A103524 - OEIS

%I #6 May 02 2015 03:58:10

%S 1,0,1,0,1,1,0,1,1,1,0,4,1,1,1,0,9,4,1,1,1,0,49,9,4,1,1,1,0,256,49,9,

%T 4,1,1,1,0,4225,256,49,9,4,1,1,1,0,103041,4225,256,49,9,4,1,1,1,0,

%U 20666116,103041,4225,256,49,9,4,1,1,1,0,11574962569,20666116,103041

%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]=t, P[2,t]=t^2, P[n,t]=tP[n-1,t]+t^2*P^2[n-2,1].

%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 from the root to the first branch node (k edges if there are no branch nodes). 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.

%F T(n, k)=0 for k>n; T(n, n)=1; T(n, 1)=0 for n>=2; T(n, k)=A000278(n-k)^2 for 2<=k<=n-1.

%e P[3,t]=t^2+t^3; therefore T(3,1)=0, T(3,2)=1, T(3,3)=1.

%e Triangle begins:

%e 1;

%e 0,1;

%e 0,1,1;

%e 0,1,1,1;

%e 0,4,1,1,1;

%e 0,9,4,1,1,1;

%e 0,49,9,4,1,1,1;

%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*subs(t=1,P[n-2])^2)) od: for n from 1 to 12 do seq(coeff(P[n],t^k),k=1..n) od;# yields sequence in triangular form

%Y Cf. A000278.

%K nonn,tabl

%O 1,12

%A _Emeric Deutsch_, Mar 21 2005