Let $\Phi(n) = \varphi(n)/n = \prod_{p|n}(p-1)/p$ be the "normalized totient" of $n$.
Some facts:
$\Phi(p) = (p-1)/p < 1$ for prime numbers with $\lim_{p\rightarrow \infty}\Phi(p) = 1$
$\Phi(n) = 1/2$ iff $n$ is a power of $2$
$\Phi(n) < 1/2$ for all even $n$ that are not powers of $2$ and some odd $n$
if $\Phi(n) > 1/2$ then $n$ is odd
I have some questions concerning numbers with $\Phi(n) < 1/2$:
Are there numbers with arbitrary small $\Phi(n)$? Or is there a lower bound $\Phi_{\text{min}} > 0$?
Are there odd numbers with arbitrary small $\Phi(n)$?
How can this astonishing regularity been explained when displaying in a square spiral only those numbers with $\Phi(n) < 1/3$ – a regular pattern of triples pointing right, down, left, up clockwise (with some irregularily distributed defects of course):
Note that the regular background pattern vanishes when choosing values other than 1/3, e.g. 0.3 (left) or 0.4 (right):
Since the cases $\Phi(n) < 1/2$ and $\Phi(n) < 1/3$ display regular patterns, one might suspect that also $\Phi(n) < 1/5$ gives rise to some regularity. But the numbers envolved in creating that pattern are too big, so I cannot visualize it.
- Supposed one would visualize $\Phi(n) < 1/5$ which regular pattern would emerge (if any)?
The answer seems simple: Having a closer look at the numbers in the spiral reveals that most of them are - not surprisingly - multiples of $6 = 2\cdot 3$:
And the spiral forces the multiples of $6$ to arrange in triples (except along the lower right diagonal):
But not all multiples of $6$ have $\Phi(n) < 1/3$, e.g. $n = 2^k\cdot 3$, and not all $n$ with $\Phi(n) < 1/3$ are multiples of $6$, the smallest one being $770 = 2\cdot 5\cdot 7 \cdot 11$.
To answer partly the last of my questions: This is how the integers divisible by 8 and 10 are distributed (the second picture giving the blueprint for the case $\Phi(n) < 1/5$):
Note that in the right picture the "arrows" (5-tuples) go counter-clockwise.