## Efficient exponentiation

### Problem 122

Published on Friday, 2nd June 2006, 06:00 pm; Solved by 4590; Difficulty rating: 40%The most naive way of computing *n*^{15} requires fourteen multiplications:

*n* × *n* × ... × *n* = *n*^{15}

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

*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}

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

*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}

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 ∑ m(*k*).