{'fraction_non_alphanumeric': 0.053168960822870974, 'fraction_numerical': 0.01767256592547774, 'mean_word_length': 3.703225041498696, 'pattern_counts': {'":': 0, '<': 9, '<?xml version=': 0, '>': 7, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 25, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Frieze patterns with coefficients

Frieze patterns, as introduced by Coxeter in the 1970's, are closely related to cluster algebras without coefficients. A suitable generalization of frieze patterns, linked to cluster algebras with coefficients, has only briefly appeared in an unpublished manuscript by Propp. In this paper we study these frieze patterns with coefficients systematically and prove various fundamental results, generalizing classic results for frieze patterns. As a consequence we see how frieze patterns with coefficients can be obtained from classic frieze patterns by cutting out subpolygons from the triangulated polygons associated to classic Conway-Coxeter frieze patterns. We address the question of which frieze patterns with coefficients can be obtained in this way and solve this problem completely for triangles. Finally, we prove a finiteness result for frieze patterns with coefficients by showing that for a given boundary sequence there are only finitely many (non-zero) frieze patterns with coefficients with entries in a discrete subset of the complex numbers.


Introduction
Frieze patterns have been introduced by Coxeter [3] and a beautiful theory for frieze patterns over positive integers has been developed subsequently by Conway and Coxeter [2]. Only some three decades later the importance of frieze patterns for other areas of mathematics has become clearer when Fomin and Zelevinsky invented cluster algebras [8]. Namely, the exchange condition in cluster algebras mimics the local condition defining frieze patterns. In this way, starting with a set of indeterminates, the entries in a frieze pattern (over the field of rational functions) are precisely the cluster variables of the corresponding cluster algebra of Dynkin type A with trivial coefficients. Via cluster algebras, frieze patterns are now connected to many areas of mathematics and therefore, frieze patterns are currently a very active area of research; see the survey by Morier-Genoud [10] for more details.
such that all neighbouring 2 × 2-matrices have determinant 1. From the cluster algebra viewpoint the two bounding diagonals of 1's mean to set all the coefficients (or frozen variables) in the cluster algebra equal to 1. Basically, the entire extensive recent literature on frieze patterns deals with such classic frieze patterns.
In the theory of cluster algebras the coefficients are very important, so it would be natural to consider more general frieze patterns where the bounding diagonals can have arbitrary entries. Such frieze patterns have appeared briefly already in an unpublished manuscript by Propp [11]. In particular, one finds there the modified local condition on 2 × 2-determinants, now involving the boundary entries. This is set up in such a way that, again, the entries in a certain frieze pattern (over the field of rational functions) are precisely the cluster variables of the corresponding cluster algebra with coefficients.
Surprisingly, it seems that this topic has so far not been considered further in the literature. One aim of the present paper is to develop the general theory of these more general frieze patterns, which we call frieze patterns with coefficients; see Definition 2.1 for a precise definition. As in the classic case, a satisfactory theory can only be expected for tame frieze patterns with coefficients, i.e. imposing that all adjacent 3 × 3-determinants are 0. (There are far too many wild frieze patterns, see [4]. ) In Sections 2 and 3 we develop a theory for tame frieze patterns with coefficients, generalizing several of the well-known results for classic frieze patterns.
We introduce certain 2 × 2-matrices, called µ-matrices, which govern the propagation along the rows and the columns of a frieze pattern with coefficients, that is, multiplication with such matrices transforms two consecutive entries in a row or column to the next two consecutive entries. See Definition 2.7 and Proposition 2.9. Moreover, we present in Corollary 2.10 formulae for how every entry in the frieze pattern appears in a certain product of µ-matrices.
As another application of the propagation formulae we prove that every tame frieze pattern with coefficients satisfies a glide reflection, see Theorem 2.11. A fundamental domain for the entries of the tame frieze pattern with coefficients under the glide reflection can be indexed in such a way that it corresponds bijectively to the edges and diagonals of a regular (n + 3)-gon (where n is the height of the frieze pattern). This is the starting point in Section 3. It allows a useful alternative viewpoint for tame frieze patterns with coefficients: the entries in such a frieze pattern become the labels for the edges and diagonals of a polygon (where the boundary entries become the labels for the edges). We show in Theorem 3.3 that these labels have to satisfy many more than the defining local conditions, namely the so-called Ptolemy relations.
For distinguishing the two viewpoints we speak of a tame frieze with coefficients if we consider a map from the edges and diagonals of a polygon satisfying all Ptolemy relations.
In Section 4 we present a construction for obtaining friezes with coefficients from classic friezes: given a classic frieze one can cut out any subpolygon and get a frieze with coefficients. If this frieze with coefficients is a classic frieze, then it is a summand in the sense of [5] or [9].
