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 $$s_m(n)=\exp(-m)\sum\limits_{k=0}^{\infty}(k + 1)^n\frac{m^k}{k!}$$ Is there a way to prove it?

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?