For each positive integer n we may define the convergent sum $$ s(n)=\sum\_{p}\frac{(n,p-1)}{p^2} $$
where the summation is over primes p and $(a,b)$ denotes the greatest common divisor of a,b.


It is immediate to deduce that s(n) is bounded on average: 


Using $\sum \limits\_{d|a, d|b}\phi(d)=(a,b)$ and inverting the order of summation we get


$s(n)=\sum\_{d|n}\phi(d) a\_d$ where $a\_d=\sum\_{p \equiv 1 (mod \ d)}p^{-2} $


Ignoring the fact that we sum over primes we get the bound $a\_d \ll \frac{1}{d^2}$ which leads to $$\sum\_{n \leq x}s(n) \ll x \sum\_{d \geq 1}\frac{\phi(d)a\_d}{d}=O(x) $$
and $$s(n) \ll \exp(\sum\_{p|n}\frac{1}{p}).$$ The last inequality means that $s(n)$ stays bounded if $\omega(n)$ is 
bounded. Towards the other direction, it seems fair to expect 
that $s(n)$ grows to infinity if $\omega(n)$ is large in some quantitative sense, say 
$\omega(n) \geq (1+\epsilon) \log \log n$. Taking into account that the contribution to the sum $s(n)$ of the primes $p$ that satisfy $(p-1,n) \leq \frac{p}{\log p}$ is bounded, since $\sum\_{p}\frac{1}{p \log p}$ converges, we see that $ s(n)=s'(n)+O(1)$ where $s'(n)=\sum\_{ (p-1,n)>\frac{p}{\log p}} \frac{(n,p-1)}{p^2}$ We are therefore led to the question as to whether a condition of the form $\frac{\omega(n)}{\log \log n}-1 \gg 1$ can guarantee that $s'(n) \to +\infty$ Are there any non-trivial techniques that can be used to answer this question ?



A question on the prime divisors of p-1
Your guess that $s(n)$ gets large if $\omega(n)$ is large is not correct.  

It is possible for $n$ to have many primes, and for $s(n)$ still to be small. 


This can be seen from some of the work in your question. As you note $s(n) =\sum\_{d|n} \phi(d) a\_d$ where $a\_d =\sum\_{p\equiv 1\pmod d} p^{-2} \ll 1/d^2$. Therefore 
$$ 
s(n) \ll \sum\_{d|n} \frac{1}{d} \le \prod\_{p|n} \Big(1-\frac 1p\Big)^{-1}. 
$$ 
If now every prime factor of $n$ exceeds $\log n$, then (since $\omega(n) \le \log n$ 
trivially) we have 
$$ 
s(n) \ll \Big(1-\frac{1}{\log n} \Big)^{-\log n} \ll 1.
$$ 


Thus $n$ can have about $\log n/\log \log n$ prime factors, all larger than $\log n$ 
and still $s(n)$ would be $\ll 1$. 


