Distribution of integral values for the ratio of two linear recurrences

TL;DR

This paper investigates the distribution of integral values of the ratio of two linear recurrences over a number field, providing bounds on their density and assuming conjectures for optimality.

Contribution

It establishes explicit bounds on the count of integers where the ratio of two linear recurrences lies in a finitely generated subring, extending prior zero-density results.

Findings

Bound on the number of such integers: x imes (rac{\u2212 ext{log log x}}{ ext{log x}})^h

Result holds under mild hypotheses and is nearly optimal under Hardy-Littlewood conjecture

Provides a quantitative measure of the distribution of ratios of linear recurrences

Abstract

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 $\log \log x$.

Equations

$$F(n+k)=\sum _{j=0}^{k-1}{c}_{j}F(n+j),$$

$$F(n)=\sum _{i=1}^r{f}_i(n){\alpha }_i^n,$$

$$X^k-c_{k-1}{X}^{k-1}-\cdots -c_{1}X-c_0.$$

$$\mathcal{N}:=\left\{n\in \mathbb{N}:G(n)\ne 0,F(n)/G(n)\in \mathfrak{R}\right\}.$$

$$\#\mathcal{N}(x)\ll \frac{x}{(\log x{)}^{\delta }},$$

$$\#\mathcal{N}(x)\ll \frac{x}{\log x},$$

$$\#\mathcal{N}(x)\le x^{1-\left(\frac{1}{2}+o(1)\right)\frac{\log \log \log x}{\log \log x}},$$

$$\#\mathcal{N}(x){\ll }_{F,G}x\cdot {\left(\frac{\log \log x}{\log x}\right)}^h,$$

$$T_{\mathbf{h}}(x)\sim C_{\mathbf{h}}\cdot \frac{x}{(\log x{)}^h},$$

$$\#\mathcal{N}(x)\ge T_{\mathbf{h}}(x)\gg \frac{x}{(\log x{)}^h},$$

$$\sum _{p\le x}{\eta }_f(p)\cdot \frac{\log p}{p}=h\log x+O_f(1),$$

$${\pi }_C(x)=\frac{\#C}{\#\mathcal{G}}\cdot \mathrm{Li}(x)+O_{\mathbb{L}}\left(\frac{x}{\exp (C\sqrt{\log x})}\right),$$

$$\sum _{g\in \mathcal{G}}\#X^g=\#\mathcal{G},$$

$$\sum _{p\le x}{\eta }_f(p)=\mathrm{Li}(x)+O_{\mathbb{L}}\left(\frac{x}{\exp (C\sqrt{\log x})}\right),$$

$$\sum _{d\mid n}g(n/d){\Lambda }_g(d)=g(n)\log n,$$

$$\sum _{n\le x}{\Lambda }_g(n)=h\log x+O(L)$$

$$\sum _{n\le x}|g(n)|\ll (\log x{)}^h,$$

$$\sum _{n\le x}g(n)=(\log x{)}^h\cdot \left(c_g+O_h\left(\frac{L}{\log x}\right)\right),$$

$$c_g:=\frac{1}{\Gamma (h+1)}\prod _p\left(1+g(p)+g(p^2)+\cdots \right){\left(1-\frac{1}{p}\right)}^h$$

$$\sum _{p\le x}\#{\Omega }_p\cdot \frac{\log p}{p}=h\log x+O(c_2),$$

$$\#\left\{n\le x:(n\mathrm{mod}p)\notin {\Omega }_p,\forall p\in ]y,z]\right\}{\ll }_{c_1,c_2,h,{\delta }_1,{\delta }_2}x\cdot {\left(\frac{\log y}{\log x}\right)}^h,$$

$$\#\left\{n\le x:(n\mathrm{mod}p)\notin {\Omega }_p,\forall p\in ]y,z]\right\}\ll x\cdot {\left(\sum _{m\le w}{g}_y(m)\right)}^{-1},$$

$$g_y(p)=\frac{\#{\Omega }_p}{p-\#{\Omega }_p},$$

$$\sum _{p\le w}{g}_y(p)\log p=h\log w+O(\log y),$$

$$\sum _{n\le w}{\Lambda }_{g_y}(n)=h\log w+O(\log y),$$

$$\prod _{p\le t}{\left(1-\frac{\#{\Omega }_p}{p}\right)}^{-1}\asymp (\log t{)}^h,$$

