Let $F$ and $G$ be linear recurrences over a number field $\mathbb{K}$, and
let $\mathfrak{R}$ be a finitely generated subring of $\mathbb{K}$.
Furthermore, let $\mathcal{N}$ be the set of positive integers $n$ such that
$G(n) \neq 0$ and $F(n) / G(n) \in \mathfrak{R}$. Under mild hypothesis,
Corvaja and Zannier proved that $\mathcal{N}$ has zero asymptotic density. We
prove that $\#(\mathcal{N} \cap [1, x]) \ll x \cdot (\log\log x / \log x)^h$
for all $x \geq 3$, where $h$ is a positive integer that can be computed in
terms of $F$ and $G$. Assuming the Hardy-Littlewood $k$-tuple conjecture, our
result is optimal except for the term $\log \log x$

Sanna, Carlo

English

arXiv

Institutional Research Information System University of Turin

Distribution of integral values for the ratio of two linear recurrences

Let F and G be linear recurrences over a number field K, and
let R be a finitely generated subring of K. Furthermore, let
N be the set of positive integers n such that G(n) = 0 and
F(n)/G(n) ∈ R. Under mild hypothesis, Corvaja and Zannier
proved that N has zero asymptotic density. We prove that
#(N ∩ [1, x])  x · (log log x/ log x)h for all x ≥ 3, where
h is a positive integer that can be computed in terms of F
and G. Assuming the Hardy–Littlewood k-tuple conjecture,
our result is optimal except for the term log log 

PORTO@iris (Publications Open Repository TOrino - Politecnico di Torino)

