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A121873
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Number of non-crossing plants in the (n+1)-sided regular polygon (contains non-crossing trees).
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0
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1, 3, 14, 80, 510, 3479, 24848, 183465, 1389090, 10726452, 84150858, 668825768, 5373971036, 43580383095, 356234802952, 2932097981824, 24279982680870, 202134854855973, 1690839212784240, 14204198452365180, 119784707913644598, 1013675671656956976
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OFFSET
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1,2
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LINKS
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FORMULA
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x = (y - y^2 - y^3)/(1+y)^2 is the inverse of the generating function.
a(n) = sum(j=0..n-1, (sum(i=0..n-j-1, binomial(i+n-1,n-1) *binomial(i+n,n-j-i-1))) *binomial(n,j))/n, n>0, a(0)=0. - Vladimir Kruchinin, Oct 12 2011
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EXAMPLE
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a(2)=3 because the non-crossing plants in a triangle are the three non-crossing trees, made of two sides.
G.f. = x + 3*x^2 + 14*x^3 + 80*x^4 + 510*x^5 + 3479*x^6 + 25848*x^7 + ...
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MATHEMATICA
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a[n_] := If[n==0, 0, Sum[Sum[Binomial[i+n-1, n-1]*Binomial[i+n, n-j-i-1], {i, 0, n-j-1}]*Binomial[n, j], {j, 0, n-1}]/n]; Table[a[n], {n, 1, 16}] (* Jean-François Alcover, Feb 20 2017, after Vladimir Kruchinin *)
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PROG
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(Maxima)
a(n):=if n=0 then 0 else sum((sum(binomial(i+n-1, n-1)*binomial(i+n, n-j-i-1), i, 0, n-j-1))*binomial(n, j), j, 0, n-1)/n; /* Vladimir Kruchinin, Oct 12 2011 */
(PARI) {a(n) = if( n<1, 0, polcoeff( serreverse( (x - x^2 + x^3) / (1 + x)^2 + x * O(x^n)), n))}; /* Michael Somos, Dec 31 2014 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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