Tribonacci and Tribonacci-Lucas Numbers via the Determinants - Hikari

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Keywords: Tribonacci numbers, Tribonacci-Lucas numbers, tridiagonal matrix, determinant ... (2) as the form t3 − t2 − t − 1 = 0 with having the roots α = 1 +. 3. √.
Applied Mathematical Sciences, Vol. 8, 2014, no. 39, 1947 - 1955 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2014.4270

Tribonacci and Tribonacci-Lucas Numbers via the Determinants of Special Matrices Nazmiye Yilmaz and Necati Taskara Department of Mathematics, Science Faculty Selcuk University, 42075, Campus, Konya, Turkey c 2014 Nazmiye Yilmaz and Necati Taskara. This is an open access article Copyright  distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this paper, by using determinants of special matrices, it has been mainly obtained Tribonacci and Tribonacci-Lucas numbers.

Mathematics Subject Classification: 11B39, 11C20 Keywords: Tribonacci numbers, Tribonacci-Lucas numbers, tridiagonal matrix, determinant

1

Introduction and Preliminaries

For n ≥ 2, the Tribonacci sequence {Tn }n∈ and the Tribonacci-Lucas sequence {Kn }n∈ are defined by the recursive equations: Tn+1 = Tn + Tn−1 + Tn−2 Kn+1 = Kn + Kn−1 + Kn−2

(T0 = 0, T1 = T2 = 1), ( K0 = 3, K1 = 1, K2 = 3).

(1) (2)

Also, for n ≥ 2, Tribonacci sequence {Bn }n∈ and the Tribonacci-Lucas sequence {Cn }n∈ with negative indices are defined by the recursive equations: Bn = −Bn−1 − Bn−2 + Bn−3

(B−1 = 1, B0 = B1 = 0),

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Nazmiye Yilmaz and Necati Taskara

Cn = −Cn−1 − Cn−2 + Cn−3

(C−1 = 1, C0 = 3, C1 = −1).

Considering [1-4], one can clearly obtain the characteristic equations of (1) and (2) as the form t3 − t2 − t − 1 = 0 with having the roots     √ √ √ √ 3 3 3 3 1 + 19 + 3 33 + 19 − 3 33 1 + w 19 + 3 33 + w 2 19 − 3 33 α = , β= , 3 3   √ √ √ 3 3 1 + w 2 19 + 3 33 + w 19 − 3 33 −1 + i 3 γ = , w= . 3 2 Hence the Binet formulas αn+1 β n+1 γ n+1 + + , (α − β) (α − γ) (β − α) (β − γ) (γ − α) (γ − β) = αn + β n + γ n ,

Tn = Kn

can be thought as solutions of the recursive equations in (1) and (2). Additionally, it is hold the equalities T−n = Bn , K−n = Cn . Recently, Fibonacci and Lucas numbers have investigated very largely and authors tried to developed and give some directions to mathematical calculations using these type of special numbers. One of these directions goes through to the Tribonacci and the Tribonacci-Lucas numbers. In fact Tribonacci numbers have been firstly defined by Feinberg in 1963 and then some important properties over this numbers have been created by [1-4]. On the other hand, Tribonacci-Lucas numbers have been investigated by Catalani in [5]. In addition, there exists another mathematical term, namely to be tridiagonal matrix, on Fibonacci and Lucas numbers. As a brief background, the Fibonacci numbers as tridiagonal matrix determinants were present by Strang [7,8], and further authors extended results to construct families of tridiagonal matrices whose determinants generate any arbitrary linear subsequence Fαk+β or Lαk+β , of the Fibonacci and Lucas numbers in another important paper [6]. Also, it is given a new generalization of tridiagonal matrices whose determinants generate Fibonacci and Lucas numbers [9]. Additionally, in [10], Feng give new proof of the Fibonacci identities by using the method of Laplace expansion to evaluate the determinant tridiagonal matrices. Proposition 1 [6] Let A (k) be as following ⎡ a1,1 ⎢ a2,1 ⎢ ⎢ A (k) = ⎢ ⎢ ⎣

a family of tridiagonal matrices which is form a1,2 a2,2 a2,3 a3,2 a3,3 .. .

