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Say you have a single rhombic dodecahedron, call it "layer p0". If you then tile identical dodecahedrons around it so that it gets completely covered, the number of dodecahedrons on the outer layer (p1) would be 12. Continuing this pattern, p2=42 etc. How would you create a formula for the number of dodecahedrons on each shell layer?

Thanks!

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  • $\begingroup$ I guess you know what you mean, but I don't get it. Is the first layer 12 touching spheres, each on one face of the dodecahedron? And then another layer (shell) of spheres, all touching their neighbours? $\endgroup$
    – Weather Vane
    Commented Apr 1 at 17:08
  • $\begingroup$ If you have a dodecahedron and glue more to that one creating an air tigh tiling pattern until you cannot see the original one anymore, kinda make a shell around it, and then count how many dodecahedrons you have added that would be p1. $\endgroup$
    – TheMasterOfCubes
    Commented Apr 1 at 17:14
  • $\begingroup$ So yes, they all need to be touching neighbours. $\endgroup$
    – TheMasterOfCubes
    Commented Apr 1 at 17:15
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    $\begingroup$ If you stack same-size dodecahedrons around a dodecahedron, they dont quite touch, because their internal angle between two faces is a bit less than 120° (about 116.5°). You can't fill a 3D space with them. $\endgroup$
    – Weather Vane
    Commented Apr 1 at 17:16
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    $\begingroup$ Please excuse me, I should have specified that they are rombic dodecahedrons. $\endgroup$
    – TheMasterOfCubes
    Commented Apr 1 at 17:17

1 Answer 1

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Not sure this is the correct answer, but will post it.

This Wikipedia page about Rhombic dodecahedral honeycomb states

The rhombic dodecahedral honeycomb is a space-filling tessellation (or honeycomb) in Euclidean 3-space. It is the Voronoi diagram of the face-centered cubic sphere-packing, which has the densest possible packing of equal spheres in ordinary space.

So if the formula is the same as ...

... for close sphere packing then the number $N$ in each layer $x$ would be ($x > 0$)
$ N = 10x^2 + 2$
giving the series
$1, 12, 42, 92, 162, 252, 362, ...$
which is OEIS A005901.

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    $\begingroup$ It awnsers my question although I wonder how to get to the formula, the proof. $\endgroup$
    – TheMasterOfCubes
    Commented Apr 1 at 18:24

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