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-   -   Very (large) PRPs? (https://www.mersenneforum.org/showthread.php?t=21294)

PawnProver44 2016-05-18 04:54

Here's the truth: I had asked someone for help to organize all the scripts, and they did that for me and showed me the proper output several times. However, they refused to give the prp's decimal expansion until I show them a proper output that I did myself.

PawnProver44 2016-05-18 05:09

1 Attachment(s)
I decided to put my largest found prp in an attachemt for proof that my claim about finding a 6056 digit prp is true. (though I just realized this now.)

If anyone does a proof on it, then becomes my largest found prime.

PawnProver44 2016-05-18 05:15

Are you talking about my 6056 digit prp or the one that I asked for help on and the people who helped me find it refuse to upload the decimal expansion?

EDIT: What just happened to post #69 I saw a few seconds ago?!?!

paulunderwood 2016-05-18 05:18

I tested it with my FU algorithm and it passed. It took about 5 seconds. ECPP would take several hours (??) and would give proof of primality -- 100%. :smile:

PawnProver44 2016-05-18 05:19

What does your "FU" algorithm stand for? :smile:

paulunderwood 2016-05-18 05:20

[QUOTE=PawnProver44;434281]Are you talking about my 6056 digit prp or the one that I asked for help on and the people who helped me find it refuse to upload the decimal expansion?

EDIT: What just happened to post #69 I saw a few seconds ago?!?![/QUOTE]

I deleted it, because I thought you had found a ~200k digit PRP, which I was willing to test for you. If you have found such a PRP, send it to me at [email]paulunderwood@mindless.com[/email] :smile:

paulunderwood 2016-05-18 05:25

[QUOTE=PawnProver44;434283]What does your "FU" algorithm stand for? :smile:[/QUOTE]

It tests (x+2)^(n+1) == 5+2*a (mod n, x^2-a*x+1) where a >= 0 is minimal such that jacobiSymbol(a^2-4,n)==-1. You can find my paper on it [URL="http://www.mersenneforum.org/showthread.php?t=19844"]here[/URL]. I have done several implementations: javaScript, Ruby and GWNUM. The latter is quick and is the one I would use to test a 200k digit number :smile:

PawnProver44 2016-05-18 05:28

[QUOTE=paulunderwood;434284]I deleted it, because I thought you had found a ~200k PRP, which I was willing to test for you. If you have found such a PRP, send it to me at [email]paulunderwood@mindless.com[/email] :smile:[/QUOTE]

yes, I has asked someone find that for me, but when I asked for an attachment with the decimal expansion, they said that I should find my own prp that size. I did however get average timing, and sieving up to 2e+10, with that showed me it only took about 2 to 4 days? Sounds weird.

LaurV 2016-05-18 05:46

The ~6k digit number is either a carmichael or a prime, it passed a random, 50-digit, prime bases PRP test. Like in "for(i=1,100,print([b=nextprime(random(10^50)), lift(Mod(b,n)^(n-1))]))" (ran in 4 cores, each base runs below 10 seconds on my laptop). Say your number is prime, with a probability of "99.9..(another~800 of nines)..9%". Now next step for you is to extend the ~6000 digits to ~200000 digits... :razz:

PawnProver44 2016-05-18 15:25

For this let me assume that there are 16,814 candidates remaining each sieve (to 2e+10) for 200k digits: Prp test takes (first please what is the time increase by doubling the number of digits?) Looked at a prime 400k digits [URL="https://primes.utm.edu/primes/page.php?id=119726"]here[/URL] and took about 12 min.) :smile:

paulunderwood 2016-05-18 15:36

That timing was on a 3.5GHz core with AVX using special mod. For Generic mod you can multiply by 4. For non-AVX you can multiply by 2 (??) and I will leave it to you work out the difference for your clock and 3.5GHz. Finally you can divide by 4 for a 200k number. Let's call it 40 minutes on your 2.15 GHz if it does not have AVX. So to find a 200k PRP you will need to do ~7000*40 core minutes i.e. 195 core days plus sieving time :smile:


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