[seqfan] Analogue of A244951 in four dimensions

[Felix Fröhlich]

Dear SeqFans,

I recently thought about sequence 28. at
https://oeis.org/wiki/Suggested_Projects#Requested_sequences

I think I have the values for a(1)-a(3), but I am bit stuck with a(4)-a(6).

For n = 1: In the 5-cell, just as in the tetrahedron each face shares an
edge with every other face, each cell shares a common "ridge" with every
other cell, so no two cells can have the same color and thus a(1) = 5.

For n = 2: In the 8-cell (tesseract), each cell shares a common "ridge"
with six other cells, but shares no common "ridge" with the "opposite"
cell, i.e., the other cell lying in a 3-dimensional hyperplane parallel to
the hyperplane of the original cell. So there are four pairs of opposite
cells and thus a(2) = 4.

For n = 3: Just like the octahedron can be decomposed into two square
pyramids, the 16-cell can be decomposed into two octahedral pyramids (two
4-polytopes), each of which is bounded by one base octahedron and eight
tetrahedra. Tetrahedra that share a common edge of the "base"-octahedron
are neighboring cells when folded into the fourth dimension, so coloring
the tetrahedra, say, black and white such that at each edge of the
octahedron a white tetrahedron meets a black tetrahedron ensures that after
folding the tetrahedra into the fourth dimension each triangular
tetrahedron face is shared by a black and a white tetrahedron. Then one can
rotate the two octahedral pyramids against each other and reassemble them
into a 16-cell that is bounded by cells in two colors. So we have a(3) = 2.

The case n = 4 seems to be a bit more difficult, and I am not sure about
the value I obtained.
My approach was to look at the vertex figure of the 24-cell. At each
vertex, 6 octahedra meet in a cubic fashion, i.e., the vertex figure of the
24-cell is a cube. This readily gives a lower bound of 3 colors, because
the cube requires at least 3 colors if no two faces (square faces of the
cube) of similar color are supposed to meet at an edge.
As explained in https://en.wikipedia.org/wiki/24-cell#Visualization, the
24-cell can be thought of as consisting of 5 latitudinal layers of cells:

Layer Cells Description
-----    -----   -----------
1       1       North Pole
2       8       Northern circle of latitude
3       6       Equator
4       8       Southern circle of latitude
5       1       South Pole

I started by coloring the North and South pole cells red. The equatorial
cells touch the northern and southern circle cells at their faces and touch
the pole cells at their vertices. Since the equator has an even number of
cells, we can color the equatorial cells, say, red and green in an
alternating fashion. Lastly, we color the northern and southern circle
cells blue and yellow in an alternating fashion. So it seems a(4) = 4.

I have not yet tried to obtain values for the 120-cell and the 600-cell.

Are the values I obtained correct? Does someone have the values for the
120-cell and the 600-cell? If we have all six values, then I definitely
think this could be submitted.

Best regards
Felix