Let
- $a(n)$ be A032188 (i.e., number of labeled series-reduced mobiles (circular rooted trees) with $n$ leaves (root has degree $0$ or $\geqslant 2$)), whose exponential generating function is $A(x)$ such that $$ \log(1-A(x))+2A(x)-x=0. $$
- Start with vector $\nu$ of fixed length $m$ with elements $\nu_i := \delta_{1,i}$, reserve $A$ and for $i$ from $1$ to $m-1$, for $j$ from $i+1$ to $m$ apply $A := 0$ (at the beginning of each cycle for $i$) and also apply $$ A := A + \nu_{j-1}, \nu_j := (j-i+1)\nu_j + A. $$
I conjecture that after the whole tranform we have $\nu_n = a(n)$. Here is the PARI/GP program to check it numerically:
upto1(n) = my(x = 'x); Vec(serlaplace(serreverse(2*x+log(1-x + O(x*x^n)))))
upto2(n) = {my(A, v1); v1 = vector(n, i, 0); v1[1] = 1;
for(i=1, n-1, A = 0;
for(j=i+1, n,
A += v1[j-1];
v1[j] = (j-i+1)*v1[j] + A));
v1}
test(n) = upto2(n) == upto1(n)
It takes about $2$ seconds on my relatively old PC to compute the first $1000$ terms using upto2. In the same time, it takes about $20$ seconds on the same PC to compute first $200$ terms using upto1.
Is there a way to prove it?