04 August 2020POLITECNICO DI TORINORepository ISTITUZIONALEDistribution of integral values for the ratio of two linear recurrences / Sanna, Carlo. - In: JOURNAL OF NUMBERTHEORY. - ISSN 0022-314X. - STAMPA. - 180(2017), pp. 195-207.OriginalDistribution of integral values for the ratio of two linear recurrencesPublisher:PublishedDOI:10.1016/j.jnt.2017.04.015Terms of use:openAccessPublisher copyright(Article begins on next page)This article is made available under terms and conditions as specified in the  corresponding bibliographic description inthe repositoryAvailability:This version is available at: 11583/2722597 since: 2020-05-03T09:53:34ZAcademic Press IncorporatedDISTRIBUTION OF INTEGRAL VALUES FOR THE RATIO OFTWO LINEAR RECURRENCESCARLO SANNAAbstract. Let F and G be linear recurrences over a number field K, and let R be a finitelygenerated subring of K. Furthermore, letN be the set of positive integers n such that G(n) 6= 0and F (n)/G(n) ∈ R. Under mild hypothesis, Corvaja and Zannier proved that N has zeroasymptotic density. We prove that #(N∩[1, x]) x·(log log x/ log x)h for all x ≥ 3, where h isa positive integer that can be computed in terms of F and G. Assuming the Hardy–Littlewoodk-tuple conjecture, our result is optimal except for the term log log x.1. IntroductionA sequence of complex numbers F (n)n∈N is called a linear recurrence if there exist somec0, . . . , ck−1 ∈ C (k ≥ 1), with c0 6= 0, such thatF (n+ k) =k−1∑j=0cjF (n+ j),for all n ∈ N. In turn, this is equivalent to an (unique) expressionF (n) =r∑i=1fi(n)αni ,for all n ∈ N, where f1, . . . , fr ∈ C[X] are nonzero polynomials and α1, . . . , αr ∈ C∗ are all thedistinct roots of the polynomialXk − ck−1Xk−1 − · · · − c1X − c0.Classically, α1, . . . , αr and k are called the roots and the order of F , respectively. Furthermore,F is said to be nondegenerate if none the ratios αi/αj (i 6= j) is a root of unity, and F is said tobe simple if all the f1, . . . , fr are constant. We refer the reader to [6, Ch. 1–8] for the generaltheory of linear recurrences.Hereafter, let F and G be linear recurrences and let R be a finitely generated subring of C.Assume also that the roots of F and G together generate a multiplicative torsion-free group.This “torsion-free” hypothesis is not a loss of generality. Indeed, if the group generated bythe roots of F and G has torsion order q, then for each r = 0, 1, . . . , q − 1 the roots of thelinear recurrences Fr(n) = F (qn + r) and Gr(n) = G(qn + r) generate a torsion-free group.Therefore, all the results in the following can be extended just by partitioning N into thearithmetic progressions of modulo q and by studying each pair of linear recurrences Fr, Grseparately. Finally, define the following set of natural numbersN := {n ∈ N : G(n) 6= 0, F (n)/G(n) ∈ R} .Regarding the condition G(n) 6= 0, note that, by the “torsion-free” hypothesis, G(n) is nonde-generate and hence the Skolem–Mahler–Lech Theorem [6, Theorem 2.1] implies that G(n) = 0only for finitely many n ∈ N. In the sequel, we shall tacitly disregard such integers.Divisibility properties of linear recurrences have been studied by several authors. A clas-sical result, conjectured by Pisot and proved by van der Poorten, is the Hadamard-quotient2010 Mathematics Subject Classification. Primary: 11B37 Secondary: 11A07, 11N25.Key words and phrases. linear recurrence; divisibility.12 CARLO SANNATheorem, which states that if N contains all sufficiently large integers, then F/G is itself alinear recurrence [13, 19].Corvaja and Zannier [5, Theorem 2] gave the following wide extension of the Hadamard-quotient Theorem (see also [4] for a previous weaker result by the same authors).Theorem 1.1. If N is infinite, then there exists a nonzero polynomial P ∈ C[X] such thatboth the sequences n 7→ P (n)F (n)/G(n) and n 7→ G(n)/P (n) are linear recurrences.The proof of Theorem 1.1 makes use of the Schmidt’s Subspace Theorem. We refer thereader to [3] for a survey on several applications of the Schmidt’s Subspace Theorem inNumber Theory.Let K be a number field. For the sake of simplicity, from now on we shall assume that R ⊆ Kand that F and G have coefficients and values in K. Corvaja and Zannier [5, Corollary 2] provedalso the following theorem about the set N .Theorem 1.2. If F/G is not a linear recurrence, then N has zero asymptotic density.We recall that a set of natural numbers S has zero asymptotic density if #S(x)/x → 0, asx→ +∞, where we define S(x) := S ∩ [1, x] for