Some Results Concerning Uniqueness of Triangle Sequences

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1 Some Results Concerning Uniqueness of Triangle Sequences T. Cheslack-Postava A. Diesl M. Lepinski A. Schuyler August Abstract In this paper we will begin by reviewing the triangle iteration. We will then investigate the question of when a triangle sequence corresponds to a unique ordered pair α β. It will be shown that certain classes of triangle sequences most importantly those that are bounded do yield unique ordered pairs. As a corollary to our work on uniqueness we will also show some clean relations dealing with the geometry of the partition triangles. 1 Introduction This paper will start with an overview of the standard triangle iteration as outlined in work by Garrity [1]. We will begin with the geometrical interpretation of triangle sequences. We define an iteration T on the triangle = {x y : 1 x y > 0}. This triangle is partitioned into an infinite set of disjoint subtriangles k = {x y : 1 x ky 0 > 1 x k + 1y} where k is any nonnegative integer. Given α β k define the map T : {x 0 : 0 x 1} by β T α β = α 1 α kβ. α The triangle sequence is recovered from this iteration by keeping track of the number of the triangle that the point is mapped into at each step. In other words if T k 1 α β ak then the point α β will have the triangle sequence a 1 a We will recursively define a sequence of vectors as follows: Set C 2 = C 1 = and C 0 = Let the components of C n be denoted by C n = p n q n r n. Then let C n = C n 3 C n 2 a n C n 1. 1

2 These C n vectors can be thought of as integer vectors approximating the plane x + αy + βz = 0. We thus refer to the C n vectors as approximation vectors. We define positive numbers d n in the following manner: d n = 1 α β C n. These are interpreted as the distance from the vector C n to the plane x + αy + βz = 0. It is then easy to see that the following recursion relations hold: d n = d n 3 d n 2 a n d n 1 p n = p n 3 p n 2 a n p n 1 q n = q n 3 q n 2 a n q n 1 r n = r n 3 r n 2 a n r n 1 This paper will focus on the problem of discovering when a particular sequence a 1 a 2... corresponds to a unique pair α β which has implications in determining if α and β are cubic irrationals. We consider only the case of infinite triangle sequences because it is trivial to show that all finite triangle sequences correspond to an infinite number of points and are thus not unique. The proof of this assertion will appear in section 2 after appropriate notation has been introduced. There are two main ways to frame the question of uniqueness. The first relies on vector algebra. We know that the vectors C n are approaching the plane x + αy + βz = 0. It is also easy to see that the C n vectors are determined completely by the sequence a 1 a Thus if it can be shown that the cross product C n C n+1 of two successive approximation vectors approaches the vector 1 α β then the sequence is unique. In other words if lim C n C n+1 = 1 α β C n C n+1 then the numbers α and β will be uniquely determined by the triangle sequence a 1 a The second approach to understanding the uniqueness problem relies on the geometry of the actual triangle iteration. We consider the original partitioning of the triangle into an infinite disjoint union of subtriangles: = k=0 It is clear from the definition of triangle sequences that k consists of all points α β whose triangle sequences have first term equal to k. In other words k k = {x y : a 1 = k}. 2

3 We can further partition each k in a way similar to that in which we partitioned. Define kl in the following way: kl = {x y k : T x y l }. Thus kl is the set of all points in k that map to l. We can think of this set in the following way: kl = {x y : a 1 = k and a 2 = l}. We can further partition the triangles in the same way. We define a1 a 2... a n to be the set of points α β whose triangle sequences have a 1 a 2... a n as their first n terms. This gives us a method for investigating uniqueness. If we can say that the set of points with the infinite triangle sequence a 1 a 2... consists of only one point α β then we can say that α β is the unique point with that triangle sequence. One way of investigating the set a1a 2... is to look at the limit of the sets a1 a 2... a n as n approaches infinity. Section 2 of this paper uses a geometric argument to show that there do in fact exist certain infinite triangle sequences which correspond to an infinite number of points α β. This will be demonstrated by showing that in certain cases the set a1 a 2... consists of an entire line segment. Section 3 is devoted to a development of the vector approach to the uniqueness problem. We will define new notation for cross products of successive approximation vectors. We will then prove a new recursion relation and develop algebraic conditions that guarantee uniqueness. Section 4 uses the algebraic conditions developed in section 3 in order to prove uniqueness of certain bounded triangle sequences. Specifically we prove that an infinite triangle sequence a 1 a 2... corresponds to a unique ordered pair α β for the case a i = A for all i and some positive integer A. Section 5 explores the connections between the vector algebra and the geometry of the partitioning of the triangles. We show that there is a clean relation between the cross product vectors and the vertices of the partition triangles a1 a 2... a n. We will also show that there is an interesting relationship between the vertices of a1 a 2... a n and the vertices of an a n 1... a 1. These results are interesting in their own right but will be mainly used as technical lemmae in proving the main theorem of section 6. Section 6 deals with the case of proving the uniqueness of a larger class of triangle sequences including the entire bounded case. Specifically we show that for any infinite sequence in which the integer C appears infinitely many times that sequence corresponds to a unique pair α β. We use both geometric and algebraic arguments to prove this result. 2 Existence of Non-unique Triangle Sequences In this section we show using arguments based on the geometry of the triangle iteration that there exist infinite sets of points that all have the same infi- 3

4 nite triangle sequence. We begin with some notation to describe the partition triangles. Definition 1 Define T n a 1...a n α β to be the unique point x y a1... a n such that T n x y = α β. Proposition 1 Given any finite triangle sequence a 1... a n there exists a line segment such that all points on that segment have the sequence a 1... a n. The set Ta n 1...a n I where I is the interval 0 1 is precisely the set of all points with triangle sequence equal to a 1... a n. Definition 2 For any finite sequence of nonnegative integers a 1... a n let da 1... a n = min{lengthta n 1...a n I lengthta n 1...a n 1a n+1 I} where I is the interval 0 1. In the figure below we have A = Ta n 1...a n I a = Ta n 1...a n +1 B = Ta n 1...a n+1 I b = n T C = T n a 1...a n +1 m = T n a 1...a n +1 1 l T n a 1...a n 1 0 c = T n 1 0 a 1...a n 1 0 a 1...a n 0 0 x 0 0 n = T n a 1...a n +1 Note: 0 < 1 l+1 < x < 1 Figure 1: Triangle created by the consecutive partition lines Ta n 1...a n I and Ta n I. 1...a n+1 4

