Kepler's lattice

Paul H. Erlich [TD 230.4]

>$ A---------E---------B
> /|\ /|\ / \
> / | \ / | \ / \
> / Eb--------Bb \ / * \
> /,' `.\ /,' `.\ / \
> F---------C---------G---------D

Ray Tomes
>>This shape is interesting because it is the same as Kepler's sphere
>>stacking problem. It can be represented as a triangular lattice with
>>another one on top or a square lattice with another one on top.

>That is, one can represent it by stacking triagular lattices in _three_
>staggered positions, or square lattices in two staggered positions. For
>viewing more than three layers of this lattice at once, an oblique view is
>necessary.

That is true in the position with the square base, but in the triangular
view that you drew multiple layers can be seen at once.

$ A---------E---------B
/|\ /|\ / \
B--/-|-\--F#-/-|-\--C# / \
| / Eb--------Bb \ | / * \
|/,' `.\|/,' `.\|/ \
F---------C---------G---------D

>If we regard these as equilateral triangles, then each note is at the center
>of an imaginary sphere, and the spheres are (probably) packed as efficiently
>as possible. However, Ray, I think you and I would both prefer not to regard
>the triangles as equilateral, but as isosceles with, e.g., the 3:1 side
>shorter than the 5:1 side or the 5:3 side.

Yes, in terms of psychological space the sides are quite unequal as you
say. Has any consensus been reached on the relative sizes of these?
I stated earlier that p*ln(p) was the relative distance for prime p but
later realised that this ratio refers to something slightly different,
namely how often each prime should be present (inversely).

Thanks to all the others that mentioned that Kepler's conjecture has now
been proven. I haven't dared look yet because after the extremely
inelegant proofs of Fermat's last theorem and the four colour problem I
couldn't bear to see another such.

-- Ray Tomes -- http://www.kcbbs.gen.nz/users/rtomes/rt-home.htm --
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