Some Results Concerning Uniqueness of Triangle Sequences

T. Cheslack-Postava, A. Diesl, M. Lepinski, A. Schuyler

August 12, 1999

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 + 1)y},

where k is any nonnegative integer.

Given (α, β) ∈ △_{k}, define the map T : △ → △ ∪ {(x, 0) : 0 ≤ x ≤ 1} by

T (α, β) = (\frac{β}{α}, \frac{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) ∈ △_{a_{k}}, then the point (α, β) will have the triangle sequence (a_{1}, a_{2}, …).

We will recursively define a sequence of vectors as follows: Set C_{−2} = (1, 0, 0), C_{−1} = (0, 1, 0) and C_{0} = (0, 0, 1). 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}.

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_{2}, …). 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_{n→∞} \frac{C_{n} × C_{n+1}}{||C_{n} × C_{n+1}||} = (1, α, β),

then the numbers α and β will be uniquely determined by the triangle sequence (a_{1}, a_{2}, …).

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}^{∞} △_{k}

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} = {(x, y) ∈ △ : a_{1} = k}.

We can further partition each △_{k} in a way similar to that in which we partitioned △. Define △_{k,l} in the following way:

△_{k,l} = {(x, y) ∈ △_{k} : T (x, y) ∈ △_{l}}.

Thus △_{k,l} is the set of all points in △_{k} that map to △_{l}. We can think of this set in the following way:

△_{k,l} = {(x, y) ∈ △ : a_{1} = k and a_{2} = l}.

We can further partition the triangles in the same way. We define △_{a_{1},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 △_{a_{1},a_{2},...} is to look at the limit of the sets △_{a_{1},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 △_{a_{1},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 △_{a_{1},a_{2},... ,a_{n}}. We will also show that there is an interesting relationship between the vertices of △_{a_{1},a_{2},... ,a_{n}} and the vertices of △_{a_{n},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 infinite triangle sequence. We begin with some notation to describe the partition triangles.

Definition 1 Define T_{a_{1}...a_{n}}^{−n}(α, β) to be the unique point (x, y) ∈ △_{a_{1},... ,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}).

Proof:

The set T_{a_{1}...a_{n}}^{−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}).

QED

Definition 2 For any finite sequence of nonnegative integers, a_{1} … a_{n}, let d(a_{1}, … a_{n}) = min{length(T_{a_{1}...a_{n}}^{−n}(I)), length(T_{a_{1}...a_{n−1},a_{n}+1}^{−n}(I))}, where I is the interval (0, 1).

In the figure below, we have

\begin{aligned}A & = & T_{a_{1}...a_{n}}^{−n}(I) & a & = & T_{a_{1}...a_{n}+1}^{−n}(1, 0) \\ B & = & T_{a_{1}...a_{n}+1}^{−n}(I) & b & = & T_{a_{1}...a_{n}}^{−n}(1, 0) \\ C & = & \overline{T_{a_{1}...a_{n}+1}^{−n}(1, 0) T_{a_{1}...a_{n}}^{−n}(1, 0)} & c & = & T_{a_{1}...a_{n}}^{−n}(0, 0) \\ m & = & T_{a_{1}...a_{n}+1}^{−n}(\frac{1}{l+1}, 0) & n & = & T_{a_{1}...a_{n}+1}^{−n}(x, 0) \end{aligned}

Note: 0 < \frac{1}{l+1} < x < 1

Figure 1: Triangle created by the consecutive partition lines T_{a_{1}...a_{n}}^{−n}(I) and T_{a_{1}...a_{n}+1}^{−n}(I).

Lemma 1 For any small ϵ > 0 and any sequence a_{1} … a_{n} of nonnegative integers, there exists a nonnegative integer l such that d(a_{1} … a_{n}, l) ≥ d(a_{1} … a_{n})− ϵ.

Proof:

\underline{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:

d(a_{1} … a_{n}, l) = length(T_{a_{1}...a_{n},l+1}^{−(n+1)}(I))
≥ length(T_{a_{1}...a_{n}}^{−n}(I))
≥ d(a_{1} … a_{n}) − ϵ

Therefore the lemma holds when c is obtuse.

\underline{Case II:} Consider the case where the angle at c is acute. Consider a line segment, N, in △_{a_{1}...a_{n}} with the following properties:

(i) One endpoint is at T_{a_{1}...a_{n}}^{−n}(1, 0) (point b in Figure 1).

(ii) The other endpoint is at T_{a_{1}...a_{n}+1}^{−n}(x, 0) for some x ∈ (1, 0) (on line B in Figure 1). (iii) The length of the line segment is at least length(T_{a_{1}...a_{n}}^{−n}(I)) − ϵ. (iv) Given any m on the segment \overline{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^{+} so that \frac{1}{l+1} ≤ x. Then T_{a_{1}...a_{n}+1}^{−n}(\frac{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 length(T_{a_{1}...a_{n}}^{−n}(I)) − ϵ = A − ϵ. But this line segment is also T_{a_{1}...a_{n},l}^{−(n+1)}(I).

Then, since length(T_{a_{1}...a_{n},l+1}^{−(n+1)}(I)) > length(T_{a_{1}...a_{n},l}^{−(n+1)}(I)), we have:

d(a_{1} … a_{n}, l) = length(T_{a_{1}...a_{n},l}^{−(n+1)}(I))
≥ length(T_{a_{1}...a_{n}}^{−n}(I)) − ϵ
≥ d(a_{1} … a_{n}) − ϵ

QED Theorem 1 There exist distinct points (α_{1}, β_{1}) and (α_{2}, β_{2}) in △ with identical infinite triangle sequences.

Proof:

Consider a triangle sequence constructed in the following manner:

Choose a_{1} so that d(a_{1}) ≥ 1 − \frac{1}{4}.

Choose a_{2} so that d(a_{1}, a_{2}) ≥ d(a_{1}) − \frac{1}{8} ≥ 1 − \frac{1}{4} − \frac{1}{8}.

Choose a_{n} so that d(a_{1} … a_{n}) ≥ d(a_{1} … a_{n−1}) − \frac{1}{2^{n+1}} for n ≥ 2.

Our lemma guarantees that a sequence may be constructed in this way.

Then for all n, d(a_{1} … a_{n}) ≥ 1 − \frac{1}{4} − \frac{1}{8} − … − \frac{1}{2^{n+1}} > \frac{1}{2}. Therefore there is some line segment of length \frac{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_{2}, …).

QED

3 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_{n→∞} \frac{C_{n} × C_{n+1}}{||C_{n} × C_{n+1}||} = (1, α, β),

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.

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}

Proof:

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}

QED

Lemma 3 X_{n} × X_{n+1} = C_{n+1}

Proof:

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 det(C_{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}.

QED Corollary 1 det(X_{n−2}, X_{n−1}, X_{n}) = 1 Proof:

det(X_{n−2}, X_{n−1}, X_{n}) = X_{n−2} · (X_{n−1} × X_{n})
= X_{n−2} · C_{n}
= (C_{n−2} × C_{n−1}) · C_{n}
= det(C_{n−2}, C_{n−1}, C_{n})
= 1

QED

Corollary 2 |\frac{y_{n−1}}{x_{n−1}} − \frac{y_{n}}{x_{n}} | = \frac{|r_{n}|}{x_{n}x_{n−1}} Proof:

From Lemma 3 we know that

|x_{n}y_{n−1} − y_{n}x_{n−1}| = |r_{n}|.

