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%I #26 Feb 09 2024 11:16:35
%S 0,2,12,33,85,165,315,518,846,1260,1870,2607,3627,4823,6405,8220,
%T 10540,13158,16416,20045,24465,29337,35167,41538,49050,57200,66690,
%U 76923,88711,101355,115785,131192,148632,167178,188020,210105,234765,260813,289731,320190
%N Nonequivalent ways to place two different markers (e.g., a pair of Go stones, black and white) on an n X n grid.
%C We say two placements are equivalent if one can be obtained from the other by rotating or reflecting the grid.
%C The formula was derived by categorizing and counting grid cells into four exclusive categories: central cell (if any); other diagonal cells, other horizontal and vertical midline cells (if any), and all others (in eight triangular regions) (if any); then determining for each category, how many ways a white stone could be placed in each category, given the category in which the black stone was placed prior. The sequence was verified by another program which generated all positions, removed reflections and rotations, and tallied the residue.
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-6,0,6,-2,-2,1).
%F For odd n, a(n) = n*(n^3 + 3*n - 4)/8.
%F For even n, a(n) = n*(n^3 + n - 2)/8.
%F G.f.: -x^2*(x^5+x^4+7*x^3+5*x^2+8*x+2) / ((x-1)^5*(x+1)^3). - _Colin Barker_, May 21 2014
%F a(n) = n*(n^3 + n*3^(n mod 2) - 2*2^(n mod 2))/8. - _Wesley Ivan Hurt_, May 21 2014
%p A242709:=n->n*(n^3 + n*3^(n mod 2) - 2*2^(n mod 2))/8; seq(A242709(n), n=1..50); # _Wesley Ivan Hurt_, May 21 2014
%t f[n_] := If[OddQ[n], (n^3 + 3 n - 4), (n^3 + n - 2)] n/8;
%t Table[f[n], {n, 1, 40}]
%o (PARI) concat(0, Vec(-x^2*(x^5+x^4+7*x^3+5*x^2+8*x+2)/((x-1)^5*(x+1)^3) + O(x^100))) \\ _Colin Barker_, May 21 2014
%o (Magma) [n*(n^3 + n*3^(n mod 2) - 2*2^(n mod 2))/8: n in [1..50]]; // _Wesley Ivan Hurt_, May 21 2014
%Y Cf. A014409 (with indistinguishable checkers)
%K nonn,easy
%O 1,2
%A _James Stein_, May 21 2014
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