5
$\begingroup$

The full title for this question should be On the many patterns of Euler totient function manipulations and their tendency towards symmetry at primorials, (but didn't want to take up too much room). For the rest of the question, $\#_n$ will represent the $n$th primorial.

Question / observations

A plot of $n/\phi(n)$ for $1\leq n\leq \#_4$ groups each $n$ that has an identical $\phi(n)$ value:

enter image description here

It can be seen by drawing reflected lines in $\#_4/2$ there is a tendency towards a symmetry. For the cases where $|n/\phi(n)-(\#_4-n)/\phi(\#_4-n)|=0$ where $n$ is prime are clearly the additive Goldbach prime partitions of $\#_4.$

enter image description here

The tendency for all $|n/\phi(n)-(\#_k-n)/\phi(\#_k-n)|\rightarrow 0$ for some $k$ shows a tendency towards symmetry:

enter image description here

At the primorials, for prime values only of $|n/\phi(n)-(\#_k-n)/\phi(\#_k-n)|$ looks like this for $k=5,$ surprisingly regular in comparison to non-primmorial multiples (right):

Similarly, limiting $n$ to values of all equal $\phi(N)$ where $N$ is a primorial results in clear patterns (left), whereas for non-primorial multiples, a completely diferent behaviour is exhibited (right):

Presumably, many of these patterns have simple explanations, but is something deeper going on here? I can't explain many of the patterns occurring here - in particular, the striking differences between primorials and nion-primorials. Can someone help to explain what is going on in some of these images?

MMA code for above plots

$\endgroup$
4
$\begingroup$

If $n = \prod_k p_k^{e_k}$ is the prime factorization of $n$, then $\varphi(n) = \prod_k (p_k - 1) p_k^{e_k - 1}$, and hence

$$\frac{\varphi(n)}{n} = \prod_k \frac{p_k - 1}{p_k} = \prod_{p \mid n} \left( 1 - \frac{1}{p} \right).$$

In particular, this quotient only depends on which primes appear in the prime factorization of $n$ (and not on their multiplicities). Primorials keep getting divisible by more and more primes so this function behaves quite nicely for them, but regular ol' numbers alternate between being divisible and not being divisible by various primes so there's more variation there.

This identity has a nice probabilistic interpretation: $\frac{\varphi(n)}{n}$ is the probability that a random number between $0$ and $n-1$ is relatively prime to $n$, and the RHS says that a number is relatively prime to $n$ iff it shares no prime factors with $n$, and furthermore that each of these events is independent.

$\endgroup$
  • $\begingroup$ thanks for the answer (that was quick!) goes some way to explaining the regularity I am seeing - thanks :) $\endgroup$ – martin Nov 18 '15 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.