ON THE THIRD-ORDER JACOBSTHAL AND THIRD-ORDER JACOBSTHAL-LUCAS SEQUENCES AND THEIR MATRIX REPRESENTATIONS GAMALIEL CERDA-MORALES Abstract. In this paper, we first give new generalizations for third-order Ja- (3) (3) cobsthal {Jn }n∈N and third-order Jacobsthal-Lucas {jn }n∈N sequences for Jacobsthal and Jacobsthal-Lucas numbers. Considering these sequences, we (3) (3) define the matrix sequences which have elements of {Jn }n∈N and {jn }n∈N. Then we investigate their properties. 2010 Mathematics Subject Classification. 11B37, 11B39, 15A15. Keywords and phrases. Third-order Jacobsthal number, third-order Jacobsthal- Lucas number, matrix representation, matrix methods, generalized Jacobsthal number. 1. Introduction The Jacobsthal numbers have many interesting properties and applications in many fields of science (see, e.g., [1]). The Jacobsthal numbers Jn are defined by the recurrence relation (1.1) J =0, J =1, Jn = Jn +2Jn− , n 1. 0 1 +1 1 ≥ Another important sequence is the Jacobsthal-Lucas sequence. This sequence is defined by the recurrence relation j0 =2, j1 =1, jn+1 = jn +2jn−1, n 1. (see, [7]). ≥ In [6] the Jacobsthal recurrence relation (1.1) is extended to higher order re- currence relations and the basic list of identities provided by A. F. Horadam [7] is expanded and extended to several identities for some of the higher order cases. In (3) particular, third order Jacobsthal numbers, Jn n≥0, and third order Jacobsthal- (3) { } Lucas numbers, jn n≥ , are defined by { } 0 (1.2) J (3) = J (3) + J (3) +2J (3), J (3) =0, J (3) = J (3) =1, n 0, n+3 n+2 n+1 n 0 1 2 ≥ arXiv:1806.03709v1 [math.CO] 10 Jun 2018 and (1.3) j(3) = j(3) + j(3) +2j(3), j(3) =2, j(3) =1, j(3) =5, n 0, n+3 n+2 n+1 n 0 1 2 ≥ respectively. The following properties given for third order Jacobsthal numbers and third order Jacobsthal-Lucas numbers play important roles in this paper (see [2, 3, 6]). (3) (3) n+1 (1.4) 3Jn + jn =2 , (1.5) j(3) 3J (3) =2j(3) , n − n n−3 (3) 2 if n 1 (mod 3) (1.6) J 4J (3) = − ≡ , n+2 − n 1 if n 1 (mod 3) 6≡ 1 2 G. CERDA-MORALES 2 if n 0 (mod 3) ≡ (1.7) j(3) 4J (3) = 3 if n 1 (mod 3) , n n − −1 if n ≡ 2 (mod 3) ≡ (3) (3) (3) (1.8) jn+1 + jn =3Jn+2, 1 if n 0 (mod 3) (3) ≡ (1.9) j(3) J = 1 if n 1 (mod 3) , n n+2 − −0 if n ≡ 2 (mod 3) ≡ 2 (3) (3) (3) n (1.10) jn−3 +3Jn jn =4 , n (3) (3) Jn+1 if n 0 (mod 3) (1.11) Jk = (3) 6≡ k ( Jn+1 1 if n 0 (mod 3) X=0 − ≡ and 2 2 (1.12) j(3) 9 J (3) =2n+2j(3) . n − n n−3 Using standard techniques for solving recurrence relations, the auxiliary equa- tion, and its roots are given by 1 i√3 x3 x2 x 2 = 0; x =2, and x = − ± . − − − 2 Note that the latter two are the complex conjugate cube roots of unity. Call them ω1 and ω2, respectively. Thus the Binet formulas can be written as 2 3+2i√3 3 2i√3 1 (1.13) J (3) = 2n ωn − ωn = 2n+1 V (3) n 7 − 21 1 − 21 2 7 − n and 8 3+2i√3 3 2i√3 1 (1.14) j(3) = 2n + ωn + − ωn = 2n+3 +3V (3) , n 7 7 1 7 2 7 n (3) respectively. Here, the sequence Vn n≥ is defined by { } 0 2 if n 0 (mod 3) ≡ V (3) = 3 if n 1 (mod 3) . n −1 if n ≡ 2 (mod 3) ≡ In [4, 5], the authors defined a new matrix generalization of the Fibonacci and Lucas numbers, and using essentially a matrix approach they showed properties of these matrix sequences. The main motivation of this article is to study the matrix sequences of third-order Jacobsthal sequence and third-order Jacobsthal sequence. 