On numbers n dividing the n th term of a linear recurrence
Juan Jose Alba Gonzalez, Florian Luca, Carl Pomerance, Igor Shparlinski
aa r X i v : . [ m a t h . N T ] F e b On numbers n dividing the n th term of alinear recurrence Juan Jos´e Alba Gonz´alez
Instituto de Matem´aticas,Universidad Nacional Autonoma de M´exico,C.P. 58089, Morelia, Michoac´an, M´exico [email protected]
Florian Luca
Instituto de Matem´aticas,Universidad Nacional Autonoma de M´exico,C.P. 58089, Morelia, Michoac´an, M´exico [email protected]
Carl Pomerance
Mathematics Department, Dartmouth College,Hanover, NH 03755, USA [email protected]
Igor E. Shparlinski
Dept. of Computing, Macquarie UniversitySydney, NSW 2109, Australia [email protected]
Abstract
Here, we give upper and lower bounds on the count of positiveintegers n ≤ x dividing the n th term of a nondegenerate linearlyrecurrent sequence with simple roots. Introduction
Let { u n } n ≥ be a linear recurrence sequence of integers satisfying a homoge-neous linear recurrence relation u n + k = a u n + k − + · · · + a k − u n +1 + a k u n , for n = 0 , , . . . , (1)where a , . . . , a k are integers with a k = 0.In this paper, we study the set of indices n which divide the correspondingterm u n ; that is, the set: N u := { n ≥ n | u n } . But first, some background on linear recurrence sequences.To the recurrence (1), we associate its characteristic polynomial f u ( X ) := X k − a X k − − · · · − a k − X − a k = m Y i =1 ( X − α i ) σ i ∈ Z [ X ] , where α , . . . , α m ∈ C are the distinct roots of f u ( X ) with multiplicities σ , . . . , σ m , respectively. It is then well-known that the general term of therecurrence can be expressed as u n = m X i =1 A i ( n ) α ni , for n = 0 , , . . . , (2)where A i ( X ) are polynomials of degrees at most σ i − i = 1 , . . . , m , withcoefficients in K := Q [ α , . . . , α m ]. We refer to [6] for this and other knownfacts about linear recurrence sequences.For upper bounds on the distribution of N u , the case of a linear recurrencewith multiple roots already can pose problems (but see below). For example,the sequence of general term u n = n n for all n ≥ f u ( X ) = ( X − shows that N u may contain all the positiveintegers. So, we look at the case when f u ( X ) has only simple roots. In thiscase, the relation (2) becomes u n = k X i =1 A i α ni , for n = 0 , , . . . . (3)2ere, A , . . . , A k are constants in K . We may assume that none of themis zero, since otherwise, a little bit of Galois theory shows that the integersequence { u n } n ≥ satisfies a linear recurrence of a smaller order.We remark in passing that there is no real obstruction in reducing to thecase of the simple roots. Indeed, let D ∈ N be a common denominator of allthe coefficients of all the polynomials A i ( X ) for i = 1 , . . . , m . That is, thecoefficients of each DA i are algebraic integers. Then Du n = m X i =1 DA i (0) α ni + m X i =1 D ( A i ( n ) − A i (0)) α ni . If n ∈ N u , then n | Du n . Since certainly n divides the algebraic integer m X i =1 D ( A i ( n ) − A i (0)) α ni , it follows that n divides P mi =1 DA i (0) α ni . If this is identically zero (i.e., A i (0) = 0 for all i = 1 , . . . , m ), then we are in an instance similar to theinstance of the sequence of general term u n = n n for all n ≥ N u contains at least a positive proportion of all thepositive integers (namely, all n coprime to D ). Otherwise, we may put w n = m X i =0 DA i (0) α ni for n = 0 , , . . . . A bit of Galois theory shows that w n is an integer for all n ≥
