Binet – Type Formula For The Sequence of Tetranacci Numbers by Alternate Methods

The sequence {Tn} of Tetranacci numbers is defined by recurrence relation Tn= Tn-1 + Tn-2 + Tn-3 + Tn-4; n≥4 with initial condition T0=T1=T2=0 and T3=1. In this Paper, we obtain the explicit formulla-Binet-type formula for Tn by two different methods. We use the concept of Eigen decomposition as well as of generating functions to obtain the result.

; n ≥ 4 with initial condition T T T 0 1 2 0 = = = and T 3 1 = . In this paper, we obtain the explicit formula -Binet -type formula for T n by two different methods. We use the concept of eigen decomposition as well as of generating functions to obtain the result.

INTRODUCTION
Fibonacci sequence is a sequence of numbers defined by the recursive formula F F F n n n = + --1 2 ; for n > 2 with initial condition F 1 1 = , F 2 1 = . This sequence possesses many interesting properties which have been studied in detail ( [8], [12]). Analogous to Fibonacci sequence, many other sequences have been defined, either by changing the initial terms or the recursive relation or both, to obtain the new sequence which may possess similar properties ([9], [11]). One of the important result that is associated with Fibonacci numbers and which has been studied for centuries now, is the Binet formula given as where φ = + 1 5 2 and F n is the n th Fibonacci number.
Also, by changing the recursive formula or the initial terms, we can have a new sequence of numbers defined and correspondingly, a new Binet -type formula can be obtained ( [2], [6], [7]). In this paper we consider the sequence of Tetranacci numbers and obtain the Binet -type formula for by two different methods. Many salient features of this sequence have been studied in detail ( [4], [5], [10]).

PRELIMINARY RESULTS
In this section, we give some preliminary results which will be useful to derive the Binet -type formula for .
Proof: Above result can be proved using principal of mathematical induction. It is clearly observed that result is true for n = 1. Let it be true for some positive integer k. This gives,  Thus result is true for n = k + 1 also, and hence for all n.
We next derive the generating function for the sequence T n { }. and − ( ) x 4 successively and adding them, we get

T T T T T x
This gives, 1 1 , as required.
We use this result to derive the Binet -type formula for T n . We first use the concept of eigen decomposition of a matrix to derive this formula and then we use the theory of generating functions to obtain the explicit formula for T n .
Proof: 2 3 4 . Then for some α β γ ,� ,� and δ , we write Thus, 1 1 1 α β γ , , and 1 δ are the roots of f(x). This gives α β γ ,� ,� and δ as the This implies x x x x We now solve this equation using Ferrari's method. Consider the substitution . This converts the above equation into depressed quartic form y y y Next we introduce a new variable m on the L.H.S. of (1) by adding We note that (1) and (2) On simplification, we get Equation (3) Next we impose the condition 3 7 6 0 ij + = , , which implies ij = − 7 18 . .

This gives
i j i j Thus, we get a quadratic equation z z and .
Thus, t = i + j gives t = --+ + ( ) 1 0 − − − − = . By above discussion it is clear that the roots of this equation are α, β, γ and δ and they are the eigenvalues of P. Since P has four distinct eigenvalues, it is a diagonalizable matrix [1]. Moreover, for the eigen decomposition of P we find the eigenvectors for the corresponding eigenvalues.
Consider the matrix equation Px = ax, where x is the eigenvector corresponding to eigenvalue α. This gives P I x − ( ) = α θ. Solving this matrix equation is equivalent to finding the null -space of (P -αI). Thus, we have Similarly, for β, γ and δ, the corresponding eigenvectors are be the matrix formed Using This gives, This gives on simplification. , as required.
We next find the same result by the use of generating function for T n .
If we consider x = 1 α , we get