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Number of non-equivalent ways to choose 4 points in an equilateral triangle grid of side n.
4

%I #27 Feb 10 2024 09:22:41

%S 0,0,4,41,244,1029,3485,9926,25030,57126,120570,238330,446344,797825,

%T 1370684,2274259,3660612,5734776,8771181,13127940,19270240,27789713,

%U 39435814,55142010,76066910,103627784,139554142,185929971,245260890,320527585,415268815

%N Number of non-equivalent ways to choose 4 points in an equilateral triangle grid of side n.

%H Heinrich Ludwig, <a href="/A231653/b231653.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_19">Index entries for linear recurrences with constant coefficients</a>, signature (2,3,-5,-8,3,19,4,-24,-15,15,24,-4,-19,-3,8,5,-3,-2,1).

%F a(n) = (n^8 + 4*n^7 - 6*n^6 - 32*n^5 + 84*n^4 - 32*n^3 - 16*n^2 - 192*n + B + C)/2304

%F where

%F B = 84*n^3 - 234*n^2 + 168*n + 171 if n==1 (mod 2)

%F = 0 otherwise

%F and

%F C = 128*n^2 + 128*n - 256) if n==1 (mod 3)

%F = 0 otherwise

%F G.f.: -x^3*(x^14 +7*x^12 +26*x^11 +146*x^10 +432*x^9 +947*x^8 +1418*x^7 +1621*x^6 +1405*x^5 +932*x^4 +438*x^3 +150*x^2 +33*x +4) / ((x -1)^9*(x +1)^4*(x^2 +x +1)^3). - _Colin Barker_, Feb 15 2014

%e For n = 3 there are the following a(3) = 4 choices of 4 points (=X) (rotations and reflections ignored):

%e X . X X

%e X X X X X X . .

%e . X . X . X . . X X X X

%Y Cf. A234249, A001399, A227327, A230723, A231654, A231655.

%K nonn,easy

%O 1,3

%A _Heinrich Ludwig_, Nov 12 2013