Dividing this expression by x_{n}x_{n−1} yields

|\frac{y_{n−1}}{x_{n−1}} − \frac{y_{n}}{x_{n}} | = \frac{|r_{n}|}{x_{n}x_{n−1}} .

QED

Definition 4 Let α_{n} be defined so that d_{n−3} − d_{n−2} − α_{n}d_{n−1} = 0. Definition 5 Let C̃_{n} = C_{n−3} − C_{n−2} − α_{n}C_{n−1}. Definition 6 Let ϵ_{n} = α_{n} − a_{n}. From these definitions it should be clear that

a_{n} ≤ α_{n} < a_{n} + 1

0 ≤ ϵ_{n} < 1
C̃_{n} · (1, α, β) = 0

Additionally, if we let,

C̃_{n} × C̃_{n+1} ≡ X̃_{n} ≡ (x̃_{n}, ỹ_{n}, z̃_{n}).

Then,

α = \frac{ỹ_{n}}{x̃_{n}}

and

β = \frac{z̃_{n}}{x̃_{n}} .

Lemma 4 X̃_{n} = X_{n} + α_{n}ϵ_{n}X_{n−1} + ϵ_{n}X_{n−2} Proof:

X̃_{n} = C̃_{n} × C̃_{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})
= −X_{n−3} − a_{n}X_{n−2}

Therefore we have that,

X̃_{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−2}
+(1 + a_{n−1} − α_{n+1}a_{n}a_{n−1})X_{n−3} + (1 − 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} + ϵ_{n}X_{n−2}
= α_{n+1}ϵ_{n}X_{n−1} + X_{n−1} + a_{n}X_{n−2} + X_{n−3} + ϵ_{n}X_{n−2}
= X_{n} + α_{n+1}ϵ_{n}X_{n−1} + ϵ_{n}X_{n−2}

QED

Lemma 5 |\frac{y_{n−1}}{x_{n−1}} − α| ≤ \frac{|r_{n}|+|r_{n−1}|}{x_{n}x_{n−1}} Proof:

We know that,

α = \frac{ỹ_{n}}{x̃_{n}} = \frac{y_{n} + α_{n+1}ϵ_{n}y_{n−1} + ϵ_{n}y_{n−2}}{x_{n} + α_{n+1}ϵ_{n}x_{n−1} + ϵ_{n}x_{n−2}} .

Hence,

|\frac{y_{n−1}}{x_{n−1}} − α| = \frac{y_{n−1}}{x_{n−1}} − \frac{y_{n} + α_{n+1}ϵ_{n}y_{n−1} + ϵ_{n}y_{n−2}}{x_{n} + α_{n+1}ϵ_{n}x_{n−1} + ϵ_{n}x_{n−2}}
= \frac{(x_{n}y_{n−1} − y_{n}x_{n−1}) + (y_{n−1}x_{n−2} − x_{n−1}y_{n−2})ϵ_{n}}{x_{n−1}(x_{n} + α_{n+1}ϵ_{n}x_{n−1} + ϵ_{n}x_{n−2})}

Therefore, since α_{n} ≥ 0, ϵ_{n} ≥ 0 and x_{n} ≥ 0, then

|\frac{y_{n−1}}{x_{n−1}} − α| ≤ \frac{|r_{n−1}ϵ_{n} − r_{n}|}{x_{n}x_{n−1}}
|\frac{y_{n−1}}{x_{n−1}} − α| ≤ \frac{|r_{n}| + |r_{n−1}|}{x_{n}x_{n−1}}

QED

Lemma 6 |\frac{z_{n−1}}{x_{n−1}} − β| ≤ \frac{|q_{n}|+|q_{n−1}|}{x_{n}x_{n−1}} Proof:

The proof of this lemma is similar to the proof of the previous lemma.

QED

Lemma 7 If lim_{n→∞} \frac{|r_{n}|}{x_{n}x_{n−1}} = 0 and lim_{n→∞} \frac{|q_{n}|}{x_{n}x_{n−1}} = 0 then the triangle sequence is unique. Proof:

Since lim_{n→∞} \frac{|r_{n}|}{x_{n}x_{n−1}} = 0, we know that

lim_{n→∞} \frac{|r_{n−1}|}{x_{n−1}x_{n−2}} = 0

⇒ lim_{n→∞} \frac{|r_{n−1}|}{x_{n}x_{n−1}} = 0

⇒ lim_{n→∞} \frac{|r_{n}| + |r_{n−1}|}{x_{n}x_{n−1}} = 0

Similarily, since lim_{n→∞} \frac{|q_{n}|}{x_{n}x_{n−1}} = 0, we know that

lim_{n→∞} \frac{|q_{n−1}|}{x_{n−1}x_{n−2}} = 0

⇒ lim_{n→∞} \frac{|q_{n−1}|}{x_{n}x_{n−1}} = 0

⇒ lim_{n→∞} \frac{|q_{n}| + |q_{n−1}|}{x_{n}x_{n−1}} = 0

Therefore by the two previous lemmas,

lim_{n→∞} \frac{y_{n}}{x_{n}} = α

lim_{n→∞} \frac{z_{n}}{x_{n}} = β

Hence, the triangle sequence is unique.

QED

4 Uniqueness of Certain Bounded Triangle Sequences

4.1 Review of Key Ideas

In section 3, we demonstrated that the relations

lim_{n→∞} \frac{|r_{n}|}{x_{n}x_{n−1}} = 0

lim_{n→∞} \frac{|q_{n}|}{x_{n}x_{n−1}} = 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 x_{n}x_{n−1} 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.

Proof:

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}

|r_{n+1}| ≤ |r_{n−1}| + |r_{n−1}| + A|r_{n}|

Additionally,

q_{n+1} = q_{n−2} − q_{n−1} − a_{n+1}q_{n}

|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 + 1)^{0}
|r_{1}| = A  ≤ (A + 1)^{1}
|r_{2}| = A^{2} − 1≤ (A + 1)^{2}

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} + A(A + 1)^{n}

= (A + 1)^{n}((A + 1)^{−2} + (A + 1)^{−1} + A)

≤ (A + 1)^{n+1}

Note that the last of the above steps is justified since A ≥ 1 implies that (A + 1)^{−2} + (A + 1)^{−1} ≤ 1. 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 + 1)^{0}
|q_{1}| = 1 ≤ (A + 1)^{1}
|q_{2}| = A + 1≤ (A + 1)^{2}

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} + A(A + 1)^{n}

= (A + 1)^{n}((A + 1)^{−2} + (A + 1)^{−1} + 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 + \frac{3}{2})^{n} = (\frac{3}{2})^{n}.

|r_{0}| = 1≤ (\frac{3}{2})^{0}

|r_{1}| = 0≤ (\frac{3}{2})^{1}

|r_{2}| = 1≤ (\frac{3}{2})^{2}

Suppose that |r_{k}| ≤ (\frac{3}{2})^{k} for all k ≤ n. Then,

|r_{n+1}| ≤ |r_{n−2}| + |r_{n−1}|

≤ (\frac{3}{2})^{n−2} + (\frac{3}{2})^{n−1}

= (\frac{3}{2})^{n} ((\frac{3}{2})^{−2} + (\frac{3}{2})^{−1})

≤ (\frac{3}{2})^{n+1}

Therefore, by induction, |r_{n}| ≤ (\frac{3}{2})^{n} for all n.

