Checking primality for $2 \uparrow \uparrow n + 3 \uparrow \uparrow n$ Is there a clever way to test primality for 
$$2 \uparrow \uparrow n \quad + \quad 3 \uparrow \uparrow n$$
where $n \gt 3$? (Not surprisingly, I got stuck after that)
For $n \le 3$ we get:
$$
\begin{matrix}
2 \uparrow \uparrow 0 \quad + \quad 3 \uparrow \uparrow 0 & = &  1 + 1  & = &   2 & (\text{prime}) & (\text{...}) \\
2 \uparrow \uparrow 1 \quad + \quad 3 \uparrow \uparrow 1 & = &  2 + 3  & = &   5 & (\text{prime}) & (\text{yes}) \\
2 \uparrow \uparrow 2 \quad + \quad 3 \uparrow \uparrow 2 & = &  2^2 + 3^3  & = &   31 & (\text{prime}) & (\text{okay}) \\
2 \uparrow \uparrow 3 \quad + \quad 3 \uparrow \uparrow 3 & = &  2^{2^2} + 3^{3^3}  & = & 7625597485003 & (\text{prime}) & (\text{hmm...}) \\
\end{matrix}
$$
Example:
One simple way to rule out primality is to check if the last digit of the sum is $5$. Now we know that the last digit of $2^k$ and $3^k$ follow a pattern such that for $2^k$ with $k \equiv 0 \pmod{4}$ the last digit is $6$ and for $3^k$ with $k \equiv 3 \pmod{4}$ the last digit is $7$. From this we can state that for $n \gt 2$ the last digit of  $2 \uparrow \uparrow n$ will be $6$ and for $n \gt 1$ the last digit of $3 \uparrow \uparrow n$ will be $7$, which means we can't rule out primality this way.

For $n$ large enough ($n \ge 6$) ,$ 2\uparrow \uparrow n \equiv 2197 \pmod {4423}$ and $3 \uparrow \uparrow n \equiv 2226 \pmod {4423}$, and so $2 \uparrow \uparrow n + 3\uparrow \uparrow n$ is always a multiple of $4423$.
Assuming the limit of $a \uparrow \uparrow n$ modulo $p$ is "random" in $\Bbb F_p^*$, you can expect to get $2\uparrow \uparrow n +3\uparrow \uparrow n \to 0 \pmod p$ with probability about $1/p$.
The sum over all primes of $1/p$ diverges, so we can expect that there are infinitely many such primes. $4423$ is the smallest such prime.
So you are left with the cases $n=4$ and $n=5$ to inverstigate.
