On the Largest Prime factor of the k –generalized Lucas numbers

Let ( L ( k ) n ) n ≥ 2 − k be the sequence of k –generalized Lucas numbers for some fixed integer k ≥ 2 whose first k terms are 0 , . . . , 0 , 2 , 1 and each term afterwards is the sum of the preceding k terms. For an integer m , let P ( m ) denote the largest prime factor of m , with P (0) = P ( ± 1) = 1. We show that if n ≥ k + 1, then P ( L ( k ) n ) > (1 / 86) log log n . Furthermore, we determine all the k –generalized Lucas numbers L ( k ) n whose largest prime factor is at most 7.


Introduction 1.Background
Let k ≥ 2 be an integer.The k-generalized Lucas numbers is the sequence defined by the recurrence relation with the initial condition L −1 = 0 for all k ≥ 3. When k = 2, this sequence is the classical sequence of Lucas numbers and in this case we omit the superscript (k) in the notation.
For an integer m, let P (m) be the largest prime factor of m with the convention P (0) = P (±1) = 1.The challenge of determining lower bounds for the largest prime factor of terms in linear recurrence sequences has sparked much interest among mathematicians.Numerous studies have been conducted on this topic, see, for example [1].In this work, our focus is on the sequence of k-generalized Lucas numbers.Specifically, we aim to derive effective lower bounds for P (L (k) n ) in relation to both k and n.We prove the following results.

Main Results
Theorem 1.1.Let (L (k) n ) n≥2−k be the sequence of k-generalized Lucas numbers.Then, the inequality holds for all n ≥ k + 1.
In addition, L

Preliminaries
It is known that for all 2 ≤ n ≤ k. (2.1) In particular, P (L Next, we revisit some properties of the k-generalized Lucas numbers.They form a linearly recurrent sequence of characteristic polynomial which is irreducible over Q[x].The polynomial Ψ k (x) possesses a unique real root α(k) > 1 and all the other roots are inside the unit circle, see [9].The root α(k) as noted in [14].As in the classical case when k = 2, it was shown in [2] that Let k ≥ 2 and define We have In particular, inequality (2.3) implies that for all k ≥ 3. It is easy to check that the above inequality holds for k = 2 as well.Further, it is easy to verify that |f k (α i )| < 1, for all 2 ≤ i ≤ k, where α i are the remaining roots of Ψ k (x) for i = 2, . . ., k.
The following lemma will be useful in our applications of Baker's theory.It is Lemma 2 in [5].
Moreover, it was shown in [2] that for all k ≥ 2 and n ≥ 2 − k.This means that The left expression in (2.7) is known as the Binet-like formula for L (k) n .Furthermore, the right inequality expression in (2.7) shows that the contribution of the zeros that are inside the unit circle to In particular, for all i = 1, 2, . . .s. Lastly, for k ≥ 3 one checks that 1/ log α ≤ 2 by using α When k = 2, the number α represents the golden ratio for which 1/ log α < 2.1.Thus, the inequality (2.10)

Linear forms in logarithms
We use Baker-type lower bounds for nonzero linear forms in logarithms of algebraic numbers.There are many such bounds mentioned in the literature but we use one of Matveev from [8].Before we can formulate such inequalities, we need the notion of height of an algebraic number recalled below.
Definition 2.1.Let γ be an algebraic number of degree d with minimal primitive polynomial over the integers where the leading coefficient a 0 is positive.Then, the logarithmic height of γ is given by In particular, if γ is a rational number represented as γ = p/q with coprime integers p and q ≥ 1, then h(γ) = log max{|p|, q}.The following properties of the logarithmic height function h(•) will be used in the rest of the paper without further reference: With these properties, it was easily computed in Section 3, equation ( 12) of [1] that (2.11) A linear form in logarithms is an expression where for us γ 1 , . . ., γ t are positive real algebraic numbers and b 1 , . . ., b t are integers.We assume, Λ ̸ = 0. We need lower bounds for |Λ|.We write K := Q(γ 1 , . . ., γ t ) and D for the degree of K over Q.We give Matveev's inequality from [8].
During the calculations, upper bounds on the variables are obtained which are too large, thus there is need to reduce them.To do so, we use some results from approximation lattices and the so-called LLL-reduction method from [7].We explain this in the following subsection.