Similarily, we will show that |q_{n}| ≤ (A + \frac{3}{2})^{n} = (\frac{3}{2})^{n}.

|q_{0}| = 0≤ (\frac{3}{2})^{0}

|q_{1}| = 1≤ (\frac{3}{2})^{1}

|q_{2}| = 1≤ (\frac{3}{2})^{2}

Suppose that |q_{k}| ≤ (\frac{3}{2})^{k} for all k ≤ n. Then,

|q_{n+1}| ≤ |q_{n−2}| + |q_{n−1}|

≤ (\frac{3}{2})^{n−2} + (\frac{3}{2})^{n−1}

= (\frac{3}{2})^{n} ((\frac{3}{2})^{−2} + (\frac{3}{2})^{−1})

≤ (\frac{3}{2})^{n+1}

Therefore, by induction, |q_{n}| ≤ (\frac{3}{2})^{n} for all n.

This means that for all A ≥ 0 we have that

|r_{n+1}| ≤ (A + \frac{3}{2})^{n+1}

|q_{n+1}| ≤ (A + \frac{3}{2})^{n+1}

Next we will bound x_{n} from below. We will show that x_{n} ≥ C(A + 2)^{n/2}.

x_{0} = 1
x_{1} = 1 + A
x_{2} = 2 + 2A
x_{3} = 3 + 3A + A^{2}

Therefore,

x_{2} ≥ (A + 2)x_{0}
x_{3} ≥ (A + 2)x_{1}

Suppose that

x_{k} ≥ (A + 2)x_{k−2}
x_{k+1} ≥ (A + 2)x_{k−1}

We will show that x_{k+2} ≥ (A+2)x_{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 + 1)x_{k} + (x_{k−1} + Ax_{k−1} + x_{k−2})
≥ (A + 1)x_{k} + (x_{k−1} + Ax_{k−2} + x_{k−3})
= (A + 1)x_{k} + x_{k}
= (A + 2)x_{k}

Therefore, by induction we have that x_{n+2} ≥ (A + 2)x_{n} for all n. This implies that x_{n} ≥ C(A + 2)^{n/2} for all n.

Finally we will show that \frac{|r_{n}|}{x_{n}x_{n−1}} and \frac{|q_{n}|}{x_{n}x_{n−1}} approach 0 as n grows without bound.

Recall that,

|r_{n}| ≤ (A + \frac{3}{2})^{n}

|q_{n}| ≤ (A + \frac{3}{2})^{n}

x_{n} ≥ C(A + 2)^{n/2}

x_{n−1} ≥ C(A + 2)^{(n/2−1/2)} ≥ D(A + 2)^{n/2}

Therefore, we have that

lim_{n→∞} \frac{|r_{n}|}{x_{n}x_{n−1}} ≤ lim_{n→∞} \frac{(A + \frac{3}{2})^{n}}{CD(A + 2)^{n}} = 0

lim_{n→∞} \frac{|q_{n}|}{x_{n}x_{n−1}} ≤ lim_{n→∞} \frac{(A + \frac{3}{2})^{n}}{CD(A + 2)^{n}} = 0

Thus, by Lemma 7, the triangle sequence a_{i} = A is unique for all A ≥ 0.

QED

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.

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 △_{a_{1},... ,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 △_{a_{1},... ,a_{n}} to those of △_{a_{n},... ,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 △_{a_{1},... ,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 △_{a_{1},... ,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 △_{a_{1},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 △_{a_{1},a_{2},... ,a_{n}} are written as

T_{a_{1},a_{2},... ,a_{n}}^{−n}(0, 0) = (\frac{y_{00}}{x_{00}} , \frac{z_{00}}{x_{00}} )

T_{a_{1},a_{2},... ,a_{n}}^{−n}(1, 0) = (\frac{y_{10}}{x_{10}} , \frac{z_{10}}{x_{10}} )

T_{a_{1},a_{2},... ,a_{n}}^{−n}(1, 1) = (\frac{y_{11}}{x_{11}} , \frac{z_{11}}{x_{11}} )

Then the vertices of △_{a_{n},a_{n−1},... ,a_{1}} are equal to

T_{a_{n},a_{n−1},... ,a_{1}}^{−n}(0, 0) = (\frac{y_{00}}{y_{10}} , \frac{y_{11} − y_{10}}{y_{10}} )

T_{a_{n},a_{n−1},... ,a_{1}}^{−n}(1, 0) = (\frac{x_{00}}{x_{10}} , \frac{x_{11} − x_{10}}{x_{10}} )

T_{a_{n},a_{n−1},... ,a_{1}}^{−n}(1, 1) = (\frac{z_{00} + x_{00}}{z_{10} + x_{10}} , \frac{z_{11} − z_{10} + x_{11} − x_{10}}{z_{10} + x_{10}} )

Proof:

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 (\frac{y}{x}, \frac{z}{x}) where x,y 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.

\underline{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

(0, 0)

(1, 0)

(1, 1)

It is easy to compute that

(\frac{0}{1}, \frac{1 − 1}{1}) = (0, 0)

(\frac{1}{1}, \frac{1 − 1}{1}) = (1, 0)

(\frac{0 + 1}{0 + 1}, \frac{1 − 0 + 1 − 1}{0 + 1}) = (1, 1)

This proves the lemma for the case n = 0, since △ is the reverse of itself.

\underline{Case n = 1} The vertices of △_{a_{1}} are

(1, 0)

(\frac{1}{a_{1} + 1}, \frac{1}{a_{1} + 1})

(\frac{1}{a_{1} + 2}, \frac{1}{a_{1} + 2})

It is again easy to compute that

(\frac{1}{1}, \frac{1 − 1}{1}) = (1, 0)

(\frac{1}{a_{1} + 1}, \frac{(a_{1} + 2) − (a_{1} + 1)}{a_{1} + 1}) = (\frac{1}{a_{1} + 1}, \frac{1}{a_{1} + 1})

(\frac{0 + 1}{1 + (a_{1} + 1)}, \frac{1 − 1 + (a_{1} + 2) − (a_{1} + 1)}{1 + (a_{1} + 1)}) = (\frac{1}{a_{1} + 2}, \frac{1}{a_{1} + 2})

This proves the lemma for the case n = 1, since △_{a_{1}} 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 △_{a_{2},... ,a_{n−1}} and construct the coordinates of the vertices of both △_{a_{1},... ,a_{n}} and △_{a_{n},... ,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.

T_{k}^{−1} (\frac{y}{x}, \frac{z}{x}) = (\frac{x}{ky + z + x}, \frac{y}{ky + z + x})

This notation refers to the preimage of the point (\frac{y}{x}, \frac{z}{x}) in triangle k.

We also need to check that the operations of reversing and applying T_{k}^{−1} preserve the reduced fractional form of the vertex coordinates. This means that we need to check that, under one of these operations, a pair (\frac{y}{x}, \frac{z}{x}) with (x, y, z) = 1 will produce another pair (\frac{y′}{x′} , \frac{z′}{x′}) with (x′, y′, z′) = 1. We need a couple of propositions.

Proposition 2 If (x, y, z) = 1, then T_{k}^{−1}(\frac{y}{x}, \frac{z}{x}) = (\frac{x}{ky+x+z} , \frac{y}{ky+x+z}) is in reduced fractional form.

Proof:

Suppose (\frac{x}{ky+x+z} , \frac{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.

QED

Proposition 3 If the points

(\frac{y_{00}}{x_{00}} , \frac{z_{00}}{x_{00}} )

(\frac{y_{10}}{x_{10}} , \frac{z_{10}}{x_{10}} )

(\frac{y_{11}}{x_{11}} , \frac{z_{11}}{x_{11}} )

are in reduced fractional form, then

(\frac{y_{00}}{y_{10}} , \frac{y_{11} − y_{10}}{y_{10}} )

(\frac{x_{00}}{x_{10}} , \frac{x_{11} − x_{10}}{x_{10}} )

(\frac{z_{00} + x_{00}}{z_{10} + x_{10}} , \frac{z_{11} − z_{10} + x_{11} − x_{10}}{z_{10} + x_{10}} )

are in reduced fractional form.

