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.

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

Offset: 0

Author

Peter Kagey, May 11 2020

Keywords

• nonn

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.

Examples

			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.
		

Links

• Peter Kagey, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000292, A334581.

Formula

a(n) = A334581(n) - 4*A334581(n-1) + 6*A334581(n-2) - 4*A334581(n-3) + A334581(n-4) for n >= 4.