$\DeclareMathOperator\wt{wt}$Let $\wt(n)$ be A000120, number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).

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 $n$.

Also $$n=2^{t_1}(1+2^{t_2+1}(1+\dotsb(1+2^{t_{wt(n)}+1}))\dotsb).$$ Then we have an integer sequence given by $$a(n)=\sum\limits_{j=0}^{2^{\wt(n)}-1}(-1)^{\wt(n)-\wt(j)}\prod\limits_{k=0}^{\wt(n)-1}\left(1+f\left(\left\lfloor\frac{j}{2^k}\right\rfloor+1\right)\right)^{t_{k+1}+1},\quad a(0)=1.$$ Let $$s(n)=\sum\limits_{k=0}^{2^n-1}a(k).$$ Then I conjecture that $s(n)$ is A095989, INVERTi transform applied to the ordered Bell numbers.

I also conjecture that \begin{align} a(0)=a(1)&=1\\ a(2n+1) &= a(2n)\\ a(2n)& = a(n)+a(2n-2^{f(n)})+b(n-1)\\ b(2n+1) &= b(n)\\ b(2n) &= a(2n). \end{align} In other words \begin{align} a(2n) &= c(n)\\ c(0)&=1\\ c(n)& = c\left(\left\lfloor\frac{n}{2}\right\rfloor\right)+c\left(\left\lfloor\frac{2n-2^{f(n)}}{2}\right\rfloor\right)+c(g(n-1)) \end{align} where $g(n)$ is A025480, $g(2n) = n$, $g(2n+1) = g(n)$.

Is there a way to prove it? Is it possible to at least get a closed form for $s(n)$?