Proof:

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

We claim that

det ( \begin{matrix}x_{00} & x_{10} & x_{11} \\ y_{00} & y_{10} & y_{11} \\ z_{00} & z_{10} & z_{11} \end{matrix} ) = 1

This implies that

det ( \begin{matrix}x_{00} & x_{10} & x_{11} \\ y_{00} & y_{10} & y_{11} \\ x_{00} + z_{10} & x_{01} + z_{01} & x_{11} + z_{11} \end{matrix} ) = 1

which proves the lemma. To prove the claim, we use induction on n, the length of the sequence. We

define the vertices of △_{a_{1},... ,a_{k}} in reduced fractional form as follows:

(\frac{y_{00}^{(k)}}{x_{00}^{(k)}} , \frac{z_{00}^{(k)}}{x_{00}^{(k)}} )

(\frac{y_{10}^{(k)}}{x_{10}^{(k)}} , \frac{z_{10}^{(k)}}{x_{10}^{(k)}} )

(\frac{y_{11}^{(k)}}{x_{11}^{(k)}} , \frac{z_{11}^{(k)}}{x_{11}^{(k)}} )

and let

M^{(k)} = ( \begin{matrix}x_{00}^{(k)} & x_{10}^{(k)} & x_{11}^{(k)} \\ y_{00}^{(k)} & y_{10}^{(k)} & y_{11}^{(k)} \\ z_{00}^{(k)} & z_{10}^{(k)} & z_{11}^{(k)} \end{matrix} )

We start with the case k=0. The vertices of △ are:

(\frac{0}{1}, \frac{0}{1})

(\frac{1}{1}, \frac{0}{1})

(\frac{1}{1}, \frac{1}{1})

Then

det M^{(0)} = det ( \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ) = 1

Now, suppose that det M^{(n−1)} = 1. The matrix for T_{a_{n}^{−1}} is as follows:

[T_{a_{n}}^{−1}] = ( \begin{matrix}1 & a_{n} & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} )

Thus det[T_{a_{n}}^{−1}] = 1 Since M^{(n)} = [T_{a_{n}}^{−1}]M^{(n−1)} this implies that det M^{(n)} = 1 This proves the claim, and thus completes the proof of the proposition.

QED

Now, we are ready to construct the proof of the reversing lemma. Let △_{a_{2},... ,a_{n−1}} have vertices equal to

(\frac{y_{00}}{x_{00}} , \frac{z_{00}}{x_{00}} )

(\frac{y_{10}}{x_{10}} , \frac{z_{10}}{x_{10}} )

(\frac{y_{11}}{x_{11}} , \frac{z_{11}}{x_{11}} )

Now, if we apply the mapping T_{a_{1}}^{−1} to each of these points, we recover the vertices of △_{a_{1},... ,a_{n−1}}, which are calculated to be:

(\frac{x_{00}}{a_{1}y_{00} + z_{00} + x_{00}} , \frac{y_{00}}{a_{1}y_{00} + z_{00} + x_{00}} )

(\frac{x_{10}}{a_{1}y_{10} + z_{10} + x_{10}} , \frac{y_{10}}{a_{1}y_{10} + z_{10} + x_{10}} )

(\frac{x_{11}}{a_{1}y_{11} + z_{11} + x_{11}} , \frac{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 △_{a_{n−1},... ,a_{1}}, shown below.

(\frac{x_{00}}{x_{10}} , \frac{x_{11} − x_{10}}{x_{10}} )

(\frac{a_{1}y_{00} + z_{00} + x_{00}}{a_{1}y_{10} + z_{10} + x_{10}} , \frac{a_{1}y_{11} − a_{1}y_{10} + z_{11} − z_{10} + x_{11} − x_{10}}{a_{1}y_{10} + z_{10} + x_{10}} )

(\frac{y_{00} + a_{1}y_{00} + z_{00} + x_{00}}{y_{10} + a_{1}y_{10} + z_{10} + x_{10}} , \frac{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}} )

Applying T_{a_{n}}^{−1} to these points yields the vertices of △_{a_{n},... ,a_{1}}:

(\frac{x_{10}}{a_{n}x_{00} + x_{11}} , \frac{x_{00}}{a_{n}x_{00} + x_{11}} )

(\frac{a_{1}y_{10} + z_{10} + x_{10}}{a_{n}a_{1}y_{00} + a_{n}z_{00} + a_{n}x_{00} + a_{1}y_{11} + z_{11} + x_{11}} , \frac{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}} )

(\frac{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}} ,

\frac{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 △_{a_{1},... ,a_{n}}. First, we use the reversing lemma to calculate the vertices of △_{a_{n−1},... ,a_{2}} to

be:

(\frac{y_{00}}{y_{10}} , \frac{y_{11} − y_{10}}{y_{10}} )

(\frac{x_{00}}{x_{10}} , \frac{x_{11} − x_{10}}{x_{10}} )

(\frac{z_{00} + x_{00}}{z_{10} + x_{10}} , \frac{z_{11} − z_{10} + x_{11} + x_{10}}{z_{10} + x_{10}} )

Applying T_{a_{n}}^{−1} yields the vertices of △_{a_{n},... ,a_{2}}:

(\frac{y_{10}}{a_{n}y_{00} + y_{11}} , \frac{y_{00}}{a_{n}y_{00} + y_{11}} )

(\frac{x_{10}}{a_{n}x_{00} + x_{11}} , \frac{x_{00}}{a_{n}x_{00} + x_{11}} )

(\frac{z_{10} + x_{10}}{a_{n}z_{00} + a_{n}x_{00} + z_{11} + x_{11}} , \frac{z_{00} + x_{00}}{a_{n}z_{00} + a_{n}x_{00} + z_{11} + x_{11}} )

Now we apply the reversing lemma to calculate the vertices of △_{a_{2},... ,a_{n}}:

(\frac{y_{10}}{x_{10}} , \frac{z_{10}}{x_{10}} )

(\frac{a_{n}y_{00} + y_{11}}{a_{n}x_{00} + x_{11}} , \frac{a_{n}z_{00} + z_{11}}{a_{n}x_{00} + x_{11}} )

(\frac{a_{n}y_{00} + y_{00} + y_{11}}{a_{n}x_{00} + x_{00} + x_{11}} , \frac{a_{n}z_{00} + z_{00} + z_{11}}{a_{n}x_{00} + x_{00} + x_{11}} )

Finally, we apply T_{a_{1}}^{−1} to yield the vertices of △_{a_{1},... ,a_{n}}:

(\frac{x_{10}}{a_{1}y_{10} + x_{10} + z_{10}} , \frac{y_{10}}{a_{1}y_{10} + x_{10} + z_{10}} )

(\frac{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}} , \frac{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}} )

(\frac{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}} ,

\frac{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 △_{a_{n},... ,a_{1}}. Thus the lemma is true for sequences of length n and is thus true by induction for all sequences.

