Let $n$ be a positive integer. The Euler $\phi$-function is defined by
$$\displaystyle \phi(n) = \# \{1 \leq a \leq n-1 : \gcd(a,n) = 1\}.$$
It is in fact a multiplicative function, and one has the formula
$$\displaystyle \phi(n) = n \prod_{p | n} \left(1 - \frac{1}{p}\right).$$
Let $S_\phi(X)$ be the number of $1 \leq n \leq X$such that $\gcd(n,\phi(n)) = 1$. Does $S_\phi(X)$ have an asymptotic formula?
It is easy to see that for any prime $p$ we have $\gcd(p, \phi(p)) = 1$, so
$$\displaystyle S_\phi(X) \gg X (\log X)^{-1},$$
but whenever $n$ is not square-free, we have $\gcd(n, \phi(n)) > 1$. Therefore $S_\phi(X) \leq (1 - \zeta(2)^{-1}) X$.