Common Values of Generalized Fibonacci and Leonardo Sequences

Bibhu Prasad Tripathy and Bijan Kumar Patel Department of Mathematics School of Applied Sciences KIIT University Bhubaneswar Odisha 751024 India

                      bptbibhu@gmail.com
                   bijan.patelfma@kiit.ac.in

Abstract For an integer k ≥ 2, let F_{n}^{(k)} be the k-generalized Fibonacci sequence that starts with 0, …, 0, 1, 1 (k terms) and each term afterwards is the sum of k preceding terms. In this paper, we find all the k-generalized Fibonacci numbers that are Leonardo numbers. More explicitly, we solve the Diophantine equation F_{n}^{(k)} = Le_{m} in positive integers n, k, m with k ≥ 2.

1 Introduction

The Fibonacci and Lucas sequence are two fascinating topics in integer sequences. The Leonardo sequence (Le_{m})_{m≥0} is an integer sequence that is related to the Fibonacci and Lucas sequences. Leonardo numbers are discussed by Catarino and Borges [9]. It is the sequence \underline{A001595} in the OEIS satisfying the recurrence relation

Le_{m} = Le_{m−1} + Le_{m−2} +1  (1)

for m ≥ 2 with the initial terms Le_{0} = 1 and Le_{1} = 1. The first few terms of (Le_{m})_{m≥0} are

1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, ….

In the recent past, many aspects of Leonardo sequence have been studied such as hybrid Leonardo numbers [1], incomplete Leonardo numbers [10], Leonardo Pisano polynomials, hybrinomials [14] and q-Leonardo hybrid numbers [16].

The Fibonacci sequence (F_{n})_{n≥0} is the binary recurrence sequence given by

F_{n+2} = F_{n+1} + F_{n} for n ≥ 0

with the initial terms F_{0} = 0 and F_{1} = 1.

Let k ≥ 2 be an integer. One of numerous generalizations of the Fibonacci sequence, called the k-generalized Fibonacci sequence (F_{n}^{(k)})_{n≥−(k−2)} is given by the recurrence

F_{n}^{(k)} = F_{n−1}^{(k)} + F_{n−2}^{(k)} + ⋯ + F_{n−k}^{(k)} = ∑_{i=1}^{k} F_{n−i}^{(k)} for all n ≥ 2,  (2)

with the initial conditions F_{−(k−2)}^{(k)} = F_{−(k−3)}^{(k)} = ⋯ = F_{0}^{(k)} = 0 and F_{1}^{(k)} = 1. Here, F_{n}^{(k)} denotes the nth k-generalized Fibonacci number.

Note that for k = 2, we have F_{n}^{(2)} = F_{n}, the nth Fibonacci number. For k = 3, we have F_{n}^{(3)} = T_{n}, the nth Tribonacci number. They are followed by the Tetranacci numbers for k = 4, and so on.

A Leonardo number is called k-Fibonacci Leonardo number if it is a k-generalized Fibonacci number. The aim of this paper is to determine all the k-Fibonacci Leonardo numbers.

Finding the intersection of two recurrent sequences of positive integers is a topic that has been extensively studied in number theory. Currently, several researchers have been interested in finding the intersection of the k-generalized Fibonacci sequence with other number sequences. For instance, one can go through [4, 5, 6, 13, 17, 18].

Motivated by the above literature, we study the Diophantine equation

F_{n}^{(k)} = Le_{m} .  (3)

In particular, our main result is the following.

Theorem 1. All the solutions of the Diophantine equation (3) in positive integers with k ≥ 2 are given by

(n, k, m) ∈ {(1, k, 0), (2, k, 0), (1, k, 1), (2, k, 1), (4, 2, 2), (5, 2, 3), (6, 4, 5)}.

Thus, the only k-Fibonacci Leonardo numbers are 1, 3, 5, and 15.

2 Auxiliary results

Our proof of Theorem 1 is mainly based on linear forms in logarithms of algebraic numbers and a reduction algorithm originally introduced by Baker and Davenport [3] (and improved by Dujella and Pethö [12]). Here, we use a variant due to de Weger [19], but first, recall some basic notation from algebraic number theory.

2.1 Linear forms in logarithms

Let γ be an algebraic number of degree d with minimal primitive polynomial

f(X) := a_{0}X^{d} + a_{1}X^{d−1} + ⋯ + a_{d} = a_{0} ∏_{i=1}^{d}(X − γ^{(i)}) ∈ ℤ[X],

where the a_{i} are relatively prime integers, a_{0} > 0, and the γ^{(i)} are conjugates of γ. Then

h(γ) = \frac{1}{d}(log a_{0} + ∑_{i=1}^{d} log(max\{|γ^{(i)}|, 1\}))  (4)

is called the logarithmic height of γ.

