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A090985
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Triangle read by rows: T(n,k) is the number of dissections of a convex n-gon by nonintersecting diagonals, having exactly k triangles (n >= 2, k >= 0).
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0
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1, 0, 1, 1, 0, 2, 1, 5, 0, 5, 4, 6, 21, 0, 14, 8, 35, 28, 84, 0, 42, 25, 80, 216, 120, 330, 0, 132, 64, 309, 540, 1155, 495, 1287, 0, 429, 191, 890, 2475, 3080, 5720, 2002, 5005, 0, 1430, 540, 3058, 7788, 16302, 16016, 27027, 8008, 19448, 0, 4862, 1616, 9580, 30108, 54964, 96005, 78624, 123760, 31824, 75582, 0, 16796
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OFFSET
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2,6
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COMMENTS
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T(n,n-2) = [binomial(2n-4, n-2)]/(n-1) = Catalan(n-2) (A000108).
T(n,n-4) = binomial(2n-5, n-4) (A002054).
T(n,n-5) = binomial(2n-6, n-5) (A002694).
Row sums give the little Schroeder numbers (A001003).
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LINKS
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FORMULA
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T(n, k) = binomial(n+k-2, k)*(Sum_{i=0..floor((n-2-k)/2)} binomial(n-2+k+i, i)*binomial(n-3-k-i, i-1))/(n-1).
G.f.: G=G(t, z) satisfies (1-t)G^3 + (1+t)zG^2 - z^2*(1+z)G + z^4 = 0.
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EXAMPLE
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T(5,1)=5 because the dissections of a convex pentagon having exactly one triangle are obtained by the placement of a diagonal between any pair of non-adjacent vertices.
T(6,0)=4 because the dissections of a convex hexagon with no triangles are obtained by the null placement and by placing one diagonal between any of the 3 pairs of opposite vertices.
Triangle starts:
1;
0, 1;
1, 0, 2;
1, 5, 0, 5;
4, 6, 21, 0, 14;
8, 35, 28, 84, 0, 42;
...
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MAPLE
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T := (n, k)->binomial(n+k-2, k)*sum(binomial(n-2+k+i, i)*binomial(n-3-k-i, i-1), i=0..floor((n-2-k)/2))/(n-1): seq(seq(T(n, k), k=0..n-2), n=2..14);
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MATHEMATICA
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T [n_, k_] := Binomial[n+k-2, k] Sum[Binomial[n-2+k+i, i] Binomial[n-3-k-i, i-1], {i, 0, (n-2-k)/2}]/(n-1);
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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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