The most naive way of computing $n^{15}$ requires fourteen multiplications: $$n\times n\times \cdots \times n=n^{15}.$$

But using a "binary" method you can compute it in six multiplications:

$$\begin{array}{rl}n\times n& =n^2\\ n^2\times n^2& =n^4\\ n^4\times n^4& =n^8\\ n^8\times n^4& =n^{12}\\ n^{12}\times n^2& =n^{14}\\ n^{14}\times n& =n^{15}\end{array}$$

However it is yet possible to compute it in only five multiplications:

$$\begin{array}{rl}n\times n& =n^2\\ n^2\times n& =n^3\\ n^3\times n^3& =n^6\\ n^6\times n^6& =n^{12}\\ n^{12}\times n^3& =n^{15}\end{array}$$

We shall define $m(k)$ to be the minimum number of multiplications to compute $n^k$; for example $m(15)=5$.

Find $\sum _{k=1}^{200}m(k)$.