Exact Determinants of the Rfpгlгr Circulant Involving Jacobsthal

WSEAS TRANSACTIONS on MATHEMATICS Tingting Xu, Zhaolin Jiang

Exact Determinants of the RFPrLrR Circulant Involving Jacobsthal, Jacobsthal-Lucas, Perrin and Padovan Numbers

TINGTING XU ZHAOLIN JIANG∗ Department of Mathematics Department of Mathematics Linyi University Linyi University Shuangling Road, Linyi city Shuangling Road, Linyi city CHINA CHINA [email protected] [email protected]

Abstract: The row ﬁrst-plus-rlast r-right (RFPrLrR) circulant matrix and row last-plus-rﬁrst r-left (RLPrFrL) circulant matrices are two special pattern matrices. On the basis of the beautiful properties of famous numbers and the inverse factorization of polynomial, we give the exact determinants of the two pattern matrices involving Jacobsthal, Jacobsthal-Lucas, Perrin and Padovan numbers.

Key–Words: Determinant; RFPrLrR circulant matrix; RLPrFrL circulant matrix; Jacobsthal number; Jacobsthal- Lucas number; Perrin number; Padovan number;

1 Introduction first-plus-r last r-right (RFPrLrR) circulant matrix and row last-plus-r first r-left (RLPrFrL) circulan- Circulant matrix family occurs in various fields, ap- t matrix. Based on the characteristic polynomial of plied in image processing, communications, signal the basic RFPrLrR circulant matrix and the Binet processing, encoding and preconditioner. Meanwhile, formulae of famous numbers, we obtain the exac- the circulant matrices [1, 2] have been extended in t determinants of two patten matrices with Jacobsthal, many directions recently. The f(x)-circulant matrix Jacobsthal-Lucas, Perrin and Padoven numbers. is another natural extension of the research category, A row first-plus-rlast r-right (RFPrLrR) cir- please refer to [3, 11]. culant matrix with the first row( (a0, a1, . . . , a)n−1), Recently, some authors researched the circulant denoted by RFPrLRcircrfr a0, a1, . . . , an−1 , is type matrices with famous numbers. In [3], Shen et al. meant to be a square matrix of the form discussed the explicit determinants of the RFMLR and   RLMFL circulant matrices involving certain famous a0 a1 . . . an−1   numbers. Jaiswal [4] showed some determinants of  ran−1 a0 + ran−1 . . . an−2    circulant matrices whose elements are the generalized A =  ran−2 ran−1 + ran−2 . . . an−3  . Fibonacci numbers. Lind presented the determinants  ......  of circulant and skew circulant involving Fibonacci ra1 ra2 + ra1 . . . a0 + ran−1 numbers in [5]. Gao et al. [6] considered the deter- n minants and inverses of skew circulant and skew left Be aware that the RFPrLrR circulant matrix is a x − circulant matrices with Fibonacci and Lucas number- rx − r-circulant matrix. s. Akbulak and Bozkurt [7] proposed some properties We define Θ(r,r) as the basic RFPrLrR circulant of Toeplitz matrices involving Fibonacci and Lucas matrix, that is   numbers. In [8], authors considered circulant matri- 0 1 0 ... 0 0 ces with Fibonacci and Lucas numbers and proposed    0 0 1 ... 0 0  their explicit determinants and inverses by construct-   Θ(r,r) =  ......  ing the transformation matrices. Jiang and Hong gave   exact determinants of some special circulant matrices 0 0 0 ... 0 1 r r 0 ... 0 0 involving four kinds of famous numbers in [9]. See n×n more of the literatures in [10, 11, 12]. = RFPrLRcircrfr(0, 1, 0,..., 0). We introduce two new patten matrices, i.e. row Both the minimal polynomial and the characteris- ∗ n Corresponding author. E-mail addresses: jzh1208@sina. tic polynomial of Θ(r,r) are g(x) = x −rx−r, which com has only simple roots, denoted by πi (i = 1, 2, . . . , n).

