) dz Dr)xr r+i rdr)x r, r+i 2π r+ir zz + ) dz and likewise for t r. Here it is essential that we were using the modified version of Perron s formula to obtain a convergent integral. For the part r 7/5 t r, we need another estimate for Dz). Lemma 4. Write z = r + it, and suppose that r 7/5 t r. Then we have the estimate uniformly for 0 < r /3. log Dr) Rlog Dz)) r /5 Proof. First of all, we reduce the task to an estimate for certain exponential sums. We have Hence, log Dz) = m M = m M k= log Φ z ) m = log Dr) Rlog Dz)) = m M m M k= k Φ kz m k Φ kr m coskt log Φ m) isinkt log Φ m )). m M k Φ kr m k= coskt log Φ m)) Φ r m cost log Φ m )). By Lemma 2, Φ m < α φm)+ α m for m M, which implies Φ r m /α for m r, m M. So we obtain the following estimate: log Dr) Rlog Dz)) α cost log Φ m )), m r which allows us to restrict our attention to a somewhat simpler sum. We distinguish several cases. 7

18 If t r, we then have t log Φ m t m log α rm log α log α < π 2 for m r and thus cost log Φ m ) t 2 log Φ m ) 2. Hence, we obtain log Dr) Rlog Dz)) t 2 log Φ m ) 2 t 2 φm) 2. m r m r Now an elementary argument shows that φm) 2 M 3, m M so that using t r 7/5, we have log Dr) Rlog Dz)) t 2 r 3 r /5. If r t r /5, we can apply the same argument to obtain log Dr) Rlog Dz)) α cost log Φ m )) m r α cost log Φ m )) m t t 2 log Φ m ) 2 t 2 t 3 r /5. m t For t r /5, we need different arguments. Write X = r. Clearly, cost log Φ m )) cost log Φ m )) m X m N for any set N of integers in [,X]. We focus on the following two cases: I. Take the set N that consists of all numbers p and 2p where p is a prime in the interval [ X 4, X 2 ]. We have α p β p ) logφ p ) = log F p = log = p log α logα β) + O α X/2), α β using β/α = α 2, and logφ 2p ) = logf 2p /F p ) = log α p + β p ) = p log α + O α X/2). 8

19 Thus, m N expit log Φ m ) = + exp it logα β))) X/4 p X/2 expitp log α) + O Xα X/2). Since #N is twice the number of primes in [X/4,X/2], we now have the estimate cost log Φ m ) = R expit log Φ m ) expit log Φ m ) m N m N m N #N 2 + exp it logα β)) + o). II. Similarly, for an odd prime q < log X, we consider the set N q of all integers of the form pq or 2pq, where p [ X 4q, X 2q ] is prime. We have α pq β pq ) )α β) logφ pq ) = log α p β p )α q β q ) = pq p)log α + logα β) log α q β q ) + O α 2p) and α pq + β pq ) logφ 2pq ) = log α p + β p )α q + β q ) = pq p)log α log α q + β q ) + O α 2p). Then the same reasoning as above shows that cost log Φ m ) m N q #N q 2 α it q + exp + β q )) logα β) + it log α q β q + o). Now suppose that X is large enough that there exists a prime q in the interval [ 6 7 log Y ] 2log α, log Y 2log α 9

20 for all Y X /0, which is guaranteed by the prime number theorem. Choose q in such a way that it lies inside this interval with Y = t X /3 ; by our assumptions on t, we have Y X /5 X /3 > X /0. It follows that and thus Y α 2q Y 6/7 X /3 = t Y t α 2q t Y 6/7 = t /7 X 2/7 X /7. Using β/α = α 2, we thus have α it q exp + β q )) log α q β q X /3. 2) and Now consider the two expressions A = + exp it log α β)) 2 A 2 = α 2 it q + exp + β q )) log α β) + it log α q β q. Trivially, A,A 2, and the estimate 2) now shows that either A C X 2/3 or A 2 C X 2/3 for some positive constant C. In the first case, we obtain the estimate m N cost log Φ m )) #N C X 2/3 )#N In the second case we have, analogously, = C X 2/3 #N X/3 log X. m N q cost log Φ m )) X/3 log 2 X. In either case, the proof of the lemma is completed. Making use of this lemma, it is now easy to estimate the remaining part of the integral: r+ir Dz)x z r+ir 2πi zz + ) dz Dr)xr exp C 2 r /5 ) r+ir 7/5 zz + ) dz r+ir 7/5 log r exp C 2 r /5 )Dr)x r 20

21 for some positive constant C 2. Putting all three parts of the integral together, we obtain Gx) = n x n G n ) = Dr)xr r ) x 2 + Or /5 ) πa = exp 2 ) Alog x + Olog x) ǫ/2 ) for any fixed ǫ > 0. The quantity #G x) that we are actually interested in can be estimated in terms of Gx) as follows: trivially, On the other hand, however, we have #G x) #G x) Gx). ) n ) log x xlog x n xlog x n G = ) Gxlog x) = exp 2 ) Alog x + Olog x) ǫ/2 ), log x which proves Theorem 2 and thus also completes the proof of Theorem. We remark that it is possible to obtain an asymptotic formula for Gx) and also for #G x)) by further studying the behavior of Cs) near s = 0). Acknowledgments. F. L. and S. W. thank Arnold Knopfmacher for useful discussions. Part of this work was done during a visit of F. L. and S. W. at the School of Mathematics of the University of the Witwatersrand in February of 200. They thank the people of this institution for their hospitality. During the preparation of this paper, F. L. was also supported in part by grants SEP-CONACyT and PAPIIT and C. P. was supported in part by NSF grant DMS S. W. was supported financially by the National Research Foundation of South Africa under grant number References [] Y. Bilu, G. Hanrot, and P. M. Voutier. Existence of primitive divisors of Lucas and Lehmer numbers. J. Reine Angew. Math., 539:75 22, 200. With an appendix by M. Mignotte. 2

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