all x ≥ 1.Corvaja and Zannier also suggested [5, Remark p. 450] that their proof of Theorem 1.2 couldbe adapted to show that if F/G is not a linear recurrence then(1) #N (x) x(log x)δ,for any δ < 1 and for all sufficiently large x > 1, where the implied constant depends on K.In our main result we obtain a more precise upper bound than (1). Before state it, wemention some special cases of the problem of bounding #N (x) that have already been studied.Alba Gonza´lez, Luca, Pomerance, and Shparlinski [1, Theorem 1.1] proved the following:Theorem 1.3. If F is a simple nondegenerate linear recurrence over the integers, r ≥ 2,G(n) = n, and R = Z, then#N (x) xlog x,for all sufficiently large x > 1, where the implied constant depends only on r.For G(n) = n and R = Z, a still better upper bound can be given if F is a Lucas sequence,that is, F (0) = 0, F (1) = 1, and F (n + 2) = aF (n + 1) + bF (n), for all n ∈ N and somefixed integers a and b. In such a case the arithmetic properties of N were first investigated byAndre´-Jeannin [2] and Somer [16, 17]. Luca and Tron [11] studied the case in which F is thesequence of Fibonacci numbers (a = b = 1) and Sanna [15], using some results on the p-adicvaluation of Lucas sequences [14], generalized Luca and Tron’s result to the following upperbound.Theorem 1.4. If F is a nondegenerate Lucas sequences, G(n) = n, and R = Z, then#N (x) ≤ x1−( 12+o(1)) log log log xlog log x ,as x→ +∞, where the o(1) depends on F .Now we state the main result of this paper.Theorem 1.5. If F/G is not a linear recurrence, then#N (x) x ·(log log xlog x)h,for all x ≥ 3, where h is a positive integer that can be computed in terms of F and G, whilethe implied constant depends on F and G.The computation of h will be clear in the proof of Theorem 1.5. In particular, it leadsimmediately to the following corollary.DISTRIBUTION OF INTEGRAL VALUES FOR THE RATIO OF TWO LINEAR RECURRENCES 3Corollary 1.1. If F/G is not a linear recurrence, G ∈ Z[X], and gcd(G, f1, . . . , fr) = 1, thenh can be taken as the number of irreducible factors of G in Z[X] (counted without multiplicity).Except for the term log log x, Corollary 1.1 should be optimal. Indeed, pick a positive integerh and an admissible h-tuple h = (n1, . . . , nh), that is, n1 < · · · < nh are positive integers suchthat for each prime number p there exists a residue class modulo p which does not intersect{n1, . . . , nh}. Assuming Hardy–Littlewood h-tuple conjecture [7, p. 61], we have that thenumber Th(x) of positive integers n ≤ x such that n + n1, . . . , n + nh are all prime numberssatisfiesTh(x) ∼ Ch · x(log x)h,as x→ +∞, where Ch > 0 depends on h. Therefore, taking F (n) = (2n+n1−2) · · · (2n+nh−2)and G(n) = (n+ n1) · · · (n+ nh), we obtain#N (x) ≥ Th(x) x(log x)h,for all sufficiently large x > 1.Notation. Hereafter, the letter p always denotes a prime number. We employ the Landau–Bachmann “Big Oh” and “little oh” notations O and o, as well as the associated Vinogradovsymbols  and , with their usual meanings. If A  B and A  B, we write A  B. Anydependence of implied constants is explicitly stated or indicated with subscripts.2. PreliminariesFirst, we need a quantitative form of a result due to Kronecker [10] (see also [18, p. 32]),which states that the average number of zeros modulo p of a nonconstant polynomial f ∈ Z[X]is equal to the number of irreducible factors of f in Z[X].Theorem 2.1. Given a nonconstant polynomial f ∈ Z[X], for each prime number p let ηf (p)be the number of zeros of f modulo p. Then∑p≤xηf (p) · log pp= h log x+Of (1),for all x ≥ 1, where h is the number of irreducible factors of f in Z[X].Proof. It is enough to prove the claim for irreducible f . Let G be the Galois group of f over Q.By a quantitative version of the Chebotarev’s density theorem [12, Ch. 2, Theorem 7.2], thenumber of primes p ≤ x such that the irreducible factors of f modulo p have degrees d1, . . . , dsispiG(d1, . . . , ds)#G · Li(x) +Of(xexp(C√log x)),for all x > 1, where Li(x) is the logarithmic integral function, C > 0 is a constant dependingon f , and piG(d1, . . . , ds) is the number of g ∈ G that have cycle decomposition with lengthsd1, . . . , ds when regarded as permutations of the roots of f . Furthermore, G acts transitivelyon the roots of f , since f is irreducible, hence∑g∈G#Xg = #G,by Burnside’s lemma, where Xg is the set of roots of f which are fixed by g. Hence,∑p≤xηf (p) = Li(x) +Of(xexp(C√log x)),and the desired result follows by partial