On the problem of Pillai with k –generalized Fibonacci numbers and powers of 3

For an integer k ≥ 2


Introduction
The problem of Pillai states that for each fixed integer c ≥ 1, the Diophantine equation has only a finite number of positive solutions {a, b, x, y}.This problem is still open; however, the case c = 1 is the conjecture of Catalan and was proved by Mihȃilescu [22].In 1936 (see [23,24]), in the special case (a, b) = (3, 2) which is a continuation of the work of Herschfeld [19,20] in 1935, Pillai conjectured that the only integers c admitting at least two representations of the form 2 x − 3 y are given by This was confirmed by Stroeker and Tijdeman [25] in 1982.The general problem of Pillai is difficult to solve and this has motivated the consideration of special cases of this problem.In the past years, several special cases of the problem of Pillai have been studied.For example, see [6, 8-12, 17, 18].Let k ≥ 2 be an integer.We consider a generalization of Fibonacci sequence called the k-generalized Fibonacci sequence {F (k) n } n≥2−k defined as with the initial conditions We call F (k) n the nth k-generalized Fibonacci number.Note that when k = 2, it coincides with the Fibonacci numbers and when k = 3 it is the Tribonacci number.The first k + 1 nonzero terms in F (k) n are powers of 2, namely Furthermore, the next term is F (k) k+2 = 2 k − 1.Thus, we have that We also observe that the recursion (1.3) implies the three-term recursion for all n ≥ 3, which can be used to prove by induction on m that F (k) n < 2 n−2 for all n ≥ k + 2 (see also [5, Lemma 2]).
The generalized Fibonacci analogue of the problem of Pillai under the same conditions as in (1.1) concerns studying for fixed (k, ) all values of the integer c such that the equation F (k)  n − F ( ) m = c (1.5) has at least two solutions (n, m).We are not aware of a general treatment of equation (1.5) (namely, considering k and parameters), although the particular case when {k, } = {2, 3} was treated by Chim, Pink and Ziegler in [8].Ddamulira, Gómez and Luca [13] studied the Diophantine equation where k is also a parameter, which is a variation of Eq. (1.5).They determined all integers c such that Eq. (1.6) has at least two solutions (n, m).These c together with their multiple representations as in (1.6) turned out to be grouped into four parametric families.
In this paper, we study a related problem and we find all integers c admitting at least two representations of the form F (k) n − 3 m for some positive integers k, n, and m.This can be interpreted as solving the equation with (n, m) = (n 1 , m 1 ).The cases k = 2 and k = 3 have been solved completely by the first author in [10,11], respectively.So, we focus on the case k ≥ 4.
Our main result is the following.

Preliminary Results
In this section, we recall some general results from algebraic number theory and Diophantine approximations and properties of the k-generalized Fibonacci sequence.

Notations and terminology from algebraic number theory
We begin by recalling some basic notions from algebraic number theory.
Let η be an algebraic number of degree d with minimal primitive polynomial over the integers where the leading coefficient a 0 is positive and the η (i) 's are the conjugates of η.Then the logarithmic height of η is given by In particular, if η = p/q is a rational number with gcd(p, q) = 1 and q > 0, then h(η) = log max{|p|, q}.The following are some of the properties of the logarithmic height function h(•), which will be used in the next sections of this paper without reference:

k-generalized Fibonacci numbers
It is known that the characteristic polynomial of the k-generalized Fibonacci numbers and has just one root outside the unit circle.Let α := α(k) denote that single root, which is located between 2 1 − 2 −k and 2 (see [14]).This is called the dominant root of F (k) .To simplify notation, in our application we shall omit the dependence on k of α.We shall use α (1) , . . ., α (k) for all roots of Ψ k (x) with the convention that α (1) := α.We now consider for an integer k ≥ 2, the function With this notation, Dresden and Du presented in [14] the following "Binet-like" formula for the terms of F (k) : It was proved in [14] that the contribution of the roots which are inside the unit circle to the formula (2.2) is very small, namely that the approximation 3) It was proved by Bravo and Luca in [5] that Before we conclude this section, we present some useful lemma that will be used in the next sections on this paper.The following lemma was proved by Bravo and Luca in [5].(i) The inequalities

