option for finding multiple factors during trial factoring

[tha]

I tried undoc, readme and google but couldn't find the information I was looking for. If I got it right mprime (or prime95) stops looking for more factors when it has found a factor during trial factoring, so the line

Factor=N/A,some_exponent,0,56

returns 1 factor at most. Again, if I am correct.

Is there an option to force it to find all factors between 0 and 56 bits using only trial factoring?

[LaurV]

No. For GIMPS (i.e. finding primes) the exponent falls from grace once a factor is found (i.e. is composite).

And you should not use P95 anymore for doing TF. This because GPUs do TF from 100 to 300 times faster. Even if you have an antediluvian CPU/system which could not run P-1 or DC (i.e. slow, no memory), keeping it running for just TF is not justified economically [edit: better load yafu and work some aliquots on it]. You could get a low-range GPU and do the same TF work in less time and with lower electricity bill. My two cents. Everybody is free to do whatever work he wants with his resources.

[Stargate38]

If you want to do more than 1 factor, use factor5. That doesn't stop at the 1st factor. (Google it if you need to)

[tha]

Originally Posted by tha

Again, if I am correct.

Is there an option to force it to find all factors between 0 and 56 bits using only trial factoring?

Hmm, the data is in and prime95 does return all factors between both limits. Problem solved.

[kracker]

Yes... it's really not worth TF'ing on the CPU at all in my opinion... especially with factor5. I think it was mainly made/optimized for the very high ranges (Operation Billion Digits)

For example: TF'ing a 55M number from 58 to 59 bits takes 51 seconds on a haswell core.

[R.D. Silverman]

Originally Posted by tha

I tried undoc, readme and google but couldn't find the information I was looking for. If I got it right mprime (or prime95) stops looking for more factors when it has found a factor during trial factoring, so the line

Factor=N/A,some_exponent,0,56

returns 1 factor at most. Again, if I am correct.

Is there an option to force it to find all factors between 0 and 56 bits using only trial factoring?

What purpose is served by finding more than one factor?

GIMPS is looking for Mersenne Primes. The purpose of TF is to quickly
eliminate candidates that have small prime factors so that they need
not run a full LL test. Once a factor is found, we know the number is
composite.

Move on.

[paulunderwood]

One possible use might be prp testing the remaining cofactor

[R.D. Silverman]

Originally Posted by paulunderwood

One possible use might be prp testing the remaining cofactor

And?

[paulunderwood]

Maybe to top this The Prime Pages table for Mersenne cofactors

However, if it is beyond proof, then it might be added to Henri Lifchitz's PRP database.

[science_man_88]

Originally Posted by R.D. Silverman

What purpose is served by finding more than one factor?

I could see eliminating possible candidates for other prime exponents if the factor is still small enough to be useful, for example 7,21,35,49,... all divide by 7
and are 1 mod :

(12:44) gp > forstep(x=1,100,2,a=7*x;print(factor(a-1)))
[2, 1; 3, 1]
[2, 2; 5, 1]
[2, 1; 17, 1]
[2, 4; 3, 1]
[2, 1; 31, 1]
[2, 2; 19, 1]
[2, 1; 3, 2; 5, 1]
[2, 3; 13, 1]
[2, 1; 59, 1]
[2, 2; 3, 1; 11, 1]
[2, 1; 73, 1]
[2, 5; 5, 1]
[2, 1; 3, 1; 29, 1]
[2, 2; 47, 1]
[2, 1; 101, 1]
[2, 3; 3, 3]
[2, 1; 5, 1; 23, 1]
[2, 2; 61, 1]
[2, 1; 3, 1; 43, 1]
[2, 4; 17, 1]
[2, 1; 11, 1; 13, 1]
[2, 2; 3, 1; 5, 2]
[2, 1; 157, 1]
[2, 3; 41, 1]
[2, 1; 3, 2; 19, 1]
[2, 2; 89, 1]
[2, 1; 5, 1; 37, 1]
[2, 7; 3, 1]
[2, 1; 199, 1]
[2, 2; 103, 1]
[2, 1; 3, 1; 71, 1]
[2, 3; 5, 1; 11, 1]
[2, 1; 227, 1]
[2, 2; 3, 2; 13, 1]
[2, 1; 241, 1]
[2, 4; 31, 1]
[2, 1; 3, 1; 5, 1; 17, 1]
[2, 2; 131, 1]
[2, 1; 269, 1]
[2, 3; 3, 1; 23, 1]
[2, 1; 283, 1]
[2, 2; 5, 1; 29, 1]
[2, 1; 3, 3; 11, 1]
[2, 5; 19, 1]
[2, 1; 311, 1]
[2, 2; 3, 1; 53, 1]
[2, 1; 5, 2; 13, 1]
[2, 3; 83, 1]
[2, 1; 3, 1; 113, 1]
[2, 2; 173, 1]

respectively so for example we know 35 isn't prime but that eliminates k=1 for p=17. the lower the factor the better though since it eliminates 1/factor k in theory if it divides one.

[R.D. Silverman]

Originally Posted by paulunderwood

Maybe to top this The Prime Pages table for Mersenne cofactors

However, if it is beyond proof, then it might be added to Henri Lifchitz's PRP database.

Golllleee!