5 Lemma 1 For any small ɛ > 0 and any sequence a 1... a n of nonnegative integers there exists a nonnegative integer l such that da 1... a n l da 1... a n ɛ. Case I: Consider the case where the angle at c is obtuse. Then all line segments from b to a point on B will be longer than A. Since the endpoints of consecutive partition lines are getting closer to c we have: da 1... a n l = lengtht n+1 a 1...a n l+1 I lengtht n a 1...a n I da 1... a n ɛ Therefore the lemma holds when c is obtuse. Case II: Consider the case where the angle at c is acute. Consider a line segment N in a1...a n with the following properties: i One endpoint is at Ta n 1...a n 1 0 point b in Figure 1. ii The other endpoint is at Ta n 1...a n+1 x 0 for some x 1 0 on line B in Figure 1. iii The length of the line segment is at least lengthta n 1...a n I ɛ. iv Given any m on the segment cn M > N. Properties i and ii will clearly be able to be satisfied. Since our line segment N gets longer as the endpoint n approaches c we know that we can satisfy property iii. By the above reasoning we also know that we will satisy property iv when m gets sufficiently close to c. Choose l Z + 1 n so that l+1 x. Then Ta 1...a n +1 1 l+1 0 point m in Figure 1 lies between c and n. So the line segment from b to m has length at least lengthta n 1...a n I ɛ = A ɛ. But this line segment is also T n+1 a I. 1...a nl Then since lengtht n+1 n+1 a 1...a n l+1i > lengthta 1...a n li we have: da 1... a n l = lengtht n+1 a 1...a n l I lengthta n 1...a n I ɛ da 1... a n ɛ 5

6 Theorem 1 There exist distinct points α 1 β 1 and α 2 β 2 in with identical infinite triangle sequences. Consider a triangle sequence constructed in the following manner: Choose a 1 so that da Choose a 2 so that da 1 a 2 da Choose a n so that da 1... a n da 1... a n n+1 for n 2. Our lemma guarantees that a sequence may be constructed in this way. Then for all n da 1... a n > 1 n+1 2. Therefore there is some line segment of length 1 2 which is inside any triangle a 1...a n. Thus every point on this line segment has the infinite triangle sequence a 1 a Properties of Cross Products of Approximation Vectors 3.1 Motivation We know that the approximation vectors C n approach the plane given by x + αy+βz = 0. Thus it is reasonable to suppose that the cross-products C n C n+1 would approach the vector 1 α β which is normal to the plane. Since the approximation vectors are completely determined by the sequence it is easy to see that the relation lim C n C n+1 = 1 α β C n C n+1 implies that the pair α β is completely determined by the sequence and thus that the sequence is unique. The above limit relation is in fact equivalent to the uniqueness of the sequence. As was shown in section 2 since a triangle sequence does not necessarily uniquely determine the pair α β it is not always true that the cross product approaches the normal vector 1 α β. However it is still useful to investigate the properties of these cross products as an approach to answering questions of uniqueness. 6

7 3.2 Notation and Results Definition 3 Let X n = C n C n+1. Set X n = x n y n z n. Lemma 2 X n+1 = X n 2 + a n+1 X n 1 + X n X n+1 = C n+1 C n+2 = C n+1 C n 1 C n a n+2 C n+1 = C n+1 C n 1 + C n C n+1 = C n 2 C n 1 a n+1 C n C n 1 + C n C n+1 = C n 2 C n 1 + a n+1 C n 1 C n + C n C n+1 = X n 2 + a n+1 X n 1 + X n Lemma 3 X n X n+1 = C n+1 We know that x n = q n r n+1 q n+1 r n y n = r n p n+1 r n+1 p n z n = p n q n+1 p n+1 q n Let us refer to the third component of X n X n+1 as w n+1. Then we know that w n+1 = x n y n+1 y n x n+1 = q n r n+1 q n+1 r n r n+1 p n+2 r n+2 p n+1 r n p n+1 r n+1 p n q n+1 r n+2 q n+2 r n+1 = r n+1 p n+2 q n r n+1 q n p n+1 r n+2 r n q n+1 p n+2 + r n r n+2 q n+1 p n+1 r n+1 r n+1 p n q n+2 p n q n+1 r n+2 r n q n+2 p n+1 r n r n+2 q n+1 p n+1 = r n+1 p n+2 q n r n+1 + p n q n+1 r n+2 + r n q n+2 p n+1 r n+1 p n q n+2 q n p n+1 r n+2 r n q n+1 p n+2 Therefore since detc n C n+1 C n+2 = 1 w n+1 = r n+1. Now since X n and X n+1 are both perpendicular to C n+1 we know that X n X n+1 = kc n+1 for some k. However since w n+1 = r n+1 we know that k = 1 and hence that X n X n+1 = C n+1. 7

8 Corollary 1 detx n 2 X n 1 X n = 1 detx n 2 X n 1 X n = X n 2 X n 1 X n = X n 2 C n Corollary 2 yn 1 x n 1 yn x n = r n x nx n 1 From Lemma 3 we know that = C n 2 C n 1 C n = detc n 2 C n 1 C n = 1 x n y n 1 y n x n 1 = r n. Dividing this expression by yields y n 1 y n x n 1 = r n. x n Definition 4 Let α n be defined so that d n 3 d n 2 α n d n 1 = 0. Definition 5 Let C n = C n 3 C n 2 α n C n 1. Definition 6 Let ɛ n = α n a n. From these definitions it should be clear that a n α n < a n ɛ n < 1 Additionally if we let C n 1 α β = 0 C n C n+1 X n x n ỹ n z n. Then and α = ỹn x n β = z n x n. 8

9 Lemma 4 X n = X n + α n ɛ n X n 1 + ɛ n X n 2 X n = C n C n+1 = C n 3 C n 2 α n C n 1 C n 2 C n 1 α n+1 C n = C n 3 C n 2 C n 3 C n 1 α n+1 C n 3 C n + C n 2 C n 1 α n+1 C n 2 C n α n C n 1 C n 2 + α n α n+1 C n 1 C n We now compute the following: C n 3 C n 1 = C n 3 C n 4 C n 3 a n 1 C n 2 = X n 4 a n 1 X n 3 C n 3 C n = C n 3 C n 3 C n 2 a n C n 1 = X n 3 a n C n 3 C n 1 = X n 3 + a n X n 4 + a n a n 1 X n 3 C n 2 C n = C n 2 C n 3 C n 2 a n C n 1 Therefore we have that = X n 3 a n X n 2 X n = X n 3 + X n 4 + a n 1 X n 3 α n+1 X n 3 + a n X n 4 + a n a n 1 X n 3 X n 2 + α n+1 X n 3 a n X n 2 + α n X n 2 + α n α n+1 X n 1 = α n α n+1 X n 1 + α n + 1 α n+1 a n X n a n 1 α n+1 a n a n 1 X n α n+1 a n X n 4 = α n α n+1 X n 1 α n+1 a n X n 2 + a n 1 X n 3 + X n 4 +X n 2 + a n 1 X n 3 + X n 4 + a n X n 2 + X n 3 + ɛ n X n 2 = α n+1 ɛ n X n 1 + X n 1 + a n X n 2 + X n 3 + ɛ n X n 2 = X n + α n+1 ɛ n X n 1 + ɛ n X n 2 Lemma 5 y n 1 x n 1 α r n + r n 1 We know that α = ỹn x n = y n + α n+1 ɛ n y n 1 + ɛ n y n 2 x n + α n+1 ɛ n x n 1 + ɛ n x n 2. 9

