THE LUCAS TRIANGLE RECOUNTED. Arthur T. Benjamin Dept. of Mathematics, Harvey Mudd College, Claremont, CA Introduction

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1 THE LUCAS TRIANLE RECOUNTED Arthur T Bejami Dept of Mathematics, Harvey Mudd College, Claremot, CA bejami@hmcedu 1 Itroductio I 2], Neville Robbis explores may properties of the Lucas triagle, a ifiite triagular array with properties similar to Pascal s triagle I this paper, we provide a combiatorial explaatio for the etries of this triagle This iterpretatio results i extremely quic ad ituitive proofs of most of the properties (proved mostly by iductio i 2]) ad allows for a atural geeralizatio, with equally trasparet proofs Usig Robbis otatio, we defie the umber by the iitial coditios = 1 for 1, = 2 for 0, 0] ] ad the Pascal-lie recurrece for 1 1, ] ] 1 1 = + 1 For combiatorial coveiece, we shall defie = 0 wheever < 0 or < 0 or > This allows the recurrece to be true for all values of ad except for = = 0 ad = 1, = 0 Below we list rows 0 through 9 of the Lucas triagle: 1

2 2 THE LUCAS TRIANLE RECOUNTED Combiatorial Iterpretatio The coectio to Lucas umbers is easy to see as follows As is well-ow 1], L couts the ways to tile a bracelet of legth with squares ad domioes The cells of the bracelet are labeled from 1 through ad we defie the first tile to be the tile that covers cell 1, the cell that is to the right of the clasp For example, the Lucas umber L 4 = 7, reflects the fact that there are seve bracelets of legth four, amely dd (two ways), dss (two ways), sds, ssd, ad ssss where d deotes a domio, s deotes a square; the first two tiligs have two represetatios, depedig o whether the iitial domio covers the clasp (cells 4 ad 1) or covers cells 1 ad 2 Theorem 1 For 1, couts bracelets cotaiig tiles with exactly domioes (ad therefore squares) Proof Let B(, ) deote the umber of bracelets with domioes ad squares We prove that B(, ) = by showig that B(, ) satisfies the same iitial coditios ad recurrece Clearly, there is oe bracelet with squares ad o domioes ad there are two bracelets with domioes ad o squares

3 THE LUCAS TRIANLE RECOUNTED 3 (depedig o the locatio of the iitial domio) Thus B(, 0) = 1 = ad 0] B(, ) = 2 = For the recurrece, otice that a bracelet with domioes ] ad squares will either ed with a square (preceded by domioes ad 1 squares) or ed with domio (preceded by 1 domioes ad squares) That is, B(, ) = B( 1, ) + B( 1, 1) satisfies the same recurrece as, ad therefore B(, ) = Usig this combiatorial iterpretatio, most of the idetities preseted i 2] ca be proved by ispectio (more precisely, by ispectig the size of a set i two differet ways) We ote that a bracelet with tiles cosistig of domioes ad squares will have legth + It follows that for 1, = + 1 1] sice a bracelet with tiles with just oe domio has legth + 1 ad that domio ca start o ay of the + 1 cells Liewise, for 1, ] = sice a bracelet with 1 domioes ad oe square has legth 2 1 ad therefore has 2 1 places to put its loe square Similarly, for 2, ] = ( 1) 2 2 sice a bracelet of legth 2 2 with two squares ca be obtaied by isertig the squares i ay two cells of opposite parity

4 4 THE LUCAS TRIANLE RECOUNTED The followig sums are also easy to see For 1, = 3(2 1 ) =0 sice the left side couts bracelets with tiles with a arbitrary umber of domioes There are three choices for the first tile (oe possible square ad two possible domioes) followed by two choices for each of the followig 1 tiles By the same reasoig, we have, for 2, = 2 ] = 3( ) 0 0 sice if we specify the parity of the umber of domioes, the the last tile is ] i forced Fially otice that couts bracelets of legth with i domioes i ad therefore i 0 ] i = L i Notice that whe p is prime, we have for all 1 i (p 1)/2, ] p p i i sice every bracelet of prime legth p with at least 1 domio has p distict bracelets i its orbit, obtaied by shiftig each tile clocwise uits as varies from 0 to p 1 The bracelets are distict sice p is prime Eve Robbis s followig extesio ca be appreciated combiatorially Propositio: Let p be a odd prime, the for all j such that 1 j p 2 ad ] for all i such that j + 1 i p 1, we have p p + j i Proof Let X be a bracelet with p + j tiles, where 1 j p 2, with i domioes where j + 1 i p 1 Here we obtai the orbit of X by eepig the first j tiles fixed, ad the rotatig the remaiig p tiles oe tile at a time Notice that the

