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 2021-06-30, 07:53 #12 LaurV Romulan Interpreter     Jun 2011 Thailand 72·199 Posts Good choice! Thanks to the mod colleague who did it.
2021-06-30, 13:03   #13
Dr Sardonicus

Feb 2017
Nowhere

3·1,609 Posts

Quote:
 Originally Posted by LaurV Good choice! Thanks to the mod colleague who did it.
I agree, good decision.

But we mere Supermods can't ban a blogger. That requires an Act of God. And God has heard the prayers of the afflicted, and has done an Act of Mercy.

Our kvetching kibbitzer has temporarily been quietened.
Quote:
 Originally Posted by Humble Pie Thirty days in the hole That's what they give you now Thirty days in the hole Oh, yeah Thirty days in the hole All right, all right

 2021-06-30, 13:12 #14 paulunderwood     Sep 2002 Database er0rr 381410 Posts I'm amazed by Sweety's SNR and the amount ensuing noise from posters, especially in this thread about Ryan's discovery of a huge PRP. Is nothing sacred? I hope Ryan goes to 30M bits on this one. I am not saying he should do it because I say so Last fiddled with by paulunderwood on 2021-06-30 at 13:14
 2021-06-30, 14:01 #15 Dr Sardonicus     Feb 2017 Nowhere 10010110110112 Posts After my tour de force proving that (2^p + 1)/3 automatically "passes" the Rabin-Miller test to the base 2 when p > 3 is prime (and stupidly failing to point out that I had actually proven that), it occurred to me that a simple Fermat PRP test for (2^p + 1)/3 to any base is easily refined to a Rabin-Miller test in this case since (N-1)/2 is odd, and for small bases the results are predictable using pencil-and-paper calculations. We have N = (2^15135397+1)/3 Clearly N == -1 (mod 4). For base b = 3, 5, 7, and 11, reducing the exponent mod 6, 4, 6, and 10 respectively, we find that 3*N == 3 (mod 9) so that N == 1 (mod 3), N == 1 (mod 5), N == 1 (mod 7), and N == -1 (mod 11). Assuming N is prime, applying the law of quadratic reciprocity tells us that (3/N) = (-1/N)(-3/N) = (-1)(N/3) = -1 (5/N) = (N/5) = +1 (7/N) = (-1/N)(-7/N) = (-1)(N/7) = -1 (11/N) = (-1/N)(-11/N) = (-1)(N/11) = (-1)(-1) = +1, so if N is prime, 3^((N-1)/2) == -1 (mod N) 5^((N-1)/2) == +1 (mod N) 7^((N-1)/2) == -1 (mod N), and 11^((N-1)/2) == +1 (mod N). If any of these congruences fail to hold, that would prove that N is composite. Further, if any of these residues were anything other than 1 or -1 (mod N), it would give a proper factorization of N. I am confident that ryanp et al know all this, and checked all this. Someone who natters that "you should run a Rabin-Miller test" without even bothering to do the above paper-and-pencil calculations (and perhaps doesn't even know how), or considering that they might be addressing their quibbling to people who know all this and have almost certainly already done the calculations, should IMO thank God for His infinite mercy in not issuing a permanent ban.
2021-07-01, 00:20   #16
ryanp

Jun 2012
Boulder, CO

313 Posts

Quote:
 Originally Posted by paulunderwood I hope Ryan goes to 30M bits on this one. I am not saying he should do it because I say so
I may, but most likely it will be later this summer.

Also, LLR with "all the switches" reports:

Code:
(2^15135397+1)/3 is BPSW and Frobenius PRP! (P = 1, Q = 3, D = -11)  Time : 51568.343 sec.

 2021-07-12, 17:21 #17 ryanp     Jun 2012 Boulder, CO 313 Posts I've now searched the whole 13M...19M range and should finish up to 20M this week.

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