20211110, 05:09  #1 
"Sam"
Nov 2016
101001111_{2} Posts 
Mersenne prime exponents that are also Sophie Germain primes
So I stumbled on A065406 by curiosity, and somewhere in the comments said this list was expected to be finite.
Basically, the number of primes p such that 2*p+1 and 2^p1 are both prime is expected to be finite. We know that for p > 3, p = 1 mod 4. While it has been conjectured that the list of Mersenne exponents is infinite, it has also been conjectured that they behave in a specific manner. Specifically that the sequence of primes p such that 2^p1 is also prime (roughly) inhibit exponential growth. So (assuming this is the case), we have a sequence of (somewhat) random primes in which two successive terms (on average) tend to share a common ratio. Now consider a new sequence of odd integers 2*p+1. So essentially speaking, under the assumption that 2*p+1 behaves like a random number, and that the sequence (roughly) inhibits some form of exponential growth, the expected number of primes of the form 2*p+1 (where p is a Mersenne Prime exponent) less than n should be S = C*(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/n) where C is a constant. This observation is clearly noticeable in the conjectured infinitude of Mersenne Primes themselves. And as n goes to infinity, so does S because this series diverges. Thus, this should imply the number of primes p such that 2*p+1 and 2^p1 are both prime is infinite. To see an example of this, consider a sequence of known Mersenne Prime Exponents p: [2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941] now the sequence for 2*p+1 is [5, 7, 11, 15, 27, 35, 39, 63, 123, 179, 215, 255, 1043, 1215, 2559, 4407, 4563, 6435, 8507, 8847, 19379, 19883] Assuming each of these behave like a random odd integer, the expected number of primes is: 2/log(5) + 2/log(7) + 2/log(11) + ... + 2/log(19883) = 9.949. However, there are only 5. And out of 51 known Mersenne prime exponents, only 8 are Sophie Germain Primes. The modular restrictions on p can make such an occurrence even rarer than usual, however, it is only a matter of a constant that differs: The fact remains that such primes should still be infinite (provided the Mersenne Prime Exponents are). A similar argument could be applied to show there are an infinite number of twin primes p such that 2^p1 is also prime (A346645). Though interestingly enough, the expected number of twin primes (p and p+2) such that both 2^p1 and 2^(p+2)1 are prime, is finite. In short, the expected number of primes would be C*(1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/49 + ... + 1/n) Such a series converges, so it is reasonable to expect that such primes are finite. A twin prime pair (p, p+2) with both 2^p1 and 2^(p+2)1 being prime is pretty darn rare, but it's not the rarest occurrence that could happen (Fermat Primes would be MUCH rarer). Any thoughts or observations? My rationale was quite simple, I'm sure there are more sophisticated answers. 
20211110, 23:13  #2 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
111001100010_{2} Posts 
Conjectures:
* For any finite set {f_1, f_2, f_3, ..., f_m} of irreducible polynomial sequences f_i(n) = a_{i,0}+a_{i,1}*n+a_{i,2}*n^2+a_{i,3}*n^3+... (all a_{i,j} are (positive or negative) integers, the leading coefficient of all f_i(n) are positive integers) which does not have covering congruence, there are infinitely many n such that all f_i(n) are primes. (this is Schinzel's hypothesis H) * For any exponential sequence (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1)) which does not have covering set (covering congruence, full algebraic covering set, or partial algebraic/partial numerical covering set), there are infinitely many n such that (a*b^n+c)/gcd(a+c,b1) is prime. * For any polynomial sequence a_0+a_1*n+a_2*n^2+a_3*n^3+...+a_m*n^m (all a_i are (positive or negative) integers, the leading coefficient (a_m) is positive integer) and any exponential sequence (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1)), there are only finitely many n such that both a_0+a_1*n+a_2*n^2+a_3*n^3+...+a_m*n^m and (a*b^n+c)/gcd(a+c,b1) are primes. * For any two exponential sequences (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1)), there are only finitely many n such that both sequence are primes. References: prime number theorem https://oeis.org/wiki/User:Charles_R...special_primes https://docs.google.com/document/d/e...W67N_vn96J/pub Last fiddled with by sweety439 on 20211110 at 23:17 
20211111, 03:10  #4  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·7·263 Posts 
Quote:
Last fiddled with by sweety439 on 20211111 at 03:11 

20211112, 00:10  #5  
"Sam"
Nov 2016
5·67 Posts 
Quote:
The probability of Mp = 2^p1 being prime is C/p where C is a constant. By the prime number theorem, p = n log(n) for some number n. The comparable series C/(n log(n)) diverges as n goes to infinity, suggesting the infinitude of Mersenne Primes. Whereas if p is restricted to say a twin prime, or Sophie Germain prime, then by the Prime number Theorem, p = n log(n)^2, and the series is C/(n log(n)^2) which converges, so finitude is suggested instead. So now two different (reasonable) arguments for either side. Interesting. 

20221019, 01:44  #6  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×7×263 Posts 
Quote:


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