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

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 polynomials and Jacobsthal polynomials are recently generalized by the authors to two variables in the form of hybrid Fibonacci polynomials which exhibit many interesting 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

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 polynomial 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. Rangarajan, Honnegowda C.K., and Shashikala P.

An interesting special case For x = y = 1 in the Theorem 2.3, we deduce the following: 1 3 l(H )(1, 1) = 2 T 1 + = 2 T 2n n 2 n 2 1 3 l(H ) (1, 1) = W 1 + = W 2n+1 n 2 n 2

3. Combinatorial Properties of Hybrid Lucas Polynomials In this section, the combinatorial properties such as generating function, matrix identities, and determinant formula are derived for the two variable hybrid Lucas polynomials. Theorem 3.1. The generating function for generalized hybrid polynomials in two variable is ∞ 2 − xt l(H )(x, y)tn = . n 1 − tx − yt2 n=0 (H ) Proof. Keeping in the mind the three term recurrence relation for ln (x, y), we proceed ∞ = (H ) n with the derivation. Put l(x,y,t) ln (x, y)t . n=0 We write = (H ) + (H ) +···+ (H ) n+1 +··· l(x,y,t) l0 (x, y) l1 (x, y)t ln+1(x, y)t − =−(H ) − (H ) 2 −···− (H ) n+1 −··· xtl(x,y,t) xl0 (x, y)t xl1 (x, y)t xln (x, y)t

− 2 =−(H ) 2 − (H ) 3 −···+ (H ) n+1 −··· yt l(x,y,t) l0 (x, y)yt l1 (x, y)yt ln−1(x, y)yt .

Summing all the three expressions on both sides, we get (1 − tx − yt2)l(x, y, t) = 2 + xt − tx(2) − yt2.2 + (x2 + 2y)t2 +··· 2 − xt l(x,y,t) = 1 − tx − yt2

Put y = 1, in the above Theorem 3.1, it becomes generating function for Lucas- Catalan polynomials in terms of x ∞ 2 − xt l(C)(x)tn = . n 1 − tx − t2 n=0 Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials 1677

Put x = 1, in the above Theorem 3.1, it becomes generating function for Lucas- Jacobsthal polynomials in terms of y

∞ 2 − t l(J )(y)tn = . n 1 − t − yt2 n=0

When x = 1 and y = 1 in the above Theorem 3.1, the generating function for hybrid polynomials becomes generating function of Lucas numbers

∞ 2 − t L tn = . n 1 − t − t2 n=0

Theorem 3.2. The generalized polynomials in two variable can be expressed in matrix form in odd and even functions are as follows: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ (H ) (H ) (H ) (H ) 2 + n l2n+4(x, y) l2n+2(x, y) l4 (x, y) l2 (x, y) x 2y 1 (1) ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ (H ) (H ) (H ) (H ) − 2 l + (x, y) l (x, y) l (x, y) l (x, y) y 0 ⎡ 2n 2 2n ⎤ 2 0 ⎡ ⎤ ⎡ ⎤n l(H ) (x, y) l(H ) (x, y) (H ) (H ) 2 + ⎢ 2n+5 2n+3 ⎥ l5 (x, y) l3 (x, y) x 2y 1 (2) ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ (H ) (H ) (H ) (H ) −y2 l2n+3(x, y) l2n+1(x, y) l3 (x, y) l1 (x, y) 0

Proof. (1) The theorem is proved by using the principle of mathematical induction: For n = 1 the result is true. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ H H H H 2 + l6 (x, y) l4 (x, y) l4 (x, y) l2 (x, y) x 2y 1 ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ H H H H − 2 l4 (x, y) l2 (x, y) l2 (x, y) l0 (x, y) y 0 ⎡ ⎤ H H l6 (x, y) l4 (x, y) = ⎣ ⎦ . H H l4 (x, y) l2 (x, y)

We assume the result is true for n = k. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤k l(H ) (x, y) l(H ) (x, y) (H ) (H ) 2 + ⎢ 2k+4 2k+2 ⎥ l4 (x, y) l2 (x, y) x 2y 1 ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ . (H ) (H ) (H ) (H ) −y2 l2k+2(x, y) l2k (x, y) l2 (x, y) l0 (x, y) 0

Now we prove the result is true for n = k + 1. 1678 R. Rangarajan, Honnegowda C.K., and Shashikala P.

Consider ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ l(H ) (x, y) l(H ) (x, y) 2 + (H ) (H ) ⎢ 2k+4 2k+2 ⎥ x 2y 1 l4 (x, y) l2 (x, y) ⎣ ⎦ . ⎣ ⎦ = ⎣ ⎦ (H ) (H ) −y2 (H ) (H ) l2k+2(x, y) l2k (x, y) 0 l2 (x, y) l0 (x, y) ⎡ ⎤k ⎡ ⎤ x2 + 2y 1 x2 + 2y 1 × ⎣ ⎦ . ⎣ ⎦ . −y2 0 −y2 0

On simpliﬁcation by using the three term recurrence relation for hybrid polynomials we get ⎡ ⎤ ⎡ ⎤ (H ) (H ) (H ) (H ) l (x, y) l (x, y) k+1 ⎢ 2k+6 2k+4 ⎥ l4 (x, y) l2 (x, y) 2 + = ⎣ ⎦ x 2y 1 ⎣ ⎦ . 2 (H ) (H ) (H ) (H ) −y 0 l2k+4(x, y) l2k+2(x, y) l2 (x, y) l0 (x, y) The proof of 2 is similar to that of 1. For y = 1, in the Theorem 3.2, we obtain matrix identities for Lucas-Catalan polynomials in the following form: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ (C) (C) (C) (C) 2 + n l2n+4(x) l2n+2(x) l4 (x) l2 (x) x 21 (3) ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ (C) (C) (C) (C) −10 l2n+2(x) l2n (x) l2 (x) l0 (x) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ l(C) (x) l(C) (x) (C) (C) 2 + n ⎢ 2n+5 2n+3 ⎥ l5 (x) l3 (x) x 21 (4) ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ (C) (C) (C) (C) −10 l2n+3(x) l2n+1(x) l3 (x) l1 (x)

