A257238 Triangle T(n, k) = n^3 - k^3, 0 <= k < = n.

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

Offset: 0

Author

Wolfdieter Lang, May 12 2015

Keywords

• nonn
• easy
• tabl

Comments

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.

Examples

			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
  ...
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Crossrefs

Cf. A126274, A103532, A025581, A079904, A335821.

Programs

Maple

for n from 0 to 10 do seq(n^3-k^3,k=0..n) od; # Robert Israel, May 10 2018

Mathematica

Table[n^3-k^3,{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Jan 02 2021 *)

Formula

T(n, k) = A025581(n, k)*(A025581(n, k)^2 + 3* A079904(n, k)) (see the identity mentioned in a comment).

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