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Does there exist an unbounded convex polyhedron with faces that have 3, 5, 7, 11, 13, ... edges, i.e., such that the number of edges of each face realize exactly the odd primes, with each prime realized exactly once (i.e., there is just one face with that number of edges)? If so, can this be achieved with adjacent primes realized by adjacent faces? Here is a start. :-)
                 Prime Polyhedron

Update. Now answered negatively by Ed Pegg! For a related question, which does have a positive solution, see the MO question, "Polyhedra that combinatorially shadow a sequence": There is a polyhedron whose shadows combinatorially mimic the primes when the polyhedron is appropriately continuously reoriented.

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    @JoeM: Yes, not quite what I had in mind, but still your construction is clever, even if it undermines the intent I couldn't quite formulate precisely. :-) Thanks!2011-10-01

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It's impossible.

Consider up to the first 20 primes, giving 20 faces. By Euler's formula V=2+E-F

337 vertices = 2 + (Total@Prime[Range[2, n]]/2) - (n - 2 + 1)

Now consider the dual object, with 20 vertices. A maximally triangularized polyhedron with n vertices has 2v − 4 faces. Or 2(20)-4=76 faces. There is no way to support 337 faces.

As the number of primes increases, the dual polyhedron becomes increasingly impossible. Consider taking a slice that has 500 settled faces. That slice would also go through a minimum of 206821 other faces for the dual of this slice off piece to be possible. Pulling in two more faces fully requires that a minimum of 1793 additional faces get cut.

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    Brilliant, Ed! $\mbox{}$2013-01-13