The Sum and Product of Fibonacci Numbers and Lucas Numbers, Pell Numbers and Pell-Lucas Numbers Representation by Matrix Method
WSEAS TRANSACTIONS on MATHEMATICS Fuliang Lu, Zhaolin Jiang
The sum and product of Fibonacci numbers and Lucas numbers, Pell numbers and Pell-Lucas numbers representation by matrix method
Fuliang Lu Zhaolin Jiang Linyi University Linyi University School of Sciences School of Sciences Linyi Shandong, 276005 Linyi Shandong, 276005 China China fl[email protected] [email protected]
Abstract: Denote by {Fn} and {Ln} the Fibonacci numbers and Lucas numbers, respectively. Let Fn = Fn × Ln and Ln = Fn + Ln. Denote by {Pn} and {Qn} the Pell numbers and Pell-Lucas numbers, respectively. Let Pn = Pn × Qn and Qn = Pn + Qn. In this paper, we give some determinants and permanent representations of Pn, Qn, Fn and Ln. Also, complex factorization formulas for those numbers are presented.
Key–Words: Fibonacci number; Lucas number; Pell numbers; Pell-Lucas number; matrix.
1. Introduction [12] considered certain generalizations of the well- known Fibonacci and Lucas numbers, the general- The Fibonacci and Lucas sequences are defined by the ized Fibonacci and Lucas p-numbers. They also de- following recurrence relations: termined certain matrices whose permanents generate the Lucas p-numbers and their sums. Fn+2 = Fn+1 + Fn where F0 = 0,F1 = 1; By the determinant of tridiagonal matrix, an iden- L = L + L where L = 2,L = 1. n+2 n+1 n 0 1 tity of Pell identities is presented [19]. Yilmaz and The Pell and Pell-Lucas sequences are defined by the Bozkurt [20] derived some relationships between Pell following recurrence relations, respectively: sequence and permanents and determinants of one type of Hessenberg matrices. Li [14] gave another Pn+2 = 2Pn+1 + Pn where P0 = 0,P1 = 1; proofs of two results relative to the Pell and Perrin numbers by constructing new Fibonacci-Hessenberg Qn+2 = 2Qn+1 + Qn where Q0 = 2,Q1 = 2. matrices. Fu and Zhou [6] derived the relation be- tween Pell numbers and its companion sequence by Let A = [ai,j] be an n×n matrix. The permanent matrix representations of them. Gulec and Taskara of A is defined by [7] gave new generalizations for (s, t)- Pell and (s, t)- Pell-Lucas sequences for Pell and Pell-Lucas num- ∑ ∏n bers, and defined the matrix sequences which have per A = a , iσ(i) elements of them and investigated their properties. ∈ σ Sn i=1 Kilic [11] gave the definition of generalized Pell − where the summation extends over all permutations σ (p, i) numbers and then presented their generating of the symmetric group S [15]. matrix. He obtained relationships between the gen- n eralized Pell (p, i)−numbers and their sums and per- It is known that there are lots of relations be- manents of certain matrices. Civciv and Turkmen¨ tween determinants or permanents of matrices and the [1] defined a new matrix generalization of the Lucas famous sequences. By the determinant of tridiago- numbers, and using essentially a matrix approach they nal matrix, an identity of Fibonacci number is proved showed properties of these matrix sequences. [5]. Demirturk¨ [4] derived some Fibonacci and Lucas sums by matrix method. In [9], the authors present a result involving the permanent of an (−1, 0, 1)-matrix A. Nalli [17] present a family of tridiagonal ma- and the Fibonacci number Fn+1. Kilic and Stakhov trices given by
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The Jacobsthal and Jacobsthal-Lucas sequences are defined by the following recurrence relations, re- 3 1 0 ··· 0 0 spectively: 1 3 1 0 ··· 0 0 1 3 1 0 ··· Jn+2 = Jn+1 + 2Jn where J0 = 0,J1 = 1; A1 = , n ...... . . . . . 0 jn+2 = jn+1 + 2jn where j0 = 2, j1 = 1. 0 ··· 0 1 3 1 Let 0 0 ··· 0 1 3 n×n 3 2 0 ··· 0 0 | 1 | such that the determinant An is the Fibonacci num- 1 1 2 0 ··· 0 ber F2n+2. Another example in [2] is the family of 0 1 1 2 0 ··· A3 = . tridiagonal matrices given by: n ...... . . . . . 0 1 i 0 ··· 0 0 0 ··· 0 1 1 2 i 1 i 0 ··· 0 0 0 ··· 0 1 1 0 i 1 i 0 ··· . be an n-square matrix. ...... . . . . . 0 F. Yilmaz and D. Bozkurt [21] showed that 0 ··· 0 i 1 i 3 0 0 ··· 0 i 1 per An = Jn+2 n×n Let Strang [18] presents the tridiagonal matrix 1 2 0 ··· 0 0 1 −1 0 ··· 0 0 ··· 1 3 2 0 0 1 1 −1 0 ··· 0 ··· 0 1 1 2 0 0 1 1 −1 0 ··· A4 = 2 n ...... An = . . . . . , . . . . . 0 ...... . . 0 0 ··· 0 1 1 2 ··· − 0 0 1 1 1 0 0 ··· 0 1 1 0 0 ··· 0 1 1 n×n be an n-square matrix. | 2 | the determinant An is the Fibonacci number Fn+1. F. Yilmaz and D. Bozkurt [21] showed that It can be checked that per A4 = j . n n 1 1 0 ··· 0
.. Also, they gave a complex factorization formulas for 1 1 1 . 0 Jacobsthal numbers. That is Fn+1 = ...... , . . . . 0 ∏n √ kπ 0 1 1 1 Jn = (1 + 2 2i cos ). n + 1 0 ··· 0 1 1 k=1 n×n
Let 2 1 0 ··· 0 .. 1 1 −1 ··· 0 0 0 0 1 2 1 . 0 1 1 1 1 ··· 0 0 0 Pn+1 = ...... , . . . . 0 0 1 1 1 −1 ··· 0 0 0 1 2 1 ...... 5 ...... 0 ··· 0 1 2 A = n×n n ··· 0 0 1 1 1 1 0 1 1 0 ··· 0 0 0 0 ··· 0 1 1 1 −1 −2 1 1 0 ··· 0 0 0 0 ··· 0 1 1 1
0 −1 1 1 0 ··· 0 0 0 0 ··· 0 1 1 L = . n ...... 0 be an n-square matrix. F. Yilmaz and D. Bozkurt [20]
0 ··· 0 −1 1 1 showed that
0 0 ··· 0 −1 1 A5 = P . n×n per n n
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N. D. Cahill, D. A. Narayan [3] showed where a ∈ Z+ and b ∈ N, has determinants that the symmetric tridiagonal family of matrices | | Ma,b(k)(k = 1, 2, ···), whose elements are given by: Ta,b(k) = Lak+b. ⌈ ⌉ F For example, m = F , m = 2a+b , 1,1 a+b 2,2 Fa+b 3 0 0 ··· 0 0
m = L , 3 ≤ i ≤ k, 0 6 −1 0 ··· 0 i,i a √ 0 −1 7 −1 0 ··· m = m = m F − F , L − = , 1,2 2,1 2,2 a+b 2a+b 4k 2 ...... √ . . . . . 0 a mj,j+1 = mj+1,j = (−1) , 2 ≤ j < k, 0 ··· 0 −1 7 −1