$$\sum _{n\le w}|g_y(n)|\le \prod _{p\le w}(1+g_y(p))\le \prod _{p\le w}{\left(1-\frac{\#{\Omega }_p}{p}\right)}^{-1}\ll (\log w{)}^h.$$

$$\sum _{n\le w}g(n)=(\log w{)}^h\cdot \left(c_{g_y}+O\left(\frac{\log y}{\log w}\right)\right),$$

$$c_{g_y}=\frac{1}{\Gamma (h+1)}\prod _p(1+g_y(p)){\left(1-\frac{1}{p}\right)}^h.$$

$$c_{g_y}=\frac{1}{\Gamma (h+1)}\prod _p{\left(1-\frac{\#{\Omega }_p}{p}\right)}^{-1}{\left(1-\frac{1}{p}\right)}^h\prod _{p\le y}\left(1-\frac{\#{\Omega }_p}{p}\right)\gg \frac{1}{(\log y{)}^h}.$$

Full text

1. Introduction

A sequence of complex numbers $F(n)_{n\in\mathbb{N}}$ is called a linear recurrence if there exist some $c_{0},\ldots,c_{k-1}\in\mathbb{C}$ ($k\geq 1$), with $c_{0}\neq 0$, such that

[TABLE]

for all $n\in\mathbb{N}$. In turn, this is equivalent to an (unique) expression

[TABLE]

for all $n\in\mathbb{N}$, where $f_{1},\ldots,f_{r}\in\mathbb{C}[X]$ are nonzero polynomials and $\alpha_{1},\ldots,\alpha_{r}\in\mathbb{C}^{*}$ are all the distinct roots of the polynomial

[TABLE]

Classically, $\alpha_{1},\ldots,\alpha_{r}$ and $k$ are called the roots and the order of $F$, respectively. Furthermore, $F$ is said to be nondegenerate if none the ratios $\alpha_{i}/\alpha_{j}$ ($i\neq j$) is a root of unity, and $F$ is said to be simple if all the $f_{1},\ldots,f_{r}$ are constant. We refer the reader to [7, Ch. 1–8] for the general theory of linear recurrences.

Hereafter, let $F$ and $G$ be linear recurrences and let $\mathfrak{R}$ be a finitely generated subring of $\mathbb{C}$. 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 by the roots of $F$ and $G$ has torsion order $q$, then for each $r=0,1,\ldots,q-1$ the roots of the linear recurrences $F_{r}(n)=F(qn+r)$ and $G_{r}(n)=G(qn+r)$ generate a torsion-free group. Therefore, all the results in the following can be extended just by partitioning $\mathbb{N}$ into the arithmetic progressions of modulo $q$ and by studying each pair of linear recurrences $F_{r},G_{r}$ separately. Finally, define the following set of natural numbers

[TABLE]

Regarding the condition $G(n)\neq 0$, note that, by the “torsion-free” hypothesis, $G(n)$ is nondegenerate and hence the Skolem–Mahler–Lech Theorem [7, Theorem 2.1] implies that $G(n)=0$ only for finitely many $n\in\mathbb{N}$. In the sequel, we shall tacitly disregard such integers.

Divisibility properties of linear recurrences have been studied by several authors. A classical result, conjectured by Pisot and proved by van der Poorten, is the Hadamard-quotient Theorem, which states that if $\mathcal{N}$ contains all sufficiently large integers, then $F/G$ is itself a linear recurrence [13, 20].

Corvaja and Zannier [6, Theorem 2] gave the following wide extension of the Hadamard-quotient Theorem (see also [5] for a previous weaker result by the same authors).

Theorem 1.1**.**

If $\mathcal{N}$ is infinite, then there exists a nonzero polynomial $P\in\mathbb{C}[X]$ such that both the sequences $n\mapsto P(n)F(n)/G(n)$ and $n\mapsto G(n)/P(n)$ are linear recurrences.

The proof of Theorem 1.1 makes use of the Schmidt’s Subspace Theorem. We refer the reader to [3] for a survey on several applications of the Schmidt’s Subspace Theorem in Number Theory.

Let $\mathbb{K}$ be a number field. For the sake of simplicity, from now on we shall assume that $\mathfrak{R}\subseteq\mathbb{K}$ and that $F$ and $G$ have coefficients and values in $\mathbb{K}$. Corvaja and Zannier [6, Corollary 2] proved also the following theorem about the set $\mathcal{N}$.

