20201217, 17:02  #540 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
23×211 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!
"đ’‰ºđ’ŒŒđ’‡·đ’†·đ’€"
May 2003
Down not across
11,027 Posts 

20201217, 21:25  #542 
Feb 2017
Nowhere
5133_{10} 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
3×29×59 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
111101011111_{2} Posts 

20210224, 09:16  #546  
May 2004
FRANCE
245_{16} 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
5×787 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 Ă la BPSW before embarking on a lengthy ECPP path. 
20210224, 09:51  #548  
May 2004
FRANCE
7×83 Posts 
Quote:
Jean 

20210224, 14:46  #549  
"James Heinrich"
May 2004
exNorthern Ontario
2·5·353 Posts 
Quote:


20210224, 20:17  #550  
"Robert Gerbicz"
Oct 2005
Hungary
7^{2}·31 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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