Generalization of the 2 -Fibonacci sequences and their Binet formula

: We will explore the generalization of the four different 2 -Fibonacci sequences defined by Atanassov. In particular, we will define recurrence relations to generate each part of a 2 -Fibonacci sequence, discuss the generating function and Binet formula of each of these sequences


Introduction
The Fibonacci sequence {F n } is a sequence that satisfies the recurrence relation F n = F n−1 +F n−2 and has the initial conditions F 0 = 0 and F 1 = 1.The Fibonacci sequence and its generalizations have been studied extensively by various researchers [7, 10, 12, 13, 15, 16, 18, 20-22, 27, 28].Authors have generalized the sequence by altering the initial conditions [10,15,16,18,20], altering the coefficients in the recurrence relation [7,12,13,21,22] or both [3,7].In 2009, Edson and Yayenie introduced the concept of a bi-periodic Fibonacci sequence q n = a n−1 q n−1 + a n−2 q n−2 where the coefficients are periodic [12].Vernon, Arora, and Unnithan, in 2020, proved that a linear recurrence relation with periodic coefficients can sometimes be split into relations with constant coefficients [25].In another paper, Vernon, Arora, and Unnithan proved that all such linear recurrence relations with periodic coefficients can be split, regardless of the coefficients, and provided a simple formula to compute the new constant coefficients [26].
In Section 2, we will explore a generalization of the four 2-Fibonacci sequences that Atanassov defined but with arbitrary initial conditions and coefficients.In addition, we will define new recurrence relations to construct these generalized 2-Fibonacci sequences, discuss the generating function of each of these sequences, obtain the Binet formula for each case, and provide necessary and sufficient conditions for each type of Binet formula obtained.In Section 3, we explore the effects of using period-p coefficients when constructing 2-Fibonacci sequences.In Section 4, we provide examples of using each type of Binet formula to define the first few terms of the sequences for each applicable case.

Main results
An order-k linear recurrence relation is an equation of the form a n+k = f (a n+k−1 , . . ., a n ), where f is a linear function.We will define an order-2 linear recurrence relation in two variables as a pair of equations If f 1 and f 2 are as expressed below, we will refer to the relation as a 2-Fibonacci recurrence relation.A pair of sequences ({a n }, {b n }) that satisfy a 2-Fibonacci recurrence relation will be referred to as a 2-Fibonacci sequence.
There are four 2-Fibonacci recurrence relations in consideration: Case 1: We will assume γ i and δ i are real constants for each i such that γ i ̸ = 0 and δ i ̸ = 0.The initial conditions for a corresponding 2-Fibonacci sequence are a 0 , a 1 , b 0 , and b 1 .There has been extensive literature examining each of these cases where 3,6,7].These results have also inspired discussion from others [9-11, 17, 19, 24].The goal of this section will be to find the Binet formula for each case where all of the initial conditions and coefficients are arbitrary.We will omit the discussion of Case 4 since it is trivial.

Recurrence relation
In each of the cases of the 2-Fibonacci sequences, we can separate the recurrence relations into two new recurrence relations each using only one of the sequences and obtain the following: Theorem 2.1.If ({a n }, {b n }) is a 2-Fibonacci sequence, then there exists a single order-4 linear recurrence relation such that {a n } and {b n } are each solutions to the relation.
Proof.We will prove the result for {a n } in Case 1.The proofs for Cases 2 and 3 and for {b n } in Case 1 are similar and will be omitted.Note that in each case, the relation we obtain for {b n } will have the same coefficients as the corresponding relation for {a n }.Let γ 1 , γ 2 , δ 1 , δ 2 be real numbers.Let {a n }, {b n } be sequences such that Using the same technique as above, for the cases given at the beginning of Section 2 we can obtain the following linear recurrence relations for which {a n } and {b n } are each solutions.

Generating function
In each of the cases of the 2-Fibonacci sequences, we can establish a generating function to obtain a closed form for a n or b n .By Theorem 2.1, in each case we have {a n } and {b n } satisfy the same recurrence relation, so the only difference in the generating functions will be the initial conditions.Therefore, we will simply focus on obtaining a generating function for {c n }, where {c n } is equal to either {a n } or {b n }.That is, define We can determine the rational function form of g(x), decompose g(x) into its partial fractions, and determine the coefficients of the partial fractions to establish a Binet formula for c n .Consider a sequence {c n } satisfying the order-4 linear recurrence relation: This corresponds to the following generating function. (2)

Case 1
From Theorem 2.1, a = c = 0, and we have that Thus from (2), we have

Case 2
From Theorem 2.1, we have that Thus from (2), can write g(x) with these values.

