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#650 | |
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
132728 Posts |
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i proved it with: http://www.alpertron.com.ar/ECM.HTM |
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#651 | |
May 2003
3·7·11 Posts |
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? 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 non-square 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 no-hoper? |
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#652 |
Nov 2003
1D2416 Posts |
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#653 | |
May 2003
3·7·11 Posts |
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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 trivially-calculable Gauss sums, in the same way that simple n-th root-of-unity cyclotomics' factors can be, rather than having to resort to heavy, but certainly tractible, polynomial factorisation?" |
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#654 | |
Nov 2003
1D2416 Posts |
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the coefficients simply by solving some linear simultaneous equations. |
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#655 | |
May 2003
3·7·11 Posts |
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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. |
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#656 |
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
245338 Posts |
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An update is in progress as I type. There are 166 composites left in the main tables now.
Paul |
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#657 | |
Jun 2003
Ottawa, Canada
7·167 Posts |
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#658 | |
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3×3,529 Posts |
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#659 |
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3·3,529 Posts |
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I've just posted the last update of 2008.
Paul |
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#660 |
Oct 2004
Austria
2·17·73 Posts |
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Can you please put a link to the reservation page at the tables-page? 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 2008-12-29 at 15:38 Reason: inserted missing words |
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