A361226 Square array T(n,k) = k*((1+2*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.
0, 0, 0, 0, 1, 1, 0, 2, 5, 3, 0, 3, 9, 12, 6, 0, 4, 13, 21, 22, 10, 0, 5, 17, 30, 38, 35, 15, 0, 6, 21, 39, 54, 60, 51, 21, 0, 7, 25, 48, 70, 85, 87, 70, 28, 0, 8, 29, 57, 86, 110, 123, 119, 92, 36, 0, 9, 33, 66, 102, 135, 159, 168, 156, 117, 45
Offset: 0
Examples
The rows are 0, 0, 1, 3, 6, 10, 15, 21, ... = A161680 0, 1, 5, 12, 22, 35, 51, 70, ... = A000326 0, 2, 9, 21, 38, 60, 87, 119, ... = A005476 0, 3, 13, 30, 54, 85, 123, 168, ... = A022264 0, 4, 17, 39, 70, 110, 159, 217, ... = A022266 ... . Columns: A000004, A001477, A016813, A017197=3*A016777, 2*A017101, 5*A016873, 3*A017581, 7*A017017, ... (coefficients from A026741). Difference between two consecutive rows: A000290. Hence A143844. This square array read by antidiagonals leads to the triangle 0 0 0 0 1 1 0 2 5 3 0 3 9 12 6 0 4 13 21 22 10 0 5 17 30 38 35 15 ... .
Crossrefs
Programs
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Mathematica
T[n_, k_] := k*((2*n + 1)*k - 1)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 05 2023 *) -
PARI
a(n) = { my(row = (sqrtint(8*n+1)-1)\2, column = n - binomial(row + 1, 2)); binomial(column, 2) + column^2 * (row - column) } \\ David A. Corneth, Mar 05 2023 -
Python
# Seen as a triangle: from functools import cache @cache def Trow(n: int) -> list[int]: if n == 0: return [0] r = Trow(n - 1) return [r[k] + k * k if k < n else r[n - 1] + n - 1 for k in range(n + 1)] for n in range(7): print(Trow(n)) # Peter Luschny, Mar 05 2023
Formula
Take successively sequences n*(n-1)/2, n*(3*n-1)/2, n*(5*n-1)/2, ... listed in the EXAMPLE section.
G.f.: y*(x + y)/((1 - y)^3*(1 - x)^2). - Stefano Spezia, Mar 06 2023
E.g.f.: exp(x+y)*y*(2*x + y + 2*x*y)/2. - Stefano Spezia, Feb 21 2024
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