Case 3
From Theorem 2.1, we have that Similarly, we can write g(x) as we have done for the two previous cases.

Decompositions
Suppose we have the following rational function: where h(x) is a third degree polynomial and a, b, c, d are constants and d ̸ = 0. Then g(x) can be decomposed into one of the following, depending on the multiplicity of each root of the denominator.

Distinct roots
Here we will calculate the coefficients for the case of the distinct roots (5) of the partial fraction decomposition of (4).We can see from ( 4) and (5) that By setting x = 1 ϵ i for i = 1, 2, 3, 4, we obtain the following.
Once we calculate the coefficients, we can represent each term of (5) as a geometric series.

Multiplicity 2
Here we will explore the scenario in which the denominator of g(x) has two roots each with multiplicity 2 and the necessary and sufficient conditions to obtain this scenario in each case.Consider the factorization of where ϵ 1 , ϵ 2 are distinct.If we expand (13), then we have We can see from each of the cases of the 2-Fibonacci sequences that a = 0 or c = 0.This implies that ϵ 1 + ϵ 2 = 0 or ϵ 1 ϵ 2 2 + ϵ 2 1 ϵ 2 = 0, both of which imply ϵ 2 = −ϵ 1 .Since ϵ 2 = −ϵ 1 , we have from ( 4) and (6) that If we let x = 0, then And if we let Then we can solve for A and C from ( 18) and ( 19) to obtain As before, we can rewrite each term of (6) as a power series: Thus from (1), we have Case 1 In this case, it is necessary from ( 14) that δ 1 .An example of the choices of the parameters for this case is outlined in Example 5 with a table of values for (23).Furthermore, this is a sufficient condition for obtaining two distinct roots, because if there exists some number ϵ 1 such that b = 2ϵ 2  1 and d = −ϵ 4 1 , then the denominator of Case 2 In this case, it is necessary from ( 14) that 1 .An example of the choices of the parameters for this case is outlined in Example 6 with a table of values for (23).As in Case 1, we can see that this is also a sufficient condition.
Case 3 In this case, it is necessary from ( 14) that 1 .An example of the choices for the parameters for this case is outlined in Example 7 with a table of values for (23).As in Case 1, we can see that this is also a sufficient condition.

Multiplicity 2 and 1
Here we will explore the scenario in which the denominator of g(x) has three distinct roots and the necessary and sufficient conditions to obtain this scenario in each case.Consider the factorization of where ϵ 1 , ϵ 2 , ϵ 3 are distinct.If we expand (24), then we have Next, we calculate the coefficients in (7).We can see from ( 4) and ( 7) that And if we let x = 0, then As with the previous cases, once we calculate the coefficients, we can represent each term in (7) as a power series: Thus from (1), we have ϵ 2 +ϵ 3 , implying ϵ 2 = ϵ 3 , which contradicts our assumption that ϵ 2 and ϵ 3 are distinct.Thus the scenario of Multiplicity 2 and 1 is impossible for Case 1.
(ϵ 2 +ϵ 3 ) 2 for some distinct numbers ϵ 2 , ϵ 3 .This also implies that ϵ 2 ̸ = −ϵ 3 .Additionally, this means ϵ 1 is distinct from ϵ 2 and ϵ 3 , since otherwise we would have a root of multiplicity 3, and as we will show in the subsequent section, no real coefficients γ 1 , γ 2 , δ 1 , δ 2 exist to yield this case.An example of the choices of the parameters for this case is outlined in Example 8 with a table of values for (32).As in Case 1 of the Multiplicity 2 scenario, we can show that this is also a sufficient condition.
Case 3 It is necessary from (34) that ϵ 2 + 3ϵ 1 = a = 0, which means that ϵ 2 = −3ϵ 1 .Thus we have that γ which implies that γ 1 is complex or ϵ 1 = 0, both of which are contradictions.Thus the scenario of Multiplicity 3 is impossible for Case 3, as well.

Multiplicity 4
We will prove that the scenario outlined in ( 9) is impossible in each of our cases.
Lemma 2.3.If ({a n }, {b n }) is a 2-Fibonacci sequence and g is a generating function for {a n } or {b n }, and 1 ϵ i is a root of the denominator of g(x), then 1 ϵ i does not have multiplicity 4.
Proof.Consider the factorization of where ϵ 1 is not 0. If we expand (35), then we have Since for each of our cases we have a = 0 or c = 0, we can see that ϵ 1 = 0, a contradiction.

