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%I #32 May 14 2020 19:13:12
%S 0,0,4,8,12,16,32,36,28,32,60,100,80,84,64,80,96,88,116,132,172,188,
%T 144,208,128,228,112,188,156,268,212,312,196,224,288,328,296,324,232,
%U 344,324,412,260,384,244,512,420,364,296,492,316,452,432,556,404,588
%N 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.
%C A regular tetrahedral grid with n points on each side contains a total of A000292(n) points.
%C 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.
%C a(n) is divisible by 4 for all n.
%C Conjecture: a(n) - 4*(n-1) is divisible by 12 for n > 0.
%H Peter Kagey, <a href="/A334386/b334386.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = A334581(n) - 4*A334581(n-1) + 6*A334581(n-2) - 4*A334581(n-3) + A334581(n-4) for n >= 4.
%e 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:
%e {(0, 0, 0, 1), (0, 0, 1, 0), ( 0, 1, 0, 0)},
%e {(0, 0, 1/5, 4/5), (0, 1/5, 4/5, 0), ( 0, 4/5, 0, 1/5)},
%e {(0, 0, 2/5, 3/5), (0, 2/5, 3/5, 0), ( 0, 3/5, 0, 2/5)}, and
%e {(0, 0, 2/5, 3/5), (0, 3/5, 2/5, 0), (3/5, 1/5, 0, 1/5)},
%e where two triangles are considered equivalent if the coordinates of one are permutations of the other.
%e The equivalence classes contain 4, 8, 8, and 8 elements respectively.
%Y Cf. A000292, A334581.
%K nonn
%O 0,3
%A _Peter Kagey_, May 11 2020
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