For x = 1 in the Theorem 3.2, we obtain matrix identities for Lucas-Jacobsthal polynomials in the following form: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ (J ) (J ) (J ) (J ) + n l2n+4(y) l2n+2(y) l4 (y) l2 (y) 1 2y 1 (5) ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ (J ) (J ) (J ) (J ) −10 l2n+2(y) l2n (y) l2 (y) l0 (y) ⎡ ⎤ ⎡ ⎤ l(J ) (y) l(J ) (y) (J ) (J ) ⎢ 2n+5 2n+3 ⎥ l5 (y) l3 (y) + n = ⎣ ⎦ 1 2y 1 (6) ⎣ ⎦ . − (J ) (J ) (J ) (J ) 10 l2n+3(y) l2n+1(y) l3 (y) l1 (y) Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials 1679

For x = 1 and y = 1 in the Theorem 3.2, we obtain Lucas matrix identities as follows: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ n L2n+4 L2n+2 L4 L2 31 (7) ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ L2n+2 L2n L2 L0 −10 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ n L2n+5 L2n+3 L5 L3 31 (8) ⎣ ⎦ = ⎣ ⎦ . ⎣ ⎦ L2n+3 L2n+1 L3 L1 −10

Determinants Formulas A direct application of the following three term recurrence relation

[ (H ) − n]= 2 + [ (H ) − n−1]− [ (H ) − n−2] (i) l2n (x, y) 2y (x 2y) l2n−2(x, y) 2y y l2n−4(x, y) 2y

(H ) =[(H ) − n+1]− [ (H ) − n] (ii) x l2n+1(x, y) l2n+2(x, y) 2y y l2n (x, y) 2y will yield the two determinant formulas stated in the following theorem with out proof. Theorem 3.3. The determinants formulas for generalized hybrid polynomials are x2 + 2y −y 0 ··· 0 −y −yx2 + 2y −y 0 ··· 0 H n . l (x, y) − 2y = 0 −yx2 + 2y .. −y 0 2n ...... ...... −y 000−yx2 + 2y nXn x2 + 2y −y 0 ··· 0 −y −yx2 + 2y −y 0 ··· 0 (H ) 2 .. xl + (x, y) = 0 −yx+ 2y . −y 0 2n 1 ...... ...... −y 000−yx2 + 2y (n+1)X(n+1) x2 + 2y −y 0 ··· 0 −y −yx2 + 2y −y 0 ··· 0 . − y 0 −yx2 + 2y .. −y 0 ...... ...... −y 000−yx2 + 2y nXn for n = 3, 4, 5, ··· 1680 R. Rangarajan, Honnegowda C.K., and Shashikala P.

Special cases (1) For y = 1 and n = 3, 4, 5, ··· we deduce x2 + 2 −10··· 0 −1 −1 x2 + 2 −10··· 0 (C) . l (x) − 2 = 0 −1 x2 + 2 .. −10 2n ...... ...... −10 00−1 x2 + 2 nXn x2 + 2 −10··· 0 −1 −1 x2 + 2 −10··· 0 − 1 (C) 2 .. xl + (x) = 0 −yx+ 2y . −y 0 2n 1 ...... ...... −10 00−1 x2 + 2 (n+1)X(n+1) x2 + 2 −10··· 0 −1 −1 x2 + 2 −10··· 0 . − 0 −1 x2 + 2 .. −10 ...... ...... −10 00−1 x2 + 2 nXn (2) For x = 1 and n = 3, 4, 5, ··· we deduce 1 + 2y −y 0 ··· 0 −y −y 1 + 2y −y 0 ··· 0 . (J ) − n = − + .. − l2n (x, y) 2y 0 y 1 2y y 0 ...... ...... − − + y 000y 1 2y nXn 1 + 2y −y 0 ··· 0 −y −y 1 + 2y −y 0 ··· 0 (C) = − + ... − l2n+1(y) 0 y 1 2y y 0 ...... ...... − − + y 000y 1 2y (n+1)X(n+1) 1 + 2y −y 0 ··· 0 −y −y 1 + 2y −y 0 ··· 0 − − + ... − y 0 y 1 2y y 0 ...... ...... − − + y 000y 1 2y nXn Lucas Polynomials Corresponding to Hybrid Fibonacci Polynomials 1681

(3) For x = 1,y= 1 and n = 3, 4, 5, ··· we deduce 3 −10··· 0 −1 −13−10··· 0 − = − ... − L2n 2 0 13 10 ...... ...... − − 10 0 0 13nXn

3 −10··· 0 −1 −13−10··· 0 = − − ... − L2n+1 1 13 10 ...... ...... − − 10 0 0 13(n+1)X(n+1) 3 −10··· 0 −1 −13−10··· 0 . − − .. − 0 13 10 ...... ...... − − 10 0 0 13nXn

Acknowledgement The second author would like to thank both UGC-SWRO, F.No. FIP/12th Plan/KADA018 TF02 and Govt. of Karnataka (DCE) and the third author would like to thank UGC, Govt. of India for encouraging this work under Post Doctoral Fellowship For SC/ST Candidates Order No. F./PDFSS - 2014 -15-ST-KAR-10116.

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