For all prime $p \ge 7$, the decimal representation of $1/p$ repeats. Let $l_p$ be the length of the period or the non-repeating part of the decimal representation of $1/p$. It can be shown that $l_p \le p-1$. Thus $\frac{l_p}{p-1}$ is a measure of the length of $l_p$ relative to the magnitude of $p$.
Definition: We define the relative magnitude of the length of the period of a prime $p$ as $\frac{l_p}{p-1}$.
Question 1: What is the limiting value of
$$ c_n = \frac{1}{n}\sum_{k = 1}^n\frac{l_{p_n}}{p_n-1} $$
Question 2: Is it true that
$$ \lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^n\frac{l_{p_n}}{p_n} = \lim_{n \to \infty}\frac{\sum_{k = 1}^n l_{p_n}}{\sum_{k = 1}^n p_n} = \prod_{p} \Big (1 - \frac{p}{p^3-1}\Big ) \approx 0.57596 $$
The plot of $c_n$ vs. $n$ for $n \le 2460000$ and for this value of, $c_n \approx 0.5763$.
Source code
p = 7
s1 = s2 = s3 = k = 0
step = target = 10^4
while True:
k = k + 1
d = divisors(p-1)
l = len(d)
i = 1
while i < l:
e = d[i]
if (10^e)%p == 1:
s1 = s1 + 1/e.n()
s2 = s2 + (e/(p-1)).n()
s3 = s3 + e
break
i = i + 1
if k >= target:
print(k, s1, s2/k, s3)
target = target + step
p = next_prime(p)