10 Hence y n 1 α x n 1 = y n 1 y n + α n+1 ɛ n y n 1 + ɛ n y n 2 x n 1 x n + α n+1 ɛ n x n 1 + ɛ n x n 2 = x ny n 1 y n x n 1 + y n 1 x n 2 x n 1 y n 2 ɛ n x n 1 x n + α n+1 ɛ n x n 1 + ɛ n x n 2 Therefore since α n 0 ɛ n 0 and x n 0 then y n 1 α x n 1 r n 1ɛ n r n y n 1 α x n 1 r n + r n 1 Lemma 6 zn 1 x n 1 β qn + qn 1 x nx n 1 The proof of this lemma is similar to the proof of the previous lemma. r Lemma 7 If lim n q x nx n 1 = 0 and lim n sequence is unique. x nx n 1 = 0 then the triangle r Since lim n = 0 we know that lim lim lim r n 1 x n 1 x n 2 = 0 r n 1 = 0 r n + r n 1 = 0 q Similarily since lim n = 0 we know that lim lim lim q n 1 x n 1 x n 2 = 0 q n 1 = 0 q n + q n 1 = 0 10

11 Therefore by the two previous lemmas y n lim lim Hence the triangle sequence is unique. = α x n z n = β x n 4 Uniqueness of Certain Bounded Triangle Sequences 4.1 Review of Key Ideas In section 3 we demonstrated that the relations lim lim r n = 0 q n = 0 are equivalent to uniqueness. Now we will use this formulation to prove the uniqueness of certain bounded infinte triangle sequences in the two main theorems of this section. 4.2 Proof of the Theorem We prove the following theorems by finding lower bounds on the growth of and upper bounds on the growth of r n and q n in order to show that the appropriate limit relations hold. Theorem 2 The infinite triangle sequence given by a i = A for all i is unique. First we shall bound r n and q n from above. We know that r n+1 = r n 2 r n 1 a n+1 r n Additionally r n+1 r n 1 + r n 1 + A r n q n+1 = q n 2 q n 1 a n+1 q n 11

12 q n+1 q n 1 + q n 1 + A q n Consider the case where A 1. We will show that r n A + 1 n. r 0 = 1 A r 1 = A A r 2 = A 2 1 A Suppose that r k A + 1 k for all k n. Then r n+1 r n 2 + r n 1 + A r n A + 1 n 2 + A + 1 n 1 + AA + 1 n = A + 1 n A A A A + 1 n+1 Note that the last of the above steps is justified since A 1 implies that A A Therefore by induction r n A + 1 n for all n. Similarily we will show that q n A + 1 n. q 0 = 0 A q 1 = 1 A q 2 = A + 1 A Suppose that q k A + 1 k for all k n. Then q n+1 q n 2 + q n 1 + A q n A + 1 n 2 + A + 1 n 1 + AA + 1 n = A + 1 n A A A A + 1 n+1 Therefore by induction q n A + 1 n for all n. Consider the case where A = 0. We will show that r n A n = 3 2 n. 3 r 0 = r 1 = r 2 =

13 Suppose that r k 3 2 k for all k n. Then r n+1 r n 2 + r n 1 n n = n n Therefore by induction r n 3 2 n for all n. Similarily we will show that q n A n = 3 2 n. 0 3 q 0 = q 1 = q 2 = 1 2 Suppose that q k 3 2 k for all k n. Then q n+1 q n 2 + q n 1 n n = n n 1 Therefore by induction q n 3 2 n for all n. This means that for all A 0 we have that r n+1 A + 3 n+1 2 q n+1 A + 3 n Next we will bound x n from below. We will show that x n CA + 2 n/2. x 0 = 1 x 1 = 1 + A x 2 = 2 + 2A x 3 = 3 + 3A + A 2 13

14 Therefore x 2 A + 2x 0 x 3 A + 2x 1 Suppose that x k A + 2x k 2 x k+1 A + 2x k 1 We will show that x k+2 A+2x k. Since the x n are increasing we know that x k+2 = x k+1 + Ax k + x k 1 = x k + Ax k 1 + x k 2 + Ax k + x k 1 = A + 1x k + x k 1 + Ax k 1 + x k 2 A + 1x k + x k 1 + Ax k 2 + x k 3 = A + 1x k + x k = A + 2x k Therefore by induction we have that x n+2 A + 2x n for all n. This implies that x n CA + 2 n/2 for all n. r Finally we will show that n q and n approach 0 as n grows without bound. Recall that Therefore we have that r n q n A + 3 n 2 A + 3 n 2 x n CA + 2 n/2 x n 1 CA + 2 n/2 1/2 DA + 2 n/2 lim lim r n lim q n lim A n CDA + 2 n = 0 A n CDA + 2 n = 0 Thus by Lemma 7 the triangle sequence a i = A is unique for all A 0. This establishes uniqueness for the very particular case where a i = A for all i. Although we have made only partial progress toward an extension of this method we do believe that it can be extended to prove uniqueness of all bounded triangle sequences. 14

