Growth rate of $(p-1)/\phi(p-1)$ for large primes

Question

Was looking into things regarding primitive roots of primes, and wanted to know roughly what fraction of residues are primitive roots. This led me to consider $S_n= (p_n-1)/\phi(p_n-1)$. $S_n$ is the reciprocal of the primitive root fraction, chosen because "unbounded" is easier to think about than "not bounded away from zero". And since number theory isn't my forte, I decided to investigate numerically.

For a lower bound, numerics suggest $\liminf_{n\rightarrow\infty}S_n = 2$. I can easily show this follows from the conjecture that there are infinitely safe primes, but I'm not sure about whether the converse is true.

More interesting to me is the upper bound. There seems to be no upper bound to $S_n$, but it grows very slowly. It first exceeds $4$ at $p_{47} = 211$, $5$ at $p_{4568} = 43891$, and I gave up searching at $p_{856690} = 13123111$ where it first reaches $5.75$. It appears that $S_n = O(\ln p_n)$ and that this bound isn't tight, but it's hard to tell since my range is so small for such slow-growing functions. (Also, it amuses me that in experimental number theory, seven orders of magnitude is "too small".)

Anyways, I'm sure I'm not the first person to consider $S_n$, but I wanted to take a bit of a stab at it myself first. What results, if any, already exist about $S_n$?

Answer 1

To the best of my understanding there is nothing special about the fact that $p$ is a prime when considering $p-1.$ What you are observing is probably the result that $$ \varphi(n) <\frac{n}{e^\gamma \log \log n}, $$ for infinitely many $n$ but you are looking along the subsequence $n_i=p_i-1.$ See for example the Wikipedia page on the Euler totient function.

Answer 2

We have the following result by I. Katai: http://archive.numdam.org/item/CM_1968__19_4_278_0/

$$ \frac1{\pi(x)}\sum_{p\leq x, \ \phi(p-1)/(p-1)\leq y} 1 \rightarrow F(y) $$ where $\pi(x)$ is the number of primes up to $x$ and $F(y)$ is a continuous function. The result was originally stated with $p+1$. The argument can be suitably adjusted with $p-1$.

The method of the paper builds on to a result of Erdos who proved the existence of limiting distribution of arithmetic functions under certain conditions (listed in the paper as 1, 2, 3), and Bombieri-Vinogradov theorem.

On the other hand, we have Lemma 1 of P. J. Stephens 'An average result for Artin's conjecture': $$ \frac1{\pi(x)}\sum_{p\leq x} \frac{\phi(p-1)}{p-1}\rightarrow A $$ where $A$ is Artin's constant $$ A=\prod_p\left(1-\frac1{p(p-1)}\right) \approx 0.37395. $$

Combining the results together, we cannot have $F(y)=1$ if $y<A$. Also, for any $\epsilon>0$, we have an arithmetic progression of primes whose value $\phi(p-1)/(p-1)$ is less than $\epsilon$. The density function $F$ has the mean value $A$, and $F(y)>0$ if $y>0$.

Thanks to Jyrki Lahtonen connecting another post of the similar topic, I was able to google the following result by Halberstam and Rotkiewicz (a weaker form of their Lemma 5): http://matwbn.icm.edu.pl/ksiazki/aa/aa13/aa13124.pdf

Let $k$ be an odd positive integer. For sufficiently large $x$, there is a prime $p$ such that $$ p\equiv 1 \ \mathrm{mod} \ k, $$ $$ x\leq p< x\left( 1+\frac1{\log x}\right),$$ $$ c\left(1-\frac 4{x^{1/5}}\right)\leq \frac{\phi(p-1)}{p-1}\leq c, $$ $$ c=\frac{\phi(2k)}{2k}. $$

Therefore, by density of $\phi(k)/k$ on $[0,1]$, for any $c\in [0,1/2]$, there is a sequence of prime numbers satifying $\phi(p-1)/(p-1)\rightarrow c$.