20201217, 17:02  #540 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
2×7×13×29 Posts 
As I get older I notice 2 things starting to happen:
1. I repeat myself 2. I repeat myself 
20201217, 17:40  #541 
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
10110110001110_{2} Posts 

20201217, 21:25  #542 
Feb 2017
Nowhere
6,229 Posts 

20201220, 18:15  #543 
Nov 2018
Poland
3×5 Posts 
The 350th fullyfactored or probablyfullyfactored Mersenne number with prime exponent
The 350th fullyfactored or probablyfullyfactored Mersenne number with prime exponent (not including the Mersenne primes themselves) is M1399.
The most recent factor (61 digits) was found by Ryan Propper on December 19 (UTC) and the PRP test was done by mikr and myself. There are 3 factors in all, plus the cofactor. 
20201220, 19:48  #544  
Feb 2017
Nowhere
1100001010101_{2} Posts 
Quote:
Code:
? n=(2^13991)/28875361/4320651071020341609502042221583629017824960697/9729831901051958663829453004687723271026191923786080297556081; ? isprime(n) %2 = 1 The manual entry says Quote:
Last fiddled with by Dr Sardonicus on 20201221 at 21:03 Reason: Add code tags 

20210224, 06:42  #545 
Sep 2002
Database er0rr
5·29·31 Posts 

20210224, 09:16  #546  
May 2004
FRANCE
1001100111_{2} Posts 
Congrats for this nice result!
Quote:
Jean P.S. : How did you do the PRP test before the certification using Primo ? 

20210224, 09:46  #547 
Sep 2002
Database er0rr
1000110001111_{2} Posts 
Thanks, Jean.
I merely got the candidate from www.mersenne.ca. I might have run a 3PRP to be sureish. Anyway, Primo does a quick Fermat+Lucas ร la BPSW before embarking on a lengthy ECPP path. 
20210224, 09:51  #548  
May 2004
FRANCE
3·5·41 Posts 
Quote:
Jean 

20210224, 14:46  #549  
"James Heinrich"
May 2004
exNorthern Ontario
4098_{10} Posts 
Quote:


20210224, 20:17  #550  
"Robert Gerbicz"
Oct 2005
Hungary
5×17×19 Posts 
Quote:
Quote:
https://www.mersenne.org/report_expo...exp_hi=&full=1 Notice that for N=(k*2^n+c)/d we're using a Fermat test using base^d as base, then (base^d)^N=base^d mod N should hold for a prp number. So base^(k*2^n+c)==base^d mod N, to help a lot we're using reduction mod (d*N)=mod (k*2^n+c). Then do only one big division at the end of the test, in real life d is "small", at most ~1000 bits. And you can build in a strong check in the routine like for the normal prp test for k*2^n+c numbers. There is only a very small slow down at error check, because here our base is "large". ps. so actually p95 has done a Fermat test using 3^d as base, and not 3. The reason is that we have a check only for 3^d [or base^d]. Last fiddled with by R. Gerbicz on 20210224 at 20:18 

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