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A row last-plus-rfirst r-left (RLPrFrL) circulant and n n matrix with the first( row (a0, a1, . . . ,) an−1), denoted jn = α + β , by RLPrFLcirc fr a , a , . . . , a − , is meant to be r 0 1 n 1 where α, β are the roots of the equation x2−x−2 = 0. a square matrix of the form While, recurrences (5) and (6) involve the charac-   teristic equation x3 − x − 1 = 0, its roots are denoted a0 . . . an−2 an−1   by r1, r2, r3. Then we obtain:  a1 . . . an−1 + ra0 ra0  B =  a . . . ra + ra ra  .   2 0 1 1   r + r + r = 0,  ......  1 2 3 r1r2 + r1r3 + r2r3 = −1, (7) an−1 + ra0 . . . ran−3 + ran−2 ran−2  r1r2r3 = 1. Let A=RLPrFLcirc fr(a , a , . . . , a − ) and r 0 1 n 1 In addition, the Binet form for the Perrin se- B = RFPrLRcirc fr(a − , a − , . . . , a ). By ex- r n 1 n 2 0 quences is plicit computation, we find n n n Rn = r1 + r2 + r3 , (8) b A = BIn, (1) and the Binet form for Padovan sequences is where Ib is the backward identity matrix of the form P n n n n n = a1r1 + a2r2 + a3r3 , (9)   1 where     ∏3 r − 1  0 1  a = j , i = 1, 2, 3. b  ..  i − In =  .  . (2) ri rj   j=1   j≠ i 1 0 1 2 Main Results The Jacobsthal sequences {Jn}, Jacobsthal- By Proposition 5.1 in [19] and properties of RFPrLrR Lucas sequences {jn}[13, 14, 15], Perrin sequences circulant matrices. we deduce the following lemma. {Rn} and Padovan sequences {Pn} [16, 17, 18] are defined by the following recurrence relations, respec- Lemma 1. Let A = RFPrLRcirc fr(a , a ,..., tively: r 0 1 an−1). Then the eigenvalues of A are given by J = J − + 2J − , n ≥ 2, (3) n n 1 n 2 n∑−1 ≥ j − jn = jn−1 + 2jn−2, n 2, (4) λi = f(πi) = ajπi , i = 0, 1, 2, . . . , n 1, j=0 Rn = Rn−2 + Rn−3, n ≥ 3, (5) ( ) P P P ≥ n = n−2 + n−3, n 3, (6) where, πi i = 1, 2, . . . , n are the eigenvalues of Θ(r,r), and the determinant of A is given by with the initial condition J0 = 0, J1 = 1, j0 = 2, P j1 = 1, R0 = 3, R1 = 0, R2 = 2 and 0 = 1, ∏n ∏n n∑−1 P1 = 1, P2 = 1 . j det A = λi = ajπi . The first few members of these sequences are giv- i=1 i=1 j=0 en as follows: n 0 1 2 3 4 5 6 Lemma 2. Assume πi (i = 1, 2, . . . , n) are the roots of the characteristic polynomial of Θ . If a = 0, J 0 1 1 3 5 11 21 (r,r) n then jn 2 1 5 7 17 31 65 ∏n Rn 3 0 2 3 2 5 5 3 2 (aπ + bπ + cπi + d) Pn 1 1 1 2 2 3 4 i i i=1 ∏n = (bπ2 + cπ + d) The sequences {J } and {j } are given by the i i n n i=1 Binet formulae [ ( ) ( ) =dn − bn−1 dr sn−1 + tn−1 + br sn + tn αn − βn ( )] J = − r2 d − c + b , n α − β

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∈ R n 2 2n 2n 2n n n where a, b, c and x3 ) = x1 + x2 + x3 + 2[(x1x2) + (x1x3) + n √ √ (x2x3) ]. Taking the relation of roots and coefﬁcients −c + c2 − 4bd −c − c2 − 4bd s = ; t = .  2b 2b  − b x1 + x2 + x3 = a x x + x x + x x = c If a ≠ 0, then  1 2 1 3 2 3 a x x x = − d ∏n 1 2 3 a (aπ3 + bπ2 + cπ + d) i i i into account, we get that i=1 − − n 1 ( ) ∏n n ( a) − 2 2 =d + d rX2(n−1) rXn−1 + 2r Xn−1 3 2 2 (aπi + bπi + cπi + d) − n ( ) i=1 ( a) 2 − 2 − 2 + rX2n + 2r Xn rXn 2r Xn+1 − n−1 ( ) 2 n ( a) 2 2 ( ( ) ) − − n−1 3 2 = d + d rX2(n−1) rXn−1 + 2r Xn 1 + (−a) r a − b + c − d − br Xn , 2 − n ( ) ( a) 2 2 2 n n n + rX2n + 2r Xn − rX − 2r Xn+1 where Xn = x1 + x2 + x3 , and x1, x2, x3 are the n 3 2 2 ( ( ) ) roots of the equation aπi + bπi + cπi + d = 0. n−1 3 2 + (−a) r a − b + c − d − br Xn , Proof. Since πi (i = 1, 2, . . . , n) are the roots of the n n n n characteristic polynomial of Θ(r,r), g(x) = x −rx− where Xn = x1 + x2 + x3 , and x1, x2, x3 are the 3 2 ⊓⊔ r can be factored as roots of the equation aπi + bπi + cπi + d = 0. ∏n In the following, we show the exact determinants xn − rx − r = (x − π ). of the RFPrLrR and RLPrFrL circulant matrices in- i volving some related famous numbers. i=1

