20151024, 17:52  #34  
Aug 2002
Buenos Aires, Argentina
31·43 Posts 
Quote:


20151024, 19:45  #35 
"Forget I exist"
Jul 2009
Dumbassville
10000010110001_{2} Posts 
Too late, but it's not really the newest idea. Or the fastest.edit never mind that pm was a test of my programming skills.
Last fiddled with by science_man_88 on 20151024 at 19:46 
20151024, 20:06  #36  
"Curtis"
Feb 2005
Riverside, CA
23×191 Posts 
Quote:
My educated guess is that your method would take longer than our lifetime to calculate an actual 1 million digit prime. Your claims so far amount to "I have a formula, but I don't know how complicated it is for big numbers. Computers are fast, so I just need a programmer to make my formula on a computer, and it'll turn out primes fast. Easy!" 

20151024, 20:06  #37 
"Rashid Naimi"
Oct 2015
Remote to Here/There
3·643 Posts 

20151025, 12:17  #38  
Aug 2002
Buenos Aires, Argentina
31·43 Posts 
Quote:


20151026, 00:12  #39  
"Rashid Naimi"
Oct 2015
Remote to Here/There
3611_{8} Posts 
Hi,
Since someone let the cat out of the bag (with good intentions no doubt) I might as well let it out properly and 1st hand. However, the following statement is likely to be overlooked by a few members and in order of not having to repeat myself, please forgive the text format: The following is not what I had in mind when I decided to tackle the onebillion+ prime challenge. That theorem and its associated approach is still safe and will only be disclosed to a party with the right computing credentials which would be willing to assist me in this endeavor. Anyone who is not pleased with that then, though. One more note: This is a first draft of the proof and I have not proofread it. So should you find an i not dotted or a t not crossed, then good for you. A real mathematician would not need a proof and would instantly see that the statement is true (I know there are a few of them around). Theorem 1: For the set A of n consecutive prime numbers staring from 2 and ending at Pn. Let: B⊂A & C=AB Let the number b be equal to multiples of anyintegerpower greaterthan 0 of alltheelements of B. Let the number c be equal to multiples of anyintegerpower greaterthan 0 of alltheelements of C. Then d = ±b±c is a prime number, ∀d: d < Pn2 Proof: d is not divisible by any elements of A, because (I) if there are any a such that: a ∈ A & (a ∈ B ⊕a ∈ B) &a∣d ⇒ ( d/a = m & m ∈ ℤ) Then d/a = m ⇒ ±b±c/a = m ⇒ b/a ± c/a = m (ii) ∵ (a ∈ B ⊕a ∈ C) ∴ (b/ a ∈ ℤ⊕ c/ a ∈ ℤ ) Since b/a ± c/a is an integer and either b/a or c/a is an integer, then the other has to be an integer since the sum of an integer number with a non integer number cannot be an integer. This conflicts with statement (ii) which states that either b/a or c/a is an integer exclusively. Thus statement (i) is false. Since d < Pn2 and d is not divisible by any prime number less than or equal to Pn , then d is a prime number. ∎ The following is the routine that I have forwarded to the 2 people who have offered to assist me. Quote:
https://www.physicsforums.com/thread...ormula.838617/ Thank you to all for your time and replies. Last fiddled with by a1call on 20151026 at 00:29 

20151026, 00:49  #40 
Aug 2002
Buenos Aires, Argentina
31·43 Posts 
Although the method generates primes as expected, the minuend and subtrahend are almost as big as in the previous version. For more interesting ranges, you should subtract extremely huge numbers and the timing would be worse than trial division.
You could use modular arithmetic in order to speed up your algorithm, but again this will be worse than trial division. Why? The union of the sets A and B must include all prime numbers below n. So you need to perform about n/log(n) multiplications mod prime greater than n^{2}. So the number of steps is the same as the worst case of trial division. Notice also that to find a prime p<n^{2} first you will need to compute a prime larger than n^{2} in order to use it as the modulus. So again this is not useful at all. 
20151026, 00:57  #41 
"Kieren"
Jul 2011
In My Own Galaxy!
97×103 Posts 
Is this Phase
Last fiddled with by kladner on 20151026 at 01:01 
20151026, 01:11  #42 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
so that wasn't just a test of my programming skills ? edit: is that better Batalov.
Last fiddled with by science_man_88 on 20151026 at 01:26 
20151026, 01:16  #43  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·7·163 Posts 
Quote:
Good luck with that. This universe is not equipped to store all primes up to 10^{1000}, is it? 

20151026, 01:22  #44 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·7·163 Posts 

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