With the established notation, Matveev (see [15] or [8, Theorem 9.4]) proved the following result.

Theorem 2. Assume that η_{1}, …, η_{t} are positive real algebraic numbers in a real algebraic number field 𝕂 of degree D, b_{1}, …, b_{t} are rational integers, and

Λ := η_{1}^{b_{1}} ⋯ η_{t}^{b_{t}} − 1,

is not zero. Then

|Λ| ≥ exp(−1.4 · 30^{t+3} · t^{4.5} · D^{2}(1 + log D)(1 + log B)A_{1} ⋯ A_{t}),

where

B ≥ max{|b_{1}|, …, |b_{t}|},

and

A_{i} ≥ max{Dh(η_{i}), | log η_{i}|, 0.16}, for all i = 1, …, t.

2.2 The de Weger reduction algorithm

Here we present a variant of the reduction method of Baker and Davenport [3] (and improved by Dujella and Pethö [12]) due to de Weger [19].

Let ϑ_{1}, ϑ_{2}, δ ∈ ℝ be given and let x_{1}, x_{2} ∈ ℤ be unknowns. Let

Λ = δ + x_{1}ϑ_{1} + x_{2}ϑ_{2}.  (5)

Set X = max{|x_{1}, |x_{2}|}. Let X_{0}, Y be positive. Assume that

|Λ| < c exp (−ρY)  (6)

and

Y ≤ X ≤ X_{0},  (7)

where c, ρ be positive constants. When δ = 0 in (5), we get

Λ = x_{1}ϑ_{1} + x_{2}ϑ_{2}.

Put ϑ = −ϑ_{1}/ϑ_{2}. We assume that x_{1} and x_{2} are coprime. Let the continued fraction expansion of ϑ be given by

[a_{0}, a_{1}, a_{2}, …],

and let the kth convergent of ϑ be p_{k}/q_{k} for k = 0, 1, 2, . . .. We may assume without loss of generality that |ϑ_{1}| < |ϑ_{2}| and x_{1} > 0. We have the following results.

Lemma 3. [19, Lemma 3.1] If (6) and (7) hold for x_{1}, x_{2} with X ≥ 1 and δ = 0, then (−x_{2}, x_{1}) = (p_{k}, q_{k}) for an index k that satisfies

k ≤ −1 + \frac{log(1 + X_{0}\sqrt{5})}{log (\frac{1+\sqrt{5}}{2})} := Y_{0}.

Lemma 4. [19, Lemma 3.2] Let

A = max_{0≤k≤Y_{0}} a_{k+1}.

If (6) and (7) hold for x_{1}, x_{2} with X ≥ 1 and δ = 0, then

Y < \frac{1}{ρ} log (\frac{c(A + 2)}{|ϑ_{2}|}) + \frac{1}{ρ} log X < \frac{1}{ρ} log (\frac{c(A + 2)X_{0}}{|ϑ_{2}|} ) .

When δ ≠ 0 in (5), put ϑ = −ϑ_{1}/ϑ_{2} and ψ = δ/ϑ_{2}. Then we have

\frac{Λ}{ϑ_{2}} = ψ − x_{1}ϑ + x_{2}.

Let p/q be a convergent of ϑ with q > X_{0}. For a real number x, we let ∥x∥ = min{|x−n| : n ∈ ℤ} be the distance from x to the nearest integer. We have the following result.

Lemma 5. [19, Lemma 3.3] Suppose that

∥qψ∥ > \frac{2X_{0}}{q} .

Then the solutions of (6) and (7) satisfy

Y < \frac{1}{ρ} log (\frac{q^{2}c}{|ϑ_{2}|X_{0}} ) .

2.3 Properties of the Leonardo sequence

The characteristic equation of (Le_{m})_{m≥0} is x^{3} − 2x^{2} − 1 = 0, which has roots α = \frac{1+\sqrt{5}}{2} and β = \frac{−1}{α} (see [2]). The Binet formula for Le_{m} is

Le_{m} = 2 (\frac{α^{m+1} − β^{m+1}}{α − β} ) − 1 = \frac{α(2α^{m} − 1) − β(2β^{m} − 1)}{α − β} for all m ≥ 0.  (8)

Lemma 6. The inequality

α^{m} ≤ Le_{m} ≤ α^{m+1},  (9)

holds for all positive integers m ≥ 2.

