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A014409
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Number of inequivalent ways (mod D_4) a pair of checkers can be placed on an n X n board.
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27
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0, 2, 8, 21, 49, 93, 171, 278, 446, 660, 970, 1347, 1863, 2471, 3269, 4188, 5356, 6678, 8316, 10145, 12365, 14817, 17743, 20946, 24714, 28808, 33566, 38703, 44611, 50955, 58185, 65912, 74648, 83946, 94384, 105453, 117801, 130853, 145331, 160590, 177430, 195132
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OFFSET
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1,2
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REFERENCES
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Computed by Fred Hallden.
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LINKS
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FORMULA
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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).
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
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
For odd n: a(n) = (n^4 + 8*n^2 - 8*n - 1)/16;
For even n: a(n) = n*(n^3 + 6*n - 4)/16. (End)
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
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
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MATHEMATICA
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LinearRecurrence[{2, 2, -6, 0, 6, -2, -2, 1}, {0, 2, 8, 21, 49, 93, 171, 278}, 40]
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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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Borghard, William (bogey(AT)hostare.att.com)
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EXTENSIONS
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STATUS
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approved
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