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A227970
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Triangular arithmetic on half-squares: b(n)*(b(n) - 1)/2 where b(n) = floor(n^2/2).
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2
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0, 0, 1, 6, 28, 66, 153, 276, 496, 780, 1225, 1770, 2556, 3486, 4753, 6216, 8128, 10296, 13041, 16110, 19900, 24090, 29161, 34716, 41328, 48516, 56953, 66066, 76636, 87990, 101025, 114960, 130816, 147696, 166753, 186966, 209628, 233586, 260281, 288420, 319600, 352380, 388521, 426426
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OFFSET
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0,4
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COMMENTS
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A "mirrored" repeating pattern of cycle length 20 exists in the last digit.
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LINKS
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FORMULA
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Let b(n) = floor(n^2/2), for n => 0, then a(n) = b(n)*(b(n)- 1)/2.
G.f. -x^2*(1+4*x+14*x^2+4*x^3+x^4) / ( (1+x)^3*(x-1)^5 ). - R. J. Mathar, Aug 14 2013
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MAPLE
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MATHEMATICA
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Table[Binomial[Floor[n^2/2], 2], {n, 0, 50}] (* Wesley Ivan Hurt, Sep 27 2013 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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