Proof. This can be easily proved by the method of induction on m.

Lemma 7. [9, Lemma 2.1] For all m ≥ 0, the m-th Leonardo number Le_{m} is an odd number.

2.4 Properties of the k-generalized Fibonacci sequence

In this subsection, we recall some facts and properties of the k-generalized Fibonacci sequence which will be used later. The characteristic polynomial of the k-generalized Fibonacci sequence is

Ψ_{k}(x) = x^{k} − x^{k−1} − ⋯ − x − 1.

The polynomial Ψ_{k}(x) is irreducible over ℚ[x] and has just one root outside the unit circle. It is real and positive, so it satisfies φ(k) > 1. The other roots are strictly inside the unit circle. Throughout this paper, φ := φ(k) denotes that single root, which is located between 2(1 − 2^{−k}) and 2 (see [20]). To simplify the notation, in general, we omit the dependence of k on φ.

Dresden and Du [11] gave the following simplified Binet-like formula for F_{n}^{(k)}:

F_{n}^{(k)} = ∑_{i=1}^{k} f_{k}(φ_{i})φ_{i}^{n−1} = ∑_{i=1}^{k} \frac{φ_{i} − 1}{2 + (k + 1)(φ_{i} − 2)}φ_{i}^{n−1},  (10)

where φ := φ_{1}, φ_{2}, …, φ_{k} are the roots of the characteristic polynomial Ψ_{k}(x). It was also proved in [11] that the contribution of the roots that lie inside the unit circle to the formula (10) is very small, namely that the approximation

|F_{n}^{(k)} − f_{k}(φ)φ^{n−1}| < \frac{1}{2} holds for all n ≥ 2 − k.  (11)

Furthermore, it was shown by Bravo and Luca in [7] that the inequality

φ^{n−2} ≤ F_{n}^{(k)} ≤ φ^{n−1} holds for all n ≥ 1 and k ≥ 2.  (12)

The first direct observation is that the first k + 1 non-zero terms in F_{n}^{(k)} are powers of 2, namely

F_{1}^{(k)} = 1, F_{2}^{(k)} = 1, F_{3}^{(k)} = 2, F_{4}^{(k)} = 4, …, F_{k+1}^{(k)} = 2^{k−1},

while the next term in the above sequence is F_{k+2}^{(k)} = 2^{k} − 1. Thus, we have that

F_{n}^{(k)} = 2^{n−2} holds for all 2 ≤ n ≤ k + 1.

We also observe that the recursion (2) implies the three-term recursion

F_{n}^{(k)} = 2F_{n−1}^{(k)} − F_{n−k−1}^{(k)} for all n ≥ 3,

which shows that the k-generalized Fibonacci sequence grows at a rate less than 2^{n−2}. In fact, the inequality F_{n}^{(k)} < 2^{n−2} holds for all n ≥ k + 2 (see [7, Lemma 2]).

The following result was proved by Bravo and Luca [7].

Lemma 8. Let k ≥ 2, φ be the dominant root of (F_{n}^{(k)})_{n≥−(k−2)}, and consider the function defined in (10). Then the inequalities

\frac{1}{2} < f_{k}(φ) < \frac{3}{4} and |f_{k}(φ^{(i)})| < 1 hold for all 2 ≤ i ≤ k.

In addition, they proved that the logarithmic height of f satisfies

h(f_{k}(φ)) < log(k + 1) + log 4 for all k ≥ 2.  (13)

We finish this subsection with the following estimate due to Bravo, Gómez, and Luca [5], which will be used later.

Lemma 9. Let k ≥ 2 and suppose that n < 2^{k/2}. Then

F_{n}^{(k)} = 2^{n−2}(1 + ξ) where |ξ| < \frac{1}{2^{k/2}} .

3 Proof of Theorem 1

Proof. First, note that F_{1}^{(k)} = F_{2}^{(k)} = 1 = Le_{0} = Le_{1}. Therefore, we may assume that n ≥ 3. For 3 ≤ n ≤ k + 1, we have that F_{n}^{(k)} = 2^{n−2}, but Le_{m} is an odd number for m ≥ 0. Thus, there is no solution of (3) in this range. From now, we assume that n ≥ k + 2 and k ≥ 2.

Combining the inequalities (9) and (12) together with equation (3), we have

φ^{n−2} ≤ α^{m+1} and α^{m} ≤ φ^{n−1}.

