Claremot Colleges Scholarship @ Claremot All HMC Faculty Publicatios ad Research HMC Faculty Scholarship 5-1-2015 Coutig o r-fiboacci Numbers Arthur Bejami Harvey Mudd College Curtis Heberle Harvey Mudd College Recommeded Citatio Coutig o r-fiboacci Numbers (with Curtis R. Heberle*) The Fiboacci Quarterly, Vol. 52, Number 2, pp. 121-128, May 2014. This Article is brought to you for free ad ope access by the HMC Faculty Scholarship at Scholarship @ Claremot. It has bee accepted for iclusio i All HMC Faculty Publicatios ad Research by a authorized admiistrator of Scholarship @ Claremot. For more iformatio, please cotact scholarship@cuc.claremot.edu.
COUNTING ON r-fibonacci NUMBERS ARTHUR T. BENJAMIN AND CURTIS R. HEBERLE Abstract. We prove the r-fiboacci idetities of Howard ad Cooper [4] usig a combiatorial tilig approach. 1. Itroductio I a recet issue of this joural [4], Fred Howard ad Curtis Cooper establish may iterestig idetities for the r-geeralized Fiboacci sequece defied, by G = 0 for 0 <, G = 1, ad for r, G = G 1 + G 2 + + G r, where r is a positive iteger. Thus whe r = 2, G is the traditioal Fiboacci umber F. At the ed of their article, they say that may of their idetities ca udoubtedly be proved usig combiatorial argumets i the maer of Bejami ad Qui [1]. It would be iterestig to see differet approaches to our theorems. I this paper, we provide such proofs for all of their idetities. For combiatorial simplicity, we defie g = G + so that g = 0 for <0, g 0 = 1 ad for r 1, g = g 1 + g 2 + + g r. We call this the r-fiboacci sequece. It s easy to see (as oted i [1] ad [4]) that g couts tiligs of a -board with tiles of legth at most r. (A -board is a 1 grid with cells labeled 1, 2,...,.) This fact is verified by iductio sice g 0 = 1 couts the empty tilig ad for >0, ad 1 r, the umber of tiligs edig with a legth tile is g. Tiles of legth 1 ad 2 are called squares ad domioes respectively. I this paper, a -tilig shall deote a tilig of a -board where all tiligs have legth at most r. 2. Combiatorial Idetities All of the idetities i this sectio appear i [4] (although ot i the same order as preseted here) with G replaced by g r+1. Idetity 1. For r +1, 2g = g +1 + g r. Proof. We prove the idetity by fidig a 1:2 correspodece betwee the set of -tiligs (couted by g ) ad the uio of the set of ( + 1)-tiligs with the set of ( r)-tiligs (couted by g +1 + g r ). First, give a -tilig, we ca produce a ( + 1)-tilig simply by addig a square. This maps every -tilig to a ( + 1)-tilig that eds with a square. Secod, for each -tilig, we ca do oe of two thigs. If the last tile has legth ad <r, the we ca replace it with a tile of legth +1 r to get a ( + 1)-tilig that does ot ed with a square. Otherwise, = r ad we ca simply remove the last tile to get a tilig of legth r. 1
2 ARTHUR T. BENJAMIN AND CURTIS R. HEBERLE Thus each -tilig maps to two uique tiligs o the right-had side of the equatio, which may have legth either + 1 or r. Coversely, each ( + 1)-tilig ad ( r)-tilig is achieved i a uique way by this costructio, ad the theorem is proved. Idetity 2. For 1 r, g =2 1. Proof. Begi by tilig the -board with squares. There are the 1 dividigliesbetwee squares. Sice r, we ca obtai ay legth tilig by selectively removig or leavig these dividig lies. Thus g is equal to the umber of ways to remove dividig lies. There are 1 such lies, ad two choices for each: eep or remove. This gives 2 1 possibilities i all. I the proof above, it was essetial that we be able to remove dividig lies without restrictio. This was possible because the legth of the board was at most r, so that there was o ris of removig dividig lies i a way that created a illegal tile of legth r + 1 or greater. The ext theorem cosiders boards that are log eough to cotai exactly oe illegal tile. Idetity 3. For r +1 2r +1, g =2 1 ( r + 1)2. Proof. The left-had side couts the umber of legth tiligs. We obtai the right-had side by coutig the total umber of possible legth tiligs (usig tiles of ay legth), the subtractig away the tiligs that cotai a tile of legth r + 1 or greater to get the umber of legth tiligs with tiles of legth at most r. We obtai the total umber of tiligs by startig with the all-squares tilig ad coutig the umber of ways to remove dividig lies. There are agai 1 such lies, so the total umber of tiligs is 2 