Let $wt(n)$ be [A000120][1], number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).

Let  $f(n)$ be [A007814][2], 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
$$n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_{wt(n)}+1}))\dots)$$
Then we have an integer sequence given by
$$a(n)=\sum\limits_{j=0}^{2^{wt(n)}-1}(-1)^{wt(n)-wt(j)}\prod\limits_{k=0}^{wt(n)-1}(1+f(\left\lfloor\frac{j}{2^k}\right\rfloor+1))^{t_{k+1}+1}, a(0)=1$$
Let
$$s(n)=\sum\limits_{k=0}^{2^n-1}a(k)$$
then I conjecture that $s(n)$ is [A095989][3], 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\lfloor\frac{n}{2}\right\rfloor)+c(\left\lfloor\frac{2n-2^{f(n)}}{2}\right\rfloor)+c(g(n-1))
\end{align}
where $g(n)$ is [A025480][4], $g(2n) = n, g(2n+1) = g(n)$.

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


  [1]: https://oeis.org/A000120
  [2]: https://oeis.org/A007814
  [3]: https://oeis.org/A095989
  [4]: https://oeis.org/A025480