Then, we deduce that

(n − 2) (\frac{log φ}{log α}) − 1 ≤ m ≤ (n − 1) (\frac{log φ}{log α}) .

Using the fact 2(1 − 2^{−k}) < φ(k) < 2 for all k ≥ 2, it follows that

0.8n − 2.6 < m < 1.5n − 1.5.  (14)

3.1 An inequality for n and m in terms of k

By using (3), (8), (11) and taking absolute value, we obtain

|f_{k}(φ)φ^{n−1} − \frac{2α^{m+1}}{\sqrt{5}} | < \frac{1}{2} + \frac{2|β|^{m+1}}{\sqrt{5}} + \frac{| − α|}{\sqrt{5}} + \frac{|β|}{\sqrt{5}}.  (15)

Dividing both sides of the above inequality by \frac{2α^{m+1}}{\sqrt{5}} , we conclude that

|(\frac{\sqrt{5}f_{k}(φ)}{2}) φ^{n−1}α^{−(m+1)} − 1| < 4α^{−m}.  (16)

Let

Λ_{1} := (\frac{\sqrt{5}f_{k}(φ)}{2}) φ^{n−1}α^{−(m+1)} − 1,  (17)

and inequality (16) becomes

|Λ_{1}| < 4α^{−m}.  (18)

Before applying Theorem 2, we need to prove that Λ_{1} ≠ 0. Assume that Λ_{1} = 0, then we get

f_{k}(φ) = \frac{2}{\sqrt{5}}φ^{−(n−1)}α^{(m+1)},

and so f_{k}(φ) is an algebraic integer, which is impossible. Thus, Λ_{1} ≠ 0. Therefore, we apply Theorem 2 to get a lower bound for Λ_{1} given by (17) with the parameters:

η_{1} := \frac{\sqrt{5}f_{k}(φ)}{2}, η_{2} := φ, η_{3} := α,

and

b_{1} := 1, b_{2} := n − 1, b_{3} := −(m + 1).

Note that η_{1}, η_{2}, η_{3} are positive real numbers and belong to the field 𝕂 := ℚ(φ, \sqrt{5}). So we can take D := [𝕂 : ℚ] ≤ 2k. Since h(η_{2}) = (log φ)/k < (log 2)/k and h(η_{3}) = (log α)/2, we choose

max{2kh(η_{2}), | log η_{2}|, 0.16} = 2 log 2 := A_{2}

and

max{2kh(η_{3}), | log η_{3}|, 0.16} = k log α := A_{3}.

By using the estimate (13) and the properties of logarithmic height, we get that for all k ≥ 2

h(η_{1}) ≤ h(f_{k}(φ)) + h (\frac{\sqrt{5}}{2})
< log(k + 1) + log 4 + log (2\sqrt{5})
< 5.8 log k.

Thus, we obtain

max{2kh(η_{1}), | log η_{1}|, 0.16} = 11.6k log k := A_{1}.

In addition, by (14) we take B := 1.5n. Then by Theorem 2, we have

|Λ_{1}| > exp (−1.432 × 10^{11}(2k)^{2}(1 + log 2k)(1 + log 1.5n)(11.6k log k)(2 log 2)(k log α)) . (19)

Comparing (18) and (19), taking logarithms and then performing the respective calculations, we get that

m log α − log 4 < 4.43 × 10^{12}k^{4} log k(1 + log 2k)(1 + log 1.5n).

Taking into consideration the facts 1 + log 2k < 3.5 log k for all k ≥ 2 and 1 + log 1.5n < 2.1 log n for all n ≥ 4, we conclude that

m < 6.77 × 10^{13}k^{4} log^{2} k log n.

Using (14), the last inequality becomes

\frac{n}{log n} < 8.47 × 10^{13}k^{4} log^{2} k.  (20)

Since the function x ↦ x/ log x is increasing for all x > e, it is easy to check that

\frac{x}{log x} < S ⟹ x < 2S log S whenever S ≥ 3.  (21)

Then, taking x := n and S := 8.47 × 10^{13}k^{4} log^{2} k, inequality (21) together with 32.07 + 4 log k + 2 log log k < 49.3 log k for all k ≥ 2, yields

n < 2 (8.47 × 10^{13}k^{4} log^{2} k) log (8.47 × 10^{13}k^{4} log^{2} k)
< (1.69 × 10^{14}k^{4} log^{2} k)(32.07 + 4 log k + 2 log log k)
< 8.35 × 10^{15}k^{4} log^{3} k.

By combining the above results, we obtain the following lemma.