0 0 ··· 0 −1 7 where a ∈ Z+ and b ∈ N, has determinants k×k √ · ·· |Ma,b(k)| = Fak+b. √18 14 0 0 0
14 5 i 0 · ·· 0
For example, 0 i 4 i 0 ··· L = , 3k+3 ...... 1 0 0 ··· 0 0 . . . . . 0
0 8 1 0 ··· 0 0 ··· 0 i 4 i
··· 0 0 ··· 0 i 4 0 1 7 1 0 k×k F − = , 4k 2 ...... 0 and
0 ··· 0 1 7 1 √ · ·· √29 11 0 0 0 0 0 ··· 0 1 7 k×k 11 3 1 0 ··· 0
0 1 3 1 0 ··· √ L2k+5 = ...... 8 6 0 · ·· 0 0 ...... √ . . 0 ··· 6 5 i 0 · ·· 0 0 0 1 3 1 0 0 ··· 0 1 3 0 i 4 i 0 ··· k×k
F3k+3 = . . . . . , ...... 0 N. D. Cahill and D. A. Narayan [3] also derived 0 ··· 0 i 4 i the following factorization 0 0 ··· 0 i 4 k×k n∏−1 and kπ F2mn = F2m (L2m − 2cos ). √ n k=1 13√ − 5 0 · ·· 0 0
− 5 3 −1 0 ··· 0 A. Nalli and H. Civciv [17] generalized the result
0 −1 3 −1 0 ··· of N. D. Cahill and D. A. Narayan [3]. The tridiagonal F = . 2k+5 . . . . . family of matrices M− − (k), k = 1, 2, ··· whose el- ...... 0 a, b ements are given by: 0 ··· 0 −1 3 −1 ⌈ ⌉ 0 0 ··· 0 −1 3 k×k F−2a−b m1,1 = F−a−b, m2,2 = , F− − In the same paper, they proved that the symmetric a b tridiagonal family of matrices Ta,b(k)(k = 1, 2, ···), mi,i = L−a, 3 ≤ i ≤ k, whose elements are given by: √ ⌈ ⌉ m1,2 = m2,1 = m2,2F−a−b − F−2a−b, L2a+b √ t1,1 = La+b, t2,2 = , a La+b mj,j+1 = mj+1,j = (−1) , 2 ≤ j < k, where a ∈ Z+ and b ∈ N. Then ti,i = La, 3 ≤ i ≤ k, √ (1) if a and b are odd, then t1,2 = t2,1 = t2,2La+b − L2a+b, { √ −Fak+b k is odd − a ≤ det(M−a,−b(k)) = ; tj,j+1 = tj+1,j = ( 1) , 2 j < k, Fak+b k is even
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(2) if a is odd and b is even, then (2) if a is odd and b is even, then { { Fak+b k is odd −Lak+b k is odd det(M−a,−b(k)) = ; det(T−a,−b(k)) = ; −Fak+b k is even Lak+b k is even (3) if a and b are even, then (3) if a and b are even, then
det(M−a,−b(k)) = −Fak+b; det(T−a,−b(k)) = Lak+b; (4) if a is even and b is odd, then (4) if a is even and b is odd, then − det(M−a,−b(k)) = Fak+b. det(T−a,−b(k)) = Lak+b. For example, For example, − ··· −8 i 0 ··· 0 0 29 6i 0 0 0 − ··· i 7 1 0 ··· 0 6i 3 i 0 0 − ··· 0 1 7 1 0 ··· 0 i 4 i 0 T− − (k) = , M− − (k) = , 3, 4 . . . . . 4, 2 . . . . . ...... ...... 0 . . 0 ··· − 0 ··· 0 1 7 −1 0 0 i 4 i ··· − ··· 0 0 0 i 4 k×k 0 0 0 1 7 k×k | | − |M− − (1)| = −F , T−3,−4(1) = L7, 4, 2 6 | | |M− − (2)| = −F , T−3,−4(2) = L10, 4, 2 10 | | − |M− − (3)| = −F , T−3,−4(3) = L13, 4, 2 14 | | |M− − (4)| = −F , T−3,−4(4) = L{16, 4, 2 18 − |M−4,−2(k)| = −F4k+2. L3k+4 k is odd |T−3,−4(k)| = . L3k+4 k is even −1 0 0 ··· 0 0 − ··· √ 0 2 i 0 0 − ·· · − ··· √ 29 11i 0 0 0 0 i 1 i 0 M− − (k) = , ··· 1, 1 ...... 11i 3 1 0 0 . . . . . 0 ··· 0 ··· 0 i −1 i 0 1 3 1 0 T− − (k) = , 0 0 ··· 0 i −1 2, 5 . . . . . k×k ...... 0 0 ··· 0 1 3 1 | | − M−1,−1(1) = F2, 0 0 ··· 0 1 3 k×k |M−1,−1(2)| = −F3, |M−1,−1(3)| = −F4, |T−2,−5(1)| = −L7, |M− − (4)| = −F , 1, 1 { 5 |T− − (2)| = −L , − 2, 5 9 Fk+1 k is odd |T− − (3)| = −L , |M−1,−1(k)| = . 