## Efficient exponentiation

### Problem 122

Published on Friday, 2nd June 2006, 06:00 pm; Solved by 3681

The most naive way of computing n15 requires fourteen multiplications:

n n ... n = n15

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

n n = n2
n2 n2 = n4
n4 n4 = n8
n8 n4 = n12
n12 n2 = n14
n14 n = n15

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

n n = n2
n2 n = n3
n3 n3 = n6
n6 n6 = n12
n12 n3 = n15

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

For 1 k 200, find m(k).