Polygonal Chain Sequences in the Space of Compact Sets

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1 Journal of Integer Sequences, Vol , Article Polygonal Chain Sequences in the Space of Compact Sets Steven Schlicker Grand Valley State University Allendale, MI USA schlicks@gvsu.edu Lisa Morales University of California, Riverside Riverside, CA 951 USA lmora008@student.ucr.edu Daniel Schultheis University of California, San Diego La Jolla, CA 9093 USA dschulth@math.ucsd.edu Abstract Configurations in the hyperspace of all non-empty compact subsets of n-dimensional real space provide a potential wealth of examples of familiar and new integer sequences. For example, Fibonacci-type sequences arise naturally in this geometry. In this paper, we introduce integer sequences that are determined by polygonal chain configurations. 1 Introduction The Hausdorff metric h provides a way of measuring the distance between nonempty closed and bounded subsets of R n using the standard Euclidean metric d E. The metric h imposes a 1

2 geometry on HR n, the hyperspace of all non-empty, compact subsets of R n. The concept of betweenness in this geometry gives rise to interesting examples of integer sequences. In this context we will introduce polygonal chains and new integer sequences related to them. Throughout this article, we let F k denote the k th Fibonacci number with F 0 = 0,F 1 = 1 and L k the k th Lucas number with L 0 =,L 1 = 1. Recall that the Lucas numbers L k satisfy L k = F k 1 + F k+1 = F k 1 + F k. The Hausdorff Metric The Hausdorff metric h was introduced by Felix Hausdorff in the early 0th century as a way to measure the distance between compact sets. The Hausdorff metric imposes a geometry on the space HR n, which will be the subject of our study. To distinguish between R n and HR n, we will refer to points in R n and elements in HR n. Definition 1. Let A and B be elements in HR n. The Hausdorff distance, ha,b, between A and B is ha,b = max{da,b,db,a}, where da,b = max a A {da,b} and da,b = min b B {d E a,b}. As an example, let A be the two point set { 1, 0, 1, 0} and B the circle centered at 0,0 of radius 1 in R. Since A B, we have da,b = 0. However, db,a = d E 0, 1, 1, 0 =. Note that da,b db,a, so the function d is not symmetric. That is why we need the maximum of da,b and db,a to obtain a metric. In our example, we have ha,b = db,a =. See Barnsley [1] for a proof that h is a metric on HR n. The corresponding metric space, HR n,h, is then itself a complete metric space [1]. 3 Betweenness in HR n Betweenness will play a major role in this paper, allowing us to find new integer sequences. In this section we define betweenness in HR n, mimicking the idea of betweenness in R n under the Euclidean metric. It is in this context that we will later define polygonal chains and introduce an infinite family of new integer sequences. Before we discuss betweenness, we define the dilation of a set, which will play a critical role in what follows. Definition. the set Let A HR n and let s > 0 be a real number. The dilation of A by s is A s = {x R n : d E x,a s for some a A}. As a simple example, the dilation of a single point set A = {a} by s is the closed ball centered at a of radius s. In general, the s-dilation of the set A is the union of all closed s balls centered at points in A. Using dilations, we can alternatively define ha,b as the minimum value of s so that the s-dilation of A encloses B and the s-dilation of B encloses A.

3 A C C* B A Figure 1: Infinitely many elements at the same location between A and B. An important and useful result about dilations is the following Theorem 4 from the article [6]. Theorem 3. Let A HR n and let s > 0. Then A s is a compact set with ha, A s = s. Moreover, if C HR n and ha,c s, then C A s. In other words, the element A s is the largest element in the sense of containment that is a distance s from A. Now we discuss betweenness. In the standard Euclidean geometry, a point x lies between the points a and b if and only if d E a,b = d E a,x + d E x,b. We extend this idea to HR n. Definition 4. Let A,B HR n with A B. The element C HR n lies between A and B if ha,b = ha,c + hc,b. Notation: We write ACB to signify that C is between A and B as in Blumenthal [5]. Betweenness in R n is quite different than betweenness in HR n. Consider the following examples. Example 5. Let A the union of the origin with the circle centered at 0,0 with radius 10 and B the circle centered at 0,0 with radius 5 in HR as shown in Figure 1. Then ha,b = da,b = db,a = 5. The union C C of the two dashed circles is the element A B 3. Theorem 3 shows that any element X satisfying AXB with ha,x = s and hb,x = ha,b s must be a subset of A s B ha,b s. It is not difficult to see that if X c = {c} C where c is any point on the circle C, then X satisfies AXB with ha,x =. So in this example, there are infinitely many different elements X satisfying AXB and ha,x =. Example 6. Let A = {, 0, 0, 0,, 0} and B = { 1, 0, 1, 0} in R as shown in Figure. Let 0 < s < ha,b and let t = ha,b s. In this case, the set C = A s B t is the four point set C = {c 1,c,c 3,c 4 }. Since any element C that lies between A and B at the distance s from A is a subset of C, there are at most 16 elements C that could satisfy AC B with ha,c = s. However, not every subset of C has this property. For example, if C = {c 1 }, then ha,c = 3 + t and C does not satisfy AC B. We leave it to the reader to show that the sets C, {c 1,c,c 4 } and {c 1,c 3,c 4 } are the only sets that lie between A and B at the distance s from A. It is this situation in which we have only finitely many elements 3

