A103525 - OEIS

%I #9 Jun 14 2019 00:28:01

%S 1,1,1,1,2,1,3,3,1,7,7,2,16,25,17,6,1,65,123,94,34,5,321,923,1263,

%T 1076,626,254,70,12,1,4546,16913,28612,28620,18476,7876,2166,352,26,

%U 107587,609479,1691387,3050910,4001833,4044516,3255042,2126032,1138124,500806

%N Triangle read by rows: T(n,k) is the coefficient of t^k (k>=0) in the polynomial P[n,t] defined by P[1,t] = P[2,t] = 1, P[3,t] = 1+t, P[n,t] = P[n-1,t] + P^2[n-2,1] for n >= 4.

%C T(n,k) is the number of certain types of trees (see the Duke et al. reference) of height n and having k branch nodes at level n-1. Row n has 2^(ceiling(n/2)-2)+1 terms (n >= 3). Row sums yield A000278. T(n,0) = A000278(n-1) for n >= 2.

%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(1,0)=1; T(2,0)=1; T(3,0)=T(3,1)=1; T(n,k)=0 for k >= ceiling(n/2); T(n,k) = T(n-1, k) + Sum_{j=0..k} T(n-2, j)*T(n-2, k-j) for n >= 4.

%e P[5,t] = 3 + 3*t + t^2; therefore T(3,0)=3, T(3,1)=3, T(3,2)=1.

%e Triangle begins:

%e    1;

%e    1;

%e    1,  1;

%e    2,  1;

%e    3,  3,  1;

%e    7,  7,  2;

%e   16, 25, 17,  6,  1;

%p P[1]:=1:P[2]:=1:P[3]:=1+t:for n from 4 to 13 do P[n]:=sort(expand(P[n-1]+P[n-2]^2)) od:for n from 1 to 11 do seq(coeff(t*P[n],t^k),k=1..2^(ceil(n/2)-2)+1) od;# yields sequence in triangular form

%Y Cf. A000278.

%K nonn,tabf

%O 1,5

%A _Emeric Deutsch_, Mar 21 2005