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).