colorings of n^3 cubes

[franktaw at netscape.net]

I have some problems with this.

First, I would change "can be assembled into a monochromatic nxnxn
cube of any of the n colors" to "can be assembled in different ways
into a monochromatic nxnxn cube of each of the n colors".  It wasn't
until I went to the web site that I realized that each color should be
obtainable, not just one of them.

Second, your conditions for two arrangements being equivalent seem
to me to be too weak.  For example, with 4 or more colors, if there
are only 2 3-3 cubes, with one being blue-green, I would consider it
different if the other is blue-red instead of yellow-red.

Third, shouldn't that second formula sum to 12n(n-2), not 4n(n-2)?
(I assume that your program got this right, since as written there
are no solutions for n = 1 (mod 3).)  Also, the third one should
total 6n(n-2)^2, not 6(n-2)^2.

Franklin T. Adams-Watters

-----Original Message-----
From: Joshua Zucker <joshua.zucker at gmail.com>

Hi folks,
I'm wondering a few things, before I submit this sequence:
Is the description I've written here clear enough?
Is this sequence interesting enough to be submission-worthy?
What are good techniques for counting the number of nonnegative
solutions to a system of linear diophantine equations like this?  I've
just written a program to search the space by brute force, actually
listing every single solution as it counts.  If you have any
references to better algorithms I'd love to learn!

Thanks,
--Joshua Zucker

%I A000001
%S A000001 
1,1,1,289,16743,2455061,9069797,25617192,60158319,122758692,227734800,398
364425,
664106324,1063255428,1644644641,2469420200,3612955276,5166901813
%N A000001 Number of colorings of n^3 unit cubes in n colors such that
they can be assembled into a monochromatic nxnxn cube of any of the n
colors
%C A000001 Classify the n^3 cubes according to how many sides are
colored with each color.  In general, we can have 3-3, 3-2-1, 3-1-1-1,
2-2-2, 2-2-1-1, 2-1-1-1-1, and 1-1-1-1-1-1 cubes.  The set of cubes is
considered to be colored in the same way if the number of cubes in
each of these seven categories is the same.

Thus, also number of solutions of
2A + B + C = 8n (corners)
B + 3D + 2E + F = 4n(n-2) (edges)
B + 3C + 2E + 4F + 6G = 6(n-2)^2 (face interiors)
A+B+C+D+E+F+G = n^3 (total cubes; but this is redundant with the
previous equations, since every side of every cube must be painted)
%H A000001 Denise at <a
href="http://letsplaymath.wordpress.com/2008/03/19/leonhards-block-puzzle
s/">Let's
Play Math</a> blog asks how to color in the n=2 and n=3 cases
%e A000001 a(2) = 1 because all 8 blocks must be colored with 3 of one
color and 3 of another, so there's only 1 way to color the cubes.
%O A000001 1
%K A000001 ,nonn,
%A A000001 Joshua Zucker (joshua.zucker at stanfordalumni.org), May 22 2008