4 B t B t A s A s A s A c 1 B c A c 3 B c 4 A Figure : Exactly 3 different elements at the same location between A and B. between two fixed elements at a specific distance from one of them that will be of interest to us. The next definition formalizes what is meant when we refer to two elements at the same location between sets A and B. Definition 7. Let A,B HR n with A B and let C,C HR n satisfy ACB and AC B. The elements C,C are said to be at the same location between A and B if ha,c = ha,c = s for some 0 < s < ha,b. 4 Finding Points Between A and B Let A,B HR n. If a set C HR n satisfies ACB with ha,c = s, then it follows from Theorem 3 that C A s B ha,b s. As we saw in Examples 5 and 6, for some elements A and B there are infinitely many different elements at the same location between A and B and for other A and B there are only finitely many elements at the same location between them. The latter situation is of interest to us. In article [4], the authors show that in order for a pair of elements A and B in HR n to have a finite number of elements in HR n at a given location between A and B, every point in A must be a common distance r = ha,b away from B that is, da,b = r for all a A and every point in B must be that same distance r from A db,a = r for all b B. If this condition is not met, there will be infinitely many elements in HR n at each location between A and B. If a pair of elements A,B HR n satisfies this condition, we call the pair a configuration. Definition 8. A finite configuration is a pair [A,B] of finite sets A,B HR n where da,b = db,a = ha,b for all a A and b B. In article [4], the authors also show that if [A,B] is a finite configuration, then the number of elements C satisfying ACB and ha,c = s is independent of s for all 0 < s < ha,b. Thus, we denote by #[A,B] the number of elements between A and B in a finite configuration [A,B] at any distance s. Note that if [A,B] is a finite configuration, then C = A s B ha,b s will be a finite set for any 0 < s < ha,b, so there are only finitely many subsets of C and #[A,B] must be finite by Theorem 3 see Figure for an example. There are infinite configurations, but in this paper we will only be concerned with finite ones. 4

5 Figure 3: The P 4 configuration. 5 String and Polygonal Configurations Two basic configurations that will be relevant to our work are string and polygonal configurations. A string configuration is any finite set of points equally spaced on a line segment, with sets A and B consisting of alternate points on the line [9]. Figure is an example. If A B = k, then we will call the string configuration [A,B] a k string and denote it as S k. So Figure shows a 5-string or S 5. In article [9] Theorem 6.1 the authors show #S k = F k 1. In Figure, the elements {c 1,c,c 3,c 4 }, {c 1,c,c 4 }, and {c 1,c 3,c 4 } are the elements that satisfy ACB at the indicated location in this example. This is as expected, since #S 5 = F 4 = 3. Another basic family of configurations is the collection of polygonal configurations. Let A and B be the alternate vertices of a regular m-gon with m N. We denote the configuration [A,B] in this case by P m and call this a polygonal configuration [9]. An example of P 4 is shown in Figure 3, with the set A as the filled points and the set B as the open points. Line segments are drawn between points to indicate the pairs of points a A and b B satisfying d E a,b = ha,b. These points will be important in what follows. Definition 9. Let [A,B] be a finite configuration. The points a A and b B are adjacent if d E a,b = ha,b. If a A, the adjacency set of a, [a] B, is defined as [a] B = {b B : d E a,b = ha,b}. Again in the paper [9] Theorem 6.3 the authors show #P m = F m + F m 1 = L m. String and polygonal configurations form the building components of polygonal chains. 6 Polygonal Chains A polygonal chain PC k is a configuration obtained by connecting k copies of P m at antipodal points with string configurations of a fixed length l + 1. When k = 1, we define PC 1 to be P m. Examples of polygonal chains are shown in Figure 4. In this section we determine recursive formulas to find #PC k. 5

