A125166 Triangle R(n,k), companion to A125165, left column n^3.

1, 8, 1, 27, 9, 1, 64, 36, 10, 1, 125, 100, 46, 11, 1, 216, 225, 146, 57, 12, 1, 343, 441, 371, 203, 69, 13, 1, 512, 784, 812, 574, 272, 82, 14, 1, 729, 1296, 1596, 1386, 846, 354, 96, 15, 1, 1000, 2025, 2892, 2982, 2232, 1200, 450, 111, 16, 1

Offset: 0

Author

Gary W. Adamson, Nov 22 2006

Keywords

• nonn
• easy
• tabl

Comments

Riordan array ((1 + 4*x + x^2)/(1 - x)^4, x/(1 - x)). - Philippe Deléham, Dec 09 2013

Examples

			With other offset (k >= 1) from first formula: R(5,3) = 146 = R(4,3) + R(4,2) = 46 + 100.
The Riordan triangle R begins:
  n\k|     0    1    2    3    4    5   6   7   8  9
  --------------------------------------------------
   0 |     1
   1 |     8    1
   2 |    27    9    1
   3 |    64   36   10    1
   4 |   125  100   46   11    1
   5 |   216  225  146   57   12    1
   6 |   343  441  371  203   69   13   1
   7 |   512  784  812  574  272   82  14   1
   8 |   729 1296 1596 1386  846  354  96  15   1
   9 |  1000 2025 2892 2982 2232 1200 450 111  16  1
... refomatted and extended by _Wolfdieter Lang_, Mar 25 2025.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Crossrefs

Cf. A000578 (column 0), A000537 (column 1), A024166 (column 2), A101094 (column 3).

Cf. A257448 (row sums).

Cf. A008292, A028246, A097805, A125165, A050488, A382057.

Programs

Mathematica

A125166[n_, k_] := A125166[n, k] = Switch[k, 0, (n + 1)^3, n, 1, _, A125166[n - 1, k - 1] + A125166[n - 1, k]];
Table[A125166[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Apr 08 2025 *)

SageMath

y = polygen(QQ, 'y')
x = y.parent()[['x']].gen()
gf = (1 + 4*x + x^2)/((x - 1)^3*(y*x + x - 1))
[list(u) for u in list(gf.O(11))]  # Peter Luschny, Apr 02 2025

Formula

Binomial transform of an infinite matrix M with diagonal D, subdiagonal (D-1)..., etc; as follows: (D) = (1,1,1...); (D-1) = (7,7,7...); (D-2) = (12,12,12...); (D-3) = (6,6,6...). Alternatively, given left border n^3: (1, 8, 27, 64...); for k>1, R(n,k) = R(n-1,k) + R(n-1,k-1).

From Wolfdieter Lang, Mar 27 2025: (Start)

Riordan triangle (see a comment above):

R(n, 0) = (n+1)^3, R(n, k) = R(n-1, k-1) + R(n-1, k), for k >= 1. (from the (finite) A-sequence {1, 1} with offset 0),

R(n, 0) = Sum_{k=0..n-1} Z(j)*R(n-1, k), for n >= 1, and R(0, 0) = 1, with the Riordan Z-sequence A382057. For Riordan A- and Z-sequences see the first W. Lang link in A006232.

R(n, k) = Sum{j=0..n} (j+1)^3*A097805(n-j, k) (convolution property).

R(n, k) = Sum{j=0..3} A028246(4, j+1)*binomial(n, k+j). (Proof for k=0 with a standard (n+1)^3 formula, and for k >= 1 using the Pascal type recurrence for the triangle.)

O.g.f. column k (with leading 0s): ((1 + 4*x + x^2)/(1 - x)^4)*(x/(1-x))^k. (Numerator polynomial from row 3 of the triangle A008292.)

O.g.f. row polynomials P(n, y) = Sum_{k=0..n} R(n, k)*y^k:

G(y, x) = (1 + 4*x + x^2)/((1 - x)^3*(1 - (1+y)*x)). (End)

Extensions

a(27) corrected by Georg Fischer, Feb 18 2020