20210628, 20:14  #1 
Jun 2012
Boulder, CO
5·61 Posts 
Announcing a new Wagstaff PRP
(2^15135397+1)/3 is a Fermat Probable prime! (4556209 decimal digits)
Also submitted to PRPTop. I am searching the range n=13M .. 17M currently, and nearly done. No other discoveries as of yet. 
20210628, 21:19  #2 
Sep 2002
Database er0rr
7341_{8} Posts 

20210628, 23:38  #3 
Sep 2006
The Netherlands
3^{2}×83 Posts 
That's a very lucky find! Congrats on that one!
Had you asked me i would've guessed next one might've lurked at 30M earliest and 70M latest. 
20210629, 11:29  #4 
Feb 2017
Nowhere
2^{2}·3·401 Posts 

20210629, 13:24  #5 
Jun 2012
Boulder, CO
5·61 Posts 

20210629, 15:32  #6  
Feb 2017
Nowhere
4812_{10} Posts 
Quote:
Silly me, I failed to consider that you had tested multiple bases. Of course, these numbers automatically "pass" the test to base 2. Paper and pencil suffices for this one. If p > 3 is prime, N = (2^p + 1)/3, then (N1)/2 = (2^(p1)  1)/3 is odd and divisible by p, so N = (2^p + 1)/3 divides 2^p + 1, and 2^p + 1 divides 2^((N1)/2) + 1, so N divides 2^((N1)/2) + 1. Now 2^((N1)/2) + 1 divides 2^(N1)  1, so N divides 2^(N1) 1, but does not divide 2^((N1)/2)  1. 

20210629, 16:59  #7  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
5^{2}·11^{2} Posts 
Quote:


20210629, 17:02  #8  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
5^{2}·11^{2} Posts 
Quote:
Last fiddled with by sweety439 on 20210629 at 17:02 

20210629, 20:23  #9  
Mar 2019
C4_{16} Posts 
Quote:
Maybe a mod can ban this guy until he stops flooding the forum with requests for other people to do things. 

20210629, 20:37  #10  
Sep 2002
Database er0rr
13·293 Posts 
Quote:
Last fiddled with by paulunderwood on 20210629 at 20:38 

20210629, 21:33  #11 
"Curtis"
Feb 2005
Riverside, CA
5·23·43 Posts 
We gave him some time off to consider his behavior, and it hasn't changed much. I suppose your suggestion and this reply might be considered yet another warning to Sweety before the banhammer falls again.

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