A063723 Number of vertices in the Platonic solids (in the order tetrahedron, cube, octahedron, dodecahedron, icosahedron).
4, 8, 6, 20, 12
Offset: 1
Examples
a(2) = 8 since a cube has eight vertices.
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a(2) = 8 since a cube has eight vertices.
a(2) = 12 since a cube has twelve edges.
a(2) = 8 since a 4D hypercube contains eight cubes.
a(2) = 16 since a 4D hypercube contains sixteen vertices.
a(0) = 1 because the 0-D regular polytope is the point. a(1) = 2 because the only regular 1-D polytope is the line segment, with 2 vertices, one at each end. a(2) = 0, indicating infinity, because the regular k-gon has k vertices. a(3) = 50 (4 for the tetrahedron, 6 for the octahedron, 8 for the cube, 12 for the icosahedron, 20 for the dodecahedron) = the sum of A053016. a(4) = 773 = 5 + 8 + 16 + 24 + 120 + 600 = sum of A063924. For n>4 there are only the three regular n-dimensional polytopes, the simplex with n+1 vertices, the hypercube with 2^n vertices and the hyperoctahedron = cross polytope = orthoplex with 2*n vertices, for a total of A086653(n) + 1 = 2^n + 3*n + 1 (again restricted to n > 4).
LinearRecurrence[{4, -5, 2}, {1, 2, 0, 50, 773, 48, 83, 150}, 32] (* Georg Fischer, May 03 2019 *)
For n = 4, there are five finite groups, denoted A(4) (symmetry group of the simplex), B(4) (= C(4)) (symmetry group of the tesseract and the 4-dimensional cross polytope), D(4) (symmetry group of the demitesseract), F(4) (symmetry group of the 24-cell) and H(4) (symmetry group of the 120-cell and the 600-cell).
PadRight[{1, -1, 3, 5, 3, 4, 4, 4}, 100, 3] (* Paolo Xausa, Dec 07 2024 *)
a(n)=if(n>8,3,[1,-1,3,5,3,4,4,4][n]) \\ Charles R Greathouse IV, Jul 15 2024
a(2)=3 because the plane can be tiled by equilateral triangles, squares or regular hexagons. a(3)=1 since the only platonic solid that can tile 3-dimensional space is the cube. a(4)=3 because the 4-dimensional space can be tiled by hypercubes (tesseracts), hyperoctahedra or 24-cell polytopes.
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