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A Two Variable Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials

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A Two Variable Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials Global Journal of Pure and Applied Mathematics. ISSN 0973-1768 Volume 13, Number 6 (2017), pp. 1669–1682 © Research India Publications http://www.ripublication.com/gjpam.htm A Two Variable Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials R. Rangarajan Department of Studies in Mathematics, University of Mysore, Manasagangotri, Mysuru - 570 006, INDIA. Honnegowda C.K.1 Department of Studies in Mathematics, University of Mysore, Manasagangotri, Mysuru - 570 006, INDIA. Shashikala P. Department of Studies in Mathematics, University of Mysore, Manasagangotri, Mysuru - 570 006, INDIA. Abstract Fibonacci numbers, Fibonacci polynomials in one variable, namely Catalan poly- nomials and Jacobsthal polynomials are recently generalized by the authors to two variables in the form of hybrid Fibonacci polynomials which exhibit many interest- ing combinatorial properties useful for research workers in combinatorics. In the present paper, two variable Lucas polynomials corresponding to two variable hybrid Fibonacci polynomials are defined and studied their combinatorial properties. AMS subject classification: 05A19, 30B70 and 33C70. Keywords: Combinatorial identities, Fibonacci and Lucas numbers and hyper ge- ometric functions in one or more variables. 1Corresponding author. 1670 R. Rangarajan, Honnegowda C.K., and Shashikala P. 1. Introduction (C) (J ) Two Lucas polynomials denoted ln (x) and ln (x) are defined in the literature corre- (C) sponding to Catalan polynomials and Jacobsthal polynomials denoted by fn (x) and (J ) fn (x) respectively which are natural extension of Fibonacci numbers Fn to one variable polynomials with many interesting combinatorial properties relevant to current literature [5, 7, 8, 9, 11, 12, 15]. The natural and the most beautiful three term recurrence relation is Fn+1 = Fn + Fn−1,F1 = 1,F2 = 1,n= 1, 2, 3,...[1, 3, 4, 16] given by Fibonacci numbers. Closely connected to this is one more such relation Ln+1 = Ln + Ln−1, L1 = 1,L2 = 3,n= 1, 2, 3,... given by Lucas numbers. Motivated by the rela- (C) tions, the sequence ln (x) is defined by the following three term recurrence relation (C) = (C) + (C) (C) = (C) = = [5]: ln+1(x) xln (x) ln−1(x), with l1 (x) 1,l2 (x) x, n 1, 2, 3,... (J ) Also the sequence ln (x) is defined by the following three term recurrence relation [5]: (J ) = (J ) + (J ) (J ) = (J ) = + = ln+1(x) ln (x) xln−1(x), with l1 (x) 1,l2 (x) 1 2x, n 1, 2, 3,... (H ) The main aim of the paper is to define a two variable Lucas polynomial ln (x, y) cor- (H ) responding to two variable hybrid Fibonacci polynomials [10] fn (x, y) which contain (C) (J ) naturally both ln (x) and ln (x). The properties of both types of Lucas polynomials in one variable will be extended to hybrid polynomial in two variables. In the present paper, the two variable hybrid Lucas polynomials are shown to be directly connected to Tchebychev polynomials [6, 7, 8, 9, 11, 12, 14, 15] and the direct formula using Pascal like table is derived. The combinatorial properties such as generating function, matrix identities and determinant formula are derived. 2. Generalization of Lucas polynomials to two variables Definition 2.1. The generalized hybrid polynomials in two variables x and y of degree (H ) n is denoted by ln (x, y) is x + x2 + 4y n x − x2 + 4y n l(H )(x, y) = + . (2.1) n 2 2 It is a two variable polynomial of degree n in x and y. When y = 1in(2.1), we get back Lucas-catalan polynomials in terms of x (H ) = (H ) = (C) ln (x, y) ln (x, 1) ln (x). When x = 1in(2.1), the hybrid polynomials becomes Lucas-Jacobsthal polynomials in terms of y (H ) = (H ) = (J ) ln (x, y) ln (1,y) ln (y). When x = 1 and y = 1in(2.1), the hybrid polynomials becomes Lucas numbers (H ) = ln (1, 1) Ln. Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials 1671 Three Term Recurrence Relations The hybrid polynomials satisfy the following three term recurrence relations: (H ) = (H ) + (H ) ln+1(x, y) xln (x, y) yln−1(x, y). (2.2) (H ) = (H ) = = l0 (x, y) 2,l1 (x, y) x, n 1, 2, 3,... The hybrid polynomials can also satisfies the following three term recurrence relation connected to hybrid Fibonacci polynomials [10] as follows: (H ) = (H ) + (H ) ln (x, y) xfn (x, y) 2yfn−1(x, y). (2.3) When y = 1in(2.2), the three term recurrence relation computes the Lucas-catalan polynomial in terms of x (C) = (C) + (C) ln+1(x) xln (x) ln−1(x), (C) = C = = l0 (x) 2,l1 (x) x, n 1, 2, 3,... When x = 1in(2.2), the three term recurrence relation computes the Lucas-Jacobsthal polynomial in terms of y (J ) = (J ) + (J ) ln+1(y) ln (y) yln−1(y), (J ) = (J ) = = l0 (y) 2,l1 (y) 1,n 1, 2, 3,... When x = 1 and y = 1in(2.2), the hybrid polynomials becomes the formula for Lucas numbers Ln+1 = Ln + Ln−1,L0 = 2,L1 = 1,n= 1, 2, 3,... 1672 R. Rangarajan, Honnegowda C.K., and Shashikala P. Initial Polynomials The first eight initial polynomials of hybrid Lucas polynomials in two variable, Lucas- Catalan in terms of x and Lucas-Jacobsthal polynomial in terms of y is (H ) (C) (J ) nln (x, y) ln (x, 1)ln (1,y) 02 2 2 1 xx1 2 x2 + 2yx2 + 21+ 2y 3 x3 + 3xy x3 + 3x 1 + 3y 4 x4 + 4x2y + 2y2 x4 + 4x2 + 21+ 4y + 2y2 5 x5 + 5x3y + 5xy2 x5 + 5x3 + 5x 1 + 5y + 5y2 6 x6 + 6x4y + 9x2y2 + 2y3 x6 + 6x4 + 9x2 + 21+ 6y + 9y2 + 2y3 7 x7 + 7x5y + 14x3y2 + 7xy3 x7 + 7x5 + 14x3 + 7x 1 + 7y + 14y2 + 7y3. When x = 1 and y = 1, the hybrid polynomials becomes the sequence of Lucas numbers Ln = 2, 1, 3, 4, 7, 11, 18, 29,... Theorem 2.2. The explicit formula for hybrid polynomial in two variable is n 2 n n − k − l(H )(x, y) = xn 2kyk. (2.4) n n − k k k=0 Proof. The result is proved by using mathematical induction method. For n = 1, 1 1 − 0 l(H )(x, y) = x1−0y0 = x. 1 1 − 0 0 For n = 2, 2 2 − 0 − 2 2 − 1 − l(H )(x, y) = x2 0y0 + x2 2y1 = x2 + 2y. 2 2 − 0 0 2 − 1 1 We assume the result is true for n = m. i.e., m 2 m m − k l(H )(x, y) = xm−2kyk. m m − k k k=0 Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials 1673 We prove the result is true for n = m + 1. By applying the three term recurrence relation (2.2), for the hybrid polynomials (H ) = (H ) + (H ) lm+1(x, y) xlm (x, y) ylm−1(x, y) m 2 m m − 1 − k =x xm−2kyk m − k k k=0 m−1 2 m − k m − 1 − k + y xm−1−2kyk m − 1 − k k k=0 m m − 0 m m − 1 =x xm−0y0 + xm−2y1 m − 0 0 m − 1 1 m m − 2 m − 1 m − 1 + xm−4y2 + y xm−1y0 m − 2 2 m − 1 0 m − 1 m − 2 m − 1 m − 3 + xm−3y1 + xm−5y2 m − m − 2 1 3 2 m + 1 m − 0 m + 1 m − 1 = xmy0 + xm−2y − − m 0 0 m 1 1 m + 1 m − 2 + xm−4y2 +··· m − 2 2 m+1 2 m + 1 m + 1 − k = xm+1−2kyk m + 1 − k k k=0 When y = 1 in (2.4), it becomes the explicit formula for the Lucas-Catalan polyno- mial in terms of x n 2 n n − k l(H )(x, y) = l(C)(x) = xn−2k. n n n − k k k=0 When x = 1 in (2.4), it becomes the explicit formula for the Lucas-Jacobsthal polynomial in terms of y n 2 n n − k l(H )(x, y) = l(J )(y) = yk. n n n − k k k=0 When x = 1 and y = 1 in (2.4), the hybrid polynomials becomes the explicit formula for Lucas numbers n 2 n n − k L = . n n − k k k=0 1674 R. Rangarajan, Honnegowda C.K., and Shashikala P. Theorem 2.3. The hybrid polynomial can be connected to Tchebychev polynomials of first kind Tn(x) and fourth kind Wn(x) as follows: x2 1. lH (x, y) = 2yn T 1 + 2n n 2y 2 H n x 2. l + (x, y) = xy W 1 + 2n 1 n 2y Proof. (1) Consider x + x2 + 4y 2n x − x2 + 4y 2n l(H )(x, y) = − 2n 2 2 ⎡⎛ ⎞ n x2 x2 2 = ⎣⎝ 1 + y + y 1 + − 1⎠ 2y 2y ⎤ x2 x2 2 n − 1 + y − y 1 + − 1 ⎦ 2y 2y ⎡ 1 x2 x2 2 n = yn2. ⎣ 1 + + 1 + − 1 2 2y 2y ⎤ x2 x2 2 n − 1 + − 1 + − 1 ⎦ 2y 2y x2 = 2ynT 1 + n 2y (2) Consider + + + 2 + 2n 1 − 2 + 2n 1 H x x 4y x x 4y l + (x, y) = + 2n 1 2 2 + 2 + 2n + 2 + 2n = x x x 4y + x x 4y 2 2 2 2 + + 2 + 2n + 2 + 2n + x 4y x x 4y − x x 4y 2 2 2 Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials 1675 ⎡ x.yn x2 x2 2 n = ⎣ 1 + + 1 + − 1 2 2y 2y ⎤ x2 x2 2 n + 1 + − 1 + − 1 ⎦ 2y 2y ⎡ ⎤ x2 + 4y x2 x2 2 n + yn ⎣ 1 + + 1 + − 1 ⎦ 2 2y 2y ⎡ ⎤ x2 + 4y x2 x2 2 n − yn ⎣ 1 + − 1 + − 1 ⎦ 2 2y 2y 2 2 2 n x x x = xy 2 1 + U − 1 + − U − 1 + 2y n 1 2y n 2 2y 2 n x +xy U − 1 + n 1 2y 2 2 n x x = xy U 1 + + U − 1 + n 2y n 1 2y x2 = xynW 1 + n 2y When y = 1 in the Theorem 2.3, we deduce the following: x2 l(H )(x, y) = l(C)(x) = 2T 1 + 2n 2n n 2 x2 l(H ) (x, y) = l(c) (x) = xW 1 + 2n+1 2n+1 n 2 When x = 1 in the Theorem 2.3, we deduce the following: 1 l(H )(x, y) = l(J )(y) = 2ynT 1 + 2n 2n n 2y 1 l(H ) (x, y) = l(J ) (y) = ynW 1 + 2n+1 2n+1 n 2y 1676 R.
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