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A289230 Number of nonequivalent ways to select 3 disjoint point triples from an n X n X n triangular point grid, each point triple forming a 2 X 2 X 2 triangle.

%I A289230 #10 Jul 01 2017 07:38:43
%S A289230 0,2,19,127,536,1688,4357,9789,19844,37172,65397,109335,175214,270934,
%T A289230 406329,593463,846934,1184212,1625979,2196509,2924050,3841240,4985531,
%U A289230 6399647,8132044,10237410,12777167,15820007,19442436,23729352,28774625,34681717,41564304,49546932
%N A289230 Number of nonequivalent ways to select 3 disjoint point triples from an n X n X n triangular point grid, each point triple forming a 2 X 2 X 2 triangle.
%C A289230 Rotations and reflections of a selection are not counted. If they are to be counted see A289224.
%H A289230 Heinrich Ludwig, <a href="/A289230/b289230.txt">Table of n, a(n) for n = 3..100</a>
%H A289230 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,6,0,0,0,-6,9,-5,1).
%F A289230 a(n) = (n^6 -6*n^5 -24*n^4 +220*n^3 -153*n^2 -1488*n +2592)/36 + IF(MOD(n, 2) = 1, -1)/2 + IF(MOD(n, 3) = 1, -2)/9.
%F A289230 G.f.: x^4*(2 + 9*x + 50*x^2 + 60*x^3 + 37*x^4 - 21*x^5 - 20*x^6 - 4*x^7 + 7*x^8) / ((1 - x)^7*(1 + x)*(1 + x + x^2)). - _Colin Barker_, Jun 30 2017
%e A289230 There are two nonequivalent ways to choose three 2 X 2 X 2 triangles (aaa, bbb, ccc) from a 4 X 4 X 4 point grid:
%e A289230       a           a
%e A289230      a a         a a
%e A289230     b c c       b . c
%e A289230    b b c .     b b c c
%e A289230 Note: aaa, bbb, ccc are not distinguishable, they are denoted differently for a better perception of the 2 X 2 X 2 triangles only.
%o A289230 (PARI) concat(0, Vec(x^4*(2 + 9*x + 50*x^2 + 60*x^3 + 37*x^4 - 21*x^5 - 20*x^6 - 4*x^7 + 7*x^8) / ((1 - x)^7*(1 + x)*(1 + x + x^2)) + O(x^40))) \\ _Colin Barker_, Jun 30 2017
%Y A289230 Cf. A289224, A289229, A117662, A289231, A289232.
%K A289230 nonn,easy
%O A289230 3,2
%A A289230 _Heinrich Ludwig_, Jun 30 2017