A360665
Square array T(n, k) = k*((2*n-1)*k+1)/2 read by rising antidiagonals.
Original entry on oeis.org
0, 0, 0, 0, 1, -1, 0, 2, 3, -3, 0, 3, 7, 6, -6, 0, 4, 11, 15, 10, -10, 0, 5, 15, 24, 26, 15, -15, 0, 6, 19, 33, 42, 40, 21, -21, 0, 7, 23, 42, 58, 65, 57, 28, -28, 0, 8, 27, 51, 74, 90, 93, 77, 36, -36, 0, 9, 31, 60, 90, 115, 129, 126, 100, 45, -45
Offset: 0
By rows:
0, 0, -1, -3, -6, -10, -15, -21, -28, ... = -A161680
0, 1, 3, 6, 10, 15, 21, 28, 36, ... = A000217
0, 2, 7, 15, 26, 40, 57, 77, 100, ... = A005449
0, 3, 11, 24, 42, 65, 93, 126, 164, ... = A005475
0, 4, 15, 33, 58, 90, 129, 175, 228, ... = A022265
0, 5, 19, 42, 74, 115, 165, 224, 292, ... = A022267
0, 6, 23, 51, 90, 140, 201, 273, 356, ... = A022269
... .
Cf. Antidiagonal sums:
A034827(n+1).
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T[n_, k_] := ((2*n - 1)*k^2 + k)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 31 2023 *)
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T(n, k) = ((2*n-1)*k^2+k)/2 \\ Thomas Scheuerle, Mar 17 2023
A361226
Square array T(n,k) = k*((1+2*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.
Original entry on oeis.org
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
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
... .
Cf.
A000004,
A000290,
A000326,
A001477,
A002414,
A005476,
A016777,
A016813,
A016873,
A017017,
A017101,
A017197,
A017581,
A022264,
A022266,
A026741,
A034827,
A160378,
A161680,
A360962.
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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 *)
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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
-
# 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
Showing 1-2 of 2 results.
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