QED

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 △_{a_{1},... ,a_{n}} are given by:

T_{a_{1},... ,a_{n}}^{−n}(0, 0) = (\frac{y_{n−1}}{x_{n−1}} , \frac{z_{n−1}}{x_{n−1}} )

T_{a_{1},... ,a_{n}}^{−n}(0, 1) = (\frac{y_{n}}{x_{n}} , \frac{z_{n}}{x_{n}} )

T_{a_{1},... ,a_{n}}^{−n}(1, 1) = (\frac{y_{n−2} + y_{n}}{x_{n−2} + x_{n}} , \frac{z_{n−2} + z_{n}}{x_{n−2} + x_{n}} )

Proof:

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 △_{a_{1}} are:

(1, 0)

(\frac{1}{a_{1} + 1}, \frac{1}{a_{1} + 1})

(\frac{1}{a_{1} + 2}, \frac{1}{a_{1} + 2})

We also know that:

X_{−1} = (x_{−1}, y_{−1}, z_{−1}) = (1, 0, 0)
X_{0} = (x_{0}, y_{0}, z_{0}) = (1, 1, 0)
X_{1} = (x_{1}, y_{1}, z_{1}) = (a_{1} + 1, 1, 1)

We check that:

(\frac{y_{0}}{x_{0}} , \frac{z_{0}}{x_{0}} ) = (\frac{1}{1}, \frac{0}{1}) = (1, 0)

(\frac{y_{1}}{x_{1}} , \frac{z_{1}}{x_{1}} ) = (\frac{1}{a_{1} + 1}, \frac{1}{a_{1} + 1})

(\frac{y_{−1} + y_{1}}{x_{−1} + x_{1}} , \frac{z_{−1} + z_{1}}{x_{−1} + x_{1}} ) = (\frac{0 + 1}{(a_{1} + 1) + 1}, \frac{0 + 1}{(a_{1} + 1) + 1}) = (\frac{1}{a_{1} + 2}, \frac{1}{a_{1} + 2})

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 △_{a_{1},... ,a_{n−1}} are

(\frac{y_{n−2}}{x_{n−2}} , \frac{z_{n−2}}{x_{n−2}} )

(\frac{y_{n−1}}{x_{n−1}} , \frac{z_{n−1}}{x_{n−1}} )

(\frac{y_{n−3} + y_{n−1}}{x_{n−3} + x_{n−1}} , \frac{z_{n−3} + z_{n−1}}{x_{n−3} + x_{n−1}} )

We will then construct the vertices of △_{a_{1},... ,a_{n}} by applying the reversing lemma and the map T_{a_{n}}^{−1}.

First, the reversing lemma gives us the vertices of △_{a_{n−1},... ,a_{1}}

(\frac{y_{n−2}}{y_{n−1}} , \frac{y_{n−3}}{y_{n−1}} )

(\frac{x_{n−2}}{x_{n−1}} , \frac{x_{n−3}}{x_{n−1}} )

(\frac{z_{n−2} + x_{n−2}}{z_{n−1} + x_{n−1}} , \frac{z_{n−3} + x_{n−3}}{z_{n−1} + x_{n−1}} )

Next, we apply the map T_{a_{n}}^{−1} to give the vertices of △_{a_{n},... ,a_{1}}

(\frac{y_{n−1}}{y_{n−1} + a_{n}y_{n−2} + y_{n−3}} , \frac{y_{n−2}}{y_{n−1} + a_{n}y_{n−2} + y_{n−3}} )

(\frac{x_{n−1}}{x_{n−1} + a_{n}x_{n−2} + x_{n−3}} , \frac{x_{n−2}}{x_{n−1} + a_{n}x_{n−2} + x_{n−3}} )

(\frac{x_{n−1} + z_{n−1}}{x_{n−1} + a_{n}x_{n−2} + x_{n−3} + z_{n−1} + a_{n}z_{n−2} + z_{n−3}} , \frac{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}} )

By using the recursion relation X_{n} = X_{n−1}+a_{n}X_{n−2}+X_{n−3} we can simplify this

(\frac{y_{n−1}}{y_{n}} , \frac{y_{n−2}}{y_{n}} )

(\frac{x_{n−1}}{x_{n}} , \frac{x_{n−2}}{x_{n}} )

(\frac{x_{n−1} + z_{n−1}}{x_{n} + z_{n}} , \frac{x_{n−2} + z_{n−2}}{x_{n} + z_{n}} )

Finally, we apply the reversing lemma to these points to yield the vertices of △_{a_{1},... ,a_{n}}. They are

(\frac{y_{n−1}}{x_{n−1}} , \frac{z_{n−1}}{x_{n−1}} )

(\frac{y_{n}}{x_{n}} , \frac{z_{n}}{x_{n}} )

(\frac{y_{n−2} + y_{n}}{x_{n−2} + x_{n}} , \frac{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.

QED

Corollary 3 The area of △_{a_{1},... ,a_{n}} is equal to \frac{1}{2x_{n}x_{n−1}(x_{n}+x_{n−2})}.

Proof:

We know from the previous theorem that the vertices of △_{a_{1},... ,a_{n}} are:

(\frac{y_{n−1}}{x_{n−1}} , \frac{z_{n−1}}{x_{n−1}} )

(\frac{y_{n}}{x_{n}} , \frac{z_{n}}{x_{n}} )

(\frac{y_{n−2} + y_{n}}{x_{n−2} + x_{n}} , \frac{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:

E_{1} = (\frac{y_{n−2} + y_{n}}{x_{n−2} + x_{n}} − \frac{y_{n}}{x_{n}} , \frac{z_{n−2} + z_{n}}{x_{n−2} + x_{n}} − \frac{z_{n}}{x_{n}} , 0)

E_{2} = (\frac{y_{n−1}}{x_{n−1}} − \frac{y_{n}}{x_{n}} , \frac{z_{n−1}}{x_{n−1}} − \frac{z_{n}}{x_{n}} , 0)

Combining the fractions yields:

E_{1} = (\frac{y_{n−2}x_{n} − x_{n−2}y_{n}}{x_{n}(x_{n−2} + x_{n})} , \frac{z_{n−2}x_{n} − x_{n−2}z_{n}}{x_{n}(x_{n−2} + x_{n})} , 0)

E_{1} = (\frac{y_{n−1}x_{n} − x_{n−1}y_{n}}{x_{n}x_{n−1}} , \frac{z_{n−1}x_{n} − x_{n−1}z_{n}}{x_{n}x_{n−1}} , 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:

E_{1} = (\frac{−r_{n−1} + r_{n−2}}{x_{n}(x_{n−2} + x_{n})}, \frac{q_{n−1} − q_{n−2}}{x_{n}(x_{n−2} + x_{n})}, 0)

E_{2} = (\frac{−r_{n}}{x_{n}x_{n−1}} , \frac{q_{n}}{x_{n}x_{n−1}} , 0)

E_{1} × E_{2} = (0, 0, \frac{(q_{n})(−r_{n−1} + r_{n−2}) − (−r_{n})(q_{n} − q_{n−2})}{x_{n}^{2}x_{n−1}(x_{n} + x_{n−2})})
= (0, 0, \frac{(r_{n}q_{n−1} − q_{n}r_{n−1}) + (q_{n}r_{n−2} − r_{n}q_{n−2})}{x_{n}^{2}x_{n−1}(x_{n} + x_{n−2})})
= (0, 0, \frac{(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_{n}^{2}x_{n−1}(x_{n} + x_{n−2})})
= (0, 0, \frac{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_{n}^{2}x_{n−1}(x_{n} + x_{n−2})})
= (0, 0, \frac{x_{n−1} + a_{n}x_{n−2} + x_{n−3}}{x_{n}^{2}x_{n−1}(x_{n} + x_{n−2})} )
= (0, 0, \frac{x_{n}}{x_{n}^{2}x_{n−1}(x_{n} + x_{n−2})})
= (0, 0, \frac{1}{x_{n}x_{n−1}(x_{n} + x_{n−2})})

Thus, Area of△_{a_{1},... ,a_{n}} = \frac{1}{2}||E_{1} × E_{2}|| = \frac{1}{2x_{n}x_{n−1}(x_{n}+x_{n−2})}

QED

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.

