Eventually monotonic solutions of the generalized Fibonacci equations

Ahmet Yaşar Özban; Halim Özdemir

doi:10.1515/math-2025-0198

Article Open Access

Eventually monotonic solutions of the generalized Fibonacci equations

Ahmet Yaşar Özban and Halim Özdemir

Published/Copyright: November 19, 2025

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From the journal Open Mathematics Volume 23 Issue 1

Abstract

The purpose of this work is, for the first time, to introduce some results about the eventually monotonic solutions of the generalized Fibonacci equations. In addition, some concrete examples are given to illustrate the theoretical results.

Keywords: generalized Fibonacci equation; monotonic solution; eventually monotonic solution

MSC 2020: 11B39; 39A06

1 Introduction

For the Fibonacci sequence F n n = 1 defined by the difference equation

F n + 2 = F n + 1 + F n , n = 1,2 , … ,

where F ₁ = 0 and F ₂ = 1, Binet’s formula states that (see, for example, [1])

F n = φ n − 1 − 1 − φ n − 1 5 , n = 1,2 , … ,

where φ = 1 + 5 / 2 is the golden ratio, the positive solution of 1 + φ = − φ / 1 − φ . Meanwhile, note the equalities φ 2 / 1 + φ = 1 , φ = − 1 / 1 − φ and φ ² = 1 + φ in relation to the golden ratio.

Consider the generalized Fibonacci sequence defined by the difference equation

(1) G n + 2 = G n + 1 + G n , n = 1,2,3 , … ,

where G ₁ = a and G ₂ = b with a , b ∈ R [2] (for more in-depth historical information, see, for instance, the references in [3]). It is clear that G n n = 1 = F n n = 1 if a = b = 1. Recall that; if a sequence c n n of real numbers is strictly increasing (decreasing) for all n > N for some positive integer N, it is said to be eventually strictly monotonic, that is eventually strictly increasing (decreasing).

In the literature, many studies have been carried out on the monotonic, eventually monotonic, periodic, eventually periodic, and oscillatory solutions of difference equations. Moreover, although there are many studies about the solutions of some difference equations in terms of Fibonacci and generalized Fibonacci numbers (see, for example, [4], [5], [6], [7], [8], [9], [10], [11], [12]), to the best of the authors’ knowledge, there is no study on the eventual monotonicity of the generalized Fibonacci sequences or, equivalently, eventually monotonic solutions of the difference equation (1).

In the first part of Section 2, we determine the conditions for the sequence to become eventually monotonic when a > 0 and b < 0 and give some examples. In the last part of Section 2, we discuss the eventual monotonicity of the sequence under the condition a < 0 and b > 0 and give relevant examples.

2 Main results

It is clear that if a, b > 0 (a, b < 0), then the sequence G n n = 1 is strictly increasing and positive (strictly decreasing and negative). Here we will investigate the behavior of the sequence G n n = 1 for the initial values a, b with ab < 0. We first consider two cases out of the four possible:

(i) If a > 0 and b < 0 such that b ≥ a , then a + b ≤ 0. So, G ₁ = a > 0, G ₂ = b < 0, G ₃ = a + b ≤ 0 and G ₄ = a + 2b < 0, that is, G n is eventually strictly decreasing and negative.
(ii) If a < 0 and b > 0 such that a ≤ b , then a + b ≥ 0. Hence G ₁ = a < 0, G ₂ = b > 0, G ₃ = a + b ≥ 0 and G ₄ = a + 2b > 0, that is, G n is eventually strictly increasing and positive.

The remaining two cases are:
(iii) If a > 0 and b ∈ − a , 0 , then a + b > 0. So, G ₁ = a > 0, G ₂ = b < 0, G ₃ = a + b > 0, G ₄ = a + 2b, G ₅ = 2a + 3b, G ₆ = 3a + 5b, and so on.
(iv) If a < 0 and b ∈ 0 , − a , then G ₁ = a < 0, G ₂ = b > 0, G ₃ = a + b < 0, G ₄ = a + 2b, G ₅ = 2a + 3b, G ₆ = 3a + 5b, and so on.

