Solving recurrence $T_k(n) = T_k\left(\left \lfloor \frac{n}{2} \right \rfloor\right) + T_k\left(\left \lceil \frac{n}{2} \right \rceil\right)$
Observe that $T_k(2^n k) = 2^n$. Furthermore, it is easy to see that either $T_k(\alpha) = T_k(\alpha-1)$ or $T_k(\alpha) = 1+T_k(\alpha-1)$. Using the previous two results, it can also be shown that $T_k(2^n k - 2^{n-1}) = T_k(2^{n-1} k)$. 

Now, consider 
$$\delta_k(n) = T_k(2^n k - 2^{n-1} + 1) - T_k(2^n k - 2^{n-1} )$$

It should be easy to prove by induction that for all $k, n$ it holds that $\delta_k(n) = 1$. This should be sufficient to uniquely determine the value of $T_k(n)$ for all $n$.