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2020-11-02, 02:37   #23
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

89·113 Posts

Quote:
 Originally Posted by Gelly ... the Unique Primes list, so I'll be proving Phi(79710,10) (PRP21248)
Quote:
 Originally Posted by sweety439 There is a much smaller PRP factor of ...
...and is it a unique number?

2020-11-02, 06:54   #24
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

10110000010002 Posts

Quote:
 Originally Posted by Batalov ...and is it a unique number?
Sweety doesn't let knowledge or relevance get in the way of telling people what to do with their time. Shame, really.

2020-11-02, 13:49   #25
Dr Sardonicus

Feb 2017
Nowhere

11000010100102 Posts

Quote:
Originally Posted by Batalov
Quote:
 Originally Posted by Gelly ... the Unique Primes list, so I'll be proving Phi(79710,10) (PRP21248)
Quote:
 Originally Posted by sweety439 There is a much smaller PRP factor of ...
...and is it a unique number?
No. This is easily inferred from the latter post. The factor in the denominator is obviously larger than (107 - 1)/9. We find that

$\Phi_{6881}(10)\text{ has the prime factors 13763 and 3213467855281944242613907639214901656160438225853}$

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 2020-11-02 at 13:51 Reason: Forgot to say "the reciprocals of"

2020-11-02, 15:58   #26
paulunderwood

Sep 2002
Database er0rr

5·29·31 Posts

Quote:
 Originally Posted by sweety439 There is a much smaller PRP factor of Phi(n,10), (10^6881-1)/(10^983-1)/49141059632832877096172610809992897380296624365337454176129, see https://stdkmd.net/nrr/repunit/prpfactors.htm
For one who is obsessed with all things small, why don't you do it yourself? If you have Win10 you can install Linux in WSL and run Primo. What is your excuse?

2020-11-23, 22:11   #27
Gelly

May 2020

2F16 Posts

Quote:
 Originally Posted by Gelly Someone mentioned to me another list that is outdated in terms of compute power is the Unique Primes list, so I'll be proving Phi(79710,10) (PRP21248)
Done! Took longer than it should have, as I'm running double duty on a bunch of silly projects with too few computers, but it's done. As an added bonus, this one managed to get onto FactorDB, so no need to rely on having it be a top 20 Primo result for proof!

Next project is a PRP22234 (thanks, Thomas Ritschel!), which will be the largest proven irregular prime by a wide margin once it's done.

 2020-12-25, 02:11 #28 paulunderwood     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.
 2020-12-25, 04:46 #29 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 89×113 Posts I am nearly done with a 'small' Euler irregular (21k digits) and working on a new Bern irregular (26k digits).
2021-01-03, 05:55   #30
Gelly

May 2020

47 Posts

Quote:
 Originally Posted by Gelly Next project is a PRP22234 (thanks, Thomas Ritschel!), which will be the largest proven irregular prime by a wide margin once it's done.
Got the wrong one. I evidently don't understand how to order candidates in Primo right, so instead I proved 798*Bern(8766)/(487*4657*6468702182951641) (23743 digits).

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:
 Originally Posted by paulunderwood I am 5.5 months into certifying R49081. Expect the result in August or September 2021 assuming all goes well.
I'm extremely excited to hear about how fast that is, considering how large the candidate is! I may have to splurge on a 3990x myself, assuming that's your current machine, or just wait until the next major development in thread technology or whatever.

Quote:
 Originally Posted by Batalov I am nearly done with a 'small' Euler irregular (21k digits) and working on a new Bern irregular (26k digits).
Beans. Looks like my Bern irregular record will be terribly ephemeral. Nice work!

 2021-01-03, 06:30 #31 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 1005710 Posts Summarizing:798*Bern(8766)/(487*4657*6468702182951641) (23743 digits) Done 348054*Bern(8286)/(103*409*1381*123997*217757687456069039) (22234 digits) Reserved; in progress ? numerator(Bern(9702)) cofactor (26709 digits) Reserved; 60% done Re: order: the candidates will be done in the order in which you shift-clicked. 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 bit-size and then quickly see that size in the Bits dialog window: Attached Thumbnails
 2021-01-03, 06:45 #32 Batalov     "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
2021-01-03, 06:52   #33
Gelly

May 2020

578 Posts

Quote:
 Originally Posted by Batalov [*]348054*Bern(8286)/(103*409*1381*123997*217757687456069039) (22234 digits) Reserved; in progress ?
Yup! Using the formula you provided in the second post, it's 3.6% done.

I recall the split being between 4 and 5 being something you could determine using heuristic arguments - on Wikipedia, they give the big-O 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 wishy-washy 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 2021-01-03 at 07:03 Reason: i'm a bean and i can't read

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