Let $\phi(x)$ be the Euler totient function. Is $\dfrac{n}{\phi(n)}$ bounded if $\phi(n) = \phi(n+1)$? My experimental data for $n \le 10^{13}$ suggests that:
Claim: If $\phi(n) = \phi(n+1)$ then, $$\lim \inf \dfrac{n}{\phi(n)} \ge 2
\text{, } \lim \sup \dfrac{n}{\phi(n)} < 3$$
Given below if the plot of all the $1,014$ solutions below $4.7 \times 10^9$
Related Question: Conjecture on the gap between integers having the same number of co-primes
Best Answer
For the lower bound, One of $n$ or $n+1$ is even.
If $n$ is even, then $\phi(n)/n\le 1/2$.
If $n+1$ is even, then $\phi(n+1)/(n+1)\le 1/2$.
Note that if $\phi(n)=\phi(n+1)$, then $$ \left|\frac{\phi(n)}n -\frac{\phi(n+1)}{n+1}\right| = \frac{\phi(n)}{n(n+1)}. $$
These prove the lower bound of the limit infimum.