2, 5 11 Fk+1 k is even |T−2,−5(4)| = −L13, Also, they proved if the symmetric tridiagonal |T− − (k)| = −L . ··· 2, 5 2k+5 family of matrices T−a,−b(k), k = 1, 2, whose el- Let A = [ai,j] be an n × n matrix with row vec- ements are given by: tors r1, r2, ··· , rn. Suppose column k contains ex- ⌈ ⌉ actly two nonzero elements aik ≠ 0 ≠ ajk and i ≠ j. L−2a−b − × − t1,1 = L−a−b, t2,2 = , Then the (n 1) (n 1) matrix Aij:k obtained from L− − a b A by replacing row i with ajkri + aikrj and deleting row j and column k is called the contraction of A on ti,i = L−a, 3 ≤ i ≤ k, √ column k relative to rows i and j, and we say A is t1,2 = t2,1 = t2,2L−a−b − L−2a−b, contractible on column k. Similarly, if row k contains √ exactly two nonzero elements, aki ≠ 0 ≠ akj and a tj,j+1 = tj+1,j = (−1) , 2 ≤ j < k, ̸ T T i = j, then the matrix Ak:ij = [Aij:k] is called the where a ∈ Z+ and b ∈ N. Then contraction of A on row k relative to columns i and j. (1) if a and b are odd, then If A is a nonnegative matrix and B is a contraction of { A, then per A = per B [10]. Lak+b k is odd A directed pseudo graph G = (V,E), with set det(T−a,−b(k)) = ; { ··· } −Lak+b k is even of vertices V (G) = 1, 2, , n and set of edges
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n 0 1 2 3 4 Let An = [aij]n×n be the adjacency matrix of the Fn 0 1 3 8 21 graph given by Figure 1, in which L 2 2 4 6 10 n a11 = at,t+1 = 1, ass = 3, al,l−1 = 1 Pn 0 2 12 70 408 for t = 1, 2, ··· , n − 1, s = 2, 3, ··· , n and l = Q 2 3 8 19 46 n 3, 4, ··· , n and otherwise 0. That is Table 1: The first few values of the sequences. 1 1 0 ··· 0 0 0 3 1 0 ··· 0 0 1 3 1 0 ··· A = . n ...... . . . . . 0 0 ··· 0 1 3 1 0 0 ··· 0 1 3 Figure 1. Let Sn be a (1, −1) matrix of order n, defined as 1 1 ··· 1 1 E(G) = {e , e , ··· , e }, is a graph in which loops 1 2 m −1 1 ··· 1 1 and multiple edges are allowed. A directed graph rep- 1 −1 1 ··· 1 resented with arrows on its edges, each arrow pointing Sn = . . . . . . towards the head of the corresponding arc. The adja- ...... × cency matrix A(G) = [ai,j] is n n matrix, defined 1 ··· 1 −1 1 by the rows and the columns of A(G) are indexed by Denote the matrices A ◦ S by H , where A ◦ S V (G), in which ai,j is the number of edges jointing vi n n n n n denotes Hadamard product of An and Sn. Thus and vj [13]. Let Fn = Fn × Ln and Ln = Fn + Ln. Then we 1 1 0 ··· 0 0 can get the following recurrence relations: 0 3 1 0 ··· 0 0 −1 3 1 0 ··· Fn+2 = 3Fn+1 − Fn where F0 = 0, F1 = 1; H = . (1) n ...... . . . . . 0 ··· − Ln+2 = Ln+1 + Ln where L0 = 2, L1 = 2. 