OFFSET
1,1
FORMULA
a(n) = 2 * (n!)^(n+1)
EXAMPLE
For n=2, construct a triangle from 3 labeled line segments with labeled endpoints. Solutions which differ by a rotation are considered equivalent, but solutions which differ by a reflection are considered distinct (assume the triangle we are constructing will be embedded in a plane, so we cannot flip it over to convert a left-handed solution to a right-handed solution). There are 2 ways to order the line segments, and each line segment can be oriented in 2 ways, so the total number of solutions is 2 * 2^3 = 16.
For n=3, construct a tetrahedron from 4 labeled triangles with labeled vertices. Assume the tetrahedron is confined to 3-space where we cannot flip it around in 4-space to convert left-handed solutions to right-handed solutions, so we consider reflections to be distinct. Without loss of generality, we can pick one labeled triangle to serve as our face of reference. For this face, we must pick which side of the triangle will face the interior of the tetrahedron, but we do not care about which rotation we pick as these just translate into rotations of the tetrahedron. From this reference triangle, there are 6 (=3!) ways to assign the remaining triangles to the faces of the tetrahedron, and each triangle can be oriented in 6 (=3!) ways (we can pick which side of the triangle will face the interior of the tetrahedron, and we can pick from 3 rotations). This gives 2 * 6^4 solutions. The factor of "2" comes from the choice of which side of the reference triangle faces the interior (a choice that goes away if we consider reflections to be equivalent).
CROSSREFS
KEYWORD
nonn
AUTHOR
Andrew Weimholt, Sep 23 2009, Sep 25 2009
STATUS
approved