Is there a Divisibility Metric for Numbers? Both prime numbers and highly divisible numbers have a common characteristic: divisibility. The former are divisible by as few lower numbers as possible, and the latter by as many as possible, like two poles on a scale.
I'm interested into fitting all the other non-prime and non-highly-divisible whole numbers into such a scale too.
Any suggestions for creating a formula that translates whole numbers to the range of [0,1], where prime numbers result in 0 and highly divisible numbers in 1, and all the other numbers in between?
Have there already been attempts to do this?
 A: This is an interesting question. I'm not sure of a fully detailed answer. But if you fix a prime $p$, one can study the so called $p$-adic numbers, which is the rationals, but with a metric that depends on divisibility by $p$. See  https://en.m.wikipedia.org/wiki/P-adic_number for more details. I know it doesn't totally answer what you want, but as far as studying goes it's a good place to start. 
FYI, start with the analytic approach, not the algebraic one. 
A: The numbers with the most prime factors are the powers of 2.
So one approach would be:
$$Number\ between\ 1\ and\ 0\ =\dfrac{\sigma_0(n)}{log_2(n)}$$
If you ment unique prime factors, then it's a little more complicated. Again, primes have the least, but the numbers with the greatest values for $\dfrac{\sigma_0(rad(n))}{n}$ are the primorials.
Since the n'th primorial $p_n\#$ is about:
$$p_n\#\approx\prod_{k=1}^{n}k*log\ k=n!*log\ n!$$
there is no way that I know of to reverse factorials, so you can't check which primorial is closest to a given integer $n$.
A: Well, you can look at the Prime Omega function : $\Omega(n)$ maps to the number of prime factors of n. This is not a function that goes from 0 to 1, but such a function is not possible I think.
It is impossible for one major reason : Take, for exemple the number $n!$, it has as many prime factors as you want (you just have to increase $n$). So, if you create a number $\psi \in \mathbb{N}$ such that $f(\psi)=1$ where $f$ is the function of your dreams. Then what is $f(\psi!)$ ? Knowing that $\psi!$ has much, much more prime factors than $\psi$, we will have $f(\psi!)>f(\psi)=1$ so your function isn't bounded.
Now, if what you want is a function that say "how much" a prime is divisible, you can take $k(n) = (\Omega(n)-1)/n$ which is very interesting too ! This function measure how much a number is divisible compared to the value of this number.
If $p$ is a prime number, then $k(p)=0$ and if $k(n)=1$ for some $n$, then it means $n$ is extremely divisible. In fact, this number is so divisible, that it has more factors than its own value, which I think is totally impossible.
The more $n$ is divisible compared to its own value, the more $k(n)$ will be high.
Edit : notice that $k(1) = -1$ because $1$ is not a prime number but it has many properties of it, for exemple, it is divisible by $1$ and itself, but no other number. $1$ is a very special case.
