20151023, 22:46  #12  
Nov 2003
2^{6}×113 Posts 
Quote:


20151023, 23:18  #13  
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 
Quote:
However, I am perfectly willing to share the theorems along with their proofs to any largeintegercomputationsavvy entities that would be willing to assist me. 

20151023, 23:23  #14  
Aug 2002
Buenos Aires, Argentina
1333_{10} Posts 
Quote:
For polynomials this is very easy to see: suppose that f(n) = p where n is an integer, p a prime number and f() is not constant. Then f(n+kp) is always multiple of p. At most deg(f) values of k will make f(n+kp) equal to p, so there are infinite values for the integer k for which the value of that polynomial f(n+kp) is not prime. If you want someone to program your algorithm, this person should be completely sure that the algorithm is sound. He will not necessarily believe you, so you will need to publish it somewhere with peer review before the coding start. In general this problem does not exist, because the mathematician can also program the algorithms. Last fiddled with by alpertron on 20151023 at 23:51 

20151023, 23:46  #15 
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 
Hello alpertron,
I am aware of the proof for impossibility of regular as well as irregular polynomials generating prime numbers for all values of n. However there are no proofs that other formulas will fail as well. An example would be Mills' formula, which is generally accepted to generate primes if the constant can be calculated. To be clear my theorem has nothing to do with Mills' formula. My theorem is very similar to the example I posted above and you quoted(but can be made to diverge for large n). It obviously does work for n= 6 and n= 7 and it can be proven that it would work for any results less than 49 which happen to be 41 and 23. A quick proof would be that the two sides do not have a common divisor and thus their sum can not divide any of their prime factors which are all the positive primes less than or equal to 7. Thus any sum that is less than 7^{2} has to be prime. No primality testing is required. Last fiddled with by a1call on 20151024 at 00:03 
20151024, 00:10  #16 
Aug 2002
Buenos Aires, Argentina
535_{16} Posts 
With respect to Mills, it is clear that the constant was computed from prime numbers, so it cannot be used to generate prime numbers.
It is like saying that the sequence N(10^{n}) generates always prime numbers (where N(k) means the next prime after k). The sequence exists because of Bertrand's postulate. But it is not easily computable, especially in the billiondigit range. Your example does not generalize to higher primorials: for example 11*7*5*32^{n} = 1155  2^{n}. You get 11551024 > 121, so you get no primes using your method. Last fiddled with by alpertron on 20151024 at 00:29 
20151024, 00:30  #17  
"Rashid Naimi"
Oct 2015
Remote to Here/There
11110001011_{2} Posts 
Quote:
There are other formulas as well. Regardless my theorem as mentioned has nothing to do with Mills' formula. Your sum should be 131 and 131 is not less than 11^{2} and does not apply to the proof I made. Your examples fails to diverge to less than 121. But I already pointed that out before. The example is not my theorem but is similar. The difference is that my theorem can be made to diverge to less than the largest prime factor squared. Last fiddled with by a1call on 20151024 at 00:44 

20151024, 00:31  #18  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3×2,917 Posts 
Quote:


20151024, 00:42  #19 
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 
That is true and I planned to submit the theorems for publication. Until I came across the EFF prize. The official rules explicitly disqualify any theorems from winning the prize and require a specific prime. I find it idiotic to publish a thereon which will almost immediately enable someone with the right computing power to claim the EFF prize. Still if there are no takers I will go with the submit for publication option.
Last fiddled with by a1call on 20151024 at 00:43 
20151024, 00:45  #20  
Aug 2002
Buenos Aires, Argentina
2465_{8} Posts 
Quote:
Since your prime is bounded by n^{2}, in order to find the billiondigit prime number, you will need to find the difference of two numbers of about 10^{5*10^8}/log(10) digits. Forget it. Last fiddled with by alpertron on 20151024 at 00:54 

20151024, 00:54  #21  
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 
Quote:
The example that I provided is not only difficult to diverge, but virtually impossible. That is why I posted it on the open forum, since it will not make anyone claim any prizes. However, my theorem can result in a formula that will diverge to a value where the associated proof confirms the output number is prime. 

20151024, 00:57  #22 
"Rashid Naimi"
Oct 2015
Remote to Here/There
1,931 Posts 
My bad alpertron,
I meant converge. Sorry about that 
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