This naturally leads to the question which friezes with coefficients can be obtained from classic friezes by cutting out subpolygons. Small examples already show that not every frieze with coefficients can be obtained from classic friezes in this way.
The rest of Section 4 and also Section 5 deal with this fundamental question for friezes with coefficients over positive integers. Our results not only give insight into friezes with coefficients but also shed new light on the classic Conway-Coxeter friezes, i.e. friezes over positive integers with all boundary entries equal to 1.
In Lemma 4.2 we show that any triangle cut out of a classic Conway-Coxeter frieze has the property that the greatest common divisor of any two of the labels on the triangle divides the third label. Conversely, we show in Lemma 4.4 that given two coprime natural numbers a, b there exists a classic Conway-Coxeter frieze containing a triangle with labels 1, a, b. The proof of this result exhibits a close connection between friezes and the Euclidean Algorithm, which we think might be of interest in its own right.
Note that the results just mentioned do not completely settle the question which labelled triangles can actually appear as subpolygons of classic Conway-Coxeter friezes. It turns out that this problem is rather subtle and we address it in Section 5. As a main result we give a complete classification of the problem, that is, we describe the triples (a, b, c) ∈ N 3 such that there exists a triangle in some classic Conway-Coxeter frieze with labels a, b, c: Theorem 1.1 (Thm. 5.12). Let a, b, c ∈ N. Then the triple (a, b, c) appears as labels of a triangle in a classic Conway-Coxeter frieze if and only if gcd(a, b) = gcd(b, c) = gcd(a, c) and where ν 2 (n) is the 2-valuation of n.
In Section 6 we go back to frieze patterns with coefficients having entries in any subset R ⊆ C (not necessarily integers as in the previous section). We address another fundamental question: for a given boundary sequence, how many frieze patterns with coefficients exist over R? Our main interest is whether there are finitely or infinitely many such frieze patterns. Easy examples show that allowing entries to be 0 rather quickly leads to infinitely many frieze patterns. So we restrict to friezes patterns with non-zero entries.
In Lemma 6.1 we prove a rather general finiteness result, which for a given boundary sequence yields an upper bound for the values in the quiddity cycle of the corresponding frieze pattern (i.e. for the entries in the diagonal next to the boundary diagonal, or the labels of the length 2 diagonals in the frieze). This generalizes a similar result for classic frieze patterns from [6,Theorem 3.6]. As a consequence we can conclude that for every discrete subset R of the complex numbers and every boundary sequence there exist only finitely many frieze patterns over R \{0} with the given boundary sequence; see Proposition 6.3.

Definition and fundamental properties
In this section we introduce frieze patterns with coefficients. This concept goes back to an unpublished manuscript by Propp [11]. A large amount of research has been carried out on classic frieze patterns, i.e. where all coefficients are set to 1. A systematic treatment of the far more general frieze patterns with coefficients does not seem to be contained in the literature so far. Therefore, we give a detailed account here, in particular proving several fundamental properties of such frieze patterns with coefficients, generalizing well-known properties of the classic frieze patterns.
A frieze pattern with coefficients of height n over R is an infinite array of the form . . . . . .
where we also set c i,i = 0 = c i,n+i+3 for all i ∈ Z, such that the following holds: (i) c i,j ∈ R for all i ∈ Z and i < j < n + i + 3.
(iii) For every (complete) adjacent 2 × 2-submatrix we have (1) The diagonals with entries c i,i+1 and c i,n+i+2 are called the boundary of the frieze pattern with coefficients. Part (iii) of the definition says that the determinant of every (complete) adjacent 2 × 2-submatrix is given by the product of two specific entries on the boundary of the frieze pattern, see the blue numbers in Figure 1.
Since c i+1,i+2 = 0 by part (ii) and R ⊆ C has no zero divisors we conclude that This means that there is a glide symmetry on the boundary entries. In particular, applying (2.1) twice we get c i,i+1 = c n+i+3,n+i+4 for all i ∈ Z. That is, for a frieze pattern with coefficients of height n the boundary is periodic of period n + 3. Hence, the entire boundary is determined by a sequence of n + 3 consecutive boundary entries (c i,i+1 , . . . , c n+i+2,n+i+3 ). (3) From the glide symmetry in (2.1) it also follows that the local condition (E i,j ) in Definition 2.1 could as well be stated in terms of two other boundary entries, namely the red numbers instead of the blue numbers in Figure 1. It is easy to check from the local conditions that x = z and y = w. Then this is a frieze pattern with coefficients if and only if xy = ac + bd. In other words, given a, b, c, d ∈ N we obtain for each divisor of ac + bd such a frieze pattern with coefficients.
As an explicit example we have the following frieze pattern with coefficients over N: . . . For general subsets R ⊆ C there are many frieze patterns with coefficients having entries in R, and many of them do not exhibit nice symmetry properties. This is already the case for classic frieze patterns, see [4]. As in the classic case, a satisfactory theory can only be expected for tame frieze patterns.
