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Certain Combinatorial Results on Two Variable Hybrid Fibonacci Polynomials

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Certain Combinatorial Results on Two Variable Hybrid Fibonacci Polynomials International Journal of Computational and Applied Mathematics. ISSN 1819-4966 Volume 12, Number 2 (2017), pp. 603–613 © Research India Publications http://www.ripublication.com/ijcam.htm Certain Combinatorial Results on Two variable Hybrid Fibonacci Polynomials R. Rangarajan, Honnegowda C.K.1, Shashikala P. Department of Studies in Mathematics, University of Mysore, Manasagangotri, Mysuru - 570 006, INDIA. Abstract Recently many researchers in combinatorics have contributed to the development of one variable polynomials related to hypergeometric family. Here a class of two variable hybrid Fibonacci polynomials is defined and proved certain combinatorial results. AMS subject classification: 05A19, 11B39 and 33C45. Keywords: Combinatorial Identities, Fibonacci Numbers and Hyper Geometric Functions in one or more Variables. 1. Introduction The natural and the most beautiful three term recurrence relation is Fn+1 = Fn + Fn−1, F1 = 1,F2 = 1, where Fn is the nth Fibonacci number [1, 2, 3, 13]. There are many one variable generalizations of Fn related to hypergeometric family [3, 4]. Two important such generalizations are Catalan polynomials and Jacobsthal polynomials denoted by (C) (J ) fn (x) and fn (x) which exhibits many interesting combinatorial properties [4, 6, 7, 8, 9, 10, 12]. 1Coressoponding author. 604 R. Rangarajan, Honnegowda C.K., and Shashikala P. Two hypergeometric polynomials, namely, Un(x) and Vn(x) called Tchebyshev poly- nomials of second and third kind are given by n+1 n+1 1 2 2 Un(x) = √ x + x − 1 − x − x − 1 2 x2 − 1 3 1 − x = (n + 1) F −n, n + 2; ; 2 1 2 2 and = − Vn(x) Un+1(x) Un(x) 1 1 − x = F −n, n + 1; ; 2 1 2 2 which have many applications in both pure and applied mathematics [5, 6, 7, 8, 9, 10, 3 3 11, 12]. It is interesting to note that U = F + and V = F + [12]. n 2 2n 2 n 2 2n 1 In the present paper, a two variable hybrid Fibonacci polynomials are defined which (C) naturally generalizes Fibonacci-Catalan polynomials fn (x) [3] and Fibonacci-Jacobsthal (J ) polynomials fn (y) [3] by a nontrivial hybridization. Further the defined hybrid Fi- bonacci polynomials are shown to be directly connected to Tchebyshev polynomials of second and third kinds. Certain combinatorial results such as generating function, matrix identities, determinant formula and the direct formula using Pascal like table are proved for the hybrid Fibonacci polynomials. 