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Apr 24, 2020 at 22:43 comment added ReverseFlowControl @GerhardPaseman, looking at the vector representation and counting those with properties that place them on the boundaries is the only "reasonable" way I know of looking at it. A face of the hypercube has all point lying on some hyperplane at $x_i=0$ or $x_i=d$. Using higher level geometry, or fancier math, seems unnecessary. Just like in 3D $y=0$ means zx-plane, or $z=0$ means xy-plane, and so on.
Apr 24, 2020 at 22:39 comment added Gerhard Paseman He is counting a different quantity, which may (or may not) bear some relation to your quantity. I suggested using Euler's formula for d=2. For higher d, one might start out with his formula to count related quantities. By the way, boundary is not clear yet in higher dimensions for me. I can see counting only lattice points on edges or on edges and faces as qualifying, but you might mean something more general. Gerhard "Not Used To High-Dimensional Combinatorics" Paseman, 2020.04.24.
Apr 24, 2020 at 22:39 comment added ReverseFlowControl I think we can easily see the first case is indeed 4: 3 triangles all with one corner at the origin and then the entire square. For $n=3$, that is the right count if you realize that polytopes need not be a triangle, you could choose 4 points including the origin and get a polypote with 4 sides where an edge would be part of the boundary, there are many such cases that only counting triangles leaves out..
Apr 24, 2020 at 22:06 comment added skd Thanks. Again, apologies for possibly being silly, but it seems like the sequence γ(L(n,2)) goes 4, 120, 2036, ... It's not clear to me how this sequence relates to the question.
Apr 24, 2020 at 20:57 history edited ReverseFlowControl CC BY-SA 4.0
Update.
Apr 24, 2020 at 20:48 history edited ReverseFlowControl CC BY-SA 4.0
Fixed notation. again.
Apr 24, 2020 at 20:38 comment added ReverseFlowControl Fixed. Thanks! (Also, the 15 character comment length minimum is silly.)
Apr 24, 2020 at 20:37 history edited ReverseFlowControl CC BY-SA 4.0
Fixed minor notation issue.
Apr 24, 2020 at 19:56 comment added skd Thanks for the response. I'm not sure I fully understand; I don't parse the string "L(n,d) := [0,n]^d" and then the equations for L(n,d). Perhaps I've not calculated/understood your answer correctly, but it doesn't seem like any of the equations listed recover the sequence 4, 56, 340 for an n-by-n square in two dimensions.
Apr 24, 2020 at 9:18 history edited ReverseFlowControl CC BY-SA 4.0
Fixed sum.
Apr 24, 2020 at 7:00 history answered ReverseFlowControl CC BY-SA 4.0