6 Figure 4: Top: PC 3,3 Bottom: PC 3,. 6.1 Adjoining Configurations To compute #PC k, we will ultimately reduce the problem to one of determining #X for smaller configurations X created from adjoins of polygonal and string configurations. Before discussing adjoins, we need to understand equivalent configurations. The idea here is that in the computation of #[A,B] the distance between adjacent points is irrelevant - only adjacencies matter. Definition 10. The finite configuration [X, Y ] is equivalent to the finite configuration [A,B] if there are bijections f : A X and g : B Y such that for all a A and b B, we have a adjacent to b if and only if fa is adjacent to gb. When [X,Y ] is equivalent to [A,B] we write [X,Y ] [A,B]. It is easy to show that the relation is an equivalence relation on the set of finite configurations. One important result involving equivalent configurations is that if [X, Y ] and [A,B] are equivalent configurations, then #[X,Y ] = #[A,B] [4] Theorem 4.1. Now we adjoin a point to a configuration. Definition 11. Let [A, B] be a finite configuration. The configuration [A, B]a, y obtained by adjoining a point y A B to [A,B] at the point a A is the configuration [A,B ], where B = B {y}, d E y,a = ha,b, and d E y,x > ha,b for all x A B with x a. For example, Figure 5 shows the configuration obtained by adjoining a point y to P 4 at point a. One tool for computing #[A,B]a,y for a finite configuration [A,B] is the following [9] Theorem 6.. a y Figure 5: Adjoining a point to P 4. 6

7 a c 0 y Figure 6: Adjoining P and P 3. Theorem 1. Let [A,B] be a finite configuration. Define a new configuration X by adjoining a point y to [A,B] at the point a A. Furthermore, assume a is adjacent to k points in B, each of which is adjacent to at least one more point in A. Then #X = #[A,B] + #[A {a},b]. As an example, if [A,B] = P 4 and X = [A,B]a,y as shown in Figure 5, then #X = #P 4 + #S 7 = L 8 + F 6 = 55. We can also adjoin two configurations. Definition 13. Let [A,B] and [X,Y ] be finite configurations and let a A and y Y. Let [X,Y ] be a configuration equivalent to [X,Y ] through bijections f : X X and g : Y Y so that 1. hx,y = ha,b,. d E gy,a = ha,b, 3. dz,a B > ha,b for all z X Y, z gy, and 4. dw,x Y > ha,b for all w A B, w a. The adjoin [A,B][a] [X,Y ][y] of the configurations at the points a and y is the configuration [A X,B Y ]. As an example, the adjoin of P with P 3 at the points a and y is shown in Figure 6. Notation: In this paper, we will consider configurations obtained by adjoining string configurations at their endpoints to polygonal configurations. If we adjoin a single string configuration to a polygonal configuration, the point of adjoin does not matter due to symmetry. The notation we use for adjoining a string configuration S l to a polygonal configuration P m is P m S l. There are cases in which we will adjoin more than one string configuration to Figure 7: P 3 {S 1 [1]; S 1 [4]}. 7

