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 $S\_\phi(X) \leq (1 - \zeta(2)^{-1}) X$.



Density of numbers $n$ which are co-prime with their $\phi$-value
Oeis has this as sequence [A003277](https://oeis.org/A003277). Erdos proved in ["Some asymptotic formulas in number theory"](https://users.renyi.hu/~p_erdos/1948-11.pdf) that
$$S\_{\phi}(X)=(1+o(1))\frac{Xe^{-\gamma}}{\log\log\log X}$$


