20081201, 18:41  #650  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,969 Posts 
Quote:
i proved it with: http://www.alpertron.com.ar/ECM.HTM 

20081201, 19:27  #651  
May 2003
3·7·11 Posts 
Quote:
? factor(hf(11,9,9)) params=[11, 9, 11/9, 1/3, 11, 1, 198] [[5750932177, 1; 104907806276834514493, 1], [100981, 1; 251857, 1; 26343084072127026078073, 1]] I currently have 2 simple scripts. One works for half the candidates, another works for about a half, but with a reasonable overlap. Many of the cases that you listed still require further tweaking to one of my scripts, but that's mostly just sign fiddling. However, I'm hoping to unify the logic so that there's just one script and it works out exactly what needs to be done. Presently I'm using polynomialfactors = factor(polsubst(polcyclo(n),x,aur*x^2)))[,1]~ numericfactors = subst(polynomialfactors, x, sq)*fudge for suitably chosen n, aur, sq, fudge dependent on the two bases and the index. sq's basically the squared part, and aur is the nonsquare part. This is a bit slow as it relies on polynomial factoring, which is at least pretty swift compared to integer factoring. Dr. Sardonicus might investigate if there's a Gauss sum representation which would be practically instant to evaluate. Bob  do you think Gauss sums are a worthwhile thing to look at, or do you think it's a nohoper? 

20081201, 23:23  #652 
Nov 2003
7460_{10} Posts 

20081201, 23:46  #653  
May 2003
11100111_{2} Posts 
Quote:
If so, why did you keep the first paragraph, and only the final sentence of my final paragraph, thereby chopping out the immediately preceding context for the final question and giving it an entirely different context? In particular, what on earth made you think that '99' was in any related to that final question?! And oddly, I think you'll find that I'm the one (along with Sun et al., and Wagstaff), who is saying that 11^99+9^99 admits an Aurefeuillian factorisation, and you, Gauss, and Kraitchik (and probably Brent too), who are avoiding that number with your "not 99 because 9 is already a square" and "squarefree n" comments. Anyway, the question, this time containing all necessary context in one sentence was: "Do you think that there is a way to express the Aurefeuillian factorisations of homogeneous cyclotomics in terms of triviallycalculable Gauss sums, in the same way that simple nth rootofunity cyclotomics' factors can be, rather than having to resort to heavy, but certainly tractible, polynomial factorisation?" 

20081203, 17:54  #654  
Nov 2003
2^{2}×5×373 Posts 
Quote:
the coefficients simply by solving some linear simultaneous equations. 

20081203, 18:48  #655  
May 2003
3·7·11 Posts 
Quote:
It seems as if you final sentence is referring to: http://wwwmaths.anu.edu.au/~brent/pub/pub127.html http://wwwmaths.anu.edu.au/~brent/pub/pub135.html I'll grab them and give them a read. They're quite possible more efficient than polynomial factorisation, and worth looking into. Thanks. 

20081214, 19:19  #656 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
41×269 Posts 
An update is in progress as I type. There are 166 composites left in the main tables now.
Paul 
20081215, 14:58  #657  
Jun 2003
Ottawa, Canada
3·17·23 Posts 
Quote:


20081215, 19:16  #658  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
41×269 Posts 
Quote:


20081229, 14:57  #659 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
41·269 Posts 
Last of the year
I've just posted the last update of 2008.
Paul 
20081229, 15:36  #660 
Oct 2004
Austria
100110110010_{2} Posts 
Can you please put a link to the reservation page at the tablespage? It is otherwise quite tricky to find it (digging through a HUGE number of posts in this thread...), when you don't have a bookmark (yet).
Last fiddled with by Andi47 on 20081229 at 15:38 Reason: inserted missing words 
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