Prime number expressible as $2^2+3^3+5^5+7^7+11^{11}+\cdots$ I found that $2^2+3^3=31$ and $31$ is a prime, also $2^2+3^3+5^5+7^7$ is a prime. After these the only prime I found is $2^2+3^3+5^5+7^7+11^{11}+\cdots+83^{83}+89^{89}$. And I've checked $p(n)$ up to $1009$, but I couldn't find another prime of the form 
$$
2^2+3^3+5^5+7^7+11^{11}+\cdots+p(n-1)^{p(n-1)}+p(n)^{p(n)}.
$$
Is there any other prime with that form?
 A: The standard heuristic for such things is this. 
Given a (growing) sequence $(a_n)_{n\in \mathbb N}$ where there is no apparent reason for a number in there to be prime or not consider (as the probability that a number of a certain rough size $N$ is prime is $(\log N)^{-1}$) 
$$\sum_{n=1}^{\infty} \frac{1}{ \log a_n} $$
If this converges there should be only finitely many primes in your sequence. 
Let us do this for you sequence. We have that $a_n \ge  p_n^{p_n}$ and as $p_n \sim n \log n $ let us estimate $a_n$ as   $(n \log n)^{n \log n}$ so $\log a_n$ is about $n  (\log n)^2 $ (I drop all lower order terms). 
The series $$\sum_{n =2}^{\infty} \frac{1}{ n (\log n)^2 } $$ 
converges, so we expect only finitely many primes in that sequence. 
It thus is conceivable you found all of them, but there is likely no way to prove this with current "technology." 
Note that there is also no proof yet that there are only finitely many Fermat primes, which should be rather easier since that sequence grows faster and is generally easier to handle.   
