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-   -   how to get F118 factor 1527888802614951*2^120+1? (https://www.mersenneforum.org/showthread.php?t=26166)

 bbb120 2020-11-05 05:45

how to get F118 factor 1527888802614951*2^120+1?

1527888802614951*2^120 + 1
1527888802614951 has 16 digits ,
if we calculate and try 10^5 per second ,
it will cost 10^15/(365*24*60*60)/10^5=317.1years for us to find this factor ,

Peter Strasser found this factor of F118,
and F118 approximate has 10^35digits , so it is impossible to find this factor by ECM .
who can tell me how to find this factor of F118

F118=2^(2^118)+1
[url]http://www.prothsearch.com/fermat.html[/url]

 Batalov 2020-11-05 05:52

[QUOTE=bbb120;562248]1527888802614951*2^120 + 1
1527888802614951 has 16 digits ,
[STRIKE]if we calculate and try 10^5 per second ,
it will cost 10^15/(365*24*60*60)/10^5=317.1years for us to find this factor ,[/STRIKE]
[/QUOTE]
That's irrelevant. (If all you have is a hammer, everything looks like a nail.)

Peter Strasser searched really hard with software [URL="http://www.fermatsearch.org/productivity.html"]shown here[/URL] and found this factor of F118, and that's the end of this story.

It says it right there - he [URL="http://www.fermatsearch.org/news.html"]found it with mmff[/URL]. That means - using a GPU.

Now, how did I find this one?
[QUOTE]June 26th, 2014
mmff discovers a new Fermat prime!

48595346636925 * 2^197+1 is a Factor of F195!!!
Serge Batalov found his fifth Fermat factor using a version of mmff extended by himself, for this discovery.
Congratulations to Serge from FermatSearch, for the third factor of the year![/QUOTE]
...not just [I]by running[/I] mmff. First, I was studying mmff's source for a week and then modified and experimented with it for a few weeks and when I was content with the new extended range (n in range high hundred-ish to low-200-ish) passing all tests, I ran it for several more months (and I paid a bunch for ~20 GPUs on AWS for several months) -- but I did get it. It was not easy, I will tell you that.

 bbb120 2020-11-05 06:23

[QUOTE=Batalov;562249]That's irrelevant. (If all you have is a hammer, everything looks like a nail.)

Peter Strasser searched really hard with software [URL="http://www.fermatsearch.org/productivity.html"]shown here[/URL] and found this factor of F118, and that's the end of this story.

It says it right there - he [URL="http://www.fermatsearch.org/news.html"]found it with mmff[/URL]. That means - using a GPU.

Now, how did I find this one?

...not just [I]by running[/I] mmff. First, I was studying mmff's source for a week and then modified and experimented with it for a few weeks and when I was content with the new extended range (n in range high hundred-ish to low-200-ish) passing all tests, I ran it for several more months (and I paid a bunch for ~20 GPUs on AWS for several months) -- but I did get it. It was not easy, I will tell you that.[/QUOTE]

so it is impossible for me to get f118 factor without GPU at present ?
I have no GPU ,I only have CPU

 Uncwilly 2020-11-05 06:34

[QUOTE=bbb120;562250]so it is impossible for me to get f118 factor without GPU at present ?
I have no GPU ,I only have CPU[/QUOTE]
And without an F1 car you won't win the Eifel Grand Prix.

You can buy or rent GPU's for a much more reasonable rate than buying an F1 car.

 Batalov 2020-11-05 09:01

[QUOTE=bbb120;562250]so it is impossible for me to get f118 factor without GPU at present ?
I have no GPU ,I only have CPU[/QUOTE]
There is an important consideration: a factor can only be found if it exists in the "range of possible to find". (You seem to think that there are factors everywhere, that programs just produce them like pancakes if you run them for a while. But they don't, the factors are very rare. People who find them invariably use extra knowledge - they know [I]where [/I]to search to improve chances of "winning the lottery".)

