The Fibonacci Quilt Sequence

Transcription

1 The Sequence Minerva Catral, Pari Ford*, Pamela Harris, Steven J. Miller, Dawn Nelson AMS Section Meeting Georgetown University March 7, 2015

2 1

3 1 2

4 1 2 3

5

6

7 Fibonacci sequence Theorem ( s Theorem) Every positive integer can be written uniquely as the sum of non-consecutive Fibonacci numbers where and F 1 = 1, F 2 = 2. F n = F n 1 + F n 2

8 Fibonacci sequence Theorem ( s Theorem) Every positive integer can be written uniquely as the sum of non-consecutive Fibonacci numbers where and F 1 = 1, F 2 = 2. F n = F n 1 + F n 2 Thus, if we create an increasing sequence of positive integers such that any positive number can be written uniquely as a sum of non-consecutive terms, we construct the sequence 1, 2, 3, 5, 8, 13, 21, 34,...

9

10 Expanding to a 2-Dimensional Construction

11 Expanding to a 2-Dimensional Construction Beginning in the center with the 1 by 1 square and spiraling out counter clockwise, we construct an increasing sequence of positive integers where every positive integer can be expressed as a sum of terms that do not share an edge in the.

12 The Sequence

13 The Sequence Theorem ( Numbers) The terms of the Sequence {a n } satisfy the recurrence relation a n+1 = a n 1 + a n 2 for n 4 and a i = i otherwise.

14 The Sequence Lemma (Recurrence Relation) The terms of the Sequence {a n } satisfy the recurrence relation a n+1 = a n + a n 4 for n 6 and a i = i for i 5 and a 6 = 7.

15 Proof of Lemma: a n+1 = a n + a n 4 for n 6 Proof. We can verify the recurrence for 6 n 10.

16 Proof of Lemma: a n+1 = a n + a n 4 for n 6 Proof. We can verify the recurrence for 6 n 10. If a n+1 is an element of the FQ Sequence where n 10, then a n+1 cannot be legally decomposed from any previous elements of the FQ Sequence. Thus, a n+1 1 is the largest positive integer that can be legally decomposed using the elements from the set {a 1 < a 2 < < a n }.

17 Proof of Lemma: a n+1 = a n + a n 4 for n 6 Proof. We can verify the recurrence for 6 n 10. If a n+1 is an element of the FQ Sequence where n 10, then a n+1 cannot be legally decomposed from any previous elements of the FQ Sequence. Thus, a n+1 1 is the largest positive integer that can be legally decomposed using the elements from the set {a 1 < a 2 < < a n }. Then (a n+1 1) (a n ) must be the largest number that can be legally decomposed using elements from the set {a 1 < a 2 < < a n 5 }, which equals a n 4 1 for n 10 by construction. Thus a n+1 = a n + a n 4 for n 6.

18 Proof of Theorem: a n+1 = a n 1 + a n 2 for n 4 Proof. Using the lemma and an inductive strategy, we first confirm that the recursive rule is satisfied for n = 4 and n = 5.

19 Proof of Theorem: a n+1 = a n 1 + a n 2 for n 4 Proof. Using the lemma and an inductive strategy, we first confirm that the recursive rule is satisfied for n = 4 and n = 5. Then a n+1 = a n + a n 4 = (a n 1 + a n 5 ) + a n 4 = a n 1 + (a n 4 + a n 5 ) = a n 1 + a n 2.

20 Not a Positive Linear Recurrence a n+1 = a n 1 + a n 2 for n 4 is not a Positive Linear Recurrence as the leading coefficient in the recurrence relation is zero.

21 Not a Positive Linear Recurrence a n+1 = a n 1 + a n 2 for n 4 is not a Positive Linear Recurrence as the leading coefficient in the recurrence relation is zero. a n+1 = a n + a n 4 for n 6 is also not a Positive Linear Recurrence because the initial conditions are not met (would have had to have a 6 = 7, but we actually have a 6 = 7).