1. To cout the umber of illegal tiligs, we cosider the positio of the illegal tile s left edge. If its left edge coicides with the left edge of the board (the edge before cell 1), the there are r dividig lies iteral to the illegal tile (amely the dividig lies to the right of cells 1, 2,...,r) which must remai abset for it to have legth at least r + 1. This leaves 1 r dividig lies elsewhere o the board, each of which ca be either removed or ept, so that there are a total of 2 illegal tiligs with the illegal tile flush with the left edge of the board. Otherwise, there are r 1 places to place the illegal tile s left edge (ad leave room to the right for it to have legth at least r + 1). Oce the left edge has bee established, the status of that dividig lie is fixed, as is the status of the first r lies iteral to the illegal tile, so that there are i this case 2 r dividig lies which we ca choose to either eep or remove. There are thus ( r 1)2 such tiligs. The total umber of illegal tiligs is thus ( r 1)2 +2 =( r + 1)2. Subtractig this from the total umber of tiligs, we obtai g =2 1 ( r +1)2. We ow cosider tiligs where the umber of illegal tiles is at most. Idetity 4. For (r + 1) <( + 1)(r + 1), g =2 1 + ( 1) i a,i 2 1 i(r+1), where a,0 = 1 for 1, a,1 = 0 for r, a r+1,1 = 2, ad for r +2, i 1, a,i = a 1,i + a (r+1),i 1.
COUNTING ON r-fibonacci NUMBERS 3 Proof. Cosider a board of legth. The umber of ways to tile this board with tiles of legth at most r is of course g. We ca also cout tiligs by coutig all those with tiles of ay legth, the subtractig away the illegal tiligs i which a tile of legth r + 1 or greater appears. The umber of urestricted tiligs is 2 1. We first subtract, for each cell j, theumber of tiligs with a illegal tile startig at cell j. However, this over-subtracts tiligs with more tha oe illegal tile, so by the priciple of iclusio/exclusio we have to add bac i tiligs with two illegal startig poits, the subtract tiligs with three illegal startig poits, ad so o. Next we cout the umber of tiligs with i desigated startig poits (where these startig poits are at least r + 1 apart). For such a tilig, i of the board s 1 dividers are fixed, as are the r dividers that must be removed for each illegal tile to esure it is sufficietly log. If the set of i dividers does ot iclude the left edge of the board, the there are 1 i(r + 1) choices to mae about the remaiig dividers, so there are 2 1 i(r+1) ways to tile the rest of the board. If the set does cotai the left edge of the board, the we have oe additioal divider choice so there are 2 i(r+1) ways to complete the tilig. Thus whe the set of desigated startig poits of illegal tiles cotais the left edge of the board, the there are twice as may ways to complete the tilig tha would otherwise be true. All that remais to cosider is the umber of ways to desigate the left edges. Let a,i deote the umber of ways to desigate left edges for i illegal tiles o a legth board, where we give weight 2 to those desigatios that use the left edge of the board. The by the Priciple of Iclusio-Exclusio, the total umber of legal tiligs is g =2 1 + ( 1) i a,i 2 1 i(r+1). We must show that a,i behaves as we claim. There is of course exactly oe way to desigate left edges for zero such tiles, so a,0 = 1. Whe <r+ 1, there s o way to desigate the left edge of a illegal tile, sice the board is t log eough to accommodate oe, so a,1 = 0. Whe = r +1, there is exactly oe way to desigate the left edge of a illegal tile (sice such a tile fills the board), but sice this puts the illegal tile flush with the left edge of the board, we give it weight 2 so that a r+1,1 = 2. Fially, we ca obtai a recurrece by cosiderig whether or ot the last desigated illegal tile has legth r +1 ad cosists of the last r +1 tiles of the board. If so, the the remaiig i 1 desigated illegal startig poits ca be chose a (r+1),i 1 ways. If ot, the there as may ways to choose i desigated illegal startig poits from a ( 1)-board as with a -board, so there are a 1,i such desigatios. Summig over both cases, we have a,i = a 1,i + a (r+1),i 1. This completes the proof. Note that the expressio for g foud above could have bee writte a little more compactly. Specifically, g = ( 1) i a,i 2 1 i(r+1). i=0 Idetity 5. For (r + 1) <( + 1)(r + 1), g = i=0 ( 1) i ri i ri 1 + i 1 2 1 i(r+1).
