Can we deduce that there is infinitely many indices $n$ such that the period length of $1/(2^{2^n}+1)$ is strictly less then $2^{2^n}$.

Question

In this page (http://mathworld.wolfram.com/FermatPrime.html) we have the following result:

$2^{2^n}+1$ is a Fermat prime if and only if the period length of $1/(2^{2^n}+1)$ is equal to $2^{2^n}$. In other words, Fermat primes are full reptend primes.

Thus, we get:

$2^{2^n}+1$ is not a Fermat prime if and only if the period length of $1/(2^{2^n}+1)$ is strictly less then $2^{2^n}$. See: Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$

My question is: Can we deduce that there is infinitely many indices $n$ such that the period length of $1/(2^{2^n}+1)$ is strictly less then $2^{2^n}$.

Answer 1

Showing existence of infinitely many such $n$'s would imply infinitely many $n$'s for which Fermat numbers are composite, which is an open problem (and as suggested in comments, period of $1/n$ cannot be larger than $n-1$). See also Composite Fermat's numbers and Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$.