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A360962 Square array T(n,k) = k*((3+6*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards. 2
0, 0, 1, 0, 4, 5, 0, 7, 17, 12, 0, 10, 29, 39, 22, 0, 13, 41, 66, 70, 35, 0, 16, 53, 93, 118, 110, 51, 0, 19, 65, 120, 166, 185, 159, 70, 0, 22, 77, 147, 214, 260, 267, 217, 92, 0, 25, 89, 174, 262, 335, 375, 364, 284, 117, 0, 28, 101, 201, 310, 410, 483, 511, 476, 360, 145 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,5
COMMENTS
The main diagonal is A024394.
The antidiagonals sums are A000537.
LINKS
FORMULA
Take successively sequences n*(3*n-1)/2, n*(9*n-1)/2, n*(15*n-1)/2, n*(21*n-1)/2, ... listed in the EXAMPLE section.
From Stefano Spezia, Feb 21 2024: (Start)
G.f.: y*(1 + 2*y + x*(2 + y))/((1 - x)^2*(1 - y)^3).
E.g.f.: exp(x+y)*y*(2 + 3*y + 6*x*(1 + y))/2. (End)
EXAMPLE
The rows are:
0 1 5 12 22 35 51 70 ... = A000326
0 4 17 39 70 110 159 217 ... = A022266
0 7 29 66 118 185 267 364 ... = A022272
0 10 41 93 166 260 375 511 ... = A022278
0 13 53 120 214 335 483 658 ... = A022284
... .
Columns: A000004, A016777, A017581, A154266=3*A017209, 2*A348845, 5*A161447, 3*A158057(n+1), ... (coefficients from A026741).
Difference between two consecutive rows are: A033428.
This square array read by antidiagonals leads to the triangle
0
0 1
0 4 5
0 7 17 12
0 10 29 39 22
0 13 41 66 70 35
0 16 53 93 118 110 51
... .
MAPLE
T:= (n, k)-> k*(k*(3+6*n)-1)/2:
seq(seq(T(d-k, k), k=0..d), d=0..10); # Alois P. Heinz, Feb 28 2023
MATHEMATICA
T[n_, k_] := ((6*n + 3)*k - 1)*k/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Feb 27 2023 *)
PROG
(PARI) T(n, k) = k*((3+6*n)*k-1)/2; \\ Michel Marcus, Feb 27 2023
CROSSREFS
Cf. A033428.
Sequence in context: A318740 A240160 A249860 * A320162 A354068 A200013
KEYWORD
nonn,tabl
AUTHOR
Paul Curtz, Feb 27 2023
STATUS
approved

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Last modified July 25 18:29 EDT 2024. Contains 374612 sequences. (Running on oeis4.)