0,3

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 (bdrake(AT)brandeis.edu), Jul 28 2008

a(n) = (1/2^n)* Sum_{k=0..n} binomial(n,k)*(n+k+1)^(n-1) - Vladeta Jovovic (vladeta(AT)eunet.yu), 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! = ln(A(x)) then b(n) = (1/2^n)* Sum_{k=0..n} binomial(n,k)*(n+(m-1)*k)^(n-1). - Paul Hanna and Vladeta Jovovic (vladeta(AT)eunet.rs), 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).

Let A(x) = e.g.f. of A138860, 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)).

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

5766*x^5/5! +...

(PARI) (PARI) /* Formula Due to Vladeta Jovovic: */ 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))}

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

Cf. A138764.

Sequence in context: A016036 A000314 A128709 this_sequence A145087 A005046 A141827

Adjacent sequences: A138857 A138858 A138859 this_sequence A138861 A138862 A138863

nonn

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 01 2008, Apr 02 2008, Apr 03 2008