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A090985
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Triangle read by rows: T(n,k) = 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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 2,6
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COMMENTS
| T(n,n-2)=[binom(2n-4,n-2)]/(n-1)=Catalan(n-2) (A000108); T(n,n-4)=binom(2n-5,n-4) (A002054); T(n,n-5)=binom(2n-6,n-5) (A002694); T(n,0)=A046736(n); Row sums give the little Schroeder numbers (A001003).
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REFERENCES
| P. Flajolet and M. Noy, Analytic combinatorics of non-crossing configurations, Discrete Math., 204, 1999, 203-229.
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FORMULA
| 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). 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
| 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.
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
| 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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CROSSREFS
| Cf. A000108, A002054, A002694, A046736.
Sequence in context: A104505 A175958 A021469 * A011131 A058241 A021827
Adjacent sequences: A090982 A090983 A090984 * A090986 A090987 A090988
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 28 2004
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