a(n) ~ c * n^n / (exp(n) * (log(2))^n), where c = 1.16670181891916121...
G.f. satisfies A(x) = 1 + Sum_{n>=0} B^n(x) / 2^(n+1), where B(x) = x*A(x) and B^n(x) denotes the n-th iteration of B(x) with B^0(x) = x. -
Paul D. Hanna, Apr 05 2026
Let b(n,k) = [x^n] A(x)^k.
b(0,1) = b(1,1) = 1; b(n,1) = Sum_{j=1..n-1} b(j,1) * b(n-j,j).
For k > 1, b(0,k) = 1; b(n,k) = (1/n) * Sum_{j=1..n} ((k+1)*j-n) * b(j,1) * b(n-j,k).
a(n) = b(n,1). (End)