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Thread: Factor bragging

  1. #121
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    Originally posted by MikeH
    hc_grove,

    I don't see this program or the source on the linked page. If it's still available (in particular the source) I'd be very interested. Thanks.
    I don't remember if it's susposed to be available on the page, but here's the source.

    Code:
    /* Written by Henrik Christian Grove */
    /* Released as beerware */
    
    #include <stdlib.h>
    #include <stdio.h>
    
    int main(int argc, char *argv[])
    {
      unsigned long long number;
      char *endptr[10];
      unsigned long div = 2;
      unsigned int exp = 0;
      
      if (argc != 2) {
        printf("Error!\n");
        exit(-1);
      }
      number = strtoll(argv[1],endptr,10);
      if ((number % 2) == 1) {
        number--;
      }
      printf("Factoring %llu\n",number);
    
      printf("%llu = ",number);
      while ((number % 2) == 0) {
        exp++;
        number /= 2;
      }
      if (exp > 1) {
        printf("2^%u*",exp);
      } else {
        printf("2*");
      }
    
      div = 3;
      while (div < number) {
        exp = 0;
        while ((number % div) == 0) {
          exp++;
          number /= div;
        }
        switch (exp) {
        case 0: 
          break;
        case 1: 
          printf("%lu*",div);
          break;
        default: 
          printf("%lu^%u*",div,exp);
        }
        div +=2;
      }
      printf("%llu\n",number);
    
      exit(0);
    }

  2. #122
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    Originally posted by garo
    It is unfortunate that you wasted your time on finding this factor but the solution is to filter out numbers that have already been factored and not change the P-1 limits (though that may be necessary for other reasons).

    Bottomline: Sieving and P-1 find factors in different ways so messing with limits will not ensure NO overlap. P-1 should NOT be done on numbers that have alerady been factored.
    I of course filter out regularly, but to be honest, only befor I dump my worktodo.ini, so once every week or so.

    645.161T 4847 8799543 6.452 Sat 26-Feb-2005 156112.697 (2)

    There is only five days between engracio submitted his factor and the date I commented on bad luck on my side. So, you decide...


    Still, I do think it would be better to use decimals. This is simply because, it gives a better estimate on what to expect and the relative feasibility of doing factoring for that k/n pair.

  3. #123
    Well, I guess this was just sheer bad luck for you.

  4. #124
    First, I just wanted to post these two because I think they are beautiful; simple coincidence, though, I guess.

    2400062667363983| 33661*2^8786592+1 (2 109 10193 10429 103567)
    2448491069125963| 4847*2^8786607+1 (2 3 3 7 6947 28807 97103)

    And then I realized that I was logged out while submitting ; It's because of my browser, I guess, if I log in, and put the sieve submission page in the command line, I'm logged out again, even with 'remember my computer'. I have to click me until there. Sorry so much... If you decide that it's too much work, I will not be angry either.
    (But if you want to do it, it's hh, not hhh, the account name)
    See you, H.

  5. #125
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    I finally have a factor that's really worth bragging about:

    191074725996848743974112872853597 | 67607*2^9999491+1

    If I'm not complete mistaken, this is the second largest factor for a n>10000 we've ever found, to make it a little bit funnier, I'm the one who found the largest one too (36 digits, this one only has 33).

    Unfortunately this number is to large for my trial factorer, and I don't have java here, so I can't use the online ECM factoring thing Louie has told about, so I'm neither able to factor this p (to see if it's prime) nor p-1 (to see how smooth it is).

    I was using quite (approaching the absurd) large bounds (B1=135000, B2=1991250) to find this, but without knowing how smooth it is, I can't tell whether that actually made a difference or just was a waste of time.

  6. #126
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    ???

    191 074725 996848 743974 112872 853596 = 2 ^ 2 x 3 ^ 2 x 7 x 758233 039670
    034698 309971 717673

    Number of divisors: 36

    Sum of divisors: 551 993652 879785 260369 659410 466672

    Euler's Totient: 54 592778 856242 498278 317963 672384

    Moebius: 0

    Sum of squares: a^2 + b^2 + c^2 + d^2
    a = 11057 584279 459218
    b = 7388 520780 436290
    c = 3769 851063 803796
    d = 50 393782 390434


    Side note:

    758233 039670 034698 309971 717673 is prime

    Number of divisors: 2

    Sum of divisors: 758233 039670 034698 309971 717674

    Euler's Totient: 758233 039670 034698 309971 717672

    Moebius: -1

    Sum of squares: a^2 + b^2
    a = 619 909546 902227
    b = 611 510583 170488

  7. #127
    Moderator vjs's Avatar
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    Hold on a sec did you factor this using p-1???