2. The third-order Jacobsthal, third-order Jacobsthal-Lucas sequences and their matrix sequences Now, considering these sequences, we define the matrix sequences which have elements of third-order Jacobsthal and third-order Jacobsthal-Lucas sequences. (3) Definition 2.1. Let n 0. The third-order Jacobsthal matrix sequence JMn n∈N ≥ (3) { } and third-order Jacobsthal-Lucas matrix sequence jMn n∈N are defined respec- tively by { } (3) (3) (3) (3) (2.15) JMn+3 = JMn+2 + JMn+1 +2JMn , (3) (3) (3) (3) (2.16) jMn+3 = jMn+2 + jMn+1 +2jMn , ON THE THIRD-ORDER JACOBSTHAL MATRIX SEQUENCES 3 with initial conditions 1 0 0 1 1 2 1 3 2 (3) (3) (3) JM0 = 0 1 0 , JM1 = 1 0 0 , JM2 = 1 1 2 0 0 1 0 1 0 1 0 0 and 1 4 4 5 5 2 10 7 10 (3) (3) (3) jM0 = 2 1 2 , jM1 = 1 4 4 , jM2 = 5 5 2 . 1− 1 2 2 1 2 1 4 4 − − In the rest of this paper, the third-order Jacobsthal and third-order Jacobsthal- (3) (3) (3) Lucas matrix sequences will be denoted by MJ,n and Mj,n instead of JMn and (3) jMn , respectively. Theorem 2.2. For n 0, we have (2.17) ≥ (3) (3) (3) (3) (3) (3) M + M + M M (2 + ω2)M +2ω2M M (3) = J,2 J,1 J,0 2n J,2 − J,1 J,0 ωn J,n (2 ω )(2 ω ) − (2 ω )(ω ω ) 1 − 1 − 2 ! − 1 1 − 2 ! (3) (3) (3) M (2 + ω1)M +2ω1M + J,2 − J,1 J,0 ωn. (2 ω )(ω ω ) 2 − 2 1 − 2 ! (2.18) (3) (3) (3) (3) (3) (3) M + M + M M (2 + ω2)M +2ω2M M (3) = j,2 j,1 j,0 2n j,2 − j,1 j,0 ωn j,n (2 ω )(2 ω ) − (2 ω )(ω ω ) 1 − 1 − 2 ! − 1 1 − 2 ! (3) (3) (3) M (2 + ω1)M +2ω1M + j,2 − j,1 j,0 ωn. (2 ω )(ω ω ) 2 − 2 1 − 2 ! Proof. (2.17): The solution of Eq. (2.15) is (3) n n n (2.19) MJ,n = c12 + c2ω1 + c3ω2 . (3) (3) (3) 2 Then, let MJ,0 = c1 + c2 + c3, MJ,1 =2c1 + c2ω1 + c3ω2 and MJ,2 =4c1 + c2ω1 + c ω2. Therefore, we have (2 ω )(2 ω )c = M (3) (ω + ω )M (3) + ω ω M (3), 3 2 − 1 − 2 1 J,2 − 1 2 J,1 1 2 J,0 (2 ω )(ω ω )c = M (3) (2 + ω )M (3) +2ω M (3), (2 ω )(ω ω )c = − 1 1 − 2 2 J,2 − 2 J,1 2 J,0 − 2 1 − 2 3 M (3) (2 + ω )M (3) +2ω M (3). Using c , c and c in Eq. (2.19), we obtain J,2 − 1 J,1 1 J,0 1 2 3 (3) (3) (3) (3) (3) (3) M + M + M M (2 + ω2)M +2ω2M M (3) = J,2 J,1 J,0 2n J,2 − J,1 J,0 ωn J,n (2 ω )(2 ω ) − (2 ω )(ω ω ) 1 − 1 − 2 ! − 1 1 − 2 ! (3) (3) (3) M (2 + ω1)M +2ω1M + J,2 − J,1 J,0 ωn. (2 ω )(ω ω ) 2 − 2 1 − 2 ! (2.18): The proof is similar to the proof of (2.17). The following theorem gives us the n-th general term of the sequence given in (2.15) and (2.16). Theorem 2.3. For n 3, we have ≥ (3) (3) (3) (3) Jn+1 Jn +2Jn−1 2Jn M (3) (3) (3) (3) (3) (2.20) J,n = Jn Jn−1 +2Jn−2 2Jn−1 J (3) J (3) +2J (3) 2J (3) n−1 n−2 n−3 n−2 4 G. CERDA-MORALES (3) (3) (3) (3) jn+1 jn +2jn−1 2jn M (3) (3) (3) (3) (3) (2.21) j,n = jn jn−1 +2jn−2 2jn−1 j(3) j(3) +2j(3) 2j(3) n−1 n−2 n−3 n−2 Proof. (2.20): Let use the principle of mathematical induction on n. Let us consider n = 0 in (1.2). We have J (3) = 0, J (3) = 1 and J (3) = 1 . Then we write −1 −2 2 −3 − 4 (3) (3) (3) (3) J1 J0 +2J−1 2J0 1 0 0 (3) M = J (3) J (3) +2J (3) 2J (3) = 0 1 0 . J,0 0 −1 −2 −1 J (3) J (3) +2J (3) 2J (3) 0 0 1 −1 −2 −3 −2 By iterating this procedure and considering induction steps, let us assume that the equality in (2.20) holds for all n = k N. To finish the proof, we have to show that (2.20) also holds for n = k + 1 by considering∈ (1.2) and (2.15). Therefore we get (3) (3) (3) (3) MJ,k+2 = MJ,k+1 + MJ,k +2MJ,k−1 (3) (3) (3) (3) (3) (3) (3) (3) Jk+2 + Jk+1 +2Jk Jk+2 +2Jk+1 2Jk+1 +2Jk +4Jk−1 = (3) (3) (3) (3) (3) (3) (3) (3) Jk+1 + Jk +2Jk−1 Jk+1 +2Jk 2Jk +2Jk−1 +4Jk−2 J (3) + J (3) +2J (3) J (3) +2J (3) 2J (3) +2J (3) +4J (3) k k−1 k−2 k k−1 k−1 k−2 k−3 (3) (3) (3) (3) Jk+3 Jk+2 +2Jk+1 2Jk+2 = (3) (3) (3) (3) .