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
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
Programs
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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 *)
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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
Comments