First, consider the number of divisors of a given number $m$. This is a product of numbers derived from the exponents of the prime factorization of $m$. The only way this is a prime is if $m$ is itself a prime to a power one less than some (likely different) prime. In general, think of the divisor lattice of $m$ as a parallelpiped of grid points, with the total number almost always composite.

Now consider throwing away a part of this block of divisors. Divisors $d$ of $m$ do not divide some other integer $f$ if and only if they do not divide a special divisor of $m$ which is $d'=\gcd(m,f)$. So out of our divisor block, we carve out a smaller block out of a corner, and consider how many are left.

If $d'$ and $m$ contain all the powers of a prime $p$ that divides $m$ (i.e. $p$ divides $m$ and $p$ does not divide $m/d'$), then the number of remaining divisors is composite. One way to see this is if $d$ divides $m$, $d$ does not divide $d'$ and $p$ does not divide $d$, then $dp^i$ also does not divide $d'$ for $i$ from 0 up to the appropriate power of $p$. (A situation where this fails is if $m=q^jp^i$ for a prime $q$, and then $d' has to be $m/q$. But the end point is to consider what usually happens, so this is like a measure 0 exception.)

Now to your situation. If $n$ is even, $n-1$ is odd and $\phi(n)$ has an odd divisor which may or may not relate to a divisor of $n-1$, and in general we do not expect $(n-1)/\gcd(n-1,\phi(n))$ to be coprime to any prime divisors of $n-1$. Not much to say without further analysis.

If $n$ is odd however, then both $n-1$ and $\phi(n-1)$ share some powers of $2$. I believe the effect you are seeing is when $\phi(n-1)$ has as many or more powers of $2$ than does $n-1$, which leads to the number of selected non-divisors of $\phi(n)$ being composite. $\phi(n)$ is divisible by 4 more often than $n$ is, for example.

Interesting data, but I don't see how much further you can take it, especially if you are looking at the conjecture $\phi(n)$ divides $n-1$ implies $n$ is prime.

Gerhard "But Enjoy The Journey Anyway" Paseman, 2016.08.25.

First, consider the number of divisors of a given number $m$. This is a product of numbers derived from the exponents of the prime factorization of $m$. The only way this is a prime is if $m$ is itself a prime to a power one less than some (likely different) prime. In general, think of the divisor lattice of $m$ as a parallelpiped of grid points, with the total number almost always composite.

Now consider throwing away a part of this block of divisors. Divisors $d$ of $m$ do not divide some other integer $f$ if and only if they do not divide a special divisor of $m$ which is $d'=\gcd(m,f)$. So out of our divisor block, we carve out a smaller block out of a corner, and consider how many are left.

If $d'$ and $m$ contain all the powers of a prime $p$ that divides $m$ (i.e. $p$ divides $m$ and $p$ does not divide $m/d'$), then the number of remaining divisors is composite. One way to see this is if $d$ divides $m$, $d$ does not divide $d'$ and $p$ does not divide $d$, then $dp^i$ also does not divide $d'$ for $i$ from 0 up to the appropriate power of $p$. (A situation where this fails and such a number is prime is if $m=q^jp^i$ for a prime $q$, and $i+1$ is prime, and then $d'$ has to be $m/q$. But the end point is to consider what usually happens, so this is like a measure 0 exception.)

Now to your situation. If $n$ is even, $n-1$ is odd and $\phi(n)$ has an odd divisor which may or may not relate to a divisor of $n-1$, and in general we do not expect $(n-1)/\gcd(n-1,\phi(n))$ to be coprime to any prime divisors of $n-1$. Not much to say without further analysis.

If $n$ is odd however, then both $n-1$ and $\phi(n-1)$ share some powers of $2$. I believe the effect you are seeing is when $\phi(n-1)$ has as many or more powers of $2$ than does $n-1$, which leads to the number of selected non-divisors of $\phi(n)$ being composite. $\phi(n)$ is divisible by 4 more often than $n$ is, for example.

Interesting data, but I don't see how much further you can take it, especially if you are looking at the conjecture $\phi(n)$ divides $n-1$ implies $n$ is prime.

Gerhard "But Enjoy The Journey Anyway" Paseman, 2016.08.25.