cp's OEIS Frontend

A361388 Number of orders of distances to vertices of n-dimensional cube.

%I A361388 #19 Mar 27 2023 07:53:00
%S A361388 1,2,8,96,5376,1981440,5722536960,138430238607360
%N A361388 Number of orders of distances to vertices of n-dimensional cube.
%C A361388 Let C be an n-dimensional cube and p be a point in R^n such that the distances from p to the 2^n vertices of C are all different. List the vertices in order of their distance from p. The number of different orders of vertices is given by a(n).
%C A361388 Equality of any two distances defines a hyperplane in R^n, although different pairs of distances may define the same hyperplane. All these hyperplanes partition the space into cells, and the interior of each (n-dimensional) cell corresponds to a particular strong order of the differences. Hence, a(n) equals the number of cells in the partition of R^n by the hyperplanes. The given SageMath code implements this approach. - _Max Alekseyev_, Mar 10 2023
%C A361388 Computing the sequence is slow. The Sage program took 20 minutes to compute a(5) on Lucas Brown's box; the C++ program took 3.5 seconds to compute a(5) on Pierre Abbat's box, a 12-thread Ryzen. The C++ program took 6 hours to compute a(6). Neither of us has computed a(7) with the program; that's from A009997.
%C A361388 For n >= 4 the frequencies of the orders appear to vary widely.
%H A361388 Pierre Abbat, <a href="https://github.com/phma/cubeorders">Cubeorders</a>
%F A361388 a(n) = 2^n*n!*A009997(n).
%e A361388 For n=3, a 3-dimensional cube has 8 corners, numbered 0 to 7. A point can be closest to any of the 8 corners. A point closest to 0 can have distances to corners 1, 2, and 4 in any of 6 orders. A point whose distances to corners 0, 1, 2, and 4 are in increasing order can be closer to 3 than to 4, or closer to 4 than to 3. So the total number of orders is 8*6*2=96.
%o A361388 (Sage)
%o A361388 def a(n):
%o A361388     x = polygens(QQ,n,'x')
%o A361388     dist2 = [sum((xi - ti)^2 for xi,ti in zip(x,t)) for t in Tuples(range(2),n)]    # squared distances
%o A361388     diffs = {p[0]-p[1] for p in Combinations(dist2,2)}     # set of pairwise differences of squared distances
%o A361388     H = HyperplaneArrangements(QQ, tuple(map(str,x)))
%o A361388     A = H([[[d.coefficient({xi:1}) for xi in x], d.constant_coefficient()] for d in diffs])
%o A361388     return A.n_regions()
%o A361388 print( [a(n) for n in (1..4)] ) # _Max Alekseyev_, Mar 10 2023
%o A361388 (C++) // See Cubeorders link.
%o A361388 (PARI) A361388(n) = A009997(n)*n!<<n \\ _M. F. Hasler_, Mar 10 2023
%Y A361388 Cf. A009997.
%K A361388 nonn,hard,more
%O A361388 0,2
%A A361388 _Pierre Abbat_, Mar 10 2023