⎤ .. ..

.

. ak−1,k ak,k−1 ak,k

⎥ ⎥ ⎥ ⎥. ⎥ ⎦

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Tribonacci and Tribonacci-Lucas numbers

The determinant of A (k) is |A (1)| = a1,1 , |A (2)| = a2,2 a1,1 − a2,1 a1,2 , |A (k)| = ak,k |A (k − 1)| − ak,k−1ak−1,k |A (k − 2)| ,

k ≥ 3.

Proposition 2 [5] Tribonacci-Lucas numbers are hold the following equality, Kk+n = Kk Kn − Kk−n Cn + C2n−k , where Cn = αn β n + αn γ n + β n γ n . Under these circumstances, the main goal of this paper is to present Tribonacci and Tribonacci-Lucas numbers in the meaning of determinants of special matrices.

2

The determinants of tridiagonal matrices with Tribonacci numbers

In the following lemma, we reveal the relationship between Tribonacci and Tribonacci-Lucas numbers. Lemma 3 Tribonacci numbers is satisfied the following equality, Tk+n = Tk Kn − Tk−n Cn + Tk−2n , where Cn = αn β n + αn γ n + β n γ n . Proof. For P = (α − β) (α − γ) , R = (β − α) (β − γ) , S = (γ − α) (γ − β) , by using the Binet formula, we have k+1

α β k+1 γ k+1 + + Tk Kn − Tk−n Cn + Tk−2n = (αn + β n + γ n ) P R S

k−n+1 β k−n+1 γ k−n+1 α + + (αn β n + αn γ n + β n γ n ) − P R S αk−2n+1 β k−2n+1 γ k−2n+1 + + + . P R S By considering αβγ = 1, if we rewrite this last equality, then we get Tk Kn − Tk−n Cn + Tk−2n = Tk+n .

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In this section, for Tribonacci numbers, the tridiagonal matrix ⎡

Tr+s

⎢ t ⎢ 2,1 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

t1,2 T2r+s Tr+s

Cr −

Ts Tr+s

⎤ Cr − Kr .. .

Ts Tr+s

..

.

..

.

Cr −

T(n−3)r+s T(n−2)r+s

Cr −

T(n−3)r+s T(n−2)r+s

Kr

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ n×n

 will be denoted by Tr,s (n) = (tij )i,j∈ , where t1,2 = t2,1 = t2,2 Tr+s − T2r+s and r, s ∈ Z. The following main result establishes that the Tribonacci numbers can be obtained as a linear combination of the indices of det (Tr,s (n)) . Theorem 4 Let Tr,s (k) be the symmetric tridiagonal family of matrices with r ∈ Z+ , s ∈ N. That is,   T2r+s , t1,1 = Tr+s , t2,2 = Tr+s tj,j = Kr , 3 ≤ j ≤ k,  t1,2 = t2,1 = t2,2 Tr+s − T2r+s ,  T(j−2)r+s tj,j+1 = tj+1,j = Cr − , 2 ≤ j < k. T(j−1)r+s The determinant of Tr,s (k) is det(Tr,s (k)) = Tkr+s .

(3)

Proof. Let us use the principle of mathematical induction on k to prove (3). While, for k = 1, det(Tr,s (1)) = Tr+s , it is easy to see that for k = 2, ⎤ ⎡  t2,2Tr+s −T2r+s Tr+s ⎦ = T2r+s . T2r+s det(Tr,s (2)) = det ⎣  t2,2 Tr+s − T2r+s Tr+s As the usual next step of inductions, let us assume that it is true for all positive integers k. That is, det(Tr,s (k)) = Tkr+s .

