Modulo $2$ binomial transform of A243499 applied $k$ times Let $m \geqslant 1$ be a fixed integer.
Let $f(n)$ be A007814, 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, product of parts of integer partitions as enumerated in the table A125106.
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?
 A: The definition of $a_1$ given in OEIS is based on a bijection between integer partitions and natural numbers. A partition $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_m>0$ with exactly $m$ parts corresponds to the number
$$2^{\lambda_1+m-2}+2^{\lambda_2+m-1}+\dots+2^{\lambda_m-1}.$$
The definition of $a_1$ can then be written as
\begin{equation}\label{a1d}(1)\qquad  a_1\left(2^{\lambda_1+m-2}+2^{\lambda_2+m-1}+\dots+2^{\lambda_m-1}\right)=\lambda_1\dotsm\lambda_m.\end{equation}
When $n$ is a natural number, I will write $\|n\|$ for the number of ones in the binary expansion of $n$, and $n\preceq m$ if all binary digits in $n$ are smaller than or equal to those in $m$. It is well-known that $\binom nk\,\operatorname{mod}\,2=1$ if and only if $k\preceq n$. It is easy to see from this that
$$a_m(n)=\sum_{k\preceq n}(m-1)^{\|n-k\|}a_1(k).$$
Roughly speaking, we go from $k$ to $n$ by changing zeroes to ones, and for each of the $\|n-k\|$ zeroes we can choose to change it in any one of $m-1$ binomial transforms.
The final ingredient is the identity
$$\left\{\array{k+l\\l}\right\}=\sum_{l\geq \lambda_1\geq\lambda_2\geq\dots\geq\lambda_k> 0}\lambda_1\lambda_2\dotsm\lambda_k.$$
Probably this is well-known. I verified it using induction; just let me know if you need more explanation. By (1), this can be written
\begin{equation}\label{sa}(2)\qquad\left\{\array{k+l\\l}\right\}=\sum_{0\leq j<2^{l+k-1},\,\|j\|=k}a_1(j).\end{equation}
We now have all ingredients we need. We have
$$s_m(n)=\sum_{0\leq k<2^n}a_m(k)=\sum_{0\leq k<2^n}\sum_{l\preceq k}(m-1)^{\|k-l\|}a_1(l).$$
For fixed $l$ and $j=\|k-l\|$, we obtain $k$ by choosing $j$ from $n-\|l\|$ zeroes in $l$ and changing them to ones. Thus, we can write
$$s_m(n)=\sum_{0\leq l<2^n}a_1(l)\sum_{j}\binom{n-\|l\|}{j}(m-1)^{j}
=\sum_{0\leq l<2^n}a_1(l)m^{n-\|l\|}.$$
Writing this as a sum over $k=n-\|l\|$ and using (2) gives indeed
$$s_m(n)=\sum_{k=0}^n m^k\left\{\array{n+1\\k+1}\right\}.
$$
