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A046736
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Number of ways to place non-intersecting diagonals in convex n-gon so as to create no triangles.
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14
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1, 0, 1, 1, 4, 8, 25, 64, 191, 540, 1616, 4785, 14512, 44084, 135545, 418609, 1302340, 4070124, 12785859, 40325828, 127689288, 405689020, 1293060464, 4133173256, 13246527139, 42557271268, 137032656700, 442158893833, 1429468244788
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OFFSET
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2,5
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LINKS
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FORMULA
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G.f.: A(x)=sum_{n>0} a(n)x^(n-1) satisfies A(x)-A(x)^2-A(x)^3 = x*(1-A(x)).
Let g=(1-x)/(1-x-x^2); then a(m) = coeff. of x^(m-2) in g^(m-1)/(m-1).
D-finite with recurrence: 5*(n-1)*n*(37*n-95)*a(n) = 4*(n-1)*(74*n^2-301*n+300)*a(n-1) + 8*(2*n-5)*(74*n^2-301*n+297)*a(n-2) - 2*(n-3)*(2*n-7)*(37*n-58)*a(n-3). - Vaclav Kotesovec, Aug 10 2013
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EXAMPLE
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a(4)=a(5)=1 because of null placement; a(6)=4 because in addition to not placing any, we might also place one between any of the 3 pairs of opposite vertices.
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MAPLE
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a := n->1/(n-1)*sum(binomial(n+k-2, k)*binomial(n-k-3, k-1), k=0..floor(n/2-1)); seq(a(i), i=2..30);
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MATHEMATICA
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a[2]=1; a[n_] := Sum[Binomial[n+k-2, k]*Binomial[n-k-3, k-1], {k, 0, Floor[n/2]-1}]/(n-1);
(* 2nd program: *)
x*InverseSeries[Series[(y-y^2-y^3)/(1-y), {y, 0, 29}], x]
(* 3rd program: *)
a[2]=1; a[3]=0; a[n_] := HypergeometricPFQ[{2-n/2, 5/2-n/2, n}, {2, 4-n}, -4]; Table[a[n], {n, 2, 30}] (* Jean-François Alcover, Apr 14 2017 *)
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PROG
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(PARI) a(n)=if(n<2, 0, polcoeff(serreverse((x-x^2-x^3)/(1-x)+x*O(x^n)), n-1))
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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