    Or P+1 or ecm

  8. #128
    191074725996848743974112872853597 = 9750606540977083 * 19596188728757959

  9. #129
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    Originally posted by vjs
    Hold on a sec did you factor this using p-1???
    Yes.

  10. #130
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    Originally posted by prime95
    191074725996848743974112872853597 = 9750606540977083 * 19596188728757959
    And:
    9750606540977082 = 2*3*11^2*139*112361*859933
    19596188728757958 = 2*3*83*4159*8669*1091401

  11. #131
    Originally posted by hc_grove
    I don't have java here, so I can't use the online ECM factoring thing.
    Dario Alpern's factoring applet at http://www.alpertron.com.ar/ECM.HTM

    says

    9750606540977083 x 19596188728757959
    Poohbah of the search for Odd Perfect Numbers
    http://OddPerfect.org

  12. #132
    Senior Member Frodo42's Avatar
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    just found myself a 24 digit factor ... my second largest factor so far

    P-1 found a factor in stage #2, B1=40000, B2=440000.
    55459*2^9155398+1 has a factor: 267877184000769088280263

    P-1 factorized:
    267877 184000 769088 280262 = 2 x 3 ^ 3 x 7 x 13 x 613 x 643 x 15199 x 27427 x 331769

  13. #133
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    Originally posted by biwema
    Sorry for not reserving, but I thought, that my range does not fit into the main range.

    At the moment, I work on 67607 starting from 9million. Depending on how much time it takes I will continue until 9.3..10 million.

    At the moment I use B1= 60000

    Reason why I factor in that area:

    No special reason. I like to play around with numbers.
    I this case I want to bring the entry 67607 at 9..10M in MikeH's Database below 1000.
    It's just for fun



    I will keep the factoring details, so you will have that information, when the main factoring effort reaches that area.

    Reto
    Does anyone know if biwema did go through all k/n pairs for k=67607 and 9m<n<10m?

    There are two reasons I'm asking that.

    First of all, if he did so, may be P-1 factorers might consider skipping k=67607 and 9m<n<10m pairs.

    And secondly, I would like to see that particular statistic to have a number <1000, meaning 36 more factors to be found. Just for the fun of course. But, to be honest, I dunno how much work it would require. If it would need to much work, may be I'll simply save it for later (i.e. at least a couple of months from now). It looks like, the figure will drop to 1000- before sieving reaches 2^52. Long way to go... So, any feedback would be appreciated.

  14. #134
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    Hi
    Sorry for not answering for such a long time.
    I was quite busy all the time.

    Finally, I worked on 67607 until 9.3 million and stopped then. On one machine I changed to proth numbers, the other has a mainboard failure and is not running now. therefore I have to change the drive first, before I can tell the exact bounds, but I think it was B1 =60000 from 9.1 to 9.3

    Regards,
    Reto

  15. #135
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    Once again

    Well, it seems that I have some special gift for finding really large factors, I just got this one:
    5767642985639413495402269732331|28433*2^9985657+1

    It's only 31 digits and thus only comes third among my factors in size, but it still 5 digits more than MikeH's largest, which is now number four on the list of large factors (for n>10000 - if we only consider n>100000, MikeH is still in fourth place but a digit further behind).

    And I still don't have Java on this machine, so I can't use the online ECM factoring thing for factoring p (to see if it's prime) or p-1 (to see how smooth it is).

    I think this calls for some
    , and
    in short:

  16. #136
    Senior Member Frodo42's Avatar
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    Re: Once again

    Originally posted by hc_grove
    Well, it seems that I have some special gift for finding really large factors, I just got this one:
    5767642985639413495402269732331|28433*2^9985657+1

    It's only 31 digits and thus only comes third among my factors in size, but it still 5 digits more than MikeH's largest, which is now number four on the list of large factors (for n>10000 - if we only consider n>100000, MikeH is still in fourth place but a digit further behind).

    And I still don't have Java on this machine, so I can't use the online ECM factoring thing for factoring p (to see if it's prime) or p-1 (to see how smooth it is).

    Congrats again ... you do have a certain nack for these factors.

