--description--

The most naive way of computing $n^{15}$ requires fourteen multiplications:

$$n × n × \ldots × n = n^{15}$$

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

$$\begin{align} & n × n = n^2\\ & n^2 × n^2 = n^4\\ & n^4 × n^4 = n^8\\ & n^8 × n^4 = n^{12}\\ & n^{12} × n^2 = n^{14}\\ & n^{14} × n = n^{15} \end{align}$$

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

$$\begin{align} & n × n = n^2\\ & n^2 × n = n^3\\ & n^3 × n^3 = n^6\\ & n^6 × n^6 = n^{12}\\ & n^{12} × n^3 = n^{15} \end{align}$$

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

For $1 ≤ k ≤ 200$, find $\sum{m(k)}$.

--hints--

efficientExponentiation() should return 1582.

assert.strictEqual(efficientExponentiation(), 1582);

--seed--

--seed-contents--

function efficientExponentiation() {

  return true;
}

efficientExponentiation();

--solutions--

// solution required