For an integer $n \ge 2$, let $\omega (n)$ denote the number of distinct prime divisors of $n$ and $\pi (n) $ be number of primes not exceeding $n$.
Let $a_1, \ldots, a_k$ be integers greater than $1$ and less than $n$ such that they are mutually relatively prime i.e. $\gcd(a_i, a_j) = 1$ for $i \ne j$ and $\gcd(a_i, n) = 1, \forall i = 1(1)k$; then of course any prime $p$ dividing one of the $a_i $'s cannot divide any other $a_j$ and not also $n$ and any such prime divisor of course does not exceed $n$ ; hence we can write $\omega(n) + \sum\omega(a_i)\le\pi(n)$.
It is also clear that $k \le \phi(n) - 1$; what can be the maximum possible value of $k$ depending on each $n$ i.e. what is the maximum number of integers greater than $1$ and less than a given $n$ that are relatively prime to $n$ and are also mutually co-prime? Can it ever reach $\phi(n) - 1$? When does the equality hold in $\omega(n) + \sum\omega(a_i)\le\pi(n)$?
In particular, when do we have $\omega(n-1) + \omega(n) = \pi(n)$ that is when does $p$ a prime $p \le n , \implies p|n$ , or $p|n-1 $? Is there anything about this in the known literature?