Is there a holomorphic funcion $f\colon B(0,1)\to\Bbb C$, $f({1\over2n})=f({1\over2n+1})={1\over n} $ for all $n$ positive integers?

Question

I used the following reasoning to try to prove that there isn't such function:

$f({1\over2z})=f({1\over 2z+1})$ on infinite points and both functions are analytics on $B(0,1) \setminus \{0,-\frac12\}$, then $f({1\over2z})=f({1\over 2z+1})$ on $B(0,1) \setminus \{0,-\frac12\}$.

I think that implies that both functions are constants, and that would be absurd. But is there a way to prove they are constants? Or is the problem solved in other way? Any advice is welcome.

Answer 1

I would have concluded $f(z)\equiv 2z$ from the fact that this holds for all $z=\frac1{2n}$. But then $f(\frac13)\ne\frac11$.