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?

• One option, which however doesn't quite reach the extremes you wany, would be $1-\frac{\varphi(n)}{n}$. Aug 16 '15 at 16:03
• For every $n>1$ there exists an integer that has more than $n$ divisors. So the concept of “highly divisible number” is not well defined. Aug 16 '15 at 16:14
• Isn't a highly divisible number n defined as having more divisors than all numbers smaller than n? Should be well defined, but probably hard to scale.
– fabb
Aug 16 '15 at 17:03
• See my comments here. This gives you way of computing the norm of any natural number. Then its just a matter of squishing that norm into the interval $[0,1]$. Aug 19 '15 at 14:05

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.

• Interesting, thank you for this direction!
– fabb
Aug 16 '15 at 17:04

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 $n$.

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.