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%I #19 Dec 24 2017 09:49:23
%S 1,4,20,104,553,2984,16272,89440,494681,2749772,15348372,85967112,
%T 482927985,2719787856,15351385152,86816721792,491819758417,
%U 2790451952660,15854070902964,90187514559208,513619224125657,2928073006131704,16708228671139600,95423104768226144,545408567460801513
%N Total number of triangles in all the dissections of a convex (n+3)-gon by nonintersecting diagonals.
%H Vincenzo Librandi, <a href="/A089382/b089382.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: (3-z+q)*(1+z-q)^2/(64*q*z^2), where q = sqrt(1-6*z+z^2).
%F Recurrence: n*(n+2)*a(n) = (8*n^2 + 7*n - 3)*a(n-1) - (13*n^2 - 17*n - 6)*a(n-2) + 2*(n-3)*(n+1)*a(n-3). - _Vaclav Kotesovec_, Oct 14 2012
%F a(n) ~ (1 + sqrt(2))^(2*n+3) / (2^(11/4) * sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 14 2012, simplified Dec 24 2017
%F a(n) = Sum_{j=0..n}((-1)^j*2^(n-j)*binomial(n+3,j)*binomial(2*n-j+2,n+2)). - _Vladimir Kruchinin_, Apr 08 2016
%e a(1)=4 because in the three dissections of a square we have altogether four triangles: no triangle in the "no-diagonals" dissection and two triangles in each of the dissections by one of the two diagonals of the square.
%t Table[SeriesCoefficient[(3-x+Sqrt[1-6*x+x^2])*(1+x-Sqrt[1-6*x+x^2])^2/(64*Sqrt[1-6*x+x^2]*x^2),{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 14 2012 *)
%o (PARI) x='x+O('x^66); q = sqrt(1-6*x+x^2); Vec((3-x+q)*(1+x-q)^2/(64*q*x^2)) \\ _Joerg Arndt_, May 10 2013
%o (Maxima)
%o a(n):=sum((-1)^j*2^(n-j)*binomial(n+3,j)*binomial(2*n-j+2,n+2),j,0,n); /* _Vladimir Kruchinin_, Apr 08 2016 */
%Y Cf. A001003.
%K nonn
%O 0,2
%A _Emeric Deutsch_, Dec 28 2003
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