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-   -   Gratuitous factors thread (https://www.mersenneforum.org/showthread.php?t=10029)

KyleAskine 2011-12-13 00:12

[QUOTE=Mini-Geek;281999]I've just found my first Mersenne P-1 factor (IIRC), discovered as part of GPU to 72. :smile: It's 76 bits (~2^75.22, or 23 digits) and with k=83*9029*15359*40433 just fit into the stage 1 bounds.
[CODE]P-1 found a factor in stage #1, B1=445000.
M47403971 has a factor: 44122627934020297446119[/CODE]Curiously, GPU to 72 credited me 3.020 GHz-days, though PrimeNet gave me 1.1202 GHz-days (a result which mersenne-aries [URL="http://mersenne-aries.sili.net/credit.php?worktype=P-1&exponent=47403971&f_exponent=&b1=445000&b2=445000&numcurves=&factor=44122627934020297446119&frombits=&tobits=&submitbutton=Calculate"]agrees on[/URL]). Does GPU to 72 lack the information for properly crediting factors, or do they use different criteria?[/QUOTE]

I would guess it credited you with a stage 2 find.

Ask chalsall.

chalsall 2011-12-13 01:35

[QUOTE=KyleAskine;282000]I would guess it credited you with a stage 2 find.

Ask chalsall.[/QUOTE]

Yes, you are correct. I don't have access to the knowledge which would tell me in which stage the factor was found, nor what the bounds might have been.

Keep in mind the PrimeNet stats are authoritative; the G72 stats might occasionally be a little off.

fivemack 2011-12-26 20:29

aliquot 1920:2385 = 2^2 *
[code]
Mon Dec 26 11:07:27 2011 prp79 factor: 1080971570084750386114055097550469938365602220050226861147118886537318654934991
Mon Dec 26 11:07:27 2011 prp83 factor: 42412151161131291621417440650540020062317130151681212879891510543747189551240271449
[/code]

10metreh 2012-01-01 20:08

Aliquot 895480:1808 had a c85 which was almost a perfect triple nice-split:
[code]c85 = p28 * p29 * p29
p28 = 9521823302237403992799998219
p29 = 16820732007128558763841201367
p29 = 17078868238455374562369008041[/code]

schickel 2012-01-09 07:04

Nice split
 
[URL="http://factordb.com/index.php?id=1000000000018244084"]3^345-2[/URL] = 505027 * 25631055313 * 620565502016389534342436351929 * c119

c119:[code]prp60 factor: 212176376860378077776315019478089127085008105128842176485863
prp60 factor: 237283278978015820127013108788399507515644303307564184722133[/code]So the 3 largest factors are p30 * p60 * p60!

Batalov 2012-02-01 18:46

From Aliquout 5448 i1366, by P-1

[CODE]GMP-ECM 6.2.3 [powered by GMP 4.3.0] [P-1]
Input number is 52364077819281691479821647122905290170177880534370243423677325607770912673020572598391633958558086124535515922715712863098724652052797240076435998369 (149 digits)
Using B1=1000000000, B2=205702371522480, polynomial x^1, x0=1156318219
Step 1 took 241118ms
Step 2 took 153645ms
********** Factor found in step 2: 10659456725594355643693545042984455938999705295539
Found probable prime factor of 50 digits: 10659456725594355643693545042984455938999705295539
[/CODE]

It is unreserved. (I took 50 largest factors from open sequences and ran P-1, just to see if P-1 can do anything these days...)

Mr. Odd 2012-02-02 13:55

ECM score!

ecm: 1290/2352 curves on C135 input, at B1 = 3M, B2 = gmp-ecm default
ecm: found prp51 factor = 276763067825903334965164807383096632510038738916593

swishzzz 2012-02-03 18:57

2^574+5 c142:
r1=11025276101966306994999199806300967370241421 (pp44)
r2=58348218048475237797758620061263725083658323 (pp44)
r3=2487541752115632972745724234914170833139290349268646101 (pp55)

fivemack 2012-02-25 10:34

Lucas(1701) (29-bit large primes, sieve 5M-15M with alim=rlim=15M, took 500 CPU-hours to sieve)

[code]
Sat Feb 25 03:55:38 2012 prp95 factor: 22053198539174195948748486948346798956430614869939319268405502791755896077736394742050587488171
Sat Feb 25 03:55:38 2012 prp109 factor: 6201756859165816632234222132664550414494970621916970706979309159192424400764254045924913302931130250212889091
[/code]

Thanks to Random Poster for saying lucas(N)^2 is a good number to do linear-dependency calculations against to get the sextic for lucas(21N).

Batalov 2012-02-25 11:11

...or simply lucas(2N)

Random Poster 2012-02-25 12:51

[QUOTE=Batalov;290854]...or simply lucas(2N)[/QUOTE]
That works too, since L(2n)=L(n)^2-2(-1)^n, but then you lose the beautiful duality: if you express either F(kn)/F(n) or L(kn)/L(n) as a polynomial in L(n)^2 and the other one as a polynomial in 5F(n)^2, the coefficients of those polynomials will have the same absolute values. The polynomials will also depend only on k, so you could just make a table of them, as only a few values of k give useful SNFS polynomials.


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