20201102, 02:37  #23 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
89·113 Posts 

20201102, 06:54  #24 
"Curtis"
Feb 2005
Riverside, CA
1011000001000_{2} Posts 

20201102, 13:49  #25  
Feb 2017
Nowhere
1100001010010_{2} Posts 
Quote:
in addition to the PRP cofactor. The decimal expansions of the reciprocals of these two primes have period 6881. Last fiddled with by Dr Sardonicus on 20201102 at 13:51 Reason: Forgot to say "the reciprocals of" 

20201102, 15:58  #26  
Sep 2002
Database er0rr
5·29·31 Posts 
Quote:


20201123, 22:11  #27  
May 2020
2F_{16} Posts 
Quote:
Next project is a PRP22234 (thanks, Thomas Ritschel!), which will be the largest proven irregular prime by a wide margin once it's done. 

20201225, 02:11  #28 
Sep 2002
Database er0rr
5×29×31 Posts 
I am 5.5 months into certifying R49081. Expect the result in August or September 2021 assuming all goes well.

20201225, 04:46  #29 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
89×113 Posts 
I am

20210103, 05:55  #30  
May 2020
47 Posts 
Quote:
Currently being verified by FactorDB and will post with a new proof code soon enough. It beats the old record by almost 10k digits and it was an accident. Figures, haha. Next is the actual PRP I meant to work on, which didn't actually list the candidate, defeating the purpose of the "primo reservation" bit. The PRP is 348054*Bern(8286)/(103*409*1381*123997*217757687456069039) (22234 digits). Quote:
Quote:


20210103, 06:30  #31 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10057_{10} Posts 
Summarizing:
Re: order: the candidates will be done in the order in which you shiftclicked. If you select/drag a range of numbers, then primo will "decide itself". Hint: always have the size estimates at hand and (as needed) convert them to bitsize and then quickly see that size in the Bits dialog window: 
20210103, 06:45  #32 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
89×113 Posts 
Btw, a few people asked before, how to estimate how far are you down the path of the proof. A number of years ago, I'd splined the runtimes to "Bits" size and the power slope was not quite "4" and not quite "5". (Probably due to backtracks. If descent was ideal, then one could expect nearly "4".)
I use "4.5". That is: if the progress in bits is (using example shown above): 70889 / 88724, then the percent remaining work is : Code:
? (70889 / 88724) ^ 4.5 0.36426 \\ this much remaining, approximately ? 1  % 0.63573 \\ this much done, approximately 
20210103, 06:52  #33  
May 2020
57_{8} Posts 
Quote:
I recall the split being between 4 and 5 being something you could determine using heuristic arguments  on Wikipedia, they give the bigO as definitely O(log(n)^(5+epsilon)), with some versions of heuristic arguments bringing it down to O(log(n)^(4+epsilon)). Since Wikipedia is very wishywashy about it, I would put it somewhere between those two estimates, giving what you found, 4.5. Edit: Think you got your numbers backward  (80000/80000)^4.5 = 1, meaning 100% remaining, not 100% done. At least, when you read the number from Primo directly. Last fiddled with by Gelly on 20210103 at 07:03 Reason: i'm a bean and i can't read 

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