- Let $a(n)$ be A000311 (i.e., Schroeder's fourth problem; also series-reduced rooted trees with $n$ labeled leaves; also number of total partitions of $n$). Here exponential generating function is $A(x)$ such that $$ e^{A(x)} = 2A(x)-x+1. $$
- Let $b(n)$ be an integer sequence such that $$ b(n) = \sum\limits_{i=0}^{n-1} {n+i-1 \brace i}(-1)^{n+i-1}[x^{n+i-1}]\frac{1+(x+1)^{2n-1}}{x+2} $$
I conjecture that $$ b(n) = a(n). $$
Here is the PARI/GP program to compute $b(n)$:
b(n) = my(x = 'x, A = (1+(x+1)^(2*n-1))/(x+2)+x*O(x^(2*(n-1)))); sum(i=0, n-1, stirling(n+i-1,i,2)*(-1)^(n+i-1)*polcoeff(A,n+i-1,x)) Is there a way to prove it? Is there a way to independently compute polynomial for $k$-th diagonal of Stirling numbers of the second kind? If so, then it will significantly speed up the calculation of $b(n)$.
