Notes on Number Theory and Discrete Mathematics Print ISSN 30 532, Online ISSN 2367 8275 Vol 24, 208, No, 03 08 DOI: 07546/nntdm2082403-08 Fibonacci and Lucas numbers via the determinants of tridiagonal matrix Taras Goy Department of Mathematics and Informatics Vasyl Stefanyk Precarpathian National University 57 Shevchenko Str, 7608 Ivano-Frankivsk, Ukraine e-mail: tarasgoy@yahoocom Received: 2 June 206 Accepted: 3 January 208 Abstract: Applying the apparatus of triangular matrices, we proved new recurrence formulas for the Fibonacci and Lucas numbers with even (odd) indices by tridiagonal determinants Keywords: Fibonacci numbers, Lucas numbers, Horadam sequence, Triangular matrix, Parapermanent of triangular matrix 200 Mathematics Subject Classification: B39, C20 Triangular matrix and parapermanents of triangular matrix The functions of triangular matrices are widely used in algebra, combinatorics, number theory and other branches of mathematics [9,, 2] Definition [] A triangular number table is called a nth-order triangular matrix a a A n = 2 a 22 a n a n2 a nn Note that a matrix () is not a triangular matrix in the usual sense of this term as it is not a square matrix 03 ()
The product a ij a i,j+ a ii is denoted by {a ij } and is called a factorial product of the element a ij Definition 2 [] The parapermanent pper(a n ) of a triangular matrix () is the number a a pper(a n ) 2 a 22 a n a n2 a nn n = n r= p ++p r=n s= r {a p ++p s, p ++p s +}, (2) where p, p 2,, p r are positive integers, {a ij } is the factorial product of the element a ij Example 3 The parapermanent of a 4-th order matrix: a a 2 a 22 pper(a 4 ) = a 3 a 32 a 33 = a 4 a 42 a 43 a 44 = a 4 a 42 a 43 a 44 + a 3 a 32 a 33 a 44 + a 2 a 22 a 43 a 44 + a 2 a 22 a 33 a 44 + +a a 42 a 43 a 44 + a a 32 a 33 a 44 + a a 22 a 43 a 44 + a a 22 a 33 a 44 To each element a ij of a matrix () we associate the triangular table of elements of matrix A n that has a ij in the bottom left corner We call this table a corner of the matrix and denote it by R ij (A n ) Corner R ij (A n ) is a triangular matrix of order (i j + ), and it contains only elements a rs of matrix () whose indices satisfy the inequalities j s r i Theorem 4 [] (Decomposition of a parapermanent pper(a n ) by elements of the last row) The following formula are valid: n pper(a n ) = {a ns } pper(r s, (A n )), (3) where pper(r 0, (A n )) s= Example 5 Decomposition of a parapermanent pper(a 4 ) by elements of the last row: pper(a 4 ) = a 44 pper(a 3 ) + a 43 a 44 pper(a 2 ) + a 42 a 43 a 44 pper(a ) + a 4 a 42 a 43 a 44 pper(a 0 ), where pper(a ) = a, ppera 0 R Zatorsky and I Lishchynskyy [0, 3] established connection between the paradeterminats and the lower Hessenberg determinants by formula {a } 0 0 0 {a 2 } {a 22 } 0 0 {a pper(a n ) = 3 } {a 32 } {a 33 } 0 0, (4) {a n, } {a n,2 } {a n,3 } {a n,n } {a n } {a n2 } {a n3 } {a n,n } {a nn } where {a ij } is factorial product of the element a ij 04