Definition 2.5. Let C be a frieze pattern with coefficients as in Definition 2.1. Then C is called tame if every complete adjacent 3 × 3-submatrix of C has determinant 0.
We aim at showing that every tame frieze pattern with coefficients satisfies a glide symmetry. We have already seen in Remark 2.2 that the boundary entries have a glide symmetry, namely we have c i,i+1 = c i+1,n+i+3 for all i ∈ Z.
For extending this to all entries of the frieze patterns with coefficients we need formulae describing how entries are 'propagated' along rows and columns, that is, how to obtain two consecutive entries in a row or column from the previous two entries. To this end it is useful to slightly extend the definition of a frieze pattern with coefficients by introducing two extra diagonals.
Remark 2.6. Let C be a frieze pattern with coefficients over R ⊆ C as in Definition 2.1. The corresponding extended frieze pattern with coefficientsĈ is the infinite array of the form That is, in addition to Definition 2.1 we set We mention some fundamental properties of these extended frieze patterns with coefficients.
(1) All local conditions (E i,j ) also hold inĈ. In fact, the only new equations to check are (E i,i ) and (E i,n+i+3 ) for i ∈ Z. The former one reads which is true since c i,i = 0 by Definition 2.1, c i+1,i = −c i,i+1 by the above definition and c i+1,n+i+3 = c i,i+1 by (2.1). Similarly, one checks that (E i,n+i+3 ) holds.
(2) C is tame if and only ifĈ is tame (cf. Definition 2.5). In fact, the new complete adjacent 3 × 3-submatrices inĈ are of the form The matrix on the left can be shown to have determinant 0 by expanding along the first column, using equation (E i−1,i+1 ) and the glide symmetry formula (2.1). Similarly, the matrix on the right has determinant 0.
Instead of only adding two further diagonals to a frieze pattern with coefficients C one could extend C to an entire SL 2 -tiling of the plane, still satisfying all local conditions and tameness. This is well-known for classic frieze patterns and can easily be transferred to frieze patterns with coefficients. We do not introduce this viewpoint here since it is not used in the present paper.
We can now start to develop the propagation formulae. It will turn out that the following matrices are crucial for this.
The following result provides the propagation formulae. Note that we write the entries as they appear in the frieze pattern with coefficients in Definition 2.1, that is as row vectors for the propagation along rows and as column vectors for the propagation along columns. Proposition 2.9. Let R ⊆ C be a subset. Let C = (c i,j ) be a tame frieze pattern with coefficients over R of height n, andĈ the corresponding extended frieze pattern (see Remark 2.6). We write Proof. We prove part (1), the proof of part (2) is similar. We first consider the cases j = i and j = i + 1 separately. For j = i we have as claimed. For j = i + 1 we similarly get Then we consider the following complete adjacent 3 × 3submatrix of the extended frieze pattern The first two columns of M cannot be linearly dependent because the upper left 2×2-submatrix has determinant d i d j−1 = 0, by Definition 2.1 and Remark 2.2. But C is tame by assumption, hence the determinant of M is zero, so there are suitable numbers s, t such that and we conclude that Thus we see from the shape of the matrix M that for fixed j − 1, there is a t j−1 = t such that for all i we have In particular, this equation holds for i = j − 1 and we get As a consequence we can give a useful formula for determining the entries of the frieze pattern with coefficients from the corresponding µ-matrices. Corollary 2.10. Let R ⊆ C be a subset. Let C = (c i,j ) be a tame frieze pattern with coefficients over R of height n, and as before set We consider the extended frieze pattern with coefficients as in Remark 2.6, in particular we set (1) From the first part of Proposition 2.9 we know that and the claim follows.
As another main result of this section we can now prove that the entries of a tame frieze pattern with coefficients are invariant under a glide reflection. This will become crucial in the next sections.
Theorem 2.11. Let R ⊆ C be a subset. Let C = (c i,j ) be a tame frieze pattern with coefficients over R of height n. Then for all entries of C we have Proof. We consider the i th row and the (n + i + 3) rd column in the extended frieze pattern corresponding to C: . . .
We propagate along the (n + i + 3) rd column using Proposition 2.9 (2) and get This implies that where the last equality holds by Corollary 2.10.

Ptolemy relations
In the previous section we have seen that every tame frieze pattern with coefficients over some subset R ⊆ C satisfies a glide symmetry, see Theorem 2.11. More precisely, Theorem 2.11 states that in Figure 2 the red area forms a fundamental domain for the action of the glide symmetry; that is, the blue row is the same as the rightmost column of the red area and the green column is the same as the top row of the red area. Note that the indices of the entries in the red fundamental area are in bijection with the edges and diagonals of a regular (n + 3)-gon (viewed as pairs of vertices), with vertices labelled 1, 2, . . . , n + 3. This means that we can view every tame frieze pattern with coefficients of height n over R as a map on the edges and diagonals of a regular (n + 3)-gon with values in R.

