I'd been meaning to look at this for a while. I finally got around to it.
In the present situation, we want p and q = 2*k*p + 1 both to be prime. The BatemanHorn Conjecture says there is a constant C = C(k) such that the number of such p up to X is asymptotically
C*x/log^{2}(X)
We also require q == 1 or 7 (mod 8). This introduces a "fudge" factor, giving
C*x/log^{2}(X), for k == 0 (mod 4),
(1/2)*C*x/log^{2}(X), for k == 1 (mod 4),
0, for k == 2 (mod 4), and
(1/2)*C*x/log^{2}(X), for k == 3 (mod 4).
The restriction q == 1 or 7 (mod 8) means that 2 is a quadratic residue (mod q). Also, since k(q1), the finite field F_{q} contains q qth roots of unity. We would expect that, among primes q having 2 as a quadratic residue, and q == 1 (mod k), 2 would be a 2kth power residue of 1/k of them.
If we assume this proportion holds under the restriction that (q1)/(2*k) is prime, we obtain a conjectural formula for the number of q up to X for which q = 2*k*p + 1 is a prime which divides M_{p}. Obviously, the factor 1/k reduces the constant multiplier if k > 1.
The formula for C is a bit complicated to go into here, but there is one aspect that may pertain to the value k = 12 being "favored."
If l is a prime not dividing 2*k, about 1/(l1) of the primes p will be in the residue class (2*k)^{1} (mod l); l divides q = 2*k*p + 1 for these p. Thus, if 3 does not divide k, we can expect about half the values of 2*k*p + 1 to be divisible by 3. If 3 divides k, however, this possibility is eliminated.
The value k = 12 is thus thrice blessed  it is divisible by 4, making the "fudge factor" as large as possible; the factor 1/k is not terribly small; and, the factor 3 eliminates the possibility of q being divisible by 3, which helps make the constant C bigger than it otherwise would be.
Someone more ambitious than I am might like to compute values of C for some given k's, and perhaps check how well the assumed factor 1/k fits the data.
Last fiddled with by Dr Sardonicus on 20200913 at 13:41
Reason: grammar errors
