20120510, 20:38  #1 
"Åke Tilander"
Apr 2011
Sandviken, Sweden
236_{16} Posts 
Ten Billion Digits Mersenne Numbers
Well, I got a bit carried away and trial factored the smallest 100 Ten Billion Digits Mersenne Numbers with prime exponents (M33,219,280,951  M33,219,283,529) up till ^70 bits and found 62 factors Here is the result:
Code:
M33219280951 has 0 factors in [2^1, 2^701]. M33219281003 has a factor: 61455669855551  Program: L5.0x M33219281003 has 1 factors in [2^1, 2^701]. M33219281023 has 0 factors in [2^1, 2^701]. M33219281027 has 0 factors in [2^1, 2^701]. M33219281063 has 0 factors in [2^1, 2^701]. M33219281111 has a factor: 2391788239993  Program: L5.0x M33219281111 has a factor: 294465063646236102041  Program: L5.0x M33219281111 has a factor: 6976049033311  Program: L5.0x M33219281111 has 3 factors in [2^1, 2^701]. M33219281159 has a factor: 66438562319  Program: L5.0x M33219281159 has 1 factors in [2^1, 2^701]. M33219281201 has 0 factors in [2^1, 2^701]. M33219281213 has 0 factors in [2^1, 2^701]. M33219281243 has 0 factors in [2^1, 2^701]. M33219281267 has 0 factors in [2^1, 2^701]. M33219281269 has a factor: 797262750457  Program: L5.0x M33219281269 has 1 factors in [2^1, 2^701]. M33219281279 has a factor: 66438562559  Program: L5.0x M33219281279 has 1 factors in [2^1, 2^701]. M33219281297 has 0 factors in [2^1, 2^701]. M33219281299 has 0 factors in [2^1, 2^701]. M33219281327 has 0 factors in [2^1, 2^701]. M33219281371 has a factor: 16894663979942936777  Program: L5.0x M33219281371 has 1 factors in [2^1, 2^701]. M33219281383 has 0 factors in [2^1, 2^701]. M33219281393 has a factor: 234042327863488769  Program: L5.0x M33219281393 has a factor: 150788962099105601  Program: L5.0x M33219281393 has 2 factors in [2^1, 2^701]. M33219281399 has a factor: 1594525507153  Program: L5.0x M33219281399 has a factor: 34880245468951  Program: L5.0x M33219281399 has 2 factors in [2^1, 2^701]. M33219281411 has 0 factors in [2^1, 2^701]. M33219281437 has a factor: 492321512315809895599  Program: L5.0x M33219281437 has 1 factors in [2^1, 2^701]. M33219281441 has 0 factors in [2^1, 2^701]. M33219281449 has a factor: 730824191879  Program: L5.0x M33219281449 has 1 factors in [2^1, 2^701]. M33219281453 has a factor: 5919675954924601  Program: L5.0x M33219281453 has 1 factors in [2^1, 2^701]. M33219281467 has 0 factors in [2^1, 2^701]. M33219281479 has 0 factors in [2^1, 2^701]. M33219281489 has a factor: 116996795601600546271  Program: L5.0x M33219281489 has 1 factors in [2^1, 2^701]. M33219281497 has a factor: 80462680625025497  Program: L5.0x M33219281497 has 1 factors in [2^1, 2^701]. M33219281507 has 0 factors in [2^1, 2^701]. M33219281521 has a factor: 16476763634417  Program: L5.0x M33219281521 has a factor: 689012466796645582273  Program: L5.0x M33219281521 has 2 factors in [2^1, 2^701]. M33219281531 has 0 factors in [2^1, 2^701]. M33219281549 has a factor: 486304386450814108031  Program: L5.0x M33219281549 has 1 factors in [2^1, 2^701]. M33219281569 has 0 factors in [2^1, 2^701]. M33219281587 has a factor: 20728831710289  Program: L5.0x M33219281587 has 1 factors in [2^1, 2^701]. M33219281629 has 0 factors in [2^1, 2^701]. M33219281653 has a factor: 8629771426379647  Program: L5.0x M33219281653 has 1 factors