Just realized, the question needs a little more precision, when you say partition into several distinct polytopes....are you counting polytopes that are solids, facets, ridges, planes, lines....is there any restriction to the kind of polytopes? I've been making assumptions. Also, for some reason the phrasing with lines instead of hyperplanes made a lot more sense.

There is a combinatorial formula for this, it gets cumbersome to generalize for higher dimension. Its easy for a(2,2), it also validates 56 as the answer. Its a double sum, inner sum over outer lattice points - 2, and outer sum over increasingly fewer lattice points as the starting point.

@SimonHenry: I found counter examples. It is subtle, but this problem is not exactly equivalent to the Collatz Conjecture; at least not as you describe.

I think you have a typo, did you mean $2^{a_k}$ instead of $a^{a_k}$?

So...finding orbits with 2 odd integers is trivial. Just take $b_n=\frac{4^n-1}{3}$, which is exactly what you have listed there. It is also easy to prove that this is the only form there is for 2 odd integers in an orbit containing s and 1.The interesting part is taking it beyond that using explicit formulas without using recursive constructs. That is much harder.