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.

1, 48, 3932160, 2958824445050880, 65878553108096586952810168320, 106856067198182393582849337977733120000000000000, 26299193579608484719502346688357440131301651853401391104000000000000000

Offset: 1

Author

Andrew Weimholt, Sep 23 2009

Keywords

• nonn

Examples

			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.
		

Crossrefs

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

Formula

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

Extensions

Example reformatted by Andrew Weimholt, Sep 25 2009