Is there an analytic function (on the open unit disk) satisfying $\forall k \in \mathbb{Z}^+\left[f(\frac1{2k})=f(\frac1{2k+1})=\frac1{2k}\right]$?
analytic function on $\mathbb{C}$ such that $f\Bigl(\frac{1}{2n}\Bigr)=f\Bigl(\frac{1}{2n+1}\Bigr)=\frac{1}{2n}$