
Re: Combining Primes
Posted:
Aug 11, 2013 6:28 PM


On 11/08/2013 8:51 AM, Pubkeybreaker wrote: > On Saturday, August 10, 2013 11:23:49 AM UTC4, jim wrote: >> Is there any way to combine 2 primes to get a larger prime, either guarantying the primality of the result or with a quick test for the primality? Thanks, Jim > > In the hope of actually introducing some real mathematics into the discussion I > put forth the following:
I'm fine with "real mathematics" where there's a standard assumption/intuition of what the natural numbers or the real numbers (your "log" function mentioned below) be.
Let me then offer another analysis to Jim's question. Please note that I'll make some use of the definition of Major number M(x) and Minor number m(x), per the post:
https://groups.google.com/d/msg/sci.math/MMwHTdMFBt0/CZwOoo9ahb0J
> Let p,q, be prime. > > By strict computer science definitions, "quick test" means polynomial time. > The answer, in this case is yes. Consider N = 2k pq + 1, for k = 1,2,3,.... > By Cramer's conjecture, if N is prime, k is polynomially bounded in N, and > since the full factorization of N1 is known, one can apply the BaillieBrillhartLehmerSelfridgeWagstaff methods to prove primality in time O(log^3 N). QED.
[So, is Cramer's _conjecture proven_ ?]
Now, I'll define _two specific constant primes P1, P2_ below and would expect you or anyone to use your algorithm to clearly, as Jim put it, "get a larger prime".
======> Assumption 1: There are finitely many (nonzero) counter examples of Goldbach Conjecture.
Then let Pa = Pb be defined as the least prime number greater than this number of counter examples.
======> Assumption 2: There is _no_ counter example of Goldbach Conjecture.
For each even number e >= 4, there corresponds the leftmost prime Pe. And in this case of the Goldbach Conjecture being true, there are infinitely many distinct of such leftmost primes; let K = the set of those primes.
Let a real number X be defined as:
.Pe1Pe2Pe3....
where each Pei is in K and (i < j) => (Pei < Pej).
Then let the real number Y be defined as:
Y = M(x).
If Y is an irrational, then Pa = Pb be defined as the prime number next to the right of the number formed by the first googleplex numbers of digits of Y.
If Y is rational then Y = n1/n2 where n1, n2 are 2 natural numbers and n2 is the least possible number for Y. Then Pa = Pb be defined as the prime number next to the right of n2
======> Assumption 3: There are infinitely many counter examples of Goldbach Conjecture.
This is a similar case to "Assumption 2", except instead of "For each even number" we'd state "For each even counter example".
In summary, let our P1 be Pa and P2 be Pb (where Pa, Pb is defined in each of these 3 mutually exclusive assumption (which is a trichotomy).
If your algorithm work for any 2 general primes p, q, how would it work for these 2 _specific constants_ P1, P2?
How would you "QED" that?
  There is no remainder in the mathematics of infinity.
NYOGEN SENZAKI

