Some one know something about the prime numbers of the form $p = 2^k + 2^n +1$, I would like to know if is it possible give conditions to $n$ and $k$ for $p$ will be a prime.

1$\begingroup$ oeis.org/A081091 $\endgroup$– CharlesJun 10, 2012 at 6:38

$\begingroup$ At least one of k and n should be odd. You might find it of interest to rule out divisibility by small primes. Gerhard "Ask Me About System Design" Paseman, 2012.06.09 $\endgroup$– Gerhard PasemanJun 10, 2012 at 6:49

$\begingroup$ Too vague and no motivation, vote to close. $\endgroup$– user9072Jun 10, 2012 at 8:02
1 Answer
Hello,
For each odd prime $q$, there should be conditions on $n,k$ mod (the order of $2$ mod $q$) that are necessary conditions for $q$ to not divide $2^n+2^k+1$. For $q=3$, the idea is implied in @Charles' comment on @joro's answer to this question.
For example, for $q=3$, $n,k$ should not both be even. For $q=5$, $n,k$ should not both be $1 \bmod 4$, and it should not be the case that one of $n,k$ is $3\bmod 4$ while the other is $0 \bmod 4$. For $q=7$, it should not be the case that one of $n,k$ is $1 \bmod 3$ while the other is $2 \bmod 3$. For $q=11$, it should not be the case that one of $n,k$ is $1 \bmod 10$ while the other is $3 \bmod 10$, and it should not be the case that one of $n,k$ is $6 \bmod 10$ while the other is $0 \bmod 10$, and it should not be the case that one of $n,k$ is $7 \bmod 10$ while the other is $8 \bmod 10$, and it should not be the case that one of $n,k$ is $2 \bmod 10$ while the other is $9 \bmod 10$, and it should not be the case that $n,k$ are both $4 \bmod 10$.
In general, if Artin's conjecture on primitive roots is true, there are infinitely many necessary conditions; one gets necessary conditions for each such $q$ where $2^{(q1)/2} \equiv 1 \bmod q$, because this translates to $q2$ values that $(n,k)$ cannot take in $(\mathbb{Z}/(q1)\mathbb{Z})^2$, one of which is the condition that $n,k$ should not both be $(q3)/2 \bmod q1$.