5 THE LUCAS TRIANLE RECOUNTED 5 bouds o i ad j esure that the fixed part has at least oe tile, ad that the rotatig part has at least oe square ad at least oe domio (If the rotatig part of X were all squares or all domioes, the its orbit would have size oe or two, respectively) Thus, sice p is prime, every orbit of X has p elemets, as desired Alteratig sums ca also be easily hadled combiatorially For example, for 2, the idetity ] ( 1) = 0 =0 says that, amog bracelets with tiles, there are as may bracelets with a eve umber of domioes as with a odd umber of domioes This is easy to see by the simplest of ivolutios: togglig the last tile I other words, if the last tile of your bracelet is a square the tur it ito a domio; if the last tile is a domio, the tur it ito a square (Note that the coditio that 2 assures that the last tile is ot the same as the first tile) Either way, the parity of the umber of domioes has chaged I this way, every bracelet with tiles ad a eve umber of domioes holds hads with a bracelet with tiles ad a odd umber of domioes The precedig idetity ad argumet ca be geeralized to give: m ] ( 1) =0 ] 1 = ( 1) m m For this idetity, we apply the same togglig argumet as before, but ow we are restricted to bracelets with tiles that have at most m domioes The oly bracelets that are ot matched up are those that have m domioes ad ed with

6 6 THE LUCAS TRIANLE RECOUNTED a square (These bracelets are umatched because togglig the last square would create a bracelet with m + 1 domioes, which exceeds our upper boud) Sice ] 1 there are bracelets of this type, all of which have m tiles, the idetity m follows Next, we ote that etries of the Lucas triagle ca be expressed i terms of the classical biomial coefficiets that ihabit Pascal s triagle Specifically, for 2 ad 0, = ( ) + ( ) 1 1 For ay bracelet with tiles ad domioes, it either has a domio coverig the clasp (resultig i a out-of-phase bracelet) or it does ot (resultig i a i-phase bracelet) The first biomial coefficiet couts the i-phase bracelets, sice we simply choose which of the tiles are domioes The umber of out-of-phase bracelets is ( 1 1) sice the first tile must be a domio coverig the clasp, ad the we ca freely choose 1 of the remaiig 1 tiles to be domioes I fact ca be expressed eve more directly i terms of biomial coefficiets, amely or equivaletly, = + = ( + ) ( ), The left side couts the ways to create a bracelet with tiles, cotaiig ( ) domioes, ad placig a star o oe of the tiles (If the selected tile is a domio, we shall place the star o its left half) The right side couts i-phase bracelets with tiles, cotaiig domioes where we place a star o oe of its

7 THE LUCAS TRIANLE RECOUNTED 7 + cells We show that these two sets have the same size by creatig a bijectio betwee the two sets Let X be a bracelet from the first set The X has legth + ad has a star o a tile that begis o some cell j, where 1 j + If X is i-phase, the we leave the bracelet uchaged, ad simply move the star from the tile to the cell j If X is out-of-phase, the we rotate X couterclocise j 1 uits so that it becomes a i-phase bracelet begiig with the tile that origially had the star The we trasfer the star to the cell below the right half of the domio that used to cover the clasp (This would put the star o cell j) Sice this process is easily reversed by examiig the tile that covers the cell with the star, we have our desired bijectio The ext two idetities mae use of the complemet of a tilig ive a tilig X, we obtai its complemet X by togglig each tile For example, if X = ssdds the X = ddssd Recallig that F m couts the set of tiligs (or i-phase bracelets) of legth m 1, we obtai the ext idetity (misstated i 2]) For 1, F 2+2 = i=0 ] + i 2i To see this, let X be a tilig of legth If X begis with a square, the we create a i-phase bracelet Y by deletig the first square, the taig the complemet of the rest of X If we let X 1 be the tilig X with its first tile missig, the our mappig ca be deoted by sx 1 X 1 (i-phase) Notice that sice X has odd legth, the the umber of squares i X must be a odd umber 2i + 1 for some 0 i It follows that X has + i + 1 tiles

