mersenneforum.org PArtial factorization of 2^852348659-1
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 2018-05-01, 07:37 #1 ET_ Banned     "Luigi" Aug 2002 Team Italia 25·149 Posts PArtial factorization of 2^852348659-1 When This thread came out, I started some trivial trial factoring with Factor5, in the hope that a factor would appear and the claims of primality of the OP could be dismissed. Meanwhile, the OP changed many times his claim, and the thread was finally locked. After that, I concluded my trial-factoring, but had no place to propagate its results, so I am publishing my results here in the faint hope to avoid the same waste of work (just about 9 GHz/day) to someone else (maybe either Aaron, James, Chris or George might be interested to update the general DB). Code: Trial-factoring M852348659 in [2^1, 2^60-1] M852348659 has 0 factors in [2^1, 2^60-1]. Trial-factoring M852348659 in [2^60, 2^62-1] M852348659 has 0 factors in [2^60, 2^62-1]. Trial-factoring M852348659 in [2^62, 2^65-1] M852348659 has 0 factors in [2^62, 2^65-1]. Trial-factoring M852348659 in [2^65, 2^75-1] M852348659 has 0 factors in [2^65, 2^75-1]. HTH Last fiddled with by ET_ on 2018-05-01 at 07:46
 2018-05-01, 07:50 #2 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
2018-05-01, 12:35   #3
axn

Jun 2003

3×112×13 Posts

Quote:
 Originally Posted by Dubslow PrimeNet already shows no factors < 2^80
If you had used the "Show full details", you would have seen the below information.

Quote:
 Originally Posted by Dubslow Ditto mersenne.ca, which admittedly says that 70-80 was done by the user in question. Perhaps it's worth double checking that anyways.

Last fiddled with by axn on 2018-05-01 at 12:36

 2018-05-01, 15:33 #4 Mark Rose     "/X\(‘-‘)/X\" Jan 2013 B3C16 Posts I'll take it to 84 over the next 10 days or so.
2018-05-01, 20:56   #5
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×2,399 Posts

Quote:
 Originally Posted by axn If you had used the "Show full details", you would have seen the below information.
Gah! Back in my day, we didn't need to click no fancy checkboxes to get detailed info!

2018-05-03, 17:19   #6
Mark Rose

"/X\(‘-‘)/X\"
Jan 2013

22×719 Posts

Quote:
 Originally Posted by Mark Rose I'll take it to 84 over the next 10 days or so.
No factors < 2^81. I double checked everything below 2^80.

Work continues.

Last fiddled with by Mark Rose on 2018-05-03 at 17:20

2018-05-10, 15:37   #7
Mark Rose

"/X\(‘-‘)/X\"
Jan 2013

22·719 Posts

Quote:
 Originally Posted by Mark Rose I'll take it to 84 over the next 10 days or so.
No factors found.

2018-05-10, 16:28   #8
CRGreathouse

Aug 2006

22×1,483 Posts

Quote:
Originally Posted by Mark Rose
Quote:
 Originally Posted by Mark Rose I'll take it to 84 over the next 10 days or so.
No factors found.
So the Bayesian probability that M852348659 is prime rises to

$\frac{84e^\gamma}{852348659} \approx 175/10^9$

from

$\frac{e^\gamma\log(2\cdot852348659)}{852348659\log 2} \approx 64/10^9$

before we did any work on the number.

That's 1/8 the chance of being dealt a royal flush (in 5 cards). Or for the hold 'em fans, 40% the chance of being dealt a full house four times in a row.

 2018-05-10, 18:27 #9 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23×7×163 Posts Four times in a row? That's easy. Attached Thumbnails
2018-05-31, 10:34   #10
LaurV
Romulan Interpreter

Jun 2011
Thailand

53×71 Posts

Quote:
 Originally Posted by CRGreathouse That's 1/8 the chance of being dealt a royal flush (in 5 cards). Or for the hold 'em fans, 40% the chance of being dealt a full house four times in a row.
What? You missed few zeros here, or so... There are 52 cards in a pack and you randomly pick 5, you get one in C(52,5) chances to get the royal flush. This is like one in half million (edit: a bit less than 2 in a million (10^6)) or so... We played poker in our life and had royal flushes occasionally, but mersenne primes with exponents in 850M didn't find yet....

Last fiddled with by LaurV on 2018-05-31 at 10:37

2018-05-31, 12:15   #11
CRGreathouse

Aug 2006

10111001011002 Posts

Quote:
 Originally Posted by LaurV What? You missed few zeros here, or so... There are 52 cards in a pack and you randomly pick 5, you get one in C(52,5) chances to get the royal flush.
That's the chance to get (for example) a royal flush in hearts, about 385 in 10^9. The chance to get any royal flush is 4 times larger, about 1539 in 10^9. You'd have to divide by 8 or 9 to be the same size as the probability I computed for M852348659.

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