The table looks great, I don't know if I can complete it though. Deforming the shape into wedge sums is easy for low dimensional shapes, but it gets really hard in higher dimensions, especially for frames in the middle. I'll edit it a bit after I answer your questions. It's worth keeping, but a table of homology groups might be more useful and easier to calculate.
1. Is that 1-frame hypercube guess correct?
Not quite, but the 5,7,9 sequence is in there. The 3-frame tesseract is 3^7. You can get this by drawing the net for the tesseract and counting all the cubes but one. Similarly the 4-frame penteract is 4^7.
2. How can we find the other missing values?
The others are hard to calculate but I found some of them. I can tell you that the 1-frame tesseract has 17 lines, and a few other things about hypercubes that I'll put into the table. All of my work on this is currently in a stack of unordered loose pages
.
3. Since it's apparently not correct to just count the numbers in the expanded rotatope, how is the hole-sequence found? I see a pattern that appending 1 to the expanding rotatope multiplies the elements in the hole-sequence by two, and I've seen a few other patterns, but I can't see anything completely general.
It's very difficult
. I do have recursive formulas for the following:
1. Cartesian product of two hyperspheres (it's basically what you'd expect)
2. The product of a hypersphere with many lines (in any frame)
3. The n-torus T^n.
4. 3T^n, 33T^n and 4T^n.
I think when I get a few more formulas I'll start seeing the pattern. I'll post all of these formulas in a separate thread later so we have them all in one place.
4. Can we generate an expanded rotatope from a hole-sequence? Does every hole-sequence have an associated shape, even if it isn't a rotatope?
Good question. I think we'll have to wait till we have a general formula to find out. Certainly not every hole-sequence has a shape. For instance you can't have H
0X = 0, unless X is the empty set. I don't know if there are less trivial counterexamples though.