PARI's commands

[CRGreathouse]

Originally Posted by science_man_88

Digital root of the lucas lehmer residue ... really I didn't know that, Thanks. I'm just trying to limit it to a list of digital root 9 because we know the exponents must be in that group to work.

Actually, I'm not sure about my claim anymore. It would certainly be easy to compute mod 9 rather than mod but that won't give you what you want since

Of course, to me, that suggests that this is not really going to be meaningful... but that's your issue, not mine.

[science_man_88]

3,5,7,13,17,19,23,31,33,51,61,71,89,101,107,127,139,191,271,273,305,331,347,351,367,397,405,407,427,435,457,467,489,521,525,539,543,549,559,565,577,583,589,597,601,607,611,613,617,619,641,643,661,693,717,729,787,793,809,817,819,837,871,879,891,899,983,987,991,

this is the sequence I got for lucaslehmer2()

[3.14159]

Originally Posted by science_man_88

Digital root of the lucas lehmer residue ... really I didn't know that, Thanks. I'm just trying to limit it to a list of digital root 9 because we know the exponents must be in that group to work.

Exponents? 0 mod 9? 9 does not divide any prime, so this condition is impossible to meet.

You can try 1 mod 9, 2 mod 9, 4 mod 9, 5 mod 9, 7 mod 9, and 8 mod 9.

[science_man_88]

Originally Posted by 3.14159

Exponents? 0 mod 9? 9 does not divide any prime, so this condition is impossible to meet.

You can try 1 mod 9, 2 mod 9, 4 mod 9, 5 mod 9, 7 mod 9, and 8 mod 9.

I'm not saying anything about a prime dividing by 9 lol. If 0=9 mod 9 then you know the mersenne prime exponent residues all fit in the group of exponents with residues that are 9 mod 9, they are a subgroup.

[3.14159]

Originally Posted by science_man_88

I'm not saying anything about a prime dividing by 9 lol. If 0=9 mod 9 then you know the mersenne prime exponent residues all fit in the group of exponents with residues that are 9 mod 9, they are a subgroup.

Okay.. what about it?

[3.14159]

And;

Adding Fermat's factoring algorithm was rather easy;

fmtfcs(a, x, m) = {
for(n=a,x,
if(issquare(n^2+m), print(n))
);
}

Okay; A few edits to that..

Revision:

fmtfcs(a,x,m)=for(n=a,x,if(issquare(n^2+m),print(sqrtint(n^2+m)+n, " * ", sqrtint(n^2+m)-n)));

[CRGreathouse]

Bad Pi! Functions should return, not print. What if you wanted to use that data from another program?

[3.14159]

Originally Posted by CRGreathouse

Bad Pi! Functions should return, not print. What if you wanted to use that data from another program?

Okay, fine. I've changed it to "return". Nope, that's unable to return the factors. Back to printing it goes.

I'm busy collecting relations for a c106. I'm only about ... 31% done. 159k relations are needed. So, I estimate that this will take two or three days. (I've only collected 49105 relations, so far, given eight hours.)

[3.14159]

If adding Fermat's was easy.. How would one go about in adding an amateur QS?

(Yes, yes, I know that PARI already uses MPQS.)

My knowledge of it so far is;

1. Set bound.
2. Set factor base.
3. b-smooth congruence hunting.
4. Try to set ≥2 congruences into a congruence of squares.
5. gcd.
6. Factors.

[3.14159]

I did a comparison, and it seems that trial division by far outspeeds the Fermat script.

I think I didn't have the most efficient script there is.. But I'll try improving on it.

I'm now 2/3 done with the collection of relations for the c106 I'm trying to split (Hopefully it does not crash.)|

Update: 72% done.

[3.14159]

Record split; Part 3!

starting SIQS on c106: 4475950135356613778937951617143741209311215873146646798431494279597781200242083780419779316018232365322301

==== sieving in progress ( 4 threads):  159064 relations needed ====
====            Press ctrl-c to abort and save state            ====
159188 rels found: 35794 full + 123394 from 2418802 partial, ( 48.30 rels/sec)

SIQS elapsed time = 51034.7770 seconds.


***factors found***

PRP53 = 59032375916416525973964206982565204661985227334884669
PRP53 = 75821954747240330698482729033441429926359516470800129