0 0 1 3 1 0 0 ··· 0 −1 3 The graph gotten by join a single vertex to every ver- tices of a path is called a fan. By [16], the number of Theorem 1 Let Hn be an matrix as in (1). Then spanning tree of a fan with n + 1 vertices is Fn. (n−2) F per Hn = per Hn = n. Let Pn = Pn × Qn and Qn = Pn + Qn. Then we can get the following recurrence relations: Proof. By definition of the matrix Hn, it can be con- (r) tracted on column 1. Let Hn be the rth contraction P = 6P − P where P = 0, P = 2; of H . If r = 1, then n+2 n+1 n 0 1 n 3 1 0 ··· 0 0 Q = 2Q + Q where Q = 2, Q = 3. n+2 n+1 n 0 1 −1 3 1 0 ··· 0 The first few values of the sequences are in Table 1. 0 −1 3 1 0 ··· (1) Hn = ...... In this paper, we investigate relationships be- ...... 0 F tween adjacency matrices of graphs and the n, the 0 ··· 0 −1 3 1 L , the P and the Q sequences. We also give com- n n n 0 0 ··· 0 −1 3 plex factorization formulas for the numbers. (1) Note that Hn also can be contracted by the first col- umn, then 1 Determinant representations of Fn 8 3 0 ··· 0 0 L and n −1 3 1 0 ··· 0 0 −1 3 1 0 ··· In this section, we consider a class of pseudo graph H(2) = . n ...... given in Figure 1 and Figure 2, respectively. Then we . . . . . 0 investigate relationships between permanents of the 0 ··· 0 −1 3 1 adjacency matrices of the graphs and Fn and Ln. 0 0 ··· 0 −1 3
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(1) Note that Kn also can be contracted by the first col- umn, then 6 4 0 ··· 0 0 1 1 1 0 ··· 0 Figure 2. 0 1 1 1 0 ··· (2) Kn = ...... ...... Similarly, . . 0 0 ··· 0 1 1 1 21 8 0 ··· 0 0 0 0 ··· 0 1 1 −1 3 1 0 ··· 0 Similarly, 0 −1 3 1 0 ··· H(3) = . n ...... 10 6 0 ··· 0 0 . . . . . 0 1 1 1 0 ··· 0 0 ··· 0 −1 3 1 0 1 1 1 0 ··· 0 0 ··· 0 −1 3 K(3) = . n ...... . . . . . 0 Going with this induction process, we have 0 ··· 0 1 1 1 [ ] 0 0 ··· 0 1 1 F − F − H(n−2) = n 1 n 2 , n −1 3 Going with this induction process, we have [ ] L − L − so K(n−2) = n 1 n 2 , (n−2) F n 1 1 per Hn = per Hn = n. ⊓⊔ so (n−2) L per Kn = per Kn = n. Let Kn = [kij]n×n be the adjacency matrix of the pseudo graph given in Figure 2, with ⊓⊔ Denote the matrices Kn ◦ Sn by Bn. That is k11 = k12 = kt,t−1 = 2, kss = 1, kl,l+1 = 1 2 2 0 ··· 0 0 for t = 2, ··· , n, s = 2, 3, ··· , n and l = 2, 3, ··· , n, −2 1 1 0 ··· 0 and otherwise 0. That is 0 −1 1 1 0 ··· B = . n ...... 2 1 0 ··· 0 0 . . . . . 0 0 ··· 0 −1 1 1 2 1 1 0 ··· 0 0 0 ··· 0 −1 1 0 1 1 1 0 ··· K = . (2) n ...... . . . . . 0 Then we have 0 ··· 0 1 1 1 det An = per Hn = Fn 0 0 ··· 0 1 1 and L Theorem 2 Let Kn be a matrix as in (2). Then det Bn = per Kn = n. (n−2) per Kn = per Kn = Ln. Let Cn+1 be an (n + 1) × (n + 1) matrix defined as Proof. By definition of the matrix K , it can be con- n 2 0 0 ··· 0 0 tracted on column 1. Let K(r) be the rth contraction n 0 1 1 0 ··· 0 of K . If r = 1, then n 0 −1 1 1 0 ··· C = . n+1 ...... 4 2 0 ··· 0 0 . . . . . 0 1 1 1 0 ··· 0 0 ··· 0 −1 1 1 0 1 1 1 0 ··· 0 0 ··· 0 −1 1 K(1) = . n ...... . . . . . 0 Similar to Theorem 2, it can be proofed that 0 ··· 0 1 1 1 ◦ L 0 0 ··· 0 1 1 per (Cn+1 Sn+1) = det Cn+1 = n.
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Figure 3. Figure 4.
