(Not a) Formula for prime with 10^8 and 10^9 digits

[jense 2185]

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

?

[Dobri]

Is Y = 107 out of the spotlight?

[Dobri]

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.

...

[jense 2185]

X=7(9^Y)+1
?

[Dobri]

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.

...

[VBCurtis]

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?

[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.

Thanks

[kruoli]

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.

[retina]

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.

[Dr Sardonicus]

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.