%I
%S 1,2,8,36,160,692,2928,12200,50304,205940,838928,3405496,13788736,
%T 55723592,224863712,906365136,3649978880,14687731572,59067989072,
%U 237424661016,953914608320,3831159414552,15381896102432,61739966366256,247750559632640,993955865320392,3986890331450528
%N Number of nedge ordered trees with bicolored boundary edges.
%H G. C. Greubel, <a href="/A228197/b228197.txt">Table of n, a(n) for n = 0..1000</a>
%H D. E. Davenport, L. K. Pudwell, L. W. Shapiro, L. C. Woodson, <a href="http://faculty.valpo.edu/lpudwell/papers/treeboundary.pdf">The Boundary of Ordered Trees</a>, 2014.
%H Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, Leon C. Woodson, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Davenport/dav3.html">The Boundary of Ordered Trees</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.8.
%H E. Deutsch, L. W. Shapiro, <a href="http://dx.doi.org/10.1016/S0012365X(02)003412">A bijection between ordered trees and 2Motzkin paths and its many consequences</a>, Disc. Math. 256 (2002) 655670.
%F G.f.: (1+4*x^2*B^2*C)/(12*x), C is the Catalan g.f. (see A000108) and B =(14*x)^(1/2) is the g.f. for the central binomial coefficients (A000984).
%F a(n) ~ 4^n * (11/(sqrt(Pi*n))).  _Vaclav Kotesovec_, Aug 23 2013
%F Conjecture: (n+1)*a(n) +2*(5*n8)*a(n1) +4*(8*n+17)*a(n2) +16*(2*n5)*a(n3)=0.  _R. J. Mathar_, Aug 25 2013
%F a(n) = 2^(2*n)2^n*JacobiP(n1,1/2,n,3) = 2^(2*n)2*A082590(n1), which satisfies the above conjecture.  _Benedict W. J. Irwin_, Sep 16 2016
%e When n=3 edges there are A000108(3)= 5 ordered trees. Four of these consist of three boundary edges each contributing 2^3 trees to the count. The last, UDUDUD, has two boundary edges giving the last 2^2 trees for a total of 36.
%t CoefficientList[Series[(12*x2*x*Sqrt[14*x])/((4*x1)*(2*x1)), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Aug 23 2013 *)
%t Table[2^(2n)2^n*JacobiP[n1,1/2,n,3],{n,0,20}] (* _Benedict W. J. Irwin_, Sep 16 2016 *)
%o (PARI)
%o x = 'x + O('x^66);
%o C = serreverse( x/( 1/(1x) ) ) / x; \\ Catalan A000108
%o B = (14*x)^(1/2); \\ central binomial coefficients
%o gf = (1+4*x^2*B^2*C)/(12*x);
%o Vec(gf) \\ _Joerg Arndt_, Aug 21 2013
%Y Cf. A000108, A000984, A228178, A228180.
%K nonn
%O 0,2
%A _Louis Shapiro_, Aug 20 2013