P (1) 2 Determinant representations of n Note that Un also can be contracted by the first col- and Qn umn, then ··· In this section, we consider a class of pseudo graph 70 12 0 0 0 given in Figure 1 and Figure 2, respectively. Then we −1 6 1 0 ··· 0 investigate relationships between permanents of the 0 −1 6 1 0 ··· U (2) = . adjacency matrices of the graphs and Pn and Qn. n ...... ′ . . . . . 0 Let A = [a ] × be the adjacency matrix of the n ij n n 0 ··· 0 −1 6 1 graph given by Figure 1, in which 0 0 ··· 0 −1 6 ′ ′ ′ ′ a11 = 2, att+1 = all−1 = 1, ass = 6, Similarly, ··· − ··· for t = 1, 2, , n 1, s = 2, 3, , n and l = 3, 4, ··· , n and otherwise 0. That is 408 70 0 ··· 0 0 ··· −1 6 1 0 ··· 0 2 1 0 0 0 0 −1 6 1 0 ··· 0 6 1 0 ··· 0 (3) Un = ...... 0 1 6 1 0 ··· ...... 0 ′ An = ...... ··· − ...... 0 0 0 1 6 1 ··· − 0 ··· 0 1 6 1 0 0 0 1 6 ··· 0 0 0 1 6 Going with this induction process, we have ′ ◦ ◦ [ ] Denote the matrices An Sn by Un, where Un Sn (n−2) Pn−1 Pn−2 denotes Hadamard product of Un and Sn. Thus U = , n −1 6 2 1 0 ··· 0 0 so 0 6 1 0 ··· 0 − per U = per U (n 2) = P 0 −1 6 1 0 ··· n n n U = . (3) n ...... . ⊓⊔ . . . . . 0 Let V = [v ] × be the adjacency matrix of the 0 ··· 0 −1 6 1 n ij n n pseudo graph given in Figure 2, with 0 0 ··· 0 −1 6
v11 = 3, v21 = vss = 2, vtt−1 = vll+1 = 1
for Theorem 3 Let Un be an matrix as in (3). Then ··· ··· (n−2) t = 3, , n, s = 2, 3, , n per Un = per Un = Pn. and Proof. By definition of the matrix Un, it can be con- l = 1, 2, ··· , n − 1, U (r) r tracted on column 1. Let n be the th contraction and otherwise 0. That is of Un. If r = 1, then 3 1 0 ··· 0 0 12 2 0 ··· 0 0 2 2 1 0 ··· 0 −1 6 1 0 ··· 0 ··· − ··· 0 1 2 1 0 0 1 6 1 0 V = . (4) U (1) = . n ...... n ...... . . . . . 0 . . . . . 0 0 ··· 0 1 2 1 0 ··· 0 −1 6 1 0 0 ··· 0 1 2 0 0 ··· 0 −1 6
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Theorem 4 Let Vn be an matrix as in (4). Then and (n−2) ′ Q per Vn = per Vn = Qn. det Bn = per Vn = n. ′ × Let Cn+1 be an (n + 1) (n + 1) matrix defined Proof. By definition of the matrix Vn, it can be con- (r) as tracted on column 1. Let Vn be the rth contraction of Vn. If r = 1, then 2 0 0 ··· 0 0 0 3/2 1 0 ··· 0 8 3 0 ··· 0 0 0 −1 2 1 0 ··· ··· ′ 1 2 1 0 0 Cn+1 = ...... ...... 0 1 2 1 0 ··· . 0 V (1) = . ··· − n ...... 0 0 1 2 1 . . . . . 0 0 0 ··· 0 −1 2 0 ··· 0 1 2 1 0 0 ··· 0 1 2 Similar to Theorem 2, it can be proofed that
(1) (C′ ◦ S ) = C′ = Q . Note that Vn also can be contracted by the first col- per n+1 n+1 det n+1 n umn, then 19 8 0 ··· 0 0 3 Complex factorization formulas 1 2 1 0 ··· 0 0 1 2 1 0 ··· In this section, we give complex factorization formu- (2) F L P V = . las for n, n and n. n ...... . . . . . 0 ∏ F n−1 kπ 0 ··· 0 1 2 1 Theorem 5 n = k=1(3 + cos n ). 0 0 ··· 0 1 2 Proof. The characteristic equation of An is Similarly, − 0 = det(An λI) 46 19 0 ··· 0 0 − ··· 1 2 1 0 ··· 0 1 λ 1 0 0 0 − ··· 0 1 2 1 0 ··· 0 3 λ 1 0 0 (3) − ··· Vn = ...... 0 1 3 λ 1 0 ...... 0 = ...... ··· . . . . . 0 0 0 1 2 1 ··· 0 ··· 0 1 3 − λ 1 0 0 0 1 2 0 0 ··· 0 1 3 − λ