Theorem 1.2**.**

If $F/G$ is not a linear recurrence, then $\mathcal{N}$ has zero asymptotic density.

We recall that a set of natural numbers $\mathcal{S}$ has zero asymptotic density if $\#\mathcal{S}(x)/x\to 0$, as $x\to+\infty$, where we define $\mathcal{S}(x):=\mathcal{S}\cap[1,x]$ for all $x\geq 1$.

Corvaja and Zannier also suggested [6, Remark p. 450] that their proof of Theorem 1.2 could be adapted to show that if $F/G$ is not a linear recurrence then

[TABLE]

for any $\delta<1$ and for all sufficiently large $x>1$, where the implied constant depends on $\mathbb{K}$.

In our main result we obtain a more precise upper bound than (1). Before state it, we mention some special cases of the problem of bounding $\#\mathcal{N}(x)$ that have already been studied.

Alba González, 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, $r\geq 2$, $G(n)=n$, and $\mathcal{R}=\mathbb{Z}$, then

[TABLE]

for all sufficiently large $x>1$, where the implied constant depends only on $r$.

For $G(n)=n$ and $\mathcal{R}=\mathbb{Z}$, 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 $n\in\mathbb{N}$ and some fixed integers $a$ and $b$. In such a case the arithmetic properties of $\mathcal{N}$ were first investigated by André-Jeannin [2] and Somer [17, 18]. Luca and Tron [12] studied the case in which $F$ is the sequence of Fibonacci numbers ($a=b=1$) and Sanna [15], using some results on the $p$-adic valuation of Lucas sequences [14], generalized Luca and Tron’s result to the following upper bound.

Theorem 1.4**.**

If $F$ is a nondegenerate Lucas sequences, $G(n)=n$, and $\mathcal{R}=\mathbb{Z}$, then

[TABLE]

as $x\to+\infty$, 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

[TABLE]

for all $x\geq 3$, where $h$ is a positive integer depending on $F$ and $G$.

Both the positive integer $h$ and the implied constant in the bound of Theorem 1.5 are effectively computable, we give the details in the last section. In particular, we have the following corollary.

Corollary 1.1**.**

If $F/G$ is not a linear recurrence, $G\in\mathbb{Z}[X]$, and $\gcd(G,f_{1},\ldots,f_{r})=1$, then $h$ can be taken as the number of irreducible factors of $G$ in $\mathbb{Z}[X]$ (counted without multiplicity).

Except for the term $\log\log x$, Corollary 1.1 should be optimal. Indeed, pick a positive integer $h$ and an admissible $h$-tuple $\mathbf{h}=(n_{1},\ldots,n_{h})$, that is, $n_{1}<\cdots<n_{h}$ are positive integers such that for each prime number $p$ there exists a residue class modulo $p$ which does not intersect $\{n_{1},\ldots,n_{h}\}$. Assuming Hardy–Littlewood $h$-tuple conjecture [8, p. 61], we have that the number $T_{\mathbf{h}}(x)$ of positive integers $n\leq x$ such that $n+n_{1},\ldots,n+n_{h}$ are all prime numbers satisfies

[TABLE]

as $x\to+\infty$, where $C_{\mathbf{h}}>0$ depends on $\mathbf{h}$. Therefore, taking $F(n)=(2^{n+n_{1}}-2)\cdots(2^{n+n_{h}}-2)$ and $G(n)=(n+n_{1})\cdots(n+n_{h})$, we obtain

[TABLE]

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 Vinogradov symbols $\ll$ and $\gg$, with their usual meanings. If $A\ll B$ and $A\gg B$, we write $A\asymp B$. Any dependence of implied constants is explicitly stated or indicated with subscripts.

2. Preliminaries

First, we need a quantitative form of a result due to Kronecker [11] (see also [19, p. 32]), which states that the average number of zeros modulo $p$ of a nonconstant polynomial $f\in\mathbb{Z}[X]$ is equal to the number of irreducible factors of $f$ in $\mathbb{Z}[X]$.

Lemma 2.1**.**

Given a nonconstant polynomial $f\in\mathbb{Z}[X]$, for each prime number $p$ let $\eta_{f}(p)$ be the number of zeros of $f$ modulo $p$. Then

[TABLE]

for all $x\geq 1$, where $h$ is the number of irreducible factors of $f$ in $\mathbb{Z}[X]$.

Proof.

