20210919, 08:55  #1 
Sep 2021
3 Posts 
(Not a) Formula for prime with 10^8 and 10^9 digits
X=3(9^Y) +1
(10^X 1) ÷ 4.5 + ((10^X ) ×7)+5 While Y=108 or While Y=109 Result is prime with 100,000,000 digits or 1,000,000,000 digits ? 
20210919, 15:05  #2 
May 2018
C2_{16} Posts 
Is Y = 107 out of the spotlight?

20210919, 15:55  #3 
May 2018
2×97 Posts 
For Y = 1, n = 72,222,222,222,222,222,222,222,222,227 is a prime.
For Y = 2, n = 72,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222, 222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222, 222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222, 222,222,222,222,222,222,222,222,227 = 251 × 5,426,180,741 × ... is composite. For Y = 3, n = 72,222,...,222,227 = 1,322,011 × ... is composite. ... Last fiddled with by Dobri on 20210919 at 16:01 
20210919, 17:02  #4 
Sep 2021
3 Posts 
X=7(9^Y)+1
? 
20210919, 17:19  #5 
May 2018
2×97 Posts 
For Y = 1 and X=7(9^Y)+1, n = 72,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,227 is a prime.
For Y = 2 and X=7(9^Y)+1, n = 72,222,...,222,227 = 12,391 × 173,429 × 9,639,563 × 53,804,376,883 × 72,037,926,256,979 × ... is composite. For Y = 3 and X=7(9^Y)+1, n = 72,222,...,222,227 = 1,352,773 × ... is composite. ... Last fiddled with by Dobri on 20210919 at 17:24 
20210919, 18:05  #6 
"Curtis"
Feb 2005
Riverside, CA
3^{3}·5·37 Posts 

20210919, 20:48  #7 
Sep 2021
3 Posts 
Never claimed to be smart, but I hoped to be lucky. Also I don't have the hardware to factor the large numbers currently.
Thanks 
20210920, 08:12  #8 
"Oliver"
Sep 2017
Porta Westfalica, DE
1010001101_{2} Posts 
Everything what Dobri computed can be done on a Android phone from 2012. Yes, I tested it. So I assume you have the hardware, but you need to look up how to do these computations. Hint: Google Alpertron ECM.

20210920, 09:05  #9 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
11·571 Posts 
If a prime formula was really just that simple then there is a very strong possibility that it would have been discovered 400 years ago.
Today you could write a simple BASIC program to generate millions of formulae and hope to get "lucky" that one of them "works". I recommend you try it to see how easy it is to generate formulae. But I don't recommend you try posting all of them here and hoping others will verify/disprove all those formulae. That would be your job, to show it works, not our job to debunk endless lists of random formulae. Last fiddled with by retina on 20210920 at 09:05 
20210920, 13:27  #10  
Feb 2017
Nowhere
2^{3}×7×89 Posts 
Quote:
N = 2*(10^X  1)/9 + 7*10^X + 5 This formula can be expressed more compactly as follows: N = (65*10^X + 43)/9 1) Given a prime p, p = 2, 3, 5, 7, etc are there any values of X for which p divides N? 2) If so, characterize such X (this will be a congruence class) 3) Are there any such X of a particular form, say X = 3^(2*Y + 1) + 1? It is easy to see "by formula" that p = 2, 5, 13, and 43 can never divide N. I supply the following table for when p divides N: p = 2: impossible "by formula" p = 3: X == 0 (mod 3) p = 5: impossible "by formula" p = 7: X == 1 (mod 6) p = 11: X == 1 (mod 2) p = 13: impossible "by formula" p = 17: X == 11 (mod 16) Up to p = 17, there are no Y for which 3^(2*Y + 1) + 1 can satisfy the congruence condition for X. I am giving you the following homework assignment: A) If you don't know how to do (2), learn how. B) If you don't know how to do (3), learn how. C) Find the smallest prime p which can divide N = (65*10^X + 43)/9 if X = 3^(2*Y + 1) + 1, and the congruence condition on Y for which this p divides N. The kind of checking indicated above only involves modulo arithmetic to very small moduli. 

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