A combinatorial proof of multiple angle formulas involving Fibonacci and Lucas numbers

Transcription

1 Notes on Number Theory and Dscrete Mathematcs ISSN Vol. 20, 2014, No. 5, A combnatoral proof of multple angle formulas nvolvng Fbonacc and Lucas numbers Fernando Córes 1 and Dego Marques 2 1 Department of Mathematcs, Unversty of Brasla Brasla, DF, Brazl e-mal: fccores@gmal.com 2 Department of Mathematcs, Unversty of Brasla Brasla, DF, Brazl e-mal: dego@mat.unb.br Abstract: Let F n and L n be the nth Fbonacc and Lucas number, respectvely. In ths note, we gve a combnatoral proof for the followng dentty F kn+p = F k n F p + n (F n 1 F (k )n+p + F n F (k )n+p 1 ). Keywords: Fbonacc, Lucas, Multple angle, Combnatoral proof. AMS Classfcaton: 11B39. 1 Introducton Let (F n ) n 0 be the Fbonacc sequence gven by F n+2 = F n+1 + F n, for n 0, where F 0 = 0 and F 1 = 1. A few terms of ths sequence are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584,.... We cannot go very far n the lore of Fbonacc numbers wthout encounterng ts companon Lucas sequence (L n ) n 0, whch follows the same recursve pattern as the Fbonacc numbers, but wth ntal values L 0 = 2 and L 1 = 1: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571,

2 The Fbonacc numbers are well-known for possessng wonderful and amazng propertes (consult [1] together wth ts very extensve annotated bblography for addtonal references and hstory). In 1963, the Fbonacc Assocaton was created to provde enthusasts an opportunty to share deas about these ntrgung numbers and ther applcatons. Also, n the ssues of The Fbonacc Quarterly we can fnd many new facts, applcatons, and relatonshps about Fbonacc numbers. The search for denttes nvolvng Fbonacc numbers has always been a popular area of research. Among the several denttes, we are nterested n multple angle formulas. The most famous among them s the very useful dentty F 2n = F n L n. Ths dentty can be easly rewrtten as F 2n = F 2 n + 2F n F n 1 (recently, ths dentty appeared as Theorem 2.5 of [2]). For the Lucas case, we have L 2n = (5F 2 n + L 2 n)/2. Below we present some general multple angle formulas and F kn = L k F k(n 1) ( 1) k F k(n 2) (k 1)/2 1 = 2 5 F k 1 n 2+1 L k 1 2 n (k 1)/2 1 = F n ( 1) (n+1) L k 1 2 n = F F nf n 1 k = F F nfn+1 k We also have the general formula L kn = L k L k(n 1) ( 1) k L k(n 2) k/2 1 = 2 5 F k 1 n 2 Ln k 2 2 = L F nfn 1. k F kn+p = F p F nfn+1. k In ths note, we shall gve combnatoral proofs for new (to the best of authors knowledge) multple angle formulas. More precsely, we have the followng Theorem 1.1. For all postve ntegers n, k and p, t holds that F kn+p = F k n F p + n (F n 1 F (k )n+p + F n F (k )n+p 1 ). (1) 36