3 2 Let x1, x2, x3 be the roots of the equation aπi +bπi + cπi + d = 0. If a = 0, please see [11] for details of 3 Determinants of the RFPrLrR and the proof. If a ̸= 0, then RLPrFrL Circulant Matrices In- ∏n 3 2 volving the Jacobsthal Numbers (aπi + bπi + cπi + d) i=1 Theorem 3. If A = RFPrLRcirc fr (J ,...,J − ), ∏n r 0 n 1 b c d then =an (π3 + π2 + π + ) i a i a i a i=1 detA ∏n ( ) ( ) [ ( )] n − − − − n − n−1 2 n−1 n−1 =a (πi x1)(πi x2)(πi x3) rJn + 2rJn−1 r Jn s1 + t1 i=1 = 1 − rjn−1 − rjn ∏n ∏n ∏n ( ) − [ ( ) ] n n 1 2 n n 2 =(−a) (x − π ) (x − π ) (x − π ) − 2rJn−1 2r Jn−1 s + t − r 1 i 2 i 3 i + 1 1 , i=1 i=1 i=1 1 − rjn−1 − rjn − n n − − n − − n − − =( a) (x1 rx1 r)(x2 rx2 r)(x3 rx3 r) { [ where n n n−1 =(−a) (x1x2x3) − rx1x2x3 (x1x2) ] [ rJn + 2rJn−1 − 1 n−1 n−1 n n s1 = + (x1x3) + (x2x3) − r (x1x2) + (x1x3) − ] √ 4rJn−1 n 2 n−1 n−1 n−1 ( ) + (x2x3) + r x1x2x3(x1 + x2 + x3 ) 2 2 [ ] 1 − rJn − 2rJn−1 − 8r JnJn−1 2 n n n + r x1 (x2 + x3) + x2 (x1 + x3) + x3 (x1 + x2) + , −4rJn−1 2 n n n − 3 + r (x1 + x2 + x3 ) r x1x2x3 3 − r (x1x2 + x1x3 + x2x3) } rJn + 2rJn−1 − 1 3 3 t1 = − r (x1 + x2 + x3) − r . −4rJn−1 √( ) − − 2 − 2 Let X = xn + xn + xn, we obtain (x x )n + 1 rJn 2rJn−1 8r JnJn−1 n 1 2 3 1 2 − . X2 −X n n n 2n n n −4rJn−1 (x1x3) + (x2x3) = 2 from (x1 + x2 +

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′ Proof. The matrix A can be written as Theorem 4. If A = RFPrLRcirc fr (J − ,...,J ),   r n 1 0 then J0 J1 ...Jn−1   ( ) ( ) ( )  rJn−1 J0 + rJn−1 ...Jn−2  − n − − n−1 −  . . . .  ′ r 2Jn−1 r Jn r r 2Jn−1 A =  . . .. .  . detA =  . . .  (−2)n + 2rj − − rj   ( ) n 1 n rJ2 rJ3 + rJ2 ...J1 n r Jn − r rJ rJ + rJ ...J + rJ − − 1 2 1 0 n 1 n . (−2) + 2rjn−1 − rjn Using the Lemma 1, the determinant of A is detA Theorem 5. If J = RLPrFLcircrfr (J0,...,Jn−1), then we have ∏n ( ) ··· n−1 [( ) ( ) ( ) = J0 + J1πi + + Jn−1πi n n−1 r − 2Jn−1 − r Jn − r r − 2Jn−1 i=1 ( detJ = n (−2)n + 2rj − − rj ∏ − 2 − 2 ( ) n 1 n α β α β 2 n ] = π + π + ··· − − − i − i r Jn r n(n 1) α β α β − − 2 i=1 n ( 1) . ) (−2) + 2rjn−1 − rjn αn−1 − βn−1 + πn−1 α − β i Proof. The matrix J can be written as ( ) ( )   ∏n − 2 − − − − 2rJn−1 πi 1 rJn 2rJn−1 πi rJn J0 ...Jn−2 Jn−1 = 2 .   1 −πi − 2π  J1 ...Jn−1 + rJ0 rJ0  i=1 i   J =  . .. . .  According to Lemma 2, we can get  . . . .    Jn−2 . . . rJn−4 + rJn−3 rJn−3 ∏n [( ) ( ) ] Jn−1 + rJ0 . . . rJn−3 + rJn−2 rJn−2 2   − 2rJn−1 π − 1− rJn − 2rJn−1 πi − rJn i Jn−1 Jn−2 ...J0 i=1   ( ) ( ) [ ( )  rJ0 Jn−1 + rJ0 ...J1  n n−1 2 n−1 n−1 = − rJ − − 2rJ − − r J s + t  . . . .  n n 1( ) n ]1 1 =  . . .. .  2 n n 2   − 2r J − s + t + r .   n 1 1 1 rJn−3 rJn−4 + rJn−3 ...Jn−2 rJ − rJ − + rJ − ...J − + rJ Then we obtain that  n 2 n 3 n 2  n 1 0 detA 0 0 ... 0 1 ( ) ( ) [ ( )]  0 ... 0 1 0  n n−1 2 n−1 n−1   − rJ + − 2rJ − r J s + t   n n 1 n 1 1 ×  ......  =  . . . . .  , 1 − rjn−1 − rjn ( ) [ ( ) ]  0 1 0 ... 0  n−1 2 n n 2 − 2rJn−1 2r Jn−1 s + t − r + 1 1 , 1 0 0 ... 0 1 − rjn−1 − rjn where then we can get rJ + 2rJ − − 1 n n 1 detJ = detA′detΓ, s1 = − √4rJn−1 ( ) ′ 2 2 which matrix A = RFPrLRcirc fr (J − ,...,J ), 1 − rJn − 2rJn−1 − 8r JnJn−1 r n 1 0 + , its determinant could be obtained through Theorem 2. −4rJn−1 − ( ) ( ) ( ) rJn + 2rJn−1 1 − n − − n−1 − t1 = ′ r 2Jn−1 r Jn r r 2Jn−1 −4rJ − detA = √ n 1 − n − ( ) ( 2)( + 2rj) n−1 rjn 2 2 n 1 − rJn − 2rJn−1 − 8r JnJn−1 r J − r − . − n n , −4rJn−1 (−2) + 2rjn−1 − rjn ⊓⊔ and Using the method in Theorem 3 similarly, we also n(n−1) have detΓ = (−1) 2 .