in [2^1, 2^701]. M33219281677 has 0 factors in [2^1, 2^701]. M33219281687 has 0 factors in [2^1, 2^701]. M33219281713 has 0 factors in [2^1, 2^701]. M33219281737 has a factor: 531508507793  Program: L5.0x M33219281737 has 1 factors in [2^1, 2^701]. M33219281741 has a factor: 7972627617841  Program: L5.0x M33219281741 has 1 factors in [2^1, 2^701]. M33219281863 has a factor: 3986313823561  Program: L5.0x M33219281863 has 1 factors in [2^1, 2^701]. M33219281917 has a factor: 54940373784068791  Program: L5.0x M33219281917 has a factor: 6909610638737  Program: L5.0x M33219281917 has 2 factors in [2^1, 2^701]. M33219281933 has a factor: 797262766393  Program: L5.0x M33219281933 has a factor: 7175364897529  Program: L5.0x M33219281933 has 2 factors in [2^1, 2^701]. M33219281957 has 0 factors in [2^1, 2^701]. M33219282001 has 0 factors in [2^1, 2^701]. M33219282013 has 0 factors in [2^1, 2^701]. M33219282059 has a factor: 18602797953041  Program: L5.0x M33219282059 has a factor: 4392386350969217  Program: L5.0x M33219282059 has 2 factors in [2^1, 2^701]. M33219282067 has 0 factors in [2^1, 2^701]. M33219282079 has a factor: 12035375927452699417  Program: L5.0x M33219282079 has 1 factors in [2^1, 2^701]. M33219282097 has 0 factors in [2^1, 2^701]. M33219282101 has a factor: 36076140361687  Program: L5.0x M33219282101 has 1 factors in [2^1, 2^701]. M33219282121 has a factor: 731938037166137002199  Program: L5.0x M33219282121 has 1 factors in [2^1, 2^701]. M33219282137 has 0 factors in [2^1, 2^701]. M33219282143 has a factor: 5479771195752336793  Program: L5.0x M33219282143 has 1 factors in [2^1, 2^701]. M33219282203 has a factor: 470385035994481  Program: L5.0x M33219282203 has a factor: 42938659516231027201  Program: L5.0x M33219282203 has 2 factors in [2^1, 2^701]. M33219282211 has 0 factors in [2^1, 2^701]. M33219282241 has 0 factors in [2^1, 2^701]. M33219282287 has a factor: 151937821815764009  Program: L5.0x M33219282287 has 1 factors in [2^1, 2^701]. M33219282409 has 0 factors in [2^1, 2^701]. M33219282451 has a factor: 92473513137322231  Program: L5.0x M33219282451 has 1 factors in [2^1, 2^701]. M33219282493 has a factor: 829153291025281  Program: L5.0x M33219282493 has 1 factors in [2^1, 2^701]. M33219282521 has 0 factors in [2^1, 2^701]. M33219282541 has 0 factors in [2^1, 2^701]. M33219282589 has 0 factors in [2^1, 2^701]. M33219282617 has 0 factors in [2^1, 2^701]. M33219282623 has a factor: 16742518441993  Program: L5.0x M33219282623 has a factor: 2458226914103  Program: L5.0x M33219282623 has 2 factors in [2^1, 2^701]. M33219282631 has 0 factors in [2^1, 2^701]. M33219282661 has a factor: 921569339581463  Program: L5.0x M33219282661 has 1 factors in [2^1, 2^701]. M33219282721 has a factor: 178586863908097  Program: L5.0x M33219282721 has 1 factors in [2^1, 2^701]. M33219282727 has a factor: 4489735680203452409  Program: L5.0x M33219282727 has 1 factors in [2^1, 2^701]. M33219282773 has 0 factors in [2^1, 2^701]. M33219282779 has 0 factors in [2^1, 2^701]. M33219282781 has 0 factors in [2^1, 2^701]. M33219282821 has a factor: 79104679550253449  Program: L5.0x M33219282821 has 1 factors in [2^1, 2^701]. M33219282847 has 0 factors in [2^1, 2^701]. M33219282869 has 0 factors in [2^1, 2^701]. M33219282883 has 