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.

$$34,276,387$$ is a prime factor of $2\uparrow\uparrow 5+3\uparrow\uparrow 5$ ,found by Sheldon L
(See the link above).
I do not know a prime factor of $2\uparrow\uparrow 4+3\uparrow\uparrow 4$, but the number is very probably composite, and if it is prime, even a probable-prime
test will be out of reach.