It is enough to prove the claim for irreducible $f$. Let $\mathbb{L}$ be the splitting field of $f$ over $\mathbb{Q}$ and let $\mathcal{G}:=\operatorname{Gal}(\mathbb{L}/\mathbb{Q})$. For any coniugacy class $C$ of $\mathcal{G}$, let $\pi_{C}(x)$ be the number of primes $p\leq x$ which do not ramified in $\mathbb{L}$ and such that their Frobenius substitutions $\sigma_{p}$ belong to $C$. A quantitative version of the Chebotarev’s density theorem [16, Theorem 3.4] states that

[TABLE]

for $x\to+\infty$, where $\operatorname{Li}(x)$ is the logarithmic integral function and $C>0$ is a constant depending on $\mathbb{L}$. If the elements of $C$ have cycle pattern $d_{1},\ldots,d_{s}$, when regarded as permutations of the roots of $f$, then $\pi_{C}(x)$ is the number of primes $p\leq x$ not dividing the discriminant of $f$ and such that the irreducible factors of $f$ modulo $p$ have degrees $d_{1},\ldots,d_{s}$.

Furthermore, $\mathcal{G}$ acts transitively on the roots of $f$, since $f$ is irreducible, hence

[TABLE]

by Burnside’s lemma, where $X^{g}$ is the set of roots of $f$ which are fixed by $g$. Therefore,

[TABLE]

and the desired result follows by partial summation. ∎

The following lemma [6, Lemma A.2] regards the minimum of the multiplicative orders of some fixed algebraic numbers modulo a prime ideal.

Lemma 2.2**.**

Let $\beta_{1},\ldots,\beta_{s}\in\mathbb{K}$ such that none of them is zero or a root of unity. Then, for all $x\geq 1$, the number of prime numbers $p\leq x$ such that some $\beta_{i}$ has order less than $p^{1/4}$ modulo some prime ideal of $\mathcal{O}_{\mathbb{K}}$ lying above $p$ is $O(x^{1/2})$, where the implied constant depends only on $\beta_{1},\ldots,\beta_{s}$.

Given a multiplicative function $g$, let $\Lambda_{g}$ be its associated von Mangoldt function, that is, the unique arithmetic function satisfying

[TABLE]

for all positive integers $n$ (see [9, p. 17]). It is easy to prove that $\Lambda_{g}$ is supported on prime powers.

Theorem 2.3**.**

Let $g$ be a multiplicative arithmetic function such that

[TABLE]

and

[TABLE]

for all $x\geq 2$, where $h,L>0$ are some constants. Then

[TABLE]

where

[TABLE]

and $\Gamma$ is the Euler’s Gamma function.

Proof.

The proof proceeds exactly as the proof of [9, Theorem 1.1], but using the error term $O(L)$ instead of $O(1)$. ∎

Now we state a technical lemma about the cardinality of a sieved set of integers.

Lemma 2.4**.**

For each prime number $p$, let $\Omega_{p}\subsetneq\{0,1,\ldots,p-1\}$ be a set of residues modulo $p$. Suppose that there exist constants $c_{1},c_{2},h>0$ such that $\#\Omega_{p}\leq c_{1}$ for each prime number $p$ and

[TABLE]

for all $x>1$. Then we have

[TABLE]

for all $\delta_{1},\delta_{2}>0$, $x>1$, $2\leq y\leq(\log x)^{\delta_{1}}$, and $z\geq x^{\delta_{2}}$.

Proof.

All the constants in this proof, included the implied ones, may depend on $c_{1}$, $c_{2}$, $h$, $\delta_{1}$, $\delta_{2}$. Clearly, we can assume $\delta_{2}\leq 1/2$. By the large sieve inequality [9, Theorem 7.14], we have

[TABLE]

where $w:=x^{\delta_{2}}$ and $g_{y}$ is the multiplicative arithmetic function supported on squarefree numbers with all prime factors $>y$ and such that

[TABLE]

for any prime number $p>y$.

For sufficiently large $x$, we have $y\leq w$, and it follows from (4) and $\#\Omega_{p}\leq c_{1}$ that

[TABLE]

which in turn implies that

[TABLE]

since $\Lambda_{g_{y}}$ is supported on prime powers $p^{s}$, with $p>y$, and $\Lambda_{g_{y}}(p^{s})=-(-g_{y}(p))^{s}\log p$.

