%I
%S 1,1,1,1,2,1,1,3,2,1,1,4,4,2,0,1,5,7,4,1,0,1,6,11,8,3,0,0,1,7,16,15,7,
%T 1,0,0,1,8,22,26,15,4,0,0,0,1,9,29,42,30,11,1,0,0,0,1,10,37,64,56,26,
%U 5,0,0,0,0,1,11,46,93,98,56,16,1,0,0,0,0,1,12
%N Array T read by rows: T(i,0)=1 for i >= 0; T(i,i)=0 for i=0,1,2,3; T(i,i)=0 for i >= 4; T(i,j)=T(i1,j)+T(i2,j1) for 1<=j<=i1.
%C T(i+j,j)=number of strings (s(1),...,s(i+1)) of nonnegative integers s(k) such that 0<=s(k)s(k1)<=1 for k=2,3,...,i+1 and s(i+1)=j.
%C T(i+j,j)=number of compositions of j consisting of i parts, all of in {0,1}.
%H C. Kimberling, <a href="https://www.fq.math.ca/Scanned/404/kimberling.pdf">Pathcounting and Fibonacci numbers</a>, Fib. Quart. 40 (4) (2002) 328338, Example 1B.
%e Triangle begins:
%e 1;
%e 1,1;
%e 1,2,1;
%e 1,3,2,1;
%e 1,4,4,2,0;
%e ...
%e T(7,4) counts the strings 3334, 3344, 3444, 2234, 2334, 2344, 1234.
%e T(7,4) counts the compositions 001, 010, 100, 011, 101, 110, 111.
%Y Row sums: A000032 (Lucas numbers, 1, 2, 4, 7, 11, 18, ...).
%Y T(2n, n)=A000125(n) (Cake numbers, 1, 2, 4, 8, 15, 26, ...).
%Y T(2n+2, n)=A027660(n).
%K nonn,tabl
%O 0,5
%A _Clark Kimberling_, May 28 2000
