E.g.f. is reversion of 2*log(1+x)-x.
Also exponential transform of
A000311, define b by 1+sum b_n x^n / n! = exp ( 1 + sum a_n x^n /n!).
E.g.f.: A(x), B(x)=x*A(x) satisfies the differential equation B'(x)=(1+B(x))/(1-B(x)). -
Vladimir Kruchinin, Jan 18 2011
The generating function A(x) satisfies the autonomous differential equation A'(x) = (1 + A)/(1 - A) with A(0) = 0. Hence the inverse function A^(-1)(x) = Integral_{t = 0..x} (1 - t)/(1 + t) dt = 2*log(1 + x) - x, which yields A(x) = -1 - 2*W(-1/2*exp((x - 1)/2)), where W is the Lambert W function.
The expansion of A(x) can be found by inverting the above integral using the method of [Dominici, Theorem 4.1] to arrive at the result a(n) = D^(n-1)(1) evaluated at x = 0, where D denotes the operator g(x) -> d/dx((1 + x)/(1 - x)*g(x)). Compare with
A032188.
Applying [Bergeron et al., Theorem 1] to the result x = Integral_{t = 0..A(x)} 1/phi(t) dt, where phi(t) = (1 + t)/(1 - t) = 1 + 2*t + 2*t^2 + 2*t^3 + ..., leads to the following combinatorial interpretation for the sequence: a(n) gives the number of plane increasing trees on n vertices where each vertex of outdegree k >= 1 can be in one of 2 colors. An example is given below. (End)
A134991 gives (b.+c.)^n = 0^n , for (b_n)=
A000311(n+1) and (c_0)=1, (c_1)=-1, and (c_n)=-2*
A000311(n) = -
A006351(n) otherwise. E.g., umbrally, (b.+c.)^2 = b_2*c_0 + 2 b_1*c_1 + b_0*c_2 =0. -
Tom Copeland, Oct 19 2011
G.f.: 1/S(0) where S(k) = 1 - x*(k+1) - x*(k+1)/S(k+1); (continued fraction). -
Sergei N. Gladkovskii, Dec 18 2011
a(n) = ((n-1)!*sum(k=1..n-1, C(n+k-1,n-1)*sum(j=1..k, (-1)^(j)*C(k,j)* sum(l=0..j, (C(j,l)*(j-l)!*2^(j-l)*(-1)^l*stirling1(n-l+j-1,j-l))/ (n-l+j-1)!)))), n>1, a(1)=1. -
Vladimir Kruchinin, Jan 24 2012
G.f.: -1 + 2/Q(0), where Q(k)= 1 - k*x - x*(k+1)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, May 01 2013
a(n) ~ sqrt(2)*n^(n-1)/((2*log(2)-1)^(n-1/2)*exp(n)). -
Vaclav Kotesovec, Jul 17 2013
G.f.: Q(0)/(1-x), where Q(k) = 1 - x*(k+1)/( x*(k+1) - (1 -x*(k+1))*(1 -x*(k+2))/Q(k+1) ); (continued fraction). -
Sergei N. Gladkovskii, Oct 10 2013
a(1) = 1; a(n) = a(n-1) + Sum_{k=1..n-1} binomial(n-1,k) * a(k) * a(n-k). -
Ilya Gutkovskiy, Aug 28 2020
