20171229, 06:53  #78 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2745_{16} Posts 
Subjectively, they can feel as if they are still alive for a long time.
Ambrose Bierce described that in 'An Occurrence at Owl Creek Bridge'. 
20171229, 09:33  #79  
Feb 2017
3·5·11 Posts 
Quote:
You are correct, that is a false positive (prime) result returned by the algorith. There are others as well. If you run the Code that I posted #73, you can identify many more. False positives are not uncommon with Primality algorithms, only the percentage of false positives w.r.t total number of "primes" tested Refer Fermat Primality test which have "......21853 pseudoprimes base 2 that are less than 2.5×10^10 (see page 1005 of [4]). This means that, for n up to 2.5×10^10, if 2n−1 (modulo n) equals 1, then n is prime, unless n is one of these 21853 pseudoprimes"......Source https://en.wikipedia.org/wiki/Primality_test Indications are that this algorithm will have less false positives at 2.5x10^10, currently sitting at +750 at 10^7 and decreasing trend with increasing magnitude. Interestingly the smallest false positive for the Fermat primality test is also the smallest false positive for this algorithm!!! 341 Last fiddled with by gophne on 20171229 at 09:39 Reason: Adding exponent sign for Fermat Pseudoprime figure 

20171229, 09:52  #80  
Feb 2017
3·5·11 Posts 
Quote:
Fermat Primality Test has 21,853 pseudoprimes in 2.5x10^10....refer my post #79 for reference for this figure. Do you consider the Fermat Primality Test fake as well based on your reasoning/reference? You refer to 3 examples, I have listed (and have provided Code) for 245 false primes up to 1,000,000 (magnitude) Last fiddled with by gophne on 20171229 at 09:54 Reason: Correct spellin/grammar errors 

20171229, 15:56  #81 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}·1,117 Posts 
There is no reason for this selfflagellation to go on.
Please meditate in silence. Or ask for your own logorrhea thread. The thread is locked. 
20171229, 23:17  #82 
∂^{2}ω=0
Sep 2002
República de California
5×2,351 Posts 
Briefly reopening the thread for the benefit of our lessmathematical readers who may not see why the test in question is simply a base2 Fermat pseudoprime test in disguise  start with the OP's version:
(2^n1) == (n+1)/2 mod (n+2) Replace the silly 'n+2' modulus by n, i.e. everywhere in the formula subtract 2 from n: 2^(n2)  1 == (n1)/2 (mod n) Multiply by 2: 2^(n1)  2 == (n1) (mod n) == 1 (mod n) Add 2 to both sides: 2^(n1) == 1 (mod n) . QED Last fiddled with by ewmayer on 20171229 at 23:17 
20171230, 18:36  #83  
Feb 2017
3×5×11 Posts 
Quote:
Thanks for your post. I don't understand why you say my algorithm is; (2^n1) == (n+1)/2 mod (n+2) I thought it was (2^n1) mod (n+2) == (n+1)/2 Retracing your steps would then give [1] Replacing (n+2) with n 2^(n2)1 mod n == (n1)/2 [2] Multiplying by 2 2^(n1)2 mod n == (n1) [3] Add 2 to both sides 2^(n1) mod n == (n+1) This is different from your result and therefore not the algorith you reduced. Similar, but very different, like 1 is different from 2 Am I mistaken with this? If you would concur it would still mean my algorithm is original. 

20171230, 18:42  #84  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}·1,117 Posts 
Quote:
Do you always read only the second explanation? First is not enough? It has to be repeated twice? What Ernst wrote I already spelled out for you two days ago. 

20171230, 18:57  #85  
"Dana Jacobsen"
Feb 2011
Bangkok, TH
3^{2}·101 Posts 
Quote:
Quote:


20171230, 19:01  #86  
Feb 2017
10100101_{2} Posts 
Quote:
I respect your opinions as an expert and a Mod. However I have a right to defend myself as well. Please look at my reply to ewmayer in post #83 and correct me where I am wrong/mistaken. According to my humble understanding I do not think that the two are the same. Please explain how you can say the two algoriths are the same when ewmayer appears to have tweaked my algorithm, possibly unintentionally. Just juxtapose the two. They are not the same look at the formulas, not the same as what ewmayer reworked. But thanks for taking me to task, but I will try to defend my position as I understand sometimes ignorance is bliss. I will also see how other users understand the issues, but for now I cannot agree that what ewmayer "reworked" is the algorithm I posted. Thanks Last fiddled with by gophne on 20171230 at 19:03 Reason: grammar 

20171230, 19:49  #87  
Einyen
Dec 2003
Denmark
2·17·101 Posts 
Quote:
Quote:
If something is equal to (n+1) mod n, then it is also equal to 1 mod n, and equal to 2n+1 mod n and 3n+1 mod n and so on. If say n=5: Then: 31 mod n = 1 because 31=6*n + 1 31 mod n = 6 = n+1 because 31= 5*n + 6 31 mod n = 11 = 2n+1 because 31 = 4*n + 11 and so on. Last fiddled with by ATH on 20171230 at 19:50 

20171230, 21:42  #88  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}×1,117 Posts 
I wrote a groundbreaking new poem and I shall go around telling everyone that I've truly made something out of myself. Here is small bit of it. I have more, much more! 
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