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A257238
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Triangle T(n, k) = n^3 - k^3, 0 <= k < = n.
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4
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0, 1, 0, 8, 7, 0, 27, 26, 19, 0, 64, 63, 56, 37, 0, 125, 124, 117, 98, 61, 0, 216, 215, 208, 189, 152, 91, 0, 343, 342, 335, 316, 279, 218, 127, 0, 512, 511, 504, 485, 448, 387, 296, 169, 0, 729, 728, 721, 702, 665, 604, 513, 386, 217, 0, 1000, 999, 992, 973, 936, 875, 784, 657, 488, 271, 0
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OFFSET
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0,4
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COMMENTS
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See the comments in A025581 and A079904 on a problem by François Viète (Vieta)(1593). Regarding that problem, note the simple identity: n^3 - k^3 = (n - k)^3 + 3*n*k*(n - k), for n > = k >= 0.
Row sums give A126274(n-1) for n >= 1, and 0 for n=0.
Alternating row sums are ars(2*n) = ars(2*n-1) = (4*n-3)*n^2 = A103532(n-1), for n >= 1, and ars(0) = 0.
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LINKS
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FORMULA
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Columns (with one leading zero and offset 0): k=0: l^3 = A000578(l), k=1: (l+1)^3 - 1 = A068601(l+1), k=2: l*(l^2 + 6*l + 12), k=3: l*(l^2 + 9*l + 27), k=4: l*(l^2 + 12*l + 48), k=5: l*(l^2 + 15*l + 75), ...
G.f. for T(n,k): (1+4*x+4*x*y+x^2-14*x^2*y+x^2*y^2-2*x^3*y-2*x^3*y^2+7*x^4*y^2)*x/((1-x*y)^3*(1-x)^4). - Robert Israel, May 10 2018
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EXAMPLE
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The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10
0: 0
1: 1 0
2: 8 7 0
3: 27 26 19 0
4: 64 63 56 37 0
5 125 124 117 98 61 0
6: 216 215 208 189 152 91 0
7: 343 342 335 316 279 218 127 0
8: 512 511 504 485 448 387 296 169 0
9: 729 728 721 702 665 604 513 386 217 0
10: 1000 999 992 973 936 875 784 657 488 271 0
...
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MAPLE
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for n from 0 to 10 do seq(n^3-k^3, k=0..n) od; # Robert Israel, May 10 2018
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MATHEMATICA
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Table[n^3-k^3, {n, 0, 10}, {k, 0, n}]//Flatten (* Harvey P. Dale, Jan 02 2021 *)
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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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