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

0,3

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

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

Formula

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

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

Cf. A000292, A334581.

Sequence in context: A261650 A178731 A381699 * A071072 A175670 A355031

Adjacent sequences: A334383 A334384 A334385 * A334387 A334388 A334389

Keyword

nonn

Author

Peter Kagey, May 11 2020

Status

approved