15 5 Relationships Between Cross Product Vectors and Vertices of Partition Triangles 5.1 Overview In order to understand the uniqueness question through the geometry of the iteration we need to investigate the partition triangles a1... a n. In particular we need to study the vertices of these triangles. In the following section we begin by stating and proving a lemma that relates the vertices of a1... a n to those of an... a 1. This result is interesting in its own right but it is most useful to us as a tool in proving the main theorem of the section. This main theorem shows that there is a clean relation between the vertices of a1... a n and the coordinates of the vectors X n associated with the sequence a 1... a n. Finally as a corollary to this theorem we will construct a formula for the area of a1... a n. 5.2 Proofs of the Relations The main theorem of this paper concerns a direct relationship between the vertices of the partition triangles and the coordinates of the cross product vectors in three dimensions. Lemma 8 Reversing Let a1 a 2... a n be the set of points whose triangle sequences have first n terms equal to a 1 a 2... a n. Suppose the vertices of a1a 2... a n are written as Then the vertices of an a n 1... a 1 T n a 1 a 2... a n 0 0 = y 00 x 00 z 00 x 00 T n a 1 a 2... a n 1 0 = y 10 x 10 z 10 x 10 T n a 1a 2... a n 1 1 = y 11 x 11 z 11 x 11 are equal to T n a n a n 1... a = y 00 y 10 y 11 y 10 y 10 T n a n a n 1... a = x 00 x 10 x 11 x 10 x 10 T n a n a n 1... a = z 00 + x 00 z 10 + x 10 z 11 z 10 + x 11 x 10 z 10 + x 10 15

16 In order to prove this lemma it is necessary to ensure that this notation for the vertices is well-defined. It is easy to see that the vertices of the partition triangles will always be of the form u v where u and v are rational. Write u v in the form y x z x where xy and z are integers and x is as small as possible. This will ensure that x y z = 1 unless of course either u or v is zero. The proof of the reversing lemma will proceed by induction on n the length of the sequence. We will begin by establishing the base cases where n = 0 and n = 1. Case n = 0 It is natural to consider the case n = 0 to be the entire triangle. We will denote this triangle by. The vertices of are It is easy to compute that = = = This proves the lemma for the case n = 0 since is the reverse of itself. 16

17 Case n = 1 The vertices of a1 are a a a a It is again easy to compute that = a a a = a a a a a a = 1 + a a a This proves the lemma for the case n = 1 since a1 is the reverse of itself. Now we assume that the theorem holds true for all triangle sequences of length k with 0 k n 1 and show that the theorem holds true for sequences of length n. The structure of the proof will be as follows. We will begin with the coordinates of the vertices of a2... a n 1 and construct the coordinates of the vertices of both a1... a n and an... a 1. We will then demonstrate that the appropriate relation occurs between the coordinates of the vertices of the two triangles. In order to construct the vertices we need the following relation which follows easily from the definition of the mapping T. y T 1 k x x z x = ky + z + x y ky + z + x This notation refers to the preimage of the point y x z x in triangle k. We also need to check that the operations of reversing and applying T 1 k preserve the reduced fractional form of the vertex coordinates. This means 17

18 that we need to check that under one of these operations a pair y x z x with x y z = 1 will produce another pair y couple of propositions. x z y ky+x+z is in re- Proposition 2 If x y z = 1 then T 1 k y x z x = x duced fractional form. x with x y z = 1. We need a ky+x+z x Suppose ky+x+z y ky+x+z is not in reduced fractional form. Then there is an positive integer m 1 such that m x m y and m ky + x + z. This implies that m z which contradicts the assumption that x y z = 1. This proves the proposition. Proposition 3 If the points y00 z 00 x 00 x 00 y10 z 10 x 10 x 10 y11 z 11 x 11 x 11 are in reduced fractional form then y00 y 11 y 10 y 10 y 10 x00 x 11 x 10 x 10 x 10 z00 + x 00 z 11 z 10 + x 11 x 10 z 10 + x 10 z 10 + x 10 are in reduced fractional form. It suffices to show that y 00 y 10 y 11 y 10 = x 00 x 10 x 11 x 10 = z 00 +x 00 z 10 +x 10 z 11 z 10 +x 11 x 10 = 1 This is equivalent to showing that y 00 y 10 y 11 = x 00 x 10 x 11 = z 00 + x 00 z 10 + x 10 z 11 + x 11 = 1 18

19 We claim that This implies that det det x 00 x 10 x 11 y 00 y 10 y 11 z 00 z 10 z 11 = 1 x 00 x 10 x 11 y 00 y 10 y 11 x 00 + z 10 x 01 + z 01 x 11 + z 11 x k 00 = 1 which proves the lemma. To prove the claim we use induction on n the length of the sequence. We define the vertices of a1... a k in reduced fractional form as follows: y k 00 zk 00 x k and let M k = y k 10 x k 10 y k 11 x k zk 10 x k 10 zk 11 x k 11 x k 00 x k 10 x k 11 y k 00 y k 10 y k 11 z k 00 z k 10 z k 11 We start with the case k=0. The vertices of are: Then det M 0 = det = 1 19

20 Now suppose that det M n 1 = 1. The matrix for T a 1 n [T 1 a n ] = 1 a n is as follows: Thus det[t 1 a n ] = 1 Since M n = [T 1 a n ]M n 1 this implies that det M n = 1 This proves the claim and thus completes the proof of the proposition. Now we are ready to construct the proof of the reversing lemma. have vertices equal to y00 z 00 x 00 a2... a n 1 x 00 y10 z 10 x 10 x 10 y11 z 11 x 11 x 11 Now if we apply the mapping Ta 1 1 to each of these points we recover the vertices of a1... a n 1 which are calculated to be: x 00 y 00 a 1 y 00 + z 00 + x 00 a 1 y 00 + z 00 + x 00 x 10 y 10 a 1 y 10 + z 10 + x 10 a 1 y 10 + z 10 + x 10 x 11 a 1 y 11 + z 11 + x 11 y 11 a 1 y 11 + z 11 + x 11 These are the vertices of a triangle whose sequence has n 1 terms. We then apply the reversing lemma to the vertices to yield the vertices of an 1... a 1 shown below. x00 x 11 x 10 x 10 x 10 a1 y 00 + z 00 + x 00 a 1 y 10 + z 10 + x 10 a 1y 11 a 1 y 10 + z 11 z 10 + x 11 x 10 a 1 y 10 + z 10 + x 10 y00 + a 1 y 00 + z 00 + x 00 y 11 y 10 + a 1 y 11 a 1 y 10 + z 11 z 10 + x 11 x 10 y 10 + a 1 y 10 + z 10 + x 10 y 10 + a 1 y 10 + z 10 + x 10 Let 20

