20210630, 07:53  #12 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×3×5×163 Posts 
Good choice! Thanks to the mod colleague who did it.

20210630, 13:03  #13  
Feb 2017
Nowhere
4988_{10} Posts 
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:


20210630, 13:12  #14 
Sep 2002
Database er0rr
5·773 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 20210630 at 13:14 
20210630, 14:01  #15 
Feb 2017
Nowhere
2^{2}×29×43 Posts 
After my tour de force proving that (2^p + 1)/3 automatically "passes" the RabinMiller 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 RabinMiller test in this case since (N1)/2 is odd, and for small bases the results are predictable using pencilandpaper 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^((N1)/2) == 1 (mod N) 5^((N1)/2) == +1 (mod N) 7^((N1)/2) == 1 (mod N), and 11^((N1)/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 RabinMiller test" without even bothering to do the above paperandpencil 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. 
20210701, 00:20  #16  
Jun 2012
Boulder, CO
2^{2}×3^{4} Posts 
Quote:
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. 

20210712, 17:21  #17 
Jun 2012
Boulder, CO
2^{2}·3^{4} Posts 
I've now searched the whole 13M...19M range and should finish up to 20M this week.

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