From Veit Elser, April 13 2020: Here are the counts up to dimension d = 4. Starting with the simplex, e.g. for d=3, 000 001 011 111 I iteratively generate sets of convex pieces by slicing through all the pieces with another hyperplane to get the pieces in the next iteration. I checked my hyperplane generator against A007847 (you can drop the 2d hyperplanes that bound the hypercube). Here are the counts for the number of pieces having d+1, d+2, … vertices: d=2 {{3, 2}} total pieces: 2 d=3 {{4, 12}, {5, 4}} total pieces: 16 d=4 {{5,6784},{6,4024},{7,4936},{8,2704},{9,1912},{10,936},{11,824},{12,496},{13,360},{14,352},{15,256},{16,72},{17,48},{18,16},{19,16},{20,16}} total pieces: 23752 To get the number of pieces in the hypercube just multiply these by d!. This gives the following values: a(1) = 1*1! = 1 a(2) = 2*2! = 4 a(3) = 16*3! = 96 a(4) = 23752*4! = 570048