If $n$ is composite then $\phi(n) < n-1$, hence there is at least one divisor $d$ of $n-1$ which does not divide $\phi(n)$. We call $d$ as the totient divisor of $n$. Trvially, if $n$ is prime then it has no totient divisor and if $n-1$ is prime then $n$ has exactly 1 totient divisor. The number of such integers $\le x$ is $\pi(x)$.

I counted how many integers $\le x$ have exactly $2,3,4,5, ...$ totient divisors. I observed nothing interesting. Then I counted how many even integers $\le x$ have exactly $2,3,4,5, ...$ totient divisors. I observed nothing interesting either. Finally I counted how many odd integers $\le x$ have exactly $2,3,4,5, ...$ totient divisors. I found something which looked interesting.

Let $T_{o}(n,x)$ be the number of odd integers $\le x$ which have $n$ totient divisors. I plotted the graph of $T_{o}(n,x)$ vs. $x$ for different values of $x$ and found a consistent pattern in them as shown below.

The red dots are the spikes and the green dots are the crests or local minima. We observe that every primes $>2$ appears on a green dot i.e. odd primes seem to appear only at the crests. This suggests odd numbers prefer to have a composite number of totient divisors i.e.somehow odd numbers do not like having a prime number of totient divisors.

Question: I see no obvious reason why $T_{o}(p,x) < T_{o}(p \pm 1,x)$, for a prime $p >2$. So I would like to ask what is the phenomenon that is driving primes to appear on the local minimas?

If this observation is true then we can claim that

Odd numbers prefer not to have a prime number of totient divisors.

Note: Every prime $> 2$ is green but the converse is not true. We have a crest at 25.