4 ARTHUR T. BENJAMIN AND CURTIS R. HEBERLE Proof. It suffices to show that i the previous theorem, ri ri 1 a,i = +. i i 1 That is, we must cout the ways to desigate the left edges of i illegal tiles o a board of legth, where we give weight 2 to desigatios that use the first cell. First we cout the ways to choose left edpoits, whe we igore the weightig coditio. There are cells o the board, r of which ca be desigated the leftmost cell of a illegal tile (cells r+1 through are too close to the right edge of the board to permit a sufficietly log tile to begi at them). Thus we wish to choose i cells x 1,...,x i from the set {1,..., r} to serve as edge cells for our illegal tiles. The cells must be spaced far eough apart for the illegal tiles to fit, so we require that x j x j 1 r + 1 for all j. To do this, we first choose y 1,...,y i from the set {1,..., ri}, thesetx 1 = y 1, x 2 = y 2 + r, x 3 = y 3 +2r,...,x i = y i +(i 1)r. (Notice that sice 1 y 1 <y 2 < <y i ri, the1 x 1 <x 2 < <x i r ad that x j x j 1 = y j y j 1 + r r + 1.) The umber of ways to choose the y j is ri i.sice the equatios above provide a bijectio betwee the x i ad the y i, this is also the umber of ways to choose the x i ad thus the umber of ways to desigate leftmost cells of illegal tiles. Now to give tiligs with a illegal tile o the left edge of the board weight 2, we simply cout those tiligs agai ad add them to the total, so that they get couted twice. If oe illegal tile has its left edge flush with the left edge of the board, the what remais is to choose i 1 cells to serve as left edpoits for the remaiig illegal tiles, where we choose from {r+2,..., r}(sice the first illegal tile will cotai at least cells 1 through r+1.) Applyig the formula from the last paragraph, the umber of ways to do this is (r+1) r(i 1) i 1 = ri 1 i 1. Thus i total a,i = ri i + ri 1 i 1, as desired. The ext theorem allows us to cosider much more geeral, buttheidetityisrecursive. Idetity 6. For 2r 1, r g =2 g r + 2 i g r. =1 Proof. To show that the right-had side couts all -tiligs, we start with a board of legth. Assumig o tile crosses the iterface betwee cell r ad cell r + 1, we ca tile its first r cells ad its last r cells separately. These ca be legally tiled i 2 ad g r ways, respectively, givig a total of 2 g r such tiligs. We must add i the -tiligs that do have a tile crossig the lie after cell r. Cosider tiligs with a tile of legth i + crossig the r, r + 1 iterface, where i is the umber of cells the tile exteds past the iterface to the left, ad is the umber of cells the tile exteds past the iterface to the right. To the right of the crossig tile there are the g r possible tiligs, ad to the left of the crossig tile there are 2 r i 1 possible tiligs. Note that ca rage from 1 to r 1(if were ay larger, the iterface-crossig tile would be illegally log, as i must have legth at least 1), while for fixed, i ca rage from 1 to r. Thus the total umber of tiligs with a iterface-crossig tilig is r =1 2 i g r. Thus the total umber of tiligs of a legth board with tiles of legth at most r is r g =2 g r + 2 i g r. =1
COUNTING ON r-fibonacci NUMBERS 5 The ext theorem follows a similar approach but, lie i Idetity 5, it begis by overcoutig the umber of legal tiligs. Idetity 7. For r, g =2 g +1 r r 2 2 g r +1. Proof. As usual, the left-had side couts the set of -tiligs. To obtai the right-had side, we divide the board ito two parts. The first r 1 cells ca be tiled i 2 ways, ad the remaiig (r 1) cells ca be tiled i g r+1 ways. We have ot yet addressed what happes at the iterface betwee cell r 1 ad cell r. We ca either eep or remove the dividig lie here, givig us a extra factor of 2 ad a total of 2 g r+1 tiligs. However, removig the dividig lie betwee r 1 ad r will occasioally result i the creatio of a tile of illegal legth. We ow cout the umber of such illegal tiligs. First cosider the rightmost r +1 cells (i.e., cells r through.) If the tile begiig at cell r has legth, the remaiig cells to its right ca be tiled i g r +1 ways. Now remove the lie betwee cells r 1 ad r. For this to create a tile of illegal legth, we must have had a tile of legth at least r +1 edig o cell r 1. The r iteral lies directly to the left of cell r 1 are removed, so that we have (r 2) (r ) = 2 choices to mae about the remaiig lies. Thus