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So where √ ′ ( ) detJ = detA detΓ 2 2 jn − r + r − jn + 8r + 16rjn−1 [( ) ( ) − ( ) n n 1 s3 = , r − 2Jn−1 − r Jn − r r − 2Jn−1 4r = √( ) (−2)n + 2rj − − rj 2 ( ) n 1 n j − r − r − j + 8r2 + 16rj − n ] n n n 1 r Jn − r n(n−1) t3 = . − − 2 4r n ( 1) . (−2) + 2rjn−1 − rjn ′ Theorem 8. If J = RLPrFLcirc fr (j , . . . , j − ), ⊓⊔ r 0 n 1 then we have ( ) [ n − r − 2jn−1 4 Determinants of the RFPrLrR and detJ ′ = (−2)n + 2rj − − rj ( ) n 1 (n ) RLPrFrL Circulant Matrix In- n−1 2 n−1 n−1 2r (r + 2rjn−1) s + t + 3 3 volving Jacobsthal-Lucas Number- − n − ((2) + 2rjn−1 rjn ) s ( ) ( ) n−1 2 n n − 2 − ] 2r 2r s3 + t3 r (jn 2jn−1) Theorem 6. If B = RFPrLRcircrfr (j0, . . . , jn−1), − (−2)n + 2rj − − rj then n 1 n ( ) n(n−1) n × (−1) 2 , 2 − rjn detB = 1 − rj − − rj ( n) 1 [ ( n )( )] where n−1 n−1 n−1 √ − 2rjn−1 r rjn − 2 s + t ( ) + 2 2 2 j − r + r − j + 8r2 + 16rj − 1 − rjn−1 − rjn n n n 1 ( ) [ ( ) ] s3 = , − n−1 2 n n 2 4r 2rjn−1 2r s2 + t2 + 3r √( ) + , 2 2 1 − rjn−1 − rjn jn − r − r − jn + 8r + 16rjn−1 t = . 3 4r where Proof. J ′ 1 + 2rjn−1 + rjn The matrix can be written as s2 =   −4rjn−1 √( ) j0 . . . jn−2 jn−1 2   1 + 2rj − + rj + 16rj − − 8r2j j −  j1 . . . jn−1 + rj0 rj0  n 1 n n 1 n n 1   + , J ′ =  . .. . .  −4rjn−1  . . . .    1 + 2rjn−1 + rjn jn−2 . . . rjn−4 + rJn−3 rjn−3 t2 = −4rjn−1 jn−1 + rj0 . . . rjn−3 + rjn−2 rjn−2 √( )   2 − 2 jn−1 jn−2 . . . j0 1 + 2rjn−1 + rjn + 16rjn−1 8r jnjn−1   − .  rj0 jn−1 + rj0 . . . j1  −4rj −   n 1  . . .. .  =  . . . .    Similarly, we also get the following Theorem 7. rjn−3 rjn−4 + rjn−3 . . . jn−2 ′ rjn−2 rjn−3 + rjn−2 . . . jn−1 + rj0 Theorem 7. If B = RFPrLRcircrfr (jn−1, . . . , j0),   then 0 0 ... 0 1   ( ) 0 ... 0 1 0 n   − r − 2j −  . . . . .  ′ n 1 ×  ......  , detB = n   (−2) + 2rjn−1 − rjn   ( ) ( ) 0 1 0 ... 0 n−1 2 n−1 n−1 2r (r + 2rjn−1) s + t 1 0 0 ... 0 + 3 3 (−2)n + 2rj − − rj ( ) ( n) 1 n n−1 2 n n 2 then we can get 2r [2r s + t − r (jn − 2jn−1)] − 3 3 , − n − ′ ′ ( 2) + 2rjn−1 rjn detJ = detB detΓ,

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′ with matrix B = RFPrLRcircrfr (jn−1, . . . , j0), its 5 Determinants of the RFPrLrR and determinant could be obtained through Theorem 7. RLPrFrL Circulant Matrix In-

