

A268408


Triangle T(d,v) read by rows: the number of hypertetrahedra with volume v/d! defined by selecting d+1 vertices of the ddimensional unithypercube.


1



0, 1, 0, 4, 12, 56, 2, 1360, 2672, 320, 16, 350000, 431232, 107904, 12864, 3872, 320, 255036992, 234667968, 98251776, 19523136, 10633728, 1615552, 1182720, 163520, 127360, 13440
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OFFSET

1,4


COMMENTS

The unit hypercube in dimension d has 2^d vertices, conveniently expressed by their Cartesian coordinates as binary vectors of length d of 0's and 1's. Hypertetrahedra (simplices) are defined by selecting a subset of 1+d of them. The (signed) volume V of a tetrahedron is the determinant of the d vectors of the edges divided by d!. (The volume may be zero if some edges in the tetrahedron are linearly dependent.) The triangle T(d,v) is a histogram of all A136465(d+1) tetrahedra classified by absolute (unsigned) volume V=v/d!.
The number of nonflat simplices (row sums without the leftmost column) are tabulated by Brandts et al. (Table 1, column beta_n).  R. J. Mathar, Feb 06 2016


LINKS



EXAMPLE

In d=2, 4 tetrahedra (triangles) are defined by taking subsets of d+1=3 vertices out of the 2^2=4 vertices of the unit square. Each of them has the same volume (area) 1/2!, so T(d=2,v=1)=4.
In d=3, 12 = T(d=3,v=0) tetrahedra with zero volume are defined by taking subsets of d+1=4 vertices out of the 2^3=8 vertices of the unit cube. These are the cases of taking any 4 vertices on a common face. (There are 6 faces and two different edge sets for each of them; one with edges along the cube's edges, and one with edges along the face diagonals.)
The triangle starts in row d=1 as follows:
0 1;
0 4;
12 56 2;
1360 2672 320 16 ;
350000 431232 107904 12864 3872 320;


CROSSREFS



KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



