# Given a number $k$(for large $k$) find $f(k)$ such that for all $x \geq f(k)$, $\phi(x) > k$

As the title says, my question is: Given a number $k$(for large $k$) find $f(k)$ such that for all $x \geq f(k)$, $\phi(x) > k$ where $\phi$ is the Euler totient function.

This question came up while I was proving that $\phi(x) = N$ has finitely many solutions. I noticed that as $x$ became larger and larger $\phi(x)$ also became larger so it must have finitely many solutions. But a thing to notice was that $\phi(x)$ also oscillates.

For example $\phi(89) = 88$ as $89$ is a prime, but $\phi(x) = 88$ is also true for $x = 178,230,276$. In this case the last $x$ such that $\phi(x) = 88$ is almost $2x$ apart from the first occurrence(that is $x = 89$).

As $k$(mentioned in the title) increases this interval between the first value to satisfy the equation and the last value keeps on increasing. So can the length of the interval be written as some $f(k)$ with suitable error terms?

In the Wikipedia article about the totient function you may find, among others, the following inequality: $$\phi(n)>\frac{n}{e^\gamma\log\log n+\frac{3}{\log\log n}}.$$