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A360962
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Square array T(n,k) = k*((3+6*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.
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2
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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
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OFFSET
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0,5
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COMMENTS
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The antidiagonals sums are A000537.
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LINKS
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FORMULA
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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.
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)
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EXAMPLE
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The rows are:
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
... .
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
... .
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MAPLE
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T:= (n, k)-> k*(k*(3+6*n)-1)/2:
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MATHEMATICA
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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 *)
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PROG
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(PARI) T(n, k) = k*((3+6*n)*k-1)/2; \\ Michel Marcus, Feb 27 2023
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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