2. Two Variable Hybrid Fibonacci Polynomials Definition 2.1. The generalized hybrid Fibonacci polynomials in two variables x and y (H ) of degree n, denoted by fn (x, y) is + 2 + n − 2 + n (H ) = 1 x x 4y − x x 4y fn (x, y) (2.1) x2 + 4y 2 2 When y = 1 in (2.1), the hybrid Fibonacci polynomials become Fibonacci-Catalan (H ) = (C) = polynomials in terms of x, that is fn (x, 1) fn (x). When x 1 in (2.1), they (H ) = (J ) become Fibonacci-Jacobsthal polynomials in terms of y, that is fn (1,y) fn (y). When x = 1 and y = 1 in (2.1), they become Fibonacci numbers. Three Term Recurrence Relations By direct verification using the definition, one can show that the following recurrence relation is satisfied by the generalized hybrid fibonacci polynomials in two variables: (H ) = (H ) + (H ) fn+1(x, y) xfn (x, y) yfn−1(x, y), (2.2) (H ) = (H ) = (H ) = 2 + = f1 (x, y) 1,f2 (x, y) x,f3 (x, y) x y, n 1, 2, 3,... Combinatorial Results on Two variable Hybrid Fibonacci Polynomials 605 Illustration The following table with first five initial polynomials of hybrid Fibonacci polynomials in two variables, Fibonacci-Catalan polynomial and Fibonacci-Jacobsthal polynomials illustrates the nontrivial hybridization. (H ) (C) (J ) nfn (x, y) fn (x, 1)fn (1,y) 11 1 1 2 xx1 3 x2 + yx2 + 11+ y 4 x3 + 2xy x3 + 2x 1 + 2y 5 x4 + 3x2y + y2 x4 + 3x2 + 11+ 3y + y2 Theorem 2.2. The hybrid polynomial can be connected to Tchebychev polynomials of second kind Un(x) and third kind Vn(x) as follows: x2 1. f (H ) (x, y) = xynU 1 + 2n+2 n 2y x2 2. f (H ) (x, y) = ynV 1 + . 2n+1 n 2y Proof. (1) Consider + 2 + 2n+2 − 2 + 2n+2 (H ) = 1 x x 4y − x x 4y f2n+2(x, y) . x2 + 4y 2 2 ⎡⎡ ⎤ n+1 2 2 x ⎢⎣ x x2 ⎦ = ⎣ 1 + y + y 1 + − 1 2 2y 2y + x2 − 2y 1 2y 1 ⎤ x2 x2 2 n+1 − 1 + y − y 1 + − 1 ⎦ . 2y 2y 606 R. Rangarajan, Honnegowda C.K., and Shashikala P. ⎡ n+1 2 + x.y ⎣ x x2 2 n 1 = 1 + + 1 + − 1 2 2y 2y + x2 − 2y 1 2y 1 ⎤ x2 x2 2 n+1 − 1 + − 1 + − 1 ⎦ 2y 2y x2 = xynU 1 + . n 2y (2) Consider + 2 + 2n+1 + 2 + 2n+1 (H ) = 1 x x 4y − x x 4y f2n+1(x, y) x2 + 4y 2 2 + 2 + 2n + 2 + 2n = 1 x x 4y − x x 4y 2 x2 + 4y 2 2 + 2 + 2n + 2 + 2n + 1 x x 4y − x x 4y 2 2 2 ⎡ 2 n 2 x .y ⎣ x x2 2 n = 1 + + 1 + − 1 2 2y 2y + x2 − 4y 1 2y 1 ⎤ x2 x2 2 n − 1 + − 1 + − 1 ⎦ 2y 2y ⎡ yn x2 x2 2 n + ⎣ 1 + + 1 + − 1 2 2y 2y ⎤ x2 x2 2 n − 1 + − 1 + − 1 ⎦ 2y 2y 2 2 2 x n x n x = y U − 1 + + y T 1 + . 2y n 1 2y n 2y But Tn(x) satisfies x2 x2 x2 x2 T 1 + = 1 + U − 1 + − U − 1 + . n 2y 2y n 1 2y n 2 2y Combinatorial Results on Two variable Hybrid Fibonacci Polynomials 607 2 2 2 2 (H ) n x x x x f (x, y) = y 2 1 + Un− 1 + − Un− 1 + − Un− 1 + 2n+1 2y 1 2y 2 2y 1 2y 2 2 n x x = y Un 1 + − Un− 1 + 2y 1 2y 2 n x = y Vn 1 + . 