Prime divisors of Lucas sequences and a conjecture of Ska lba

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1 Prime divisors of Lucas sequences and a conjecture of Ska lba Florian Luca, Instituto de Matemáticas Universidad Nacional Autónoma de México C.P , Morelia, Michoacán, México fluca@matmor.unam.mx Pantelimon Stănică Auburn University Montgomery Department of Mathematics Montgomery, AL , USA pstanica@mail.aum.edu October 3, 2005 Abstract Ihis paper, we give some heuristics suggesting that if (u n ) n 0 is the Lucas sequence given by u n = (a n )/(a ), where a > is an integer, then ω(u n ) ( + o()) log n log log n holds for almost all positive integers n. Introduction If n is a positive integer, we write ω(n) and Ω(n) for the number of distinct prime factors of n and total prime factors of n; i.e, including multiplicities, 2000 Mathematics Subject Classification: B39,N37,N60 2 Keywords: Lucas sequences, prime numbers, divisibility.

2 ihe latter case. In what follows, for a real number x > we write log x for the natural logarithm of x. Hasse proved in [5] that the set of primes p dividing 2 n + for some positive integer n has relative density 7/24. Inspired by Hasse s result, Ska lba proved the following results in [7]. Theorem. Let p be a prime. If ord p (2) p 0.8, then p divides a number of the form 2 a + 2 b + for some positive integers a and b. Here, ord p (2) stands for the multiplicative order of 2 modulo p. Theorem 2. If Ω(2 m ) < log m/ log 3, thehere exists a prime divisor q of 2 m such that q is not a divisor of 2 a + 2 b + for any positive integers a and b. We point out that the inequality in Theorem 2 in Ska lba s paper [7] should be strict, since otherwise the assertion is not true. Moreover, he proposed two conjectures: Conjecture.. (i) The number of primes p x that are divisors of some number of the form 2 a + 2 b + is ( + o())x/log x as x. (ii) There are infinitely many primes q such that q does not divide any number of the form 2 a + 2 b +. Regarding (i) above, we point out that a result of Pappalardi (see Theorem 2.3 in [6]) implies that ord p (2) > p 0.8 holds for almost all primes p under the Generalized Riemann Hypothesis, which, via Theorem, supports (i) above. We cannot comment on (ii) above, but ihis paper we look at the condition Ω(2 m ) < log m/ log 3, which, via Theorem 2, would support (ii) above. In [2], Bugeaud et al., proved that for the Fibonacci sequence (F n ) n 0 given by F 0 = 0, F = and F n+2 = F n+ + F n for all n 0, if ω(f n ) 2, then either n =, 2, 4, 8, 2, or n = l, 2l, l 2 for some odd prime l. Clearly, these are only necessary conditions for F o have at most two distinct prime factors but not sufficient. They also showed that the inequality ω(f n ) (log n) log 2+o() holds for almost all positive integers n, and offered an heuristic to support that the inequality ω(f n ) log n holds for all composite positive integers n. Here and in what follows, we use the Vinogradov symbols, and and the Landau symbols O and o with their usual meanings. 2

3 We recall that A B, B A and A = O(B) are all equivalent and mean that A < cb holds with some constant c, while A B means that both A B and B A hold. The constants implied by such symbols may depend on our data a, ε, etc. Throughout, a property holds for almost all natural numbers if it holds for a set of asymptotic density. Acknowledgements. The authors are grateful to the referee whose suggestions have led to an improvement ihe presentation of this paper. During the preparation of this paper, F. L. was supported in part by projects PAPIIT 04505, SEP-CONACyT and a Guggenheim Fellowship. 2 The Results Let a > be an integer. Put u n = an for n = 0,,.... Ihis paper, we a offer the following conjecture, which complements the heuristics made in [2] and, if true, suggests that the positive integers m which fulfill the hypothesis of Theorem 2 are not typical ones. Conjecture 2.. The inequality ω(u n ) ( + o()) log n log log n holds for almost all positive integers n. The same conjecture can be made for the sequence (u n ) n 0 replaced by any nondegenerate Lucas sequence. In what follows, we offer an heuristic in support of the above conjecture. Let ε > 0 be fixed. Let n be a positive integer from a set of asymptotic density one. Let p > p 2 >... > p t be all the prime factors of n ihe interval I n = [ log n, exp ( log n log log n We shall assume that n fulfills various conditions such as: (i) If p > log n is a prime factor of n, then p n. )]. () (ii) There do not exist primes q > p > log n dividing n such that q (mod p). 3

