Lower Growth Rate of Euler Totient Function

Question

Let $\phi(x)$ denote Euler's Totient function. What is the slowest growing function $f(x)$ such that $$\phi(x)=f(x)$$ occurs infinitely often for integers $x≥1$?

Answer 1

Your request for equality makes this impossible to answer in your exact words. However, from Rosser and Schoenfeld (1962) we get $$ \varphi(n) \geq \frac{n}{e^\gamma \log \log n + \frac{2.50637}{\log \log n}} $$ where the constant $2.50637$ is chosen to give equality only at one number, the primorial $n= 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 = 223092870. $

Rosser and Schoenfeld asked whether, if the inequality sign is reversed, it might be possible to erase the 2.50637 term for infinitely many numbers. Jean-Louis Nicolas (1983) showed that this was possible, and that the question of saying "always" for primorials was equivalent to the Riemann Hypothesis. ALL of the numbers Nicolas uses are primorials. Primorials are the numbers with extremely low $\varphi,$ in exactly the same manner as Ramanujan's Superior Highly Composite Numbers are those with extremely large number of divisors. I gave the brief proof at Is the Euler phi function bounded below?

In symbols, Nicolas showed that there are infinitely many numbers, primorials, such that $$ \frac{n}{e^\gamma \log \log n + \frac{2.50637}{\log \log n}} < \varphi(n) < \frac{n}{e^\gamma \log \log n } $$

See references 33-35 in the Wikipedia article; R+S can be downloaded. The paper of Nicolas, in French, is item 84 HERE An interesting discussion in English HERE