2y 3. Combinatorial Properties of Hybrid Polynomials In this section, the combinatorial properties such as generating function, matrix identities and determinant formula and are direct formula using Pascal like table for the two variable hybrid Fibonacci polynomials are stated with proof. Theorem 3.1. The generating function for generalized hybrid polynomials in two vari- able is ∞ t f (H )(x, y)tn = n 1 − tx − yt2 n=0 (H ) Proof. Keeping in the mind the three term recurrence relation for fn (x, y). We proceed ∞ = (H ) n with the derivation. Put f(x,y,t) fn (x, y)t . We write n=0 = (H ) + (H ) +···+ (H ) n+1 +··· f(x,y,t) f0 (x, y) f1 (x, y)t fn+1(x, y)t − =− H − H 2 −···− H n+1 −··· xtf(x,y,t) xf0 (x, y)t xf1 (x, y)t xfn (x, y)t − 2 =−(H ) 2 − (H ) 3 −···+ (H ) n+1 −··· yt f(x,y,t) f0 (x, y)yt f1 (x, y)yt fn−1(x, y)yt . Summing all the three expressions on both sides, we get (1 − tx + yt2)f (x, y, t) = 0 + 1t − tx.0 t f(x,y,t) = . 1 − tx − yt2 Theorem 3.2. The generalized polynomials in two variable can be expressed in matrix 608 R. Rangarajan, Honnegowda C.K., and Shashikala P. form in odd and even functions are as follows: ⎡ ⎤ ⎡ ⎤ (H ) (H ) 2 + n f2n+2(x, y) yf2n (x, y) x 2yy (1) ⎣ ⎦ = x ⎣ ⎦ . − (H ) − 2 (H ) −y 0 yf2n (x, y) y f2n−2(x, y) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ (H ) (H ) 2 + 2 + n f2n+3(x, y) yf2n+1(x, y) x yy x 2yy (2) ⎣ ⎦ = ⎣ ⎦ ⎣ ⎦ . (H ) (H ) 11 −y 0 f2n+1(x, y) yf2n−1(x, y) Proof. (1) The theorem is proved by using the principle of mathematical induction: For n = 1, the result is obvious. For n = 2, ⎡ ⎤ ⎡ ⎤ (H ) (H ) 2 + 2 f6 (x, y) yf4 (x, y) x 2yy ⎣ ⎦ = x ⎣ ⎦ . − (H ) − 2 (H ) −y 0 yf4 (x, y) y f2 (x, y) ⎡ ⎤ ⎡ ⎤ (H ) (H ) 5 + 3 + 2 3 + 2 f6 (x, y) yf4 (x, y) x 4x y 3xy x y 2xy ⎣ ⎦ = ⎣ ⎦ − (H ) − 2 (H ) − 3 + 2 − 2 yf4 (x, y) y f2 (x, y) (x y 2xy ) xy is true. We assume the result is true for n = k, ⎡ ⎤ ⎡ ⎤ f (H ) (x, y) yf (H )(x, y) 2 + k ⎢ 2k+2 2k ⎥ x 2yy ⎣ ⎦ = x ⎣ ⎦ . − (H ) − 2 (H ) −y 0 yf2k (x, y) y f2k−2(x, y) Now we prove the result is true for n = k + 1. Consider ⎡ ⎤ ⎡ ⎤ f (H ) (x, y) yf (H )(x, y) 2 + ⎢ 2k+2 2k ⎥ x 2yy ⎣ ⎦ ⎣ ⎦ − (H ) − 2 (H ) −y 0 yf2k (x, y) y f2k−2(x, y) ⎡ ⎤k ⎡ ⎤ x2 + 2yy x2 + 2yy = x ⎣ ⎦ ⎣ ⎦ −y 0 −y 0 On simplification by using the three term recurrence relation for hybrid polyno- mials. We get ⎡ ⎤ ⎡ ⎤ + f (H ) (x, y) yf (H ) (x, y) 2 + k 1 ⎢ 2k+4 2k+2 ⎥ x 2yy ⎣ ⎦ = x ⎣ ⎦ . − (H ) − 2 (H ) −y 0 yf2k+2(x, y) y f2k (x, y) The proof of matrix identity 2 is similar to that of the identity 1. Combinatorial Results on Two variable Hybrid Fibonacci Polynomials 609 For y = 1, in the Theorem 3.1, we obtain matrix identities for Fibonacci–Catalan polynomials in the following form: ⎡ ⎤ ⎡ ⎤ (C) (C) 2 + n f2n+2(x) yf2n (x) x 21 (3) ⎣ ⎦ = x ⎣ ⎦ . − (C) (C) −10 f2n (x) f2n−2(x) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ (C) (C) 2 + 2 + n f2n+3(x) f2n+1(x) x 11 x 21 (4) ⎣ ⎦ = ⎣ ⎦ ⎣ ⎦ . (C) (C) 11 −10 f2n+1(x) yf2n−1(x) For x = 1 in the Theorem 3.1, we obtain matrix identities for Fibonacci-Jacobsthal polynomials in the following form: ⎡ ⎤ ⎡ ⎤ (J ) (J ) + n f2n+2(y) yf2n (y) 1 2yy (5) ⎣ ⎦ = ⎣ ⎦ .
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