# Why are primes of the form p^2 - 2 for prime p seemingly unusually likely to be factors of prime-exponent Mersenne numbers?

The sequence A049002 (primes of form $q^2 - 2$, where $q$ is prime) appears to contain a high proportion of elements that are factors of prime-exponent Mersenne numbers (see below). I wonder why?

Factors of Mersenne numbers must obey certain well-known properties (e.g., Wikipedia), but as a layperson I can't think of any obvious way to relate the above empirical result to those properties.

• 2, 7, 23, 47, 167, 359, 839, 1367, 1847, 2207, 3719, 5039, 7919, 10607, 11447, 16127, 17159, 19319, 29927, 36479, 44519, 49727, 54287, 57119, 66047, 85847, 97967, 113567, 128879, 177239, 196247, 201599, 218087, 241079, 273527, 292679, 323759, 344567, 368447, 426407, 458327, 516959, 528527, 537287, 552047, 563999, 573047, 579119, 635207, 657719, 744767, 776159, 1026167, 1042439, 1104599, 1142759, 1190279, 1329407, 1495727, 1515359, 1661519, 1682207, 1708247, 1745039, 1868687, 1985279, ...

Of the 66 elements of A049002 less than 2 million, 32 of them are factors of prime-exponent Mersenne numbers. Of these, 30 of them are of the form $2p+1$; the other two are of the form $2kp+1$ ($11447=2*59*97+1$ is a factor of $M_{97}$ and $1329407=2*13*51131+1$ is a factor of $M_{51131}$).

Since all elements but the first are odd, we can write $q^2 - 2 = 2n + 1$ for some integer $n$. Empirically over this limited dataset, it seems that $n$ is prime almost half the time, which seems surprisingly high; also, if $n=p$ is prime, then empirically over this dataset it seems that $2p+1$ is always a factor of $M_p$ (recall that $2p+1$ is always prime by the construction of the sequence).

In the general case of $p$ and $2p+1$ both being prime, it is not necessarily true that $2p+1$ is a factor of $M_p$ (counterexamples include $p$ = 29, 41, 53, 83, etc). However, for any prime $p$ such that $2p+1$ is an element of A049002, it seems that $2p+1$ is always a factor of $M_p$ (at least over the limited dataset).

Is there any interesting math here, or at least some way to prove something or derive a result?

1. If $p$ and $2p + 1$ are both prime (meaning that $p$ is a Sophie Germain prime), and $p$ is congruent to 3 (mod 4), then $2p + 1$ divides $2^p − 1$.