Going with this induction process, we have [ ] 3 − λ 1
− Qn−1 Qn−2 1 3 − λ 1 V (n 2) = , n 1 2 − ...... = (1 λ) . . . , so 1 3 − λ 1 (n−2) Q per Vn = per Vn = n. 1 3 − λ ⊓⊔ here I is the identity matrix. By [22], the eigenvalues ◦ ′ Denote the matrices Vn Sn by Bn. That is of the matrix 3 1 0 ··· 0 0 3 − λ 1 −2 2 1 0 ··· 0 1 3 − λ 1 − ··· . . . ′ 0 1 2 1 0 ...... B = . n ...... . . . . . 0 1 3 − λ 1 0 ··· 0 −1 2 1 1 3 − λ 0 0 ··· 0 −1 2 kπ · ·· − are 3 + cos n (k = 1, 2, , n 1). Then we have So the result follows. ⊓⊔ ∏ ′ P L n kπ detAn = per Un = n Theorem 6 n = 2 k=1(1 + i cos n+1 ).
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kπ · ·· − Proof. The characteristic equation of Cn+1 is are 6 + cos n (k = 1, 2, , n 1). So the result follows. ⊓⊔ 0 = det(Cn+1 − λI)
2 − λ 0 0 ··· 0 0
0 1 − λ 1 0 ··· 0 4 Conclusion
0 −1 1 − λ 1 0 ··· = We give some determinant and permanent representa- ...... 0 tions of F , L , P and Q and complex factorization n n n n 0 ··· 0 −1 1 − λ 1 formulas for F , L and P . So, it is natural to ask the n n n 0 0 ··· 0 −1 1 − λ question: what is the complex factorization formula
for Qn? 1 − λ 1
−1 1 − λ 1
− ...... = (2 λ) . . . . References:
−1 1 − λ 1 [1] H. Civciv and R. Turkmen,¨ Notes on the (s, t)- −1 1 − λ Lucas and Lucas matrix sequences, Ars Combina- By [22], the eigenvalues of the matrix toria, 89, 2008, pp. 271–285. 1 − λ 1 [2] N. D. Cahill, J. R. DErrico, J. P. Spence, Complex factorizations of the Fibonacci and Lucas num- −1 1 − λ 1 bers, Fibonacci Quart., 41, 2003, pp. 13–19. ...... . . . [3] N. D. Cahill, D. A. Narayan, Fibonacci and Lu- −1 1 − λ 1 cas numbers as tridiagonal matrix determinants, −1 1 − λ Fibonacci Quart., 42,3 2004, pp. 216–221. kπ ··· [4] B. Demirturk,¨ Fibonacci and Lucas sums by ma- are 1 + i cos n+1 (k = 1, 2, , n). So the result fol- lows. ⊓⊔ trix methods, International Mathematical Forum, Then, we give complex factorization formulas for 5, 3, 2010, pp. 99–107. Pn. [5] J. Feng, Fibonacci identities via the determinant ∏ − Theorem 7 P = 2 n 1(6 + cos kπ ). of tridiagonal matrix, Appl. Math. Comput. , 217, n k=1 n 2011, pp. 5978–5981. Proof. The characteristic equation of A′ is n [6] X. Fu and X. Zhou, On matrices related with Fi- 0 = det(A′ − λI) bonacci and Lucas numbers, Appl. Math. Comput. n , 200, 2008, pp. 96–100. 2 − λ 1 0 ··· 0 0 [7] H. H. Gulec and N. Taskara, On the (s, t)−Pell 0 6 − λ 1 0 ··· 0 and (s, t)−Pell−Lucas sequences and their ma- 0 1 6 − λ 1 0 ··· = trix representations, Applied Mathematics Letters, ...... 0 2012, doi:10.1016/j.aml.2012.01.014.
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