    As with the others this one is also not prime though ... the Alpatrons jave applet took some time to find this (1 minute 38 seconds), it came out with:

    5 767642 985639 413495 402269 732331 = 1255 267090 431443 x 4594 753602 324617

  17. #137
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    Re: Re: Once again

    Originally posted by Frodo42
    5 767642 985639 413495 402269 732331 = 1255 267090 431443 x 4594 753602 324617
    Thanks, it's quote far from being smooth:

    1255267090431442 = 2*11*5779*10529*937721
    4594753602324616 = 2^3*29*257*61511*1252819

  18. #138
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    Re: Once again

    Originally posted by hc_grove
    Well, it seems that I have some special gift for finding really large factors, I just got this one:
    5767642985639413495402269732331|28433*2^9985657+1
    The sad part is, that we found a prime for k=28433 back in january
    Seems I forgot to update the dat.

  19. #139
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    Which program are you using to find these factors second what type of bounds are you using?

    Nice find and wow thats a huge factor!!!!

    Your B1 bound must be quite large if it's p-1

  20. #140
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    Originally posted by vjs
    Which program are you using to find these factors second what type of bounds are you using?

    Nice find and wow thats a huge factor!!!!

    Your B1 bound must be quite large if it's p-1
    The first and largest factor (the one with 36 digits - found before Louie started this thread) was found with sbfactor using optimal bounds (sieve depth = <whatever was relevant at the time>, factor value = 1.5 and RAM=512 i think).

    The two new was found with mprime using some parameters that are chosen for getting high bounds, and the values actually used is: B1=135000 and B2=1991250.
    I'm perfectly aware that this is not optimal for the project, but I'm also in this for the fun, and I like finding the less smooth factors (which seems to increase the probablity of finding large factors too).

    As the factor isn't prime, you should actually think of it as being two factors found simultaneously. To find both factors (and thus the product)
    B1 only had to be >= 61511 and
    B2 only had to be >= 1252819,
    to find the smoother (which would have been enough to eliminate the test - if we didn√łt already have a prime for k=28433) all that was needed was:
    B1 >= 10529 and B2 >= 937721.

    What's really clear is that my B1 is way larger than needed, while B2 is only a factor 2 larger than needed.

  21. #141
    Moderator vjs's Avatar
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    I was just curious, it was obvious using alpertron that it was the multiple of two fairly large smooth factors.

    And I'm all for using large B1 B2 bounds wondering if you would really find that many more if you were using optimal bounds anyways.

  22. #142
    Read my writeup on hc_grove's site. He definitely would have found more factors in the same time with smaller bounds but he probably wouldn't have found factors as large.

  23. #143
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    3551338609857586 | 19249*2^9660686+1

    p-1 = 2 * 31 * 71 * 421 * 8209 * 233437

    Submitted on webpage however I'm still waiting to get my lost password issue sorted but I'd like to make sure I get credit for it (especially as it's in the upcoming window :-)

  24. #144
    Senior Member Frodo42's Avatar
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    I am doing some fast factoring with pretty low bounds ... I sure didn't expect this one

    P-1 found a factor in stage #1, B1=20000.
    22699*2^9782902+1 has a factor: 23397089200776608159181803

    23 397089 200776 608159 181802 = 2 x 107 x 109 x 313 x 2657 x 2713 x 4241 x 5591 x 18749

    For a factor of 26 digits this seems to be very smooth.

  25. #145
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    1005612431486479611763 | 21181*2^9668132+1

    Woohoo.

    p-1 is 2 * 3 * 11 * 43 * 181 * 449 * 823 * 39829 * 133013

  26. #146
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    I just found a quite smooth factor over 2^64:

    316782611247570464197 | 33661*2^9674568+1

    316782611247570464196 = 2 ^ 2 x 3 ^ 4 x 79 x 607 x 1423 x 1787 x 2081 x 3853

  27. #147
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    Just curious what people are using for B1 B2 bounds lately.

    Looking at the numbers in the threads, etc.

    - There are a little less than 250 (~244) tests in a 10000 range
    - People are getting about 4 factors per 10000.

    This boils down to one factor every ~60 tests.

    I'm assuming people are using the latest version of Prime95 with default settings???

  28. #148
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    I'm using sbfactor v1.25.5

    With 1GB of mem on a 3.0GHz Xeon it does B1=40000 B2=430000

    With 256MB of mem on a 1.7GHz P4 it uses B1=40000 B2=370000.

