mersenneforum.org (Not a) Formula for prime with 10^8 and 10^9 digits
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 2021-09-19, 08:55 #1 jense 2185   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 ?
 2021-09-19, 15:05 #2 Dobri   "Καλός" May 2018 44010 Posts Is Y = 107 out of the spotlight?
 2021-09-19, 15:55 #3 Dobri   "Καλός" May 2018 23×5×11 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 2021-09-19 at 16:01
 2021-09-19, 17:02 #4 jense 2185   Sep 2021 112 Posts X=7(9^Y)+1 ?
 2021-09-19, 17:19 #5 Dobri   "Καλός" May 2018 6708 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 2021-09-19 at 17:24
2021-09-19, 18:05   #6
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

555110 Posts

Quote:
 Originally Posted by jense 2185 X=7(9^Y)+1 ?
How is it that you think you are smart enough to come up with a formula that turns out primes very often, but aren't smart enough to actually check if the (majority of) outputs are prime?

 2021-09-19, 20:48 #7 jense 2185   Sep 2021 316 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
 2021-09-20, 08:12 #8 kruoli     "Oliver" Sep 2017 Porta Westfalica, DE 122110 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.
2021-09-20, 09:05   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

664010 Posts

Quote:
 Originally Posted by jense 2185 ... I hoped to be lucky.
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 2021-09-20 at 09:05

2021-09-20, 13:27   #10
Dr Sardonicus

Feb 2017
Nowhere

2·11·277 Posts

Quote:
 Originally Posted by jense 2185 Never claimed to be smart, but I hoped to be lucky. Also I don't have the hardware to factor the large numbers currently.
Instead of just hoping to "be lucky," it would be a good idea to at least try to check your formulas for divisibility by small primes. Your formula is

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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