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A361226
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Square array T(n,k) = k*((1+2*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.
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1
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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
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OFFSET
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0,8
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COMMENTS
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The first upper diagonal is A160378(n+1).
The antidiagonals sums are A034827(n+2).
b(n) = (A034827(n+3) = 0, 2, 10, 30, 70, ...) - (A002414(n) = 0, 1, 9, 30, 70, ...) = 0, 1, 1, 0, 0, 5, 21, 56, ... .
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LINKS
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FORMULA
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Take successively sequences n*(n-1)/2, n*(3*n-1)/2, n*(5*n-1)/2, ... listed in the EXAMPLE section.
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EXAMPLE
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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
... .
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
... .
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MATHEMATICA
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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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PROG
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(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)]
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CROSSREFS
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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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KEYWORD
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AUTHOR
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STATUS
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approved
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