Consider the primitive roots of a prime $p$ in the range $1...p$ which are not primitive roots mod $p^2$. Let $n(p)$ be this number. While looking for an answer to this question, it seems that the number of primitive roots mod $p$ that are not primitive roots mod $p^2$ (call it $n(p)$), follows a Poisson distribution with $\lambda = -\log\log 2$, in the sense that the proportion of primes up to $x$ which have $n(p)= k$ seems to be near $$ \frac{e^{-\lambda}\lambda^k}{k!} $$
Here is un updated count up to $p=711539$:
observed expected
0 39 717 39 801
1 14 517 14 588
2 2 804 2 673
3 341 327
4 39 30
5 2 2
6 1 0
This seems reasonable to me by the following reason: if you pick $g$ a primitive root mod $p^2$, and $t$ is an integer coprime with $p$, then $g^t$ is a primitive root mod $p$, it will be also a primitive root mod $p^2$ if and only if $p \not|\, t$.
So if you consider the integers $g^{pt}$ ($t=1,\dots,p$, $(t,p-1)=1$), reduce them mod $p^2$ and count how many fall in the interval $(1,p)$ then you have the numbers we are looking for. If you think that $g^{pt}$ can fall "randomly" with equal probability in any set of $p$ incongruent integers mod $p^2$ then it is natural to expect this number to follow a Poisson distribution.
The parameter $\lambda$ of the Poisson distribution should be something like the mean value of $\phi(p-1)/p$. It seems that $\lambda = -\log \log 2$ works fine but it could be any number around.
My question is if there is a way to compute the real value of $\lambda$ (if it exists) and what might be it's real value, or a reason why $\lambda = -\log\log 2$ might or not be a reasonable guess. By the way computing the mean value of $\phi(p-1)/p$ seems to be something around $0.374\dots$ while $-\log\log 2 = 0.3665...$ but the approximation using this constant is worse than using $-\log\log 2$.
