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%I #5 May 02 2015 03:58:14
%S 1,0,1,0,0,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,1,2,1,0,0,0,0,0,1,1,1,1,3,3,
%T 3,2,1,0,0,0,0,0,0,1,1,1,1,3,4,5,5,5,7,8,7,6,6,4,1,0,0,0,0,0,0,0,1,1,
%U 1,1,3,4,6,7,8,11,17,21,25,28,31,31,33,32,27,18,10,4,1,0,0,0,0,0,0,0,0,1,1
%N Triangle read by rows: T(n,k) is the coefficient of t^k (k>=1) in the polynomial P[n,t] defined by P[1,t]=P[2,t]=t^2, P[n,t]=tP[n-1,t]+t^2*P^2[n-2,t] (n>=3).
%C T(n,k) is the number of certain types of trees (see the Duke et al. reference) of height n and having k edges. Row n contains A027383(n-1) terms, the first n-1 of which are 0. Row sums yield A000278.
%H W. Duke, Stephen J. Greenfield and Eugene R. Speer, <a href="https://cs.uwaterloo.ca/journals/JIS/green2/qf.html">Properties of a Quadratic Fibonacci Recurrence</a>, J. Integer Sequences, 1998, #98.1.8.
%e P[3,t]=t^3+t^4; therefore T(3,1)=0, T(3,2)=0, T(3,3)=1, T(3,4)=1.
%e Triangle begins:
%e 1;
%e 0,1;
%e 0,0,1,1;
%e 0,0,0,1,1,1;
%e 0,0,0,0,1,1,1,1,2,1;
%e 0,0,0,0,0,1,1,1,1,3,3,3,2,1;
%p P[1]:=t:P[2]:=t^2:for n from 3 to 12 do P[n]:=sort(expand(t*P[n-1]+t^2*P[n-2]^2)) od: d[1]:=1: d[2]:=2: for n from 3 to 20 do d[n]:=2*d[n-2]+2 od: for n from 1 to 9 do seq(coeff(P[n],t^k),k=1..d[n]) od;# yields sequence in triangular form
%Y Cf. A000278, A027383.
%K nonn,tabf
%O 1,22
%A _Emeric Deutsch_, Mar 21 2005
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