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 2016-05-18, 19:37 #89 PawnProver44     "NOT A TROLL" Mar 2016 California 19710 Posts I already upgraded to one, but have to set up Wifi connections. And there are still easier ways to find such prps on Windows? Great!
 2016-05-20, 14:30 #90 PawnProver44     "NOT A TROLL" Mar 2016 California 197 Posts I had just come up with the most efficient, but useless method to generate random primes: Pick a random integer a. Now find a prime such that (a+1)^p-a^p is prime. Now ignore that large prime, then take the exponent p, and of course it is prime. Work well here, see: A random 5 digit prime: Find a expoent p such that 8^p-7^p is prime. Got it! 8^76213-7^76213! Result = 76213 And very RANDOM distribution too, now does anyone know how I use these same methods for larger prines? If you thought this was 'useless' to compute, well I already pointed that out. Spoiler Alert: I am working right now on prps sizes 12022, 50337, 142367, and 200059 digits. I will attach them here with a similar file name as I did with the last prp. Last fiddled with by PawnProver44 on 2016-05-20 at 14:52
 2016-05-20, 14:55 #91 paulunderwood     Sep 2002 Database er0rr DCD16 Posts 8^76213-7^76213 was found in 2003 by Ananda Tallur & Jean-Louis Charton. Move along: nothing new to see here
2016-05-20, 16:00   #92
danaj

"Dana Jacobsen"
Feb 2011
Bangkok, TH

2·3·151 Posts

I believe he was using that known result to show that 76213 is prime.

Quote:
 Originally Posted by PawnProver44 And very RANDOM distribution too
You are asserting something without evidence and which seems very likely to be false. The distribution of exponents in the top prime tables is almost certainly not uniform.

 2016-05-20, 16:16 #93 PawnProver44     "NOT A TROLL" Mar 2016 California 197 Posts If there was a pattern distribution, then why does G.I.M.P.S exist? Also given that there should be infinitely many primes of the form a^(p+1)-a^p for fixed p (only prime p) concludes that if a is chosen randomly, then any p value if possible would make a "random" distribution. Correction: This only holds for values of p that are prime, (I think that's obvious however). Last fiddled with by PawnProver44 on 2016-05-20 at 16:28
 2016-05-20, 16:45 #94 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100100001011112 Posts It's time to fix an overlooked problem. User PawnProver44 = Trejack = Treject = JoeBroe. One cannot have one "mulitple personality" banned and another still spewing the same nonsense. Perhaps it is time to ban by IP, that's for Xyzzy to decide.
2016-05-20, 17:06   #95
pepi37

Dec 2011
After milion nines:)

101011011102 Posts

Quote:
 Originally Posted by Batalov It's time to fix an overlooked problem. User PawnProver44 = Trejack = Treject = JoeBroe. One cannot have one "mulitple personality" banned and another still spewing the same nonsense. Perhaps it is time to ban by IP, that's for Xyzzy to decide.

AVE CEZARE!
At least some good news to defeat this spam

2016-05-20, 18:24   #96
danaj

"Dana Jacobsen"
Feb 2011
Bangkok, TH

2·3·151 Posts

Quote:
 Originally Posted by PawnProver44 If there was a pattern distribution, then why does G.I.M.P.S exist? Also given that there should be infinitely many primes of the form a^(p+1)-a^p for fixed p (only prime p) concludes that if a is chosen randomly, then any p value if possible would make a "random" distribution. Correction: This only holds for values of p that are prime, (I think that's obvious however).
You're saying that if you randomly select p using a random distrbution, then p is selected with a random distribution? What's the point? What I meant was that your method seems to be:

1) select an entry from the top proven prime lists of the form (a+1)^p-a^p,

2) p is prime.

Step 1 does not yield a uniform distribution of p values, as one is selecting from a very small set. Without knowing which small primes are prime (the point of this), how do you make sure every small prime is represented in step 1, and an equal number of times as the other small primes?

Let's say that none of the tables had an entry for 76243, but did for 76231 and 76249. You'd never select 76243, which doesn't sound like a good distribution (some values never appear).

If instead you first select a random small prime p, with uniform distribution, then go find it on the table, then of course it works since you start out assuming what you're concluding. But it leaves one wondering what the point is -- if you knew all the small primes so you could do this, what's the point of this circular exercise?

Now I realize that this is just an example showing tiny numbers, and we should be thinking of some other method that gets much larger values. But:

1) if your method relies on looking at previously found large primes, then I claim this isn't a good distribution. People search in patterns.

2) if your method relies on finding larger forms that happen to be easy to prove, then this is clearly not random. You would need to show that this restricted set of large primes that are amenable to your chosen proof method results in a uniform and complete distribution of smaller primes. Two common methods for this, Maurer and Shawe-Taylor, both have a large part (if not the majority) of their papers devoted to explaining how the distribution is flawed within tolerable bounds for their applications.

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