Let $m\geqslant1$ be a fixed integer.
Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Let $g(n)$ be A025480, $g(2n) = n$, $g(2n+1) = g(n)$.
Then we have an integer sequences given by \begin{align} a(0)&=1\\ a(n)& = a\left(\left\lfloor\frac{n}{2}\right\rfloor\right)+a\left(\left\lfloor\frac{2n-2^{f(n)}}{2}\right\rfloor\right)+ma(g(n-1)) \end{align} I conjecture that $$a(2^{k-1}-1)=T(m,k), T(m,0)=1$$ where $T(n,k)$ is A111528, square table, read by antidiagonals, where the g.f. for row $n+1$ is generated by: $$xR_{n+1}(x) = \frac{1+nx - \frac{1}{R_n(x)}}{n+1}$$ with $$R_0(x) = \sum\limits_{n=0}^{\infty}n!x^n$$ Is there a way to prove it?