21 Applying Ta 1 n to these points yields the vertices of an... a 1 : x 10 x 00 a n x 00 + x 11 a n x 00 + x 11 a 1 y 10 + z 10 + x 10 a 1 y 00 + z 00 + x 00 a n a 1 y 00 + a n z 00 + a n x 00 + a 1 y 11 + z 11 + x 11 a n a 1 y 00 + a n z 00 + a n x 00 + a 1 y 11 + z 11 + x 11 y 10 + a 1 y 10 + z 10 + x 10 a n y 00 + a n a 1 y 00 + a n z 00 + a n x 00 + y 11 + a 1 y 11 + z 11 + x 11 y 00 + a 1 y 00 + z 00 + x 00 a n y 00 + a n a 1 y 00 + a n z 00 + a n x 00 + y 11 + a 1 y 11 + z 11 + x 11 Now we will similarly construct the vertices of a1... a n. First we use the reversing lemma to calculate the vertices of an 1... a 2 to be: y00 y 11 y 10 y 10 y 10 x00 x 10 x 11 x 10 x 10 z00 + x 00 z 10 + x 10 z 11 z 10 + x 11 + x 10 z 10 + x 10 Applying Ta 1 n yields the vertices of an... a 2 : y 10 y 00 a n y 00 + y 11 a n y 00 + y 11 x 10 x 00 a n x 00 + x 11 a n x 00 + x 11 z 10 + x 10 z 00 + x 00 a n z 00 + a n x 00 + z 11 + x 11 a n z 00 + a n x 00 + z 11 + x 11 Now we apply the reversing lemma to calculate the vertices of a2... a n : y10 z 10 x 10 x 10 an y 00 + y 11 a n x 00 + x 11 a nz 00 + z 11 a n x 00 + x 11 an y 00 + y 00 + y 11 a n x 00 + x 00 + x 11 a nz 00 + z 00 + z 11 a n x 00 + x 00 + x 11 21

22 Finally we apply Ta 1 1 to yield the vertices of a1... a n : x 10 y 10 a 1 y 10 + x 10 + z 10 a 1 y 10 + x 10 + z 10 a n x 00 + x 11 a n y 00 + y 11 a 1 a n y 00 + a 1 y 11 + a n z 00 + z 11 + a n x 00 + x 11 a 1 a n y 00 + a 1 y 11 + a n z 00 + z 11 + a n x 00 + x 11 x 00 + a n x 00 + x 11 a 1 y 00 + a 1 a n y 00 + a 1 y 11 + x 00 + a n x 00 + x 11 + z 00 + a n z 00 + z 11 y 00 + a n y 00 + y 11 a 1 y 00 + a 1 a n y 00 + a 1 y 11 + x 00 + a n x 00 + x 11 + z 00 + a n z 00 + z 11 But this is what you get if you apply the reversing lemma to an... a 1. Thus the lemma is true for sequences of length n and is thus true by induction for all sequences. Now we are ready to prove the main theorem of this exposition: Theorem 3 Let X n = C n C n+1 = x n y n z n. Then the vertices of a1... a n are given by: Ta n yn a n 0 0 = z n 1 x n 1 x n 1 Ta n yn 1... a n 0 1 = z n x n x n Ta n yn 2 + y n 1... a n 1 1 = z n 2 + z n x n 2 + x n x n 2 + x n This theorem is proven by induction on n the number of terms in the sequence. We consider first the case n = 1. The vertices of a1 are: a a a a

23 We also know that: X 1 = x 1 y 1 z 1 = X 0 = x 0 y 0 z 0 = We check that: X 1 = x 1 y 1 z 1 = a y0 z 0 1 = x 0 x = y1 z 1 1 = x 1 x 1 a a y 1 + y 1 z 1 + z = x 1 + x 1 x 1 + x 1 a a = 1 a a Now we assume that the theorem holds for the vertices of all triangles with sequences of length k with 1 k n 1. Suppose the vertices of a1... a n 1 are yn 2 z n 2 x n 2 x n 2 yn 1 z n 1 x n 1 x n 1 yn 3 + y n 1 x n 3 + x n 1 z n 3 + z n 1 x n 3 + x n 1 We will then construct the vertices of a1... a n by applying the reversing lemma and the map Ta 1 n. First the reversing lemma gives us the vertices of an 1... a 1 yn 2 y n 3 y n 1 y n 1 xn 2 x n 1 x n 3 x n 1 zn 2 + x n 2 z n 1 + x n 1 z n 3 + x n 3 z n 1 + x n 1 23

24 Next we apply the map Ta 1 n to give the vertices of an... a 1 y n 1 y n 2 y n 1 + a n y n 2 + y n 3 y n 1 + a n y n 2 + y n 3 x n 1 x n 1 + a n x n 2 + x n 3 x n 2 x n 1 + a n x n 2 + x n 3 x n 1 + z n 1 x n 2 + z n 2 x n 1 + a n x n 2 + x n 3 + z n 1 + a n z n 2 + z n 3 x n 1 + a n x n 2 + x n 3 + z n 1 + a n z n 2 + z n 3 By using the recursion relation X n = X n 1 +a n X n 2 +X n 3 we can simplify this yn 1 y n 2 y n y n xn 1 x n 2 x n x n xn 1 + z n 1 x n 2 + z n 2 x n + z n x n + z n Finally we apply the reversing lemma to these points to yield the vertices of a1... a n. They are yn 1 z n 1 x n 1 x n 1 yn z n x n x n yn 2 + y n x n 2 + x n z n 2 + z n x n 2 + x n This shows that the theorem is true for sequences of length n. This proves the theorem for all n. Corollary 3 The area of a1... a n is equal to 1 2x nx n 1x n+x n 2. We know from the previous theorem that the vertices of a1... a n yn 1 z n 1 x n 1 x n 1 are: 24

25 yn z n x n x n yn 2 + y n x n 2 + x n z n 2 + z n x n 2 + x n Therefore we can calculate the area as one-half of the length of the crossproduct of two edge vectors. The edge vectors are calculated to be: yn 2 + y n E 1 = y n z n 2 + z n z n 0 x n 2 + x n x n x n 2 + x n x n E 2 = yn 1 y n z n 1 z n 0 x n 1 x n Combining the fractions yields: yn 2 x n x n 2 y n E 1 = x n x n 2 + x n z n 2x n x n 2 z n x n x n 2 + x n 0 yn 1 x n x n 1 y n E 1 = z n 1x n x n 1 z n 0 Now we know that y n 1 x n x n 1 y n = r n and that z n 1 x n x n 1 z n = q n. We can also calculate the other two numerators by using the recursion relation X n = X n 1 + a n X n 2 + X n 3 y n 2 x n x n 2 y n = y n 2 x n 1 + a n x n 2 + x n 3 x n 2 y n 1 + a n y n 2 + y n 3 = y n 2 x n 1 x n 2 y n 1 + y n 2 x n 3 x n 2 y n 3 = r n 1 + r n 2 Similarly z n 2 x n x n 2 z n = q n 1 q n 2. Thus we can rewrite E 1 and E 2 in the following way: rn 1 + r n 2 E 1 = x n x n 2 + x n q n 1 q n 2 x n x n 2 + x n 0 E 2 = rn q n 0 25

