A138860 E.g.f. satisfies: A(x) = exp( x*(A(x) + A(x)^2)/2 ).

1, 1, 4, 31, 364, 5766, 115300, 2788724, 79197040, 2583928360, 95256535936, 3916137470664, 177651980724160, 8815348234689920, 474993826614917632, 27619367979975064576, 1723821221240101984000, 114948301218300412117632

Offset: 0

Author

Paul D. Hanna, Apr 01 2008, Apr 02 2008, Apr 03 2008

Keywords

• nonn

Comments

The related sequence A007889 enumerates the number of intransitive (or alternating) trees.

a(n+1) is the number of incomplete ternary trees on n labeled vertices in which each left child has a larger label than its parent and each middle child has a smaller label than its parent. - Brian Drake, Jul 28 2008

Examples

			E.g.f.: A(x) = 1 + x + 4*x^2/2! + 31*x^3/3! + 364*x^4/4! + 5766*x^5/5! + ...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..363

Crossrefs

Cf. A007889, A088789, A058014, A036778, A138903.

Cf. A138764.

Programs

Mathematica

Table[1/2^n * Sum[Binomial[n,k]*(n+k+1)^(n-1),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 15 2013 *)

PARI

a(n)=(1/2^n)*sum(k=0,n,binomial(n,k)*(n+k+1)^(n-1))

PARI

/* Series Reversion: */
a(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(serreverse(2*x/(exp(X)+exp(2*X)) )),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*(n+k+p)^(n-1))}

Formula

a(n) = (1/2^n)* Sum_{k=0..n} binomial(n,k)*(n+k+1)^(n-1) - Vladeta Jovovic, Mar 31 2008.

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

E.g.f.: A(x) = inverse function of 2*log(x)/(x + x^2).

E.g.f.: A(x) = exp( Series_Reversion[ 2*x/(exp(x) + exp(2*x)) ] ).

E.g.f.: A(x) = G(x/2) where G(x) = e.g.f. of A138764.

More generally, if A(x) = Sum_{n>=0} a(n)*x^n/n! = exp( x*[A(x) + A(x)^m]/2 ) then a(n) = (1/2^n)* Sum_{k=0..n} binomial(n,k)*(n+(m-1)*k+1)^(n-1) and if B(x) = Sum_{n>=0} b(n)*x^n/n! = log(A(x)) then b(n) = (1/2^n)* Sum_{k=0..n} binomial(n,k)*(n+(m-1)*k)^(n-1). - Paul D. Hanna and Vladeta Jovovic, Apr 02 2008

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*(n+k+p)^(n-1).

Given e.g.f. A(x), let B(x) = e.g.f. of A007889, then

. A(x) = B(x*A(x)) = (1/x)*Series_Reversion(x/B(x)) and

. B(x) = A(x/B(x)) = x/Series_Reversion(x*A(x)).

a(n) ~ n^(n-1)*(1+r)^n*r^(n+1)/(sqrt(1+3*r)*(1-r)^(2*n+1)*exp(n)*2^n), where r = 0.6472709258412625... is the root of the equation (r/(1-r))^(1+r) = e. - Vaclav Kotesovec, Jun 15 2013