Proof:

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.

QED

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

Figure 2: Triangle corresponding to Lemma 9

triangles △_{a_{1},... ,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 \overrightarrow{AB} be the vector from A to B.

Definition 9 Let the distance between two points A and B be defined as ∥ \overrightarrow{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 ∥ \overrightarrow{AM} ∥≥∥ \overrightarrow{AN} ∥⇒∥ \overrightarrow{AM} ∥≤∥ \overrightarrow{AB} ∥.

Proof:

Since B, M and N are colinear and since M lies between N and B, then \overrightarrow{BM} = a\overrightarrow{MN} for some a ≥ 0.

Let m_{1} be the largest component of the vector \overrightarrow{AM}. Without loss of generality, assume let m_{1} be non-negative. Then ∥ \overrightarrow{AM} ∥= m_{1}. Now let x_{1}, n_{1} and b_{1} be the corresponding components of \overrightarrow{MN}, \overrightarrow{AN} and \overrightarrow{AB} respectively. Then,

∥ \overrightarrow{AM} ∥≥∥ \overrightarrow{AN} ∥⇒ m_{1} ≥ n_{1}

Therefore, since \overrightarrow{AN} = \overrightarrow{AM} + \overrightarrow{MN}, we know that x_{1} ≤ 0. Additionally, since \overrightarrow{AB} = \overrightarrow{AM} − a\overrightarrow{MN}, we know that b_{1} ≥ m_{1}. This implies that ∥ \overrightarrow{AB} ∥≥∥ \overrightarrow{AM} ∥.

QED Lemma 10 Let △ABC be a triangle with ∥ \overrightarrow{AC} ∥≤∥ \overrightarrow{AB} ∥, then for any point M on the line segment between B and C, ∥ \overrightarrow{AM} ∥≤∥ \overrightarrow{AB} ∥.

Proof:

Case1: ∥ \overrightarrow{AM} ∥≥∥ \overrightarrow{AC} ∥

This case follows immediately by letting N = C in the previous lemma. Case2: ∥ \overrightarrow{AM} ∥<∥ \overrightarrow{AC} ∥ Since ∥ \overrightarrow{AC} ∥≤∥ \overrightarrow{AB} ∥ and ∥ \overrightarrow{AM} ∥<∥ \overrightarrow{AC} ∥, then ∥ \overrightarrow{AM} ∥≤∥ \overrightarrow{AB} ∥ as

desired.

QED

Definition 10 Let s_{0}(n) be the length of the side connecting T_{a_{1}...a_{n}}^{−n}(0, 0) and T_{a_{1}...a_{n}}^{−n}(1, 0).

Lemma 11 For any n ≥ 3 one of the following is true:

s_{0}(n) ≤ s_{0}(n − 1)
s_{0}(n) ≤ s_{0}(n − 2)
s_{0}(n) ≤ s_{0}(n − 3)

Proof:

Let

A = T_{a_{1}...a_{n−1}}^{−n}(1, 0)
B = T_{a_{1}...a_{n−1}}^{−n}(0, 0)
C = T_{a_{1}...a_{n−1}}^{−n}(1, 1)

Analogously let,

A′ = T_{a_{1}...a_{n−2}}^{−n}(1, 0)
B′ = T_{a_{1}...a_{n−2}}^{−n}(0, 0)
C′ = T_{a_{1}...a_{n−2}}^{−n}(1, 1)
A′′ = T_{a_{1}...a_{n−3}}^{−n}(1, 0)
B′′ = T_{a_{1}...a_{n−3}}^{−n}(0, 0)
C′′ = T_{a_{1}...a_{n−3}}^{−n}(1, 1)

Where T_{a_{1}...a_{0}}^{−n} is the identity transformation. Note that B = A′ and B′ = A′′. Case 1: ∥ \overrightarrow{AB} ∥≥∥ \overrightarrow{AC} ∥ By Lemma 10,

s_{0}(n) ≤∥ \overrightarrow{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.

Case 2: ∥ \overrightarrow{AC} ∥≥∥ \overrightarrow{AB} ∥ and ∥ \overrightarrow{BC} ∥≥∥ \overrightarrow{AC} ∥.

Since \overrightarrow{BC} is the longest side of △ABC and s_{0}(n) lies within △ABC, then s_{0}(n) ≤∥ \overrightarrow{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) =∥ \overrightarrow{A′B′} ∥≥∥ \overrightarrow{A′C} ∥=∥ \overrightarrow{BC} ∥≥ s_{0}(n)

Case 3: ∥ \overrightarrow{BC} ∥≤∥ \overrightarrow{AC} ∥, ∥ \overrightarrow{AB} ∥≤∥ \overrightarrow{AC} ∥ and ∥ \overrightarrow{B′A′} ∥≥∥ \overrightarrow{B′C′} ∥.

Recall that the side between B′ and C′ contains the points A and C. Therefore,

s_{0}(n − 2) =∥ \overrightarrow{B′A′} ∥≥∥ \overrightarrow{B′C′} ∥≥∥ \overrightarrow{AC} ∥

We know that \overrightarrow{AC} is the longest side of △ABC, in which s_{0}(n) lies. Therefore,

s_{0}(n − 2) ≥∥ \overrightarrow{AC} ∥≥ s_{0}(n)

Case 4: ∥ \overrightarrow{BC} ∥≤∥ \overrightarrow{AC} ∥, ∥ \overrightarrow{AB} ∥≤∥ \overrightarrow{AC} ∥ and ∥ \overrightarrow{B′A′} ∥≤∥ \overrightarrow{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 ∥ \overrightarrow{B′C′} ∥≥∥ \overrightarrow{B′A′} ∥,

s_{0}(n − 3) =∥ \overrightarrow{A′′B′′} ∥≥∥ \overrightarrow{A′′C′} ∥=∥ \overrightarrow{B′C′} ∥

Since \overrightarrow{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) ≥∥ \overrightarrow{B′C′} ∥≥∥ \overrightarrow{AC} ∥≥ s_{0}(n)

QED

Definition 11 Let F_{n} = max(\frac{|r_{n−4}|}{x_{n−4}x_{n−5}} , \frac{|r_{n−3}|}{x_{n−3}x_{n−4}} , \frac{|r_{n−2}|}{x_{n−2}x_{n−3}} ).

Definition 12 Let G_{n} = max(\frac{|q_{n−4}|}{x_{n−4}x_{n−5}} , \frac{|q_{n−3}|}{x_{n−3}x_{n−4}} , \frac{|q_{n−2}|}{x_{n−2}x_{n−3}} ).

Lemma 12 If a_{n} = C and a_{n−1} > C then \frac{|r_{n}|}{x_{n}x_{n−1}} ≤ \frac{C^{2}+3C+2}{C^{2}+4C+4}F_{n}.

Proof:

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 x_{n}x_{n−1} from below in terms of x_{n−2}x_{n−3}. We will then use these two bounds fact that F_{n} ≤ \frac{max(|r_{n−2}|,|r_{n−3}|,|r_{n−4}|}{x_{n−2}x_{n−3}} to bound \frac{|r_{n}|}{x_{n}x_{n−1}} from above in terms of F_{n}.

