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I recently found a different method to compute prime number in $\mathcal O(\log(\log n))$ complexity. At present, that logic working fine for $300$ digits prime number, which I found on websites.I need to validate whether that logic will be working fine for a higher number of digits. At present, I have computed a prime number of $300\ 000$ digits(but I am not sure whether this would be valid),

My questions are:

  • Where can I find a prime number of higher digits i.e., more than $300\ 000$ digits?
  • Where can I validate $300\ 000$ digit prime number is valid one?
Peter
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ideano1
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  • A method with this complexity would be sensational! Is $\log(\log(n))$ actually the complexity of the method ? – Peter Apr 04 '18 at 13:13
  • With PFGW , you can check your number – Peter Apr 04 '18 at 13:14
  • Find a new prime of the form given in this question : https://math.stackexchange.com/questions/2635516/a-conjecture-about-numbers-of-the-form-10m2k%e2%88%9212k-1%e2%88%921-where-m-is/2636195#2636195 – Peter Apr 04 '18 at 13:23
  • Or first check the ones that have already been calculated. Not the number of digits you want, but already quite large. – Peter Apr 04 '18 at 13:29
  • Finding a prime of that magnitude is a big task! If you actually found a (new) prime of this magnitude, no matter how you did it, congratulations! – Peter Apr 04 '18 at 13:32
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    I hate to sound overly skeptical, but just printing $n$ out has complexity $O(\log n)$. How is your algorithm outputting the candidate prime numbers? – Jyrki Lahtonen Apr 04 '18 at 13:45
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    Oh, it's actually not clear to me whether you claim to have an algorithm that produces large primes, or whether the algorithm tests a given integer for primeness. Could you please clarify? – Jyrki Lahtonen Apr 04 '18 at 13:49
  • My algorithm is not producing large primes, but it can validate whether number is prime or not. But now, i have to verify whether that logic works for larger numbers. For now, i validating number of 300277 digits as "NOT A PRIME" and its divisor is 11 .. Similarly, i have to verify this output is correct or not – ideano1 Apr 05 '18 at 03:21
  • You claim your algorithm works, yet you need to "validate it" for large numbers. Something is amiss here - the algorithm either works and is proven or it does not. This sounds like hundreds of other papers on Vixra that claim to have found something significant yet fail to provide any proof beyond philosophical argumentation. – Klangen Apr 05 '18 at 06:10
  • @Pickle : Actually, i am a software developer from programming background, i dont have much knowledge in maths. In case, if I will find largest prime number, how can claim that it is the largest prime number .? That is my question... Do we need to prove it mathematically that it is largest prime number or just submitting the new prime number with the council who can approve .? Kindly help me on this – ideano1 Apr 05 '18 at 06:49
  • @ideano1 The largest prime number currently known is $2^{77,232,917} − 1$, a number with $23,249,425$ digits. In comparison, your $3\times 10^{6}$ digits are quite pale. – Klangen Apr 05 '18 at 07:59

3 Answers3

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Check it with a table of Mersenne primes (e.g. http://oeis.org/A000043) or use provable primes (e.g. Maurer's method, see Alg. 4.62 in the Handbook of Applied Cryptography).

gammatester
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Primo does this. It only runs on Linux, if you don't have one at home you can just create a virtual machine and run it as guest within your current OS. There are various tests that Primo can perform, please read the documentation on the website in order to adapt it to your special case.

Please note that it will take very long to certify that a $3\times10^{6}$-digit number is prime. If you want to test your method on large numbers, take a much smaller one to begin with, i.e., with about $10^{3}$ or $10^{4}$ digits.

Klangen
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  • Thanks for your info. but as i have primo only supports 40000 digits, In the release document they have mentioned that as follows,

    v4.3.0 (February 21, 2018) Maximal size of candidates increased up to 132,928 bits (~ 40,000 decimal digits).

     – ideano1 Apr 05 '18 at 03:16
  • @ideano1 Oh, right, I didn't see that. I guess I never needed to go beyond 40k digits before... – Klangen Apr 05 '18 at 06:11
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From what you write I understand that you want to prove that your method is fine. It probably is if you checked up to 300, digits. But the only way to validate a method is by analysing it step by step and actually prove that it works. No matter how many digits you try, that will not be a proof that your method has no flaws.

Arnaud Mortier
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    A method with this complexity would be fantastic , if it actually works in most cases. Too nice that I can believe it. – Peter Apr 04 '18 at 13:27