Let $P$ be a polyhedron in $\mathbb{R}^3$.
Say that $P$ *combinatorially shadows* a sequence of natural numbers $S$ if
there is a continuous rotation of $P$ such that its orthogonal-projection
shadows are polygons whose number of sides coincide with the elements of $S$
in order. For example, $S$ might be the odd primes: $S=(3,5,7,11,13,\ldots)$,
and we want the shadows to be a triangle, then a pentagon, then a septagon,
then a hendecagon, etc.

**Q0**. Let $S$ be an increasing sequence of natural numbers,
whose first element is $\ge 3$.
Does there exist a polyhedron $P$ that combinatorially shadows $S$?

I think the answer here is *Yes*, illustrated just for $S=(3,5)$ with this
example:

The generalization is that the needed increase above the previous element
in the sequence is achieved by bumping out near the centroid of an
appropriate face (the centroid $c$
of face $(1,2,3)$ in the
above example is bumped out to $\lbrace a, b \rbrace$),
shallow enough to be hidden for the previous element (middle image), but sufficent so
that a rotation will simultaneously expose the additional vertices (right image).
So if I am correct here, there is a *prime polyhedron* that realizes the odd primes—either
up to any given prime, or all odd primes if an infinite number of faces are countenanced.

*Correction* (*9Dec12*): I now think the above sketch fails to allow many vertices to appear in the shadow
simultaneously. ~~Better is to split existing vertices into two ...~~
[remaining bad idea deleted]. *23Dec12*: Now I believe the construction posted in a separate
answer settles **Q0** (positively).

My question concerns arbitrary—not necessarily increasing—sequences:

**Q1**. Let $S$ be an *arbitrary* sequence of natural numbers,
each $\ge 3$.
Does there exist a polyhedron $P$ that combinatorially shadows $S$?

Ideas, even half-baked, or pointers to relevant literature welcomed! Thanks!