Mersenne Cousins (numbers primes)

Question

What are the prime numbers obtained by: $$2^{2^{2^{2^{2^{2^{\vdots^{2}}}}}}} + 9$$

The numbers are $~11~$ and $~13~$. I think I would have to apply Mersenne's cousins, but how?

Answer 1

Let $$a_1=2$$ $$a_{i+1}=2^{a_i}$$ I am assuming that you are trying to find all $i\in\mathbb N$ such that $a_i+9$ a prime number.

Evidently, $i=1,2$ satisfy the given condition. By induction, $4|a_i\forall i\geq 2$ and therefore, $$\forall i\geq 3 \exists m_i\in\mathbb N \ \ such\ \ that \ \ a_i=2^{4m_i}$$ Therefore, for all $i\geq 3, a_i$ is a power of $16$, and hence, have $6$ as the unit digit, showing that $$5|(a_i+9)\forall i\geq 3$$

Hence, $11$ and $13$ are the only primes of required type.