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%I #11 Dec 28 2016 17:05:20
%S 1,0,1,1,0,1,0,3,0,2,2,0,8,0,4,0,11,0,22,0,9,5,0,45,0,61,0,21,0,41,0,
%T 166,0,171,0,51,14,0,226,0,580,0,483,0,127,0,154,0,1050,0,1962,0,1373,
%U 0,323,42,0,1070,0,4430,0,6496,0,3923,0,835,0,582,0,6005,0,17570,0,21184,0,11257,0,2188
%N Triangle read by rows: T(n,k) is the number of ordered trees with n edges and having k branches of odd length (n>=0, 0<=k<=n).
%C Row n has n+1 terms.
%C Column 0 yields the Catalan numbers (A000108) alternating with 0's.
%C The diagonal entries are the Motzkin numbers (A001006).
%C T(n,n-2) = A025566(n) for n>=2.
%H Emeric Deutsch, <a href="http://dx.doi.org/10.1016/j.disc.2003.10.021">Ordered trees with prescribed root degrees, node degrees and branch lengths</a>, Discrete Math., 282, 2004, 89-94.
%H J. Riordan, <a href="http://dx.doi.org/10.1016/S0097-3165(75)80010-0">Enumeration of plane trees by branches and endpoints</a>, J. Comb. Theory (A) 19, 1975, 214-222.
%F G.f. G = G(t,z) satisfies z(t+z)G^2-(1+tz)G+1+tz=0.
%e T(3,3)=2 because we have (i) a tree with 3 edges hanging from the root and (ii) a tree with one edge hanging from the root, at the end of which 2 edges are hanging.
%e Triangle starts:
%e 1;
%e 0,1;
%e 1,0,1;
%e 0,3,0,2;
%e 2,0,8,0,4;
%p G:=1/2/(z^2+t*z)*(t*z+1-sqrt(1-2*t*z-3*t^2*z^2-4*z^2-4*t*z^3)): Gserz:=simplify(series(G,z=0,14)):P[0]:=1: for n from 1 to 12 do P[n]:=sort(expand(coeff(Gserz,z^n))) od: for n from 0 to 12 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields the sequence in triangular form
%Y Cf. A000108, A001006, A025566, A102004.
%K nonn,tabl
%O 0,8
%A _Emeric Deutsch_, Dec 23 2004
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