What's the greatest common divisor of $\phi(n)$ and $n$, where $\phi(n)$ is the Euler Totient Function?

Question: Is there any formula for finding the $$\operatorname{gcd}(\phi(n), n)$$?

I'm not sure if this is a dumb question, but I couldn't find one myself and not on Wikipedia.

EDIT: To clarify what I'm trying to do:

I'm trying to solve another problem, where I have to plug the greatest common divisor into the Totient function again, and it would be fun if there was an expression for that so it maybe would simplify.

• OEIS sequence A009195. Mar 15, 2020 at 17:15
• @RobertIsrael Pardon me, I'm quite new to OEIS, the formula section leads to other sequences, does this mean the problem has only been partially solved or? Mar 15, 2020 at 17:31
• Depends on what you really want to do. What is wrong with just $\text{gcd}(\phi(n),n)$? Mar 15, 2020 at 17:43
• If there were a simpler expression the OEIS would probably have it. This is as simple as it gets unless I am very mistaken. Please edit your question to include your comment in the body of it. Mar 15, 2020 at 17:49
• @Peter Sorry I was in a hurry yesterday and on phone. Coincidentally, I'm trying to give this problem a try, which you have also worked on. Basically I'm just "doodling" around, trying to simplify the 3rd equation given in the problem and the term $\phi(n\phi(n))$ appeared, and since $n$ and $\phi(n)$ are not necessarily relatively prime, I used the formula $$\phi(nm)=\phi(n)\phi(m)\frac{d}{\phi(d)}$$ where $d$ is $\operatorname{gcd}(m,n)$. Mar 16, 2020 at 12:48

However we can do a little bit better than that if we know the factorization of the integer $$n \ = \ p_1^{k_1}...p_r^{k_r}$$
$$\phi(n) \ = \ p_1^{k_1}...p_r^{k_r}(\frac{p_1 - 1}{p_1})...(\frac{p_r - 1}{p_r}) \ = \ p_1^{k_1-1}...p_1^{k_r-1}(p_1-1)...(p_r-1)$$
$$\gcd(\phi(n), n)\ =\ \gcd (p_1^{k_1-1}...p_1^{k_r-1}(p_1-1)...(p_r-1),\ p_1^{k_1}...p_r^{k_r})\ =\\ p_1^{k_1-1}...p_1^{k_r-1}\gcd((p_1-1)...(p_r-1), p_1...p_r)$$