# The number of infinite frieze patterns

This blog entry was inspired by the seminar of Prof. Karin Baur on Frieze patterns. Infinite frieze patters are examined in the paper Infinite Friezes.

Definition 1
An infinite frieze is a sequence of rows $\{r_k\}_{k \in \mathbb{N}}$ where each row is a function $r_k:\mathbb{Z} \to \mathbb{N}$ and the sequence satisfies:

1. $\forall i \in \mathbb{Z} [r_0(i)=0 \land r_1(i)=1]$.
2. $\forall k \in \mathbb{N} \forall i \in \mathbb{Z} [r_{k+1}(i)r_{k+1}(i+1)-r_k(i+1)r_{k+2}(i) = 1]$

From the definition it is clear that the whole infinite frieze is determined from the row $r_2$. However, some rows $r_2$ may not give a rise to an infinite frieze or a frieze at all. For example $r_2(i):=1$ if $i \equiv 0 \mathrm{mod} 3$ and $r_2(i):=2$ otherwise. Therefore a question arises how many friezes there are.

By calculating the first rows of several friezes, which can be done quickly by writing a program and running it online, one can observe a general polynomial formula(thus not containing any division operation) defining a row from the previous two rows.

Lemma 2

Let $\{r_k\}_{k \in \mathbb{N}}$ be a sequence of rows where each row is a function $r_k:\mathbb{Z} \to \mathbb{R}$ and the sequence satisfies the 2 conditions of an infinite frieze. Let there be a positive integer $n$ such that for all $i,j \in \mathbb{Z}$ if $i \equiv j \mathrm{mod} n$ then $r_2(i)=r_2(j)$,i.e. $_2$ is given by a period of a length $n$. Then $\forall k \in \mathbb{N}-\{0,1\} \forall i \in \mathbb{Z}. r_{k+2}(i)=r_{k+1}(i) \cdot r_2(i+k)-r_k(i)$.

Proof:
The formula is proved by the induction. For the base case $r_4(i)=\frac{r_3(i) r_3(i+1)-1}{r_2(i+1)}=\frac{(r_2(i)r_2(i+1)-1)(r_2(i+1)r_2(i+2)-1)-1}{r_2(i+1)}$
$=r_2(i)r_2(i+1)r_2(i+2)-r_2(i)-r_2(i+2)=r_3(i)r_2(i+2)-r_2(i)$.
For the inductive case assume $r_{k+2}(i)=r_{k+1}(i)r_2(i+k)-r_k(i)$. Then $r_{k+3}=\frac{r_{k+2}(i)r_{k+2}(i+1)-1}{r_{k+1}(i+1)}$
$=\frac{(r_{k+1}(i)r_2(i+k)-r_k(i))(r_{k+1}(i+1)r_2(i+k+1)-r_k(i+1))-1}{r_{k+1}(i+1)}$
$=r_2(i+k+1)r_{k+2}(i)+\frac{-r_k(i+1)(r_{k+1}(i)r_2(i+k)-r_k(i))-1}{r_{k+1}(i+1)}$
$=r_2(i+k+1)r_{k+2}(i)-\frac{r_k(i+1)r_{k+2}(i)-1}{r_{k+1}(i+1)}=r_{k+2}(i)r_2(i+k+1)-r_{k+1}(i)$.
QED.

Note that if $ad-bc=1$, $a>b>0$ and $d>b>0$, then $c>a$ and $c > d$. This guarantees that if $r_2$ is given by the numbers greater than 1, then every next row will have greater entries than the preceding one and the formula from the lemma will hold as the division by 0 never occurs. This gives:

Theorem 3
Let $r_2:\mathbb{Z} \to \mathbb{N}-\{0,1\}$ be given by the period $r_2(0), r_2(1), ..., r_2(n)$ of a length $n+1$, i.e. $\forall i, j \in \mathbb{Z}. i \equiv j \mathrm{mod} n+1 \implies r_2(i) = r_2(j)$. Then $r_2$ defines an infinite frieze ${r_k}_{k \in \mathbb{N}}$.

If the row $r_2$ is not periodic, but its entries are integers greater than 1, then it still gives a rise to an infinite frieze for one may consider it as if defined by periods of an arbitrarily large length.

Corollary 4
Every row $r_2:\mathbb{Z} \to \mathbb{N}-\{0,1\}$ gives a rise to an infinite frieze.

Corollary 5
The number of infinite friezes is $\aleph_1$.

# The number of m-faces of an n-cube

$n$-cube is a generalization of a cube where 0-cube is a vertex, 1-cube is a line segment, 2-cube is a square, 3-cube is a cube and 4-cube is a hypercube. $n$-cube has its $m$-faces where 0-faces are vertexes, 1-faces edges and 2-faces are faces.

Ever wondered how many $m$-faces does an $n$-cube have? It is easy to calculate with the following hint as a starting point. Consider a cube with points $\{(0,0,0),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)\}$. What is the necessary and sufficient condition on the coordinate values of the four points of a cube to form a face?

Afterwards, you should obtain a formula:

The number of $m$-faces of $n$-cube is ${n \choose m} 2^{n-m}$.