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A165643 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 equivalent. 3
1, 48, 3932160, 2958824445050880, 65878553108096586952810168320, 106856067198182393582849337977733120000000000000, 26299193579608484719502346688357440131301651853401391104000000000000000 (list; graph; refs; listen; history; text; internal format)



Table of n, a(n) for n=1..7.


a(n) = ((2n-2)!!)^(2n-1) * (2n-1)!


For n=2, we are constructing a square from 4 labeled line-segments with labeled endpoints. Solutions which differ by a rotation or a reflection are considered equivalent. There are 3 ways to order the line-segments, and each line-segment can be oriented in 2 ways, so the total number of solutions is 3 * 2^4 = 96. (equivalently, 6 * 2^3, which more closely resembles the formula provided.) For n=3, we are constructing a cube from 6 labeled squares with labeled vertices. Without loss of generality, we can pick one labeled square to serve as our face of reference. For this face, we do not care which side of the square will face the interior of the cube as this just translates into a reflection of the cube, nor we do not care about which rotation we pick as these just translate into rotations of the cube. From this reference square, there are 5! ways to assign the remaining squares to the faces of the cube, and each square can be oriented in 8 ways (we can pick which side of the square will face the interior of the cube, and we can pick from 4 rotations). This gives 8^5 * 5! solutions.


Cf. A165642 (same idea, but reflections are distinct). A165644 and A091868 are the corresponding sequences for simplices instead of cubes.

Sequence in context: A051235 A282403 A254625 * A165047 A272096 A115480

Adjacent sequences:  A165640 A165641 A165642 * A165644 A165645 A165646




Andrew Weimholt, Sep 23 2009


Example reformatted by Andrew Weimholt, Sep 25 2009



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Last modified May 27 05:27 EDT 2017. Contains 287189 sequences.