    Should I be using something else?

  29. #149
    Senior Member Frodo42's Avatar
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    I think that is the same bounds that I am using at the moment.

    I am using mprime v. 24.13 and with the with 49.5,1.5 at the end of my lines in worktodo.ini I get the following bounds
    B1=40000, B2=430000
    on my P4 3 GHz with 512 MB ram available for factoring.

  30. #150
    Moderator vjs's Avatar
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    Greenbank,

    I think sbfactor is slower than Prime95, you probably have a feeling already of how long it take to factor one test.

    Give Prime95 a try and see how long it takes using that program.

    Creating a worktodo.ini isn't as confusing as it initally looks when usings Mike's make program.

    Check the link at the top of the page.

    I personally prefer larger bounds but prime95 was written by people who really know what they are doing. So the defaults that frodo mentioned are probably best. Yours are pratically identical.

    If we had a shortage of numbers I.E. we ranges were not being passed by etc I'd suggest larger bounds.

  31. #151
    Senior Member Frodo42's Avatar
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    He this is efficient ... I wish the factors were always this dense ... two in a row

    Code:
    [Thu Aug 11 18:04:31 2005]
    P-1 found a factor in stage #2, B1=40000, B2=430000.
    21181*2^9802124+1 has a factor: 17993672443299947
    [Thu Aug 11 19:04:50 2005]
    P-1 found a factor in stage #2, B1=40000, B2=430000.
    19249*2^9802202+1 has a factor: 27491241436006571

  32. #152
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    We don't really have alot of people factoring...

    If people have time perhaps they can post the factors that they found, and do the following.

    Go to this site... http://www.alpertron.com.ar/ECM.HTM

    Enter the factor in frodos case

    27491241436006571 press factor --> proves that the factor is prime to begin with.

    Then subtract 1 and factor it again see what the factors are

    27491 241436 006570 = 2 x 5 x 499 x 1579 x 12197 x 286061

    12197 would have been the minimum B1 required
    286061 would have been the minimum B2 required

    17993672443299947

    17993 672443 299946 = 2 x 23 x 241 x 419 x 33749 x 114781

    33749 would have been the minimum B1 required
    114781 would have been the minimum B2 required

    ------------------------------------------------------------------------

    I know we use to do this in the past.

  33. #153
    Moderator vjs's Avatar
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    Example only three lines,

    P-1 found a factor in stage #2, B1=40000, B2=430000.
    19249*2^9802202+1 has a factor: 27491241436006571

    27491 241436 006570 = 2 x 5 x 499 x 1579 x 12197 x 286061

    Would be great.

  34. #154
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    (B1=40000 B2=370000)
    1596714224499631 | 22699*2^9750934+1
    p-1 = 2*3*3*5*31*37*1597*2647

  35. #155
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    (B1=40000,B2=370000)
    112967067451648889 | 24737*2^9754231+1
    p-1 = 2 * 2 * 2 * 19 * 19 * 97 * 179 * 181 * 1033 * 12049

  36. #156
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    3521346658647360362333 | 33661*2^9872640+1

    3521346658647360362332 = 2 ^ 2 x 23 x 31 x 11003 x 18217 x 39839 x 154619

    Found with optimal bounds for sieve depth = 49.7, factor value = 1.5, and RAM = 512 MB (B1=40000, B2=400000).

  37. #157
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    That was really a very close hit. Congratz, hc_grove!

  38. #158
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    Just found this one:
    18196089861334501 | 10223*2^10101065+1

    18196089861334500 = 2^2*3^3*5^3*37*47*701*997*1109

    Now, that's smooth! (But, I've seen even smoother)

  39. #159
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    Nice, and that was just above where my range stopped! Nothing on my first one with B1=600000,B2=1M. Bah.

    The lowest B2 for a factor > 2^49 was found by sieving:-

    512099950585939 (49 bits)

    512099950585938 = 2 x 3 x 11^2 x 13^2 x 29 x 41 x 103 x 173 x 197

    (B1 = 173, B2 = 197)

    The lowest B1 for a factor > 2^49 was:-

    582971265917041 (50 bits)

    582971265917040 = 2^4 x 3 x 5 x 7 x 11 x 13 x 19 x 37 x 61 x 71 x 797

    (B1 = 71, B2 = 797)

  40. #160
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    Woot!!!

    Looks like that was me and engracio, respectively.


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