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A131638 Increasing binary trees having exactly two vertices with outdegree 1.

%I A131638 #35 Aug 05 2023 23:39:11
%S A131638 1,11,180,4288,141584,6213288,350400832,24718075136,2133652515072,
%T A131638 221311262045440,27166907582280704,3895974311462313984,
%U A131638 645512064907811491840,122381396964887716078592,26325690425815766552887296,6377608610246241663568248832
%N A131638 Increasing binary trees having exactly two vertices with outdegree 1.
%H A131638 M. P. Develin and S. P. Sullivant, <a href="http://dx.doi.org/10.1007/s00026-003-0196-9">Markov Bases of Binary Graph Models</a>, Annals of Combinatorics 7 (2003) 441-466.
%H A131638 Christiane Poupard, <a href="http://dx.doi.org/10.1016/S0195-6698(89)80009-5">Deux propriétés des arbres binaires ordonnés stricts</a>, Europ. J. Combin., vol. 10, 1989, pp. 369-374.
%F A131638 E.g.f.: (3*sec(x/sqrt(2))^2*tan(x/sqrt(2))^2-x*sec(x/sqrt(2))^2*tan(x/sqrt(2))/(sqrt(2)))/2. - _Michel Marcus_, Mar 03 2013
%F A131638 a(n) ~ (2*n)! * 2^(n+6)*n^3/Pi^(2*n+4). - _Vaclav Kotesovec_, Sep 25 2013
%F A131638 From _Klaus K Haverkamp_, Jul 02 2023: (Start)
%F A131638 a(n) = (A002105(n+2) - (n+1)*A002105(n+1))/2.
%F A131638 a(n) = A094503(2n+1,n). (End)
%t A131638 Table[n!*SeriesCoefficient[1/2*(-((x*Sec[x/Sqrt[2]]^2 *Tan[x/Sqrt[2]]) /Sqrt[2]) + 3*Sec[x/Sqrt[2]]^2 *Tan[x/Sqrt[2]]^2), {x, 0, n}], {n, 2, 40, 2}] (* _Vaclav Kotesovec_ after _Michel Marcus_, Sep 25 2013 *)
%o A131638 (PARI) lista(m) = { default(realprecision, 30); x = y + O(y^m); egf = (3*tan(x/sqrt(2))^2/cos(x/sqrt(2))^2-x*tan(x/sqrt(2))/(sqrt(2)*cos(x/sqrt(2))^2))/2; forstep (n=2, m, 2, print1(round(n!*polcoeff(egf, n, y)), ", "));}  \\ _Michel Marcus_, Mar 03 2013
%Y A131638 Cf. A002105, A094503.
%K A131638 nonn
%O A131638 1,2
%A A131638 _Wenjin Woan_, Oct 03 2007
%E A131638 More terms from _Michel Marcus_, Mar 03 2013