speeding up lucas lehmer test with p^2
 

I've been looking in Pari recently and what if I told you just by checking one thing I may have a way to use p^2 instead of 2^p-1 for the mod step. and that's because 0 mod (2^p-1) should in theory mean the end number for mod p^2 should be a multiple of 0.6931471805599453094172321215*p (I think) , if it isn't then 2^p-1 is not prime. the problem is how many value there can be along these lines : p/0.6931471805599453094172321215 possible values.

[QUOTE=science_man_88;256589]I've been looking in Pari recently and what if I told you just by checking one thing I may have a way to use p^2 instead of 2^p-1 for the mod step. and that's because 0 mod (2^p-1) should in theory mean the end number for mod p^2 should be a multiple of 0.6931471805599453094172321215*p (I think) , if it isn't then 2^p-1 is not prime. the problem is how many value there can be along these lines : p/0.6931471805599453094172321215 possible values.[/QUOTE]

okay i see a slight flaw again. oh well I was close. this idea came from [CODE](21:09)>(2^p-1)%(p^2)
%1 = 0.6931471805599453094172321215*p[/CODE]

That's the series around 0. Mersenne exponents aren't close to 0.

[QUOTE=science_man_88;256592]okay i see a slight flaw again. oh well I was close. this idea came from [CODE](21:09)>(2^p-1)%(p^2)
%1 = 0.6931471805599453094172321215*p[/CODE][/QUOTE]

the flaw I see is if the multiple originally exceeds p^2 it's not guaranteed to be of that form unless p^2 is of that form.

[QUOTE=CRGreathouse;256595]That's the series around 0. Mersenne exponents aren't close to 0.[/QUOTE]

all I was saying is because:

[CODE](21:09)>(2^p-1)%(p^2)
%1 = 0.6931471805599453094172321215*p[/CODE]

and the fact that the last one is supposed to be 0 mod 2^p-1 that it's ((s/(2^p-1))* (0.6931471805599453094172321215*p)) % p^2 . I was originally thinking that the %p^2 didn't change things much but it does.

Even if you look at the number without taking it mod p^2 it won't work since you took the series around 0 and you're giving it numbers not close to 0.

p = 1: 2^p - 1 = 1*1
p = 0.5: 2^p - 1 ~= 0.828427125p
p = 0.1: 2^p - 1 ~= 0.717734625p
p = 0.01: 2^p - 1 ~= 0.695555006p
p = 0.001: 2^p - 1 ~= 0.693387463p
p = 0.0001: 2^p - 1 ~= 0.693171204p
p = 0.00001: 2^p - 1 ~= 0.693149583p

but

p = 10: 2^p - 1 = 102.3p
p = 100: 2^p - 1 ~= 10[SUP]28[/SUP]p
p = 1000: 2^p - 1 ~= 10[SUP]298[/SUP]p

etc.

the 24m+7 thread was nice but I think I've found one better:

1) mersenne primes are 4 7 or 1 mod 9
2) all mersenne numbers >=7 are 24x+7

so according to math I did with pari ( because I wasn't thinking how to prove when they intersect on my own that means that they are either 72c+7 72c+31 or 72c+55 ( and yes I know it complicates things, but it also uses a lower variable value to calculate higher numbers).

and none of these form are in the OEIS sequences as far as I've searched ( though I've found ways of using the 24m+7 idea before).

got something new to think on:

1) is there a way to find when a lucas sequence factors into another
2) can you use algebra to assign values to p and q for the lucas lehmer sequence, ( I get algebraically, p=a(x) and q=2/a(x-1)).

I looked at the Lucas Lehmer under modular arithmetic with my pocket calculator and brain and have realized what they say would be necessary to have infinitely many Mersenne primes. [TEX]S_{p-2}= 0 (mod 2^p-1)[/TEX] is what it states however. continuing on shows 0^2-2 = -2 (mod 2^p-1) continuing more you'll find that all of the rest S_{x} with x>=p have a 2 modulo so we'd have to prove if 2^p-1 is prime there's a prime q such that (2^q-1) | [2][SUB]2^p-1[/SUB]

it would be nice if I could get this hypothesis into a exponent finding form I think:

the next Mersenne prime z has a power that is = x^y mod (Mersenne prime (z-1))

my supporting evidence ( I have more done before this):

[CODE](20:12)>for(y=0,100000,if(ispower(((2147483647^y)%524287)),print(y);break()))
1445
(20:12)>for(y=0,100000,if(ispower(((524287^y)%131071)),print(y);break()))
2
(20:13)>for(y=0,100000,if(ispower(((524287^2)%131071)),print(y);break()))
0
(20:13)>for(y=0,100000,if(ispower(((524287^y)%131071)),print(y);break()))
2
(20:14)>for(y=0,100000,if(ispower(((2305843009213693951^y)%2147483647)),print(y);break()))
2
(20:14)>for(y=0,100000,if(ispower(((618970019642690137449562111^y)%2305843009213693951)),print(y);break()))
2
(20:16)>for(y=0,100000,if(ispower(((162259276829213363391578010288127^y)%618970019642690137449562111)),print(y);break()))
2
(20:19)>for(y=0,100000,if(ispower(((170141183460469231731687303715884105727^y)%162259276829213363391578010288127)),print(y);break()))
2[/CODE]

Let's see:
You are hypothesizing that log2(M48+1) = ( X^Y )mod(M47) for some X and Y?

How would I effectively computer X and Y? Can you show me X and Y for M2 through M47?