Integers representable as differences of linear recurrence sequences

Let \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{U_n\}_{n \ge 0}$$\end{document}{Un}n≥0 and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{V_m\}_{m \ge 0}$$\end{document}{Vm}m≥0 be two linear recurrence sequences. We establish an asymptotic formula for the number of integers c in the range \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$[-x, x]$$\end{document}[-x,x] which can be represented as differences \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ U_n - V_m$$\end{document}Un-Vm. In particular, the density of such integers is 0.


Introduction
Pillai's Conjecture [10] states that for any given positive integer c the Diophantine equation has only finitely many positive integer solutions (a, b, n, m) with n, m ≥ 2. Pillai's conjecture is a corollary of the abc conjecture. For c = 1, it coincides with Catalan's conjecture, which has been proved by Mihȃilescu [8]. For all c > 1, Pillai's conjecture is still open. For fixed integers a, b Pillai [10,11] proved that for sufficiently large c there is at most one solution (n, m) with n, m ≥ 2 to Eq. (1). Pillai [9] also proved the following asymptotic result on the number of integers c in the range [1, x] which can be expressed in the form c = a n − b m : #{c ∈ [1, x]: c = a n − b m for some (n, m) ∈ N 2 } ∼ (log x) 2 2(log a) (log b) , as x → ∞. (2) We denote by N the set of all non-negative integers. In recent years, there have been several papers studying a generalised version of equation (1), that is where {U n } n≥0 and {V m } m≥0 are linear recurrence sequences of integers. For instance, in [6] the authors considered the case where {U n } n≥0 are the Fibonacci numbers and {V m } m≥0 are the powers of two. In [2] the authors considered the Tribonacci numbers and powers of two and in [4] the authors considered the Fibonacci numbers and the Tribonacci numbers. In each paper, the authors found all integers c having at least two different representations of the form c = U n − V m for the respective sequences.
Chim et al. [5] proved, that this is possible for general linear recurrence sequences (with a few subtle restrictions), i.e. there exists an effectively computable finite set C such that Eq. (3) has at least two distinct solutions (n, m) if and only if c ∈ C. This can be seen as the generalisation of Pillai's result in [10,11].
What has not been established properly yet, is for how many integers c there exists a solution to (3) at all. In other words, Pillai's result (2) has not been extended yet. This is what we aim to do in this paper. In fact, we will find an asymptotic formula analogous to (2) for the number of integers c ∈ [−x, x] which can be represented as c = U n − V m for given linear recurrence sequences {U n } n≥0 and {V m } m≥0 . Our proof is based on ideas from [5] and [12] 1 and in particular on lower bounds for linear forms in logarithms. A weaker version of this result has been proved by the third author in [12].
In order to state our result, we recall some definitions. Let {U n } n≥0 be a linear recurrence sequence of integers given by for all n ≥ 0, for some given k ≥ 1, some given integers c 1 , . . . , c k with c k = 0 and some given integers U 0 , . . . , U k−1 . Then the characteristic polynomial of {U n } n≥0 is defined by where α 1 , . . . , α t are distinct complex numbers and σ 1 , . . . , σ t are positive integers whose sum is k. It is known that for any such sequence {U n } n≥0 there exist poynomials a 1 (X), . . . , a t (X) with coefficients in Q(α 1 , . . . , α t ) and degrees deg a i (X) ≤ σ i − 1 for i = 1, . . . , t, such that the formula holds for all n ≥ 0. We call α = α 1 a dominant root, if |α 1 | > |α 2 | ≥ · · · ≥ |α t | and a 1 (X) is not the zero polynomial. In this case the sequence {U n } n≥0 is said to satisfy the dominant root condition. Now we state our main result.
Then we have the asymptotic behaviour More precisely, we have for x large enough. The implied constants are effective.
Chim et al. [5] proved for the situation of Theorem 1 that if an integer c has a representation c = U n − V m , then in the "generic" case this representation is unique. Therefore, instead of counting integers c ∈ [−x, x] which are representable as c = U n − V m , one can count the solutions (n, m) to the Diophantine inequality |U n − V m | ≤ x. We will see that both ways of counting yield the same result. In fact, Theorem 1 is equivalent to the following.
Theorem 2 Let {U n } n≥0 and {V m } m≥0 be two linear recurrence sequences of integers satisfying the dominant root condition with dominant roots α and β respectively. Suppose that |α| > 1 and |β| > 1 and that α and β are multiplicatively independent. Let Then we have for x large enough. The implied constants are effective.
The paper is structured as follows. In Sect. 2 we state some preliminary results, in particular results on linear forms in logarithms, heights, the result from [5] and some elementary inequalities. We also prove the equivalence of Theorems 1 and 2. In Sect. 3 we use elementary arguments to prove the lower bound for T (x). In Sect. 4 we prove the upper bound for T (x) using linear forms in logarithms. Finally, in Sect. 5 we put some further problems.

Preliminaries
In this section we present the tools for our proof. The most powerful one is certainly lower bounds for linear forms in logarithms. Moreover, will need some estimates for heights. Next, we state some facts on linear recurrence sequences and the result from [5], which will show the equivalence of Theorems 1 and 2. Finally, we check some simple relations between inequalities, which will be important for the proofs of the lower and the upper bound in Theorems 1 and 2.

