%I
%S 1,1,2,9,70,795,11961,224504,5051866,132523155,3969912160,
%T 133678842902,4997280555576,205320100093953,9195224163850830,
%U 445775353262707365,23255990676521697670,1299028117862237432959,77348967890083608924045
%N Column 1 of triangular matrix T=A101479, in which row n equals row (n1) of T^(n1) followed by '1'.
%C Number of Dyck paths whose ascent lengths are exactly {n, n1, .. 1}, for example the a(2) = 2 paths are uududd and uuddud.  _David Scambler_, May 30 2012
%F Equals column 0 of array A136730.
%e This sequence can also be generated in the following manner.
%e Start a table with the all 1's sequence in row 0; from then on, row n+1 can be formed from row n by dropping the initial n terms of row n and taking partial sums of the remaining terms to obtain row n+1.
%e The following table (A136730) illustrates this method:
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
%e [1], 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...;
%e [2, 5], 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, ...;
%e [9, 23, 43], 70, 105, 149, 203, 268, 345, 435, 539, 658, ...;
%e [70, 175, 324, 527], 795, 1140, 1575, 2114, 2772, 3565, ...;
%e [795, 1935, 3510, 5624, 8396], 11961, 16471, 22096, 29025, ...;
%e [11961, 28432, 50528, 79553, 117020, 164672], 224504, ...; ...
%e In the above table, drop the initial n terms in row n (enclosed in square brackets) and then take partial sums to obtain row n+1 for n>=0;
%e this sequence then forms the first column of the resultant table.
%e Note: column k of the above table equals column 1 of matrix power T^(k+1) where T=A101479, for k>=0.
%o (PARI) {a(n)=local(A=Mat(1),B);for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=(A^(i1))[i1,j]);));A=B);return(A[n+1,1])}
%Y Cf. A101479, A101481, A101483, A136730.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 21 2005