8 a polygonal configuration. In these situations, we will label the vertices of a polygonal configuration counterclockwise in order the starting point does not matter due to symmetry. The notation we will use to adjoin string configurations of length s and t to a polygonal configuration P m at the points labeled i and j, respectively is P m {S s [i],s t [j]}. An illustration of P 3 {S 1 [1];S 1 [4]} is shown in Figure 7. The next theorem shows us a method for computing #X by breaking X into smaller disconnected configurations. A configuration [A,B] is connected if, given a A and b B, there is a sequence {x k } m k=1 A B such that x 1 = a, x m = b and x i is adjacent to x i+1 for each i [4]. In other words, there is a string in A B connecting a to b. A configuration is disconnected if it is not connected. The proof of Theorem 15 depends on the next result from the paper [4] Theorem 6.1. Theorem 14. Let [A,B] be a configuration and let r = ha,b. If [X,Y ] is a configuration with ha,b = hx,y, d E a,y > ha,b for all a A and y Y, and d E b,x > ha,b for all b B and x X, then #[A X,B Y ] = #[A,B]#[X,Y ]. In other words, if [U,V ] is a disconnected configuration constructed from configurations [A,B] and [X,Y ], with U = A X and V = B Y, then #[U,V ] = #[A,B]#[X,Y ]. The next theorem describes another important computation technique. Theorem 15. Let [A,B] and [X,Y ] be finite configurations and let [U,V ] = [A,B][a] [X,Y ][y] for some a A and y Y. Then #[U,V ] = #[A,B] #[X,Y ] + #[A S 1 [a],b] #[X,Y S 1 [y]]. Proof. Without loss of generality, we assume ha,b = hx,y. Let s,t > 0 with s + t = ha,b. Let C = A s B t be the largest element that lies between A and B at a distance s from A, and let Z = X s Y t be the largest element that lies between X and Y at a distance s from X. Then in the configuration [U,V ], the largest element between A X and B Y will be C Z {c 0 }, where c 0 is the point which lies between the newly adjacent points a and y. See Figure 6 for an illustration. We now consider which subsets D of C Z {c 0 } satisfy UDV. Since D satisfies UDV, we know that [a ] D, [b ] D, [x ] D, and [y ] D for all a A,b B,x X, and y Y. We consider two cases. Case 1 c 0 D. In this case, let C = {d D : d [a ] D for some a A} and let Z = D C. By construction, if c C, then c c 0 and so c C. Thus C C. If z Z, then z c 0 and z Z. Now, if a A and b B, we know [a ] D and [b ] D. Therefore, [a ] C and [b ] C. So C satisfies AC B. Similarly, Z satisfies XZ Y. By Theorem 14, the number of such sets D = C Z is #[A,B] #[X,Y ]. Case c 0 D. In this case, we have c 0 [a] D and c 0 [y] D. The existence of the point c 0 in D allows for the potential removal of other points in D that are adjacent to a or y. In other words, D = C {c 0 } Z, where C satisfies AC B S 1 [a] and Z satisfies X S 1 [y]z Y. The number of such sets D is #[A S 1 [a],b] #X,Y S 1 [y]]. By combining the results of the two cases here, we arrive at the expected formula #[U,V ] = #[A,B] #[X,Y ] + #[A S 1 [a],b] #[X,Y S 1 [y]]. 8

9 Figure 8: PC 3,3 S 1. As an example, consider the configuration X obtained by adjoining P 3 and P as shown in Figure 6. Theorem 15 shows #X = #P 3 #P + #P 3 S 1 #P S 1. Now Theorem 1 gives us #P 3 S 1 = #P 3 + #S 5 = = 1 and #P S 1 = #P + #S 3 = = 8. So 7 Finding #PC k #X = = 94. In this section, we determine a pair of recursive formulas to find #PC k. As the first step in our computations, we determine #X S t for t for any finite configuration X in terms of #X and #X S 1. Theorem 16. Let X be a finite configuration and t. Then #X S t = F t #X S 1 + F t 1 #X. Proof. Using Theorem 1, we see that our theorem is certainly true with t = and t = 3. We proceed by induction and suppose our theorem is true for all s, 1 s t. Then #X S t+1 = #X S t + #X S t 1 = [F t #X S 1 + F t 1 #X] + [F t 1 #X S 1 + F t #X] = F t + F t 1 #X S 1 + F t 1 + F t #X = F t+1 #X S 1 + F t #X. By PC k S t we mean the adjoin of S t with PC k at a point antipodal to the connected strings. An example of PC,3 3 S 1 is shown in Figure 8. A direct consequence of Theorem 16 is Corollary 17. For m, k 1, l 1, and t, #PC k S t = F t #PC k S 1 + F t 1 #PC k Now we add in the case when t = 1 for polygonal configurations. Note first that L m + F m = F m + F m 1 + F m+1 = F m + F m+1 = F m+. 9

10 Corollary 18. For m and t 1, #P m S t = F t F m+ + F t 1 L m. Proof. When t = 1, a direct application of Theorem 1 gives us #P m S 1 = #P m + #S m 1 = L m + F m = F m+ = F 1 F m+ + F 0 L m. For t, we use Theorem 16 to obtain #P m S t = F t #P m S 1 + F t 1 #P m = F t F m+ + F t 1 L m. We need one more preinary calculation before we determine #PC k. Lemma 19. For m, #P m {S 1 [1];S 1 [m + 1]} = F m+. Proof. Figure 7 shows an example of the configurations in question. By Theorem 1 and Corollary 18, #P m {S 1 [1];S 1 [m + 1]} = #P m S 1 + #S m 1 S 1 [m] We use the identity = F m+ + #S m 1 + #S m 1 = F m+ + F m + F m. F m = F m+1 F m 1 for m 1 see [7], Corollary 5.4 or [3], Identity 14 note that f m = F m+1 to obtain F m+ + F m + F m = F m+1 + F m 1 + F m = F m+ F m + F m F m + F m = F m+. Now we derive recursive formulas for #PC k and #PCk S 1 in terms of #PC k 1 and #PC k 1 S 1. Remember that #PC 1 = #P m = L m and #PC 1 S 1 = #P m S 1 = F m+. To include all cases together, we extend the Fibonacci sequence in the negative direction by defining F 1 = 1. We also define X S 0 to be X for any finite configuration X. Theorem 0. For m, k, and l 1, #PC k = L m F l + F m+ F l 1 #PC k 1 + L mf l 1 + F m+ F l #PC k 1 S 1 where #PC k S 1 = F m+ F l + F l 1 Fm+ #PC k 1 + F m+ F l 1 + F l Fm+ #PC k 1 S 1. 10