( ) volving Perrin Numbers n ( ′ − r − 2jn−1 Theorem 9. Let C = RFPrLRcircrfr R1,R2,..., detB = n ) (−2) + 2rjn−1 − rjn ( ) ( ) Rn . Then n−1 2 n−1 n−1 2r (r + 2rjn−1) s + t ( ) + 3 3 2 − rR − rR n − n − n+1 n+2 ( 2) + 2rjn−1 rjn det C = ′ ( ) − ( ) Y n 1 2 n n − 2 − [( ) − ( ) 2r [2r s3 + t3 r (jn 2jn−1)] n 1 − , rRn rRn+1 + rRn+2 − n − + ( 2) + 2rjn−1 rjn Y′ ] ( ) × 2 − − 2 rXn−1 rX2(n−1) 2r Xn−1 where ( ) ( ) rR n rX2 − rX − 2r2X + 2r2X − n n 2n n n+1 Y′ √( ) [ ( ) ( ) − − 2 2 r3 1 − 2rR + 3 − rR − rR r2X jn r + r jn + 8r + 16rjn−1 − n+2 n n+2 n s3 = , Y′ √ 4r ] ( ) ( ) − − − 2 2 n−1 jn r r jn + 8r + 16rjn−1 × 2 rRn , t = , 3 4r where Y′ − 2 − 2 − 2 = rYn rYn−1 +rY2n +rY2(n−1) 2r Yn+1 + − 2 3 n(n 1) 2r Y − − 2r + 2. − 2 n 1 and detΓ = ( 1) . n n n Xn = x1 + x2 + x3 , and x1, x2, x3 are the roots( of − 3 − − 2 − So the equation) ( rRnx +(3 rRn) rRn+2)x + 2 rRn+1 x + − rRn+1 − rRn+2 = 0. n n n Yn = y + y + y , y1, y2, y3 are the roots of the ′ ′ 1 2 3 detJ = detB detΓ equation y3 + y2 − 1 = 0. ( ) [ n − r − 2jn−1 Proof. Obviously, the matrix C has the form = n   (−2) + 2rjn−1 − rjn ( ) ( ) R1 R2 ...Rn n−1 2 n−1 n−1   2r (r + 2rjn−1) s3 + t3  rRn R1 + rRn ...Rn−1  +  . . .  (−2)n + 2rj − − rj  . . .. .  ( n 1 n ) C =  . . . .  . ( ) ( )   n−1 2 n n − 2 − ] rR3 rR4 + rR3 ...R2 2r 2r s3 + t3 r (jn 2jn−1) − rR2 rR3 + rR2 ...R1 + rRn − n − ( 2) + 2rjn−1 rjn Owing to Lemma 1, the Binet form (8) and (7), n(n−1) × (−1) 2 , ∏n ( ) ··· n−1 det C = R1 + R2πi + + Rnπi i=1 where ∏n ∑n ∑3 k k−1 = rj πi i=1 k=1 j=1 √( ) 2 ∏n ∑3 n n j − r + r − j + 8r2 + 16rj − rj(1 − r π ) n n n 1 = j i s3 = , 1 − r π √ 4r i=1 j=1 j i ( ) [ 2 2 ∏n jn − r − r − jn + 8r + 16rjn−1 −rR π3 + (3 − rR − rR )π2 t = . = n i n n+2 i 3 − 3 − 2 4r πi πi + 1 i=1 ( ) ( )] 2 − rRn+1 πi − rRn+1 + rRn+2 + . ⊓⊔ − 3 − 2 πi πi + 1

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( ′ By Lemma 2 and the recurrence (5), we obtain Theorem) 10. Let C = RFPrLRcircrfr Rn,Rn−1, [ ...,R1 . Then ∏n [ − rR π3 + (3 − rR − rR )π2 (R − 3r)n r(R − 3r)(xn−1 + xn−1) n i n n+2 i det C′ = n − n 1 2 i=1 ] Y′′ Y′′ ( ) ( ) n n r(Rn+1 − 2r)(x + x ) + 2 − rRn+1 πi − rRn+1 + rRn+2 1 2 + ′′ ( ) ( Y ) n ] = − rR − rR r2 R + R − R [ n+1 n+2 − n n+1 n+2 · − n−1 ( ) ( ) ′′ (Rn+1 2r) , 1 n−1 Y + rRn rRn+1 + rRn+2 2 ] where ( ) 2 2 × rX − − rX − − 2r X − ′′ 3 1 2 2 n 1 2(n 1) n 1 Y =1 − r − [rY − rY − − 2r Y − ] 2 n−1 2(n 1) n 1 1( ) ( ) − n 2 − − 2 2 1 2 2 2 rRn rXn rX2n 2r Xn + 2r Xn+1 − [rY − rY2n − 2r Yn + 2r Yn+1] 2 [ 2 n ( ) ( ) √ n−1 5r − R + χ − rR r3 1 − 2rR n+2 n n+2 x1 = − , ] 2Rn+1 4r√ ( ) − − − − 2 5r Rn+2 χ + 3 rRn rRn+2 r Xn , x2 = , 2Rn+1 − 4r χ =(R − 5r)2 − 4(R − 2r)(R − 3r), n n n n+2 n+1 n where Xn = x1 +x2 +x3 , and x1, x2, x3 are the roots 3 2 n n n (of the equation) −rR( nx + (3 − rRn −) rRn+2)x + Yn = y1 + y2 + y3 , and y1, y2, y3 are the roots of the 3 − − 2 − rRn+1 x + − rRn+1 − rRn+2 = 0. And equation y y 1 = 0. Proof. The matrix C′ has the form ∏n   3 2 (−π − π + 1) Rn Rn−1 ...R1 i i   i=1  rR1 Rn + rR1 ...R2  ( )  . . .  1 2 2  . . .. .  . = − rY − rY + rY + rY −  . . . .  2 n n−1 2n 2(n 1)   rR − rR − + rR − ...R − 2 2 3 n 2 n 3 n 2 n 1 − r Yn+1 + r Yn−1 − r + 1, rRn−1 rRn−2 + rRn−1 ...Rn + rR1 n n n According to Lemma 1, the Binet form (8) and (7), we where Yn = y1 + y2 + y3 , and y1, y2, y3 are the roots of the equation y3 + y2 − 1 = 0. have Consequently, ∏n ( ) ′ n−1 det C = R + R − π + ··· + R π ( ) n n 1 i 1 i n i=1 2 − rRn+1 − rRn+2 det C = ∏n n∑−1 ∑3 ∏n ∑3 n+1 n Y′ r − rjπ ( ) ( ) = rn−kπk = j i [ n−1 j i − rRn rRn+1 + rRn+2 rj πi + i=1 k=0 j=1 i=1 j=1 Y′ n ] ∏ (R − 2r)π2 + (R − 5r)π + R − 3r ( ) = n+1 i n+2 i n . × 2 − − 2 −π3 + π + 1 rXn−1 rX2(n−1) 2r Xn−1 i=1 i i ( ) ( ) rR n rX2 − rX − 2r2X + 2r2X Using Lemma 2 and the recurrence (5), we obtain − n n 2n n n+1 ′ ( ) (Y ) ∏n [ ] [ − 2 − − r3 1 − 2rR + 3 − rR − rR r2X (Rn+1 2r)πi + (Rn+2 5r)πi + Rn 3r − n+2 n n+2 n Y′ i=1 [ ] ( ) − n − − n−1 n−1 n−1 =(Rn 3r) r(Rn 3r)(x1 + x2 ) × 2 rRn . + r(R − 2r)(xn + xn) n+1 1 2 ] ( ) ⊓⊔ 2 n−1 − r Rn + Rn+1 − Rn+2 (Rn+1 − 2r) ,