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Therefore, we have to show that it is true for k + 1. In other words, we need to check det(Tr,s (k + 1)) = T(k+1)r+s . Considering Proposition 1, we have

T(k−2)r+s det(Tr,s (k − 1)) det(Tr,s (k + 1)) = Kr det(Tr,s (k)) − Cr − T(k−1)r+s T(k−2)r+s T(k−1)r+s = Kr Tkr+s − Cr T(k−1)r+s + T(k−1)r+s = Kr Tkr+s − Cr T(k−1)r+s + T(k−2)r+s . From Lemma 3, we get det(Tr,s (k + 1)) = T(k+1)r+s which ends up the induction and the proof. Corollary 5 In Theorem 4, if we take r = s = 1, then the matrix becomes ⎤ ⎡ 1 0 √ ⎥ ⎢ 0 2 2i ⎥ ⎢ √ ⎥ ⎢ . . ⎥ ⎢ . 2i 1 T1,1 (n) = ⎢ . ⎥

.. .. ⎢ Tn−2 ⎥ . . 1 + Tn−1 i ⎥ ⎢ ⎦ ⎣

1 + TTn−2 i 1 n−1 n×n

Then det(T1,1 (n)) = Tn+1 . Furthermore, by a different approximation, we can also define the Tribonacci numbers with the determinant of the tridiagonal matrix ⎡ ⎤ 1 i ⎢ i 1 2i ⎥ ⎢ ⎥ ⎢ ⎥ . . ⎥ . i 1 T1 (n) = ⎢ . (4) ⎢   ⎥ . . ⎢ ⎥ Tn−2 +Tn−1 . . . . i ⎦ ⎣ Tn−1 i 1 n×n For the matrix T1 (n) in (4), we get det(T1 (n)) = Tn+1 .For example, for n = 1, det(T1 (1)) = T2 = 1.

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3

Nazmiye Yilmaz and Necati Taskara

The determinants of tridiagonal matrices with Tribonacci-Lucas numbers

In this section, for Tribonacci-Lucas numbers, the tridiagonal matrix ⎡ Kr+s k1,2

K2r+s ⎢ k −s Cr − KCr+s ⎢ 2,1 Kr+s ⎢

.. ⎢ −s . Cr − KCr+s Kr ⎢ ⎢

⎢ .. C .. . . ⎢ Cr − K−(n−3)r−s (n−2)r+s ⎣

C−(n−3)r−s Cr − K(n−2)r+s Kr

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ n×n

 will be denoted by Kr,s (n) = (kij )i,j∈ , where k1,2 = k2,1 = k2,2 Kr+s − K2r+s and r, s ∈ Z. The following main result establishes that the Tribonacci-Lucas numbers can be obtained as a linear combination of the indices of det (Kr,s (n)) . Theorem 6 For 1 ≤ m ≤ n, the elements of Kr,s (m) are   K2r+s k1,1 = Kr+s , k2,2 = , Kr+s kj,j = Kr , 3 ≤ j ≤ m,  k1,2 = k2,1 = k2,2 Kr+s − K2r+s ,  C−(j−2)r−s , 2 ≤ j < m, kj,j+1 = kj+1,j = Cr − K(j−1)r+s where r ∈ Z+ , s ∈ N. In addition, we have det (Kr,s (m)) = Krm+s .

(5)

Proof. Let us use the principle of mathematical induction on m to prove (5). While, for m = 1, det(Kr,s (1)) = Kr+s , it is easy to see that for m = 2, ⎤ ⎡  k2,2Kr+s −K2r+s Kr+s ⎦ = K2r+s . K2r+s det(Kr,s (2)) = det ⎣  k2,2 Kr+s − K2r+s Kr+s

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Tribonacci and Tribonacci-Lucas numbers

As the usual next step of inductions, let us assume that it is true for all positive integers m. That is, det(Kr,s (m)) = Kmr+s . Therefore, we have to show that it is true for m + 1. In other words, we need to check det(Kr,s (m + 1)) = K(m+1)r+s . Considering Proposition 1, we have