11 * * Figure 9: Using Theorem 15 to compute #PC k and #PCk S 1. Proof. We use Theorem 15 applied at the adjacent points indicated with an asterisk as illustrated in Figure 9, and Corollary 17: #PC k = #P m #PC k 1 S l 1 + #P m S 1 #PC k 1 S l = L m Fl 1 #PC k 1 S 1 + F l #PC k 1 + F m+ Fl #PC k 1 S 1 + F l 1 #PC k 1 = L m F l 1 + F m+ F l #PC k 1 S 1 + L m F l + F m+ F l 1 #PC k 1. Finally, we determine the recursive formula for #PC k S 1 by using Theorem 15, applied at the adjacent points indicated with an asterisk in Figure 9. This gives us #PC k S 1 = #PC k 1 S l 1 #P m S 1 + #PC k 1 S l #P m {S 1 [1];S 1 [m + 1]}. Therefore, Lemma 19 and Corollary 17 yield #PC k S 1 = F m+ #PC k 1 S l 1 + #PC k 1 S l Fm+ = F m+ Fl 1 #PC k 1 S 1 + F l #PC k 1 + F l #PC k 1 S 1 + F l 1 #PC k 1 Fm+ = F m+ F l 1 + F l Fm+ #PC k 1 S 1 + F m+ F l + F l 1 Fm+ #PC k 1. As examples of the use of Theorem 0, #PC 3,1 = L 6 #PC 1 3,1 + F 8 #PC 1 3,1 S 1 = 18#P 3 + 1#P 3 S 1 = = 765, #PC, = F m+ #PC k 1 + [L m + F m+ ] #PC k 1 S 1 = 8#P + [7 + 8]#P S 1 = = 176, 11

12 Sloane A-number i A1597 #P,1 i A1598 #Pi, A1599 #P,i A15930 #P, i A15931 #Pi, A1593 #P3,i Table 1: Some polygonal chain sequences and #PC,3 = [L 4 F 1 + F 6 F ] #PC 1,3 + [L 4 F + F 6 F l ] #PC 1,3 S 1 = [ ]#P + [71 + 8]#P S 1 = = Polygonal Chain Sequences Theorem 0 gives us recursive formulas for finding #PC k. By fixing any two of m,k, or l and letting the other parameter vary, we create infinite families of integer sequences. Examples of the first few terms of several of these sequences are given in Table 1. We suspect that it is possible to define many new types of configurations in this geometry that provide more interesting and previously unknown integer sequences. 9 Asymptotic Behavior Polygonal chain sequences appear to exhibit asymptotically exponential behavior as the sequences of successive quotients in Table illustrate. These sequences are natural extensions of string and polygonal sequences, both of which involve Fibonacci and Lucas numbers, so we might expect the ratios of polygonal chain sequences to do so as well. To understand this behavior, we investigate these quotients. 9.1 Fixing m and l In this section we consider the asymptotic behavior of the sequences obtained by letting k vary for fixed values of m and l. For easier notation, let X k, = #PC k and Y k, = #PC k S 1. Theorem 0 shows that X k, = a X k 1, + b Y k 1, and Y k, = c X k 1, + d Y k 1, 1 1