0, and thesequence { w n } n ≥ satisfies a linear recurrence relation of order ℓ := { ≤ i ≤ m : A i (0) = 0 } with integer coefficients, which furthermore has onlysimple roots. Hence, N u ⊆ N w , and therefore there is indeed no loss ofgenerality when proving upper bounds in dealing only with linear recurrentsequences with distinct roots.We put ∆ u := Y ≤ i 6∈ { , ± } . Its characteristic polynomial is f u ( X ) = X − a . It is easy to see that in this case N u ( x ) = O ((log x ) ω ( | a | ) ),where for an integer m ≥ ω ( m ) for the number of distinct primefactors of m . So, from now on, we assume that k ≥ u n = 2 n − n ≥ f u ( X ) = ( X − X − N u , so that the Prime NumberTheorem and estimates for the distribution of pseudoprimes show that itis possible for the estimate N u ( x ) = (1 + o (1)) x/ log x to hold as x → ∞ .However, we show that N u ( x ) cannot have a larger order of magnitude. Theorem 1. For each k ≥ , there is a positive constant c ( k ) dependingonly on k such that if the characteristic polynomial of a nondegenerate lin-ear recurrence sequence { u n } n ≥ of order k has only simple roots, then the A pseudoprime is a composite number n which divides 2 n − 2. The paper [14] showsthat there are few odd pseudoprimes compared with primes, while [10] (unpublished) doesthe same for even pseudoprimes. stimate N u ( x ) ≤ c ( k ) x log x holds for x sufficiently large. In case of a Lucas sequence, we have a better bound. To simplify no-tations, for a posititive integer ℓ we define log ℓ x iteratively as log x :=max { log x, } and for ℓ > ℓ x := log (log ℓ − x ). When ℓ = 1 we omitthe index but understand that all logarithms are ≥ 1. Let L ( x ) := exp (cid:16)p log x log x (cid:17) . (6) Theorem 2. Assume that { u n } n ≥ is a Lucas sequence. Then the inequality N u ( x ) ≤ xL ( x ) o (1) (7) holds as x → ∞ . It follows from a result of Somer [17, Theorem 8] that N u is finite if andonly if ∆ u = 1, and in this case N u = { } .For Lucas sequences with a = ± 1, we also have a rather strong lowerbound on N u ( x ). Our result depends on the current knowledge of thedistribution of y -smooth values of p − p , that is values of p − y . We use Π( x, y ) to denote thenumber of primes p ≤ x for which p − y -smooth. Since the numbers p − p prime are likely to behave as “random” integers from thepoint of view of the size of their prime factors, it seems reasonable to expectthat behavior of Π( x, y ) resembles the behavior of the counting function forsmooth integers. We record this in a very relaxed form of the assumptionthat for some fixed real v ≥ y v , y ) ≥ y v + o (1) (8)as y → ∞ . In fact, a general result from [3, Theorem 1.2] implies that (8)holds with any v ∈ [1 , / Theorem 3. There is a set of integers L such that L ⊂ N u for any Lucassequence u with a = ± , and such that if (8) holds with some v > , wehave N u ( x ) ≥ L ( x ) ≥ x ϑ + o (1) as x → ∞ , where ϑ := 1 − /v. 5n particular, since as we have already mentioned, any value of v < / ϑ = 1 / . Furthermore, since (8) is expected to hold for any v > 1, it is very likely thatthe bound of Theorem 3 holds with ϑ = 1.Finally, we record a lower bound on N ( x ) when a = ± u = 1. Theorem 4. Let { u n } n ≥ be any Lucas sequence with ∆ u = 1 . Then thereexist positive constants c and x depending on the sequence such that for x > x we have N u ( x ) > exp( c (log x ) ) . Throughout the paper, we use x for a large positive real number. We usethe Landau symbol O and the Vinogradov symbol ≪ with the usual meaningin analytic number theory. The constants implied by them may depend onthe sequence { u n } n ≥ , or only on k . We use c , c , . . . for positive constantswhich may depend on { u n } n ≥ . We label such constants increasingly as theyappear in the paper. As in the proof of [6, Theorem 2.6], put D u ( x , . . . , x k ) := det( α x j i ) ≤ i,j ≤ k . For a prime number p not dividing a k , let T u ( p ) be the maximal nonnegativeinteger T with the property that p ∤ Y ≤ x ,...,x k ≤ T max { , | N K/ Q ( D u (0 , x , . . . , x k )) |} . It is known that such T exists. In the above relation, x , . . . , x k are integersin [1 , T ], and for an element α of K we use N K/ Q ( α ) for the norm of α over Q . Since α , . . . , α k are algebraic integers in K , it follows that the numbers N K/ Q ( D u (0 , x , . . . , x k )) are integers.Observe that T u ( p ) = 0 if and only if k = 2 and p is a divisor of ∆ u =( α − α ) . 6ore can be said in the case when { u n } n ≥ is a Lucas sequence. In thiscase, we have | N K/ Q ( D u (0 , x )) | = | α x − α x | = | ∆ u | | u x | , x = 1 , , . . . . Thus, if p does not divide the discriminant ∆ u = ( α − α ) = a + 4 a ofthe sequence { u n } n ≥ , then T u ( p ) + 1 is in fact the minimal positive integer ℓ such that p | u ℓ . This is sometimes called the index of appearance of p in { u n } n ≥ and is denoted by z u ( p ). The index of appearance z u ( m ) canbe defined for composite integers m in the same way as above, namely asthe minimal positive integer ℓ such that m | u ℓ . This exists for all positiveintegers m coprime to a , and has the important property that m | u n if andonly if z u ( m ) | n . For any γ ∈ (0 , P u,γ = { p : T u ( p ) < p γ } . Lemma 1. For x γ , y ≥ , the estimates { p : T u ( p ) ≤ y } ≪ y k log y , P u,γ ( x ) ≪ x kγ γ log x hold, where the implied constants depend only on the sequence { u n } n ≥ .Proof. It is clear that the second inequality follows immediately from thefirst with y = x γ , so we prove only the first one. Suppose that T u ( p ) ≤ y .In particular, there exists a choice of integers x , . . . , x k all in [1 , y + 1] suchthat p | max { , | N K/ Q ( D u (0 , x , . . . , x k )) |} . This argument shows that Y T u ( p ) ≤ y p | Y ≤ x ,...,x k ≤ y +1 max { , | N K/ Q ( D u (0 , x , . . . , x k )) |} . (9)There are at most ( y + 1) k − = O ( y k − ) possibilities for the ( k − x , . . . , x k ). For each one of these ( k − | N K/ Q ( D u (0 , x , . . . , x k )) | = exp( O ( y )) . Hence, the right hand side in (9) is of size exp( O ( y k )). Taking logarithms inthe inequality implied by (9), we get that X T u ( p ) ≤ y log p = O ( y k ) . 7f there are a total of n primes involved in this sum and if p i denotes the i thprime, then n X i =1 log p i = O ( y k ) , so that in the language of the prime number theorem, θ ( p n ) ≪ y k . It followsthat p n ≪ y k and n ≪ y k / ( k log y ), which is what we wanted to prove.The parameter T u ( p ) is useful to bound the number of solutions n ∈ [1 , x ] of the congruence u n ≡ p ). For example, the following is [6,Theorem 5.11]. Lemma 2. There exists a constant c ( k ) depending only on k with the follow-ing property. Suppose that { u n } n ≥ is a linearly recurrent sequence of order k satisfying recurrence (1) . Suppose that p is a prime coprime