Mobius Function Find all integers $n, 1 ≤ n ≤ 100$ such that $µ(n) = 1$. My initial idea was to find all the primes between 1 and 100 using techniques of finding prime numbers. If p is prime then $µ(p)=-1$ so with the exception of $µ(1)=1$ then in order for $µ(n)=1, n$ must be composite with a prime factor expansion with an even number of unique primes. Is there an easier way to do this?
 A: The easiest way is to look up the sequence A030229 , which does what you propose, i.e., lists the product of an even number of distinct primes (and $1$). Up to $100$ they are
$$
1, 6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95.
$$
I don't think that every other method is significantly faster since we only talk about the range $1\le n\le 100$.
A: Since $2\cdot3\cdot5\cdot7=210\gt100$, you only need to consider squarefree semiprimes (and $1$). These are listed in OEIS sequence A006881, which of course coincides up to $210$ and except for $1$ with the sequence given by Dietrich Burde. You only need to find the primes up to $50$ for this, not up to $100$, since the other factor is at least $2$. There are $15$ such primes. If you arrange them along the $x$ and $y$ axis, the squarefree semprimes correspond to the pairs below the diagonal and below the hyperbola $xy=100$. They are marked in red in the following diagram; there are $30$ of them, in agreement with the OEIS sequences.

