'Lines' in the Euler phi graph

Solution 1:

In the following $\mathbb{P}$ denotes the set of primes and $p\in\mathbb{P}$ a prime.

We know $\phi(p)=p-1$, so that $\{(p,p-1):p\in\mathbb{P}\}$ is contained in the graph of $\phi$. This set lies on the line $y=x-1$, and corresponds to the steepest line in the graph.

Fix $q\in\mathbb{N}$. Then $\phi(q\cdot p)=\phi(q)(p-1)$ if $p\not\mid q$. The set of points $\{(q\cdot p,\phi(q)(p-1)):p\not\mid q\}$ is contained in the graph of $\phi$ and lies on the line $$y=\phi(q)\Bigl(\frac{x}{q}-1\Bigr).$$ The other lines you see correspond to $q=2$, $3$ and $4$.

Solution 2:

Some of the "dotted lines" that you're looking at are actually points from lines close to each other, but yes there are some lines.

Most obviously, the upper bound is the line $y=x-1$, which is hit by all the primes.

Going along that same line of thought, think of $\phi(3p)$ for primes $p$ other than $3$. You can compute that $\phi(3p)=2\phi(p)$. This says that if you went and plotted all the values $3p$ for primes other than $3$, you would get the dotted line of points $(3p, 2\phi(p))$ lying on the line $y=2(\frac{x}{3}-1)$.

I'm not sure this can be developed into a complete answer about lines, but it certainly seems to paint a picture for the $n$ which are divisible by a prime $q$ but not by $q^2$. I would suspect that numbers which do not have prime factors with multiplicity 1 are the "stray" points.