A105125 Triangle read by rows: T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.

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

Offset

0,3

Links

Table of n, a(n) for n=0..58.

Formula

T(n,k) = n^3 + k^3, n >= 0, 0 <= k <= n.

T(n, k) = A051162(n, k)*(A051162(n, k)^2 + 3* A025581(n, k)^2)/4. See the comment on A051162 for this identity. - Wolfdieter Lang, May 15 2015

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

Example

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

Maple

seq(seq(n^3+k^3, k=0..n), n=0..10); # Robert Israel, May 15 2015

Mathematica

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]

Crossrefs

Cf. A069011. Different from A004999. A257238, A025581, A051162.

Sequence in context: A367984 A046679 A004999 * A230314 A220263 A033492

Adjacent sequences: A105122 A105123 A105124 * A105126 A105127 A105128

Keyword

nonn,easy,tabl

Author

Roger L. Bagula, Apr 09 2005

Status

approved