Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let $a(n)$ be A014307. Here
$$ A(x) = \sum\limits_{k=0}^{\infty} \frac{a(k)}{k!}x^k = \sqrt{\frac{e^x}{2-e^x}} $$
- Let $b(n)$ be an integer sequence such that
$$ b(2^m(2^n(2k+1)-1)) = \sum\limits_{i=1}^{m+1} b(2^i k) (-2)^{m-i+1}\sum\limits_{j=i}^{m+1} j^n s(j,i) {m+1 \brace j}, \\ b(0) = 1 $$
- Let
$$ s(n) = \sum\limits_{k=0}^{2^n-1}b(k) $$
I conjecture that
$$ s(n) = a(n+1). $$
UPD:
It looks like we can also change $(-2)^{m-i+1}$ to $(-q)^{m-i+1}$. In this case, advanced search for $q=3$ in OEIS gives us the following result:
$$ 1+\sum\limits_{k=1}^{\infty}\frac{s_3(k-1)}{k!}x^k = \exp\left(\sum\limits_{n=1}^{\infty}\frac{1}{n!}x^n\sum\limits_{k=0}^{n-1}k!{n-1 \brace k}2^{n-k-1}\right) $$
I conjecture that it generalizes to
$$ 1+\sum\limits_{k=1}^{\infty}\frac{s_q(k-1)}{k!}x^k = \exp\left(\sum\limits_{n=1}^{\infty}\frac{1}{n!}x^n\sum\limits_{k=0}^{n-1}k!{n-1 \brace k}(q-1)^{n-k-1}\right) $$
Is there a way to prove it?
