These sequences must somewhere stop (in the sense of primality) , right?

Question

Suppose that we take some prime $p$ and form a sequence:

$$a_1=p,$$

$$a_2=2p+1,$$

$$a_3=2(2p+1)+1=4p+3,$$

$$\ldots$$

$$a_n=2^{n-1}p+2^{n-1}-1.$$

There is no prime $p$ for which this sequence consists only of primes, right?

What is the longest "run" of primes that you can find?

Answer 1

As @lulu points out, when $p \ge 3$, $\gcd(2,p) = 1$, so by Fermat's Little Theorem, $$a_p=2^{p-1}p+2^{p-1}-1$$ is a multiple of the prime number $p$.

To address the case $p = 2$, raised by @ArnaudMortier, by inspection, $$a_6=2^{6-1}(2)+2^{6-1}-1 = 64 + 32 - 1 = 95 = 5 \times 19,$$ so $(a_n)_n$ must contain composite numbers.