In both of the last two cases, determining the behavior of the sequences requires some deeper analysis.

Now, we will first discuss the case (iii). Let G ₁ = a > 0, G 2 = b ∈ − a , 0 and, so, G ₃ = a + b > 0. Hence, G ₁ = a and G ₃ = a + b are two positive terms of the sequence seperated by the negative term G ₂ = b. It is clear that the signs of the terms G ₄, G ₅, G ₆, … depends on a and b or the relation between a and b. Noticing that the terms of the sequence G n n = 1 can be written as; G 1 = 1 ⋅ a + F 1 a + b , G 2 = − 1 ⋅ a + F 2 a + b and G n = − F n − 2 a + F n a + b for n = 3, 4, … and inspiring from the Fibonacci sequence, we define a sequence A n n = 1 by

A n = − φ n − 3 − 1 − φ n − 3 5 , n = 1,2 , … .

Since A ₁ = 1, A ₂ = −1 and A _n = −F _n−2, n = 3, 4, …, it follows that

G n = A n a + F n a + b , n = 1,2,3 , … ,

or, equivalently,

G n = a A n + F n 1 + b a , n = 1,2,3 , … ,

that is,

G n = a 1 − φ n − 3 − φ n − 3 + φ n − 1 − 1 − φ n − 1 1 + b a / 5 .

To prove that the sequence G n n = 1 is eventually strictly increasing and positive (decreasing and negative), we need to show that for fixed values of a > 0 and b ∈ − a , 0 , there exists a positive integer N such that G _n ⋅ G _n+1 > 0 for all n > N. For n = 1, 2, …, we can write the successive terms G _n and G _n+1 as

G k = a φ k − 3 φ + 1 + φ b a − 1 − φ k − 3 − φ − 1 + φ − 2 b a / 5 ,

where k = n, n + 1. Their signs depend on the signs of φ + 1 + φ b a and 1 − φ k − 3 , since φ ^k−3 > 0 for any k, and φ − 1 + φ − 2 b a > 0 for a > 0 and b ∈ − a , 0 .

Before starting to discuss distinct cases in detail, for a fixed a > 0, let us define two functions N ₁ and N ₂ which we will use in the sequel:

N 1 ( b ) = ln φ − 1 + φ − 2 b a / φ + 1 + φ b a ln 1 + φ + 2 ,

N 2 ( b ) = ln φ − 1 + φ − 2 b a / 1 + φ φ + 1 + φ b a ln 1 + φ + 4 ,

where b a ∈ − 1 , − φ / 1 + φ ∪ − φ / 1 + φ , 0 . It is clear that N ₂(b) = N ₁(b) + 1.

Case 1. G _n > 0 and G _n+1 > 0. We need to show that

φ n − 3 φ + 1 + φ b a > 1 − φ n − 3 − φ − 1 + φ − 2 b a

or, equivalently,

(2) φ n − 2 φ + 1 + φ b a > − 1 − φ n − 4 − φ − 1 + φ − 2 b a ,

and

(3) φ n − 2 φ + 1 + φ b a > 1 − φ n − 2 − φ − 1 + φ − 2 b a .

Let

(4) β = 1 − φ n − 4 φ − 1 + φ − 2 b a ,

(5) γ = − 1 − φ n − 2 φ − 1 + φ − 2 b a .

So, combining inequalities (2) and (3), we must show that there is positive integer N such that

(6) φ n − 2 φ + 1 + φ b a > max β , γ ,

for all n > N. We should note that if b a = − φ / φ + 1 , then the inequality (6) is not satisfied for any a > 0, b ∈ − a , 0 and n since max β , γ = γ > 0 if n is odd, and max β , γ = β > 0 if n is even, and also φ − 1 + φ − 2 b a < 0 .

