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FORMULA
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a(0) = a(1) = 1; a(n) = n * F(n-1), where F(0) = F(1) = 1, F(n) = sum_{i=1}^{n} a(i) * F(n-i, i), where F(0, k) = 1; F(n, 1) = F(n), F(n, k) = sum_{i=0}^{n} F(i) * F(n-i, k-1).
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^(n-1)] 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)
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EXAMPLE
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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^(n-1)] 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)
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