# Does the sequence $x_0=8$ , $x_{n+1}=4x_n+1$ contain a prime?

Inspired by this question : Does every sequence $4^nx_0+\frac{4^n-1}{3}$ contain a prime?

If $x_0=21$, the sequence $x_{n+1}=4x_n+1$ produces no prime as shown in OEIS. What about the case $x_0=8$ ?

Is $4^n\cdot 8+\frac{4^n-1}{3}$ prime for any nonnegative integer $n$ ? Or equivalent : does the sequence $x_0=8$ , $x_{n+1}=4x_n+1$ produce a prime ?

I did not find a prime for $n\le 50\ 000$

• For the casual lurker: Just provided an answer to the more general question to which you refer: math.stackexchange.com/a/4351868/1714 It contains a list of additional cases $x_0$, where the sequences are proven to be primefree, using another argument than in PaoloLeonetti's answer. Commented Jan 8, 2022 at 20:06

No: $$4^n\cdot 8+\frac{4^n-1}{3}=\frac{(5\cdot 2^n+1)(5\cdot 2^n-1)}{3}.$$ (both factors at the numerator are $>3$).
• ..which can be extended to all $x_0$ for which $3x_0+1$ is a square Commented Mar 25, 2018 at 18:40
• Am I right in thinking "both factors are $>1$" is sufficient to prove it compound? Commented Mar 25, 2018 at 19:23
• I do not think so: the question is of the type whether $a\cdot 2^n+b$ with $a,b$ coprime odd integers is prime (or not). As particular instances, there are two of the major open problems in number theory: establish whether Mersenne primes $2^n-1$ and Fermat primes $2^n+1$ are finite or not; however, they seem to be out or reach at the moment.. Commented Mar 25, 2018 at 20:10