Recall that

|r_{n}| = |r_{n−3} − r_{n−2} − 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}| = |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,

|r_{n}| ≤ (Ca_{n−1} + 2C + 2) max(|r_{n−2}|, |r_{n−3}|, |r_{n−4}|)

Additionally,

x_{n}x_{n−1} = (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}x_{n−1} = ((x_{n−4} + a_{n−1} + 1)x_{n−3} + (a_{n} + 1)x_{n−2}) (x_{n−4} + a_{n−1}x_{n−3} + x_{n−2}) (2)

By dropping the x_{n−4} terms and using that a_{n} = C, we get

x_{n}x_{n−1} ≥ ((a_{n−1} + 1)x_{n−3} + (C + 1)x_{n−2})(a_{n−1}x_{n−3} + x_{n−2})

By dropping the x_{n−3}^{2} terms we get

x_{n}x_{n−1} ≥ (Ca_{n−1} + 2a_{n−1} + C + 2)x_{n−2}x_{n−3}

Therefore,

\frac{|r_{n}|}{x_{n}x_{n−1}} = \frac{(Ca_{n−1} + 2C + 2) max(|r_{n−2}|, |r_{n−3}|, |r_{n−4}|)}{(Ca_{n−1} + 2a_{n−1} + C + 2)x_{n−2}x_{n−3}}

≤ \frac{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,

\frac{|r_{n}|}{x_{n}x_{n−1}} ≤ \frac{C^{2} + 3C + 2}{C^{2} + 4C + 4}F_{n}

QED Lemma 13 If a_{n} = C and a_{n−1} > C then \frac{|q_{n}|}{x_{n}x_{n−1}} ≤ \frac{C^{2}+3C+2}{C^{2}+4C+4}G_{n}

Proof:

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.

QED

Lemma 14 If a_{n} = C, a_{n−1} > C and a_{n+1} ≥ C, then \frac{|r_{n+1}|}{x_{n+1}x_{n}} ≤ \frac{C^{2}+3C+2}{C^{2}+4C+4}F_{n}

Proof:

By replacing n with n + 1 in Equation 1, we get

|r_{n+1}| = | − a_{n+1}r_{n−3} + (a_{n+1} + 1)r_{n−2} + (a_{n}a_{n+1} − 1)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+1}| = |(a_{n}a_{n+1} − 1)r_{n−4} + (1 − a_{n+1} − Ca_{n+1})r_{n−3} + (a_{n+1} + 1 + 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−1} + 3) 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} + 1)x_{n−2} + (a_{n+1} + 1)x_{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} + 1)x_{4} + (a_{n−1} + a_{n−1}a_{n+1} + 1)x_{n−3} + (a_{n+1} + a_{n} + 2)x_{n−2}) (x_{n−4} + (a_{n−1} + 1)x_{n−3} + (a_{n} + 1)x_{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} + 1)x_{n−3} + (a_{n+1} + C + 2)x_{n−2})((a_{n−1} + 1)x_{n−3} + (C + 1)x_{n−2})

By dropping the x_{n−3}^{2} 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} + 5)x_{n−2}x_{n−3}

Therefore,

\frac{|r_{n}|}{x_{n+1}x_{n}} ≤ \frac{2Ca_{n+1} + a_{n−1}Ca_{n+1} + 2a_{n+1} + a_{n−1} + 3}{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} + 5}F_{n}

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,

\frac{|r_{n}|}{x_{n+1}x_{n}} ≤ \frac{3Ca_{n+1} + C^{2}a_{n+1} + C + 2a_{n+1} + 4}{C^{2}a_{n+1} + 3C^{2} + 4Ca_{n+1} + 10C + 4a_{n+1} + 8}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,

\frac{|r_{n}|}{x_{n+1}x_{n}} ≤ \frac{C^{2} + 3C + 2}{C^{2} + 4C + 4}F_{n}

QED

Lemma 15 If a_{n} = C, a_{n−1} > C and a_{n+1} ≥ C then \frac{|q_{n+1}|}{x_{n+1}x_{n}} ≤ \frac{C^{2}+3C+2}{C^{2}+4C+4}G_{n}

Proof:

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}.

QED

Lemma 16 If a_{n} = C, a_{n−1} > C, a_{n+1} ≥ C and a_{n+2} ≥ C then \frac{|r_{n+2}|}{x_{n+2}x_{n+1}} ≤ \frac{C^{2}+3C+2}{C^{2}+4C+4}F_{n}

Proof:

By replacing n with n + 1 in Equation 3, we get

|r_{n+2}| = |(a_{n+2}a_{n+1} − 1)r_{n+3} − (a_{n+2} + a_{n+2}a_{n+1} − 1)r_{n−2} + (a_{n+2} + 1 − 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+2} + 1 − a_{n+2}a_{n+1}C + C)r_{n−4} +

(a_{n+2}a_{n+1} − a_{n+2} − 2 + a_{n+2}a_{n+1}C − C)r_{n−3} −

(a_{n+2} + a_{n+2}a_{n+1} − 1 + 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+2} + 4) max(|r_{n−2}|, |r_{n−3}|, |r_{n−4}|) (5)

Additionally, by replacing n with n + 1 in Equation 4, we get

x_{n+2}x_{n+1} = ((a_{n+2}+1)x_{n−3}+(a_{n}a_{n+2}+a_{n}+1)x_{n−2}+(a_{n+1}+a_{n+2}+2)x_{n−1})

(x_{n−3} + (a_{n} + 1)x_{n−2} + (a_{n+1} + 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+2}x_{n+1} = ((a_{n+1}+a_{n+2}+2)x_{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} + 3)x_{n−2})((a_{n+1} + 1)x_{n−4} +

(1 + a_{n+1}a_{n−1} + a_{n−1})x_{n−3} + (a_{n} + 2 + 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+3)x_{n−2})

((1 + a_{n+1}a_{n−1} + a_{n−1})x_{n−3} + (C + 2 + a_{n+1})x_{n−2})

By dropping the x_{n−3}^{2} 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_{n+1}^{2}+3a_{n+2}+8a_{n+1}+7)

+ (Ca_{n+2}a_{n+1} + C^{2}a_{n+2} + C^{2} + 6Ca_{n+2} + 2Ca_{n+1} + 7C +

a_{n+1}^{2} + 2a_{n+2}a_{n+1} + 5a_{n+2} + 7a_{n+1} + 11)x_{n−2}x_{n−3} (6)

Now let

J = a_{n−1}(Ca_{n+2}a_{n+1} + C + a_{n+2} + 1) +

(2Ca_{n+2}a_{n+1} + 2C + 2a_{n+2}a_{n+1} + 3a_{n+2} + 4)

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_{n+1}^{2}+3a_{n+2}+8a_{n+1}+7)

+ (Ca_{n+2}a_{n+1} + C^{2}a_{n+2} + C^{2} + 6Ca_{n+2} + 2Ca_{n+1} + 7C +

a_{n+1}^{2} + 2a_{n+2}a_{n+1} + 5a_{n+2} + 7a_{n+1} + 11)

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

x_{n+2}x_{n+1} ≥ Kx_{n−2}x_{n−3}

Therefore,

\frac{|r_{n+2}|}{x_{n+2}x_{n+1}} ≤ \frac{J}{K} F_{n}

Since the derivative of \frac{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+1} + 4) + 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+1} + 11C + 4a_{n+1} + 8) +

(2C^{2}a_{n+1} + 4C^{2} + 2Ca_{n+1}^{2} + 12Ca_{n+1} + 17C + 3a_{n+1}^{2} + 15a_{n+1} + 18)

Therefore we have

\frac{|r_{n+2}|}{x_{n+2}x_{n+1}} ≤ \frac{J_{1}}{K_{1}} F_{n}

Since the derivative of \frac{J_{1}}{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,

\frac{|r_{n+2}|}{x_{n+2}x_{n+1}} ≤ \frac{a_{n+1}C^{2} + 3a_{n+1}C + 2a_{n+1} + C + 4}{a_{n+1}C^{2} + 4a_{n+1}C + 4a_{n+1} + 3C^{2} + 11C + 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,

\frac{|r_{n+2}|}{x_{n+2}x_{n+1}} ≤ \frac{C^{2} + 3C + 2}{C^{2} + 4C + 4}

QED

Lemma 17 If a_{n} = C, a_{n−1} > C, a_{n+1} ≥ C and a_{n+2} ≥ C then \frac{|q_{n+2}|}{x_{n+2}x_{n+1}} ≤

\frac{C^{2}+3C+2}{C^{2}+4C+4}G_{n}

Proof:

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}.

