
FORMULA

a(0) = a(1) = 1; a(n) = n * F(n1), where F(0) = F(1) = 1,
F(n) = sum_{i=1}^{n} a(i) * F(ni, i), where F(0, k) = 1; F(n, 1) = F(n),
F(n, k) = sum_{i=0}^{n} F(i) * F(ni, k1).
Contribution from Paul D. Hanna, Aug 08 2007 (revised Apr 28 2012): (Start)
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
a(n) = [x^(n1)] A(x)^n for n>=1;
a(n) = (n+1)*A132070(n+1) for n>=0;
A(x) = x / Series_Reversion(x*G(x)) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A132070. (End)


EXAMPLE

a(5)=605 because there are 605 possibilities to form 5 nodes into a rooted tree and order the nodes of this tree such that no two subtrees interleave. Two subtrees t1, t2 interleave if their roots are (tree)disjoint and there are four nodes l1, r1 from t1 and l2, r2 from t2 such that l1 < l2 < r1 < r2.
Comment from Paul D. Hanna, Aug 08 2007 (revised Apr 28 2012): (Start)
Illustrate a(n) = [x^(n1)] A(x)^n by the following generating method.
Form a table of coefficients in powers of the g.f. A(x):
A(x)^1: [(1), 1, 2, 9, 64, 605, 6996, 94556, ...];
A(x)^2: [1, (2), 5, 22, 150, 1374, 15539, 206676, ...];
A(x)^3: [1, 3, (9), 40, 264, 2346, 25937, 339294, ...];
A(x)^4: [1, 4, 14, (64), 413, 3568, 38558, 495848, ...];
A(x)^5: [1, 5, 20, 95, (605), 5096, 53840, 680365, ...];
A(x)^6: [1, 6, 27, 134, 849, (6996), 72302, 897558, ...];
A(x)^7: [1, 7, 35, 182, 1155, 9345, (94556), 1152936, ...]; ...
then the coefficients along the main diagonal form the initial terms of this sequence. (End)
