

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
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OFFSET

0,3


COMMENTS

A regular tetrahedral grid with n points on each side contains a total of A000292(n) points.
a(n) >= 4*(n1), because there are n1 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*(n1) 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(n1) + 6*A334581(n2)  4*A334581(n3) + A334581(n4) for n >= 4.


EXAMPLE

For n = 6 there are 28 equilateral triangles that touch all of the sides of the sixvertexperside 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: A316712 A261650 A178731 * A071072 A175670 A194374
Adjacent sequences: A334383 A334384 A334385 * A334387 A334388 A334389


KEYWORD

nonn


AUTHOR

Peter Kagey, May 11 2020


STATUS

approved