Linear forms in logarithms and heights
Let γ be an algebraic number of degree d ≥ 1 with the minimal polynomial where a 0 , . . . , a d are relatively prime integers and γ 1 , . . . , γ d are the conjugates of γ . Then the logarithmic height of γ is given by Since the first results by Baker, there have been many powerful results on lower bounds for linear forms in logarithms. In particular, in 1993 Baker and Wüstholz [1] obtained a very good explicit bound. In the following years, further improvements were made. At the present time, one of the most widely used results is due to Matveev [7]. The following theorem [3, Thm. 9.4] is a consequence of Matveev's result.

b t be rational integers, and let
In order to estimate the height of certain expressions, we will use the following two well known lemmas [5, Lems. 1 and 2]. Lemma 1 Let K be a number field and α, β ∈ K two multiplicatively independent algebraic numbers. Then there exists an effectively computable Lemma 2 Let K be a number field and p, q ∈ K[x] two arbitrary but fixed polynomials. Then there exists an effectively computable constant C = C(p, q) > 0 such that

Linear recurrence sequences and solutions to c = U n − V m
From now on, until the end of this paper, let {U n } n≥0 and {V m } m≥0 be two linear recurrence sequences of integers satisfying the dominant root condition with dominant roots α and β respectively and |α| > 1 and |β| > 1. Moreover, we assume that α and β are multiplicatively independent. Suppose that U n = a(n)α n + a 2 (n)α n 2 + · · · + a s (n)α n s and for all n, m ≥ 0. As in [5], we use the L-notation: Then we have U n = a(n)α n + L(a α n ) and for some 1 < α < |α|, 1 < β < |β| and a , b > 0. Suppose that deg a(X) = σ and deg b(X) = τ . Then there exist positive constants C 1 , C 2 , C 3 , C 4 such that C 1 |α| n ≤ |U n | ≤ C 2 n σ |α| n and (4) for all n, m large enough.
In order to prove Theorem 1, we will actually prove Theorem 2. The following lemma shows the equivalence of the two theorems. In other words, it allows us to switch between counting integers c which have a representation c = U n − V m and counting solutions (n, m) of the Diophantine inequality |U n − V m | ≤ x. For the second inequality we need the fact that "most" representations are unique. This was proved in [5]. We state the result as a lemma. Note that this is not the main result in [5], but it follows immediately from the proof.  Proof Inequality (6) implies n ≤ kz and log n ≤ log k + log z. Thus we get Proof Note that for r, s ≥ 2 we have log(r + s) ≤ log r + log s. By assumption, n and z are large, in particular c(log n) 2 ≥ 2 and kz ≥ 2. Thus, using the assumptions, we have

Some auxiliary inequalities
≤ log k + log z + log c + 2 log log n ≤ log z + 0.5 log n, for n ≥ N . Thus log n ≤ 2 log z and using the assumption again we get

Lower bound for T(x)
In this section we prove the lower bound for the number of solutions (n, m) to the Diophantine inequality |U n − V m | ≤ x: In fact, we show that if n ≤ log x/ log |α| + O(log log x) and m ≤ log Then Lemma 5 with k = 1/ log |α|, c = σ /log |α| + 1 and d = (log C 2 + log 2)/ log |α| yields for z = log x large enough Multiplying by log |α| and applying the exponential function we obtain which together with (4) implies Analogously, we obtain that Therefore, for all (n, m) ∈ N 2 satisfying (7) and (8) we have But the number of (n, m) ∈ N 2 satisfying (7) and (8) is larger than so for x large enough we have

Upper bound for T(x)
In this section we use linear forms in logarithms to prove the upper bound for the number of solutions (n, m) to the Diophantine inequality |U n − V m | ≤ x: In fact, we assume that |U n − V m | ≤ x, i.e.
Note that we can assume that n and m are large enough, i.e. n ≥ N and m ≥ M for some suitable N, M. This is because of the same argument as in the proof of Lemma 3: Ignoring solutions with n ≤ N or m ≤ M, we only miss O(log x) solutions, which has no impact on our result. Similarly, we may assume that c is large enough.
Dividing by log |β| we get Combined with (16) this yields This implies m log γ ≤ log C 16 + C 11 (log (max{n, m})) 2 ≤ log C 16 + C 11 (log(C 15 n)) 2 and we get This means that m is actually very small compared to n and therefore |c| ≈ |α| n , i.e. n ≈ log |c|/ log |α|, which is exactly what we need. In order to formalise this argument, we go back to (9) and use inequalities (4) and (5): Taking logarithms and noting that log(r − s) ≥ log r − log s for r ≥ s + 2 ≥ 4, we get log |c| ≥ log C 1 + n log |α| − log C 4 + τ (log C 17 + 2 log log n) + C 17 (log n) 2   This completes the proof of Theorem 2 and by Lemma 3 we have also proved Theorem 1.

Further problems
The key in our proof is the use of lower bounds for linear forms in logarithms. This tool only works if α and β are algebraic. The natural question is: What happens if α and/or β are transcendental? We pose the following problem.
In particular, is the above true for α = π and β = e? Note that it is an open conjecture that π and e are multiplicatively independent (this is equivalent to log π being irrational).
In that case it would be interesting to find out whether #{(n, m) ∈ N 2 : |π n − e m | ≤ x} ∼ (log x) 2 log π , as x → ∞.
Of course, the same question can be asked for sequences with sums of powers. For instance, we pose the following problem.
And finally, we ask what happens if we allow three different powers.