13 i #P i+1,1 #P,1 i #Pi+1,1 #Pi, #P,i+1 #P,i #P i+1, #P, i #Pi+1, 3 #Pi, #P3,i+1 3 #P3,i Table : Ratios of polygonal chain sequences to 8 decimal places. for values of a,b,c, and d that depend on only m and l. We can determine the asymptotic behavior of X k, by writing this system in matrix form: [ ] [ ] [ ] Xk, a b = Xk 1,. [ ] a b Let M =. Then c d The eigenvalues of M are Y k, [ Xk, Y k, ] c d Y k 1, [ ] = M k 1 X1,. Y 1, and λ = d + a + d a + 4b c λ = d + a d a + 4b c. Since d a + 4b c > 0, we see that [ M has ] two distinct real eigenvalues. Thus λ 0 M is diagonalizable to the matrix D =. If Q 0 λ is the matrix whose columns are the eigenvectors of M, then [ ] [ ] Xk, λ = Q k 1 [X1, ] 0 Y k 1 Q 1. k, 0 λ Y 1, Note that Q = [ 1 1 λ a b d λ b ] and Q 1 = 1 λ a +d [ λ d b λ a b ]. 13

14 Since X 1, = L m and Y 1, = F m+, we have where R = S = 1 λ a +d 1 λ a +d X k, = R λ k 1 + S λ k 1 λ d L m + b F m+ and λ a L m b F m+. So provides a non-recursive formula for computing X k,. Now λ > λ, so we have as k. Thus, X k, = R λ k 1 X k 1, = R λ k + S λ k 1 + S λ k R λ + S λ λ λ k R λ R = λ k R + S λ λ k X k, X k 1, = λ. The values of a,b,c, and d are given by Theorem 0: a = L m F l + F m+ F l 1, b = L m F l 1 + F m+ F l, c = F m+ F l + F l 1 F m+, and d = F m+ F l 1 + F l F m+. Thus the sequence of polygonal chain numbers as k varies for fixed m and l behaves asymptotically as the exponential sequence with ratio λ. For example, λ,1 = and λ, = compare to the ratios in Table. 9. Limits of Fibonacci and Lucas Sequences To determine the asymptotic behavior of the sequences {X k, } as either m or l vary, we will need some its of sequences of quotients of Fibonacci and Lucas numbers. Let ϕ = 1+ 5 be the golden ratio. The following results concerning Fibonacci and Lucas numbers are well known. F m+1 = ϕ [7] Corollary 8.5, [8] 5 F m L m+1 = ϕ [7] Corollary 8.7, [8] 5 L m ϕ m = F m ϕ + F m 1 for m 1 [7] Lemma 5.1, [3] Corollary 33 14

15 F m+j = ϕ j for j 0 [3] Corollary 31, [8] 7 F m L m+j For j 0, = ϕ j [8] 7 L m We will also need other its of ratios of Fibonacci and Lucas numbers. Lemma 1. For m and j 0, L m 1. For j 0, = 1 + ϕ F m+j ϕ j+1 F m+. Fm+ F m+ 3. Fm+3 F m+4 4. Fm+ = ϕ4 1 ϕ 4 = ϕ4 1 ϕ 6 = ϕ4 1 ϕ = 1 + ϕ ϕ 3 = 1 + ϕ ϕ 5 = 1 + ϕ ϕ 5. L m Fm+ = ϕ + 1 ϕ 6 Proof. We prove and 5, the remainder are similar. Recalling that ϕ 1 = ϕ, F m+ Fm+ Fm+ Fm 1 1 = 1 ϕ = ϕ4 1 ϕ 4 = 1 + ϕ ϕ ϕ 4 = 1 + ϕ ϕ 3. Fm+ Fm F m+ 15

16 Finally, we use Identity 13 from the book [3] F m+1 = F m + F m+1 to see L m Fm+ F m 1 + F m+1 Fm+ Fm 1 + F m + Fm + Fm+1 Fm+ Fm 1 Fm + + = 1 ϕ 3 F m+ + = 1 + ϕ + ϕ 4 ϕ 6 = ϕ + 1 ϕ 6. 1 ϕ + F m+ 1 ϕ Fm+1 F m+ 9.3 Fixing k and l In this section, we fix k and l and let m vary and investigate the asymptotic behavior of the resulting sequence X k,. We will need to know its of ratios of the terms a,b,c and d as m approaches infinity. Lemma. Let G l = 1+ϕ F l 1 +ϕ 3 F l 1+ϕ F l +ϕ 3 F l 1. Then for l 1, a m+1,l 1. = ϕ a b m+1,l. a b 3. = G l a c 4. = a c m+1,l 5. = ϕ G l ϕ3 1 + ϕ = ϕ5 a 1 + ϕ d ϕ 3 6. = a 1 + ϕ d m+1,l ϕ 5 7. = a 1 + ϕ G l. G l. 16

