Klitzing wrote:Did someone meanwhile made some progress with that seed?

Went on myself,

Here is the so far extended

axially pyritohedral seed:

1) 1

cube embedded centrically at origin in axially pyritohedral symmetry

its vertex coords then are A = ( 1/2, 1/2, 1/2; 0 ) with pyritohedral symmetry applied to the first three.

2) 6

hips to be square adjoined to that cube from (1)

it adds vertex coords B = ( (f+1)/2, 1, 1/2; (f-1)/2 ) for the nearest neighbours

resp. C = ( (2f+1)/2, 1/2, 1/2; f-1 ) for the opposites, each again with pyritohedral symmetry to be applied to the first three.

3) 12

tricues to be hexagon adjoined to the hips of (2)

as well as one lacing square each to be connected to the near lacing square of the neighbouring hip.

Thus all vertices here are already provided, just the single far vertex of the top base triangle,

which then reads D = ( (f+2)/2, 0, (f+1)/2; f-1 ), again with pyritohedral symmetry to be applied to the first three.

4) The near lacing triangles of the tricues of (3) provide around the pyritohedral rotational axes triangular dimples,

which thus can readily be filled by 8

tets. No further vertices arise here.

Thus it occurs that CD = f.

My next try thus was to use

5) 12

bilbiroes to be square attached to the 2 far lacing squares of the hips of (2) each,

resp. triangle attached to the top base of the neighbouring tricues of (3).

One then has lots of new vertices:

(here prime and double prime relate to accordingly applied symmetries)

- Code: Select all
` G H `

F

D'' I

E

B' C'

which can be calculated as

E = ( (f+1)/2, (f+2)/2, 1; f-1 )

F = ( (f+1)/2, (2f+1)/2, (f+2)/2; (2f-1)/2 )

G = ( 1/2, (f+2)/2, (f+3)/2; (2f-1)/2 )

H = ( 1/2, f+1, (f+2)/2; (3f-2)/2 )

I = ( 0, f+1, 1; (2f-1)/2 )

again all with pyritohedral symmetry to be applied to the first three.

This then allows

6) 24

squippies square attached to far lacing squares of the tricues of (3)

and triangle attached to the overhanging triangles of the bilbiroes of (5)

Again all vertices are already provided:

- Code: Select all
`B' C'`

E

B D'

7) the open face of the tets of (4) then has to be adjoined to one of 8

octs each,

the neighbouring oct faces then adjoin to the next free squippy triangle.

No new vertices.

8) The B'EFGD'' pentagons of the bilbiroes of (5) adjoin to one of 24

peppies each,

one lacing of which then adjoins to one of the other lacings of the octs of (7).

No new vertices.

9) The opposite triangle of the octs of (7) then adjoin to 8 further

octs.

Again no new vertices.

(to be continued ...)

--- rk