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Let $K$ be a quadratic field and let $\Lambda$ be a rank $2$ sublattice of $K$, with $\mathbb{Q} \Lambda =K$. We'll say that $\Lambda$ has CM if there is some non-rational $\theta$ in $K$ such that $\theta \Lambda \subseteq \Lambda$. Such a $\theta$ is necessarily an algebraic integer, and the ring of all $\theta$ with $\theta \Lambda \subseteq \Lambda$ is an order in $\mathcal{O}_K$.

$\bullet$ Let $K = \mathbb{Q}(z)$ and let $\Lambda = \langle 1, z \rangle$. Then $\Lambda$ has CM. Shifting the periodic sequence does not change the strict equivalence class of $\Lambda$.

$\bullet$ The above is a bijection between periodic sequences of $r$'s and $b$'s, up to shift, and strict equivalence classes of lattices with CM in real quadratic fields.

$\bullet$ Reversing the sequence sends $\Lambda$ to $\overline{\Lambda}$. If $\Lambda$ is a lattice with CM, and $R$ is its endomorphism ring, then $\Lambda \overline{\Lambda}$ is a strictly principcal fractional ideal for $R$. (This is a special property of quadratic fields, which I know of no generalization of in higher degree number fields.) So, with the understanding that we treat a fractional ideal as a fractional ideal for its full endomorphism ring, reversing the sequence sends $\Lambda$ to $\Lambda^{-1}$.

Let $K$ be a quadratic field and let $\Lambda$ be a rank $2$ sublattice of $K$, with $\mathbb{Q} \Lambda =K$. Let $\mathrm{End}(\Lambda)$ be the ring of $\theta \in K$ such that $\theta \Lambda \subseteq \Lambda$. This is an order in $\mathcal{O}_K$.

$\bullet$ Let $K = \mathbb{Q}(z)$ and let $\Lambda = \langle 1, z \rangle$. Shifting the periodic sequence does not change the strict equivalence class of $\Lambda$.

$\bullet$ The above is a bijection between periodic sequences of $r$'s and $b$'s, up to shift, and strict equivalence classes of lattices in real quadratic fields.

$\bullet$ Reversing the sequence sends $\Lambda$ to $\overline{\Lambda}$. If $\Lambda$ is a lattice in $K$, and $R = \mathrm{End}(\Lambda)$, then $\Lambda \overline{\Lambda}$ is a strictly principal fractional ideal for $R$. (This is a special property of quadratic fields, which I know of no generalization of in higher degree number fields.) So, with the understanding that we treat a fractional ideal as a fractional ideal for its full endomorphism ring, reversing the sequence sends $\Lambda$ to $\Lambda^{-1}$.

Fixing the endomorphism ring; working with triples $(a,b,c)$

We can obtain the same unit from two quite different looking continued fractions. For example $$\sqrt{10} = r\ r\ r\ b\ b\ b\ b\ b\ b\ r\ r\ r \cdots \quad \sqrt{10}/2 = r\ b\ r\ b\ b\ r\ b\ r \cdots$$ where I have given a full period for each fraction. Truncating to before the middle block gives $$r\ r\ r = \frac{3}{1} \quad r\ b\ r\ = \frac{3}{2}.$$ Both of these give the unit $3-\sqrt{10} = 3-2 \frac{\sqrt{10}}{2}$.

Fixing the endomorphism ring; working with triples $(a,b,c)$

Fixing the endomorphism ring

$\bullet$ There are only finitely many $(a,b,c)$ for any $R$.

Fixing the endomorphism ring; working with triples $(a,b,c)$

$\bullet$ There are only finitely many $(a,b,c)$ for any $R$.

$\bullet$ Adding an $r$ at the beginning of the sequence changes $(a,b,c)$ to $(a,a+b,c-a-2b)$. Adding a $b$ at the beginning changes $(a,b,c)$ to $(a-c-2b,b+c,c)$. Reversing the sequence changes $(a,b,c)$ to $(a,-b,c)$; color switching the sequence sends $(a,b,c)$ to $(c,-b,a)$.