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%I #17 Dec 30 2016 02:31:19
%S 1,0,3,9,0,1,27,6,0,3,90,18,27,0,2,297,135,81,24,0,6,1053,648,351,72,
%T 90,0,5,3888,2889,1377,756,270,90,0,15,14742,12150,6723,3888,1485,270,
%U 315,0,14,56619,51273,32805,19224,6480,3645,945,336,0,42,219429,218700
%N Triangle read by rows: number of hex trees with n edges and k branches of length 1 (0<=k<=n).
%C A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a middle child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read reference).
%H F. Harary and R. C. Read, <a href="https://doi.org/10.1017/S0013091500009135">The enumeration of tree-like polyhexes</a>, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.
%F Sum of terms in row n = A002212(n+1).
%F T(n,0) = A126322(n). T(n,n) = A126324(n).
%F T(2n,2n) = Cat(n); T(2n+1,2n+1) = 3*Cat(n), where Cat(n) are the Catalan numbers (A000108).
%F Sum_{k=0..n} k*T(n,k) = A126323(n).
%F G.f.: (1+P)C(z^2*Q^2), where C(z)=(1-sqrt(1-4z))/(2z) is the Catalan function, P=3tz + 9z^2/(1-3z) and Q=t+3z/(1-3z).
%e T(2,2)=1 because among the 10 hex trees with two edges only the tree V has 2 branches of length 1.
%e Triangle starts:
%e 1;
%e 0, 3;
%e 9, 0, 1;
%e 27, 6, 0, 3;
%e 90, 18, 27, 0, 2;
%e 297, 135, 81, 24, 0, 6;
%p C:=z->(1-sqrt(1-4*z))/2/z: G:=(1+3*t*z+9*z^2/(1-3*z))*C(z^2*(t+3*z/(1-3*z))^2): Gser:=simplify(series(G,z=0,15)): for n from 0 to 10 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 10 do seq(coeff(P[n],t,j),j=0..n) od; # yields sequence in triangular form
%t n = 10; c[z_] = (1 - Sqrt[1 - 4*z])/2/z; g[t_, z_] = (1 + 3*t*z + 9*(z^2/(1 - 3*z)))* c[z^2*(t + 3*(z/(1 - 3*z)))^2]; Flatten[CoefficientList[#, t] & /@ CoefficientList[Series[g[t, z], {z, 0, n}], z]] (* _Jean-François Alcover_, Jul 22 2011, after Maple prog. *)
%Y Cf. A000108, A002212, A126322, A126323, A126324.
%K nonn,tabl
%O 0,3
%A _Emeric Deutsch_, Dec 25 2006
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