0 factors in [2^1, 2^701]. M33219282907 has a factor: 797262789769  Program: L5.0x M33219282907 has 1 factors in [2^1, 2^701]. M33219282923 has a factor: 797262790153  Program: L5.0x M33219282923 has 1 factors in [2^1, 2^701]. M33219282937 has 0 factors in [2^1, 2^701]. M33219282953 has a factor: 797262790873  Program: L5.0x M33219282953 has 1 factors in [2^1, 2^701]. M33219282979 has 0 factors in [2^1, 2^701]. M33219283039 has a factor: 20320368312088457  Program: L5.0x M33219283039 has 1 factors in [2^1, 2^701]. M33219283063 has 0 factors in [2^1, 2^701]. M33219283133 has a factor: 44791286850946817  Program: L5.0x M33219283133 has 1 factors in [2^1, 2^701]. M33219283147 has a factor: 1044414262141681  Program: L5.0x M33219283147 has 1 factors in [2^1, 2^701]. M33219283151 has 0 factors in [2^1, 2^701]. M33219283183 has a factor: 794416900401712391  Program: L5.0x M33219283183 has a factor: 1582367335139023  Program: L5.0x M33219283183 has 2 factors in [2^1, 2^701]. M33219283217 has a factor: 41658849605921823383  Program: L5.0x M33219283217 has 1 factors in [2^1, 2^701]. M33219283301 has a factor: 103297620335723969  Program: L5.0x M33219283301 has 1 factors in [2^1, 2^701]. M33219283319 has 0 factors in [2^1, 2^701]. M33219283337 has a factor: 142251814421401423  Program: L5.0x M33219283337 has 1 factors in [2^1, 2^701]. M33219283351 has a factor: 20662659998588809  Program: L5.0x M33219283351 has 1 factors in [2^1, 2^701]. M33219283369 has 0 factors in [2^1, 2^701]. M33219283373 has a factor: 25682691677033999  Program: L5.0x M33219283373 has 1 factors in [2^1, 2^701]. M33219283429 has a factor: 114459561135898990799  Program: L5.0x M33219283429 has 1 factors in [2^1, 2^701]. M33219283519 has a factor: 1328771340761  Program: L5.0x M33219283519 has 1 factors in [2^1, 2^701]. M33219283529 has a factor: 35725346278427761  Program: L5.0x M33219283529 has 1 factors in [2^1, 2^701]. 
20120510, 22:42  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·7·683 Posts 
Why stop there? We should start testing trilliondigit numbers already.
Code:
M3321928096379 has a factor 6643856192759 M3321928097483 has a factor 6643856194967 M3321928097759 has a factor 6643856195519 M3321928098803 has a factor 6643856197607 M3321928100303 has a factor 6643856200607 M3321928102019 has a factor 6643856204039 M3321928104431 has a factor 6643856208863 M3321928132499 has a factor 664385626499801 M3321928147817 has a factor 664385629563401 M3321928390169 has a factor 664385678033801 M3321928439759 has a factor 664385687951801 M3321928556909 has a factor 664385711381801 M3321928559699 has a factor 664385711939801 M3321928570811 has a factor 664385714162201 M3321928639961 has a factor 664385727992201 M3321928651313 has a factor 664385730262601 M3321928799261 has a factor 664385759852201 M3321928836311 has a factor 664385767262201 M3321929068193 has a factor 664385813638601 M3321928164167 has a factor 3986313797000401 M3321928736839 has a factor 3986314484206801 M3321929056309 has a factor 3986314867570801 M3321928125443 has a factor 737016261623285753 M3321928177429 has a factor 4080603422302944737 
20120511, 00:47  #3 
"William"
May 2003
New Haven
2^{6}·37 Posts 
Let me know the web pages for these and I'll add a link to the them.

