# A sequence defined by $(2^n-1)\,a_n=\sum_{k=1}^n\binom n k\,a_{n-k}\,(k+1)^{-1}$

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 $$n$$.)

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 $$\sum _{k=0}^{n}\binom{n}{k}(-1)^ka_k=a_n.$$