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A118923
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Triangle T(n,k) built by placing T(n,0)=A000012(n) in the left edge, T(n,n)=A079978(n) on the right edge and filling the body with the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).
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3
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1, 1, 0, 1, 1, 0, 1, 2, 1, 1, 1, 3, 3, 2, 0, 1, 4, 6, 5, 2, 0, 1, 5, 10, 11, 7, 2, 1, 1, 6, 15, 21, 18, 9, 3, 0, 1, 7, 21, 36, 39, 27, 12, 3, 0, 1, 8, 28, 57, 75, 66, 39, 15, 3, 1, 1, 9, 36, 85, 132, 141, 105, 54, 18, 4, 0, 1, 10, 45, 121, 217, 273, 246, 159, 72, 22, 4, 0, 1, 11, 55, 166
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OFFSET
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0,8
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COMMENTS
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The fourth diagonal is 1, 2, 5, 11, 21, ..., which is 1 + A000292. The fifth diagonal is 0, 2, 7, 18, 39, 75, 132, 217, 338, 504, 725, 1012, ..., which is A051743.
T(n,n-k) is the (n,k)-th entry of the (1/(1-x^3), x/(1-x)) Riordan array.
Sums of antidiagonals give A079962. (End)
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LINKS
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FORMULA
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For 0 <= k < n, T(n,k) = (n-k)*Sum_{j=0..floor(k/3)} binomial(n-3*j,n-k)/(n-3*j).
G.f.: 1/((1+x*y+(x*y)^2)*(1-x-x*y)). (End)
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EXAMPLE
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The table begins
1
1 0
1 1 0
1 2 1 1
1 3 3 2 0
1 4 6 5 2 0
1 5 10 11 7 2 1
1 6 15 21 18 9 3 0
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MAPLE
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MATHEMATICA
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Flatten@Table[CoefficientList[Series[1/((1 + x*y + x^2*y^2)(1 - x - x*y)), {x, 0, 23}, {y, 0, 11}], {x, y}][[n + 1, k + 1]], {n, 0, 11}, {k, 0, n}] (* Michael A. Allen, Nov 30 2021 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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