Application of Chybeshev Polynomials in Factorizations of Balancing and Lucas-Balancing Numbers

In this paper, with the help of orthogonal polynomials especially Chybeshev polynomials of first and second kind, number theory and linear algebra intertwined to yield factorization of balancing and Lucas-balancing numbers


Introduction
As usual, see [1], the balancing number n is defined by the solution of the Diophantine equation where r is the balancer corresponding to the balancing number n.The first few balancing numbers are 1, 6, 35 with corresponding balancers 0, 2, 14.If B n is the n th balancing number, the recurrence relation for balancing numbers is given by with B 1 = 1, B 2 = 6.
In [1] it is shown that, if n is a balancing number, n 2 is a triangular number, that is, 8n 2 + 1 is a perfect square and for all n, √ 8n 2 + 1 generates a sequence called as the sequence of Lucas-balancing numbers [5], whose first few terms are given by 1, 3 and 17 and if C n is the n th Lucas-balancing number, its recurrence relation is given by with In the recent years many number theorists from all over the world are taking interest in this beautiful number system.Liptai [2] proved that the only Fibonacci number 50 Prasanta Kumar Ray in the sequence of balancing numbers is 1.In [3], he also proved that there is no Lucas number in the sequence of balancing numbers.Balancing numbers and its related sequences are available in the literature.Interested reader may follow [4], [6], [7].
In this paper, we observe that, with the help of orthogonal polynomials, number theory and linear algebra intertwined to yield factorization of balancing and Lucasbalancing numbers.In section 2 and 3 we derive the following factorization of these numbers: In order to derive (1.3) and (1.4) we present the following theorem whose proof is included for completeness.
Theorem 1.1 If the sequence of tridiagonal matrices {A n , n = 1, 2, • • • } is of the form , then the successive determinant of A n are given by the recursive formulas: Proof.Using Induction one can easily check that the theorem is true for n = 1, 2 and 3 and assume that it is true for all k, 3 ≤ k ≤ n, that is Lucas-Balancing Numbers

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Now, Thus the theorem is true for all natural number n.

Factorization of Balancing Numbers
In order to derive the factorization of balancing numbers (1.3), let us introduce the sequence of matrices {D n , n = 1, 2, • • • } where D n is am n × n tridiagonal matrix with entries , By virtue of Theorem 1.1, we find which is nothing but the sequence of balancing numbers starting with B 2 .Thus, Since the determinant of a matrix can be found by taking the product of its eigenvalues, we will now find the spectrum of D n in order to find an alternate formulation for det(D n ).
Let us introduce another sequence of matrices {S n , n = 1, 2, • • • } where S n is an n × n tridiagonal matrix with entries .
Clearly D n = 6I + S n , where I be the identity matrix same order as S n .Let λ k , k = 1, 2, 3 • • • , n, be the eigenvalues of S n with corresponding eigenvectors X k .Then for each j, Thus In order to find λ k , k = 1, 2 • • • , n, we recall that each λ k is zero of the characteristic polynomial p n (λ) = det(S n − λI).Since S n − λI is a tridiagonal matrix we have, Using Theorem 1.1, we get the following recursive formula for the characteristic polynomials: This family of polynomials can be transformed into another family {M n , n ≥ 1} by the transformation λ = −2x to get, Lucas-Balancing Numbers

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We observe that the family {M n , n ≥ 1} is the set of Chebyshev polynomials of second kind.It is well known that for x = cos θ, the Chebyshev polynomials of the second kind can be written as which when equal to zero gives Thus, Now applying the transformation λ = −2x, the eigenvalues of S n are given by Combining (2.1), (2.2) and (2.3), we get which is identical to the factorization (1.3).

Factorization of Lucas-Balancing Numbers
In a similar manner we can derive (1.4) by considering the sequence of matrices Again using Theorem 1.1, we obtain We observe that each member in this sequence is a Lucas-balancing number.Thus, we get If e j is the j th column of the identity matrix I, we see that det(I + e 1 e T 1 ) = 2. Therefore,we may write Also we observe that the right hand side of (3.2) can be expressed as where S n is the matrix defined earlier. If Therefore, In order to find α ′ k s, we recall that each α k is a zero of the characteristic polynomial q n (α) = det(S n − ie 1 e T 2 − αI).Since det(I − 1 2 e 1 e T 1 ) = 1 2 , we can express the characteristic polynomial as Since q n (α) is the twice of a tridiagonal matrix, we can use Theorem 1.1 to get the following recursive formulas: which is identical to the factorization(1.4).