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%I #42 Oct 18 2020 03:38:37
%S 1,4,29,229,1847,14974,121430,983476,7952111,64193728,517447289,
%T 4165721377,33500374796,269166095800,2161064409680,17339917293304,
%U 139060729285871,1114752741216196,8933074352513183,71564554425680839
%N Number of triangulations (by Euclidean triangles) having 3+3n vertices of a triangle with each side subdivided by n additional points.
%H Andrei Asinowski, Christian Krattenthaler, Toufik Mansour, <a href="http://arxiv.org/abs/1604.02870">Counting triangulations of some classes of subdivided convex polygons</a>, European Journal of Combinatorics 62 (2017), 92-114; arXiv:1604.02870 [math.CO], 2016. See also <a href="https://web.archive.org/web/20171109075721/http://discretemath.upc.edu/jmda16/wp-content/uploads/2015/09/JMDA2016_paper_15.pdf">preprint</a>, 2016.
%H R. Bacher, <a href="http://arxiv.org/abs/math.CO/0310206">Counting Triangulations of Configurations</a>, arXiv:math/0310206 [math.CO], 2003.
%F A formula is given in the Bacher reference.
%F It seems that a(n) = Sum_{i, j, k>=0} C(n, i+j)*C(n, j+k)*C(n, k+i)). - _Benoit Cloitre_, Oct 25 2004; proved in the article by Asinowski et al.
%F G.f.: seems to be (10*g^3 - 17*g^2 + 7*g - 1)/((1-3*g)*(2*g-1)*(4*g^2 - 6*g+1)) where g*(1-g)^2 = x. - _Mark van Hoeij_, Nov 10 2011; proved in the article by Asinowski et al.
%F Conjecture: 2*n*(2*n-1)*(5*n^2 - 29*n + 30)*a(n) + (-295*n^4 + 1926*n^3 - 3425*n^2 + 2106*n - 360)*a(n-1) + 24*(3*n-4)*(3*n-5)*(5*n^2 - 19*n + 6)*a(n-2) = 0. - _R. J. Mathar_, Apr 23 2015. Proved by Andrei Asinowski, C. Krattenthaler, T. Mansour, Counting triangulations of balanced subdivisions of convex polygons, 2016.
%e a(0)=1 since there is only one triangulation of a triangle (consisting of the triangle itself).
%e The a(1)=4 triangulations of a triangle with each side subdivided by one additional point are given by
%e .
%e O O
%e / \ /|\
%e O _ O O O
%e / \ / \ / \|/ \
%e O _ O _ O , O _ O _ O
%e .
%e and rotations by 120 degrees and 240 degrees of the last triangulation.
%t max = 19; f[x_] := Sum[ a[n]*x^n, {n, 0, max}]; a[0] = 1; g[x_] := Sum[ b[n]*x^n, {n, 0, max}]; b[0] = 0; coes = CoefficientList[ Series[ g[x]*(1 - g[x])^2 - x, {x, 0, max}], x]; solb = Solve[ Thread[ coes == 0]][[1]]; coes = CoefficientList[ Series[ f[x] - ((10*g[x]^3 - 17*g[x]^2 + 7*g[x] - 1)/((1 - 3*g[x])*(2*g[x] - 1)*(4*g[x]^2 - 6*g[x] + 1))), {x, 0, max}], x] /. solb; sola = Solve[ Thread[ coes == 0]][[1]]; Table[a[n] /. sola, {n, 0, max}] (* _Jean-François Alcover_, Dec 06 2011, after _Mark van Hoeij_ *)
%K nonn,nice
%O 0,2
%A _Roland Bacher_, Oct 16 2003
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