INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A18 TILING PROOFS OF SOME FIBONACCI-LUCAS RELATIONS Mark Shattuck Department of Mathematics, University of Tennessee, Knoxville, TN 37996-1300, USA shattuck@math.utk.edu Received: 3/6/07, Revised: 1/7/08, Accepted: 4/18/08, Published: 5/8/08 Abstract We provide tiling proofs for some relations between Fibonacci Lucas numbers, as requested by Benjamin Quinn in their text, Proofs that Really Count. Extending our arguments yields Gibonacci generalizations of these identities. 1. Introduction Let F n L n denote the Fibonacci Lucas numbers defined, respectively, by F 0 = 0, F 1 = 1 with F n = F n 1 + F n if n by L 0 =, L 1 = 1 with L n = L n 1 + L n if n. The following four relations can be shown using the Binet formulas for F n L n occur as (V85) (V88) in Vajda [: F (k+3)t = F t [( 1) (k+1)t + F (k+)t = F t ( 1) it L (k+ i)t, (1.1) ( 1) it L (k+1 i)t, (1.) L (k+3)t = L t [( 1) (k+1)(t+1) + F (k+)t = L t ( 1) i(t+1) L (k+ i)t, (1.3) ( 1) i(t+1) F (k+1 i)t, (1.4) where k t are nonnegative integers. Benjamin Quinn request combinatorial proofs of (1.1) (1.4) on page 145 of their text, Proofs that Really Count [1. In this note, we provide tiling proofs for (1.1) (1.4) as well as for a couple of closely related identities. The arguments
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A18 can be extended to yield generalizations involving Gibonacci numbers. The tiling proofs we give will also supply a combinatorial insight into the various divisibility relations implicit in (1.1) (1.4) within/between the Fibonacci Lucas sequences. The Fibonacci number F n+1 counts tilings of a board of length n with cells labeled 1,,..., n using squares dominos (termed n-tilings). The Lucas number L n is given by the simple relation L n = F n+1 + F n 1, n. (1.5) From (1.5), we see that L n counts n-tilings in which one may circle a domino covering cells n 1 n, which we ll term Lucas n-tilings. (There are F n 1 Lucas n-tilings which end in a circled domino F n+1 Lucas n-tilings which do not.) We ll say that a tiling is breakable at m (as in [1, p. 3) if cell m is covered by a square or by a second segment of a domino. The 5-tiling below is breakable at 0 (vacuously), 1, 3, 5. For m 0, let F m L m denote the sets consisting of m-tilings of Lucas m-tilings, respectively.. Tiling Proofs We first provide a combinatorial interpretation for (1.) when t is even. By (1.5), we re to show ( ) F (k+)t = Ft F (k+1 i)t+1 + F t F (k+1 i)t 1. (.1) Given i, 0 i k, let F i F (k+)t 1 comprise those tilings starting with it consecutive dominos, but not with (i + 1)t consecutive dominos. There are clearly F t F (k+1 i)t 1 members of F i that are not breakable at (i + 1)t, as such tilings can be decomposed as d it δ 1 dδ, where δ 1 δ are tilings of length t 1 (k + 1 i)t, respectively. So we need to show that there are F t F (k+1 i)t+1 members of F i that are breakable at (i + 1)t. Suppose λ F i is breakable at (i + 1)t. Let α be the t-tiling obtained by taking all the pieces in λ covering cells it + 1 through (i + 1)t β be the [(k + 1 i)t 1-tiling obtained by taking all the pieces in λ past cell (i + 1)t. We place α above β, offsetting the tilings by shifting β to the right by one cell. (Tiling α covers cells 1 through t β covers cells through (k + 1 i)t.) Look for the first fault line passing through α β. (A fault line occurs at cell j if both α β are breakable at cell j.) Exchange all the pieces in α past the first fault line with all the pieces in β past it (i.e., swap tails as shown) to obtain (A, B) F (k+1 i)t F t 1.