We will handle the subcases − φ / φ + 1 < b a < 0 and − 1 < b a < − φ / φ + 1 separately.

Case 1.1. φ + 1 + φ b a > 0 . Since a > 0 and b ∈ − a , 0 or b a ∈ − 1,0 , we have b a ∈ − φ / 1 + φ , 0 ⊂ − 1,0 . Let n > N 1 ( b ) be odd. Then

n − 2 ln 1 + φ > ln φ − 1 + φ − 2 b a / φ + 1 + φ b a

1 + φ n − 2 > φ − 1 + φ − 2 b a / φ + 1 + φ b a .

Since n − 2 is odd, it follows that

− − 1 + φ n − 2 > φ − 1 + φ − 2 b a / φ + 1 + φ b a

and using the fact that φ / 1 − φ = − 1 + φ , we get

− φ / 1 − φ n − 2 > φ − 1 + φ − 2 b a / φ + 1 + φ b a .

Then, multiplying both sides by − 1 − φ n − 2 φ + 1 + φ b a > 0 we obtain

φ n − 2 φ + 1 + φ b a > − 1 − φ n − 2 φ − 1 + φ − 2 b a = γ .

Now, since 1 − φ n − 4 < 0 for odd values of n and − φ − 1 + φ − 2 b a < 0 , β < 0 and γ > 0. So, φ n − 2 φ + 1 + φ b a > max β , γ = γ , as is required by inequality (6).

Let n > N 2 ( b ) be even. So,

n − 4 ln 1 + φ > ln φ − 1 + φ − 2 b a / 1 + φ φ + 1 + φ b a

and by 1 + φ = − φ / 1 − φ ,

ln − φ / 1 − φ n − 4 > ln φ − 1 + φ − 2 b a / 1 + φ φ + 1 + φ b a

− φ / 1 − φ n − 4 > 1 1 + φ φ − 1 + φ − 2 b a φ + 1 + φ b a .

Multiplying both sides by φ 2 − 1 − φ n − 4 φ + 1 + φ b a > 0 , we get

φ n − 2 φ + 1 + φ b a > 1 − φ n − 4 φ − 1 + φ − 2 b a = β > 0 .

Now, since 1 − φ n − 4 > 0 for even values of n and − φ − 1 + φ − 2 b a < 0 , one has γ < 0. So, φ n − 2 φ + 1 + φ b a > max β , γ = β , as is required by inequality (6).

Having obtained these inequalities and using the fact that N ₂(b) = N ₁(b) + 1, the first result, result for the case φ + 1 + φ b a > 0 , can be stated as follows.

Theorem 1.

If a > 0 and b a ∈ − φ / 1 + φ , 0 ⊂ − 1,0 , then for all values of n > N = N 2 ( b ) , 0 < G _n < G _n+1, that is, the sequence G n n = 1 is eventually strictly increasing and positive.

Example 1.

For a = 1 and b = −0.618030 we have b a = − 0.618030 ∈ − φ / 1 + φ , 0 = − 0.618 034,0 and b = 0.618030 < a = 1 . Then, N ₁(b) = 13.753558 and N ₂(b) = 14.753558. Hence N = 15. From the first column of Table 1 it is seen that the sequence G n n = 1 is eventually strictly increasing and positive.

Table 1:

The behaviour of the generalized Fibonacci sequences in Examples 1 and 2.

a = 1, b = −0.618030, b a ∈ − φ / 1 + φ , 0	a = 1, b = −0.61803398875, b a ∈ − φ / 1 + φ , 0
G ₁ = 1	G ₁ = 1
G ₂ = −0.618030	G ₂ = −0.6180339887498
G ₃ = 0.381 970	G ₃ = 0.38196601125
⋮	⋮
G ₁₂ = −0.004670	G ₃₀ = −8.209042 × 10⁻⁷
G ₁₃ = 0.003680	G ₃₁ = 6.164080 × 10⁻⁷
G ₁₄ = −0.000990	G ₃₂ = −2.044962 × 10⁻⁷
G ₁₅ = 0.002690	G ₃₃ = 4.119118 × 10⁻⁷
G ₁₆ = 0.001700	G ₃₄ = 2.074156 × 10⁻⁷
⋮	⋮

Example 2.

