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?

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    $\begingroup$ One option, which however doesn't quite reach the extremes you wany, would be $1-\frac{\varphi(n)}{n}$. $\endgroup$ – Henning Makholm Aug 16 '15 at 16:03
  • $\begingroup$ 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. $\endgroup$ – egreg Aug 16 '15 at 16:14
  • $\begingroup$ 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. $\endgroup$ – fabb Aug 16 '15 at 17:03
  • $\begingroup$ 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 $[0,1]$. $\endgroup$ – goblin 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.

FYI, start with the analytic approach, not the algebraic one.

  • $\begingroup$ Interesting, thank you for this direction! $\endgroup$ – 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 $n$.


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