Kirby Urner wrote: [snip]
Another way David talked about, to get cubes:
n**3 = 6 * Tetra + n where Tetra is a tetrahedral number 0,1,4,10,20,35...
We looked up Tetra here:
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?An um=A000292
Another relevant page (with a picture) from the same site is: http://www.research.att.com/~njas/sequences/demo3.html
(Darn, I'm breaking my own rule: no reference to python in this message ;-)
Andr�
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