Further Comments
Andrew Weimholt, Dec 19 2006
The a(4) terms is
a(4) = 1480206036768915456000 = 2^30 * 3^6 * 5^3 * 7^2 * 11 * 13 * 17 * 127
I actually found it back in early November, and was surprised to
see that the result contained a strange prime factor, 127.
I had expected that the largest prime factor would be less than
or equal to the dimension, 4n+3.
The factor, 127, can only be explained by the existence of multiple
sets of simplices (such that no symmetry operation of the 19-cube can
transform a simplex from one set into a simplex of another).
Indeed, it turns out that there are 3 sets of 19-simplices
which can be inscribed on the 19-cube, occurring in the following numbers:
a(4)*32/127, a(4)*57/127, a(4)*38/127
Set 1:
The symmetry group acting on the superposition of 1 of these simplices
with the 19-cube has order 171
From this, the number of simplices in this set is computed to be
2^35 * 3^6 * 5^3 * 7^2 * 11 * 13 * 17 = 372965300603191296000
Set 2:
The symmetry group acting on the superposition of 1 of these simplices
with the 19-cube has order 96
From this, the number of simplices in this set is computed to be
2^30 * 3^7 * 5^3 * 7^2 * 11 * 13 * 17 * 19 = 664344441699434496000
Set 3:
The symmetry group acting on the superposition of 1 of these simplices
with the 19-cube has order 144
From this, the number of simplices in this set is computed to be
2^31 * 3^6 * 5^3 * 7^2 * 11 * 13 * 17 * 19 = 442896294466289664000
These results prompted me to go back and look at the 15 dimensional case.
It turns out there are 5 sets of 15-simplices which can be inscribed on
the 15-cube. The group orders and counts are as follows:
15d-set-1: ord = 2688, count = 15941173248000
15d-set-2: ord = 2688, count = 15941173248000
15d-set-3: ord = 1536, count = 27897053184000
15d-set-4: ord = 9216, count = 4649508864000
15d-set-5: ord = 322560, count = 132843110400
For 11 and 7 dimensions, there's only 1 set each (unless you
only allow pure rotations in which case they can be divided
into left- and right-handed sets).
Additional comments added Dec 26 2005:
I have submitted two more terms to A108973, the number of regular
(4n+3)-simplices that can be inscribed on a (4n+3)-cube...
8898131405512141870083342336000 = 2^41 * 3^10 * 5^3 * 7^3 * 11 * 13 * 17 * 19 * 34603
10827543712227210782977570287648768000000 = 2^52 * 3^9 * 5^6 * 7^4 * 11^2 * 13 * 17 * 19 * 23 * 278617
The large prime factors at the end can be found in A048615, which is no
coincidence, since the problem of finding simplices on the cube
is related to Hadamard matrices.
These additional terms were computed using the orders of the automorphism
groups of Hadamard matrices which can be found at
http://www.research.att.com/~njas/hadamard/index.html
Up until last week, I didn't know that this data was so readily available,
so the previous terms of the sequence
were computed with much more effort than these last two.
The number of regular (4n-1)-simplices which can be inscribed on an
(4n-1)-cube is precisely the number of
regular 4n-orthoplexes which can be inscribed on a 4n-cube. Each Hadamard
matrix corresponds to such an orthoplex, and the automorphism group
of the Hadamard matrix is also the automorphism
group of the corresponding orthoplex.
For each "type" of orthoplex, we can compute the number appearing on the
cube by dividing the order of the symmetry group of the cube,
by the order of the automorphism group of the orthoplex.
We then sum the results for each type to get the total number of
orthoplexes on the cube.
For the A108973(4) term that I submitted last week, I commented on the
orders of the automorphism groups of the simplices for
dimensions 19 and 15, and later noticed this relationship...
Let H be the automorphism group for a Hadamard matrix of order 4n
Let G be the automorphism group of a corresponding simplex on cube
(dimension 4n-1)
for (n<6)
8n * |Aut(G)| = |Aut(H)|
This holds for n<6 because the all of the 8n cubically aligned (4n-1)-s
implices of the 4n-orthoplexes are equivalent. However,
for n>6, some of the orthoplexes have multiple
types of cubically aligned simplices. (By "cubically aligned", I mean
those (4n-1)-simplices of the 4n-orthoplex which are inscribed
on (4n-1)-cubes of the 4n-cube.)
In short, for n>6, there is no one "corresponding simplex" but a set
of inequivalent simplices.
The number of inequivalent 4n-orthoplexes on the 4n-cube is the same as the number of
inequivalent Hadamard matrices of order 4n (see A007299). For n<6, this is
also the number of inequivalent (4n-1)-simplices on the (4n-1)-cube, but for n=6, the
number of inequivalent simplices is at least 110 (it'll take some time to
determine the precise number).