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%I #58 Dec 30 2023 22:44:51
%S 0,2,8,21,49,93,171,278,446,660,970,1347,1863,2471,3269,4188,5356,
%T 6678,8316,10145,12365,14817,17743,20946,24714,28808,33566,38703,
%U 44611,50955,58185,65912,74648,83946,94384,105453,117801,130853,145331,160590,177430,195132
%N Number of inequivalent ways (mod D_4) a pair of checkers can be placed on an n X n board.
%D Computed by Fred Hallden.
%H Vincenzo Librandi, <a href="/A014409/b014409.txt">Table of n, a(n) for n = 1..1000</a>
%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 a(2*n) = n/2*(2*n^3 + 3*n - 1); a(2*n+1) = n/2*(2*n^3 + 4*n^2 + 7*n + 3).
%F a(0)=0, a(1)=2, a(2)=8, a(3)=21, a(4)=49, a(5)=93, a(6)=171, a(7)=278, a(n)=2*a(n-1)+2*a(n-2)-6*a(n-3)+0*a(n-4)+6*a(n-5)-2*a(n-6)- 2*a(n-7)+ a(n-8). - _Harvey P. Dale_, May 06 2012
%F G.f.: -x^2*(x^5+x^4+3*x^3+x^2+4*x+2) / ((x-1)^5*(x+1)^3). - _Colin Barker_, Jul 11 2013
%F From _James Stein_, May 22 2014: (Start)
%F For odd n: a(n) = (n^4 + 8*n^2 - 8*n - 1)/16;
%F For even n: a(n) = n*(n^3 + 6*n - 4)/16. (End)
%F a(n) = A054252(n, 2), n >= 0. - _Wolfdieter Lang_, Oct 03 2016
%F E.g.f.: (x*(1 + 13*x + 6*x^2 + x^3)*cosh(x) + (-1 + 3*x + 15*x^2 + 6*x^3 + x^4)*sinh(x))/16. - _Stefano Spezia_, Apr 14 2022
%F a(n) = (2*n^4+14*n^2-12*n-1-(-1)^n*(2*n^2-4*n-1))/32. - _Wesley Ivan Hurt_, Dec 30 2023
%t LinearRecurrence[{2, 2, -6, 0, 6, -2, -2, 1}, {0, 2, 8, 21, 49, 93, 171, 278}, 40]
%t CoefficientList[Series[- x (x^5 + x^4 + 3 x^3 + x^2 + 4 x + 2)/((x - 1)^5 (x + 1)^3), {x, 0, 50}], x] (* _Vincenzo Librandi_, Oct 15 2013 *)
%o (PARI) a(n)=if(n%2, n^4 + 8*n^2 - 8*n - 1, n^4 + 6*n^2 - 4*n)/16 \\ _Charles R Greathouse IV_, Feb 09 2017
%o (Magma) [(2*n^4+14*n^2-12*n-1-(-1)^n*(2*n^2-4*n-1))/32 : n in [1..60]]; // _Wesley Ivan Hurt_, Dec 30 2023
%Y Cf. A054252, A019318, A082966, A242279, A242358, A054247.
%K nonn,nice,easy
%O 1,2
%A Borghard, William (bogey(AT)hostare.att.com)
%E More terms and formula from _Hugo van der Sanden_
%E More terms from _Colin Barker_, Jul 11 2013
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