# Disproving the existence of a specific infinite sequence of Fibonacci primes

Consider the following sequence:

$$T_{1} = a,\: T_{i+1} = F_{T_{i}}$$

where $a \in \mathbb{P}$ and $F_i$ is the $i$-th Fibonacci number.

Is there a value of $a \neq 5$ such that this sequence generates only primes?

I certainly don't expect an answer in the positive—as it would prove that there are infinitely many Fibonacci primes, which is an open problem—but I would like to know if there are any straightforward/obvious reasons for which such an $a$ wouldn't exist.

• The question seems to contain a typo. – Peter May 3 '18 at 8:41
• Not the formulation is the problem. I do not understand the sequence. – Peter May 3 '18 at 9:11
• I see @Peter; the intention is that the sequence begins by taking some prime number $a$, determines the next term as the a-th Fibonacci number $F_{a}$, and then uses that Fibonacci number as the index for the following term $F_{F_{a}}$ and so on. For example, taking $a = 7$, $T_{1} = 7$, $T_{2} = F_{7} = 13$, and $T_{3} = F_{13} = 233$. – aghostinthefigures May 3 '18 at 9:32
• OK, I understand. And you want to show that every such sequence must contain at least one composite number. – Peter May 3 '18 at 9:51
• Precisely, and I exclude $a = 5$ since that sequence would consist entirely of $5$'s as a result of $F_{5} = 5$. – aghostinthefigures May 3 '18 at 10:17