17 Proof. We prove 5, the others are similar: c m+1,l a F m+4 F l + F m+3f l 1 L m F l + F m+ F l 1 Fm+4 F m+ F l + L m F m+3 F m+ F l 1 F m+ F l + F l 1 ϕ ϕ F l + 5 F ϕ = +1 l 1 1+ϕ ϕ 5 = 1 + ϕ = ϕ5 1 + ϕ. ϕ 3 F l + F l 1 1+ϕ F ϕ 3 l + F l 1 1+ϕ F ϕ 3 l + F l 1 Before we can analyze the asymptotic behavior of the polygonal chain sequence, we need to understand how the sequences {Y k, } and {X k, } are related as m varies. Lemma 3. For k 1, Y k, X k, = ϕ3 1 + ϕ. Y 1, F m+ 1 Proof. When k = 1 we have = X 1, 1 L m + 1 = ϕ3 1 + ϕ. So the claim is ϕ 3 ϕ true for k = 1. We proceed by induction and assume the claim is true for some k 1. Then 1 shows Y k+1, c X k, + d Y k, X k+1, a X k, + b Y k, c d Yk, a + a b Yk, 1 + a X k, ϕ 3 ϕ + 3 ϕ G 3 1+ϕ = 1+ϕ l 1+ϕ ϕ 1 + G 3 l = ϕ3 1 + ϕ. ϕ +1 X k, 17

18 Now we can identify the asymptotic behavior of the polygonal chain sequence when m varies. Theorem 4. For k 1 and l 1, X k,m+1,l X k, Proof. The proof will be by induction on k. L m+ ϕ 5 L m = ϕ k Y k,m+1,l and X k, = ϕ Y 1,m+1,l. Also, Y 1, = F m+ and X 1, = ϕk ϕ. X 1,m+1,l Now X 1, = L m and = X 1, 1 F m+4 = L m = ϕ 5 ϕ ϕ. Thus, our theorem is true when k = 1. Assume the theorem is true for some k 1. Then and X k+1,m+1,l X k+1, a m+1,l X k,m+1,l + b m+1,l Y k,m+1,l a X k, + b Y k, am+1,l Xk,m+1,l a m,1 X k, + b Yk, 1 + a m,1 ϕ ϕ k + ϕ ϕ G l k+3 1+ϕ = ϕ 1 + G l 3 = ϕ k+ 1+ϕ bm+1,l Yk,m+1,l a X k, X k, Y k+1,m+1,l X k+1, c m+1,l X k,m+1,l + d m+1,l Y k,m+1,l a X k, + b Y k, cm+1,l a Xk,m+1,l 1 + ϕ ϕ 5 k + 1+ϕ = 1 + G l = ϕk ϕ. X k, + dm+1,l a b Yk, a X k, ϕ 5 1+ϕ ϕ 3 1+ϕ ϕ G k+3 l 1+ϕ Yk,m+1,l X k, So the sequence of polygonal chain numbers as m varies for fixed k and l behaves asymptotically as the exponential sequence with ratio ϕ k. For example, ϕ and ϕ compare to the ratios in Table. Note that the its in these cases are independent of the value of l. 18

19 9.4 Fixing k and m In this section, we fix k and m and let l vary and investigate the asymptotic behavior of the resulting sequence X k,. Again, we begin with its of sequences of ratios of a,b,c, and d, this time as l goes to infinity. Lemma 5. Let H m = F m++fm+ ϕ L m +F m+. Then for m, ϕ 1. a +1 a = ϕ. b a = ϕ 3. b +1 a = ϕ 4. c a = H m 5. c +1 a = ϕh m 6. d +1 a = ϕ H m. Proof. We verify 5 and leave the others for the reader: c +1 a F m+ F l 1 + F m+f l L m F l + F m+ F l 1 F F m+ + F l m+ F l 1 L Fl m F l 1 + F m+ = F m+ + F m+ϕ 1 L m + F ϕ m+ Fm+ + F = ϕ m+ϕ. L m + F m+ ϕ Now we need to understand the relationship between the sequence {Y k, } and the sequence {X k, } as l goes to infinity. Lemma 6. For k, m, Y k, X k, = H m. 19