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where √ √ 6 Determinants of the RFPrLrR and − − − 5r Rn+2 + χ 5r Rn+2 χ RLPrFrL Circulant Matrix In- x1 = , x2 = , 2Rn+1 − 4r 2Rn+1 − 4r 2 volving Padovan Numbers χ =(Rn+2 − 5r) − 4(Rn+1 − 2r)(Rn − 3r). P P And Theorem 12. Let D = RFPrLRcircrfr( 1, 2,..., ∏n Pn). Then − 3 ( πi + πi + 1) detD i=1 ( ) ( ) n n−1 − 3 − 1 2 − − 2 P − rP 1 rP ( = 1 r [rYn−1 rY2(n−1) 2r Yn−1] 1 n+1 − 2 n P 2 = ′ ′ 1 V )( V ) 1 2 2 2 2 2 − [rY − rY − 2r Y + 2r Y ], − rP rU − rU − 2r U − 2 n 2n n n+1 ( ) n+1 n−1 2(n−1) n 1 1 n n n n rPn ( ) where Yn = y1 + y2 + y3 , and y1, y2, y3 are the roots − 2 rU 2 − rU − 2r2U + 2r2U of the equation y3 − y − 1 = 0. Therefore, V′ n 2n n n+1 ( ) − [ [ − − n 1 − n − n 1 n 1 rPn ( ) ′ (Rn 3r) − r(Rn 3r)(x1 + x2 ) − r3 P − P + a + a + a det C = ′′ ′′ V′ 1 2 1 2 3 Y Y ] − n n ( ) r(Rn+1 2r)(x1 + x2 ) + a + a + a − rP − rP r2U , + Y′′ 1 2 3 n n+2 n ( )] r2 R + R − R ( ) n n+1 n+2 n−1 ′ − · (Rn+1 − 2r) . V − 3− 1 2 − − 2 − Y′′ where( = 1 r 2 rVn−1 rV2()n−1) 2r Vn−1 1 2 − 2 − 2 n n ⊓⊔ 2 rVn rV2n +2r Vn+1 4r Vn ,Un = u1 +u2 + ( n − P 3 (u3 , u1, u2, u3 are the roots of the) equation( r nx + Theorem 11. Let R = RLPrFLcircrfr R1,R2,..., 2 ) a1 + a2 + a3 − rPn − rPn+2 x + P2 − rPn+1 − Rn . Then P P − P n n n r n+2x+ 1 r n+1 = 0 and Vn = v1 +v2 +v3 , v1, { [ 3 2 − n − n−1 n−1 v2, v3 are the roots of the equation y + y − 1 = 0. (Rn 3r) − r(Rn 3r)(x1 + x2 ) det R = Y′′ Y′′ Proof. The matrix D has the form r(R − 2r)(xn + xn) + n+1 1 2   Y′′ P P ... P ( )] } 1 2 n 2  P P P P  r Rn + Rn+1 − Rn+2  r n 1 + r n ... n−1  − · (R − 2r)n−1   Y′′ n+1  . . .. .  D =  . . . .  . n(n−1)  rP rP + rP ... P  × (−1) 2 , 3 4 3 2 rP2 rP3 + rP2 ... P1 + rPn where ′′ 3 1 2 2 Y =1 − r − [rY − − rY − − 2r Yn−1] 2 n 1 2(n 1) ∏n ( ) 1 P P ··· P n−1 − 2 − − 2 2 det D = 1 + 2πi + + nπi [rYn rY2n 2r Yn + 2r Yn+1], 2 √ i=1 5r − Rn+2 + χ ∏n ∑n ∑3 x1 = , k k−1 − = ajr π 2Rn+1 4r√ j i i=1 k=1 j=1 5r − Rn+2 − χ x2 = , ∏n ∑3 n n 2R − 4r ajrj(1 − r π ) n+1 = j i − 2 − − − 1 − r π χ =(Rn+2 5r) 4(Rn+1 2r)(Rn 3r), i=1 j=1 j i [ ( ) Y = yn + yn + yn, and y , y , y are the roots of the ∏n 3 2 n 1 2 3 1 2 3 −rPnπ + a1 + a2 + a3 − rPn − rPn+2 π 3 − − = i i equation y y 1 = 0. − 3 − 2 πi πi + 1 Proof. Since i=1 ( ) ] n(n−1) P2 − rPn+1 − rPn+2 πi + P1 − rPn+1 b − 2 + , det In = ( 1) , − 3 − 2 πi πi + 1 the result can be derived from Theorem 10 and the relation (1). ⊓⊔ from Lemma 1, the Binet form (9) and (7). Using