Linear forms in logarithms and continued fractions
In order to prove our main result (Theorem 1.1), we need to use several times a Baker-type lower bound for a nonzero linear form in logarithms of algebraic numbers.There are many such in the literature like that of Baker and Wüstholz from [2].We recall the result of Bugeaud, Mignotte and Siksek [7, Theorem 9.4, p. 989], which is a modified version of the result of Matveev [21].This result is one of our main tools in this paper.Theorem 2.2 (Matveev according to Bugeaud, Mignotte, Siksek).Let γ 1 , . . ., γ t be positive real algebraic numbers in a real algebraic number field K of degree D, b 1 , . . ., b t be nonzero integers, and assume that and During the course of our calculations, we get some upper bounds on our variables which are too large, thus we need to reduce them.To do so, we use some results from the theory of continued fractions.Specifically, for a nonhomogeneous linear form in two integer variables, we use a slight variation of a result due to Dujella and Pethő (see [15,Lemma 5a]), which itself is a generalization of a result of Baker and Davenport [1].
For a real number X, we write X := min{|X − n| : n ∈ Z} for the distance from X to the nearest integer.

Lemma 2.3 (Dujella, Pethő
).Let M be a positive integer, p/q be a convergent of the continued fraction of the irrational number τ such that q > 6M, and A, B, μ be some real numbers with A > 0 and B > 1.Furthermore, let ε := μq − M τq .If ε > 0, then there is no solution to the inequality in positive integers u, v, and w with u ≤ M and w ≥ log(Aq/ε) log B .
The above lemma cannot be applied when μ = 0 (since then ε < 0).In this case, we use the following criterion of Legendre, a well-known result from the theory of Diophantine approximation.

Lemma 2.4 (Legendre). Let τ be real number and x, y integers such that
Finally, the following lemma is also useful.It is [16,Lemma 7].

The Connection with the Classical Pillai Problem
Assume that (n, m) = (n 1 , m 1 ) are such that n1 and since min{n, n 1 } ≥ 2, we get that n = n 1 .Thus, (n, m) = (n 1 , m 1 ), contradicting our assumption.Hence, m = m 1 , and we may assume without loss of generality that m > m 1 ≥ 1.Since and the right-hand side of (3.1) is positive, we get that the left-hand side of (3.1) is also positive and so n > n 1 .Furthermore, since We analyze the possible situations.
Case 3.1.Assume that 2 ≤ n 1 < n ≤ k + 1.Then, by (1.4), we have so, by substituting them in (3.1), we get By comparing with the classical solutions in (1.2), and by using the fact that is a power of 2 if and only if n ≤ k + 1 (see [4]), we get the solutions The following lemma is useful.
Lemma 3.1.For n ≥ k + 2, the conditions cannot simultaneously hold.
Proof.If they do, then The sequence We show that from here on it is increasing.That is This is equivalent to and this last inequality holds true because in the right-hand side we have Int. J. Number Theory Downloaded from www.worldscientific.comby Mahadi Ddamulira on 04/08/20.Re-use and distribution is strictly not permitted, except for Open Access articles.

