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#23 | |
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Dec 2010
Monticello
179510 Posts |
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I am working on the required reading, but have come to the realisation that there are parts of comp sci (starting with Turing Machines) that were excluded from my engineering comp sci education. You have to realize it's hard work and it doesn't happen overnight, especially as I do have a full-time job amongst a bunch of engineers who don't know what a Mersenne prime is in the first place, have long since forgotton what a Laplace transform is, and wouldn't know the difference between the Eiger and an Eigenvalue. I'm also debating whether taking on a project here; I have one that has to finish first, and the coding involved is temporarily soaking up all my available math bandwidth. And while I'm at it, do let me know if I have correctly shown that the number I was asking about is divisible by 40. You will get much further if you encourage and reward correctness, just ask BF Skinner! Last fiddled with by Christenson on 2011-07-07 at 03:00 |
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#24 | |
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Dec 2010
Monticello
5×359 Posts |
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Wblipp: Current mfaktc assignments are of the form factor=43112609,72,73 which means: factor 2^43112609-1 with trial factors between 2^72 and 2^73. What's the best representation for the assignments you would want TF'ed, in general form? |
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#25 | |
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"William"
May 2003
New Haven
2·7·132 Posts |
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factor=base,exponent,lowlim,hilim meaning: trial factor base^exponent-1 with primes of the form k*exponent+1 between lowlim and hilim. I'd be ecstatic with factor=base,exponent,multiplier,lowlim,hilim meaning: trial factor base^(exponent*multiplier)-1 with primes of the form k*exponent+1 The ecstatic form makes it easy to trial factor a^n+1 by making multiplier=2. It also allow me to simultaneously search all the large algebraic factors of 719^(2^2 * 3 * 183925013)-1, one of the target groups from http://oddperfect.org/FermatQuotients3.html I have no particular preference for how represent lowlim and hilim, If powers of two are still easy, that has the advantage of familiarity. If there is something that much easier to implement and not too hard to understand, that would be fine, too. William |
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#26 |
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Dec 2010
Monticello
111000000112 Posts |
I'm wondering if we don't want to talk about a set of multipliers here, rather than a single multiplier, with the single multiplier being a special case of the set. After all, if I calculate 719^(large prime #) mod factor candidate, then squaring or cubing the result, mod factor candidate, is significantly easier than re-performing that large exponentiation, which, will on average, have two dozen squarings and a dozen multiplies by the base.
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#27 |
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"Forget I exist"
Jul 2009
Dumbassville
203008 Posts |
http://www.mersenneforum.org/showpos...postcount=2269
gives PARI code you can modify for GPU. Yes the TF allows all bases the MTF code is just base 2 but can be transformed into a code for all bases. |
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#28 | |
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Jan 2005
Caught in a sieve
5·79 Posts |
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#29 |
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Jun 2011
Henlopen Acres, Delaware
7×19 Posts |
I was wondering if there would be any "ideal bases" that would find primes with as few k's and as few n's as possible, but with a fair number of +/- constants after.
Maybe bases of the form: 1001 10001 100001 10000 .... 00001 So instead of varying the base for a trial run, fix the base, have a few "n's", more k's, and a boatload of constants. 1000000000001^101 = 1000000000101000000005050000000166650000004082925000079208745001267339920017199613200202095455102088319702719212541264998940114101232050855758460963550957197485177420325414028911439100404799285502021331095676088416183742868800643345599775151475315716866271985731991536958006272477324491587179583856827926513500592331682134232579237482724796258175554928347429243705759667435386284646565561783499781088436421274567683484763110034583156211880651084039705547277573596329704668260931674800987143837558770872396880395834070399710309414813490464221426099706416891571300436092852676414058203975930443434784401657764255585676928675700936563632656314818056403647636947150112778481979034466162277360787229204891362831139710078397045966918388927493954592422300818497143405081842346660487844922705804890338008781881274489066410808378334402491151807644536641030814549929237444717786503985982357609317621631880828590342131116476187226483491441690568696242366219581466610133513343184956453742237790345675933406095501985615649100397100815414027363537485177130960963550907092050855750458940114100059212541264842088319702700202095455100017199613200001267339920000079208745000004082925000000166650000000005050000000000101000000000001 k*(b^p)+c k = 2 b = 1000000000001 p = 101 Check constants in the range of +1 to +5001 (the next prime would be at +568 for the large number shown above) k = 3 b = 1000000000001 p = 101 Check constants in the range of +1 to +5001 etc. With a clever combination of b and p I would make the SWAG that a prime would be "in the neighborhood" and checking a few constants might pay high dividends. Last fiddled with by LiquidNitrogen on 2011-07-09 at 22:30 |
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#30 | |
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Dec 2010
Monticello
34038 Posts |
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Keep talking...remember that mfaktc is going to run out of effective work fairly soon and should be looking for new targets and improved methods. |
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#31 |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
I'm supposed to be preparing to move as far as I know but I also don't have the skill needed to do the programming.
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#32 | |
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"William"
May 2003
New Haven
2·7·132 Posts |
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William |
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#33 | |
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Dec 2010
Monticello
5·359 Posts |
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if TF factors 41^P-1, then it also factors 41^(K*P)-1, where K and P are positive integers and P is usually prime...since 41^P mod TF =1, and (41^P)^k mod TF = 1^k =1. therefore, we should be TF'ing over some large set of TFs by calculating 41^P mod TF, and then reporting if 41 ^P mod TF is a reasonably small root of unity (up to some bound) mod TF. Or can you give me a slightly more concrete example? |
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