Problem 122
Efficient exponentiation

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).

These problems are part of Project Euler and are licensed under CC BY-NC-SA 2.0 UK

http://projecteuler.net/problem=122