Lemma 10. If (n, k, m) is a solution in integers of equation (3) with k ≥ 2 and n ≥ k + 2, then the inequalities

0.6m < n < 8.35 × 10^{15}k^{4} log^{3} k  (22)

hold.

3.2 The case of small k

Suppose now that k ∈ [2, 230]. In order to apply Lemma 5, we let

Γ_{1} := (n − 1) log φ − (m + 1) log α + log (\frac{\sqrt{5}f_{k}(φ)}{2}) .  (23)

Then e^{Γ_{1}} − 1 := Λ_{1}, where Λ_{1} is defined by (17). Therefore, (18) can be written as

|e^{Γ_{1}} − 1| < 4α^{−m}.  (24)

Note that Γ_{1} ≠ 0. Since Λ_{1} ≠ 0, we distinguish the following cases. If Γ_{1} > 0, then e^{Γ_{1}}−1 > 0. Using the fact x ≤ e^{x} − 1 for all x ∈ ℝ and the inequality (24), we obtain

0 < Γ_{1} < 4α^{−m}.

If, on the contrary, Γ_{1} < 0, then 4α^{−m} < 1/2 holds for all m ≥ 5. Thus, from (24), we have |e^{Γ_{1}} − 1| < 1/2 and therefore e^{|Γ_{1}|} < 2. Since Γ_{1} < 0, we have

0 < |Γ_{1}| ≤ e^{|Γ_{1}|} − 1 = e^{|Γ_{1}|}|e^{Γ_{1}} − 1| < 8α^{−m}.

In both cases, we have

0 < |Γ_{1}| < 8α^{−m} < 8 exp(−0.48 × m).  (25)

Let

c := 8, ρ := 0.48, ψ := \frac{log (^{\frac{\sqrt{5}f_{k}(φ)}{2}})}{log φ},

and

ϑ := \frac{log α}{log φ}, ϑ_{1} := − log α, ϑ_{2} := log φ, δ := log (\frac{\sqrt{5}f_{k}(φ)}{2}) .

For each k ∈ [2, 230], we find a good approximation of φ and a convergent p_{l}/q_{l} of the continued fraction of ϑ such that q_{l} > X_{0}, where X_{0} = ⌊8.35 × 10^{15}k^{4} log^{3} k⌋, which is an upper bound of max{n − 1, m} from Lemma 10. After doing this, we use Lemma 5 on inequality (25). A computer search with Mathematica revealed that if k ∈ [2, 230], then the maximum value of ⌊\frac{1}{ρ} log (q^{2}c/|ϑ_{2}|X_{0})⌋ is 318, which is an upper bound on m according to Lemma 5. Hence, we deduce that the possible solutions (n, m, k) of the equation (3) for which k ∈ [2, 230] have m ≤ 318, therefore we use inequalities (14) to obtain n ≤ 401.

Finally, we used Mathematica to compare F_{n}^{(k)} and Le_{m} for the range 4 ≤ n ≤ 401 and 2 ≤ m ≤ 318, with m < n/0.6 and checked that the solutions of equation (3) are

F_{4}^{(2)} = 3 = Le_{2}, F_{5}^{(2)} = 5 = Le_{3}, and F_{6}^{(4)} = 15 = Le_{5} .

3.3 The case of large k

From now on, we assume that k > 230. Here, it follows from Lemma 10 that

0.6m < n < 8.35 × 10^{15}k^{4} log^{3} k < 2^{k/2}.

Using (8) and Lemma 9, we can write (3) as

2^{n−2} − \frac{2α^{m+1}}{\sqrt{5}} = 2^{n−2}ξ − \frac{2β^{m+1}}{\sqrt{5}} − \frac{α}{\sqrt{5}} + \frac{β}{\sqrt{5}}.  (26)

Taking absolute values on both sides of (26), we have that

|2^{n−2} − \frac{2α^{m+1}}{\sqrt{5}} | < \frac{2^{n−2}}{2^{k/2}} + \frac{5}{\sqrt{5}}.

Dividing both sides of the above inequality by 2^{n−2} and taking into account that 1/2^{n−2} < 1/2^{k/2} for n ≥ k + 2, we obtain

|1 − 2^{−n} \frac{8}{\sqrt{5}}α^{m+1}| < \frac{3.23}{2^{k/2}}.  (27)

Applying Theorem 2 for the left-hand side, we set

Λ_{2} := 2^{−n} \frac{8}{\sqrt{5}}α^{m+1} − 1.

