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

Laugier, Alexandre

Saikia, Manjil P.

English

Online Research @ Cardiff

DEMONSTRATIO MATHEMATICAVol. 49 No 3 2016Alexandre Laugier, Manjil P. SaikiaA COMBINATORIAL PROOF OF A RESULT ONGENERALIZED LUCAS POLYNOMIALSCommunicated by Z. LoncAbstract. We give a combinatorial proof of an elementary property of generalizedLucas polynomials, inspired by [1]. These polynomials in s and t are defined by therecurrence relation xny “ sxn´1y`txn´2y for n ≥ 2. The initial values are x0y “ 2, x1y “ s,respectively.1. Introduction and motivationIn this paper, we shall focus on giving a combinatorial proof of a resulton the generalized Lucas polynomials. But first we give some introductoryremarks and motivation. The famous Fibonacci numbers Fn are defined byF0 “ 0, F1 “ 1 and, for n ≥ 2,Fn “ Fn´1 ` Fn´2.The Lucas numbers Ln are defined by the same recurrence, with the initialconditions L0 “ 2 and L1 “ 1.One generalization of these numbers which has received much attentionis the sequence of Fibonacci polynomialsFnpxq “ xFn´1pxq ` Fn´2pxq, n ≥ 2,with initial conditions F0pxq “ 0, F1pxq “ 1. The generalized Fibonaccipolynomials depend on two variables s, t and are defined by t0us,t “ 0,t1us,t “ 1 and, for n ≥ 2,tnus,t “ stn´ 1us,t ` ttn´ 2us,t.Here and with other quantities depending on s and t, we will drop the2010 Mathematics Subject Classification: Primary 05A10; Secondary 11B39, 11B65.Key words and phrases: binomial theorem, Fibonacci number, Fibonomial coefficient,Lucas number q-analogue, Generalized Lucas Polynomials.DOI: 10.1515/dema-2016-0022c© Copyright by Faculty of Mathematics and Information Science, Warsaw University of TechnologyBrought to you by | Cardiff UniversityAuthenticatedDownload Date | 3/9/20 11:34 AMOn generalized Lucas polynomials 267subscripts as they will be clear from context. For example, we havet2u “ s, t3u “ s2 ` t, t4u “ s3 ` 2st, t5u “ s4 ` 3s2t` t2.For some historical remarks and relations of these polynomials we referthe reader to [1], [2] and [3].The main focus of our paper are the generalized Lucas polynomials definedbyxnys,t “ sxn´ 1ys,t ` txn´ 2ys,t, n ≥ 2together with the initial conditions x0ys,t “ 2 and x1ys,t “ s. The first fewpolynomials arex2ys,t “ s2 ` 2t, x3ys,t “ s3 ` 3st,x4ys,t “ s4 ` 4s2t` 2t2, x5ys,t “ s5 ` 5s3t` 5st2.When s “ t “ 1 these reduce to the ordinary Lucas numbers.2. Combinatorial interpretations of tnu and xnyFig. 1. Linear and circular tilingsIn addition to the algebraic approach to our polynomials, there is acombinatorial interpretation derived from the standard interpretation of Fnvia tiling, given in [3]. A linear tiling, T , of a row of squares is a covering ofthe squares with dominos (which cover two squares) and monominos (whichcover one square). We let,Ln “ tT : T a linear tiling of a row of n squaresu.The three tilings in the first row of Figure 1 are the elements of L3. We willalso consider circular tilings where the (deformed) squares are arranged in acircle. We will use the notation Cn for the set of circular tilings of n squares.So the set of tilings in the bottom row of Figure 1 is C3. For any type ofnonempty tiling T , we define its weight to bewtT “ s# of monominos in T t# of dominos in T .Brought to you by | Cardiff UniversityAuthenticatedDownload Date | 3/9/20 11:34 AM268 A. Laugier, M. P. SaikiaWe give the empty tiling ε of zero boxes the weight wt ε “ 1, if it is beingconsidered as a linear tiling or wt ε “ 2, if it is being considered as a circulartiling. The following proposition is immediate from the definitions of weightand of our generalized polynomials.Proposition 2.1. (Sagan and Savage, [3]) For n ≥ 0, we havetn` 1u “ÿTPLnwtTandxny “ÿTPCnwtT.From the above discussions on the combinatorial interpretations of tnuand xny we get the following.Theorem 2.2. (Sagan and Savage, [3]) For m ≥ 1 and n ≥ 0 we havetm` nu “ tmutn` 1u ` ttm´ 1utnu.Proposition 2.3. (Sagan and Savage, [3]) For n ≥ 1 we havexny “ tn` 1u ` ttn´ 1u.And for m,n ≥ 0 we havetm` nu “xmytnu ` tmuxny2.For some more interesting combinatorial interpretations, we refer thereader to [1] and [3].3. Main resultThe main aim of this paper is to give a combinatorial proof of the followingresult, inspired by [1].Theorem 3.1. For s, t P N such that1s` tă min"1|X|,1|Y |*,we have for X,Y ‰ 08ÿn“0xnys,tps` tqn`1“s` 2ttps` t´ 1q.Proof. We consider an infinite row of squares which extends to both directions.A random square is marked as the 0th place. The squares are numbered fromleft to right of 0 by the positive integers and from the right to left of 0 bythe negative