Compute the Mertens function

Mathematica, 22 20 bytes

Thanks to @miles for saving 2 bytes.

Tr@*MoebiusMu@*Range

Explanation

Range

Generate a list from 1 to input.

MoebiusMu

Find MoebiusMu of each number

Tr

Sum the result.

Python 2, 45 37 bytes

f=lambda n,k=2:n<k or f(n,k+1)-f(n/k)

Test it on Ideone.

Background

This uses the property

from A002321, which leads to the following recursive formula.

How it works

We use recursion not only to compute M for the quotients, but to compute the sum of those images as well. This saves 8 bytes over the following, straightforward implementation.

M=lambda n:1-sum(M(n/k)for k in range(2,n+1))

When f is called with a single argument n, the optional argument k defaults to 2.

If n = 1, n<k yields True and f returns this value. This is our base case.

If n > 1, n<k initially returns False and the code following or is executed. f(n/k) recursively computes one term of the sum, which is subtracted from the return value of f(n,k+1). The latter increments k and recursively calls f, thus iterating over the possible values of k. Once n < k + 1 or n = 1, f(n,k+1) will return 1, ending the recursion.

05AB1E, 16 15 bytes

LÒvX(ygmyyÙïQ*O

Explanation

L        # range [1 .. n]
Ò        # list of prime factors for each in list
v        # for each prime factor list
 X(ygm   # (-1)^len(factors)
 yyÙïQ*  # multiplied by factors == (unique factors)
 O       # sum

Try it online!