A360962 Square array T(n,k) = k*((3+6*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.
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
Offset: 0
Examples
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 ... .
Crossrefs
Programs
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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
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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 *) -
PARI
T(n,k) = k*((3+6*n)*k-1)/2; \\ Michel Marcus, Feb 27 2023
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)
Comments