summation. The following lemma [5, Lemma A.2] regards the minimum of the multiplicative orders ofsome fixed algebraic numbers modulo a prime ideal.4 CARLO SANNALemma 2.2. Let β1, . . . , βs ∈ K such that none of them is zero or a root of unity. Then,for all x ≥ 1, the number of prime numbers p ≤ x such that some βi has order less than p1/4modulo some prime ideal of OK lying above p is O(x1/2), where the implied constant dependsonly on β1, . . . , βs.Now we state a technical lemma about the cardinality of a sieved set of integers.Lemma 2.3. For each prime number p, let Ωp ( {0, 1, . . . , p− 1} be a set of residues modulop, and denote by Ω the whole family of Ωp’s. Suppose that there exist constants c, h > 0 suchthat #Ωp ≤ c for each prime number p and(2)∑p≤x#Ωp · log pp= h log x+O(1),for all x > 1. Then we have# {n ≤ x : (n mod p) /∈ Ωp, ∀p ∈ ]y, z]} Ω,δ1,δ2 x ·(log ylog x)h,for all δ1, δ2 > 0, x > 1, 2 ≤ y ≤ (log x)δ1, and z ≥ xδ2.Proof. All the constants in this proof, included the implied ones, may depend on Ω, δ1, δ2.Clearly, we can assume δ2 ≤ 1/2. By the large sieve inequality [8, Theorem 7.14], we have(3) # {n ≤ x : (n mod p) /∈ Ωp, ∀p ∈ ]y, z]}  x ·∑m≤wgy(m)−1 ,where w := xδ2 and gy is the multiplicative arithmetic function supported on squarefree num-bers with all prime factors > y and such thatgy(p) =#Ωpp−#Ωp ,for any prime number p > y.For sufficiently large x, we have y ≤ w, and it follows from (2) that−(A+ h log y) + h logw ≤∑p≤wgy(p) log p ≤ B + h logw,for some constants A,B > 0. Then from [9, Theorem 0.4.1] we obtain that∑m≤wgy(m) =S(w)Γ(h+ 1)· (logw)h ·(1 +O(log ylogw)),where Γ is the Euler’s Gamma function andS(w) :=∏p≤w(1 + gy(p))(1− 1p)h.In particular, since y ≤ (log x)δ1 , for sufficiently large x we get that(4)∑m≤wgy(m) S(w) · (logw)h.Now from (2) it follows easily that∏p≤t(1− #Ωpp)−1 (log t)h,for all t ≥ 2. Hence, also thanks to Mertens’ third theorem [8, p. 34, Eq. 2.16], we have(5) S(w) =∏p≤w(1− #Ωpp)−1(1− 1p)h/ ∏p≤y(1− #Ωpp)−1 1(log y)h.DISTRIBUTION OF INTEGRAL VALUES FOR THE RATIO OF TWO LINEAR RECURRENCES 5Putting together (3), (4), and (5), and recalling that w = xδ2 , the desired result follows. Finally, we need a lemma about the number of zeros of a simple linear recurrence in a finitefield of q elements Fq (see also [6, Theorem 5.10] for a more precise result).Lemma 2.4. Let c1, . . . , cr, a1, . . . , ar ∈ F∗q, and let N be the minimum of the orders of theai/aj (i 6= j) in F∗q. (If r = 1 then pick an arbitrary positive integer N .) Then the number ofintegers m ∈ [0, q − 1] such that(6)r∑i=1ciami = 0is at most 5(q − 1)N−1/2r−2.Proof. For r = 1 the claim is obvious since (6) never holds, hence we may assume r ≥ 2. In [5,Proposition A.1] it is stated and proved that for prime q the number of integers m ∈ [1, q − 1]satisfying (6) is at most 4(q − 1)N−1/2r−2 , and the same proof works also for not necessarilyprime q. Thus the claim follows, since 4(q − 1)N−1/2r−2 + 1 ≤ 5(q − 1)N−1/2r−2 . 3. Proof of Theorem 1.5The first part of the proof proceeds similarly to the proof of Theorem 1.2. If N is finite,then the claim is trivial, hence we suppose that N is infinite. Then, by Theorem 1.1 it followsthat F/G = H/P , for some linear recurrence H and some polynomial P . As a consequence,without loss of generality, we shall assume that G is a polynomial.Let S be a finite set of absolute values of K containing all the archimedean ones. Write OSfor the ring of S-integers of K, that is, the set of all α ∈ K such that |α|v ≤ 1 for all v /∈ S.Enlarging K and S we may assume that α1, . . . , αr are S-units, f1, . . . , fr, G ∈ OS [X], andR ⊆ OS .Since F/G is not a linear recurrence, it follows that G does not divide all the f1, . . . , fr.Moreover, factoring out the greatest common divisor (G, f1, . . . , fr) we can even assume that(G, f1, . . . , fr) = 1 and thatG is nonconstant. In particular, (G(n), f1(n), . . . , fr(n)) is boundedand, enlarging S, we may assume that it is an S-unit for all n ∈ N.Let NK(α) denotes the norm of α ∈ K over Q. It is easy to prove that there exist a positiveinteger g and a nonconstant polynomial G˜ ∈ Z[X] such that NK(G(n)) = G˜(n)/g for all n ∈ N.Let h be the number of irreducible factors of G˜ in Z[X]. Again by enlarging S, we may assumethat g is an S-unit.Let P be the set of all prime numbers p which do not make G˜ vanish identically modulo p,such that pOK has no prime ideal factor piv with v ∈ S, and such that the minimum order ofthe αi/αj (i 6= j) modulo any prime