the left side ca be tiled i 2 2 ways. Note that this oly maes sese for 2, sice if is 1, removig the lie betwee r 1 ad r caot create a illegal tile, o matter how the first r 1cells have bee tiled. Summig over all possible, ad subtractig this from the total umber of tiligs created i this way gives the umber of legal tiligs, amely r g =2 g +1 r 2 2 g r +1. =2 =2 The ext theorem geeralizes a well-ow sum of squares idetity for Fiboacci umbers. Idetity 8. For r 2, =0 g 2 + i=2 =i g g i = g g +1. Proof. Cosider a ( + 1)-board laid parallel to a -board, such that the left edges of the two boards alig ad the ( + 1)-board exteds oe cell to the right past the right edge of the -board. There are g ways to tile the shorter board, ad g +1 ways to tile the loger, so that there are g g +1 ways to tile the pair simultaeously. We ow show that the left-had side of the equatio couts the same quatity. Let s be the rightmost cell of either board that is ot covered by a domio, ad let + 1 be its positio withi its board. Note that sice ad + 1 have differet parity, s is always uiquely determied by. Suppose that s is covered by a square. The the cells to its right o its board must be covered by domioes ad the cells to its left ca be tiled i g ways. Liewise, cells + 1 ad beyod of the board ot cotaiig s must be covered by domioes, with g ways to tile the
6 ARTHUR T. BENJAMIN AND CURTIS R. HEBERLE remaiig cells. Thus the two boards ca be tiled i g 2 ways. The cell s ca be positioed aywhere from 1 to + 1, so ca rage from 0 to. The total umber of tiligs where s is covered by a square is thus =0 g2. Otherwise, s is covered by a tile of legth i + 1, where 2 i r 1. The the cells to the right of s must still be covered by domioes, the tile coverig s also covers the first i cells to its left, ad the remaiig i cells ca be tiled i g i ways. As before, cells +1 ad beyod of the board ot cotaiig s must be covered by domioes, ad the remaiig cells ca be tiled i g ways, givig g i g tiligs. Summig over all possible i ad gives i=2 =i g g i tiligs where s is ot covered by a square. (Note that i, as s must ecessarily be the rightmost cell of the tile that covers it.) Thus i total the umber of tiligs of the pair of boards is g 2 + g g i = g g +1. =0 i=2 =i We coclude with the most complicated looig idetity i [4]. Idetity 9. For 1, m 1, r 3, g +m r+1 = g r+1 g m r+1 + g r+1 g m r + g r g m r+1 i 1 + g i g m r+i+1 g i g m r+i j+1. i=2 j=1 Proof. Here, the left-had side couts all ( + m r + 1)-tiligs. To see that the right-had side also couts this, we begi by cosiderig the first r 2 potetial breas after cell r +2. That is, the r 2 gridlies startig with the right edge of cell r + 2 ad edig with the right edge of cell 1. For each such gridlie, we cout the tiligs that have a brea at that lie. I geeral, give a brea at cell i, there are g i ways to tile the leftmost i cells. This leaves m r +i+1 cells to be tiled to the right of the brea, which ca be doe i g m r+i+1 ways, so that there are g i g m r+i+1 tiligs with a brea at cell i. Summig over all r 2 potetial breapoits uder cosideratio gives g ig m r+i+1 tiligs. Sice it is possible for a tilig to have breas at more tha oe of the r 2 special lies, the sum just costructed couts may tiligs multiple times. We must subtract away each tilig the appropriate umber of times; if a tilig has breas at exactly of the r 2 special lies, it will have bee couted times, so we must subtract it 1times. To do this, we cosider, for each tilig, what happes i the regio of the board bouded by the first ad last of the r 2 special lies. If a tilig has breas at exactly of the r 2 lies, the those breas boud 1 tiles withi this regio. (We do ot cout tiles which overlap this regio but are ot wholly cotaied withi it.) Thus we ca achieve the appropriate subtractio by coutig the umber of tiligs with a particular legth tile i a particular positio, for each possible tile ad positio withi the special regio. For example, a tilig with exactly three breas withi the regio say at the right edges of cells 4, 2, ad 1 will be subtracted twice: oce for havig a domio o cells 3 ad 2, ad oce for havig a square o cell 1. Note that for tiligs with exactly oe brea withi the regio, o subtractio is eeded, ad sice o tiles are fully cotaied i the special regio, o subtractio will be performed.
