M Theory Lesson 226
What primes are of the form $2p + 1$ for $p$ prime? This is not an arithmetic progression, because the sequence of primes $p$ is not the ordinals. For the next prime, namely 7, we find that $2p + 1$ is not prime. Similarly, $2p + 1$ is not prime for $p = 13$, 17 or 19. The first five primes that work give values for $2p + 2$ of 6, 8, 12, 24 and 48.
Since all primes $p > 3$ are of the form $6n \pm 1$, for $2p + 1$ to be prime we require at least that $p = 3n$, which cannot be true for a prime, or $p = 3n - 1$ for some $n$. For example, $23 = 3 \times 8 - 1$ and $11 = 3 \times 4 - 1$. So now we are actually interested in primes in the arithmetic progression $3m + 2$. In this case, Dirichlet's theorem tells us that there are infinitely many such primes.
Note that $m$ must always be odd, but when do these larger $m$ correspond to primes $2p + 1 = 6m + 5$? Again, by Dirichlet's theorem, there are infinitely many primes in this sequence. With the handy table, one can quickly find more $p$ such that the truth values in dimension $p$ form a finite field of $2p + 1$ elements.