20120511, 02:53  #4 
"Nathan"
Jul 2008
Maryland, USA
5·223 Posts 
The question now, of course, is whether the EFF prizes are paid cumulatively, if a prime happens to be found that is both 100M and 1B digits in size. And would they institute higher prizes for higher orders of magnitude of digits. Hmm....

20120511, 03:10  #5 
Jun 2003
2×31×83 Posts 

20120511, 04:29  #6 
Romulan Interpreter
Jun 2011
Thailand
2^{4}×13×47 Posts 
Cute, haha, but what's the point? I am sure Serge's post was a joke, as he usually does (see his algorithm with adding 13). Here is a single line in pari:
Code:
(10:25:19) gp > p=nextprime(10^20*log(10)/log(2)); cnt=0; while(1,if(p%4==3&&isprime(q=2*p+1),print("M"p" has a factor "q);cnt++);p=nextprime(p++)) M332192809488736240559 has a factor 664385618977472481119 M332192809488736242551 has a factor 664385618977472485103 M332192809488736243919 has a factor 664385618977472487839 M332192809488736247123 has a factor 664385618977472494247 M332192809488736248023 has a factor 664385618977472496047 M332192809488736261391 has a factor 664385618977472522783 M332192809488736261991 has a factor 664385618977472523983 M332192809488736267871 has a factor 664385618977472535743 <...> <snip 400 lines with factors removed (about)> <...> M332192809488737637371 has a factor 664385618977475274743 M332192809488737640659 has a factor 664385618977475281319 M332192809488737642603 has a factor 664385618977475285207 M332192809488737648999 has a factor 664385618977475297999 M332192809488737649143 has a factor 664385618977475298287 <...> <snip 400 lines with factors removed (about)> <ctrl C pressed> *** at toplevel: ...cnt=0;while(1,if(p%4==3&&isprime(q=2*p+1),pri *** ^ *** isprime: user interrupt after 5,297 ms. *** Break loop: type <Return> to continue; 'break' to go back to GP break> cnt 814 break>^C (10:25:27) gp > (the second parameter is starting k, the third is ending k, in 2*k*p+1) Code:
(10:33:32) gp > \r tf (10:33:34) gp > p=nextprime(10^20*log(10)/log(2)); while(1, printf("%d: ",p); tf(p,0,10^6,0); p=nextprime(p++)) 332192809488736234933: Factor not found, try higher start/stop values. 332192809488736234949: Factor not found, try higher start/stop values. 332192809488736234993: Factor not found, try higher start/stop values. 332192809488736235033: Factor found: 2^3321928094887362350331 is divisible by 1979869144552867960796681 <...> <lots of lines with no factors scissored out, I will let only the one with factors> <...> 332192809488736235227: Factor found: 2^3321928094887362352271 is divisible by 118006180485921727784628119 332192809488736235309: Factor found: 2^3321928094887362353091 is divisible by 2657542475909889882473 332192809488736235513: Factor found: 2^3321928094887362355131 is divisible by 1993156856932417413079 332192809488736235779: Factor found: 2^3321928094887362357791 is divisible by 725509095923399938941337 332192809488736235981: Factor found: 2^3321928094887362359811 is divisible by 239178822831890089906321 332192809488736236499: Factor found: 2^3321928094887362364991 is divisible by 3774374701411021119101639 332192809488736236983: Factor found: 2^3321928094887362369831 is divisible by 725509095923399941570873 332192809488736237123: Factor found: 2^3321928094887362371231 is divisible by 13287712379549449484921 332192809488736237339: Factor found: 