Convention:
We use the notion (tame) frieze pattern with coefficients for an infinite array as in Definition 2.1 and the notion (tame) frieze with coefficients for a corresponding map from edges and diagonals of a regular polygon. The entries in a frieze (pattern) with coefficients are tightly connected by many remarkable equations, in addition to the local conditions given in equations (E i,j ) in Definition 2.1.
Definition 3.1. Let C = (c i,j ) be a tame frieze with coefficients on a regular m-gon. For any indices 1 ≤ i ≤ j ≤ k ≤ ≤ m the corresponding Ptolemy relation is the equation (1) By Definition 2.1, the entries c i,i are zero for all i ∈ Z. This implies that the equations (E i,j,k, ) always hold if there are equalities among the numbers i, j, k, .
(2) If i < j < k < then the Ptolemy relation (E i,j,k, ) can be visualized as follows The Ptolemy relation asks that in the quadrilateral, the product of the labels on the diagonals equals the sum of the products of labels of opposite sides; cf. Ptolemy's theorem from Elementary Euclidean Geometry.
(3) The local conditions for a frieze pattern from Definition 2.1 are a special case of Ptolemy relations. Namely, the local condition Theorem 3.3. Every tame frieze with coefficients over some subset R ⊆ C satisfies all Ptolemy relations.
Proof. Let C be a tame frieze on a regular m-gon, where m ≥ 3. Take any four vertices 1 ≤ i ≤ j ≤ k ≤ ≤ m of the regular m-gon. By Corollary 2.10 we have that Using that M i+1,k = M i+1,j M j+1,k , M j+1, = M j+1,k M k+1, , and M i+1, = M i+1,j M j+1,k M k+1, we get from the (2, 1)-entries of the matrices that With these three equations and (E j,k ) we conclude: Hence we obtain c i,j c k, + c j,k c i, = c i,k c j, , that is, the Ptolemy relation (E i,j,k, ) holds.

Remark 3.4.
We have now seen that a tame frieze pattern with coefficients as in Definition 2.1 is basically the same as a map on edges and diagonals of a polygon satisfying all Ptolemy relations. This means that indeed the local condition of a frieze pattern with coefficients produces the cluster variables of a Ptolemy cluster algebra with coefficients; see for instance [12, Section 1] for a description of the Ptolemy cluster algebra (aka cluster algebra of Dynkin type A with non-trivial coefficients).

Frieze patterns from subpolygons
In the previous section we have seen that a tame frieze pattern with coefficients of height n can be seen as a map from edges and diagonals of an (n + 3)-gon such that all Ptolemy relations are satisfied.
From now on we consider friezes with coefficients over positive integers. For classic friezes over N there is a beautiful combinatorial description via triangulations of polygons [2].
Convention: From now on we use the notion classic Conway-Coxeter frieze (pattern) for a classic frieze (pattern) over the positive integers N.
Our aim is to connect the more general theory of friezes with coefficients over N with the theory of classic Conway-Coxeter friezes.
From the viewpoint of polygons there is an obvious way to obtain friezes with coefficients from classic Conway-Coxeter friezes. Namely, take any classic Conway-Coxeter frieze C; this is given as a map f C from edges and diagonals on a polygon to N. Now cut out any subpolygon and restrict the map f C to this subpolygon. Clearly, the restricted map still satisfies all Ptolemy relations of the subpolygon, that is, the restriction yields a frieze with coefficients.
Example 4.1. We consider the following triangulation of a hexagon: The corresponding Conway-Coxeter frieze has for instance the following values on the subpolygon highlighted in red: The fundamental question then is: which friezes with coefficient over N can be obtained from classic Conway-Coxeter friezes by cutting out subpolygons?
The following result gives some restrictions for the smallest case of triangles.
Lemma 4.2. Let C = (c i,j ) be a classic Conway-Coxeter frieze and i ≤ j ≤ k. Then the greatest common divisor of any two of the numbers c i,j , c j,k , c i,k divides the third number. In particular, gcd(c i,j , c j,k ) = gcd(c j,k , c k,i ) = gcd(c i,j , c k,i ).
Proof. By symmetry and relabelling it suffices to show that gcd(c j,k , c i,k ) divides c i,j .
If i = j then the assertion holds since c i,i = 0. If j = k then gcd(c j,k , c i,k ) = c i,k = c i,j and the claim follows. If j = k − 1 then gcd(c j,k , c i,k ) = gcd(1, c i,k ) = 1 which clearly divides c i,j .
