Let $m \geqslant 1$ be a fixed integer.

Let $f(n)$ be [A007814][1], exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.

Then we have an integer sequence given by
\begin{align}
a_1(0)& = 1\\
a_1(2n+1)& = a_1(n)\\
a_1(2n)& = a_1(n-2^{f(n)})+a_1(2n-2^{f(n)})
\end{align}
Here $a_1(n)$ is [A243499][2], product of parts of integer partitions as enumerated in the table [A125106][3].

Let
$$a_m(n) = \sum\limits_{k=0}^{n}(\binom{n}{k}\operatorname{mod} 2)a_{m-1}(k)$$
Also
$$s_m(n)=\sum\limits_{k=0}^{2^n-1}a_m(k)$$
I conjecture that $s_m(n)$ is Stirling transform of
$$1, m, m^2, m^3, \cdots$$
In other words
$$s_m(n)=\exp(-m)\sum\limits_{k=0}^{\infty}(k + 1)^n\frac{m^k}{k!}=\sum\limits_{k=0}^{n}{n+1\brace k+1}m^k$$
Is there a way to prove it?

  [1]: https://oeis.org/A007814
  [2]: https://oeis.org/A243499
  [3]: https://oeis.org/A125106