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A18 3 Figure 1: Swapping tails past cell 8 converts (α, β) into (A, B). Note that λ F i implies that tails can always be swapped as it cannot be the case that α consists of β starts with t dominos. (If i = k, there is a single case in which tails of length zero are swapped.) Since t 1 is odd, there is always at least one square in a tiling of length t 1. Thus, the inverse on F (k+1 i)t F t 1 can always be defined, which completes the even case. Now suppose t 1 is odd. With the F i as above, let F i > := k j=i+1 F j, 0 i k ; (.) note that F i > = F (k i)t, as members of F i > must start with at least (i + 1)t consecutive dominos. There are F t F (k+1 i)t 1 members of F (k+)t 1 starting with it dominos that aren t breakable at (i + 1)t. These tilings consist of the F (k i)t members of F i > (they aren t breakable at (i+1)t since t is odd) as well as the members of F i that aren t breakable at (i + 1)t. Upon subtraction, there are then F t F (k+1 i)t 1 F (k i)t members of F i that aren t breakable at (i + 1)t. If λ F i is breakable at (i + 1)t, then use the correspondence above with F (k+1 i)t F t 1, noting that now the F (k i)t ordered pairs whose first component starts with t+1 dominos whose second component consists of t 1 dominos are missed. Thus, there are F t F (k+1 i)t+1 F (k i)t members of F i that are breakable at (i + 1)t. Upon adding the F t F (k+1 i)t 1 F (k i)t members of F i that aren t breakable at (i + 1)t, we see that there are (F t F (k+1 i)t+1 F (k i)t ) + (F t F (k+1 i)t 1 F (k i)t ) = F t L (k+1 i)t F (k i)t members of F i altogether, which implies F i + F i > = F t L (k+1 i)t. Therefore, F t ( 1) i L (k+1 i)t = ( 1) [ i F i + F i > = F i = F (k+)t 1 = F (k+)t,
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A18 4 where the second equality follows from a simple characteristic function argument: If 0 j k λ F j, then both the second third sums count λ one time, the second j 1 since ( 1) j + ( 1) i = 1. This completes the odd case the proof of (1.). A slight modification of the preceding argument gives (1.1) as well. We now turn to identity (1.3), first assuming t 1 is odd. A similar proof will apply to (1.4). By (1.5), we re to show L (k+3)t = L t + (F t+1 + F t 1 ) L (k+ i)t. (.3) Given i, 0 i k + 1, let L i L (k+3)t comprise those tilings starting with it consecutive dominos, but not with (i + 1)t consecutive dominos. Clearly, L k+1 = L t. If 0 i k, then there are F t+1 L (k+ i)t members of L i that are breakable at (i + 1)t as well, since such tilings can be decomposed as d it δ 1 δ, where δ 1 is a regular tiling of length t δ is a Lucas tiling of length (k + i)t. So we need to show that there are F t 1 L (k+ i)t members of L i that are not breakable at (i + 1)t. Suppose λ L i is not breakable at (i + 1)t. Let α be the (t 1)-tiling obtained by taking all the pieces in λ covering cells it + 1 through (i + 1)t 1 β be the Lucas [(k + i)t 1-tiling obtained by taking all the pieces in λ past cell (i + 1)t + 1. Set α above β, shifting β to the right by one cell, exchange pieces past the first fault line to obtain (A, B) L (k+ i)t F t. Note that this operation can always be performed since λ L i can always be reversed since t is odd. If t is even, then let L i > := k+1 j=i+1 L j, 0 i k, (.4) where the L i are as above. Note that L i > = L (k+1 i)t, since members of L i > must start with at least (i + 1)t consecutive dominos. If 0 i k, then there are F t+1 L (k+ i)t members of L (k+3)t which start with it consecutive dominos are breakable at (i + 1)t. Since t is even, there are among these the L (k+1 i)t members of L i >. By subtraction, we get F t+1 L (k+ i)t L (k+1 i)t members of L i that are breakable at (i + 1)t if 0 i k. If λ L i is not breakable at (i+1)t, then use the correspondence above with L (k+ i)t F t, noting that now the ordered pairs whose second component consists of t 1 dominos whose first component starts with t dominos are missed. Thus, there are F t 1L (k+ i)t L (k+1 i)t members of L i that aren t breakable at (i+1)t (F t+1 L (k+ i)t L (k+1 i)t )+ (F t 1 L (k+ i)t L (k+1 i)t ) = L t L (k+ i)t L (k+1 i)t members of L i in all. The rest of the proof goes much like the odd case of (1.) above. Two further identities can also be obtained from the arguments above. On page 146 of their text, Benjamin Quinn request combinatorial proofs for Identities V93 V94,
