2020-04-15, 10:28 | #1 |
"Karl-Heinz Hofmann"
Jan 2017
Gr.-Rohrheim Germany
2^{3}·3 Posts |
The spread of low k´s (1 up to 13) of Mersennes
I have made a statistic of the spread of low k`s.
I see that the low k´s are getting rare, when the exponents are getting bigger. |
2020-04-15, 11:12 | #2 |
Einyen
Dec 2003
Denmark
2,969 Posts |
And every k=12*c is more common than others:
https://mersenneforum.org/showpost.p...74&postcount=3 mersenne/mersennefactors1.png Last fiddled with by ATH on 2020-04-15 at 11:14 |
2020-09-13, 13:39 | #3 |
Feb 2017
Nowhere
6773_{8} Posts |
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 Bateman-Horn 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|(q-1), the finite field F_{q} contains q q-th 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 2k-th power residue of 1/k of them. If we assume this proportion holds under the restriction that (q-1)/(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/(l-1) 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 2020-09-13 at 13:41 Reason: grammar errors |
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