8 8 THE LUCAS TRIANLE RECOUNTED comprisig 2i+1 squares ad i domioes Hece the bracelet Y has +i tiles with 2i domioes, ad all i-phase bracelets of this type ca be obtaied this way O the other had, if X begis with a domio, the we obtai the out-of-phase bracelets by the mappig dx 1 dx 1 (out-of-phase) where the iitial domio of X covers the clasp of Y, ad all subsequet tiles of X are replaced by their complemet Here, if X has 2i 1 squares for some 1 i, the X will cotai + i tiles comprisig 2i 1 squares ad + 1 i domioes, ad therefore Y will cotai + i tiles with exactly 2i domioes as desired We leave it to the reader to verify that the exact same mappig leads to the idetity 1 ] + i F 2+1 = 2i + 1 i=0 3 The iboacci Triagle The Lucas umbers are a special case of the iboacci umbers defied by arbitrary iitial coditios 0 ad 1, ad for 2, = It is easy to show (as doe i 1]) that has the followig combiatorial iterpretatio Whe 0 ad 1 are oegative itegers ad 1, couts phased tiligs of legth These are just lie traditioal tiligs with squares ad domioes, but the first tile is give a phase If the iitial tile is a domio, the it ca be assiged oe of 0 phases; if the iitial tile is a square, the it ca be assiged oe of 1 phases Notice that whe 0 = 2 ad 1 = 1, that = L

9 THE LUCAS TRIANLE RECOUNTED 9 ad the 0 = 2 phases of a iitial domio idicate whether or ot a iitial domio covers the clasp Whe 0 = 0 ad 1 = 1, the = F ad we are coutig legth tiligs that must begi with a square (sice there are o iitial domioes) ad therefore couts arbitrary tiligs of legth 1 Although this defiitio requires 0 ad 1 to be oegative itegers, a similar iterpretatio ca be give for arbitrary real or complex umbers (as show i 1]) but we will ot pursue that here We defie the umber 0] ad for 1 1, Also we let = 1 for 1, by the iitial coditios = ] ] 1 + = 0 for 0, ] 1 1 = 0 whe < 0 or < 0 or > For example, the iitial terms 0 = 4 ad 1 = 9, geerate the iboacci sequece: 4, 9, 13, 22, 35, 57, 92, 149, 241, 390, The first te rows of the iboacci Triagle are:

10 10 THE LUCAS TRIANLE RECOUNTED Arguig exactly as before, we have a simple combiatorial iterpretatio of Theorem 2 For 1, 0, exactly domioes (ad therefore squares) couts phased tiligs with tiles cotaiig This combiatorial iterpretatio allows us to immediately geeralize most of the precedig theorems (It also immediately gives us a geeratig fuctio, amely is the a b coefficiet of ( 0 a + 1 b)(a + b) 1 ) Their proofs are almost exactly the same as before, so we leave most of their details to the reader I our first set of idetities, we decide o the legth ad phase of the iitial tile, the proceed to choose the remaiig 1 (uphased) tiles For 1, = 1] ( 1) = 1 + ( 0 1 ), ad for 2, ] 1 ] 2 =0 = ( 1) = 0 + ( 1 0 ), ( ) 1 = ( 1), 2 = ( )2 1 = 2 2 1, = 2 ] = , 0 0 ( ) ( ) ( ) ( ) = = 1 + ( 0 1 ) 1 1 Some idetities geeralize with virtually o chages to their statemet or logic For example, the sum of the diagoal etries of the iboacci triagle yields

11 iboacci umbers For 0, Although the prime idetity THE LUCAS TRIANLE RECOUNTED 11 i 0 ] i = i ] p p i i does ot have a phased aalog (because we caot rotate the phased iitial tile), the extesio of that result wors fie because the first j tiles are fixed Specifically, for ay odd prime p ad for ay 1 j p 2 ad j + 1 i p 1, ] p p + j i Liewise, the alteratig idetities, obtaied by togglig the last tile, are completely uaffected by the iitial tile Thus, for 2, ] ( 1) = 0, =0 m ] ] 1 ( 1) = ( 1) m m =0 Refereces 1] A T Bejami ad J J Qui, Proofs That Really Cout: The Art of Combiatorial Proof, Mathematical Associatio of America, Washigto, DC, ] N Robbis, The Lucas Triagle Revisited The Fiboacci Quarterly, Vol 43 (2005): pp AMS Subject Classificatio Numbers: 05A19, 11B39