Furthermore, again from (4) and $\#\Omega_{p}\leq c_{1}$, we have

[TABLE]

for all $t\geq 2$, so that

[TABLE]

At this point, we have proved that (2) and (3) hold with $g=g_{y}$ and $L=\log y$. Therefore, by Theorem 2.3 we have

[TABLE]

where

[TABLE]

Now using (6) we obtain

[TABLE]

Hence, recalling that $y\leq(\log x)^{\delta_{1}}$ and $w=x^{\delta_{2}}$, by (7) and (8) we find that

[TABLE]

Putting together (5) and (9), the desired result follows. ∎

Finally, we need a lemma about the number of zeros of a simple linear recurrence in a finite field of $q$ elements $\mathbb{F}_{q}$.

Lemma 2.5**.**

Let $c_{1},\ldots,c_{r},a_{1},\ldots,a_{r}\in\mathbb{F}_{q}^{*}$, and let $N$ be the minimum of the orders of the $a_{i}/a_{j}$ ($i\neq j$) in $\mathbb{F}_{q}^{*}$. (If $r=1$ then pick an arbitrary positive integer $N$.) Then the number of integers $m\in[0,q-2]$ such that

[TABLE]

is at most $4(q-1)N^{-1/2^{r-2}}$.

Proof.

In [6, Proposition A.1] the claim is stated and proved for prime $q$, but the same proof works also for not necessarily prime $q$. ∎

Note that results stronger than Lemma 2.5 can be obtained using bounds for the number of zeros of sparse polynomials in finite fields (see, e.g., [4, 10]).

3. Proof of Theorem 1.5

The first part of the proof proceeds similarly to the proof of Theorem 1.2. If $\mathcal{N}$ is finite, then the claim is trivial, hence we suppose that $\mathcal{N}$ is infinite. Then, by Theorem 1.1 it follows that $F/G=H/P$, 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 $\mathbb{K}$ containing all the archimedean ones. Write $\mathcal{O}_{S}$ for the ring of $S$-integers of $\mathbb{K}$, that is, the set of all $\alpha\in\mathbb{K}$ such that $|\alpha|_{v}\leq 1$ for all $v\notin S$. Enlarging $\mathbb{K}$ and $S$ we may assume that $\alpha_{1},\ldots,\alpha_{r}$ are $S$-units, $f_{1},\ldots,f_{r},G\in\mathcal{O}_{S}[X]$, and $\mathfrak{R}\subseteq\mathcal{O}_{S}$.

Since $F/G$ is not a linear recurrence, it follows that $G$ does not divide all the $f_{1},\ldots,f_{r}$. Moreover, factoring out the greatest common divisor $(G,f_{1},\ldots,f_{r})$ we can even assume that $(G,f_{1},\ldots,f_{r})=1$ and that $G$ is nonconstant. In particular, $(G(n),f_{1}(n),\ldots,f_{r}(n))$ is bounded and, enlarging $S$, we may assume that it is an $S$-unit for all $n\in\mathbb{N}$.

Let $N_{\mathbb{K}}(\alpha)$ denotes the norm of $\alpha\in\mathbb{K}$ over $\mathbb{Q}$. It is easy to prove that there exist a positive integer $g$ and a nonconstant polynomial $\widetilde{G}\in\mathbb{Z}[X]$ such that $N_{\mathbb{K}}(G(n))=\widetilde{G}(n)/g$ for all $n\in\mathbb{N}$. Let $h$ be the number of irreducible factors of $\widetilde{G}$ in $\mathbb{Z}[X]$. Again by enlarging $S$, we may assume that $g$ is an $S$-unit.

Let $\mathcal{P}$ be the set of all prime numbers $p$ which do not make $\widetilde{G}$ vanish identically modulo $p$, such that $p\mathcal{O}_{\mathbb{K}}$ has no prime ideal factor $\pi_{v}$ with $v\in S$, and such that the minimum order of the $\alpha_{i}/\alpha_{j}$ ($i\neq j$) modulo any prime ideal above $p$ is at least $p^{1/4}$. Furthermore, let us define

[TABLE]

for any $p\in\mathcal{P}$, and $\Omega_{p}:=\varnothing$ for any prime number $p\notin\mathcal{P}$.

