|
|
A334386
|
|
a(n) is the number of ways to choose 3 points in a size n tetrahedral grid in such a way that the three points form an equilateral triangle that touches all four sides of the tetrahedron.
|
|
1
|
|
|
0, 0, 4, 8, 12, 16, 32, 36, 28, 32, 60, 100, 80, 84, 64, 80, 96, 88, 116, 132, 172, 188, 144, 208, 128, 228, 112, 188, 156, 268, 212, 312, 196, 224, 288, 328, 296, 324, 232, 344, 324, 412, 260, 384, 244, 512, 420, 364, 296, 492, 316, 452, 432, 556, 404, 588
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
A regular tetrahedral grid with n points on each side contains a total of A000292(n) points.
a(n) >= 4*(n-1), because there are n-1 ways to choose three points on a single face that touch all four sides of the tetrahedron.
a(n) is divisible by 4 for all n.
Conjecture: a(n) - 4*(n-1) is divisible by 12 for n > 0.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
For n = 6 there are 28 equilateral triangles that touch all of the sides of the six-vertex-per-side tetahedron. In barycentric coordinates, these come in four equivalence classes:
{(0, 0, 0, 1), (0, 0, 1, 0), ( 0, 1, 0, 0)},
{(0, 0, 1/5, 4/5), (0, 1/5, 4/5, 0), ( 0, 4/5, 0, 1/5)},
{(0, 0, 2/5, 3/5), (0, 2/5, 3/5, 0), ( 0, 3/5, 0, 2/5)}, and
{(0, 0, 2/5, 3/5), (0, 3/5, 2/5, 0), (3/5, 1/5, 0, 1/5)},
where two triangles are considered equivalent if the coordinates of one are permutations of the other.
The equivalence classes contain 4, 8, 8, and 8 elements respectively.
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|