Bounding n in Terms of m and k
By the results of the previous section, we assume that n ≥ k + 2. Thus, n−1 and therefore So, from the above, (2.4) and (3.1), we have , and leading to We note that the above inequality (4.2) in particular implies that m < n < 1.6m + 4. We assume for technical reasons that n > 600.By (2.3) and (3.1), we get In the above, we have also used the fact that |f k (α)| < 1 (see Lemma 2.1).Dividing through by 3 m , we get where for the right-most inequality in (4.3) we used (4.1) and the fact that α 2 > 2.
For the left-hand side of (4.3), we apply Theorem 2.2 with the data We begin by noticing that the three numbers γ 1 , γ 2 , γ 3 are positive real numbers and belong to the field K := Q(α), so we can take To see why Λ = 0, note that otherwise, we would then have that and so f k (α) would be an algebraic integer, which contradicts Lemma 2.1(i).
Then, the left-hand side of (4.3) is bounded below, by Theorem 2.2, as Comparing with (4.3), we get Now the argument is split into two cases.In this case, we rewrite (3.1) as Dividing through by 3 m gives Now we put We apply again Theorem 2.2 with the following data: As before, we begin by noticing that the three numbers γ 1 , γ 2 , γ 3 belong to the field K := Q(α), so we can take D := [K : Q] = k.To see why Λ 1 = 0, note that otherwise, we would get the relation Conjugating this last equation with any automorphism σ of the Galois group of Ψ k (x) over Q such that σ(α) = α (i) for some i ≥ 2, and then taking absolute values, we arrive at the equality |f k (α So, we can take A 1 := 6.80 × 10 11 k 4 log 2 k(1 + log n).Further, as before, we take A 2 := log 2 and A 3 := k log 3. Finally, by recalling that m < n, we can take B := n.
We then get that log Comparing this with (4.4), we get that In this case, we write (3.1) as The above inequality (4.5) suggests once again studying a lower bound for the absolute value of We again apply Matveev's theorem with the following data: We can again take B := n and K := Q(α), so that D := k.We also note that, if ) is an algebraic integer, which is not the case.Thus, Λ 2 = 0. Now, we note that Thus, kh(γ 1 ) < 4k log k + (m − m 1 ) log 3 < 6.80 × 10 11 k 4 log 2 k(1 + log n), and so we can take A 1 := 6.80 × 10 11 k 4 log 2 k(1 + log n).As before, we take A 2 := log 2 We now finally rewrite Eq. (3.1) as We divide through both sides by 3 m − 3 m1 getting since n < 1.6m + 4. To find a lower-bound on the left-hand side of (4.7), we again apply Theorem 2.2 with the data We also take B := n and we take K := Q(α) with D := k.From the properties of the logarithmic height function, we have that where in the above chain of inequalities we used the bounds (4.6).So we can take A 1 := 8.3 × 10 22 k 8 log 3 k(1 + log n) 2 , and certainly as before we take A 2 := log 2 and A 3 := k log 3. We need to show that if we put then Λ 3 = 0. To see why Λ 3 = 0, note that otherwise, we would get the relation Again, as for the case of Λ 1 , conjugating the above relation with an automorphism σ of the Galois group of Ψ k (x) over Q such that σ(α) = α (i) for some i ≥ 2, and then taking absolute values, we get that This cannot hold true because in the left-hand side we have while in the right-hand side we have  7 .

The cutoff k
We have from the above lemma that Baker's method gives n < 4 × 10 42 k 11 (log k) 7 .
By imposing that the above amount is at most 2 k/2 , we get The inequality above holds for k > 600.
We now reduce the bounds and to do so we make use of Lemma 2.3 several times.

