I am trying to show that $n$ has $2^{\omega(n) - 1}$ coprime factor pairs. I'm pretty sure this is true but I don't see how to prove it. There is no obvious way to use induction.
Here is an example:
$30$ has $3$ distinct prime factors: $2, 3, 5$ so $\omega(30) = 3$. My conjecture predicts that $30$ has $2^{3 - 1} = 2^2 = 4$ coprime factor pairs.
We find they are: $(1, 2\cdot3\cdot5) $, $(2, 3\cdot5)$, $(3, 2\cdot 5)$, and $(5, 2\cdot3)$.
Can anyone provide a hint at how to prove this?
Let $n= p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_k^{\alpha_k}$. Then $\omega(n)=k$. Then the number of coprime factor pair has form $(x,y)$ where \begin{align} \begin{cases} x \cdot y = p_1p_2\ldots p_k\\ \gcd(x,y)=1 \end{cases} \end{align} Then for each $x$, you can determine exactly one $y$ that satisfies the above relations. Note that to form $x$, you can just pick $m$ prime in the set $\left\lbrace p_1,p_2,\ldots,p_k\right\rbrace$ and let $x$ be their product. There are $ \binom{k}{m}$ ways to choose such $m$-tuples of prime. So there are in total $\sum_{m=0}^k \binom{k}{m}=2^k$ coprime factor pairs. But there is a catch, each pair is repeated: for example, choosing $m=0$ and $m=k$ yields the same pairs, so in fact you only have $2^{k-1}$ pairs.