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Lemma 2 and the recurrence (6), we obtain ..., P1). Then [ ∏n ( ) ′ − P 3 − P − P 2 detD r nπi + a1 + a2 + a3 r n r n+2 πi [ ( )] n − [ i=1 P − r a + a + a 1 rn 1 ( ] n 1 2 3 − 2 P − ( ) = V′′ V′′ n r a1 + P − rP − rP π + P − rP ] 2 n+1 n+2 i 1 n+1 ) ( ) 2 2 ( ) ( ) ( ) · − − − n 1 n−1 + a2 + a3 rUn−1 rU2(n−1) 2r Un 1 = P1 − rPn+1 − rPn P1 − rPn+1 [ 2 ] 1 n ( ) 2 2 2 r 2 2 2 × rU − rU − 2r U − − · − − n−1 2(n−1) n 1 V′′ rUn rU2n 2r Un + 2r Un+1 ( ) [ ] 1 n n−1 [ ( ) − rP rU 2 − rU − 2r2U + 2r2U r n n 2n n n+1 − r3 P + P − P − rP (2 ) [ ( ) V′′ n n+1 n+2 2 n−1 3 ( ) ] − rPn r P1 − P2 + a1 + a2 + a3 P − P − P 2 ( ) ] + n+1 r 1 r 2 r Un , 2 + a1 + a2 + a3 − rPn − rPn+2 r Un , [ ′′ 3 1 2 where V = 1 − r − rV − rV − − and ] [ 2 n−1 2(n 1) ] 2 − 1 2 − − 2 2 2r Vn−1 2 rVn rV2n 2r Vn + 2r Vn+1 , n n n ∏n [ ] Un = u1 + u2 + u(3 , u1, u2, u3 are the) roots[ of the − 3 − 2 − P 3 P − P − P 2 P − πi πi + 1 equation( r 1)x + ]n+1 r 1 (r 2 x + )n+2 i=1 r a1 +a2 +a3 −rP2 x+Pn −r a1 +a2 +a3 = 0. [ ] n n n 3 1 2 2 and Vn = v1 + v2 + v3 , v1, v2, v3 are the roots of the = 1 − r − rV − rV − − 2r V − 2 n−1 2(n 1) n 1 equation y3 − y − 1 = 0. [ ] − 1 2 − 2 − 2 rVn rV2n + 2r Vn+1 4r Vn , 2 Proof. The matrix D′ has the form n n n where Un = u1 + u2 + u3(, u1, u2, u3 are the roots   − P 3 − P − P P P of the) equation( r nx + a1 + a2 + a3 r n n n−1 ... 1 2   rP x + P −rP −rP )x+P −rP = 0.  rP1 Pn + rP1 ... P2  n+2 2 n+1 n+2 1 n+1   We have the following results:  . . .. .   . . . .  .   rPn−2 rPn−3 + rPn−2 ... Pn−1 detD rPn−1 rPn−2 + rPn−1 ... Pn + rPn−1 ( ) ( ) − P − rP n 1 rP n 1 ( 1 n+1 − 2 n P = ′ ′ 1 V )( V ) 2 2 According to Lemma 1, the Binet form (9) and − rP rU − rU − 2r U − ( ) n+1 n−1 2(n−1) n 1 (7), we have 1 n rPn ( ) − 2 rU 2 − rU − 2r2U + 2r2U V′ n 2n n n+1 ∏n ( ) ( ) ′ − n−1 [ P P ··· P n 1 rP ( ) det D = n + n−1πi + + 1πi − n 3 P − P V′ r 1 2 + a1 + a2 + a3 i=1 ] ∏n n∑−1 ∑3 ∏n ∑3 a rn+1 − a r πn ( ) n−k k j j j j i − P − P 2 = ajrj πi = + a1 + a2 + a3 r n r n+2 r Un , rj − πi i=1 k=0 j=1 i=1 j=1 [ ( ) ∏n 3 2 ( ) −rπ + Pn+1 − 2r π ′ 3 1 2 2 i i where V = 1−r − rV −rV − −2r V − − = ( 2 n−1 2()n 1) n 1 −π3 + π + 1 1 2 − 2 − 2 n n i=1 i i 2 rVn rV2n +2r Vn+1 4r Vn ,Un = u1 +u2 + [ ( ) ] n − P 3 Pn+2 − r a1 + a2 + a3 − r πi (u3 , u1, u2, u3 are the roots of the) equation( r nx + + − P − P 2 P − P − − 3 a1 + a2 + a3 r n r n+2 x + 2 r n+1 πi + πi + 1 n n n ( )] rPn+2x+P1 −rPn+1 = 0 and Vn = v +v +v , v1, 1 2 3 Pn − r a1 + a2 + a3 v v y3 + y2 − 1 = 0 + . 2, 3 are the roots of the equation . − 3 ⊓⊔ πi + πi + 1