Periodic coefficient sequences
In [26], the authors showed that any sequence satisfying an order-k linear recurrence relation with period-p coefficients can be subdivided into p separate subsequences such that there is a single order-k linear recurrence relation with constant coefficients that each subsequence satisfies.
In the case of 2-Fibonacci sequences, a similar result arises.We will define a 2-Fibonacci recurrence relation with period-p coefficients in the same way as a 2-Fibonacci recurrence relation in Section 2, but replacing the constant coefficients γ i , δ i with coefficient sequences {γ i,n }, {δ i,n } of period p.Likewise, we will similarly define 2-Fibonacci p-sequences.
Lemma 3.1.If ({a n }, {b n }) is a 2-Fibonacci p-sequence, then there exist two order-4 linear recurrence relations with period-p coefficients such that {a n } and {b n } are each solutions to one of the relations.
Proof.We will prove the case that is analogous to the case used in the proof of Theorem 2.1.The other proofs are similar and will be omitted.Suppose we have sequences {a n }, {b n } such that for all n, the following holds: where {γ 1,n }, {γ 2,n }, {δ 1,n }, {δ 2,n } are given sequences such that γ i,n+p = γ i,n and δ i,n+p = δ i,n for all i and all n.Using the same technique as in the proof of Theorem 2.1, we can show that We can easily verify that the new coefficients in the obtained relations for {a n } and {b n } are period-p sequences.This means {a n } and {b n } are each solutions to order-4 recurrence relations with period-p coefficient sequences, assuming the original coefficient sequences have no zero elements.
Note that unlike in the case with constant coefficients, the new coefficients for a n+4 and b n+4 are not necessarily equal.By the results in our previous paper, any order-4 linear recurrence relation with period-p coefficient sequences can be split into p separate order-4 linear recurrence relations with constant coefficients [25].We will demonstrate this in the following example, the case where p = 2.For simplicity of notation, we will rename the coefficients.
We first construct matrices similar to the Fibonacci matrix.This can be applied to any of the cases of 2-Fibonacci sequences with period-2 coefficient sequences and can be easily generalized to any period.A similar result can be obtained for {b n }.
Last, we will show that although the period-p coefficient sequences in the relations from Lemma 3.1 are not necessarily the same for {a n } and {b n }, the derived constant coefficients for the subsequences are indeed the same.Theorem 3.2.If ({a n }, {b n }) is a 2-Fibonacci p-sequence, then there exists a single order-4 linear recurrence relation with constant coefficients such that both {a k+np } and {b k+np } are solutions for all k.
Proof.We will prove the result in Case 1.The other results are similar.By Lemma 3.1, there exists an order-4 linear recurrence relation with period-p coefficients such that {a n } is a solution.
Then by [26] there exist four constant coefficients α i such that for each n we have Since we assume γ i,j ̸ = 0 for all i, j, this means that for each n we can divide both sides by γ 2,n+4p to conclude b

Examples
The following are examples of the 2-Fibonacci sequences to illustrate each possible Binet formula calculated in their respective sections.In each example,we will use initial conditions a 0 = b 0 = 0, a 1 = b 1 = 1.A list of the first few a n are displayed in their respective columns of Table 1.

Conclusion
We have shown that we can express a 2-Fibonacci sequence with arbitrary coefficients and initial conditions using a single order-4 linear recurrence relation.We have also constructed a generating function to obtain a closed form or Binet formula for each of these sequences.We have found necessary and sufficient conditions on the coefficients corresponding to each possible form of the Binet formula, and also which forms are not possible, along with examples for each possible scenario.We extend these concepts to the case where the coefficients are periodic sequences of period p.
, the sum of the first principal minors is AE + F + B, the sum of the second principal minors is BF − AG − CE − H − D, the sum of the third principal minors is CG − DF − BH, and the sum of the fourth principal minors is DH.Then by the results of our previous paper, we have a n+8 = (AE +B+F )a n+6 −(BF −AG−CE −H −D)a n+4 +(CG−DF −BH)a n+2 −(DH)a n .
a n and b n values for 2-Fibonacci sequences Roots Case Coefficients ϵ i a 0 a 1 a 2 a 3 a 4 a 5 a 6 . . .b 0 b 1 b 2 b 3 b 4 b 5 b 6 . . .