Reduced Bases for Lattices and LLL-reduction methods
Let k be a positive integer.A subset L of the k-dimensional real vector space R k is called a lattice if there exists a basis We say that b 1 , b 2 , . . ., b k form a basis for L, or that they span L. We call k the rank of L. The determinant det(L), of L is defined by with the b i 's being written as column vectors.This is a positive real number that does not depend on the choice of the basis (see [3], Section 1.2).
Given linearly independent vectors b 1 , b 2 , . . ., b k in R k , we refer back to the Gram-Schmidt orthogonalization technique.This method allows us to inductively define vectors b * i (with , for 1 ≤ j < i ≤ n, and where ∥ • ∥ denotes the ordinary Euclidean length.The constant 3/4 above is arbitrarily chosen, and may be replaced by any fixed real number y in the interval 1/4 < y < 1 (see [7], Section 1).where || • || denotes the Euclidean norm on R k .It is well known that, by applying the LLL-algorithm, it is possible to give in polynomial time a lower bound for l (L, y), namely a positive constant c 1 such that l (L, y) ≥ c 1 holds (see [13], Section V.4).
In our application, we are given real numbers η 0 , η 1 , . . ., η k which are linearly independent over Q and two positive constants c 3 and c 4 such that where the integers x i are bounded as |x i | ≤ X i with X i given upper bounds for 1 ≤ i ≤ k.We write X 0 := max 1≤i≤k {X i }.The basic idea in such a situation, due to [4], is to approximate the linear form (2.13) by an approximation lattice.So, we consider the lattice L generated by the columns of the matrix , where C is a large constant usually of the size of about X k 0 .Let us assume that we have an LLL-reduced basis b 1 , . . ., b k of L and that we have a lower bound l (L, y) ≥ c 1 with y := (0, 0, . . ., −⌊Cη 0 ⌋).Note that c 1 can be computed by using the results of Lemma 2.2.Then, with these notations the following result is Lemma VI.1 in [13].
13) implies that we either have Finally, we present an analytic argument which is Lemma 7 in [6].
SageMath 9.5 is used to perform all computations in this work.
In this section, we prove Theorem 1.1.To do this, we first state and prove some preliminary results.We start with the following.
3.1 An upper bound on n in terms of s and k. n with β i ≥ 0, for all i = 1, 2, . . ., s.Then log n < 35s log s + 3s log k + 3 log(12s + k).
We apply Lemma 2.4 with x := n + 1, m := 3 and T := 1.21 • 10 13 s 4.5 k 3+s (60 log s) s > (4m 2 ) m = 46656.We get n + 1 < 2 We therefore assume that s < k for the remainder of this section.With this assumption, the conclusion of Lemma 3.1 becomes log n < 38s log k + 3 log(13k) for s ≥ 2 and k ≥ 2. We proceed by distinguishing between two cases.

The case
Here, we have that • 46s log k < 133s log k.
From the above, we have k/ log k < 133s.We apply Lemma 2.4 with the data: x := k, m := 1 and T := 133s > (4m

The case
Let λ > 0 be such that α +λ = 2. Since 2 That is, λ ∈ (0, 1/2 k−1 ).Moreover, where we used the fact that log(1 − x) > −2x for all x < 1/2 and e −x ≥ 1 − x for all x ∈ R. Furthermore, Next, consider the function f k (x) given at (2.5).By the Mean-Value Theorem, there exists some ω ∈ (α, 2) It can be checked that the same holds for k = 2. Hence, From the above, if we write then inequalities (3.10), (3.11) and (3.12) become Moreover, since f k (2) = 1/2 for all k ≥ 2, we have , we get In the above, we used that k ≤ 2 k−1 and that k < (5/4)2 k/2 for k ≥ 2. Dividing both sides above by 3 • 2 n−2 , having in mind that p 1 = 2 and p 2 = 3, we get Now, we intend to apply Theorem 2.1 on the left-hand side of (3.15).Let Notice that Λ 1 ̸ = 0, otherwise we would have The algebraic number field containing γ i 's is K := Q, so we take D = 1.Since h(γ i ) = log p i ≤ log p s for all i = 1, 2, . . ., s, we take A i := log p s for all i = 1, 2, . . ., s.

Proof of Theorem 1.2
We proceed in a way similar as in Section 3. Specifically, we prove following estimates.(b) In the second part, if k > 1000, then For this reason, we can use the same arguments from Subsection 3.3 relation (3.15) to write from which after applying Matveev's result with s = 4 as in Subsection 3.This completes the proof of Lemma 4.1.
To complete the proof of Theorem 1.2, we proceed in two cases, that is, the case k ≤ 1000 and the case k > 1000.We use similar analyses given on pages 1363 and 1364 of [1].
For each k ∈ [2, 1000], we used the LLL-algorithm to compute a lower bound for the smallest nonzero number of the form |τ 1 |, with integer coefficients x i not exceeding 1.4 • 10 27 k 7 (log k) 3 in absolute value.Specifically, we consider the approximation lattice , with C := 10 355 and choose y := (0, 0, 0, 0, 0, 0, 0).Now, by Lemma 2.2, we get l (L, y) Finally, we wrote a simple program in SageMath (see Appendix 1), to look at k-generalized Lucas numbers for 2 ≤ k ≤ 1000 and k + 1 ≤ n ≤ 1449.Instead of factoring the numbers fully, we checked if they could be divided by 2, 3, 5, and 7 until we couldn't divide them anymore.This way, we found out if each number could be written using only these primes.The numbers we got are the ones given in Theorem 1.2.This completes the analysis in the case k ∈ [2, 1000].

The case k > 1000
Lastly, we treat the case when k > 1000.At this point, we need to reduce our absolute upper bound on k, see Lemma 4.1, by using again the LLL-algorithm described in Lemma 2.3.To do this, let We then obtain that k ≤ 1106.After repeating this process 2 more times, we finally find that k < 1000, which is a contradiction.Thus, Theorem 1.2 is proved.

. 1 )
in nonnegative integers n, k, a, b, c, d with k ≥ 2 and n ≥ k + 1, are L
2 ) m = 4, for all s ≥ 2. We get for s ≥ 2. In the above, we used that log s < s.Consequently, p s > s log s > (1/86) log n in this case.