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%I #22 Nov 18 2017 00:23:22
%S 1,0,1,0,0,3,1,4,3,4,7,20,15,8,5,42,102,72,36,15,6,245,540,366,176,70,
%T 24,7,1428,2950,1944,912,355,120,35,8,8379,16524,10668,4920,1890,636,
%U 189,48,9,49588,94430,60021,27336,10405,3492,1050,280,63,10,296010
%N Triangle read by rows: T(n,k) is the number of noncrossing trees with n edges in which the maximal number of contiguous border edges starting from the root in both directions is equal to k.
%C Row sums yield the ternary numbers (A001764).
%C T(n,0) = A102594(n).
%H Andrew Howroyd, <a href="/A102595/b102595.txt">Table of n, a(n) for n = 0..1274</a>
%H P. Flajolet and M. Noy, <a href="http://dx.doi.org/10.1016/S0012-365X(98)00372-0">Analytic combinatorics of non-crossing configurations</a>, Discrete Math., 204, 203-229, 1999.
%H M. Noy, <a href="http://dx.doi.org/10.1016/S0012-365X(97)00121-0">Enumeration of noncrossing trees on a circle</a>, Discrete Math., 180, 301-313, 1998.
%F G.f.: G(t, z)=(g+zg-tz-2zg^2+t^2*(1-t)z^3*g^2-2t(1-t)z^2*g)/(1-tzg)^2, where g=1+zg^3 is the g.f. for the ternary numbers (A001764).
%e T(2,0)=T(2,1)=0, T(2,2)=3 because in all the noncrossing trees _\, /\ and /_, the maximal number of contiguous border edges starting from the root in both directions is equal to 2.
%e Triangle starts:
%e 1;
%e 0, 1;
%e 0, 0, 3;
%e 1, 4, 3, 4;
%e 7, 20, 15, 8, 5;
%e 42, 102, 72, 36, 15, 6;
%e ...
%p G:=(g+z*g-t*z-2*z*g^2+t^2*(1-t)*z^3*g^2-2*t*(1-t)*z^2*g)/(1-t*z*g)^2: z:=w^2: b:=w*sqrt(3): g:=2*sin(arcsin(3*b/2)/3)/b: Gser:=simplify(series(G,w=0,24)): P[0]:=1: for n from 1 to 10 do P[n]:=sort(coeff(Gser,w^(2*n))) od: for n from 0 to 10 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields sequence in triangular form
%t max = 20; z = w^2; b = w*Sqrt[3]; g = 2*(Sin[ ArcSin[3*(b/2)]/3]/b); gf = (g + z*g - t*z - 2*z*g^2 + t^2*(1 - t)*z^3*g^2 - 2*t*(1 - t)*z^2*g)/(1 - t*z*g)^2; se = Series[gf, {w, 0, max}]; Flatten[ Rest /@ DeleteCases[ (CoefficientList[t*#1, t] & ) /@ CoefficientList[se, w], {}]] (* _Jean-François Alcover_, Oct 05 2011, after Maple *)
%o (PARI)
%o S(n)={my(g=1+serreverse(x/(1+x)^3 + O(x*x^n))); Vec((g + x*g - y*x - 2*x*g^2 + y^2*(1-y)*x^3*g^2 - 2*y*(1-y)*x^2*g)/(1 - y*x*g)^2)}
%o my(v=S(10)); for(n=1, #v, my(p=v[n]); for(k=0, n-1, print1(polcoeff(p, k), ", ")); print); \\ _Andrew Howroyd_, Nov 17 2017
%Y Cf. A001764, A102594.
%K nonn,tabl
%O 0,6
%A _Emeric Deutsch_, Jan 22 2005
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