20 Proof. When k = we have Y, X, c X 1, + d Y 1, a X 1, + b Y 1, c a X 1, + X 1, + d a Y 1, b a Y 1, = H ml m + ϕh m F m+ L m + ϕf m+ = H m. So the lemma is true for k =. We proceed by induction and assume the lemma is true for some k. Then 1 shows Y k+1,+1 X k+1,+1 c X k, + d Y k, a X k, + b Y k, c d Yk, a + a b Yk, 1 + a X k, = H m + ϕh m H m 1 + ϕ H m = H m. X k, Now we determine the asymptotic behavior of the polygonal chain sequences if we allow l to vary. Theorem 7. For k and m, X k,+1 X k, = ϕ k 1 and Y k,+1 X k, = ϕ k 1 H m. Proof. The proof will be by induction on k. Recall that X 1, = L m and Y 1, = F m+ We first verify the k = case. Equation 1 shows X,+1 X, a +1 X 1,+1 + b +1 Y 1,+1 a X 1, + b Y 1, a+1 a X 1,+1 + X 1, + = ϕl m + ϕ F m+ L m + ϕf m+ = ϕ b+1 a Y 1,+1 b a Y 1, 0

21 and Y,+1 X, Now the general cases: c +1 X 1,+1 + d +1 Y 1,+1 a X 1, + b Y 1, c+1 a X 1,+1 + X 1, + d+1 a Y 1,+1 b a Y 1, = ϕ H ml m + ϕ H m F m+ L m + ϕf m+ = ϕh m. and X k+1,+1 X k+1, a +1 X k,+1 + b +1 Y k,+1 a X k, + b Y k, a+1 Xk,+1 a X k, b+1 Yk,+1 a X k, b Yk, a X k, = ϕ ϕ k 1 + ϕ ϕ k 1 H m 1 + ϕ H m = ϕ k Y k+1,+1 X k+1, c +1 X k,+1 + d +1 Y k,+1 a X k, + b Y k, c+1 Xk,+1 d+1 a X k, a b Yk, a X k, = ϕh m ϕ k 1 + ϕ H m ϕ k 1 H m 1 + ϕ H m = ϕ k H m. Yk,+1 X k, So the sequence of polygonal chain numbers as l varies for fixed k and m behaves asymptotically as the exponential sequence with ratio ϕ k 1. For example, ϕ and ϕ compare to the ratios in Table. Note that the its in these cases are independent of the value of m. 1

22 10 Conclusion The asymptotic growth rates of polygonal chain sequences are functions of the golden ratio ϕ. Since polygonal chains are constructed from string and polygonal configurations and #S l+1 #P m+1 = ϕ and = ϕ, this should not be surprising in hindsight. #S l #P m That the asymptotic ratios of the polygonal chain sequences have such nice closed forms is, however, somewhat surprising given the complex nature of betweenness in this geometry. Polygonal chain sequences provide a three parameter family of integer sequences previously unidentified in The On-Line Encyclopedia of Integer Sequences [10]. These sequences have interesting geometric interpretations as the number of compact sets at each location between the component sets that make up the polygonal chain. We suspect that there are many other similar families of unknown sequences just waiting to be found. 11 Acknowledgements This work was supported by National Science Foundation grant DMS The authors also thank Grand Valley State University for hosting and supporting a Research Experiences for Undergraduates program. References [1] M. Barnsley, Fractals Everywhere, Second Edition, Academic Press Professional, [] C. Bay, A. Lembcke, and S. Schlicker, When lines go bad in hyperspace, Demonstratio Math., , [3] A. T. Benjamin and J. J. Quinn, Proofs That Really Count, Mathematical Association of America, 003. [4] C. Blackburn, K. Lund, S. Schlicker, P. Sigmon, and A. Zupan, A missing prime configuration in the Hausdorff metric geometry, to appear, J. Geom. [5] L. M. Blumenthal, Theory and Applications of Distance Geometry, Oxford University Press, [6] D. Braun, J. Mayberry, A. Powers, and S. Schlicker, A singular introduction to the Hausdorff metric geometry, Pi Mu Epsilon J., 1 No. 3, 005, [7] T. Koshy, Fibonacci and Luca Numbers with Applications, John Wiley & Sons, 001. [8] E. Lucas, Théorie des fonctions numériques simplement périodiques, Amer. J. Math , , [9] K. Lund, P. Sigmon, and S. Schlicker, Fibonacci sequences in the space of compact sets, Involve, 1 No. 008,

23 [10] N. J. A. Sloane, 007, The On-Line Encyclopedia of Integer Sequences, published electronically at Mathematics Subject Classification: Primary 51F99; Secondary 11B83. Keywords: Hausdorff metric, configuration, metric geometry, polygonal chains, Fibonacci numbers, Lucas numbers. Concerned with sequences A00003, A000045, A1597, A1598, A1599, A15930, A15931, and A1593. Received August 008; revised version received December 008. Published in Journal of Integer Sequences, December 008. Return to Journal of Integer Sequences home page. 3

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