to a k ∆ u . As-sume that there exists a positive integer s such that u s is coprime to p . Thenfor any real x ≥ the number of solutions R ( x, p ) of the congruence u n ≡ p ) with ≤ n ≤ x satisfies the bound R u ( x, p ) ≤ c ( k ) (cid:18) xT u ( p ) + 1 (cid:19) . When { u n } n ≥ is a Lucas sequence, we put Q u,γ = { p : z u ( p ) ≤ p γ } . The remarks preceding Lemma 1 show that Q u,γ ( x ) = P u,γ ( x ) + O (1).Hence, Lemma 1 implies the following result. Lemma 3. For x > , the estimate Q u,γ ( x ) ≪ x γ log x holds, where the implied constant depends only on the sequence { u n } n ≥ . As usual, we denote by Ψ( x, y ) the number of integers n ≤ x with P ( n ) ≤ y . By [2, Corollary to Theorem 3.1], we have the following well-known result. Lemma 4. For x ≥ y > , the estimate Ψ( x, y ) = x exp( − (1 + o (1)) v log v ) uniformly in the range y > (log x ) as long as v → ∞ , where v := (log x ) / (log y ) . The proof of Theorem 1 We assume that x is large. We split the set N u ( x ) into several subsets. Let P ( n ) be the largest prime factor of n and let y := x / log log x . Let N ( x ) := { n ≤ x : P ( n ) ≤ y } ; N ( x ) := { n ≤ x : n 6∈ N ( x ) and P ( n ) ∈ P u, / ( k +1) } ; N ( x ) := N ( x ) \ (cid:0) ∪ i =1 N i ( x ) (cid:1) . We now bound the cardinalities of each one of the above sets.For N ( x ), by Lemma 4, we obtain N ( x ) = Ψ( x, y ) = x exp( − (1 + o (1)) v log v ) = o (cid:18) x log x (cid:19) (10)as x → ∞ , where v = log x log y = log log x. Suppose now that n ∈ N ( x ). Then n = pm , where p = P ( n ) ≥ max { y, P ( m ) } . In particular, p ≤ x/m therefore m ≤ x/y . Since we alsohave p ∈ P u, / ( k +1) ( x/m ), Lemma 1 implies that the number of such primes p ≤ x/m is O (cid:16) ( x/m ) k/ ( k +1) (cid:17) , where the implied constant depends on thesequence { u n } n ≥ . Summing up the above inequality over all possible valuesof m ≤ x/y , we get N ( x ) ≤ x k/ ( k +1) X ≤ m ≤ x/y m k/ ( k +1) ≪ x k/ ( k +1) Z x/y dtt k/ ( k +1) = (( k + 1) x k/ ( k +1) ) t / ( k +1) (cid:12)(cid:12)(cid:12) x/y ≪ xy / ( k +1) . (11)Now let n ∈ N ( x ). As previously, we write n = pm , where p = P ( n ) > y .We assume that x (hence, y ) is sufficiently large. Thus, m ≤ x/p < x/y .Since n ∈ N u , we have that n | u n , therefore p | u n . Furthermore, T u ( p ) ≥ p / ( k +1) . We fix p and count the number of possibilities for m . To this end,let { w ℓ } ℓ ≥ be the sequence defined as w ℓ = u pℓ for all ℓ ≥ 0. This is alinearly recurrent sequence of order k . We would like to apply Lemma 2 toit to bound the number of solutions to the congruence w m ≡ p ) , where 1 ≤ m ≤ x/p. 9f the conditions of Lemma 2 are satisfied, then this number denoted by R w ( x/p, p ) satisfies R w ( x/p, p ) ≤ c ( k ) (cid:18) xpT w ( p ) + 1 (cid:19) . Let us check the conditions of Lemma 2. Note first that if α , . . . , α k are thecharacteristic roots of { u n } n ≥ , then α p , . . . , α pk are the characteristic rootsof { w ℓ } ℓ ≥ . Hence, f w ( X ) = k Y i =1 ( X − α pi ) . In particular, the term a w,k corresponding to the recurrence { w ℓ } ℓ ≥ satisfies a w,k = a pk assuming that y > 2. Thus, assuming further that y > | a k | , wethen have that p does not divide a k , therefore p does not divide a w,k either.Next, note that ∆ w = Y ≤ i We divide the numbers n ∈ N u ( x ) into several classes:(i) N ( x ) := { n ∈ N u ( x ) : P ( n ) ≤ L ( x ) / } ;(ii) N ( x ) := { n ∈ N u ( x ) : P ( n ) ≥ L ( x ) } ;(iii) N ( x ) := N u ( x ) \ ( N ( x ) ∪ N ( x )).It follows from Lemma 4 that N ( x ) ≤ Ψ( x, L ( x ) / ) ≤ xL ( x ) o (1) as x → ∞ .For n ∈ N u and p | n , we have n ≡ p ) and n ≡ z u ( p )).For p not dividing the discriminant of the characteristic polynomial for u (andso for p sufficiently large), we have z u ( p ) | p ± 1, so that gcd( p, z u ( p )) = 1.Thus, the conditions n ∈ N u , p | n , and p sufficiently large jointly force n ≡ pz u ( p )). Hence, if p is sufficiently large, the number of n ∈ N u ( x ) with P ( n ) = p is at most Ψ( x/pz u ( p ) , p ) ≤ x/pz u ( p ).Thus, for large x , N ( x ) ≤ X p>L ( x ) xpz u ( p ) = X p>L ( x ) z u ( p ) ≤ L ( x ) xpz u ( p ) + X p>L ( x ) z u ( p ) >L ( x ) xpz u ( p ) . The first sum on the right has, by Lemma 1, at most L ( x ) terms for x large,each term being smaller than x/L ( x ) , so the sum is bounded by x/L ( x ).The second sum on the right has terms smaller than x/pL ( x ) and the sumof 1 /p is of magnitude log log x , so the contribution here is x/L ( x ) o (1) as x → ∞ . Thus, N ( x ) ≤ x/L ( x ) o (1) as x → ∞ .For any nonnegative integer j , let I j := [2 j , j +1 ). For N , we cover I :=[ L ( x ) / , L ( x ) ) by these dyadic intervals, and we define a j via 2 j = L ( x ) a j .We shall assume the variable j runs over just those integers with I j notdisjoint from I . For any integer k , define P j,k as the set of primes p ∈ I j with z u ( p ) ∈ I k . Note that, by Lemma 1, we have P j,k ≪ k . We have N ( x ) ≤ X j X k X p ∈P j,k X n ∈N u ( x ) P ( n )= p ≤ X j X k X p ∈P j,k Ψ (cid:18) xpz u ( p ) , p (cid:19) ≤ X j X k X p ∈P j,k xpz u ( p ) L ( x ) / a j + o (1) , x → ∞ , where we have used Lemma 4 for the last estimate. For k > j/ X p ∈P j,k pz u ( p ) ≤ − k X p ∈ I j p ≤ − k for x large. For k ≤ j/ 2, we use the estimate X p ∈P j,k pz u ( p ) ≪ k j k = 2 k − j , since there are at most order of magnitude 4 k such primes, as noted before.Thus, X k X p ∈P j,k pz u ( p ) = X k>j/ X p ∈P j,k pz u ( p ) + X k ≤ j/ X p ∈P j,k pz u ( p ) ≪ − j/ = L ( x ) − a j / . We conclude that N ( x ) ≤ X j xL ( x ) a j / / a j + o (1) as x → ∞ . Since the minimum value of t/ / (2 t ) for t > t = 1, weconclude that N ( x ) ≤ x/L ( x ) o (1) as x → ∞ . With our prior estimatesfor N ( x ) and N ( x ), this completes our proof.It is possible that using the methods of [5] and [7] a stronger estimatecan be made. Since a = ± 1, it is easy to see that the sequence u is purely periodic moduloany integer m . So, the index of appearance z u ( m ) defined in Section 2 existsfor all positive integers m . Further, by examining the explicit formula (5)one can see that for any prime power q = p k we have z u ( p k ) | z u ( p ) p k − . (14)In fact this is known in much wider generality.13ow, for any real number y ≥ M y := lcm [ m : m ≤ y ] . We say that a positive integer n is Lucas special if it is of the form n =2 sM y for some y ≥ s such thatgcd( s, M y ) = 1 and for every prime p | s we have p − | M y . Let L denotethe set of Lucas special numbers.We now show that L ⊂ N u for any Lucas sequence u with a = ± 1. Tosee this it suffices to show for any n = 2 sM y ∈ L and for any prime power q | n , we have z u ( q ) | n . This is easy for q | s , since then q = p is prime andeither z u ( p ) = p (in the case p | ∆ u ) or z u ( p ) | p ± p − | M y , we have z u ( p ) | n in either case.If q | M y , we consider the cases of odd and even q separately. • When q is odd, we have