What if there is no factor of F118 to be found with this century's technology? Then you can set up your program on a hundred computers, press some buttons and wait for a 100 years and nothing will be found - because it isn't there. Maybe the next factor of F118 is (a 49-digit number)*2^120 + 1. Or maybe (a 79-digit number)*2^120 + 1, then what?

 LaurV 2020-11-05 10:33

[QUOTE=Batalov;562260]then what?[/QUOTE]
Then, you buy a F1 car. :razz:
And profit...

 Dr Sardonicus 2020-11-05 13:05

Things any do-it-yourselfers should keep in mind

Ask yourself, "Do I really want to do this myself?" If you're sure...

1) Understand the job you are trying to do
2) Know the techniques for doing it
3) Get (borrow, rent, buy) good tools and equipment
4) Make sure you know how to use them

Now, I'm unlikely to go searching for factors of Fermat numbers myself. Even so, I believe I can understand (1) and, to some degree, (2) for this task. So when I see a new announcement, I can have some appreciation of the hard work that went into making it happen.

The Fermat numbers being looked at are so large, about the only plausible way to find factors is to search through "small" candidates and hope you get lucky.

About all I know in this regard is that, for n > 1, all factors of F[sub]n[/sub] are congruent to 1 (mod 2[sup]n+2[/sup]). So you look at N = k*2[sup]n+2[/sup] + 1 and see whether N divides F[sub]n[/sub]. This would involve repeated squaring starting with Mod(2, N).

At this point, I ask myself, "Is it worthwhile to weed out N-values with small prime factors to avoid doing this test on them?" My guess is, "Seems pretty likely."

So, I'm guessing you need something that can do some fast sieving, and something that can handle repeated squarings (mod N) for largish-to-large N quickly.

If I were seriously interested in trying it myself, I would inquire further into what the right tools for the job might be, and how I can get the use of them within my budget.

The regular contributors to the various forums and threads devoted to specific factoring efforts know this stuff as well as anyone on the planet. I'm sure they would be more than happy to help someone who asks respectfully. Part of that respect is IMO putting some effort of your own into learning (1) and (2) above before seeking help.

Just my :two cents:

 mathwiz 2020-11-05 13:14

The [URL="http://www.fermatsearch.org/index.html"]FermatSearch[/URL] website is a reasonably good starting point; it has links to lots of software you can download, information about which ones are good for which ranges, FAQ, lists of available ranges, etc.

Good luck!

 retina 2020-11-05 13:19

[QUOTE=Dr Sardonicus;562276]1) Understand the job you are trying to do
2) Know the techniques for doing it
3) Get (borrow, rent, buy) good tools and equipment
4) Make sure you know how to use them[/QUOTE]Yes, good.

5) Learn what has already been searched and don't waste time redoing the same work.

 Xyzzy 2020-11-05 13:40

[QUOTE=retina;562281]5) Learn what has already been searched and don't waste time redoing the same work.[/QUOTE]:goodposting:

 Dr Sardonicus 2020-11-05 14:40

[QUOTE=retina;562281]5) Learn what has already been searched and don't waste time redoing the same work.[/QUOTE]
Good point.

:paul:
I remember something I did when I was 5 or 6 years old. I had gotten up and dressed. I happened to look where I always put my glasses when I went to bed. They weren't there! I looked on the floor nearby. No glasses. I thought, "Where else do I set them down?" So I looked in the bathroom, by the washbowl, since I cleaned them there. My dad was shaving. He asked me what I was doing. I told him I was looking for my glasses. He got kind of a funny smile on his face. I looked in a few other places. I went back to Dad, to ask him if he knew where my glasses were. But before I could ask, I saw my face in the mirror. And I knew where my glasses were.

So, yes, I can say from experience that it is a good idea to make sure you're not trying to find something that is already found.

I don't remember recent things so well, of course...

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