26 Therefore E 1 E 2 = = = = = = = 0 0 q n r n 1 + r n 2 r n q n q n 2 x 2 nx n 1 x n + x n r nq n 1 q n r n 1 + q n r n 2 r n q n 2 x 2 nx n 1 x n + x n x n 1 + q n 3 q n 2 a n q n 1 r n 2 r n 3 r n 2 a n r n 1 q n 2 x 2 nx n 1 x n + x n x n 1 + r n 2 q n 3 r n 3 q n 2 + a n r n 1 q n 2 r n 2 q n 1 x 2 nx n 1 x n + x n x n 1 + a n x n 2 + x n 3 x 2 nx n 1 x n + x n 2 x n 0 0 x 2 nx n 1 x n + x n x n + x n 2 Thus Area of a1... a n = 1 2 E 1 E 2 = 1 2 x n +x n 2 Corollary 4 The set of all points α β with triangle sequence equal to a 1 a 2... consists of at most a line segment. It is easy to show that if you have any two points α 1 β 1 and α 2 β 2 that have the same triangle sequence then every point on the line segment connecting them will also have that same triangle sequence. Now suppose there are three non-collinear points that all have the same triangle sequence. Clearly all points on the triangular perimeter defined by these three points will have the same triangle sequence. Then given any point in the interior of this region we can construct a line segment containing this point with endpoints on the perimeter. This implies that there exists a region with nonzero area in which every point has the same triangle sequence. However this contradicts corollary 3. 6 Uniqueness of Triangle Sequences in Which the Integer C Appears Infinitely Often This section uses a geometric argument to show that in the case that any integer C appears infintely often in a triangle sequence a 1 a 2... the partition 26

27 Figure 2: Triangle corresponding to Lemma 9 triangles a1... a n converge to the unique point with that triangle sequence. We begin with a couple of definitions. Definition 7 For any vector V = v 1 v 2 let V = max v 1 v 2. Definition 8 For any two points A and B let AB be the vector from A to B. Definition 9 Let the distance between two points A and B be defined as AB. Lemma 9 Let ABC be a triangle and let M and N be two points such that M lies on the line segment between B and N. Then AM AN AM AB. Since B M and N are colinear and since M lies between N and B then BM = a MN for some a 0. Let m 1 be the largest component of the vector AM. Without loss of generality assume let m 1 be non-negative. Then AM = m 1. Now let x 1 n 1 and b 1 be the corresponding components of MN AN and AB respectively. Then AM AN m 1 n 1 Therefore since AN = AM + MN we know that x 1 0. Additionally since AB = AM a MN we know that b 1 m 1. This implies that AB AM. 27

28 Lemma 10 Let ABC be a triangle with AC AB then for any point M on the line segment between B and C AM AB. Case1: AM AC This case follows immediately by letting N = C in the previous lemma. Case2: AM < AC Since AC AB and AM < AC then AM AB as desired. Definition 10 Let s 0 n be the length of the side connecting Ta n 1...a n 0 0 and Ta n 1...a n 1 0. Lemma 11 For any n 3 one of the following is true: Let s 0 n s 0 n 1 s 0 n s 0 n 2 s 0 n s 0 n 3 Analogously let A = T n a 1...a n B = T n a 1...a n C = T n a 1...a n A = T n a 1...a n B = T n a 1...a n C = T n a 1...a n A = T n a 1...a n B = T n a 1...a n C = T n a 1...a n Where Ta n 1...a 0 is the identity transformation. Note that B = A and B = A. Case 1: AB AC By Lemma 10 s 0 n AB = s 0 n 1 Note that s 0 n is the length of a line segment connents to A to a point on the line segment between B and C. 28

29 Case 2: AC AB and BC AC. Since BC is the longest side of ABC and s 0 n lies within ABC then s 0 n BC. Note that A and C are both points on the line segment connenting B and C and that C lies between B and A. Therefore by Lemma 9 since BC AC s 0 n 2 = A B A C = BC s 0 n Case 3: BC AC AB AC and B A B C. Recall that the side between B and C contains the points A and C. Therefore s 0 n 2 = B A B C AC We know that AC is the longest side of ABC in which s 0 n lies. Therefore s 0 n 2 AC s 0 n Case 4: BC AC AB AC and B A B C. Note that A and C are points on the line segment connecting B and C with C between B and A. Therefore by Lemma 9 since B C B A s 0 n 3 = A B A C = B C Since AC is the longest side of ABC and since the side between B and C contains the points A and C then s 0 n 3 B C AC s 0 n Definition 11 Let F n = max Definition 12 Let G n = max r n 4 x n 4 x n 5 q n 4 x n 4 x n 5 Lemma 12 If a n = C and a n 1 > C then r n 3 x n 3 x n 4 q n 3 x n 3 x n 4 r n 2 x n 2 x n 3. q n 2 x n 2 x n 3. r n C2 +3C+2 C 2 +4C+4 F n. To prove this lemma we will bound r n from above in terms of max r n 2 r n 3 r n 4 and bound from below in terms of x n 2 x n 3. We will then use these two bounds fact that F n max r n 2 r n 3 r n 4 x n 2 x n 3 in terms of F n. Recall that r n = r n 3 r n 2 a n r n 1 to bound r n from above 29

30 By making the substitution r n 1 = r n 4 r n 3 a n 1 r n 2 we get r n = r n 3 r n 2 a n r n 4 + a n r n 3 + a n a n 1 r n 2 1 By using the triangle inequality and that a n = C we get Additionally r n Ca n 1 + 2C + 2 max r n 2 r n 3 r n 4 = x n 3 + a n x n 2 + x n 1 x n 1 By making the substitution x n 1 = x n 4 + a n 1 x n 3 + x n 2 we get = x n 4 + a n 1 + 1x n 3 + a n + 1x n 2 By dropping the x n 4 terms and using that a n = C we get x n 4 + a n 1 x n 3 + x n 2 2 a n 1 + 1x n 3 + C + 1x n 2 a n 1 x n 3 + x n 2 By dropping the x 2 n 3 terms we get Therefore Ca n 1 + 2a n 1 + C + 2x n 2 x n 3 r n = Ca n 1 + 2C + 2 max r n 2 r n 3 r n 4 Ca n 1 + 2a n 1 + C + 2x n 2 x n 3 Ca n 1 + 2C + 2 Ca n 1 + 2a n 1 + C + 2 F n Since the derivative of the above expression with respect to a n 1 is always negative in the region a n 1 C + 1 then the above expression achieves its maximum in the region a n 1 C + 1 when a n 1 = C + 1. Therefore r n C2 + 3C + 2 C 2 + 4C + 4 F n 30