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A18 5 which may be written as k+1 F (k+3)t = ( 1) (k+1)t F t [(k + 3) + 5 ( 1) it Fit i=1 (.5) k+1 F (k+3)t = ( 1) (k+1)t F t [ (k + 1) + ( 1) it L it. (.6) Replacing i by k + 1 i in (1.1) gives i=1 substituting k+1 F (k+3)t = ( 1) (k+1)t F t [1 + ( 1) it L it, (.7) i=1 L it = 5F it + ( 1) it (.8) L it = L it ( 1) it (.9) into (.7) yields (.5) (.6), respectively. The preceding argument for (1.1) hence (.7) can then be easily combined with the arguments for (.8) (.9), which are known involve tailswapping (see, e.g., Identities 36, 45, 53 of [1), to obtain combinatorial interpretations for (.5) (.6), as desired. 3. Generalizations Let (G n ) n 0 be the sequence given by G n = G n 1 + G n if n, where G 0 G 1 are nonnegative integers (termed Gibonacci numbers [1, p. 17 as shorth for generalized Fibonacci numbers). The G n are seen to enumerate n-tilings in which a terminal square is assigned one of G 1 possible phases a terminal domino is assigned one of G 0 possible phases (termed phased n-tilings). Note that G n reduces to F n+1 when G 0 = G 1 = 1 to L n when G 0 =, G 1 = 1. Reasoning as in the prior section with phased tilings instead of regular tilings yields the following (reindexed) generalizations of (1.1) (1.4): G (k+3)t 1 = ( 1) kt [ ( 1) t G t 1 + F t ( 1) it (G (i+)t + G (i+)t ), (3.1)
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A18 6 G (k+)t 1 = ( 1) kt [ ( 1) t (G 1 G 0 ) + F t G (k+3)t = ( 1) k(t+1) [ G (k+)t 1 = ( 1) k(t+1) [ ( 1) it (G (i+1)t + G (i+1)t ), (3.) ( 1) t+1 G t + L t ( 1) i(t+1) G (i+)t, (3.3) ( 1) t+1 (G 1 G 0 ) + L t ( 1) i(t+1) G (i+1)t 1. (3.4) Identities (3.1), (3.), (3.4) reduce to (1.1), (1.), (1.4), respectively, when G n = F n+1, (3.3) reduces to (1.3) when G n = L n. We outline the proof of (3.), leave the others as exercises for the interested reader. We adjust the argument above for (1.), first assuming t is even. Let G m denote the set consisting of phased tilings of length m let G i G (k+)t 1 comprise those tilings starting with it consecutive dominos, but not with (i+1)t consecutive dominos, 0 i k. There are then F t G (i+1)t members of G k i that are not breakable at (k i + 1)t if 0 i k, upon tailswapping as above, F t G (i+1)t members of G k i that are breakable at (k i + 1)t if 0 < i k. If i = 0, we omit the G 1 cases in which λ G k ends in a phased square preceded by t 1 dominos, since here tailswapping doesn t actually move any tiles (i.e., the first only fault occurs directly after cell t) hence the phased square ending λ fails to be moved. Similarly, we must omit from consideration the G 0 members of G t F t 1 whose first coordinate is a phased tiling which contains no squares whose second coordinate is a regular tiling which contains only a single square occurring at the end. Thus, there are F t G t + (G 1 G 0 ) members of G k that are breakable at (k + 1)t. Adding together all of the cases gives (3.) when t is even. Similar adjustments apply when t is odd. References [1 A. Benjamin J. Quinn, Proofs that Really Count: The Art of Combinatorial Proof, The Dolciani Mathematical Expositions, 7, Mathematical Association of America, 003. [ S. Vajda, Fibonacci & Lucas Numbers, the Golden Section: Theory Applications, John Wiley & Sons, Inc., 1989.