I was sure he would bring it up ... no?
Alright, well go there and get it. The feature? You can save
individual file names now. Very useful for those of us who do
different things on it.
http://www.gravitation3d.com/magiccube5d/
Out of boredom, I chose to see for myself the number of positions on
the 3x3x3x3 and I kept getting half the value!
Here's what I did:
24!x32!x16! - This is the total number of positions accounting all
n-sided stickers
(2^24)x(6^32)x(12^16) - The total number of orientations accounting
for all n-sided stickers
Now I divided numbers appropriately based on these findings:
2-I could not switch just two 2-colored stickers.
2-I could not switch just two 3-colored stickers.
2-I could not switch just two 4-colored stickers.
2-I could not flip just one 2-colored sticker.
2-I could flip just one 3-colored sticker, but only in three
orientations, not six.
3-I could flip just one 4-colored sticker, but only in four ways, not
twelve.
So my end result was:
((24!/2)x(32!/2)x(16!/2))x((2^24)/2)x((6^32)/2)x((12^16)/3)=3D
878,386,440,354,567,921,584,263,
039,540,512,529,807,242,315,074,
778,825,738,578,010,866,618,399,
485,084,275,300,137,443,825,041,
177,103,564,800,000,000,000,000
Which is half of the currently accepted value. I think the argument
lies where Eric Balandraud stated "Not all the permutations of the 24
2-colored and the 32 3-colored are possible. Only the permutations
that have the same parity on the 2-colored and the 3-colored ... So
the number of positions reachable by just the 2-colored [and
3-colored] pieces is (24!x32!)/2"
He did mention that neither 2- or 3-colored stickers could switch just
two pieces, but he only divided by two once, not compensating for both
occurrences. The real value, I believe, should have been half of what
he gave us in that part of the equation.