A Combinatorial Proof of a Result on Generalized Lucas Polynomials

Abstract We give a combinatorial proof of an elementary property of generalized Lucas polynomials, inspired by [1]. These polynomials in s and t are defined by the recurrence relation 〈n〉 = s〈n-1〉+t〈n-2〉 for n ≥ 2. The initial values are 〈0〉 = 2; 〈1〉= s, respectively.


Introduction and Motivation
In this paper, we shall focus on giving a combinatorial proof of a result on the generalized Lucas polynomials. But first we give some introductory remarks and motivation. The famous Fibonacci numbers, F n are defined by F 0 " 0, F 1 " 1 and, for n ≥ 2, The Lucas numbers L n are defined by the same recurrence, with the initial conditions L 0 " 2 and L 1 " 1.
One generalization of these numbers which has received much attention is the sequence of Fibonacci polynomials F n pxq " xF n´1 pxq`F n´2 pxq, n ≥ 2, with initial conditions F 0 pxq " 0, F 1 pxq " 1. The generalized Fibonacci polynomials depend on two variables s, t and are defined by t0u s,t " 0, t1u s,t " 1 and, for n ≥ 2, tnu s,t " stn´1u s,t`t tn´2u s,t .
Here and with other quantities depending on s and t, we will drop the subscripts as they will be clear from context. For example, we have t2u " s, t3u " s 2`t , t4u " s 3`2 st, t5u " s 4`3 s 2 t`t 2 .
For some historical remarks and relations of these polynomials we refer the reader to [1], [2] and [3].
When s " t " 1 these reduce to the ordinary Lucas numbers. In addition to the algebraic approach to our polynomials, there is a combinatorial interpretation derived from the standard interpretation of F n via tiling, given in [3]. A linear tiling, T , of a row of squares is a covering of the squares with dominos (which cover two squares) and monominos (which cover one square). We let, L n " tT : T a linear tiling of a row of n squaresu.

Combinatorial Interpretations of tnu and xny
The three tilings in the first row of Figure 1 are the elements of L 3 . We will also consider circular tilings where the (deformed) squares are arranged in a circle. We will use the notation C n for the set of circular tilings of n squares. So the set of tilings in the bottom row of Figure 1 is C 3 . For any type of nonempty tiling, T , we define its weight to be We give the empty tiling of zero boxes the weight wt " 1, if it is being considered as a linear tiling or wt " 2, if it is being considered as a circular tiling. The following proposition is immediate from the definitions of weight and of our generalized polynomials.
Proposition 2.1 (Sagan and Savage, [3]). For n ≥ 0, we have From the above discussions on the combinatorial interpretations of tnu and xny we get the following.
For some more interesting combinatorial interpretations, we refer the reader to [1] and [3].

Main Result
The main aim of this paper is to give a combinatorial proof of the following result, inspired by [1].
Proof. We consider an infinite row of squares which extends to both directions. A random square is marked as the 0 th place. The squares are numbered from left to right of 0 by the positive integers and from the right to left of 0 by the negative integers.
We now suppose that each square can be coloured with one of s shades of white and t shades of black. Let Z be the random variable which returns the box number at the end of the first odd-length block of boxes of the same black shade starting from the right of 0, and let Z 1 be the random variable which returns the box number at the end of the first odd-length block of boxes of the same black shade from the left of 0. And let W be the event be the combination of both Z and Z 1 .
For any integer n, the event W " n is the combination of Z " n and Z 1 "´n. Here Z " n is equivalent to having box n painted with one of the shades of black among the first n squares being of even length including 0 to the right of 0. So there are t choices for the colour of box n and s`t´1 choices for the colour of box n`1. Similarly Z 1 "´n is equivalent to having box´n painted with one of the shades of black among the first n squares being of even length including 0 to the left of 0. So, there are t choices for the colour of box´n and s`t´1 choices for the colour of box´pn`1q.
Each colouring of the first n squares gives a tiling where each white box is replaced by a monomino and a block of 2k boxes of the same shade of black is replaced by k dominoes. Also, the weight of the tiling is just the number of colourings attached to it. Thus, the number of the colourings of the first n boxes is xny since the n boxes both to the right and left of 0 will give rise to a circular tiling in this case. Indeed, the number of the colourings of the first n boxes to the right of 0, including the box 0 is tn`1u. Moreover, if the number of black shades boxes to the left of 0, not including the box 0, is even, then the number of the colourings of the first n´1 boxes to the left of 0, not including the box 0 is tnu. By convention, we fix the shade of the box 0 to be white. Since there are s possible white shades for the box 0, there are stnu colourings of the first n boxes to the left of 0, including the box 0. So, there are tn`1u´stnu " ttn´1u colourings of the first n´1 boxes to the left of 0, not including the box 0. This implies that the number of the colourings of the first n boxes is tn`1u`ttn´1u " xny.
Notice that if the shade of the box 0 is white, then the box 0 contributes by a factor s to the total number of circular tilings for which W " n whereas if the shade of the box 0 is black, then the box 0 contributes by a factor 2t to the total number of circular tilings since for each black shade, there are two possibilities (namely the two neighbours of box 0 in a circular tiling). It gives rise to a multiplicative factor s`2t in the expression of the total number of circular tilings for which W " n. Notice that once we count s`2t for the box 0, the other boxes (including the box n`1) contributes by a multiplicative factor s`t to the total number of circular tiling. Thus, the total number of circular tilings for which W " n is given by ps`2tqps`tq n`1 .
Hence we have, P pW " nq " tps`t´1qxny s,t ps`2tqps`tq n`1 . Summing these will give us the desired result.