So from now on we can assume that i < j < k − 1. Then the Ptolemy relation for the crossing diagonals (i, k − 1) and (j, k) implies that Therefore, any common divisor of c j,k and c i,k divides c i,j and the claim follows. (1) We have seen in Example 2.4 that any frieze with coefficients of height 0 over N is given by three numbers a, b, c; these are the values attached to the edges of the corresponding triangle. Lemma 4.2 implies that such a triangle can only be cut out of a classic Conway-Coxeter frieze if the greatest common divisor of two of a, b, c divides the third. For instance, a triangle with values 1, 2, 2 can not come from a classic Conway-Coxeter frieze.
In the next section we will consider triangles in more detail and will obtain a complete characterisation for which triangles with triples a, b, c can be cut out of a classic Conway-Coxeter frieze.
(2) Consider the following frieze with coefficients on a square: This square can not be cut out of a classic Conway-Coxeter frieze because it contains a triangle with values 1, 2, 2, which can not come from a classic Conway-Coxeter frieze by part (1).

We have seen in Examples 4.1 and 4.3 that whether a frieze with coefficients can be cut out of a classic Conway-Coxeter frieze does not only depend on the boundary sequence.
The next result shows that triangles can be cut out of a classic Conway-Coxeter frieze if one of the values on the edges is 1 and the other two values are coprime. Note that the coprimeness condition has to be satisfied in this case by Lemma 4.2. The proof of the lemma exhibits a close connection to the Euclidean Algorithm. Proof. Note that by assumption not both of a, b can be zero, since gcd(0, 0) = 0. If one of them is zero, say a = 0, then b = gcd(0, b) = gcd(a, b) = 1 and the assertion holds with k = 1; in fact, use the unique classic Conway-Coxeter frieze on a triangle.
So from now on we can assume that a, b ∈ N. We show how the Euclidean Algorithm leads to a triangulation of a polygon, and hence to a classic Conway-Coxeter frieze, with the required values.
First we perform the Euclidean Algorithm, where we can assume that a ≥ b. Set r 0 = a and r 1 = b, then Note that by assumption we have r +1 = gcd(a, b) = 1 and q +1 = r . We now describe how to obtain from this data a suitable triangulation of a polygon. Consider the usual number line, and the integral points as vertices. Start with the edge connecting 1 and 2. Then draw q +1 arcs from 2 to the left to the next available vertices, i.e. to the vertices 0, −1, . . . , −q +1 + 1. Then from vertex −q +1 + 1 draw q arcs to the right to the next available vertices. Continue in this alternating fashion by drawing q −1 , q −2 , . . . , q 3 , q 2 , q 1 arcs to the left and right, respectively.
This gives a triangulation of a polygon whose vertices are the integers which are attached to one of the arcs described above. (The rest of the number line is now disregarded.) Note that this polygon is an m := (2 + +1 j=1 q j )-gon. We keep the label for vertex 1, but then number the vertices of this polygon as usual consecutively by 1, 2, . . . , m.
We claim that the classic Conway-Coxeter frieze to this triangulation of the m-gon has the desired properties. For this, we use a well-known combinatorial algorithm for computing arbitrary frieze entries from the triangulation, see [1], starting from the vertex 1. The values attached to the vertices by this algorithm are then the frieze entries c 1,j .
The endpoint of the last of q +1 arcs to the left gets the label q +1 = r . Then the endpoints of the q arcs to the right get the labels r + 1, 2r + 1, . . . , q r + 1 = r −1 . Then the last of the endpoints of the q −1 arcs to the left gets the label q −1 r −1 + r = r −2 . Eventually, we get two consecutive vertices k + 1 and k in the polygon which get assigned the values r 1 = b and r 0 = a.
This means that we have constructed a classic Conway-Coxeter frieze C = (c i,j ) with c k,k+1 = 1, c 1,k = a and c 1,k+1 = b.

Triangles in Conway-Coxeter friezes
We have seen in Section 4 that cutting out any subpolygon of a classic Conway-Coxeter frieze yields a frieze with coefficients over N. In this section, we give an explicit description of all possible values on the edges of triangles cut out of classic Conway-Coxeter friezes. Recall that we have already obtained some results on triples (a, b, c) appearing as labels of a triangle cut out of a classic Conway-Coxeter frieze. Namely, the greatest common divisor of any two of the numbers divides the third (see Lemma 4.2), and for any pair a, b of coprime natural numbers, the triple (a, 1, b) appears as a triangle in some classic Conway-Coxeter frieze (see Lemma 4.4).
However, the complete classification of possible triples is rather subtle and will be the main topic of this section; see Theorem 5.11 below for a precise statement of our main result.
The following auxiliary result shows that given three vertices of an n-gon, any triangulation of the n-gon contains a triangle which separates the given points.
Lemma 5.1. Let C = (c i,j ) be a classic Conway-Coxeter frieze on an n-gon.
