20170221, 07:49  #1 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}·31 Posts 
Most Abundant form of Prime Numbers
Hi all,
From what I understand Mersenne primes are quite rare but relatively easy to prove deterministically to be prime. Ease of proof notwithstanding, what would be the most (or more) abundant form of Prime Numbers? Thank you in advance. 
20170221, 08:34  #2 
Jun 2003
2^{3}·607 Posts 
Loophole abuse
2n+1

20170221, 09:08  #3 
"Rashid Naimi"
Oct 2015
Remote to Here/There
11111000000_{2} Posts 
I would have thought something along the lines of:
m.n#(+)p, p>n Don't understand your title. ETA I think from a probabilistic point of view it can be argued that 2^n+1 is less likely to be prime than 2^n1, but not by very much. Last fiddled with by a1call on 20170221 at 09:30 
20170221, 12:12  #4 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
2n+1 covers all odd numbers so 2n+1 also covers all odd primes.

20170221, 13:52  #5  
Jun 2003
12F8_{16} Posts 
Quote:
Coming to 2^n+1. It isn't just "less" likely than 2^n1. It is whole lot less likely. We expect infinitely many Mersenne primes, but only finitely many 2^n+1. Because 2^n+1 can be prime only if n is itself a power of 2. These are the Fermat numbers. As to a good form  after sieving, all the forms are expected to yield similar number of primes (subject to size and sieve depth). Mersenne primes can be sieved deeper (due to special property of factor), but they grow fast. GFNs (b^2^n+1) can also be similarly sieved deeper, but they grow slowly. So this would be a good form. If we don't consider sieving, your form is a good candidate, since it gets sieve up to n for free. A better form would be m.X+/n.Y where X.Y = p# and X and Y are almost same size and gcd(m.X,n.Y)=1. 

20170221, 18:01  #6 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}×31 Posts 
Thank you gentlemen. It was 4:30 in the morning and I misread the post. The odd numbers is the correct answer to my under described question. What I am looking for is an algebraic integer form which has a high likelihood of evaluating to prime numbers for high integer values. Algorithms or forms that produce large PRPs and their probability of being prime would be of interest.

20170221, 18:28  #7  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
19·491 Posts 
Quote:


20170221, 22:22  #8  
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 
Quote:
if conditionals are counted as part of a form then specifically you can make 2m+1 be exactly the odd primes by saying 2m+1 such that m is not of form 2ij+i+j where there are conditions on i and j. https://en.wikipedia.org/wiki/Sieve_of_Sundaram Last fiddled with by science_man_88 on 20170221 at 22:33 

20170222, 05:59  #9 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}·31 Posts 
Thank you very much Batalov,
I have to be careful, what I wish for. Still studying into it. 
20170223, 01:17  #10 
∂^{2}ω=0
Sep 2002
República de California
2D57_{16} Posts 
The odd ones.

20170223, 02:41  #11 
Feb 2017
Nowhere
2^{4}×271 Posts 
For degree 1, a refinement of PNT (PNT for arithmetic progressions) says that, for a given n > 1, the phi(n) arithmetic progressions
n*x + b, with 0 < b < n and gcd(b, n) = 1 all contain, asymptotically, 1/phi(n) of the primes up to X as X increases without bound. AFAIK, the question of whether a given (monic, irreducible) polynomial in Z[x] without any intrinsic prime factors even assumes infinitely many prime values is unanswered for any polynomial of degree greater than 1. That said, in the case of quadratic polynomials particularly, one can sometimes exclude evaluations divisible by some small prime values, which may apparently increase the frequency of prime evaluations. The perennial champion is x^2 + x + 41, whose evaluations at integer x are always odd, and whose discriminant 163 is a quadratic nonresidue modulo every odd prime less than 41, a circumstance which excludes any evaluations divisible by any prime less than 41. As to expressions involving exponential functions, I offer 4 + (1)^n. Its evaluations are prime for all positive integers n. 
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