4 In particular, p i n (p i n and p 2 i n) for all i =, 2,..., t, and there do not exist i < j such that p i (mod p j ). For a positive integer m we write P (m) for the largest prime factor of m. Let d(n) be the largest divisor of n such that P (d(n)) < log n, and put n = n/d(n). Define m 0 = n and m i = n/(p...p i ) for i =,..., t. Consider the following finite sequence: v i = u mi /u mi, i =, 2,..., t. (2) We observe that v i = (a p i i )/(a i ), where a i = a n/p...p i. We also observe that v i and v j are coprime if i j. Indeed, assume that i < j and that there exists a prime q dividing v i and v j. Since j > i, we have that v j u mi, therefore ( ) umi q, u mi. u mi It is well-knowhat the above greatest common divisor divides m i. Thus, q m i n. However, since q u mi, it follows that there exists a unique minimal divisor d of n such that q u d. If d =, we then get q (a ), which is impossible if log n > a, because q n and n is free of primes log n (the case log n a need not be treated as there are only finitely many positive integers n satisfying that inequality). Thus, d > is a divisor of m i, and q is a primitive prime factor of u d. It is then well-knowhat q (mod d) (see [3]). Since d >, there exists a prime factor p of d. Clearly, p n. Hence, p (q ), contradicting (ii). It is knowhat the probability that a typical positive integer m is prime is / log m, and that a typical positive integer m has k distinct prime factors is (log log m) k (k )! log m. We now make the following heuristic: Heuristic 2.2. With the above notations, we suppose that ω(v i ) = k i happens with the probability (log log v i ) k i (k i )! log v i, (3) that this is uniform ihe k i s, and that these probabilities are independent for i =,..., t and uniformly in our range for t. 4

5 Under all these assumptions, the probability that u n has at most K prime factors will be S(K) = n We will also assume that (iii) t > ( ε/2) log log n. (iv) d(n) < n ε/4 ; (v) p i < n ε/4 ; t k +...+k t K Under these assumptions, we shall show that: (log log v i ) k i (k i )! log v i. (4) Theorem 2.3. If K < ( ε)t log n, thehe series S(K) converges. Theorem 2.3 has the following corollary. Corollary 2.4. Heuristic 2.2 implies that the inequality holds for all n satisfying (i) (v). ω(u n ) ( 2ε) log n log log n To end, we prove the following proposition. Proposition 2.5. Let ε > 0 be fixed. Thehe set of positive integers n satisfying (i) (v) has asymptotic density one. Clearly, letting ε to tend to zero, we get that Corollary 2.4 and Proposition 2.5 lead to the conclusiohat Heuristic 2.2 implies Conjecture Proofs Proof of Theorem 2.3. Fix k,..., k t, with t k i = k. Obviously, k i. Since a p i i v i a p i i, one sees immediately that log v i p i log a i m i 5 n p p i. (5)

6 Thus, t (log log v i ) k i (k i )! log v i = t (log (n/(p p i ))) k i (k i )! n/(p p i ) (p p i ) t (log m i ) k i. (k i )! Using the above inequality, we obtain t k + +k t=k (log log v i ) k i (k i )! log v i k + +k t=k (p p i ) t (log m i ) k i (k i )! t (p p i ) ( ) t k t (k t)! (log m i ) k i k,..., k t k + +k t=k ( t t ) = (p k t p i ) (k t)! log m i ( ( t t )) = (p k t p i ) (k t)! log m i t = (p ( ( p i ) k t log (k t)! t (p. p i ))) Moreover, from Stirling s formula and the fact that k i, we obtain s(n, K) = k +...+k t K k K (k t)! (p p i ) t (log log v i ) k i (k i )! log v i (p p i ) 0 j K t ( ( log ( e log (p p i ) ) Q t (p p i ) j j. )) k t (6) 6

7 As in [2], if y is fixed, thehe function x (ey/x) x is increasing for x < y. Thus, if we assume that ( ) K t + c 0 log (p, (7) p i ) where c 0 = c 0 (ε) < is a positive constant to be chosen later depending on ε, then estimate (6) leads to s(n, K) (p ( ) ( ) p i ) n c0 log Q e t t p p i log p. p i c 0 we get Furthermore, denoting S(K) = s(n, k) n= m(n) = n= log m(n) m(n) Now note that, by (iv) and (v), It now follows easily that if (p p i ), ( e c 0 ) c0 log m(n) = ( ) t n m(n) n ( ε/2)t. p... p t n= (8) log m(n) m(n) c 0 log(e/c 0 ). (9) ( ε/2)( c 0 log(e/c 0 ))t >, (0) thehe series (9) converges, and by (iii) it is clear that for fixed ε and c 0, the above inequality (0) holds for all but finitely many n. Finally, to conclude, it remains to check that if K satisfies the inequality from the hypothesis of Theorem 2.3, it then satisfies inequality (7), as well. But clearly, the double inequality t + c 0 log(m(n)) > t( ε/2)c 0 log n > t( ε) log n 7

8 holds if we choose c 0 (ε) to be ihe interval ( ) ε ε/2,. Proof of Corollary 2.4. Theorem 2.3 together with (v) shows that if K < ( ε)t log n < ( ε)( ε/2) log n log log n, thehe series S(K) converges. Since ( ε)( ε/2) > 2ε, it follows that the series S(K) converges when K < ( 2ε) log n log log n as well. Heuristic 2.2 now completes the proof. Proof of Proposition 2.5. Let A be the set of positive integers satisfying (i) (v). For a positive real number x, we let A(x) = A [, x]. It suffices to show that #A(x) = ( + o())x. Let B (x) = {n x : p 2 n for some p > log x}. Let n B (x). There exists a prime p > log x such that p 2 n. For fixed p, the number of such positive integers n is x/p 2. Hence, #B (x) p log x x p 2 x log x = o(x). () Let B 2 (x) = {n x : pq n for some primes q > p > log x with p q }. Let n B 2 (x). There exist primes q > p > log x such that pq n and p (q ). For fixed p and q, the number of such positive integers n is x/pq. Hence, #B 2 (x) x pq p log x q<x q (mod p) x p>log x p x log log x q<x q (mod p) p>log x q pφ(p) x log log x = o(x), (2) log x 8