COUNTING ON r-fibonacci NUMBERS 7 We ow perform this subtractio. Cosider tiligs with a tile of legth j coverig cells i + 1 through i + j. There are g i ways to tile the board to the left of this tile, ad g m r+i j+1 ways to tile the board to the right of this tile. Thus there are g i g m i j+1 such tiligs. We sum over all such tiles withi the regio bouded by the r 2 special lies. The leftmost of these lies is the right edge of cell r + 2; thus the first cell the special tile ca iclude is r + 3, ad so i ca be at most r 2. The last cell the tile ca iclude is 1, so i must be at least 2. A tile that starts at cell i + 1 must ed at or before cell 1, so j ca be at most i 1. Thus j rages from 1 to i 1, ad we have the double sum i 1 g i g m r+i j+1. i=2 j=1 Thus far we have show that the umber of tiligs with oe or more breas at specially desigated lies is i 1 g i g m r+i+1 g i g m r+i j+1. i=2 j=1 All that remais is to add i the tiligs that do t have breas at ay of the r 2 special lies. There are three possible cases. First, the r 2 lies ca be covered by a sigle tile of legth r 1coverigcells r + 2 through. Note that o shorter tile could cover all r 2lies, ad that this is the oly way to positio a tile of legth r 1tocoverr 2lies.Tiligthe board to the left ad the right of the tile, respectively, we see that there are g r+1 g m r+1 such tiligs. The other possibility is that the r 2 lies are covered by a tile of legth r. Sice this tile is loger tha ecessary by 1, it ca begi either at cell r + 1 or at cell r +2 ad still cover all r 2 of the special lies. I the former case the rest of the board ca be tiled i g r g m r+1 ways; i the latter case there are g r+1 g m r possible tiligs. Thus i total there are g r+1 g m r+1 + g r g m r+1 + g r+1 g m r tiligs which have o breas at ay of the special poits. The total umber of tiligs of a board of legth + m r +1isthus as desired. g r+1 g m r+1 + g r g m r+1 + g r+1 g m r i 1 + g i g m r+i+1 g i g m r+i j+1, i=2 j=1 3. r-fiboacci Numbers ad s-biomial Coefficiets Udoubtedly there are may more geeralized Fiboacci idetities that are ripe for combiatorial proof. Just as there are umerous coectios betwee Fiboacci umbers ad biomial coefficiets, such as F +1 =, 0 so it is that geeralized Fiboacci umbers have coectios with geeralized biomial coefficiets. Defiitio 1. For s 1, 0 ad 0 all itegers, we defie the s-biomial coefficiet to be the umber of ways of choosig a -elemet subset from a set of distict elemets, s where idividual elemets ca be chose for the subset a maximum of s times.
8 ARTHUR T. BENJAMIN AND CURTIS R. HEBERLE Techically, the objects beig couted by s are multi-subsets. Note that 0 = 1 ad 1 =. Bodareo [3] refers to the umber defied above as s 1. The followig idetity appears i [2]. We give a short combiatorial proof i hopes that it will ecourage readers to see out ad prove more idetities by this approach. Idetity 10. For 0, r 2, g = 0. Proof. As usual, the left side couts the ways to tile a -board with tiles of legth at most r. For the right side we show that the summad couts those -tiligs with exactly tiles. We begi with the tilig cosistig of squares. For each size subset of {1, 2,..., } couted by, we create a -tilig as follows: if the umber j appears x j times i the subset, the the jth tile is legtheed so it has legth 1 + x j. Sice, for each j, 0 x j r 1, each tile eds up with legth at most r. Ad sice j=1 x j =, the legth of the expaded tilig is ( )+ =, as desired, ad the idetity is established. 4. Acowledgmet The authors are very grateful to Kimberly Kidred ad the aoymous referee who carefully read this paper ad provided may suggestios that improved the expositio of this paper. Refereces [1] A. T. Bejami ad J. J. Qui. Proofs That Really Cout: The Art of Combiatorial Proof, The Dolciai Mathematical Expositios, 27, Mathematical Associatio of America, Washigto DC, 2003. [2] R. C. Bolliger. Fiboacci -Sequeces, Pascal-T Triagles ad -i-a-row Problems, The Fiboacci Quarterly 24.2 (1986), 140 144. [3] B.A. Bodareo. Geeralized Pascal Triagles ad Pyramids. Traslated by R.C. Bolliger. Sata Clara, Califoria: The Fiboacci Associatio, 1993. [4] F. T. Howard ad C. Cooper. Some Idetities for r-fiboacci Numbers, The Fiboacci Quarterly 49.3 (2011), 231 243. AMS Classificatio Numbers: 05A19, 11B39 Departmet of Mathematics, Harvey Mudd College, Claremot, CA 91711 E-mail address: bejami@hmc.edu 6 McLea Place, Apt 3. Cambridge, MA, 02140 E-mail address: cheberle@hmc.edu