2 A connection between the Horadam numbers with even (odd) indices and parapermanents In [5] A Horadam considered the sequence h = p, h 2 = q, h n = h n + h n 2, n 3, where p and q are arbitrary integer numbers This sequence generalized the Fibonacci sequence: and the Lucas sequence: F =, F 2 =, F n = F n + F n 2, n 3, L = 2, L 2 =, L n = L n + L n 2, n 3 Proposition 2 The following formula is valid: p h 2 h 0 4 h h 0 0 6 h 2n = h 3 0 0 0 0 0 0 0 0 0 0 0 0 0 h 2n 4 h 2n 7 h 2n 2 h 2n 5 (5) Proof Expanding the parapermanent (5) by elements of the last raw (see (3)), we have h 2n = h 2n 3 + h 2n 2 h 2n 5 h 2n 5 = h 2n 3 + h 2n 2 Obtained equality holds by definition of the sequence {h n } n Proposition 22 The following formula is valid: q h 3 h 0 5 h 2 h 0 0 7 h 2n = h 4 0 0 0 0 0 0 0 0 0 0 0 0 0 Proof Using (3), we have h 2n 3 h 2n 6 h 2n h 2n 4 h 2n = h 2n 2 + h 2n h 2n 4 h 2n 4 = h 2n 2 + h 2n (6) 05
3 Main results In this section we proved two recurrence formulas expressing the Horadam numbers h n by the determinant of tridiagonal matrix As a consequence we received the corresponding formulas for the Fibonacci and Lucas numbers Proposition 3 The following formulas are valid: p 0 0 0 0 0 0 h 2 0 0 0 0 0 0 h 4 h h 0 0 0 0 h 2n = 0 0 h 6 h 3 h 3 0 0 0, (7) h h 3 h 2n 5 0 0 0 0 0 h 2n 4 h 2n 7 h 2n 7 0 0 0 0 0 0 h 2n 2 h 2n 5 q 0 0 0 0 0 0 h 3 0 0 0 0 0 0 h 5 h 2 h 2 0 0 0 0 h 2n = 0 0 h 7 h 4 h 4 0 0 0 (8) h 2 h 4 h 2n 4 0 0 0 0 0 h 2n 3 h 2n 6 h 2n 6 0 0 0 0 0 0 h 2n h 2n 4 Proof We prove the formula (7) From (5) using (4), we have p 0 0 0 0 0 0 h 2 0 0 0 0 0 0 h 4 h 0 0 0 0 h 2n = 0 0 h 6 h 3 0 0 0 0 0 0 0 0 h 2n 4 h 2n 7 0 0 0 0 0 0 h 2n 2 h 2n 5 After obvious simple transformations, we get (7) Formula (8) can be proved similarly Example 32 Fibonacci numbers with odd indices: 0 0 0 0 0 0 F 2 0 0 0 0 0 0 F 4 F F 0 0 0 0 F 2n = 0 0 F 6 F 3 F 3 0 0 0 F F 3 F 2n 5 0 0 0 0 0 F 2n 4 F 2n 7 F 2n 7 0 0 0 0 0 0 F 2n 2 F 2n 5 06
Example 33 The Fibonacci numbers with even indices: 0 0 0 0 0 0 F 3 0 0 0 0 0 0 F 5 F 2 F 2 0 0 0 0 F 2n = 0 0 F 7 F 4 F 4 0 0 0 F 2 F 4 F 2n 4 0 0 0 0 0 F 2n 3 F 2n 6 F 2n 6 0 0 0 0 0 0 F 2n F 2n 4 Example 34 The Lucas numbers with odd indices: 2 0 0 0 0 0 0 L 2 0 0 0 0 0 0 L 4 L L 0 0 0 0 L 2n = 0 0 L 6 L 3 L 3 0 0 0 L L 3 L 2n 5 0 0 0 0 0 L 2n 4 L 2n 7 L 2n 7 0 0 0 0 0 0 L 2n 2 L 2n 5 Example 35 The Lucas numbers with even indices: 0 0 0 0 0 0 L 3 0 0 0 0 0 0 L 5 L 2 L 2 0 0 0 0 L 2n = 0 0 L 7 L 4 L 4 0 0 0 L 2 L 4 L 2n 4 0 0 0 0 0 L 2n 3 L Ln 6 L 2n 6 0 0 0 0 0 0 L 2n L 2n 4 Note, that determinants of matrices, elements of which are classical or generalized Fibonacci numbers, in particular, studied in [, 2, 3, 4, 6, 7, 8] 4 Acknowledgements The author is grateful to professor Roman Zatorsky, Department of Mathematics and Informatics, Vasyl Stefanyk Precarpathian National University (Ukraine), for constant attention to this work and for useful discussions References [] Civciv, H (2008) A note on the determinant of five-diagonal matrices with Fibonacci numbers, Int J Contemp Math Sciences, 3(9), 49 424 07
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