# Questions about cyclic numbers, repeating decimals, and full reptend primes

I have a few questions about cyclic numbers in base $$b$$ ($$b = 2$$ in particular). We are dealing here with primes $$p$$ such that the length of the period in the decimal (more precisely base $$b$$) representation of $$1/p$$ is equal to $$p-1$$. For instance, in base $$10$$, primes $$p=7, 17, 19, 23, 29, 47, 59, \cdots$$ have this property. In base $$2$$, $$p = 3, 5, 11, 13, 19, 29,\cdots$$ have this property.

My questions are:

• Does the period of $$1/p$$, for large $$p$$'s satisfying this property (having a period of length $$p-1$$), look random?
• Can you compute the first $$100,000$$ primes in base $$2$$ satisfying this property?

Final purpose

The final purpose of this exercise is to check whether a number such as $$e^{-1}$$ has its digits uniformly distributed, by approximating $$e^{-1}$$ with a sequence $$\{x_n\}$$ built as follows: $$x_n = b^n/q_n$$ where $$q_n$$ is the prime in question that minimizes the approximation error. Here $$b$$ is the base. The idea is that such a converging sequence (or sub-sequence) exists for a number such as $$e^{-1}$$ ($$x_n \rightarrow e^{-1})$$ but maybe not for a rational number. As $$e^{-1}$$ is approximated by rational numbers of increasing period $$n-1$$, and as long as the period look random, at the limit, we would conclude that the digits of $$e^{-1}$$ are uniformly distributed.

• Here are the first $10000$ such primes. What do you mean by "look random"? Jul 16, 2019 at 18:38
• Thanks, very useful! By "random", I mean as the period grows to infinity, the proportion of 0's, 1's, 2's and so on (in the period) is almost the same. Jul 16, 2019 at 19:52
• Heuristically your claim is true. I tested for various bases with the largest prime in the above list. Although I would have no idea how one could prove this. Jul 16, 2019 at 19:58
• I am wondering if the fact that $(n+1)^n / n^n \rightarrow e$ might help. Jul 16, 2019 at 23:54
• @VincentGranville Checking whether $e^{-1}$ , for example , has digits uniformly distributed, is almost surely utterly hopeless. No matter how precise we can calculate $e^{-1}$ , we do not even know whether the digits $0-9$ all appear infinite many times. Perhaps, only two digits appear infinite many times, noone knows. The situation won't be better in base $2$ or other bases. Jul 17, 2019 at 5:36