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A105125
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Triangle read by rows: T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.
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2
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0, 1, 2, 8, 9, 16, 27, 28, 35, 54, 64, 65, 72, 91, 128, 125, 126, 133, 152, 189, 250, 216, 217, 224, 243, 280, 341, 432, 343, 344, 351, 370, 407, 468, 559, 686, 512, 513, 520, 539, 576, 637, 728, 855, 1024, 729, 730, 737, 756, 793, 854, 945, 1072, 1241, 1458, 1000, 1001, 1008, 1027
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,3
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LINKS
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FORMULA
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T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.
G.f. for triangle: -(9*x^5*y^3 - 8*x^4*y^3 - x^4*y^2 + 7*x^3*y^3 - 36*x^3*y^2 - 2*x^2*y^3 + 5*x^3*y + 27*x^2*y^2 + 12*x^2*y - 8*x*y^2 - x^2 + 3*x*y - 4*x - 2*y - 1)*x/((x-1)^4*(x*y-1)^4). - Robert Israel, May 15 2015
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EXAMPLE
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Triangle begins (modulo 2 plot is a checkerboard):
{0}
{1, 2}
{8, 9, 16}
{27, 28, 35, 54}
{64, 65, 72, 91, 128}
{125, 126, 133, 152, 189, 250}
...
The identity for T(2, 1): 9 = 3*(3^2 + 3*1^2)/4 = 3*12/4 = 9. - Wolfdieter Lang, May 15 2015
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MAPLE
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MATHEMATICA
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f[n_, m_, p_] := n^p + m^p p = 3 a = Table[Table[f[n, m, p], {n, 0, m}], {m, 0, 20}] aa = Flatten[a]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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