A202617 E.g.f. satisfies: A(x) = exp( x*(1 + A(x)^2)/2 ).

1, 1, 3, 19, 185, 2441, 40747, 823691, 19564785, 534145105, 16482667091, 567343245635, 21552042260905, 895664877901145, 40422799315249275, 1968883362773653051, 102942561775293158369, 5750760587905912310177, 341848844954020959953059, 21545207157567497255044979

Offset: 0

Author

Paul D. Hanna, Dec 21 2011

Keywords

• nonn

Comments

Compare to e.g.f. F(x) of A007889: F(x) = exp(x*(1 + F(x))/2), where A007889(n) = number of intransitive (or alternating) trees: vertices are [0,n] and for no i

Related sequence: A058014(n) = number of labeled trees with n+1 nodes such that the degrees of all nodes, excluding the first node, are odd.

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 19*x^3/3! + 185*x^4/4! + 2441*x^5/5! +...
where log(A(x)) = x*(1 + A(x)^2)/2 and
log(A(x)) = x + 2*x^2/2! + 12*x^3/3! + 112*x^4/4! + 1440*x^5/5! + 23616*x^6/6! +...
A(x)^2 = 1 + 2*x + 8*x^2/2! + 56*x^3/3! + 576*x^4/4! + 7872*x^5/5! + 134656*x^6/6! +...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..372

Crossrefs

Cf. A007889, A058014, A214225, A138860, A202357.

Programs

Mathematica

CoefficientList[Series[Sqrt[-ProductLog[-E^x*x]/x], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Jan 10 2014 *)

PARI

a(n)=local(A=1+x); for(i=0, n, A=exp(x*(1+A^2)/2 +x*O(x^n))); n!*polcoeff(A, n)

PARI

/* Coefficients of A(x)^p are given by: */
{a(n, p=1)=(1/2^n)*sum(k=0, n, binomial(n, k)*p*(2*k+p)^(n-1))}

PARI

a(n)=n!*polcoeff(exp(sum(k=1,n,k^(k-1)*cosh(k*x +x*O(x^n))*x^k/k!) +x*O(x^n)),n)
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Nov 20 2012

Formula

E.g.f. A(x) equals the formal inverse of function 2*log(x)/(1+x^2).

E.g.f.: exp( Sum_{n>=1} n^(n-1) * cosh(n*x) * x^n / n! ). - Paul D. Hanna, Nov 20 2012

E.g.f.: exp(G(x)) where G(x) = x/(1 - tanh(G(x))) is the e.g.f. of A214225. - Paul D. Hanna, Nov 20 2012

E.g.f. satisfies: A(x) = G(x*A(x)) where G(x) = A(x/G(x)) is the e.g.f. of A058014.

a(n) = (1/2^n)*Sum_{k=0..n} C(n,k)*(2*k+1)^(n-1).

Powers of e.g.f.:

If A(x)^p = Sum_{n>=0} a(n,p)*x^n/n! then a(n,p) = (1/2^n)* Sum_{k=0..n} binomial(n,k)*p*(2*k+p)^(n-1).

a(n) ~ sqrt(1+c) * n^(n-1) / (2 * exp(n) * c^(n+1/2)), where c = LambertW(exp(-1)) = 0.278464542761... (see A202357). - Vaclav Kotesovec, Jan 10 2014

E.g.f.: sqrt(-LambertW(-x*exp(x))/x). - Vaclav Kotesovec, Jan 10 2014

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