# On numbers with small $\varphi(n)/n$

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)?
• 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 $\;\Psi(n)\;$ . – DonAntonio Jan 11 at 12:04
• Done, thanks for the hint. – Hans Stricker Jan 11 at 12:10
• Even for odd $n$, the value can get arbitary small since the product $$\prod_{p\ prime} \frac{p-1}{p}$$ diverges to $0$ – Peter Jan 11 at 12:14
• What might help for a deeper analyze is that we can replace $n$ by its radical (the product of the primes dividing $n$) – Peter Jan 11 at 12:18
• 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. – Peter Jan 11 at 12:52

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