OFFSET
1,2
FORMULA
E.g.f. A(x) satisfies:
(1) (1+A(x))*log(1+A(x)) = (3*A(x) - x)/2.
(2) log(1+A(x)) = Series_Reversion[(3-2*x)*exp(x) - 3].
(3) Let B(x) = 1+A(x), then: B(x) = exp( B(x)^2 * Integral 1/B(x)^3 dx ). - Paul D. Hanna, Dec 06 2013
a(n) ~ n^(n-1) / (sqrt(2) * exp(n-1/4) * (2*exp(1/2)-3)^(n-1/2)). - Vaclav Kotesovec, Dec 07 2013
EXAMPLE
E.g.f.: A(x) = x + 2*x^2/2! + 10*x^3/3! + 84*x^4/4! + 988*x^5/5! + ...
Series reversion of the e.g.f. A(x) begins:
x - x^2 + x^3/3 - x^4/6 + x^5/10 - x^6/15 + x^7/21 - x^8/28 +- ...
Series reversion of log(1+A(x)) begins:
x - x^2/2! - 3*x^3/3! - 5*x^4/4! - 7*x^5/5! - 9*x^6/6! - 11*x^7/7! - ...
MAPLE
series(exp(LambertW(-exp(-3/2)*(3+x)/2)+3/2)-1, x, 31): A:=simplify(%, symbolic): A180715:=n->n!*coeff(A, x, n): # Vladeta Jovovic, Sep 28 2010
PROG
(PARI) a(n)=if(n<1, 0, n!*polcoeff(serreverse(x-sum(k=2, n, (-x)^k*2/(k*(k-1)))+x*O(x^n)), n))
for(n=1, 25, print1(a(n), ", "))
(PARI) a(n)=if(n<1, 0, n!*polcoeff(exp(serreverse((3-2*x)*exp(x+x*O(x^n))-3))-1, n))
for(n=1, 25, print1(a(n), ", "))
(PARI) a(n)=local(B=1+x); for(i=1, n, B=exp(B^2*intformal(1/B^3+x*O(x^n)))); n!*polcoeff(B-1, n)
for(n=1, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Dec 06 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 24 2010
STATUS
approved