For a = 1 and b = −0.618033988750 we have b a = − 0.61803398875 ∈ − φ / 1 + φ , 0 = − 0.618 034,0 and b = 0.61803398875 < a = 1 . Then, N ₂(b) = 32.993413. So, N = 33. From the second column of Table 1 it is seen that the sequence G n n = 1 is also eventually strictly increasing and positive. So, comparing with the previous example, it is clear that N becomes larger and larger as b a becomes closer and closer to − φ / 1 + φ .

Case 1.2. φ + 1 + φ b a < 0 . Since a > 0 and b ∈ − a , 0 , we have b ∈ − a , − a φ / 1 + φ or b a ∈ − 1 , − φ / 1 + φ . Moreover,

− φ − 1 + φ − 2 b a < φ + 1 + φ b a < 0 .

For odd values of n, 1 − φ n − 2 < 0 and − 1 − φ n − 4 > 0 . Hence β < 0 and γ > 0. Then,

max β , γ = γ > 0 .

Since φ n − 2 φ + 1 + φ b a < 0 and γ > 0, inequality (6) is not satisfied for any odd n for φ + 1 + φ b a < 0 .

For even values of n we have 1 − φ n − 4 > 0 . Hence β > 0 and γ < 0. Then,

max β , γ = β > 0

But, since φ n − 2 φ + 1 + φ b a < 0 and β > 0, inequality (6) is not satisfied for any even n for φ + 1 + φ b a < 0 .

Case 2. G _n < 0 and G _n+1 < 0. We must show that

(7) φ n − 2 φ + 1 + φ b a < 1 − φ n − 4 φ − 1 + φ − 2 b a

and

(8) φ n − 2 φ + 1 + φ b a < − 1 − φ n − 2 φ − 1 + φ − 2 b a .

Now, recalling β and γ given by (4) and (5), respectively, and combining (7) and (8), we need to prove that

(9) φ n − 2 φ + 1 + φ b a < min β , γ ,

for all n ≥ N, where N is a sufficiently large positive integer. It is clear that if b a = − φ / 1 + φ , then inequality (9) is not satisfied for any a and n since min β , γ = β < 0 if n is odd, and min β , γ = γ < 0 if n is even. We will treat the subcases φ + 1 + φ b a > 0 and φ + 1 + φ b a < 0 separately.

Case 2.1. φ + 1 + φ b a > 0 . For odd values of n, 1 − φ n − 2 < 0 , − 1 − φ n − 4 > 0 . Then β < 0 and γ > 0 imply that

(10) min β , γ = β < 0 .

But φ n − 2 φ + 1 + φ b a > 0 , β < 0 imply that inequality (9) is not satisfied for any odd n for φ + 1 + φ b a > 0 .

For even values of n one has 1 − φ n − 2 > 0 and − 1 − φ n − 4 < 0 . Then β > 0 and γ < 0 imply that

min β , γ = γ < 0 .

Since φ n − 2 φ + 1 + φ b a > 0 and γ < 0, inequality (9) is not satisfied for any even n for φ + 1 + φ b a > 0 .

Case 2.2. φ + 1 + φ b a < 0 . For odd values of n we have 1 − φ n − 2 < 0 , − 1 − φ n − 4 > 0 . Then β < 0 and γ > 0 imply that

min β , γ = β < 0 .