22 Not a Positive Linear Recurrence a n+1 = a n 1 + a n 2 for n 4 is not a Positive Linear Recurrence as the leading coefficient in the recurrence relation is zero. a n+1 = a n + a n 4 for n 6 is also not a Positive Linear Recurrence because the initial conditions are not met (would have had to have a 6 = 7, but we actually have a 6 = 7). This leads to new complications and questions we can ask.

23 Non-Unique We find that we no-longer have unique decompositions. In particular, 11 = 9+2 and 11=7+4 are both legal decompositions.

24 Non-Unique Let d(m) be the number of FQ-legal decompositions of m. By construction of our sequence, d(a i ) = 1.

25 Non-Unique Let d(m) be the number of FQ-legal decompositions of m. By construction of our sequence, d(a i ) = 1. Let d ave be the average number of FQ-legal decompositions of the integers in I n := [0, a n+1 ). Then d ave (n) := 1 a n+1 1 d(m). a n+1 m=0

26 Non-Unique Let d(m) be the number of FQ-legal decompositions of m. By construction of our sequence, d(a i ) = 1. Let d ave be the average number of FQ-legal decompositions of the integers in I n := [0, a n+1 ). Then d ave (n) := 1 a n+1 1 d(m). a n+1 m=0 Theorem (Growth Rate of ) There is computable λ > 1 and a C > 0 such that d ave (n) Cλ n. Thus the average number of decompositions of integers in [0, a n+1 ) tends to infinity exponentially fast.

27 Legal We define the following: d n : the number of FQ-legal decompositions using only elements of {a 1, a 2,..., a n }. d n = a n+1 1 m=0 d(m) d ave (n) = d n a n+1

28 Legal We define the following: d n : the number of FQ-legal decompositions using only elements of {a 1, a 2,..., a n }. d n = a n+1 1 m=0 d(m) d ave (n) = d n a n+1 Similarly we define c n and b n where c n counts decompositions where a n is a summand, and b n counts decompositions where a n and a n 2 are both summands.

29 Legal Using brute force we can compute the first few values of the sequences. n a n d n c n b n Table: First few terms.

30 Legal Lemma For n 7 we have d n = c n + d n 1, (1) c n = d n 5 + c n 2 b n 2, and (2) b n = d n 7. (3) Thus d n = d n 1 + d n 2 d n 3 + d n 5 d n 9, implying d ave (n) C( ) n.

31 The for decompositions typically leads to unique representations by iteratively selecting the largest available summands.

32 The for decompositions typically leads to unique representations by iteratively selecting the largest available summands. For the FQ Sequence we do not have unique decompositions and the greedy algorithm does not always successfully terminate with a legal decomposition. We have 6 > a 5, but 6 = a 4 + a 2 is the only legal decomposition.

33 The for decompositions typically leads to unique representations by iteratively selecting the largest available summands. For the FQ Sequence we do not have unique decompositions and the greedy algorithm does not always successfully terminate with a legal decomposition. We have 6 > a 5, but 6 = a 4 + a 2 is the only legal decomposition. If N = a n + 6 < a n+1, then N does not have a legal decomposition by applying the.

34 In fact, if N = a n + x and N < a n+1, then N and x either both have a successful legal decomposition by applying the, or neither has one. The Greedy Fails less than a 18 = 200 are 6, 27, 34, 43, 55, 71, 92, 113, 120, 141, 148, 157, 178, 185, 194.

35 In fact, if N = a n + x and N < a n+1, then N and x either both have a successful legal decomposition by applying the, or neither has one. The Greedy Fails less than a 18 = 200 are 6, 27, 34, 43, 55, 71, 92, 113, 120, 141, 148, 157, 178, 185, 194. h n : number of integers in the interval [1, a n+1 ) where the greedy algorithm successfully terminates in a legal decomposition.

36 for n a n h n ρ n Table: First few terms, yields h n = h n 1 + h n

37 for Now we have h n = h n 1 + h n and ρ n hn a n+1. Using the Generalized Binet Formula, we find the Greedy Algorithm for yields a legal decomposition for about % of the integers.

38 Thank You Pari Ford