Proof. Let T be the triangulation of an n-gon corresponding to the classic Conway-Coxeter frieze C. Any diagonal of T divides the n-gon into two subpolygons. If each diagonal of T has the property that the three vertices i, j, k are contained in one of the two subpolygons then the triangle given by i, j, k is a triangle of the triangulation T and we are done by choosing i = i , j = j and k = k .
So we can choose a diagonal (u, v) of T which separates the three vertices. By symmetry and possibly relabelling the vertices we can assume that we have the situation as given in the following figure: u v i j k Moreover, we choose the diagonal (u, v) ∈ T such that |i − u| and |v − j| are minimal. That is, of all the diagonals of T separating the three vertices in this way we choose the one closest to the diagonal (i, j). Note that the latter need not be in T , but the cases i = u and j = v are allowed.
If j = i + 1 then the diagonal (u, v) of T is part of a triangle in the left subpolygon. By minimality of |i−u| and |j −v| the third endpoint w of this triangle must satisfy i+1 ≤ w ≤ j −1. Now set i = w, j = v, k = u, and we are done. This leaves us with the case that j = i + 1. But then the edge (i, i + 1) is part of a triangle of T and it is not hard to check that the assertion of the lemma holds in this case.
We now come to a crucial reformulation of the problem of describing the triangles which appear as subpolygons in classic Conway-Coxeter friezes. Namely, we are going to show that such triangles correspond (but not bijectively!) to certain tuples of coprime pairs of numbers.
To understand the following explanations, Figure 3 might be helpful. In particular, the Ptolemy relations required in the proof of Proposition 5.3 are easy to read from the picture.
The following result explains the relevance of the sets and maps in Definition 5.2 for our purposes, namely every triple of labels of a triangle cut out of a classic Conway-Coxeter frieze is in the image of the map ∆.  (a 1 , a 2 , b 1 1 , a 2 , b 1 , b 2 , c 1 , c 2 )).
Proof. Choose i , j , k as in Lemma 5.1, i.e.
Using the Ptolemy relations (Theorem 3.3 and also Remark 3.2 (1))) and Equation (5.1), we obtain Again by the Ptolemy relations we now obtain and hence ∆((c k,k , c j ,k , c i,i , c k ,i , c j,j , c i ,j )) = (c i,j , c j,k , c k,i ), as desired.
We now show a converse to Proposition 5.3, namely that every triple in the image of S ≥0 under the map ∆ actually appears as labels of a triangle cut out of some classic Conway-Coxeter frieze.
Proof. By Lemma 4.4, there are three classic Conway-Coxeter friezesC = (c i,j ), C = (c i,j ), C = (c i,j ) such thatc for some , , . LetT , T , T be the corresponding triangulations of some regular polygons. We glue these three triangulations together in such a way that the edges with labelsc 1,2 , c 1,2 , c 1,2 become edges of an inner triangle and obtain a new triangulation T . In the classic Conway-Coxeter frieze C corresponding to T , this inner triangle carries labels c i ,j = c j ,k = c k ,i = 1 for some i , j , k . Moreover, there are i, j, k such that In other words, our triangulation T has the shape as in Figure 3. Using the Ptolemy relations (Theorem 3.3) several times (as in the proof of Proposition 5.3) we obtain (c i,j , c j,k , c k,i ) = ∆ ((a 1 , a 2 , b 1 , b 2 , c 1 , c 2 )).
Our goal is now to describe the image of S ≥0 = S ∩ Z 6 ≥0 under the map ∆ : S → Z 3 . The strategy will be the following. For a given triple (a, b, c) ∈ ∆(S ≥0 ) ⊆ Z 3 ≥0 , imagine that the preimage of (a, b, c) under ∆ is an iceberg. We are looking for a non-negative preimage, i.e. for a place on the iceberg which is not under water. In a first step we find any place on the iceberg (Theorem 5.6), the second step (Theorem 5.9) is to move this solution (using Lemma 5.8) to the dry peak of the iceberg.
Theorem 5.10 clarifies the precise condition for (a, b, c) to be in the image, Theorem 5.11 collects all these results and states the complete classification of triangles appearing in classic Conway-Coxeter friezes.
Definition 5.5. For a prime number p and n ∈ N, we denote by ν p (n) ∈ Z ≥0 the exponent of p in the prime factorization of n, i.e. p νp(n) divides n but p νp(n)+1 does not divide n. We call ν p (n) the p-valuation of n.
Then multiplication by c yields For any k ∈ Z set a 1 := uc + kb , b 2 := vc − ka .
Then a 1 a + bb 2 = (uc + kb )a + b(vc − ka ) = c since kb a = bka by definition of a , b . It remains to find some k ∈ Z such that in addition gcd(a 1 , b) = gcd(a, b 2 ) = 1. We first claim that gcd(a 1 , b ) | c ; in fact, we have gcd(a 1 , b ) = gcd(uc + kb , b ) = gcd(uc , b ) | uc and since b and u are coprime by (5.2) the claim follows. Moreover, b and c are coprime by definition, so the above claim yields gcd(a 1 , b ) = 1 and we obtain Similarly, This means that it suffices to find suitable k ∈ Z such that gcd(a 1 , d) = 1 = gcd(b 2 , d).
