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.

  • $\begingroup$ How convex must it be? It seems easy enough if we allow "trivial" vertices where only two faces meet, and we can still make the interior of the polyhedron a convex subset of $\mathbb R^3$. $\endgroup$ – hmakholm left over Monica Sep 30 '11 at 21:51
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    $\begingroup$ This probably is not what you had in mind. Look at the parabola y2 = x but above the line y =0. Divide this region into parts with the lines x = k where k runs through the integers 1, 2, etc. In each of these regions, starting at the left, put 0 points, 1 point, 3 points, and so on, and draw the straight line "chords" between these points, to form a 3-gon, 5-gon, etc. Now form the cylinder (straight lines parallel to the plane used for the construction and perpendicular to that plane) that lies above this "construction." $\endgroup$ – Joseph Malkevitch Sep 30 '11 at 21:56
  • $\begingroup$ @Henning: I should have specified that all faces should be strictly convex. $\endgroup$ – Joseph O'Rourke Sep 30 '11 at 23:39
  • $\begingroup$ @Joe. One can modify the idea above to use both the upper and lower section of the parabola.Each "cell" will have strictly convex faces but there will be a plane that touches all of the cells, which is a convex "face" with infinitely many sides. $\endgroup$ – Joseph Malkevitch Oct 1 '11 at 2:43
  • $\begingroup$ @JoeM: Yes, so I should also specify that no two faces are coplanar. Likely your idea will still yield a solution... $\endgroup$ – Joseph O'Rourke Oct 1 '11 at 10:41

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.

  • $\begingroup$ Brilliant, Ed! $\mbox{}$ $\endgroup$ – Joseph O'Rourke Jan 13 '13 at 19:29

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