[seqfan] Number of ways to assemble a n-cube from 2n labeled (n-1)-cubes with labeled vertices
andrew at weimholt.com
Wed Sep 23 07:59:06 CEST 2009
Number of ways to assemble an n-cube from 2n labeled (n-1)-cubes with
labeled vertices, where left-handed and right-handed counterparts are
considered distinct. (assume the n-cube is confined to n-space where
we don't have the extra dimensions needed to flip it around to convert
left-handed cubes to right-handed cubes).
a(n) = 2 * ((2n-2)!!)^(2n-1) * (2n-1)!
starting with n=1, the sequence begins...
2, 96, 7864320, 46231631953920, ...
An alternate sequence would consider left and right handed
counterparts to be equivalent,
and each term in the alternate sequence would be half of the
corresponding term in the original sequence...
1, 48, 3932160, 23115815976960, ...
Which of the two sequences (if either), should be submitted?
I prefer the former.
The idea for this sequence came after thinking about the number of
ways to arrange the pips on dice if you abandon the requirement of
having oppisite sides add up to 7. If we only consider the numbers
assigned to each face of the die, then there are 30 variations,
however the arrangement of pips for "2", "3", and "6" have less than
square symmetry, so if we consider also the alignment of these 3
faces, we have 240 variations.
This lead me to consider the abstract case of assembling a cube from 6
squares with labeled vertices. Now, not only are there four ways to
rotate each face, but we can also select which side of the square will
face the interior of the cube, so for each face we have 8
Thus there are 7864320 ways to build a cube from 6 labeled squares
with labeled vertices.
From here it was only natural to extend this idea to n-cubes.
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