integers.Brought to you by | Cardiff UniversityAuthenticatedDownload Date | 3/9/20 11:34 AMOn generalized Lucas polynomials 269We now suppose that each square can be coloured with one of s shades ofwhite and t shades of black. Let Z be the random variable which returns thebox number at the end of the first odd-length block of boxes of the same blackshade starting from the right of 0, and let Z 1 be the random variable whichreturns the box number at the end of the first odd-length block of boxes ofthe same black shade from the left of 0. And let W be the combination ofboth Z and Z 1.For any integer n, the event W “ n is the combination of Z “ n andZ 1 “ ´n. Here Z “ n is equivalent to having box n painted with one of theshades of black among the first n squares being of even length including 0to the right of 0. So there are t choices for the colour of box n and s` t´ 1choices for the colour of box n ` 1. Similarly Z 1 “ ´n is equivalent tohaving box ´n painted with one of the shades of black among the first nsquares being of even length including 0 to the left of 0. So, there are tchoices 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 isreplaced by a monomino and a block of 2k boxes of the same shade of blackis replaced by k dominoes. Also, the weight of the tiling is just the numberof colourings attached to it. Thus, the number of the colourings of the firstn boxes is xny since the n boxes both to the right and left of 0 will give riseto a circular tiling in this case. Indeed, the number of the colourings of thefirst n boxes to the right of 0, including the box 0 is tn` 1u. Moreover, ifthe 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 theleft of 0, not including the box 0 is tnu. By convention, we fix the shade ofthe 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 thebox 0. So, there are tn` 1u ´ stnu “ ttn´ 1u colourings of the first n´ 1boxes to the left of 0, not including the box 0. This implies that the numberof 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 contributesby a factor s to the total number of circular tilings for which W “ n whereasif the shade of the box 0 is black, then the box 0 contributes by a factor 2t tothe total number of circular tilings since for each black shade, there are twopossibilities (namely the two neighbours of box 0 in a circular tiling). It givesrise to a multiplicative factor s` 2t in the expression of the total number ofcircular tilings for which W “ n. Notice that once we count s`2t for the box0, the other boxes (including the box n` 1) contributes by a multiplicativefactor s` t to the total number of circular tiling. Thus, the total number ofcircular tilings for which W “ n is given by ps` 2tqps` tqn`1.Brought to you by | Cardiff UniversityAuthenticatedDownload Date | 3/9/20 11:34 AM270 A. Laugier, M. P. SaikiaHence, we haveP pW “ nq “tps` t´ 1qxnys,tps` 2tqps` tqn`1.Summing these will give us the desired result.Acknowledgements. The second author was supported by DST IN-SPIRE Scholarship 422/2009 from the Department of Science and Technology,Government of India when this work was done.References[1] T. Amdeberhan, X. Chen, V. H. Moll, B. E. Sagan, Generalized Fibonacci polynomialsand Fibonomial coefficients, Ann. Comb. 18(4) (2014), 541–562.[2] S. Ekhad, The Sagan-Savage Lucas-Catalan polynomials have positive coefficients,preprint http://www.math.rutgers.edu/˜zeilberg/mamarim/mamarimhtml/bruce.html.[3] B. E. Sagan, C. D. Savage, Combinatorial interpretations of binomial coefficientanalogues related to Lucas sequences, Integers, A52, 10 (2010), 697–703.A. LaugierLYCÉE TRISTAN CORBIÈRE16 RUE DE KERVÉGUEN - BP 17149 - 29671MORLAIX CEDEX, FRANCEE-mail: laugier.alexandre@orange.frM. P. SaikiaDIPLOMA STUDENT, MATHEMATICS GROUPTHE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS11, STRADA COSTIERATRIESTE, ITALYcurrent addressFACULTY OF MATHEMATICSUNIVERSITY OF VIENNAOSKAR-MORGENSTERN-PLATZ 11090 WIEN, AUSTRIAE-mail: manjil@gonitsora.comReceived March 26, 2014.Brought to you by | Cardiff UniversityAuthenticatedDownload Date | 3/9/20 11:34 AM

A combinatorial proof of a result on generalized Lucas polynomials

AbstractWe 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.</jats:p

Alexandre Laugier

Manjil P. Saikia

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Demonstratio Mathematica

A Combinatorial Proof of a Result on Generalized Lucas Polynomials

Laugier Alexandre

Saikia Manjil P.

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A combinatorial proof of a result on generalized Lucas polynomials

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