2^3321928094887362373391 is divisible by 79726274277296696961361 332192809488736237793: Factor found: 2^3321928094887362377931 is divisible by 7972627427729669707033 332192809488736238023: Factor found: 2^3321928094887362380231 is divisible by 183370430837782403388697 332192809488736238521: Factor found: 2^3321928094887362385211 is divisible by 1993156856932417431127 332192809488736238593: Factor found: 2^3321928094887362385931 is divisible by 940105650853123555218191 332192809488736238767: Factor found: 2^3321928094887362387671 is divisible by 39863137138648348652041 332192809488736238953: Factor found: 2^3321928094887362389531 is divisible by 7308241808752197256967 332192809488736238989: Factor found: 2^3321928094887362389891 is divisible by 432515037954334583163679 Last fiddled with by LaurV on 20120511 at 04:41 
20120511, 06:19  #7  
"Åke Tilander"
Apr 2011
Sandviken, Sweden
1000110110_{2} Posts 
Quote:
Well maybe. Generally speaking it could have some value to know that the distribution of factors are as expected even very high up among the Mersenne numbers. Some things about the Mersenne numbers/primes are just taken for granted but there is no mathematical proof of it, so at least it could be good to have empirical proofs. If I remember it rightly (I am on the run just now so I don't have the time to make a proper check) there is not even a mathematical proof of wether there are infinitely many composite (prime exponent) Mersenne numbers or not? All of us take this for granted, but showing that there are composites high up confirms this at least up to that limit. So thinking more about it I really think there is a good point in having sets of Mersenne numbers trail factored way up which could be studied more closely. But that wasn't really why I did it. It was more because it was nice doing it and nice finding a bunch of factors like that and after all Operation Billion Digits is a whimsical project. Quote:
And I have googled all 62 factors without finding any of them on the internet. They are not included in Will Edgington's list here: http://www.garlic.com/~wedgingt/resu...k.split.aw.bz2 And they did not appear neither on PrimeNet nor on Mersenne aries since they don't cover these ranges. I have submitted them to Mersenne aries now and you can find them under the largest exponent covered by the database here: http://mersennearies.sili.net/expon...61455669855551 And I have, of course, searched this forum as well without finding any mention of them before. So to the best of my knowledge they were not known before. Last fiddled with by aketilander on 20120511 at 06:47 

20120511, 08:54  #8 
Banned
"Luigi"
Aug 2002
Team Italia
3·1,609 Posts 
I think Will Edgington may enjoy them...
Luigi 
20120820, 20:29  #9  
"James Heinrich"
May 2004
exNorthern Ontario
DAE_{16} Posts 
Quote:
Quote:
Incidentally, I now generate a daily exponent status report in the same format at PrimeNet, but for M(10^{9})M(2^{32}): mersennearies.sili.net/primenet_gimps_summary_1G/ PM or email me if you have any interest in doing broad TF prefactoring in this range. 

20120821, 02:45  #10  
Bemusing Prompter
"Danny"
Dec 2002
California
2,411 Posts 
A factor I found several years ago:
Quote:


20120821, 04:45  #11  
Romulan Interpreter
Jun 2011
Thailand
2^{4}×13×47 Posts 
Quote:
Code:
elimSGp3m4(start,stop,file,pflag)= { my(p,d,cnt); p=max(start,5); cnt=0; while(p<stop, until(p%4==3 && isprime(d=2*p+1), p=nextprime(p+1) ); if(p<stop, if(bitand(pflag,1), printf("...%d...%c",p,13)); if(bitand(pflag,2), print("M"p" is divisible by "d)); if(bitand(pflag,4), write(file,p,",",d)); cnt++ ) ); return(cnt) }; Here is some TF variation, which can be used to eliminate very fast exponents with small k, up to 2^30 or so. About half of them will be gone with this "filtering". There is also a function called "ratf" (random tf) which does the same but looking for k first. Ratfpro is its "pro" version which splits the output file in different files according with bit level and exponent size. For this ratfpro, you can create a file called "flags.txt" to "gracefully shut it down", you create that file and put a single line in it, containing a "0" (ascii char code 48, then end of line). When you want your program to stop, edit the file and put a "1" and it will stop after the next save checkpoint, so you can resume next time without losing candidates (no gaps). Code:
ten=log(10); two=log(2); /* p= exponent, start/stop=values of k in 8*k*p+{s*p}+1 */ tf(p,start,stop,fisout)= { if(p<3  !isprime(p),print("p has to be an odd prime"); return(0)); if(p<11,print("2^"p"1 is prime!"); return(2)); gettime(); /*m=1<<p1;*/ if(stop==0, error("No Stop"));/*stop=sqrtint(m)\(8*p)+1);*/ sp0=2*(4p%4)*p; sp1=8*psp0; k=start; if(k==0,d=sp0+1;t=1,d=8*k*p+1;t=0); /* while(k<=stop && (n=m%d) /* faster than exponentiation n=(lift(Mod(2,d)^p)1)*/ while(k<=stop && (n=lift(Mod(2,d)^p)1) !=0, if(t==0, d+=sp0, d+=sp1; k+=t); t=1t; if(bitand(k,1048575)==0 && t==0, printf("... %d bMit (k: %d)  last tested: %d, %d digits (2^%d++), %11.6f ms/iter ...%c", k>>20,k,d,ceil(log(d)/ten),floor(log(d)/two),gettime()/1048576.0,13); if(fisout<>0, write(fisout,k1)) ) ); if(n==0, /* print("Factor found: 2^"p"1 = "m" is divisible by "d); return(1), */ print("Factor found: 2^"p"1 is divisible by "d); return(1), /*else*/ /* if(stop>=sqrtint(m)\(8*p)+1, print("2^"p"1 = "m" is prime!"); return(2), /*else*/ print("Factor not found, try higher start/stop values."); return(0) /* )*/ ); } /* get the k in 8*k*p+{s*p}+1 (negativ for s!=0. else positiv) */ getk(d,p)= { if(Mod(2,d)^p!=Mod(1,d),print("Not a factor!");return); if((a=((d1)/p))%8==0,return(a/8),return((a2*(4p%4))/8)); } /*get smallest p such as n divides 2^p1*/ getp(n,limt)= { if(n<3  n%2==0, print("Only odd positive numbers!");return); rg=1; sm=0; until(rg==1  sm>=limt, rg+=n; while(rg%2==0 && sm<limt, rg>>=1; sm++; if(bitand(sm,1048575)==0, print(sm)); ) ); return(sm); } /*getp for a prime n (in this case p is a divisor of n1*/ getpfp(n)= { my(c); c=divisors(n1); for(i=1,#c, if(c[i]<1000000000 && Mod(2,n)^c[i]==1, return(c[i])) ); return(0); } /* "random" trial factoring, specify a starting possible factor and the outfput file */ ratf(a,fis)= { if(fis==0, print("Specify an output file!"); return(0)); while(1, until(a%8==1  a%8==7,a=nextprime(a+1)); c=factorint(a1)[,1]~; for(i=1,#c, if(c[i]<1000000000 && Mod(2,a)^c[i]==1, print(a","c[i]); write(fis,a","c[i])) ) ); } /*same as above, but split the output to different files depending on n and k if a=1, it will read the last_k from the file, assuming the file has ONE LINE only use the flags.txt file to gracefully stop it */ ratfpro(a,printstep)= { my(tnextpr,tmod,tfactint,telse); if(a==1, a=read("fact_last_k.txt")); print("Starting/continuing from: "a); tnextpr=tfactint=tmod=telse=0; if(printstep==0,printstep=2^20); prst=a; gettime(); while(1, telse+=gettime(); until(a%8==1  a%8==7,a=nextprime(a+1)); tnextpr+=gettime(); if(a>prst, prst=a+printstep; printf("... %d, (%d digits, 2^%d++)  Timers: %6.4f, %6.4f, %6.4f, %6.4f (seconds)\n", a,ceil(log(a)/ten),floor(log(a)/two), tnextpr/1000, tfactint/1000, tmod/1000, telse/1000); extern("del fact_last_k.bak"); extern("ren fact_last_k.txt fact_last_k.bak"); write("fact_last_k.txt",a1); shouldstop=read("flags.txt"); if(shouldstop==1,break) ); telse+=gettime(); c=factorint((a1)>>1)[,1]~; tfactint+=gettime(); for(i=1,#c, if(c[i]<10000000000, telse+=gettime(); if(Mod(2,a)^c[i]==1, tmod+=gettime(); print(a","c[i]" "); if(c[i]<1000000000,fis="fact_0B_", if(c[i]<2000000000,fis="fact_1B_", if(c[i]<3000000000,fis="fact_2B_", if(c[i]<4000000000,fis="fact_3B_", if(c[i]<5000000000,fis="fact_4B_", if(c[i]<6000000000,fis="fact_5B_", if(c[i]<7000000000,fis="fact_6B_", if(c[i]<8000000000,fis="fact_7B_", if(c[i]<9000000000,fis="fact_8B_", fis="fact_9B_"; break))))))))); expo=40; k=1<<expo; for(j=1,40, if(a<k,fis=concat(fis,expo); break, k<<=1; expo++)); write(fis,a","c[i]), /*else*/ tmod+=gettime() ), /*else*/ break ) ) ); } /*same as above, but use modular order instead of factoring k1*/ /*for bigger stuff it may become faster than factoring k1*/ ratfzno(a,printstep)= { my(tnextpr,tprchk,tznord,telse); if(a==1, a=read("fact_last_k")); tnextpr=tznord=tprchk=telse=0; gettime(); if(printstep==0,printstep=2^20); prst=a; while(1, until(a%8==1  a%8==7,a=nextprime(a+1)); tnextpr+=gettime(); if(a>prst, prst=a+printstep; printf("... %d, (%d digits, 2^%d++)  Timers: %10.4f, %10.4f, %10.4f, %10.4f seconds\n", a,ceil(log(a)/ten),floor(log(a)/two), tnextpr/1000, tznord/1000, tprchk/1000, telse/1000); write("fact_last_k",a)); telse+=gettime(); c=znorder(Mod(2,a)); tznord+=gettime(); if(isprime(c), tprchk+=gettime(); if(c<1000000000,fis="fact_0B_", if(c<2000000000,fis="fact_1B_", if(c<3000000000,fis="fact_2B_", if(c<4000000000,fis="fact_3B_", if(c<5000000000,fis="fact_4B_", if(c<6000000000,fis="fact_5B_", if(c<7000000000,fis="fact_6B_", if(c<8000000000,fis="fact_7B_", if(c<9000000000,fis="fact_8B_", if(c<10000000000,fis="fact_9B_", telse+=gettime(); next)))))))))); expo=40; k=1<<expo; for(j=1,40, if(a<k,fis=concat(fis,expo); break, k<<=1; expo++)); print(a","c" "); write(fis,a","c); telse+=gettime() ); tprchk+=gettime() ); } /* it will look for prime factors of the form 2*pr(v)+1 v can include 2, p and other odd primes (or else) call sample (copy/paste): > p=23; xx=[0,0,0]; while((xx=pptf([2,p],xx[2],xx[2]+10^4,10^6,0,0))[3]!=p,); xx immediate subsequent calls (to find other small factors): > xx=%; while((xx=pptf([2,p],xx[2],xx[2]+10^4,10^6,0,0))[3]!=p,); xx */ pptf(v,k,limt,iter,pp,pv)= { gata=0; a=0; /*not necessary*/ pr=prod(i=1,#v,v[i]); while(k<li I am currently playing with implementing P1 on cuda. For that, I started with a pari implementation, to see how the things go around. If you are interested in going "deeper" with the "filtering", then you can play with the next pari script too: [edit limit, I cut it out, let's see with attach] [grr need to rename it txt] 

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