Prime numbers primitive roots and $\Phi$?

Question

Let's $s$ be the sum of the primitive roots of a prime number $p$.

One observation is that $\dfrac{s+x}{p}$ is an integer when $x = \{-1, 0, 1\}$.

So one could categorize each $p$ in one of $3$ categories: $c_{-1}$, $c_0$ or $c_1$.

Let's call $r_0$ the ratio of number of $c_0$ and number of primes $\left(\text{i.e.}\,\dfrac{\text{number of $c_0$}}{\text{number of primes}}\right)$.

After observation of the $1650$ first primes, I am under the impression that $r_0$ converges by oscillating to $\dfrac{1}{\Phi}$, with $\Phi$ being $\dfrac{1+V_5}{2}$.

Is it true?

If yes:

1. Is it a well-known result?
2. Is there some documentation out there?
3. What about $r_{-1}$ & $r_1$? Do they also converge to a calculable value? So far, I have $r_{-1}=0.18612922347362182$ and $r_1=0.19594997022036928$.

I'm limited in my observation by using only a spreadsheet program and going till the $2000^{th}$ prime should be exhausting enough! Please excuse my bad English. Maybe I should mention also I am not a mathematics student or anything; I just like playing with numbers.

Examples:

1. Primitives roots of $29$ ($10^{th}$ prime) $= \{2,15,3,10,8,11,14,27,18,21,19,26\}$, sum $= 174$, and $\dfrac{174}{29} = 6 \implies 29$ is a $c_0$ prime.

2. Primitives roots of $43$ ($14^{th}$ prime) $= \{3,29,5,26,12,18,19,34,20,28,30,33\}$, sum $= 257$, and $\dfrac{257+1}{43} = 6 \implies 43$ is a $c_1$ prime.

3. Primitives roots of $11$ ($5^{th}$ prime) $= \{2,6,7,8\}$, sum $= 23$, and $\dfrac{23-1}{11} = 2 \implies 11$ is a $c_{-1}$ prime.

See the graph below for an illustration.

Answer 1

We claim that the sum of the primitive roots $\bmod p$ is equivalent to $\mu(p-1)\bmod p$, where $\mu$ is the Mobius function.

To prove this, we are going to do something similar to Mobius inversion. Letting $g$ be any primitive root $\bmod p$, we can represent the primitive roots $\bmod p$ as the set of values of $g^k$, where $\gcd(k,p-1)=1$. So, the sum of all of these is

$$\sum_{k=0}^{p-2} g^k[\gcd(k,p-1)=1],$$

where the Iverson bracket $[\gcd(k,p-1)=1]$ is $1$ if the condition is true, and $0$ otherwise. This can be expanded, using the property that

$$\sum_{d|n} \mu(d) = [n=1],$$

as

$$\sum_{k=0}^{p-2} g^k\sum_{d|k,\ d|p-1} \mu(d).$$

Switching the indices of summation and setting $k=dm,$

$$\sum_{d|p-1} \mu(d)\sum_{m=0}^{\left(\frac{p-1}{d}\right)-1} g^{md}.$$

The inside sum is a geometric series, and we can simply use the formula for the sum of one to get that it equals

$$\frac{g^{p-1}-1}{g^d-1}.$$

If the denominator is not $0\bmod p$, then this is $0\bmod p$ (as $g^{p-1}\equiv 1\bmod p$); so, the only term we don't need to neglect is that where $d=p-1$, at which point the sum if $1$ and the sum reduces to $\mu(p-1),$ finishing the proof.

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Edit: What follows is not terribly accurate numerical analysis - that of Peter in his answer is more accurate.

Now, it's probably pretty difficult to get exact distribution statistics on $\mu(p-1)$ as the connection between the factorizations of consecutive integers is not very well-hashed-out. However, we can get some heuristics based on the generic properties of the Mobius function:

First off, the distribution of $1$s should be the same as the distribution of $-1$s, so it suffices to find the distribution of $0$s. If $p\equiv 1\bmod 4$, then $p-1$ is not squarefree, so $\mu(p-1)=0$.

We know that the probability that $\mu(n)\neq0$ is $6/\pi^2$. However all of these numbers are not $0\bmod 4$, so the probability that a number that is $2\bmod 4$ (specifically, $p-1$) has $\mu(n)\neq0$ should be

$$\frac{4}{3}\left(\frac{6}{\pi^2}\right) = \frac{8}{\pi^2}.$$

Thus, the probability that a prime $p$ has $\mu(p-1)=0$ should be

$$\frac{1}{2}\left(1+1-\frac{8}{\pi^2}\right) = 1-\frac{4}{\pi^2},$$

and each of $\pm 1$ has probability $\frac{2}{\pi^2}$.

Answer 2

Based on Carl's criterion, the probability for residue $0$ is asymptotically

$$1-\prod_p (1-\frac{1}{\phi(p^2)})=1-\prod_p (1-\frac{1}{p(p-1)})=0.626044\cdots $$ where $p$ runs over the primes and $\phi(n)$ denotes the totient-function.

The reason is that for a large prime $q$, the probability that $q$ is not of the form $kp^2+1$, is $1-\frac{1}{\phi(p^2)}$ for every prime $p$ because $\phi(p^2)$ residues are possible modulo $p^2$

Hence the probability that $q$ has not the form $kp^2+1$ for any prime (hence is squarefree) is the product of $1-\frac{1}{\phi(p^2)}$ over the primes upto $\lceil \sqrt{q} \rceil$ if we assume indepence which should hold if $q$ is large enough.