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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):

enter image description here

Note that the regular background pattern vanishes when choosing values other than 1/3, e.g. 0.3 (left) or 0.4 (right):

enter image description here enter image description here

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)?
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    $\begingroup$ Perhaps you should change the symbol for that normalization, as it is easy, I think, to confuse with Euler's function. Perhaps something like $\;\Phi(n)\;$ or perhaps even $\;\Psi(n)\;$ . $\endgroup$
    – DonAntonio
    Commented Jan 11, 2019 at 12:04
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    $\begingroup$ Done, thanks for the hint. $\endgroup$
    – Hans-Peter Stricker
    Commented Jan 11, 2019 at 12:10
  • $\begingroup$ Even for odd $n$, the value can get arbitary small since the product $$\prod_{p\ prime} \frac{p-1}{p}$$ diverges to $0$ $\endgroup$
    – Peter
    Commented Jan 11, 2019 at 12:14
  • $\begingroup$ What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$) $\endgroup$
    – Peter
    Commented Jan 11, 2019 at 12:18
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    $\begingroup$ Sorry, you are right. We have $$\prod_{p\ prime, p\le x} \frac{p-1}{p}\approx \frac{e^{-\gamma}}{\ln(x)}$$ where $\gamma$ is the Euler-Mascheroni-constant. $\endgroup$
    – Peter
    Commented Jan 11, 2019 at 12:52

1 Answer 1

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The answer seems simple: Having a closer look at the numbers in the spiral reveals that most of them are - not surprisingly - multiples of $6 = 2\cdot 3$:

enter image description here

And the spiral forces the multiples of $6$ to arrange in triples (except along the lower right diagonal):

enter image description here

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 $\Phi(n) < 1/5$):

enter image description here enter image description here

Note that in the right picture the "arrows" (5-tuples) go counter-clockwise.

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