20071010, 00:37  #1 
Oct 2007
2×17 Posts 
An equation to generate all primes that uses 2 & 3
I've come up with this equation after working a long time with primes.
PRIME GENERATOR A +/ B= prime ex (2*3*5*711*13)=67 The numbers on the left are a series of all prime starting with 2 that go up any prime value in this case is 13 and any prime can appear more than once provided it only appears on one side fo the addition or subtraction. the other condition is that the number it produces in this case is 67 is always prime if (max. prime on left)^2 is greater than the number on the right. In this case 13^2 is greater than 67 therefore 67 is prime. Note the primes on the left can contain a larger prime not in sequence as in the example (3*5*72*29)=47 provided (max prime of sequence)^2 is greater than the number on the right. In this case 7^2 is greater than 47. THE PROOF The proof as why the number on the right is always prime is because it has a minimum of 2 potential factors if not prime. The primes on the left are present at least once in all the potential factor possibilities of the number on the right. Now if you factor out any potential odd factors from the primes on the left you now have a whole number on one side of the addition or subtraction and a nonwhole on the other side of the addition or subtraction, which gives you a nonwhole in the brackets on the left. You now have a nonwhole as a factor for every potential factor of the number on the right so therefore the number on the right must be prime. THE FISCHBACH CONJECTURE The Fischbach conjecture says that all primes can be generated by starting with 2 and 3 in the above equation and use the generated primes to further generate all possible primes. Here are the first 14 primes generated from 2 and 3 2+3=5 2*2+3=7 2*5+3=13 3*5+2=17 2*2*2*35=19 2*3*57=23 7*52*3=29 5*3*32*7=31 5*3*2+7=37 5*7+2*3=41 2*5*73*3*3=43 3*5*72*29=47 
20071010, 02:14  #2  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
2^{2}·11·97 Posts 
Quote:


20071010, 02:39  #3 
Oct 2007
2·17 Posts 
clarifying equation
In the example you have illustrated you have 2 and 5 on both sides
of the minus sign. In the defintion of the equation you can have one value of prime on only one side of the minus sign not on both. To change your example bit 2*2*2*53*7=19 which is prime. 
20071010, 03:52  #4 
"Ben"
Feb 2007
110111111010_{2} Posts 
Generate M45 for me and I'll say you have something...

20071010, 04:13  #5 
Oct 2007
42_{8} Posts 
unlimited computer power
When they develope unlimited computer power I"ll generate M45 with
this algorithm. 
20071010, 05:03  #6 
Oct 2007
2×17 Posts 
primes may run out
A lot of people on this forum are after the largest prime, I've got news for you, primes being a digital occurence can not be analysed with analog
equations as to if primes continue for infinity. I've developed a digital analysing system that tells you the exact number of primes from 0 to any number I've yet to finalize all the equations to come up with an exact number primes,it requires a lot of work and I don't have the time for it right now, but early results so that primes may run out, so maybe these searches are a waste of time. 
20071010, 06:04  #7 
"Ben"
Feb 2007
110111111010_{2} Posts 
What algorithm? I haven't seen anything that tells me systematically how to pick the primes on the left, or where to put the +/, other than possibly trial and error. In less than a second, I found 1000099999 to be prime using the sieve of erathosthenes. What is your formula for that?

20071010, 11:42  #8 
Jun 2003
7×167 Posts 

20071010, 12:03  #9  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AC_{16} Posts 
Quote:
Proof of infinite primes: http://en.wikipedia.org/wiki/Prime_n..._prime_numbers Basically, what it's saying is that if you take a set of primes (e.g. {2,3,5}), multiply them together and add 1 (e.g. 2*3*5+1=31) the resulting number is either prime (as with 31), or divisible by primes not in the original set (as with 3*5+1=16=2^{4}), meaning there is always another prime number. Quote:
EDIT: I think I found one. 211*223199=46854=2 x 3 ^ 2 x 19 x 137. 223^{2}=49729 which is higher than 46854. Also, 199, 211, and 223 are consecutive primes. Last fiddled with by MiniGeek on 20071010 at 12:15 

20071010, 13:01  #10 
Oct 2007
2×17 Posts 
To clarify my definition
The equation can contain any prime but it must contain a sequence of
primes from 2 to the prime just greater than ( the number generated)^.5 and if this condition is satisfied then the number generated is prime. 
20071010, 13:10  #11 
Oct 2007
100010_{2} Posts 
The algorithm
This is a prime generator not a prime tester. You have to randomly
arrange known primes in the equation to generate larger primes from the equation. 
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