Let $a_n$ be the sequence of rational numbers defined by the recurrence
$$a_0=1,\quad\left(2^n-1\right)a_n=\sum_{k=1}^n{\small\binom n k}\frac{a_{n-k}}{k+1}.\tag1$$
It begins
$$\small1,\,\frac12,\,\frac5{18},\,\frac16,\,\frac{143}{1350},\,\frac{19}{270},\,\frac{1153}{23814},\,\frac{583}{17010},\,…\tag2$$
and is apparently related to $\texttt{A272755}$ and the Fabius function.
It appears that the following simpler identity holds:
$$a_n\stackrel{\small\color{silver}?}=\sum_{k=0}^n{\small\binom n k}\,(-1)^k\,a_k.\tag3$$
How can we prove it?
(Note that $(3)$ could not be used as a definition of $a_n$ because the term $a_n$ is canceled for even $n$.)
Best Answer
If we call $f(x)=\sum \frac{a_n}{n!}x^n$, then the first recursion is equivalent to $$f(2x)=\frac{e^x-1}{x}f(x).$$ In this paper (page 5), one can see that $f(x)=e^{x/2}\hat{\text{up}}(\frac{ix}{4\pi}),$ where $$\hat{\text{up}}(x)=\prod _{n = 0}^{\infty}\frac{\sin(\pi x/2^n)}{\pi x/2^n}.$$ notice that $\hat{\text{up}}(x)$ is an even function and so
$$f(-x)e^x=e^{x/2}\hat{\text{up}}(\frac{ix}{4\pi})=f(x),$$ and so checking at the coefficient of $x^n$ on that equation, we get that $$\sum _{k=0}^{n}\binom{n}{k}(-1)^ka_k=a_n.$$