Let n > N 2 ( b ) be odd. Then,

n − 4 > ln φ − 1 + φ − 2 b a / − 1 + φ φ + 1 + φ b a ln 1 + φ

− − 1 + φ n − 4 > − φ − 1 + φ − 2 b a 1 + φ φ + 1 + φ b a .

Using the fact that − 1 + φ = φ / ( 1 − φ ) we get

− φ / 1 − φ n − 4 > − φ − 1 + φ − 2 b a 1 + φ φ + 1 + φ b a .

Now, multiplying both sides by 1 + φ − 1 − φ n − 4 φ + 1 + φ b a < 0 we obtain

(11) φ n − 2 φ + 1 + φ b a < 1 − φ n − 4 φ − 1 + φ − 2 b a = β < 0 ,

which is what is needed.

For even values of n one has 1 − φ n − 4 > 0 and hence − 1 − φ n − 4 < 0 . So, β > 0 and γ < 0, imply that

min β , γ = γ < 0 .

Let n > N 1 ( b ) be even. Then,

n − 2 ln 1 + φ > ln φ − 1 + φ − 2 b a / − φ + 1 + φ b a

or, since 1 + φ n − 2 = − 1 + φ n − 2 , it follows that

ln − 1 + φ n − 2 > ln φ − 1 + φ − 2 b a / − φ + 1 + φ b a

and, using the fact that − 1 + φ n − 2 = φ 1 − φ n − 2 > 0 ,

ln φ / 1 − φ n − 2 > ln φ − 1 + φ − 2 b a / − φ + 1 + φ b a

or,

φ / 1 − φ n − 2 > φ − 1 + φ − 2 b a / − φ + 1 + φ b a

and hence

(12) φ n − 2 φ + 1 + φ b a < − 1 − φ n − 2 φ − 1 + φ − 2 b a = γ < 0 ,

which is the required inequality.

Combining (11) and (12) and using the fact that N ₂(b) = N ₁(b) + 1, the second result, result for the case φ + 1 + φ b a < 0 , can be stated as follows.

Theorem 2.

If a > 0 and b a ∈ − 1 , − φ / 1 + φ ⊂ − 1,0 , then for all values of n > N = N 2 ( b ) , G _n+1 < G _n < 0, that is, the sequence G n n = 1 is eventually strictly decreasing and negative.

Example 3.

For a = 1.5 and b = −0.927060 we have b a = − 0.618 040 ∈ − 1 , − φ / 1 + φ = − 1 , − 0.618 034 and b = 0.927060 < a = 1.5 . Then, N ₂(b) = 14.327394. So, N = 15. From the first column of Table 2, it is seen that the sequence G n n = 1 is eventually strictly decreasing and negative.

Table 2:

The behaviour of the generalized Fibonacci sequences in Examples 3 and 4.

a = 1, b = −0.618030, b a ∈ − φ / 1 + φ , 0	a = 1, b = −0.618033988750, b a ∈ − φ / 1 + φ , 0
G ₁ = 1	G ₁ = 1
G ₂ = −0.618030	G ₂ = −0.618033988750
G ₃ = 0.381 970	G ₃ = 0.381966011250
⋮	⋮
G ₁₂ = −0.004670	G ₃₀ = −8.209042 × 10⁻⁷
G ₁₃ = 0.003680	G ₃₁ = 6.164080 × 10⁻⁷
G ₁₄ = −0.000990	G ₃₂ = −2.044962 × 10⁻⁷
G ₁₅ = 0.002690	G ₃₃ = 4.119118 × 10⁻⁷
G ₁₆ = 0.001700	G ₃₄ = 2.074156 × 10⁻⁷
⋮	⋮

Example 4.

For a = 1.5 and b = −0.927050983125 we have b a = − 0.61803398875 ∈ − 1 , − φ / 1 + φ = − 1 , − 0.618 034 and b = 0.927050983125 < a = 1.5 . Then, N ₂(b) = 32.886260. Hence N = 33. From the second column of Table 2, it is seen that the sequence G n n = 1 is eventually strictly decreasing and negative. So, comparing with the previous example, it is clear that N becomes larger and larger as b a becomes closer and closer to − φ / 1 + φ .

