Upper bound for Euler's totient function on composite Mersenne numbers

Question

Are there any good upper bounds for Euler's phi function on composite Mersenne numbers. That is, any good $f(n)$ such that $\phi(2^n-1) \leq f(n)$? It might be useful to know that $n \mid \phi(2^n-1)$ and $\phi(k) \leq k - k^{1/2}$ whenever $k$ is composite. Thanks!

Answer 1

A Mersenne number cannot be a perfect power, by Mihailescu's proof of Catalan's conjecture. Therefore any composite Mersenne number $k$ must have at least $2$ distinct prime factors $p$ and $q$, and so $$ \phi(k) \le k\bigg(1-\frac1p\bigg)\bigg(1-\frac1q\bigg) \le k\bigg(1-\frac1{\sqrt{pq}}\bigg)^2 \le k\bigg(1-\frac1{\sqrt{k}}\bigg)^2 = k-2\sqrt k+1. $$ (The middle inequality is because of the log-concavity of $f(x)=1-1/x$.)

On the other hand, I don't see any good reason why a Mersenne number $k$ might not be the product of two primes very close to $\sqrt k$, in which case the true size of $\phi(k)$ would be roughly $k-2\sqrt k$. So I doubt you'll be able to get any bound substantially better than the above.