# The probability that $\dfrac{p-1}2$ is square-free

Let $Q(x)$ denote the number of square-free integers between $1$ and $x$, we obtain the approximation

\eqalign{ &Q(x)\approx x\prod_{p\,{\rm prime}}\left(1-\dfrac1{p^2}\right)=x\prod_{p\,{\rm prime}}\dfrac{1}{\left(1-\tfrac1{p^2}\right)^{-1}} \\ &Q(x)\approx x\prod_{p\,{\rm prime}}\dfrac1{1+\tfrac1{p^2}+\tfrac1{p^4}+\cdots}=\dfrac x{\sum_{k=1}^\infty\tfrac1{k^2}}=\dfrac x{\zeta(2)}. }

Thus the asymptotic density of squarefree integers is $\dfrac{6}{\pi ^2}.$

Let $P(n)$ denote the number of odd primes $p$ among the first $n$ primes for which $\dfrac{p-1}2$ is square-free, see also A066651. I guess it may be true that $\dfrac{P(n)}{n}\approx \dfrac{Q(n)}{n} \approx \dfrac{1}{\zeta (2)}\approx 0.607927.$

I run a program to check my conjecture and get some results:

$$\begin{array}{c|lcr} n &\text{P(n)}& \text{P(n)/n} \\ \hline 10 & 7 & 0.7\\ 10^2 & 59 & 0.59\\ 10^3 & 567 & 0.567\\ 10^4 & 5604 & 0.5604\\ 10^5 & 56182 & 0.56182\\ 10^6 & 561104 & 0.561104\\ \end{array}$$

Now we can see that $\dfrac{P(n)}{n}$ is getting smaller when $n$ is getting greater (except for $n$ from $10^4$ to $10^5$), hence it's very likely that $$\lim_{n\to \infty}\dfrac{P(n)}{n}\lt \dfrac{6}{\pi^2}.$$

I'm not sure if this is another example of Strong Law of Small Numbers, so that $\dfrac{P(n)}{n}$ will $\to \dfrac{6}{\pi^2}$ when $n$ is big enough.

Hence, there are two problems:

• Does $\lim_{n\to \infty}\dfrac{P(n)}{n}$ exist?

• Is it true that $\lim_{n\to \infty}\dfrac{P(n)}{n}=\dfrac{6}{\pi^2}$?

Any tips or help is welcome, thanks in advance!

• For example, of the nine numbers n = 18k+1,3,5,7,9,11,13,15 and 17, there is one where (n-1)/2 is divisible by 9, but three can't be primes. So for primes, (n-1)/2 is divisible by 9 for one out of six instead of one out of 9. – gnasher729 May 24 '14 at 17:56

By Mirsky's result mentioned in this other question, the asymptotic density of primes $p$ such that $p-1$ is square free is $$\prod_p \left(1 - \frac{1}{p(p-1)} \right) = C_{\text{Artin}} = 0.3739558…$$ otherwise known as Artin's constant. To modify this to find the asymptotic density of primes $p$ such that $(p-1)/2$ is square free, we should slightly modify this formula to be: $$\left(1 - \frac{1}{2^2(2-1)} \right) \prod_{p \neq 2} \left(1 - \frac{1}{p(p-1)} \right) = \frac{3}{2}C_{\text{Artin}} = 0.5609337…$$ which matches your experimental results quite well!