QED Lemma 18 s_{0}(n) =∥ (\frac{r_{n}}{x_{n}x_{n−1}} , \frac{q_{n}}{x_{n}x_{n−1}} ) ∥

Proof:

From Lemma 3 we know that:

T_{a_{1}...a_{n}}^{−n}(0, 0) = (\frac{y_{n−1}}{x_{n−1}} , \frac{z_{n−1}}{x_{n−1}} )
T_{a_{1}...a_{n}}^{−n}(1, 0) = (\frac{y_{n}}{x_{n}} , \frac{z_{n}}{x_{n}} )

Therefore,

s_{0}(n) =∥ (\frac{y_{n−1}}{x_{n−1}} , \frac{z_{n−1}}{x_{n−1}} ) − (\frac{y_{n}}{x_{n}} , \frac{z_{n}}{x_{n}} ) ∥

=∥ (\frac{y_{n}x_{n−1} − x_{n}y_{n−1}}{x_{n}x_{n−1}} , \frac{z_{n}x_{n−1} − x_{n}z_{n−1}}{x_{n}x_{n−1}} ) ∥

=∥ (\frac{r_{n}}{x_{n}x_{n−1}} , \frac{q_{n}}{x_{n}x_{n−1}} ) ∥

QED

Lemma 19 If a_{n} = C, a_{n−1} > C, a_{n+1} ≥ C and a_{n+2} ≥ C then s_{0}(i) ≤ \frac{C^{2}+3C+2}{C^{2}+4C+4} max(s_{0}(n − 2), s_{0}(n − 3), s_{0}(n − 4)) for all i ∈ \{n, n + 1, n + 2\}.

Proof:

From Lemma 18 we know that,

max(s_{0}(n − 2), s_{0}(n − 3), s_{0}(n − 4)) = max \frac{|r_{n−2}|}{x_{n−2}x_{n−3}} , \frac{|r_{n−3}|}{x_{n−3}x_{n−4}} , \frac{|r_{n−4}|}{x_{n−4}x_{n−5}} ,

\frac{|q_{n−2}|}{x_{n−2}x_{n−3}} , \frac{|q_{n−3}|}{x_{n−3}x_{n−4}} , \frac{|q_{n−4}|}{x_{n−4}x_{n−5}} )

and

s_{0}(i) = max ((\frac{|r_{i}|}{x_{i}x_{i−1}} , (\frac{|q_{i}|}{x_{i}x_{i−1}} ))

From Lemmas 12 through 17, we know that for all i ∈ {n, n + 1, n + 2},

\frac{|r_{i}|}{x_{i}x_{i−1}} ≤ \frac{C^{2} + 3C + 2}{C^{2} + 4C + 4} max(\frac{|r_{n−2}|}{x_{n−2}x_{n−3}} , \frac{|r_{n−3}|}{x_{n−3}x_{n−4}} , \frac{|r_{n−4}|}{x_{n−4}x_{n−5}} )
\frac{|q_{i}|}{x_{i}x_{i−1}} ≤ \frac{C^{2} + 3C + 2}{C^{2} + 4C + 4} max(\frac{|q_{n−2}|}{x_{n−2}x_{n−3}} , \frac{|q_{n−3}|}{x_{n−3}x_{n−4}} , \frac{|q_{n−4}|}{x_{n−4}x_{n−5}} )

Therefore, for all i ∈ {n, n + 1, n + 2},

s_{0}(i) ≤ \frac{C^{2} + 3C + 2}{C^{2} + 4C + 4} max(s_{0}(n − 2), s_{0}(n − 3), s_{0}(n − 4))

QED

Lemma 20 If lim_{n→∞} s_{0}(n) = 0 then the triangle sequence (a_{1}, a_{2}, a_{3}, …) is unique.

Proof:

From Lemma 18, we know that s_{0}(n) = max (\frac{|r_{n}|}{x_{n}x_{n−1}} , \frac{|q_{n}|}{x_{n}x_{n−1}} ). Therefore,

s_{0}(n) ≥ \frac{|r_{n}|}{x_{n}x_{n−1}}
s_{0}(n) ≥ \frac{|q_{n}|}{x_{n}x_{n−1}}

This that lim_{n→∞} s_{0}(n) = 0 implies

lim_{n→∞} s_{0}(n) = 0
lim_{n→∞} s_{0}(n) = 0

Hence by Lemma 7 we know that the triangle sequence is unique.

QED

Theorem 4 If any number occurs infinitely many times in a triangle sequence, then that triangle sequence is unique.

Proof:

Given a triangle sequence (a_{1}, a_{2}, a_{3}, …), let C be the smallest term that occurs infinitely many times in the sequence.

\underline{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.

\underline{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_{3}, …) such that N_{1} > m + 1, and for all i, a_{N_{i}} = C and a_{N_{i}−1} > C. Note that since N_{1} > m + 1, it follows immediately that a_{N_{i}+1} ≥ C and a_{N_{i}+1} ≥ C. Let

M = max(s_{0}(N_{0} − 2), s_{0}(N_{0} − 3), s_{0}(N_{0} − 4))

We will show that for all n ≥ N_{i},

s_{0}(n) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{i} M

First consider the base case where i = 0. By Lemma 11, for all s_{0}(n) ≥ N_{0},

s_{0}(n) ≤ M

Now assume that for all n ≥ N_{k},

s_{0}(n) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{k} M

Therefore, we know that

max(s_{0}(N_{k+1} − 2), s_{0}(N_{k+1} − 3), s_{0}(N_{k+1} − 4)) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{k} M

So by Lemma 19 we know that

s_{0}(N_{k+1}) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{k+1} M

s_{0}(N_{k+1} + 1) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{k+1} M

s_{0}(N_{k+1} + 2) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{k+1} M

Hence by Lemma 11, we know that for all n ≥ N_{k+1},

s_{0}(n) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{k+1} M

Therefore, by induction we have that for all i and for all n ≥ N_{i}

s_{0}(n) ≤ (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{i} M

Therefore we know that

lim_{n→∞} s_{0}(n) ≤ lim_{i→∞} (\frac{C^{2} + 3C + 2}{C^{2} + 4C + 4})^{i} M

Therefore since \frac{C^{2}+3C+2}{C^{2}+4C+4} < 1,

lim_{n→∞} s_{0}(n) = 0

Thus, by Lemma 20, the triangle sequence is unique.

QED

References

[1] Thomas Garrity. On periodic sequences for algebraic numbers. 1999.