In Search of Combinatorial Fibonacci Identities

Transcription

1 Summer Research 2010 Duncan McGregor Mike Rowell Department of Mathematics and Computer Science Pacific University, Oregon NUMS, April 9, 2011

2 Outline 1 2

3 Fibonacci Sequence Defined Recursively: f 0 = 1, f 1 = 1 f n = f n 1 + f n 2

4 Fibonacci Sequence Defined Recursively: f 0 = 1, f 1 = 1 f n = f n 1 + f n 2 {1, 1, 2, 3, 5, 8, 13,... }

5 Fibonacci Sequence Defined Recursively: f 0 = 1, f 1 = 1 f n = f n 1 + f n 2 {1, 1, 2, 3, 5, 8, 13,... } this definition is off by one from the usual definition

6 Combinatorial Interpretation Theorem (Benjamin, Quinn) The number of ways to tile an (1 n)-board with 1 1 squares and 1 2 dominoes is f n

7 Combinatorial Interpretation Theorem (Benjamin, Quinn) The number of ways to tile an (1 n)-board with 1 1 squares and 1 2 dominoes is f n 1 way to tile the 0-board.

8 Combinatorial Interpretation Theorem (Benjamin, Quinn) The number of ways to tile an (1 n)-board with 1 1 squares and 1 2 dominoes is f n 1 way to tile the 0-board. 1 way to tile the 1-board.

9 Combinatorial Interpretation Theorem (Benjamin, Quinn) The number of ways to tile an (1 n)-board with 1 1 squares and 1 2 dominoes is f n 1 way to tile the 0-board. 1 way to tile the 1-board.

10 Some Vocabulary An n board has n cells.

11 Some Vocabulary An n board has n cells. A tiling has a fault at cell m if there is not a domino beginning in that cell.

12 Some Vocabulary An n board has n cells. A tiling has a fault at cell m if there is not a domino beginning in that cell. Theorem For n, k 1, f n+k = f n f k + f n 1 f k 1

13 Lucas Numbers

14 Lucas Numbers Defined Recursively: L 0 = 2, L 1 = 1 L n = L n 1 + L n 2

15 Lucas Numbers Defined Recursively: L 0 = 2, L 1 = 1 L n = L n 1 + L n 2 {2, 1, 3, 4, 7, 11, 18,... }

16 Combinatorial Interpretation Theorem (Benjamin, Quinn) The number of ways to tile a circularn bracelet with bent 1 1 squares and 1 2 dominoes is L n

17 Combinatorial Interpretation Theorem (Benjamin, Quinn) The number of ways to tile a circularn bracelet with bent 1 1 squares and 1 2 dominoes is L n The proof is similar except that there are two ways to tile the 0-bracelet.

18 Combinatorial Interpretation Theorem (Benjamin, Quinn) The number of ways to tile a circularn bracelet with bent 1 1 squares and 1 2 dominoes is L n The proof is similar except that there are two ways to tile the 0-bracelet.

19 A Useful Identity Theorem (Benjamin, Quinn) For n 2, L n = f n + f n 2

20 Zeckendorf Representations Theorem (Zeckendorf) For n N there exists a unique sequence {a j } M j=1 N such that a j+1 > a j + 1 and n = M f aj. j=1 Where the sum is called the Zeckendorf Representation of n.

21 Zeckendorf Representations Theorem (Zeckendorf) For n N there exists a unique sequence {a j } M j=1 N such that a j+1 > a j + 1 and n = M f aj. j=1 Where the sum is called the Zeckendorf Representation of n. We seek Combinatorial Proofs of Zeckendorf Representations. In particular of numbers like f n f k, 2f n f k, L n f k, L n L k.

22 A Useful Lemma Lemma For n 2, m 1, n m f n 2 f m f n 1 f m 1 = ( 1) m f n m

23 A Useful Lemma Lemma For n 2, m 1, n m f n 2 f m f n 1 f m 1 = ( 1) m f n m How many ways can we tile an (m + n 2) board such that there is a fault at n 2 but there is not a fault at n 1?