ideal above p is at least p1/4. Furthermore, let us defineΩp :={` ∈ {0, . . . , p− 1} : G˜(`) ≡ 0 (mod p)},for any p ∈ P, and Ωp := ∅ for any prime number p /∈ P.Let x ≥ 3, y := (log x)2rh, and z := x1/(d+1), where d := [K : Q]. We split N (x) into twosubsets:N1 := {n ∈ N (x) : (n mod p) /∈ Ωp, ∀p ∈ ]y, z]} ,N2 := N \N1.First, we give an upper bound for #N1. Hereafter, all the implied constants may depend onF and G. Clearly, #Ωp ( {0, 1, . . . , p − 1} and #Ωp ≤ deg(G˜) for all prime number p, whilefrom Theorem 2.1 and Lemma 2.2 it follows that∑p≤x#Ωp · log pp= h log x+O(1).6 CARLO SANNATherefore, applying Lemma 2.3, we obtain#N1  x ·(log ylog x)h(log log xlog x)h.Now we give an upper bound for #N2. If n ∈ N2 then there exist p ∈ P ∩ ]y, z] and ` ∈ Ωpsuch that n ≡ ` (mod p). In particular, p divides NK(G(`)) in OS and, since pOK has no primeideal factor piv with v ∈ S, it follows that there exists some prime ideal pi of OS lying abovep and dividing G(`). Let Fq := OS/pi, so that q is a power of p. Write n = ` + mp, for someinteger m ≥ 0. Since pi divides G(n) and F (n)/G(n) ∈ OS , we have that F (n) is divisible bypi too. As a consequence, we obtain that(7)r∑i=1fi(`)α`i(αpi)m ≡ r∑i=1fi(n)αni ≡ F (n) ≡ 0 (mod pi).Note that f1(`), . . . , fr(`) cannot be all equal to zero modulo pi, since pi divides G(`) and(G(`), f1(`), . . . , fr(`)) is an S-unit. Note also that the minimum order of the αpi /αpj (i 6= j)modulo pi is equal to the minimum order of the αi/αj (i 6= j) modulo pi, since (p, q − 1) = 1.Therefore, we can apply Lemma 2.4 to the congruence (7). The positive integer r maydecrease, and N can the taken ≥ p1/4, in light of the definition of P. It follows that thenumber of possible values of m modulo q − 1 is at most 5(q − 1)p−1/2r . Consequently, thenumber of possible values of n ≤ x is at most5(q − 1)p−1/2r(xp(q − 1) + 1) xp1+1/2r,since p(q − 1) < pd+1 ≤ zd+1 ≤ x. Hence, we have#N2 ∑p∈P ∩ ]y,z]xp1+1/2r∫ +∞ydtt1+1/2r xy1/2r=x(log x)h.In conclusion,#N (x) = #N1 + #N2  x ·(log log xlog x)has claimed.Acknowledgements. The author thanks Umberto Zannier for a fruitful conversation onTheorem 1.2.References1. J. J. Alba Gonza´lez, F. Luca, C. Pomerance, and I. E. Shparlinski, On numbers n dividing the nth term ofa linear recurrence, Proc. Edinb. Math. Soc. (2) 55 (2012), no. 2, 271–289.2. R. Andre´-Jeannin, Divisibility of generalized Fibonacci and Lucas numbers by their subscripts, FibonacciQuart. 29 (1991), no. 4, 364–366.3. Y. F. Bilu, The many faces of the subspace theorem [after Adamczewski, Bugeaud, Corvaja, Zannier. . .],Aste´risque (2008), no. 317, Exp. No. 967, vii, 1–38, Se´minaire Bourbaki. Vol. 2006/2007.4. P. 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Berlin (1880), 155–162.DISTRIBUTION OF INTEGRAL VALUES FOR THE RATIO OF TWO LINEAR RECURRENCES 711. F. Luca and E. Tron, The distribution of self-Fibonacci divisors, Advances in the theory of numbers, FieldsInst. Commun., vol. 77, pp. 149–158.12. M. R. Murty and V. K. Murty, Non-vanishing of L-functions and applications, Modern Birkha¨user Classics,Birkha¨user/Springer Basel AG, Basel, 1997.13. R. Rumely, Notes on van der Poorten’s proof of the Hadamard quotient theorem. I, II, Se´minaire de The´oriedes Nombres, Paris 1986–87, Progr. Math., vol. 75, Birkha¨user Boston, Boston, MA, 1988, pp. 349–382,383–409.14. C. Sanna, The p-adic valuation of Lucas sequences, Fibonacci Quart. 54 (2016), no. 2, 118–124.15. C. Sanna, On numbers n dividing the nth term of a Lucas sequence, Int. J. Number Theory 13 (2017), no. 3,725–734.16. L. Somer, Divisibility of terms in Lucas sequences by their subscripts, Applications of Fibonacci numbers,Vol. 5 (St. Andrews, 1992), Kluwer Acad. Publ., Dordrecht, 1993, pp. 515–525.17. L. Somer, Divisibility of terms in Lucas sequences of the second kind by their subscripts, Applications ofFibonacci numbers, Vol. 6 (Pullman, WA, 1994), Kluwer Acad. Publ., Dordrecht, 1996, pp. 473–486.18. P. Stevenhagen and H. W. Lenstra, Jr., Chebotare¨v and his density theorem, Math. Intelligencer 18 (1996),no. 2, 26–37.19. A. J. van der Poorten, Solution de la conjecture de Pisot sur le quotient de Hadamard de deux fractionsrationnelles, C. R. Acad. Sci. Paris Se´r. I Math. 306 (1988), no. 3, 97–102.Universita` degli Studi di Torino, Department of Mathematics, Turin, ItalyE-mail address: carlo.sanna.dev@gmail.com

Distribution of integral values for the ratio of two linear recurrences

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