9 where ihe above estimates we used the known fact that q<x q (mod p) q log log x, φ(p) and that this estimate is uniform in 2 p x (see, for example, Lemma in [] or bound (3.) in [4]). Now put B 3 (x) = {n x/ log x}. Obviously, #B 3 (x) x log x = o(x). (3) From now on, we consider only those n x not in 3 B i (x). It is clear that such integers satisfy (i) and (ii). Put y = log x. Let f(s) = exp (log s/ log log s) and put z = f(x/ log x) and z 2 = f(x). The function f(s) is increasing for s > s 0 = e e. Thus, if x > x 0 is sufficiently large, then the inequalities log n y < z < f(n) < z 2 (4) hold for all our n. Thus, by (4), we get [y, z ] I n = [log n, f(n)] [, z 2 ]. (5) For any s > and positive integer m, we write ω s (m) and Ω s (m) for the number of distinct prime factors of m which are s, and the total number of prime factors of m which are s, respectively. By the well-known Turán- Kubilius estimates (see [9], for example), we have g s (n) log log s 2 = O(x log log s), (6) n x where g {Ω, ω}. Further, the above estimates are uniform in e e < s x. We now put B 4 (x) = {n x : Ω y (n) (ε/4) log log x}, and B 5 (x) = {n x : ω z (n) ( ε/4) log log x}. 9

10 Using estimates (6) with (g, s) = (Ω, y) and (ω, z ), together with the fact that log log z = ( + o()) log log x, we immediately get that and #B 4 (x) x log log log x (log log x) 2 = o(x), (7) #B 5 (x) x log log x Assume now that n 5 B i (x). Then, = o(x). (8) t ω z (n) Ω y (n) > ( ε/2) log log x ( ε/2) log log n, so n satisfies (iii) (here, t is the number of distinct prime factors of n in I n, where I n is given as in ()). Furthermore, for large x we also have d(n) (log n) Ωy(n) exp ( (ε/4)(log log x) 2) ( ) ε x < < n ε, log x therefore n satisfies (iv) as well. It remains to deal with condition (v). For this, we note that if n does not fulfill condition (v), then n has a divisor t p i n ε/4 > ( ) ε/4 x > x ε/8 log x whose largest prime factor is z 2, by () and (5). Fix such a divisor d. Thehe number of positive integers n x which are multiples of d is x/d. Thus, writing B 6 (x) for the set of n 5 B i (x) which do not fulfill (v), we get that #B 6 (x) x d. (9) Let x ε/8 <d<x P (d)<z 2 u = log(x ε/8 )/ log z 2 = (ε/8) log log x. It is known (see, for example, Chapter III of [8]), that x ε/8 <d<x P (d)<z 2 d ρ(u) log x, (20) 0

11 where ρ is the Dickman function. Since ρ(u) = u (+o())u as u, we get, by estimates (9) and (20), that #B 6 (x) xρ(u) log x = x log x exp ( ( + o())(ε/8) log log x log log log x) = o(x). (2) Thus, we conclude that the complement of 6 B i (x) consists of positive integers n x satisfying (i) (v), and since by (), (2), (3), (7), (8) and (2), we have that # ( 6 B i (x) ) 6 #B i (x) = o(x), the conclusion of the proposition follows. References [] N. L. Bassily, I. Kátai and M. Wijsmuller, Ohe prime power divisors of the iterates of the Euler-φ function, Publ. Math. Debrecen 55 (999), [2] Y. Bugeaud, F. Luca, M. Mignotte and S. Siksek, On Fibonacci numbers with few prime divisors, Proc. Japan Acad. Sci. Ser. A 8 (2005), [3] R. D. Carmichael, Ohe numerical factors of arithmetical forms α n ±β n, Ann. Math. (2) 5 (93), [4] P. Erdős, A. Granville, C. Pomerance and C. Spiro, Ohe normal behavior of the iterates of some arithmetic functions, Analytic Number Theory, Birkhäuser, Boston, 990, [5] H. Hasse, Über die Dichte der Primzahlen p, für die eine vorgegebene ganzrationale Zahl a 0 von gerader bzw. ungerader Ordnung mod p ist, Math. Ann. 68 (966), [6] F. Pappalardi Ohe Order of Finitely Generated Subgroups of Q (mod p) and Divisors of p, J. Number Theory 57 (996),

12 [7] M. Ska lba, Two conjectures on primes dividing 2 a + 2 b +, Elem. Math. 59 (2004), [8] G. Tenenbaum, Introductioo analytic and probabilistic number theory, Cambridge U. Press, 995. [9] P. Turán, On a Theorem of Hardy and Ramanujan, J. London Math. Soc. 9 (934),