24 Proof Idea f n 2 f m f n 1 f m 1 = ( 1) m f n m

25 Proof Idea f n 2 f m f n 1 f m 1 = ( 1) m f n m Call the set of tilings with the fault at n 2 A and and the fault at n 1 B.

26 Proof Idea f n 2 f m f n 1 f m 1 = ( 1) m f n m Call the set of tilings with the fault at n 2 A and and the fault at n 1 B. Condition on the number of dominoes on each side of the fault.

27 Proof Idea f n 2 f m f n 1 f m 1 = ( 1) m f n m Call the set of tilings with the fault at n 2 A and and the fault at n 1 B. Condition on the number of dominoes on each side of the fault. Realize for most cases the size of the tails is the same for A and B. Cancellation yields the result.

28 Proof Idea f n 2 f m f n 1 f m 1 = ( 1) m f n m Call the set of tilings with the fault at n 2 A and and the fault at n 1 B. Condition on the number of dominoes on each side of the fault. Realize for most cases the size of the tails is the same for A and B. Cancellation yields the result. If m is even, A has one more member than B and vice versa.

29 A Picture to Convince You

30 A Picture to Convince You Fault at n 2 Fault at n 1

31 A Picture to Convince You Fault at n 2 Fault at n 1 Fault at n 2 Fault at n 1

32 What can we prove with this? Theorem For n 2 > 2k > 1 Proof. L 2k L n = f n+2k + f n+2k 2 + f n 2k + f n 2k 2

33 What can we prove with this? Theorem For n 2 > 2k > 1 Proof. Rewrite as: L 2k L n = f n+2k + f n+2k 2 + f n 2k + f n 2k 2 f n f 2k 2 (f n+2k f n f 2k ) + f n 2 f 2k 2 (f n+2k 2 f n 2 f 2k ) = f n 2k + f n 2k 2

34 What can we prove with this? Theorem For n 2 > 2k > 1 Proof. Rewrite as: L 2k L n = f n+2k + f n+2k 2 + f n 2k + f n 2k 2 f n f 2k 2 (f n+2k f n f 2k ) + f n 2 f 2k 2 (f n+2k 2 f n 2 f 2k ) = f n 2k + f n 2k 2 Use the above lemma twice, let m 2k and m 2k, n n 2.

35 Similarly

36 Similarly Theorem f 2k L n = f n+2k + f n 2k

37 Similarly Theorem f 2k L n = f n+2k + f n 2k Using a trick about off by every other fibonacci sums gives you odd cases of above theorems.

38 Similarly Theorem f 2k L n = f n+2k + f n 2k Using a trick about off by every other fibonacci sums gives you odd cases of above theorems. Thinking about f 2k L n yields ZR for f m f 2k L n, f m L 2k L n, L m L 2k L n.

39 Similarly Theorem f 2k L n = f n+2k + f n 2k Using a trick about off by every other fibonacci sums gives you odd cases of above theorems. Thinking about f 2k L n yields ZR for f m f 2k L n, f m L 2k L n, L m L 2k L n. OPEN PROBLEMS:

40 Similarly Theorem f 2k L n = f n+2k + f n 2k Using a trick about off by every other fibonacci sums gives you odd cases of above theorems. Thinking about f 2k L n yields ZR for f m f 2k L n, f m L 2k L n, L m L 2k L n. OPEN PROBLEMS: Do the above in a single map.

41 Similarly Theorem f 2k L n = f n+2k + f n 2k Using a trick about off by every other fibonacci sums gives you odd cases of above theorems. Thinking about f 2k L n yields ZR for f m f 2k L n, f m L 2k L n, L m L 2k L n. OPEN PROBLEMS: Do the above in a single map. What about 2k + 1?

42 Acknowledgments Pacific University Mathematics Faculty Pacific Research Institute for Science and Mathematics, Office of the Dean of Arts and Sciences of Pacific University, Involve: Journal of Mathematics, Arthur Benjamin and Jennifer Quinn for the Proofs That Really Count.