Hello!

I conjectured today that $\displaystyle s_{3}(2^{k})=2\left\lfloor \frac{k-1}{2}\right\rfloor$ is true for every $\displaystyle k > 9$ integer, where $\displaystyle s_{3}(2^{k})$ is giving the sum of $\displaystyle 2^{k}$ digits in base 3.

Has it already solved or hard to proof?

Ben