31 Lemma 13 If a n = C and a n 1 > C then q n x nx n 1 C2 +3C+2 C 2 +4C+4 G n The proof holds for q n because in the proof of the previous lemma nothing is used about r n except that it satisfies the recurrence r n = r n 3 r n 2 a n r n 1 which q n satisfies as well. Lemma 14 If a n = C a n 1 > C and a n+1 C then By replacing n with n + 1 in Equation 1 we get r n+1 x n+1x n C2 +3C+2 C 2 +4C+4 F n r n+1 = a n+1 r n 3 + a n+1 + 1r n 2 + a n a n+1 1r n 1 By making the substitution r n 1 = r n 4 r n 3 a n 1 r n 2 we get r n+1 = a n a n+1 1r n a n+1 Ca n+1 r n 3 + a n a n 1 Ca n+1 a n 1 r n 2 3 By using the triangle inequality and that a n = C we get r n+1 = 2Ca n+1 + a n 1 Ca n+1 + 2a n+1 + a n max r n 2 r n 3 r n 4 Additionally by replacing n with n + 1 in Equation 2 we get x n+1 x n = x n 3 + a n + 1x n 2 + a n+1 + 1x n 1 x n 3 + a n x n 2 + x n 1 By making the substitution x n 1 = x n 4 + a n 1 x n 3 + x n 2 we get x n+1 x n = a n+1 + 1x 4 + a n 1 + a n 1 a n+1 + 1x n 3 + a n+1 + a n + 2x n 2 x n 4 + a n 1 + 1x n 3 + a n + 1x n 2 4 By dropping the x n 4 terms and using that a n = C we get x n+1 x n a n 1 + a n 1 a n+1 + 1x n 3 + a n+1 + C + 2x n 2 a n 1 + 1x n 3 + C + 1x n 2 By dropping the x 2 n 3 terms we get x n+1 x n Ca n 1 a n+1 + 2Ca n 1 + Ca n+1 + C 2 + 5C + 2a n 1 a n+1 + 3a n 1 + 2a n+1 + 5x n 2 x n 3 Therefore r n 2Ca n+1 + a n 1 Ca n+1 + 2a n+1 + a n x n+1 x n Ca n 1 a n+1 + 2Ca n 1 + Ca n+1 + C 2 + 5C + 2a n 1 a n+1 + 3a n 1 + 2a n F n 31

32 Since the derivative of the above expression with respect to a n 1 is always negative when a n 1 C + 1 and a n+1 C then the above expression achieves its maximum in the region bounded by a n 1 C + 1 and a n+1 C when a n 1 = C + 1. Therefore r n 3Ca n+1 + C 2 a n+1 + C + 2a n x n+1 x n C 2 a n+1 + 3C 2 + 4Ca n C + 4a n F n Since the derivative of the above expression with respect to a n+1 is always positive when a n+1 C we can replace the right hand side of the above inequality with its limit as a n+1. Therefore r n C2 + 3C + 2 x n+1 x n C 2 + 4C + 4 F n Lemma 15 If a n = C a n 1 > C and a n+1 C then q n+1 x n+1 x n C2 +3C+2 C 2 +4C+4 G n The proof of this lemma is identical to the proof of the previous lemma with all references to r i replaced by references to q i. Lemma 16 If a n = C a n 1 > C a n+1 C and a n+2 C then C2 +3C+2 C 2 +4C+4 F n r n+2 x n+2 x n+1 By replacing n with n + 1 in Equation 3 we get r n+2 = a n+2 a n+1 1r n+3 a n+2 + a n+2 a n+1 1r n 2 + a n a n+2 a n+1 a n + a n r n 1 By making the substitution r n 1 = r n 4 r n 3 a n 1 r n 2 we get r n+2 = a n a n+2 a n+1 C + Cr n 4 + a n+2 a n+1 a n a n+2 a n+1 C Cr n 3 a n+2 + a n+2 a n a n+2 a n 1 + a n 1 a n+2 a n+1 a n 1 C + Ca n 1 r n 2 By using the triangle inequality and that a n = C we get r n+2 a n+2 a n+1 a n 1 C + 2a n+2 a n+1 C + Ca n 1 + 2C + 2a n+2 a n+1 + a n+2 a n 1 + a n 1 + 3a n max r n 2 r n 3 r n

33 Additionally by replacing n with n + 1 in Equation 4 we get x n+2 x n+1 = a n+2 +1x n 3 +a n a n+2 +a n +1x n 2 +a n+1 +a n+2 +2x n 1 x n 3 + a n + 1x n 2 + a n+1 + 1x n 1 By making the substitution x n 1 = x n 4 + a n 1 x n 3 + x n 2 we get x n+2 x n+1 = a n+1 +a n+2 +2x n 4 +a n+2 +1+a n+1 a n 1 +a n+2 a n 1 +2a n 1 x n 3 + a n+1 + a n a n+2 + a n+2 + a n + 3x n 2 a n+1 + 1x n a n+1 a n 1 + a n 1 x n 3 + a n a n+1 x n 2 By dropping the x n 4 terms and using that a n = C we get x n+2 x n+1 a n+2 +1+a n+1 a n 1 +a n+2 a n 1 +2a n 1 x n 3 +a n+1 +Ca n+2 +a n+2 +C+3x n a n+1 a n 1 + a n 1 x n 3 + C a n+1 x n 2 By dropping the x 2 n 3 terms we get x n+2 x n+1 a n 1 Ca n+2 a n+1 +2Ca n+1 +2Ca n+2 +3C+2a n+2 a n+1 +2a 2 n+1+3a n+2 +8a n Ca n+2 a n+1 + C 2 a n+2 + C 2 + 6Ca n+2 + 2Ca n+1 + 7C + a 2 n+1 + 2a n+2 a n+1 + 5a n+2 + 7a n x n 2 x n 3 6 Now let J = a n 1 Ca n+2 a n+1 + C + a n Ca n+2 a n+1 + 2C + 2a n+2 a n+1 + 3a n and K = a n 1 Ca n+2 a n+1 +2Ca n+1 +2Ca n+2 +3C+2a n+2 a n+1 +2a 2 n+1+3a n+2 +8a n Ca n+2 a n+1 + C 2 a n+2 + C 2 + 6Ca n+2 + 2Ca n+1 + 7C + a 2 n+1 + 2a n+2 a n+1 + 5a n+2 + 7a n Then Equation 5 can be rewritten as r n+2 J max r n 2 r n 3 r n 4 and Equation 6 can be rewritten as Therefore x n+2 x n+1 Kx n 2 x n 3 r n+2 x n+2 x n+1 J K F n 33