If d = 1 then we are clearly done. So assume d > 1 and let d = r i=1 p e i i be the prime factorization (so p i = p j for i = j). The idea now is to consider each prime divisor p i separately and find possible numbers k i such that gcd(a 1 , p i ) = 1 = gcd(b 2 , p i ). This leads to different congruences and finally an overall suitable k ∈ Z such that gcd(a 1 , p i ) = 1 = gcd(b 2 , p i ) for all i = 1, . . . , r is constructed by using the Chinese Remainder Theorem.
Consider first an i with p i > 2. If p i divides b , then p i does not divide c (by definition) and p i does not divide u by (5.2). Thus we have gcd(uc , p i ) = 1 and hence gcd(a 1 , p i ) = gcd(uc + kb , p i ) = 1 for any k. If p i does not divide b , then gcd(a 1 , p i ) = gcd(uc + kb , p i ) = 1 for all but one k ∈ {0, . . . , p i − 1} (use that b is invertible modulo p i ). Thus we get in all circumstances It follows by symmetry that But 2(p i − 1) > p i by assumption, thus we find a k i ∈ {0, . . . , p i − 1} such that gcd(a 1 , p i ) = gcd(p i , b 2 ) = 1.
If p i = 2, then d and hence all of a, b, c are even. It follows from Remark 5.7 that precisely one of a , b , c is even.
If c is even, then a , b are odd. But then a 1 and b 2 are both congruent to k modulo 2, thus it suffices to choose any k i := k odd.
The last case is when a or b is even. By symmetry we can assume that a is even (and hence b and c are odd). But then a 1 ≡ u + k (mod 2) and b 2 ≡ v ≡ 1 (mod 2), where the last congruence follows from (5.2). Thus we may choose any k i := k such that u + k is odd.
Note that in all the above cases, the choice of k i is independent of adding multiples of p i . So this yields a system of congruences k ≡ k i (mod p i ) for i = 1, . . . , r.
The Chinese Remainder Theorem now gives a solution for k as desired.
To understand the image of ∆, we still need some more preparations. The following lemma describes a special element in a nice group of transformations leaving ∆ invariant.
Proof. For the first claim note that Γ t (x) is given by a block diagonal matrix with 2×2-matrices in SL 2 (Z) on the diagonal. Therefore elements in S are mapped to elements in S. For the second claim, just evaluate ∆.
Theorem 5.9. Let a, b, c ∈ N be such that d := gcd(a, b) = gcd(b, c) = gcd(a, c), and assume that either Then there exists a tuple (a 1 , Proof. By symmetry, we may assume that c ≤ a and c ≤ b. Since we have the same assumptions as in Theorem 5.6, we get a 1 , b 2 ∈ Z such that a 1 a + bb 2 = c and gcd(a 1 , b) = gcd(a, b 2 ) = 1. At least one of a 1 , b 2 must be non-negative (since a, b, c ∈ N), so without loss of generality we can assume a 1 ≥ 0.
We now prove a converse to Theorem 5.9.
Proof. We prove the claim indirectly. Assume that ν 2 (a) = ν 2 (b) = ν 2 (c) = ≥ 1, thus we may write a = 2 ã, b = 2 b , c = 2 c withã,b,c odd. Let (a 1 , a 2 , b 1 , b 2 , c 1 , c 2 ) ∈ S ∩ Z 6 ≥0 be such that ∆ ((a 1 , a 2 , b 1 , b 2 , c 1 , c 2 )) = (a, b, c). Then by Definition 5.2 we have We claim that this implies In fact, consider equations (5.6)-(5.8). Since a, b, c are even by assumption, the two summands on the right hand side of each equation are both even or both odd. Suppose that in (5.6) both summands are even; the other case is dealt with similarly, the details are left to the reader. We distinguish two cases. If b 1 is even then b 2 is odd (by definition of S), hence b 1 + b 2 is odd. Since b 2 is odd our assumption on (5.6) implies that c 2 is even. Then c 1 is odd (again by definition of S), thus c 1 + c 2 is odd. But a 1 , a 2 can not both be even (by definition of S), so (5.7) implies that a 1 and a 2 are both odd. This proves the claim on the above congruences in case b 1 is even. If b 1 is odd then our assumption on (5.6) implies that c 1 + c 2 is even. By definition of S we conclude that c 1 and c 2 are both odd. Then the assumption on (5.6) yields that b 2 is even. Then b 1 is odd (by definition of S) and b 1 + b 2 is odd. Now (5.7) implies that a 1 is even. So a 2 is odd (by definition of S) and this completes the proof of the claim.