Now, finally consider the case (iv), where a < 0 and b > 0 such that a > b . The following results will be given without proofs, since we can obtain them following a way similar to the one we use to obtain the results in Theorems 1 and 2.

Theorem 3.

If a < 0 and b a ∈ − φ / 1 + φ , 0 ⊂ − 1,0 , then for all values of n > N = N 2 ( b ) , G _n+1 < G _n < 0, that is, the sequence G n n = 1 is eventually strictly decreasing and negative.

Example 5.

For a = −1 and b = 0.618030 we have b a = − 0.618030 ∈ − φ / 1 + φ , 0 = − 0.618 034,0 and b = 0.618030 < a = 1 . So, N ₂(b) = 14.753558. Hence N = 15. From the first column of Table 3, it is seen that the sequence G n n = 1 is eventually strictly decreasing and negative.

Table 3:

The behaviour of the generalized Fibonacci sequences in Examples 5 and 6.

a = −1, b = 0.618030, b a ∈ − φ / 1 + φ , 0	a = −1.5, b = 0.927060, b a ∈ − 1 , − φ / 1 + φ
G ₁ = −1	G ₁ = −1.5
G ₂ = 0.618030	G ₂ = 0.927060
G ₃ = −0.381 970	G ₃ = −0.572 940
⋮	⋮
G ₁₂ = 0.004670	G ₁₁ = −0.011700
G ₁₃ = −0.003680	G ₁₂ = 0.008340
G ₁₄ = 0.000990	G ₁₃ − 0.003360
G ₁₅ = −0.002690	G ₁₄ = 0.004980
G ₁₆ = −0.001500	G ₁₅ = 0.001620
⋮	⋮

Theorem 4.

If a < 0 and b a ∈ − 1 , − φ / 1 + φ ⊂ − 1,0 , then for all values of n > N = N 2 ( b ) , 0 < G _n < G _n+1, that is, the sequence G n n = 1 is eventually strictly increasing and positive.

Example 6.

For a = −1.5 and b = 0.927060 we have b a = − 0.618 040 ∈ − 1 , − φ / 1 + φ = − 1 , − 0.618 034 and b = 0.927060 < a = 1.5 . Then, N ₂(b) = 14.327394. Hence N = 15. From the second column of Table 3, it is seen that the sequence G n n = 1 is eventually strictly increasing and positive.

3 Conclusions

Properties such as periodicity, eventually periodicity, oscillation, monotonicity, and eventually monotonicity of solutions of difference equations are the subjects of many studies in the literature. However, the issue of eventually monotonicity of generalized Fibonacci sequences, or equivalently, eventually monotonic solutions of generalized Fibonacci equations has never been addressed in the literature. In this work, for the first time, we have determined the conditions required for the generalized Fibonacci equations to have eventually monotonic solutions and given examples for all cases.

Corresponding author: Ahmet Yaşar Özban, Department of Mathematics, Çankırı Karatekin University, TR18100, Çankırı, Türkiye, E-mail: ahmetyozban@karatekin.edu.tr

Acknowledgements

The authors are grateful to the anonymous referees for their invaluable suggestions and comments that greatly improve the quality and presentation of the original manuscript.

Research ethics: Not applicable.
Informed consent: Not applicable.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript.
Use of Large Language Models, AI and Machine Learning Tools: None declared.
Conflict of interest: The authors declare no conflict of interest.
Research funding: None declared.
Data availability: Data sharing is not applicable for this article since no dataset was created or analyzed during the study.

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Received: 2025-02-17

Accepted: 2025-09-14

Published Online: 2025-11-19

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Keywords for this article

generalized Fibonacci equation; monotonic solution; eventually monotonic solution

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