You don't need the bigger machinery of a segmented sieve for such a small range. Here is a simple $O(n\log\log n)$ algorithm to calculate all $\mu(i)$ up to $n$ based on the Sieve of Erostothanes. In fact, the sieve does fully factor all the square-free numbers; it just doesn't do it one at a time.
Depending on your computer, this approach is practical up to around $2^{30}$ at which point you need to start using higher-precision arithmetic and computing a range of $\mu(i)$ values in batches.
public static int[] GetMu(int max)
{
var sqrt = (int)Math.Floor(Math.Sqrt(max));
var mu = new int[max + 1];
for (int i = 1; i <= max; i++)
mu[i] = 1;
for (int i = 2; i <= sqrt; i++)
{
if (mu[i] == 1)
{
for (int j = i; j <= max; j += i)
mu[j] *= -i;
for (int j = i * i; j <= max; j += i * i)
mu[j] = 0;
}
}
for (int i = 2; i <= max; i++)
{
if (mu[i] == i)
mu[i] = 1;
else if (mu[i] == -i)
mu[i] = -1;
else if (mu[i] < 0)
mu[i] = 1;
else if (mu[i] > 0)
mu[i] = -1;
}
return mu;
}
Running GetMu(1000000)
takes about 10 msec on my computer.