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a(n) = (1/2)*number of non-degenerate triangular pyramids that can be formed using 4 distinct points chosen from an (n+1) X (n+1) X (n+1) lattice cube.
+10
4
29, 7316, 285400, 4508716, 42071257, 273611708, 1379620392, 5723597124, 20398039209, 64302648044, 183316772048, 480140522044, 1170651602665
COMMENTS
The observed growth rate of CPU time required to compute more terms is approximately ~ n^10.5.
EXAMPLE
a(1)=29: Only 58 of the A103157(1)=70 possible ways to choose 4 distinct points from the 8 vertices of a cube result in pyramids with volume > 0: 2 regular tetrahedra of volume=1/3 and 56 triangular pyramids of volume=1/6. The remaining A103658(1)=12 configurations result in objects with volume=0. Therefore a(1)=(1/2)*( A103157(1)- A103658(1))=58/2=29.
CROSSREFS
Cf. A103157 binomial((n+1)^3, 4), A103158 tetrahedra in lattice cube, A103658 4-point objects with volume=0 in lattice cube, A103426 non-degenerate triangles in lattice cube.
Number of different volumes assumed by triangular pyramids with their 4 vertices chosen from distinct points of an (n+1)X(n+1)X(n+1) lattice cube, including degenerate objects with volume=0.
+10
4
3, 13, 39, 90, 178, 309, 503, 756, 1096, 1523, 2059, 2683, 3469, 4355, 5406
EXAMPLE
a(1)=3 because 4-point objects with 3 different volumes can be built using the vertices of a cube: 2 regular tetrahedra (e.g. [(0,0,0),(0,1,1),(1,0,1),(1,1,0)]) with volume 1/3, 56 pyramids with volume 1/6 and 12 objects with volume=0, e.g. the faces of the cube.
a(2)=13: The A103157(2)=17550 4-point objects that can selected from the 27 points of a 3X3X3 lattice cube fall into 13 different volume classes (6*V,occurrences): (0,2918), (1,3688), (2,5272), (3,1272), (4,2788), (5,272), (6,684), (7,72), (8,494), (9,16), (10,48), (12,24), (16,2).
A103658(n) gives the occurrence counts of objects with V=0 (i.e. A103658(2)=2918).
A103659(n) gives 6*V of the most frequently occurring volume and A103660(n) gives the corresponding occurrence count, divided by 2. Therefore A103659(2)=2 and A103660(2)=2636.
A103661(n) gives the smallest value of 6*V not occurring in the list of 4-point object volumes, i.e. A103661(2)=11.
Number of configurations of 4 distinct points chosen from an (n+1) X (n+1) X (n+1) lattice cube resulting in objects with volume=0.
+10
3
12, 2918, 64576, 673943, 4058656, 19462319, 70636336, 224091378, 621046332, 1573877652, 3582064304, 3582064304, 15781332296
EXAMPLE
a(1)=12 because the 6 faces and 6 diagonal cuts along parallel diagonals of opposite faces through a cube are objects with 4 distinct points with volume=0.
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