The case of small k
We now treat the cases when k ∈ [4,600].First, we consider Eq. (3.1) which is equivalent to (1.7).For k ∈ [4,600] and n ∈ [3,600], consider the sets With the help of Mathematica, we intersected these two sets and found the only solutions listed in Theorem 1.1.
Next, we note that for these values of k, Lemma 4.1 gives us absolute upper bounds for n.However, these upper bounds are so large that we wish to reduce them to a range where the solutions can be easily identified by a computer.To do this, we return to (4.3) and put For technical reasons we assume that min{n − n 1 , m − m 1 } ≥ 20.In the case that this condition fails, we consider one of the following inequalities instead: We start by considering (4.3).Note that Γ = 0; thus we distinguish the following two cases.If Γ > 0, then e Γ − 1 > 0, then from (4.3) we get Next we suppose that Γ < 0. Since Λ = |e Γ −1| < 1  2 , we get that e |Γ| < 2. Therefore, Therefore, in any case, the following inequality holds: (5.1) By replacing Γ in the above inequality by its formula and dividing through by log 3, we then conclude that Then, we apply Lemma 2.3 with the following data: Next, we put M k := 4 × 10 42 k 11 (log k) 7 , which is the absolute upper bound on n by Lemma 4.1.An intensive computer search in Mathematica revealed that the maximum value of log(2α 6 q/ε)/ log α is < 600 and the maximum value of log((6/ log 3)q/ε)/ log 3 is < 375.Thus, either n − n 1 < log(2α 6 q/ε) log α < 600, or m − m 1 < log((6/ log 3)q/ε) log 3 < 375.
Therefore, we have that either n − n 1 ≤ 600 or m − m 1 ≤ 375.Now, let us assume that n − n 1 ≤ 600.In this case, we consider the inequality (4.4) and assume that m − m 1 ≥ 20.Then we put By similar arguments as in the previous step for proving (5.1), from (4.4) we get 0 < |Γ 1 | < 6 3 m−m1 , and replacing Γ 1 with its formula and dividing through by log 3 give As before, we keep the same τ k , M k , (A k , B k ) := ((6/ log 3), 3) and put We apply Lemma 2.3 to the inequality (5.2) with the above data.A computer search in Mathematica revealed that the maximum value of log(Aq/ε)/ log B over the values of k ∈ [4, 600] and l ∈ [1, 600] is < 377.Hence, m − m 1 ≤ 377.
Next, we assume that m − m 1 ≤ 375.Here, we consider the inequality (4.5) and also assume that n − n 1 ≥ 20.We put Thus, by the same arguments as before, we get . By substituting for Γ 2 with its formula and dividing through by log 3 in the above inequality, we get As before, we apply Lemma 2.3 with the same τ k , M k , (A k , B k ) := (2α 6 / log 3, α) and put Finally, we go to (4.7) and put Since n > 600, from (4.7) we can conclude that . Hence, by substituting for Γ 3 by its formula and dividing through by log 3, we get We apply Lemma 2.3 with the same and put A computer search in Mathematica revealed that the maximum value of log(1328q/ε)/ log 3 , for k ∈ [4, 600], l ∈ [1, 603] and j ∈ [1, 377] is < 378.Hence, n < 473, which contradicts the assumption that n > 500 in the previous section.

The case of large k
We now assume that k > 600.Note that for these values of k we have n < 4 × 10 42 k 11 (log k) 7 .
Since, n ≥ k + 2, we have that n ≥ 602.The following lemma is useful.
Lemma 5.1.For 1 ≤ n < 2 k/2 and k ≥ 10, we have Proof.When n ≤ k + 1, we have F (k) n = 2 n−2 so we can take ζ := 0. So, assume k + 2 ≤ n < 2 k/2 .It follows from [3, (1.8)] that By the above lemma, we can rewrite (3.1) as Next, we have Going back to (5.3), we have (5.4) We now apply Theorem 2.2 on the left-hand side of (5.4) with the data It is clear that Γ = 0, otherwise we would get 3 m = 2 n−2 which is a contradiction since 3 m is odd while 2 n−2 is even.We consider the field K = Q, in this case D = 1.
Since h(γ 1 ) = h(3) = log 3 and h(γ 2 ) = h(2) = log 2, we can take A 1 := log 3 and A 2 := log 2. We also take B := n.Then, by Theorem 2.2, the left-hand side of (5.We rewrite (3.1) as (5.5) We now apply Matveev's theorem, Theorem 2.2 on the left-hand side of (5.5) to with the data We rewrite (3.1) as (5.6) We again apply Matveev's theorem, Theorem 2.2 on the left-hand side of (5.5) which is with the data Note that Γ 2 = 0. Otherwise, 3 m − 3 m1 = 2 n−2 , which is impossible since the lefthand side is a multiple of 3 and the right-hand side is not.We use the same values, A 1 := log 3, A 2 := log 2, B := n as in the previous steps.In order to determine A 3 , note that So, we take A 3 := 5.90 × 10  Then we also note that (5.6) can be rewritten as Therefore, in all cases we found out that k < 3428 which gives that n < 7.2741 × 10 87 < 10 88 .These bounds are still too large.We repeat the above procedure
Int. J. Number Theory Downloaded from www.worldscientific.comby Mahadi Ddamulira on 04/08/20.Re-use and distribution is strictly not permitted, except for Open Access articles.