34 Since the derivative of J K F n with respect to a n 1 is always negative when a n 1 C+1 a n+1 C and a n+2 C then the above expression achieves its maximum in the region bounded by a n 1 C + 1 a n+1 C and a n+2 C when a n 1 = C + 1. By making the substitution a n 1 = C + 1 in the expression J we get J 1 = a n+2 C 2 a n+1 + 3Ca n+1 + C + 2a n C 2 + 4C + 5 By making the substitution a n 1 = C + 1 in the expression K we get K 1 = a n+2 C 2 a n+1 + 3C 2 + 4Ca n C + 4a n C 2 a n+1 + 4C 2 + 2Ca 2 n Ca n C + 3a 2 n a n Therefore we have r n+2 x n+2 x n+1 J 1 K 1 F n Since the derivative of J1 K 1 F n with respect to a n+2 is always positive when a n+1 C and a n+2 C then we can replace the right hand side of the above inequality with its limit as a n+2. Therefore r n+2 a n+1 C 2 + 3a n+1 C + 2a n+1 + C + 4 x n+2 x n+1 a n+1 C 2 + 4a n+1 C + 4a n+1 + 3C C + 8 Since the derivative of the above expression with respect to a n+1 is always positive when a n+1 C then we can replace the right hand side of the above inequality with its limit as a n+1. Therefore r n+2 C2 + 3C + 2 x n+2 x n+1 C 2 + 4C + 4 Lemma 17 If a n = C a n 1 > C a n+1 C and a n+2 C then C 2 +3C+2 C 2 +4C+4 G n q n+2 x n+2 x n+1 The proof of this lemma is identical to the proof of the previous lemma with all references to r i replaced by references to q i. 34

35 Lemma 18 s 0 n = rn q n From Lemma 3 we know that: T n a 1...a n 0 0 = T n a 1...a n 1 0 = yn 1 z n 1 x n 1 x n 1 yn z n x n x n Therefore yn 1 s 0 n = z n 1 yn z n x n 1 x n 1 x n x n yn x n 1 x n y n 1 = z nx n 1 x n z n 1 rn q n = Lemma 19 If a n = C a n 1 > C a n+1 C and a n+2 C then s 0 i C 2 +3C+2 C 2 +4C+4 maxs 0n 2 s 0 n 3 s 0 n 4 for all i {n n + 1 n + 2}. From Lemma 18 we know that maxs 0 n 2 s 0 n 3 s 0 n 4 = max r n 2 x n 2 x n 3 q n 2 x n 2 x n 3 r n 3 x n 3 x n 4 q n 3 x n 3 x n 4 r n 4 x n 4 x n 5 q n 4 x n 4 x n 5 and ri q i s 0 i = max x i x i 1 x i x i 1 From Lemmas 12 through 17 we know that for all i {n n + 1 n + 2} r i C2 + 3C + 2 x i x i 1 C 2 + 4C + 4 max r n 2 x n 2 x n 3 q i C2 + 3C + 2 x i x i 1 C 2 + 4C + 4 max q n 2 x n 2 x n 3 r n 3 x n 3 x n 4 q n 3 x n 3 x n 4 r n 4 x n 4 x n 5 q n 4 x n 4 x n 5 35

36 Therefore for all i {n n + 1 n + 2} s 0 i C2 + 3C + 2 C 2 + 4C + 4 maxs 0n 2 s 0 n 3 s 0 n 4 Lemma 20 If lim s 0 n = 0 then the triangle sequence a 1 a 2 a 3... is unique. From Lemma 18 we know that s 0 n = max rn q x nx n 1 n x nx n 1. Therefore This that lim s 0 n = 0 implies s 0 n r n s 0 n q n lim s 0n = 0 lim s 0n = 0 Hence by Lemma 7 we know that the triangle sequence is unique. Theorem 4 If any number occurs infinitely many times in a triangle sequence then that triangle sequence is unique. Given a triangle sequence a 1 a 2 a 3... let C be the smallest term that occurs infinitely many times in the sequence. Case 1: The sequence contains only finitely many terms a i such that a i C. Then there must exist an a N such that for all a i where i > N the terms in the sequence are C. By some Lemma 2 i the sequence C C C... is unique. Therefore the triangle sequence a 1 a 2 a 3... is unique. Case 2: The sequence contains infinitely many terms a i such that a i C. By assumption C is the smallest term that appears infinitely often in the sequence. Therefore there are only finitely many terms less than C in the triangle sequence. Let a m be the last term in the sequence that is less than C. Since there are infinitely many C terms in the sequence and infinitely many terms greater than C it is possible to select an infinite sequence of N 1 N 2 N

37 such that N 1 > m + 1 and for all i a Ni = C and a Ni 1 > C. Note that since N 1 > m + 1 it follows immediately that a Ni +1 C and a Ni +1 C. Let We will show that for all n N i M = maxs 0 N 0 2 s 0 N 0 3 s 0 N 0 4 s 0 n C 2 i + 3C + 2 C 2 M + 4C + 4 First consider the base case where i = 0. By Lemma 11 for all s 0 n N 0 Now assume that for all n N k Therefore we know that s 0 n s 0 n M C 2 k + 3C + 2 C 2 M + 4C + 4 maxs 0 N k+1 2 s 0 N k+1 3 s 0 N k+1 4 So by Lemma 19 we know that C 2 k+1 + 3C + 2 s 0 N k+1 C 2 M + 4C + 4 C 2 k+1 + 3C + 2 s 0 N k C 2 M + 4C + 4 C 2 k+1 + 3C + 2 s 0 N k C 2 M + 4C + 4 Hence by Lemma 11 we know that for all n N k+1 C 2 k+1 + 3C + 2 s 0 n C 2 M + 4C + 4 C 2 k + 3C + 2 C 2 M + 4C + 4 Therefore by induction we have that for all i and for all n N i C 2 i + 3C + 2 s 0 n C 2 M + 4C + 4 Therefore we know that C 2 i lim s + 3C + 2 0n lim i C 2 M + 4C + 4 Therefore since C2 +3C+2 C 2 +4C+4 < 1 lim s 0n = 0 Thus by Lemma 20 the triangle sequence is unique. 37

38 References [1] Thomas Garrity. On periodic sequences for algebraic numbers