loop for finding possible odd divisors

[scienceman88]

I may have print errors but if It's accurate 129704461, is prime so 2^129704461 could be prime if we had a part that figures out which exponents will result in a prime ( a.k.a primes but not all primes but more specific) then we can find all mersenne primes under (2^p) where p is prime and less than the limits of computing in C.

[scienceman88]

hence with enough room to store the last prime found in a notation like (2^p)-1 with a 64 bit system we could find mersenne primes up to (2^18 quintillion+)-1.

[scienceman88]

1,837,612,001 biggest prime before my files ran out of room by the looks of it pretty good but it would be cool to be able to use a pattern to figure out is (2^1,837,612,00 1)-1 prime

[JosAH]

1,837,612,001 biggest prime before my files ran out of room by the looks of it pretty good but it would be cool to be able to use a pattern to figure out is (2^1,837,612,00 1)-1 prime

Yep, that's the difficult part and until now they have been determining whether or not that (huge) number is a prime number by 'classical' means, i.e. check all proper divisors etc. Despite the fact that these numbers have a very special structure (they are all 1s in binary notation) no results are known (yet?).

kind regards,

Jos

[scienceman88]

Okay I've got a weird idea normal primes are fun but what if I could come up with a pattern in one of the types that if accurate will land us in the record books( might need to write massive strings instead of numbers though). I think I see a pattern in the double mersenne primes(might have to use every computer in the world to confirm the next one as prime).

[scienceman88]

if you look at wikipedia the double mersenne primes go:

7, 127, 2147483647, 170141183460469 231731687303715 884105727
first all of them end in 7
next if you look at the table for mersenne primes they have:

p Mp
3 7
7 127
127 170141183…88410 5727

If I'm right in assuming the last Mp is the last in this series and this pattern continues( the pattern is Mp becomes the next p) then the next one in this sub series( or what ever you want to call it) is:

p
170141183460469 231731687303715 884105727

Mp
2^170,141,183,4 60,469,231,731, 687,303,715,884 ,105,727


if this is true it smashes all records as the biggest prime to date and biggest mersenne prime is 2^46 million and something this is 2^ 170 Undecillion and something.

[JosAH]

Okay I've got a weird idea normal primes are fun but what if I could come up with a pattern in one of the types that if accurate will land us in the record books( might need to write massive strings instead of numbers though). I think I see a pattern in the double mersenne primes(might have to use every computer in the world to confirm the next one as prime).

If you find such a pattern you'll be world famous of course. If you're really interested in those Mersenne beasts read Eric Weisstein's Mathworld and google for Sloane's database of integer number series. They're interesting and fun.

kind regards.

Jos

[scienceman88]

yeah unfortunately they may have to reinvent computers and extend SI to fit it lol if I looked it up right it would take 2.12676479 × 10^37 bytes. and you thought I was crazy before lol. this would be over 21 tera x tera x terabytes lol.

[JosAH]

yeah unfortunately they may have to reinvent computers and extend SI to fit it lol if I looked it up right it would take 2.12676479 × 10^37 bytes. and you thought I was crazy before lol. this would be over 21 tera x tera x terabytes lol.

Here's a last link because I have the feeling that you don't quite realize what you're up to.

kind regards,

Jos

[scienceman88]

Let n be an odd prime. The Mersenne number M(n) = 2n-1 is prime if and only if S(n-2) = 0 (mod M(n)) where S(0) = 4 and S(k+1) = S(k)2-2. We can add a few things in to do this testing and to try for a new record all we'd have to do is set the lower limit to the current record.

[scienceman88]

we know the primes that can be exponents if we further test them to check S(n-2) = 0 (mod M(n)) where S(0) = 4 and S(k+1) = S(k)2-2 if they fit that all it's a new record

[JosAH]

we know the primes that can be exponents if we further test them to check S(n-2) = 0 (mod M(n)) where S(0) = 4 and S(k+1) = S(k)2-2 if they fit that all it's a new record

What's your definition of S(k)?

kind regards,

Jos

[scienceman88]

That's the part I don't understand but if we can pull it off we may beat gimps take the 13 years streak and destroy it lol.

[scienceman88]

one thing I see is most of the prime exponents that don't work under 61 are of the form (2^6x-1)-1

[scienceman88]

is there a way I could define a data type that will fit all data under 171 undecillion ?