q | M y so q ≤ y . Write q = p k with p prime,so that (14) implies z u ( q ) | ( p − p k − , p k or ( p + 1) p k − . We have p k − ≤ y and if p + 1 ≤ y , then z ( q ) | M y . The only case not covered is p + 1 > y (so p ∈ ( y − , y ]), k = 1, z u ( p ) = p + 1. Write p + 1 = 2 j m where m is odd. Then 2 j | M y and m | M y , so p + 1 | M y . Thus, inall cases, z u ( q ) | M y so z u ( q ) | n . • When q = 2 k is a power of 2 with q | M y , then since z u (2) ∈ { , } ,we see from (14) that either z u (2 k ) | k or z u (2 k ) | · k − . Since y ≥ z u ( q ) | M y .We now use the method of Erd˝os [4] to show that the set L is ratherlarge. For this we take y := log x log log x and z := y v . We say that q is a proper prime power if q = ℓ k for a prime ℓ and an integer k ≥ P as the set of primes p such that: • p ∈ [ y + 1 , z ]; • p − y -smooth; • p − q > y .14ote that if q is a proper prime power and q | p − 1, then q | p ± q is even, in which case q/ | p ± 1. Since trivially there are only O ( t / ) proper prime powers q ≤ t , there are only O ( zy − / ) primes p ≤ z for which p − q > y . Thus, recallingthe assumption (8), we obtain P ≥ Π( z, y ) − y + O ( zy − / ) = z o (1) , provided that x → ∞ .It is also obvious that for any squarefree positive integer s composed outof primes p ∈ P , the integer n = 2 sM y is Lucas special.We now take the set L v ( x ) of all such Lucas special integers n = 2 sM y ,where s is composed out of r := (cid:22) log x − y log z (cid:23) distinct primes p ∈ P . Since by the prime number theorem the estimate M y = exp((1 + o (1)) y ) holds as x → ∞ , we see that for sufficiently large x we have n ≤ x for every n ∈ L v ( x ).For the cardinality of L v ( x ) we have L v ( x ) ≥ (cid:18) P r (cid:19) ≥ (cid:18) P r (cid:19) r . Since r = ( v − + o (1)) log x log log x and P r = (log x ) v − o (1) as x → ∞ , we obtain L v ( x ) ≥ x − /v + o (1) as x → ∞ . Noting that L v ( x ) ⊂L ( x ) concludes the proof. Since ∆ u ≡ , u = 0 , 1, it follows that | ∆ u | > 1. Let r besome prime factor of ∆ u . Then r k ∈ N u for all k ≥ k be a large positive integer and look at u r k +4 . By Bilu,Hanrot and Voutier’s primitive divisor theorem (see [1]), u n has a primitive15rime factor for all n ≥ 31. Recall that a primitive prime factor of u n isa prime factor p of u n which does not divide ∆ u u m for any positive integer m < n . Such a primitive prime factor p always satisfies p ≡ ± n ).Since there are at most 5 values of k ≥ r k ≤ 30 for the sameinteger r > 1, and since u m | u n if m | n , we conclude that u r k +4 has atleast τ ( r k +4 ) − k distinct prime factors p = r , where τ ( m ) is the numberof divisors of the positive integer m . Let them be p < · · · < p k . Assumethat | α | ≥ | α | . For large n , we have that | α | n/ < | u n | < | α | n (see [6,Theorem 2.3]). If β , . . . , β k are nonnegative exponents such that β i ≤ log( x/r k +4 ) k log p i , then r k +4 p β · · · p β k k ≤ x is in N u (see [11, Page 210]), so it is counted by N u ( x ). Hence, N u ( x ) ≥ k Y i =1 (cid:18)(cid:22) log( x/r k +4 ) k log p i (cid:23) + 1 (cid:19) ≥ (cid:18) log( x/r k +4 ) k (cid:19) k Q ki =1 log p i ≥ (cid:18) log( x/r k +4 )2 r k +4 log | α | (cid:19) k , where the last inequality follows from the mean value inequality k Y i =1 log p i ≤ k k X i =1 log p i ! k ≤ (cid:18) log( | u r k +4 | ) k (cid:19) k < (cid:18) r k +4 log | α | + log 2 k (cid:19) k < (cid:18) r k +4 log | α | k (cid:19) k , for k ≥ 2. In the above, we have also used the fact that | u n | < | α | n holdsfor all n ≥ n := r k +4 . Let c := 2 log | α | . The above lowerbound is N u ( x ) ≥ (cid:18) log xr k +4 c + O (cid:18) kr k (cid:19)(cid:19) k = (cid:18) log xr k +4 c (cid:19) k (cid:18) O (cid:18) k log x (cid:19)(cid:19) ≫ (cid:18) log xr k +4 c (cid:19) k k = o ( p log x ) , (15)as x → ∞ , which is now assumed. So, it suffices to look at (cid:18) log xr k +4 c (cid:19) k = exp ( k log(log x/c ) − k ( k + 4) log r ) . Let A := log(log x/c ). In order to maximize the function f ( x ) := xA − x ( x + 4) log r , we take its derivative and set it equal to zero to get A − x log r − r = 0, therefore x = ( A − r ) / (2 log r ) = A/ (2 log r ) − k := ⌊ A/ (2 log r ) − ⌋ (so that (15) is satisfied), we get that f ( k ) = f ( x ) + O ( f ′ ( x )) = A / (4 log r ) + O ( A ), hence N u ( x ) ≥ exp (cid:18) (log(log x/c )) r + O (log log x ) (cid:19) = exp (cid:18) (log log x ) r + O (log log x ) (cid:19) , which implies the desired conclusion with any constant c < / (4 log r ). We end with a result showing that it is quite possible for N u ( x ) to be largeunder quite mild conditions. Observe that the sequence u n = 2 n − u = 0. Here is a more general version of this fact. Proposition 1. Let k ≥ and { u n } n ≥ be a linearly recurrent sequence oforder k satisfying recurrence (1) . Assume that there exists a positive integer n coprime to a k such that u n = 0 . Then N u ( x ) ≫ x/ log x, where the implied constant depends on the sequence { u n } n ≥ .Proof. Since n is coprime to a k , it follows that { u n } n ≥ is purely periodicmodulo n . Let t n be this period. Now, let R u be the set of primes p ≡ t n ) such that f u ( X ) splits into linear factors modulo p . Alternatively, R u is the set of primes p such that the polynomial f u ( X )( X t n − 1) splits17nto linear factors modulo p . The set of such primes has a positive densityby the Chebotarev density theorem. We claim that S u ⊆ N u , (16)where S u := { pn : p ∈ R u and p > n | ∆ u |} . The above inclusion implies the desired bound since then N u ( x ) ≥ R u ( x/n ) + O (1) ≫ x/ log x. So, let us suppose that p > n | ∆ u | is in R u . Then p ≡ t n ),therefore p = 1 + λt n for some positive integer λ . Thus, pn = n + λn t n and since { u n } n ≥ is purely periodic with period t n modulo n , we get that u pn = u n + λn t n ≡ u n ≡ n ) . (17)Next, observe that since the polynomial f u ( X ) factors in linear factors mod-ulo p , we get that α pi ≡ α i (mod p ) for all i = 1 , . . . , k . In particular, α pn i ≡ α n i (mod p ) for all i = 1 , . . . , k . Since the denominators of the co-efficient A i , i = 1 , . . . , k , in (3) are divisors of ∆ u and p > | ∆ u | , it followsthat such denominators are invertible modulo p , therefore A i α pn i ≡ A i α n i (mod p ) for all i = 1 , . . . , k . Summing up these congruences for i = 1 , . . . , k ,we get u pn = k X i =1 A i α pn i ≡ k X i =1 A i α n i ≡ u n ≡ p ) . (18)From the congruences (17) and (18), we get that both p and n divide u pn ,and since p is coprime to n , we get that pn | u pn . This completes the proofof the inclusion (16) and of the proposition.The condition that n is coprime to a k is not always necessary. The con-clusion of Propositon 1 may hold without this condition like in the exampleof the sequence of general term u n = 10 n − n − · n − n ≥ , for which we can take n = 2. Observe that k = 4, f u ( X ) = ( X − X − X − X − , n is not coprime to a = − p | u p holds for all primes p ≥ 11. We do not give further details.Let M u ( x ) be the set of integers n ≤ x with n | u n and n is not of theform pn , where p is prime and u n = 0. It may be that in the situation ofTheorem 1, we can get a smaller upper bound for M u ( x ) than for N u ( x ).We can show this in a special case. Proposition 2. Assume that { u n } n ≥ is a linearly recurrent sequence oforder k whose characteristic polynomial splits into distinct linear factorsin Z [ X ] . There is a positive constant c ( k ) depending on k such that forall sufficiently large x (depending on the sequence u ), we have M u ( x ) ≤ x/L ( x ) c ( k ) .Proof. Let y = L ( x ). We partition M u ( x ) into the following subsets: M ( x ) := { n ∈ M u ( x ) : P ( n ) ≤ y } ; M ( x ) := { n ∈ M u ( x ) : there is a prime p | n, p > y, pT u ( p ) ≤ kx } ; M ( x ) := M u ( x ) \ ( M ( x ) ∪ M ( x )) . As in the proof of Theorem 2, we see that Lemma 4 implies that M ( x ) ≤ x/L ( x ) / o (1) as x → ∞ .As in the proof of Theorem 1, M ( x ) ≪ X y
6∈ P u, / ( k +1) , respec-tively. Lemma 1 shows that P u, / ( k +1) ( t ) ≪ t k/ ( k +1) / log t . Thus, X y kx . Using asbefore the notation t p for the period of u modulo p , as well as the fact that T u ( p ) ≤ kt p and t p | p − f u splits in linear factors over Q [ X ]), wehave kx < pT u ( p ) ≤ kpt p ≤ kp , so that p > √ x . Thus, n can have at most one prime factor p with pT u ( p ) >kx . So, if n ∈ M ( x ), we may assume that n = mp where p > √ x > m , and P ( m ) ≤ y . Further, we may assume that u m = 0. Since p | u pm and t p | p − p | u m . Now the number of prime factors of u m is O ( m ). Since thenumber of n ∈ M ( x ) with such a prime p | n is O ( x/ ( pT u ( p )) + 1) = O (1),we have M ( x ) ≪ X m< √ xP ( m ) ≤ y m ≤ √ x Ψ( √ x, y ) = xL ( x ) / o (1) as x → ∞ , using Lemma 4.We conclude that the result holds with c ( k ) := min { / , / ( k + 2) } ,say.Finally, we note that for a given non-constant polynomial g ( X ) ∈ Z [ X ]one can consider the more general set N u,g := { n ≥ g ( n ) | u n } . We fix some real y < x / and note that by the Brun sieve (see [9, Theo-rem 2.3]), there are at most N ≪ x (cid:18) log y log x (cid:19) (19)values of n ≤ x such that g ( n ) does not have a prime divisor in the interval[ y, x / ]. We also note that for a prime p not dividing the content of g , thedivisibility p | g ( n ) puts n in at most deg g arithmetic progressions. Thus,using Lemma 2 as it was used in the proof of Theorem 1, the number of other n ≤ x with g ( n ) | u n can be estimated as N ≤ X p ∈ [ y,x / ] X n ≤ xp | g ( n ) p | u n ≪ X p ∈ [ y,x / ] (cid:18) xpT u ( p ) + 1 (cid:19) ≪ x X p ∈ [ y,x / ] pT u ( p ) + O ( x / ) . γ ∈ (0 , 1) and the trivial estimate T u ( p ) ≫ log p , wederive X p ∈ [ z, z ] pT u ( p ) ≤ z X p ∈ [ z, z ] T u ( p ) ≪ z (cid:18) z kγ (log z ) + z − γ log z (cid:19) . Taking γ to satisfy z γ = ( z log z ) / ( k +1) , we obtain 1 z X p ∈ [ z, z ] pT u ( p ) ≪ z − / ( k +1) (log z ) − ( k +2) / ( k +1) . Summing over dyadic intervals, we now have X p ∈ [ y,x / ] pT u ( p ) ≪ y − / ( k +1) (log y ) − ( k +2) / ( k +1) . Therefore, N ≪ xy − / ( k +1) (log y ) − ( k +2) / ( k +1) + x / . (20)Taking, for example, y := (log x ) k +1 , we obtain from (19) and (20) the esti-mate N u,g ( x ) ≤ N + N ≪ x (cid:18) log log x log x (cid:19) . 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