C−(m−2)r−s det(Kr,s (m + 1)) = Kr det(Kr,s (m)) − Cr − det(Kr,s (m − 1)) K(m−1)r+s C−(m−2)r−s K(m−1)r+s = Kr Kmr+s − Cr K(m−1)r+s + K(m−1)r+s = Kr Kmr+s − Cr K(m−1)r+s + C−(m−2)r−s From Proposition 2, we get det(Kr,s (m + 1)) = K(m+1)r+s which ends up the induction and the proof. Corollary 7 In Theorem 6, if we take r = s = 1, then the matrix becomes ⎡ ⎤ 3 i √ ⎢ i 2 ⎥ 2i ⎢ ⎥ √ ⎢ ⎥ .. ⎢ ⎥ . 2i 1 . K1,1 (n) = ⎢ ⎥

.. .. ⎢ C−n+2 ⎥ . . 1 + Kn−1 i ⎥ ⎢ ⎣ ⎦

1 + CK−n+2 i 1 n−1 n×n

Then det(K1,1 (n)) = Kn+1 . Furthermore, by a different approximation, we can also define the TribonacciLucas numbers with the determinant of the tridiagonal matrix ⎡ ⎤ 3 4i ⎢ i 1 4i ⎥ ⎢ ⎥ 3 ⎢ ⎥ .. ⎢ ⎥ . i 1 . (6) K1 (n) = ⎢   ⎥ .. .. ⎢ ⎥ Kn−2 +Kn−1 . . i ⎦ ⎣ Kn−1 i 1 n×n

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Nazmiye Yilmaz and Necati Taskara

For the matrix K1 (n) in (6), we get det(K1 (n)) = Kn+1 .For example, for n = 1, det(K1 (1)) = K2 = 3. Additionally, we can write second special matrix which is form as following ⎡ ⎤ 3 i −1 ⎢ i 2 i −1 ⎥ ⎢ ⎥ ⎢ ⎥ .. . ⎢ ⎥ i 1 i ⎥ . K2 (n) = ⎢ .. ⎢ . −1 ⎥ i 1 ⎢ ⎥ ⎢ ⎥ .. .. ⎣ . i ⎦ . i 1 n×n Determinant of the matrix is Tribonacci-Lucas numbers Kn+1 . That is, det(K2 (n)) = Kn+1 . Acknowledgements. This study is a part of Ms thesis of Nazmiye Yilmaz. Thank to the editor and reviewers for their interests and valuable comments.

References [1] T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons Inc, NY, (2001). [2] M. Feinberg, Fibonacci-Tribonacci, The Fibonacci Quarterly, 1 (3), 7074, (1963). [3] W.R. Spickerman, Binet’s formula for the Tribonacci sequence, The Fibonacci Quarterly, 20 (2), 118-120, (1982). [4] N. Taskara, K. Uslu, H.H. Gulec, On the properties of Lucas numbers with binomial coefficients, Appl. Math. Letter, 23(1), 68-72, (2010). [5] M. Catalani, Identities for Tribonacci-related Sequences, Cornell University Library, http://arxiv.org/abs/math/0209179v1, (2002). [6] N.D. Cahill, D.A. Narayan, Fibonacci and Lucas numbers as tridiagonal matrix determinants, The Fibonacci Quarterly, 42(3), 216-221, (2004). [7] G. Strang, Introduction to Linear Algebra, 2nd ed., Wellesley (MA): Wellesley-Cambridge; 1998. [8] G. Strang, K. Borre, Linear Algebra, geodesy and GPS., Wellesley (MA): Wellesley-Cambridge; 1997: p.555-7.

Tribonacci and Tribonacci-Lucas numbers

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[9] A. Nalli, H. Civciv, A generalization of tridiagonal matrix determinants, Fibonacci and Lucas numbers, Chaos, Solitons and Fractals, 40, 355-361, (2009). [10] J. Feng, Fibonacci identities via the determinant of tridiagonal matrix, Appl. Math. Comput., doi: 10.1016/j.amc.2010.12.025, (2010). Received: February 3, 2014