Complete n-types of the theory of atomless Boolean algebras

I have to answer the next questions:

• What is the number of complete 1-types of the theory of atomless Boolean algebras?

• What is the number of complete 2-types of the theory of atomless Boolean algebras?

I know that the theory of atomless Boolean algebras is countably categorical (has up to isomorphism only one countable model) and therefore has only finitely many n-types for each n. But what next?

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You could start by looking at the countable model and thinking about how its automorphisms act. –  Chris Eagle Jan 31 '13 at 16:05
Hint: (elaborating a little on Chris Eagle's comment) if a theory is $\omega$-categorical, then its only countable model is necessarily saturated.
Now, if you have any saturated model $M$, then it realizes all $n$-types without parameters for each $n$ and it is strongly homogeneous. This means that not only can you find representatives for each $n$-type, but $n$-types correspond exactly to orbits of $\operatorname{Aut}(M)$ acting on $M^n$.
@natural: in general, $\kappa$-saturation implies $\kappa$-homogeneity (this is easy), and homogeneity in a model's own cardinality is equivalent to strong homogeneity in its own cardinality (which can be done using a standard back-and-forth argument). –  tomasz Feb 5 '13 at 20:42
Thanks, I can follow the reasoning now. But I cannot get a grip on what the orbit of say an element $a$ (not 0 or 1) is. For we are dealing with a countably infinite BA. –  natural Feb 6 '13 at 13:43