3 As applcaton, we prove that Corollary 1.2. For all postve ntegers n, k and p, t holds that L kn+p = F k n L p + n (F n 1 L (k )n+p + F n L (k )n+p 1 ). (2) 2 The proof of the theorem 2.1 Auxlary facts Before the proof, for the convenence of the reader, we shall recall combnatoral nterpretatons for Fbonacc and Lucas numbers. By a bnary sequence, we call a fnte word wrtten usng only 0 s and 1 s. A bnary sequence s sad to be of type 0, f ts frst and last terms are equal to 0 and such that t does not contan two consecutve 1 s. Lemma 2.1. The number of bnary words of type 0 wth length n s equal to F n. Proof. Let p n be the number of bnary words of type 0 wth length n. Clearly, p 1 = p 2 = 1 ({0, 00}). Also, such a bnary word of type 0 ether starts wth 01 or 00. In the frst case, t remans 01 } 0.{{.. 0} havng, by defnton, p n 2 words. In the second case, we have 1 } 0.{{.. 0} whch n 2 bts n 1 bts are counted n p n 1 ways. In concluson, p n = p n 1 + p n 2 whch completes the proof. A bnary sequence s sad to be of type 1 f there s no 1 s smultaneously at frst and last postons. Lemma 2.2. The number of bnary words of type 1 wth length n s equal to L n. Proof. Let c n be the number of such words wth length n. Clearly, c 1 = 1 ({0}) and c 2 = 3 ({00, 01, 10}). Also, such a word ether starts wth 1 or 0. In the frst case, the second and the last bt are 0, so by the prevous lemma, we have F n 1 such words. In the case of startng wth 0, the last term can be 0 and so we have F n words, or t ends wth 1 and so we obtan F n 1 words. In concluson, c n = F n 1 + F n + F n 1 = F n 1 + F n+1. A straghtforward calculaton yelds c n 1 + c n = c n+1 whch completes the proof. As a consequence, we have that Lemma 2.3. For all postve nteger n, t holds that L n = F n 1 + F n+1. Now, we are ready to deal wth the proof of theorem. 37

4 2.2 The proof of Theorem 1.1 In order to smplfy our argument, we shall rewrte (1) as F p F k n = F kn+p n (F n 1 F (k )n+p + F n F (k )n+p 1 ). The left-hand sde above counts the number of juxtapostons of k bnary sequences of length n and one sequence of length p, all of type 0, whch s F p Fn k (by Lemma 2.1). On the other hand, we can count these juxtapostons excludng from the F kn+p bnary sequences of length n and type 0 those ones havng the bts 0 and 1 or 1 and 0 at postons n and n + 1, for {1,..., k}, respectvely. In the case of a sequence havng 0 and 1 at postons n and n + 1, respectvely (see below) 0 }.{{.. 0} 1 } 0.{{.. 0} n bts (k )n+p 1 bts we have F n juxtapostons of the frst n block and F (k )n+p 1 for the last block (snce 1 s fxed at poston n + 1). Thus, there exst F nf (k )n+p 1 juxtapostons. In the second case, that s, a sequence havng 1 and 0 at postons n and n + 1, respectvely (see below) }{{} 0 }.{{.. 0} ( 1)n bts n 1 bts }{{} (k )n+p bts we obtan Fn 1 for the frst block, F n 1 for the second one and F (k )n+p for the last one. In a total of Fn 1 F n 1 F (k )n+p ways. In concluson we have excluded sequences and therefore (F nf (k )n+p 1 + n F n 1 F (k )n+p ) F p F k n = F kn+p n (F n 1 F (k )n+p + F n F (k )n+p 1 ) holds. 2.3 The proof of Corollary 1.2 By Lemma 2.3, we have that L kn+p = F kn+p 1 + F kn+p+1. Thus, by Theorem 1.1, L kn+p = F kn+p 1 + F kn+p+1 = Fn k F p 1 + Fn 1 (F n 1 F (k )n+p 1 + F n F (k )n+p 2 ) +F k n F p+1 + n (F n 1 F (k )n+p+1 + F n F (k )n+p ) 38

5 = F k n (F p 1 + F p+1 ) + n (F n 1 (F (k )n+p 1 + F (k )n+p+1 ) +F n (F (k )n+p 2 + F (k )n+p )) = Fn k L p + Fn 1 (F n 1 L (k )n+p + F n L (k )n+p 1 ) The proof s then complete. Acknowledgements The authors are grateful to the referees for readng ths paper. Durng the preparaton of ths work, the frst author was supported by CAPES (Coordenaçao de Aperfeçoamento de Pessoal de Nível Superor). The second author s supported by CNPq (Conselho Naconal de Desenvolvmento Centífco e Tecnológco) and FAP-DF (Fundaçao de Apoo à Pesqusa do Dstrto Federal). References [1] Koshy, T., Fbonacc and Lucas Numbers wth Applcatons, Wley, New York, [2] Thongmoon, M., New denttes for the even and odd Fbonacc and Lucas Numbers, Int. J. Contemp. Math. Scences,Vol. 4, 2009,