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FORMULA
| E.g.f.: ((1-x)(2-x-x^2) - (2-x+x^2)sqrt{1-2x-x^2}) / (3x^3).
Comment from Brendan McKay, May 24 2010: The number of labelled trees with d[i] vertices of degree i for i=1,2,3 is (n-1)! * n! / (2^d[3] * d[3]! * d[2]! * d[1]!). Now sum over d[1]+d[2]+d[3]=n, d[1]+2*d[2]+3*d[3] = 2n-2.
Comment from Georgi Guninski, May 24 2010: This seems to hold up to 3000 terms:
a[n+1]=(-a[n-1]*a[n] - (-3*a[n]**2 + (2/3)*a[n-2]*a[n]*n+
(-4/3)*a[n-1]*a[n]*n+ (-4/3)*a[n]**2*n+ (-1/3)*a[n-2]*a[n]*n**2+
(-2/3)*a[n-1]*a[n]*n**2)) / (a[n-1]+ (-1/3)*a[n] -2*a[n-2]*n+
2*a[n-1]*n+a[n-2]*n**2) where a[n] == A003692.
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PROG
| (Sage program from Doug McNeil, May 24 2010):
sage: gf = ((1-x)*(2-x-x**2) - (2-x+x**2)*(1-2*x-x**2)**(1/2)) / (3*x**3)
sage: c = taylor(gf, x, 0, 12).coefficients()
sage: c
[[1, 0], [1, 1], [3/2, 2], [8/3, 3], [5, 4], [39/4, 5], [469/24, 6],
[40, 7], [333/4, 8], [1405/8, 9], [5995/16, 10], [807, 11], [42055/24,
12]]
sage: sq = [a*factorial(b) for a, b in cc]
sage: sq
[1, 1, 3, 16, 120, 1170, 14070, 201600, 3356640, 63730800, 1359666000,
32212857600, 839350512000]
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