20081215, 15:02  #12 
Banned
"Luigi"
Aug 2002
Team Italia
2^{6}·3·5^{2} Posts 
Nice work indeed!
10000! prints the actual known factors, ending with a 5532digits number whose status is unknown. Is it correct? No one factored it? Not a bug, just a question. A (typographical) bug is presented when I click on the 35660digits long expansion of 10000! I guess it's a bug related to the browser I use (Firefox 2.0): the related link is overwritten, I guess due to bad visualization properties of Firefox itself. Hint: would it be possible/feasible/worth trying to use a fixed length and add a [newline ] after it? Luigi 
20081215, 15:33  #13 
(loop (#_fork))
Feb 2006
Cambridge, England
2·3,191 Posts 
Thanks for writing this, it's a nice bit of work.
I see you haven't inhaled the most recent version of the Cunningham tables (I'm cheeky, I search for the factorisations I did most recently ...), but that gives the opportunity to use the submitfactor interface. I see you handle submissions of composite factors correctly (actually, it's slightly strange; I submitted a product of several largish factors of 2^23101, it recorded one of them, I submitted the same product again and it recorded a different one); I would appreciate some sort of error message if I put in a factor which doesn't divide the claimed number, rather than just getting the original page back. Do you have enough compute power on that server to do ~20 large GCDs per number submitted, so that you can submit a factor and it gets applied to every cofactor that it divides (IE you just say '903888164009442693590288077' rather than having to tell it that that divides 2^1606+1) ? Last fiddled with by fivemack on 20081215 at 15:45 
20081215, 16:31  #14  
Sep 2008
Krefeld, Germany
11100110_{2} Posts 
Quote:
Like this? 

20081215, 17:00  #15 
Banned
"Luigi"
Aug 2002
Team Italia
2^{6}×3×5^{2} Posts 

20081215, 17:07  #16 
"Robert Gerbicz"
Oct 2005
Hungary
2×7×103 Posts 
Some interesting answers:
Code:
1 Status: Unknown 11 Status: Unknown 12+1 Error: Negative 5*7/(2+3) Error: Not divisable 
20081215, 17:41  #17 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2×2,909 Posts 
have you added looking for algebraic factorizations

20081215, 17:52  #18  
Sep 2008
Krefeld, Germany
2·5·23 Posts 
Thank you
Quote:
Quote:
The Server is a duron 900/512MB Syd 

20081215, 18:29  #19 
(loop (#_fork))
Feb 2006
Cambridge, England
6382_{10} Posts 
You construct, using GMP and a product tree, twenty numbers
N1 = product {0th bit of compositeID is 1} C N2 = product {1st bit of compositeID is 1} C N3 = product {2nd bit of compositeID is 1} C N4 = product {3rd bit of compositeID is 1} C ... then, if you have a prime input, you can compute the GCDs (N1,p) through (N20,p), and that gives you the identity of the composite number that it divides. At least, provided that it divides only one composite number. (you can check which composite numbers share factors by a similar kind of process: suppose you expect the density to be pretty small: * label each number with a random integer in {1..100} * compute the hundred products of numbers with each label * compute the GCD of each pair of products * if it's 1, you know no two numbers in the sets with the given label share a factor * if it's prime, see which numbers with the given label it divided * if it's composite, check whether it equals one of the numbers in the sets, and if not then perform the whole randomlabelling process on the subsets with the two labels that appeared last time * repeat with different labelling until the answers consistently come out as all 1 ) I suppose this may still be too slow, it's 30 seconds or so on the K8/2200 I use for testing. Quite fun to implement, though. I did it with the ~10000 cofactors of partition numbers to check that there weren't any unexpected shared primes. Last fiddled with by fivemack on 20081215 at 18:36 
20081215, 18:35  #20 
(loop (#_fork))
Feb 2006
Cambridge, England
2·3,191 Posts 
Wow, that's much niftier than I expected.
I put in 10^100+6, it said 2*5000000..0003, I put in 5*10^99+3 and it listed it as fully factored, I went back to 10^100+6 and it's now recorded as fully factored! 
20081216, 01:53  #21 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,109 Posts 
It is nice.
I added to our 2^925+1 project's state. The p33 was missing. The composite that we are splitting is a c217. ...and it immediately showed up. 
20081216, 12:40  #22 
Nov 2008
2·3^{3}·43 Posts 
I notice 10^100+13 has a composite C94 factor. I did that one a few days ago by chance. Sadly I deleted my logfile and will do the whole thing again.
Edit: Just remembered I do have it. Problem solved! I have submitted the factor. P.S. How far have you factored Mersenne numbers? There is still a C78 for M277. Last fiddled with by 10metreh on 20081216 at 13:03 
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