Let $x\geq 3$, $y:=(\log x)^{2^{r}h}$, and $z:=x^{1/(d+1)}$, where $d:=[\mathbb{K}:\mathbb{Q}]$. We split $\mathcal{N}(x)$ into two subsets:

[TABLE]

First, we give an upper bound for $\#\mathcal{N}_{1}$. Hereafter, all the implied constants may depend on $F$ and $G$. Clearly, $\#\Omega_{p}\subsetneq\{0,1,\ldots,p-1\}$ and $\#\Omega_{p}\leq\deg(\widetilde{G})$ for all prime number $p$, while from Lemma 2.1 and Lemma 2.2 it follows that

[TABLE]

Therefore, applying Lemma 2.4, we obtain

[TABLE]

Now we give an upper bound for $\#\mathcal{N}_{2}$. If $n\in\mathcal{N}_{2}$ then there exist $p\in\mathcal{P}\cap{]y,z]}$ and $\ell\in\Omega_{p}$ such that $n\equiv\ell\pmod{p}$. In particular, $p$ divides $N_{\mathbb{K}}(G(\ell))$ in $\mathcal{O}_{S}$ and, since $p\mathcal{O}_{\mathbb{K}}$ has no prime ideal factor $\pi_{v}$ with $v\in S$, it follows that there exists some prime ideal $\pi$ of $\mathcal{O}_{S}$ lying above $p$ and dividing $G(\ell)$. Let $\mathbb{F}_{q}:=\mathcal{O}_{S}/\pi$, so that $q$ is a power of $p$. Write $n=\ell+mp$, for some integer $m\geq 0$. Since $\pi$ divides $G(n)$ and $F(n)/G(n)\in\mathcal{O}_{S}$, we have that $F(n)$ is divisible by $\pi$ too. As a consequence, we obtain that

[TABLE]

Note that $f_{1}(\ell),\ldots,f_{r}(\ell)$ cannot be all equal to zero modulo $\pi$, since $\pi$ divides $G(\ell)$ and $(G(\ell),f_{1}(\ell),\ldots,f_{r}(\ell))$ is an $S$-unit. Note also that the minimum order $N$ of the $\alpha_{i}^{p}/\alpha_{j}^{p}$ ($i\neq j$) modulo $\pi$ is equal to the minimum order of the $\alpha_{i}/\alpha_{j}$ ($i\neq j$) modulo $\pi$, since $(p,q-1)=1$. In particular, $N\geq p^{1/4}$, in light of the definition of $\mathcal{P}$.

Therefore, we can apply Lemma 2.5 to the congruence (10), getting that the number of possible values of $m$ modulo $q-1$ is at most $4(q-1)p^{-1/2^{r}}$. Consequently, the number of possible values of $n\leq x$ is at most

[TABLE]

since $p(q-1)<p^{d+1}\leq z^{d+1}\leq x$. Hence, we have

[TABLE]

In conclusion,

[TABLE]

as claimed.

4. Concluding remarks

Let us briefly explain the computation of $h$. First, we have an effective procedure to test if there exists a nonzero polynomial $P\in\mathbb{C}[X]$ such that the sequences $n\mapsto P(n)F(n)/G(n)$ and $n\mapsto G(n)/P(n)$ are linear recurrences, and in such a case $P$ can be determined (see [6, p. 435, Remark 1]).

On the one hand, if $P$ does not exist, then Theorem 1.1 implies that $\mathcal{N}$ is finite, hence $h$ can be any positive integer. Moreover, using any effective version of the Skolem–Mahler–Lech Theorem at the end of the proof of [6, Proposition 2.1], it is possible to bound $\#\mathcal{N}$. Therefore, the implied constant in Theorem 1.5 is effectively computable.

On the other hand, if $P$ exists, then we can write the linear recurrences $H=PF/G$ as

[TABLE]

for some $\beta_{1},\ldots,\beta_{s}\in\mathbb{C}^{*}$ and $h_{1},\ldots,h_{s}\in\mathbb{C}[X]$. Setting $Q:=P/(P,h_{1},\ldots,h_{s})$, we have that $\widetilde{Q}(n)=N_{\mathbb{K}}(Q(n))$ is a polynomial in $\mathbb{Q}[X]$ and $h$ can be taken as the number of irreducible factors of $\widetilde{Q}$. Furthermore, the implied constant in Theorem 1.5 is effectively computable, since all the implied constants of the results used in the proof of Theorem 1.5 are effectively computable.

Acknowledgements

The author thanks Umberto Zannier for a fruitful conversation on Theorem 1.2, and also the anonymous referee for useful comments which improved the quality of the paper.