A little more arithmetic derivative stuff.

Question

empty · Answer

$$n = \prod_i p_i^{k_i}$$
$$\mathrm{rad}(n) = \prod_i p_i$$
$$\gcd(n,n') = \frac{n}{\mathrm{rad}(n)} \prod_i \gcd(p_i,k_i)$$

So $n$ and $n'$ are relatively prime when:

$$n = \mathrm{rad}(n)$$

Or in other words, n is square free.

empty · Answer

Now some interesting things are, 

$\gcd(n,n')$, $\mathrm{rad}(n)$, $n$, and that last function there $f(n)=\prod_i \gcd(p_i,k_i)$ are ALL multiplicative functions.

empty · Answer

I don't know if this is useful for anything but, 

$$\gcd(n,n')=1 \iff \mu(n) \ne 0$$

parthkohli · Answer

What is $\mu(n)$ referring to?

empty · Answer

It's kinda weird, if the number is square free, then you raise -1 to the power of primes in the number, otherwise it's 0. Examples:

$$\mu(2)=-1$$$$\mu(3)=-1$$$$\mu(6)=1$$$$\mu(4)=0$$

It's multiplicative so that means you can just split it up into its prime factors too:

$$\mu(12)=\mu(4)*\mu(3) = 0$$$$\mu(70)=\mu(2)\mu(5)\mu(7)=(-1)^3 = -1$$

Also, just sorta outta necessity, $\mu(1)=1$ you can think of it as 0 prime factors so $(-1)^0$

empty · Answer

It actually ends up being like a derivative operator when you use it with the Dirichlet convolution haha...

empty · Answer

The way I think of all multiplicative functions is I build them up by just defining their behavior on primes like this:

$$f(p^k)=...$$

So in the case of the $\mu$ (Mobius function) it looks like this:

$$\mu(1)=1$$$$\mu(p)=-1$$$$\mu(p^k)=0 \text{ for k > 1}$$

So you can kind of see the the backwards finite difference in there if you squint. $$\Delta f(k) = f(k)-f(k-1)$$