TITLE: Does the sequence $x_0=8$ , $x_{n+1}=4x_n+1$ contain a prime?
QUESTION [5 upvotes]: 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 $n\le 50\ 000$

REPLY [13 votes]: 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 $>3$).