Dekany=Rap? and Eikosany Stats

I wrote,

>http://members.aol.com/Polycell/uniform.html

>These people are insane! We need to import some of them!

note -- this author uses "pentachoron" instead of "pentatope".

Well, looking a little further, I found in

http://members.aol.com/Polycell/section1.html,

>Dispentachoron [2]

>Alternative names:
>Rectified 5-cell (Norman W. Johnson)
>Rectified pentachoron
>Rectified pentatope
>Rectified [four-dimensional] simplex
>Rap (Jonathan Bowers: for rectified pentachoron)

>Schläfli symbol: r{3,3,3}, also t1{3,3,3} or t2{3,3,3}

>Elements:
>Cells: 5 octahedra, 5 tetrahedra
>Faces: 30 triangles (10 joining octahedra to octahedra, 20 joining
octahedra to tetrahedra)
>Edges: 30
>Vertices: 10 (located at the midpoints of the edges of a pentachoron, or at
the centroids of >its faces)

The 2)5 dekany has five 2)4 octahedra (hexanies) and five 1)4 tetrahedra
(tetrads). The number of edges is 30. The number of faces is the number of
consonant triads. The otonal triads each have one factor in common between
the notes. So there are 4choose3 = 4 otonal triads for each of the 5
factors, hence there are 4*5=20 otonal triads. The number of utonal triads
is the number that _lack_ two common factors, and there's only one triad for
each pair of lacking factors, hence the number of utonal triads is just
5choose2 = 10. 20+10=30 consonant triads!

We have a winner!

So what about the Eikosany?

We already know it has 20 vertices and 90 edges.
The number of otonal triads is 5choose3 = 10, times the number of factors
(6), gives 60 otonal triads. The number of utonal triads is the number that
lack two common factors, and there are 4choose3 = 4 triads for each pair of
lacking factors, and 6choose2 = 15 ways of choosing two lacking factors, so
there are 60 utonal triads. 120 consonant triads altogether. The number of
otonal tetrads is the number of ways of taking two common factors, or
6choose2 = 15. The number of utonal tetrads is the number of ways of
_lacking_ two common factors, or 6choose2 = 15. The number of hexanies is
the number of ways of leaving out two factors (6choose2) times the number of
ways of choosing a third factor (2) for 30 hexanies. There are _no_ pentads,
but there are five 2)5 dekanies _and_ five 3)5 dekanies.

Whew!

I'm e-mailing George Olshevsky, the author of that page.