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.


[![enter image description here][1]][1]

  [1]: https://i.sstatic.net/JH3WX.jpg

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.