Note that Λ_{2} ≠ 0. Indeed, if Λ_{2} = 0, then α^{2(m+1)} is a rational number, which is not possible for all positive integers m. Therefore Λ_{2} ≠ 0. We take t := 3,

η_{1} := 2, η_{2} := \frac{8}{\sqrt{5}}, η_{3} := α,

and

b_{1} := −n, b_{2} := 1, b_{3} := m + 1.

Note that 𝕂 := ℚ(α) contains η_{1}, η_{2}, η_{3} and has D := 2. Since m < 1.5n, we deduce that B := max{|b_{1}|, |b_{2}|, |b_{3}|} = 1.5n. The logarithmic heights for η_{1}, η_{2}, and η_{3} are calculated as follows:

h(η_{1}) = log 2, h(η_{2}) = log (8\sqrt{5}) and h(η_{3}) = \frac{log α}{2}.

Thus, we can take

A_{1} := 2 log 2, A_{2} := 5.8 and A_{3} := log α.

As before, by applying Theorem 2, we have

|Λ_{2}| > exp (−8.27 × 10^{12} log n) ,  (28)

where 1 + log 1.5n < 2.1 log n holds for all n ≥ 4. Comparing (27) and (28), we obtain

k < 2.39 × 10^{13} log n.

By Lemma 10 and using the fact 36.66 + 4 log k + 3 log log k < 11.7 log k for all k > 230, we get

k < 2.39 × 10^{13} log (8.35 × 10^{15}k^{4} log^{2} k)
< (2.39 × 10^{13})(36.66 + 4 log k + 3 log log k)
< 2.8 × 10^{14} log k.

Solving the above inequality by using the relation (21) gives

k < 1.87 × 10^{16}.

Again from Lemma 10, we obtain

n < 5.38 × 10^{85} and m < 8.97 × 10^{85}.  (29)

Let

Γ_{2} := (m + 1) log α − n log 2 + log (\frac{8}{\sqrt{5}}) .  (30)

Using a similar method to show the inequality (25), one can see that

0 < |Γ_{2}| < \frac{6.46}{2^{k/2}} < 6.46 × exp(−0.34 × k)  (31)

holds for all k > 230. The inequality (29) implies that we can take X_{0} := 8.97 × 10^{85}. Further, we choose

c := 6.46, ρ := 0.34, ψ := −\frac{log (\frac{8}{\sqrt{5}})}{log 2},

and

ϑ := \frac{log α}{log 2}, ϑ_{1} := log α, ϑ_{2} := − log 2, δ := log (\frac{8}{\sqrt{5}}) .

Using Lemma 3 with c := 6.46, ρ := 0.34 and X_{0} := 8.97 × 10^{85}, we get Y_{0} := 411.954 . . .. Let

[a_{0}, a_{1}, a_{2}, …] := [0, 1, 2, 3, 1, 2, 3, 2, 4, 2, 1, 2, 11, 2, 1, 11, 1, 1, 134, 2, 2, 2, 1, 4, 1, 1, 3, 1, …]

be the continued fraction expansion of log α/ log 2. With the help of Mathematica, we find that

max_{0≤k≤Y_{0}} a_{k+1} = 880 := A.

Then by Lemma 4, we have

k < \frac{1}{0.34} · log (\frac{6.46 · 882 · 8.97 · 10^{85}}{log 2} ) < 609.

With the above upper bound on k and by Lemma 10, we have

n < 3.1 × 10^{29} and m < 5.2 × 10^{29}.  (32)

We apply again Lemma 4 with X_{0} := 5.2 × 10^{29}. Hence by Lemma 3, we obtain Y_{0} := 142.863 … and A := 134. According to Lemma 4, it becomes

k < \frac{1}{0.34} · log (\frac{6.46 · 136 · 5.2 · 10^{29}}{log 2} ) < 223,

which contradicts our assumption that k > 230. Hence, we have shown that there are no solutions (n, k, m) to equation (3) with k > 230. This completes the proof of Theorem 1.

4 Acknowledgments

The authors would like to thank the anonymous referee for their useful comments and suggestions that improved the exposition of the paper.

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2020 Mathematics Subject Classification: Primary 11B39; Secondary 11J86, 11R52. Keywords: k-generalized Fibonacci number, Leonardo number, linear form in logarithms, reduction method.

(Concerned with sequences \underline{A000032}, \underline{A000045}, and \underline{A001595}.) Received January 31 2023; revised versions received February 8 2023; May 17 2023; May 26 2023. Published in Journal of Integer Sequences, June 12 2023. Return to Journal of Integer Sequences home page.