Hence, for the values of a 1 , a 2 , b 1 , b 2 , c 1 , c 2 , c 1 + c 2 modulo 2, only the following three cases are possible (recall that b 1 , b 2 cannot both be even by definition of S): Using Equations (5.9), (5.6), a = a + (k 1 − k 2 )b 1 b 2 . Hence we can choose k 1 = k 2 (for the case b 1 = 0 or b 2 = 0, this is clear a priori). More precisely, from the possible parities in the above table one sees that k has to be odd in all circumstances. Now we get from equations (5.7) and (5. Thus becauseã,c and k are odd, andb =ã(a 1 (u + v) + a 2 u) − kc is odd as well, we get that a 1 (u + v) + a 2 u is even. Using this fact, and the fact that u, v are coprime by (5.9), the above table extends to: As an immediate corollary to the Theorems 5.9 and 5.10 we obtain the main result of this section, giving a complete classification of triples appearing as labels of triangles in classic Conway-Coxeter friezes.  1 , a 2 , b 1 The claim now follows by Theorem 5.11.

Finiteness
In this section we prove finiteness results on the number of possible frieze patterns with coefficients over subsets R ⊆ C. Our main interest is in frieze patterns with coefficients over N but it will actually turn out that our finiteness result holds for arbitrary discrete subsets R of C.
If P ≥ 1 then every entry in the quiddity cycle of C has absolute value at most P 2 ((P M + (n − 1)P 2 + M ) M 2 .
Proof. We set B := P 2 ((P M +(n−1)P 2 +M ) for abbreviation. Note that B > 0 since P > 0 and M > 0 by assumption. To ease notation we also set d i := c i,i+1 for all i ∈ Z.
Suppose for a contradiction that there exists an entry x 1 in the quiddity cycle such that |x 1 | > B. Then we consider the corresponding two consecutive rows in the frieze pattern C, with notation as in the following figure: 0 d j x 1 x 2 . . . x n−1 x n d j−1 0 0 d j+1 y 1 y 2 . . . y n−1 y n d j 0 Note that the boundaries are as indicated because of the glide symmetry on the boundary entries, see Remark 2.2. The strategy of the proof is to proceed inductively along the rows. More precisely, we want to use the following two conditions as induction hypotheses: M for k = 2, . . . , i + 1. (2) |y k | < M P 2 M (M B−P 2 )−(k−2)P 4 (P 2 + M |x k+1 |) for k = 2, . . . , i. For the base of the induction, let us check that condition (1) holds for k = 2 and k = 3 and that condition (2) holds for k = 2.
Solving this inequality for |x 3 | then gives and this is equation (1) for k = 3.
Thus we have shown that we can indeed use conditions (1) and (2) as induction hypotheses for an induction on i.
For the induction step, we have to show that condition (2) holds for i + 1 and that condition (1) holds for i + 2.
From the defining condition of frieze patterns with coefficients we have and hence |y i+1 | ≤ |d j d j+i+2 | + |x i+2 | · |y i | |x i+1 | ≤ P 2 + |x i+2 | · |y i | |x i+1 | Using the induction hypothesis for |y i | we deduce Now we also plug in the induction hypothesis for |x i+1 | and get |yi+1| < P 4 M M (M B − P 2 ) − (i − 1)P 4 + half-integers, and M = 1 2 = P . Then the bound given in Lemma 6.1 is equal to 3 4 which is clearly not an upper bound for all entries in the quiddity cycle.
As a consequence of Lemma 6.1 we obtain the following finiteness result for frieze patterns with coefficients over discrete subsets of complex numbers. This generalizes [6,Corollary 3.8] from classic frieze patterns to frieze patterns with coefficients. Proposition 6.3. Let R ⊆ C be a discrete subset, that is, R has no accumulation point. Let n ∈ Z ≥0 and fix a boundary sequence in (R \ {0}) n+3 . Then there are only finitely many frieze patterns with coefficients of height n over R \ {0} with the given boundary sequence.
Proof. For n = 0 there is only one possible frieze pattern with coefficients for any given boundary sequence, see Example 2.4.
For n ≥ 1 we want to use Lemma 6.1. Recall that frieze patterns with coefficients can be scaled (see Remark 2.3). The existence of only finitely many frieze patterns and the discreteness of the set R are invariant under scaling, so we can assume that our boundary sequence satisfies the assumption P ≥ 1 in Lemma 6.1.
Since R is discrete, the origin is not an accumulation point and R satisfies the assumption of Lemma 6.1. Moreover, again by discreteness, every closed disk contains only finitely many elements from R. Then Lemma 6.1 implies that there are only finitely many elements of R which can appear in the quiddity cycle of a frieze pattern with coefficients having the given boundary sequence. However, any non-zero frieze pattern with coefficients is determined by the boundary sequence and the quiddity cycle, and this proves the claim.