′ Theorem 13. Let D = RFPrLRcircrfr(Pn, Pn−1, Using Lemma 2 and (9), we obtain

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( P P P Theorem) 14. Let = RLPrFLcircrfr 1, 2, [ ..., Pn . Then ∏n ( ) [ ( − 3 P − 2 P − rπi + n+1 2r πi + n+2 r a1 + a2 det P i=1 {[ ( )] [ ] n 1 n−1 ) ] ( ) Pn − r a1 + a2 + a3 r ( = − 2 P − r a + a3 − r πi + Pn − r a1 + a2 + a3 V′′ V′′ n 1 ] [ ( )] ( ) [ ( ) ( ) n 1 n−1 · 2 − − 2 = Pn −r a1 + a2 +a3 − rP1 Pn − r a1 + a2 + a3 rUn−1 rU2(n−1) 2r Un−1 )]( 2 ) 2 2 1 n + a2 + a3 rU − − rU2(n−1) − 2r Un−1 r ( ) n 1 − 2 · rU 2 − rU − 2r2U + 2r2U 1( ) ( ) V′′ n 2n n n+1 − rP n rU 2 − rU − 2r2U + 2r2U 2 1 n 2n n n+1 rn−1 [ ( ) ( ) [ ( ) − r3 P + P − P − rP n−1 3 V′′ n n+1 n+2 2 − rP1 r Pn + Pn+1 − Pn+2 − rP2 } ( ) ] ( ) ] − P − P − P 2 2 n(n 1) + n+1 r 1 r 2 r Un , + Pn+1 − rP1 − rP2 r Un · (−1) 2 ,

n n n [ where U = u + u + u , u , u , u are the roots ′′ 3 1 2 n 1 2 3( 1 2 3 ) where V = 1 − r − rV − rV − − − P 3 P − P − P 2 ] [ 2 n−1 2(n 1) ] of the equation r 1x + n+1 r 1 r 2 x + 2 1 2 2 2 [ ( ) ] ( 2r V − − rV − rV − 2r V + 2r V , P − − P P − n 1 2 n 2n n n+1 )n+2 r a1 +a2 +a3 r 2 x+ n r a1 +a2 + n n n Un = u1 + u2 + u(3 , u1, u2, u3 are the) roots[ of the a3 = 0. − P 3 P − P − P 2 P − equation( r 1)x + ]n+1 r 1 (r 2 x + )n+2 r a1 +a2 +a3 −rP2 x+Pn −r a1 +a2 +a3 = 0. ∏n ( ) n n n − 3 and Vn = v1 + v2 + v3 , v1, v2, v3 are the roots of the πi + πi + 1 equation y3 − y − 1 = 0. i=1 [ ] 3 1 2 2 Proof. The theorem can be proved by using Theorem = 1 − r − rV − rV − − 2r V − 2 n−1 2(n 1) n 1 13 and the relation (1). ⊓⊔ [ ] − 1 2 − − 2 2 rVn rV2n 2r Vn + 2r Vn+1 , 2 7 Conclusion n n n where Vn = v1 + v2 + v3 , v1, v2, v3 are the roots of the equation y3 − y − 1 = 0. In this paper, we introduce RFPrLrR circulant matrix and RLPrFrL circulant matrices, which are two spe- Then we have the following results: cial pattern matrices. Based on the Binet formulaes of famous numbers and the characteristic polynomial det D′ [ ( )] [ of the basic RFPrLrR circulant matrix, we give the P − r a + a + a n 1 rn−1 ( exact determinants of the two pattern matrices involv- n 1 2 3 − 2 P − = V′′ V′′ n r a1 ing Jacobsthal, Jacobsthal-Lucas, Perrin and Padovan ] ) ( ) numbers in section 3, 4, 5 and 6. 2 2 + a2 + a3 · rU − − rU2(n−1) − 2r Un−1 n 1 Acknowledgements: The research is supported by 1 rn ( ) the Development Project of Science & Technology of − 2 · 2 − − 2 2 V′′ rUn rU2n 2r Un + 2r Un+1 Shandong Province (Grant No. 2012GGX10115) and n−1 [ ( ) the AMEP of Linyi University, China. − r 3 P P − P − P ′′ r n + n+1 n+2 r 2 V ( ) ] 2 + Pn+1 − rP1 − rP2 r Un , References: [ [1] P. J. Davis, Circulant matrices, Wiley & Sons, V′′ − 3 − 1 2 − − where ] =[ 1 r 2 rVn−1 rV2(n−1) ] New York, 1979. 2 − 1 2 − − 2 2 2r Vn−1 2 rVn rV2n 2r Vn + 2r Vn+1 , [2] Z. L. Jiang and Z. X. Zhou, Circulant matri- n n n Un = u1 + u2 + u(3 , u1, u2, u3 are the) roots[ of the ces, Chengdu Technology University Publish- − P 3 P − P − P 2 P − equation( r 1)x + ]n+1 r 1 (r 2 x + )n+2 ing Company, Chengdu, 1999. r a1 +a2 +a3 −rP2 x+Pn −r a1 +a2 +a3 = 0. [3] N. Shen, Z. L. Jiang and J. Li, On explicit deter- n n n and Vn = v1 + v2 + v3 , v1, v2, v3 are the roots of the minants of the RFMLR and RLMFL circulan- equation y3 − y − 1 = 0. ⊓⊔ t matrices involving certain famous numbers,

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