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[QUOTE=gophne;475141]Hi jnml
Using the algorithm (sagemath), I get positive result for M3, M5 & M17. I could not test beyound M29 (dividend) due to unknown run time). e.g. M3 (2^7-1) #1......a=2^7-1..................Divisor under test~127....[COLOR="Red"]n-2[/COLOR] <-- what the hell is this? #2......b=(2^7-1)-2.............[COLOR="Red"]n~125[/COLOR] #3......c=1/2*(b+1).............[COLOR="Red"]Target Congruant~63[/COLOR] #4......d=2^b-1...................Dividend~42,535,295,865,117,307,932,921,825,928,971,026,431 #5.......e=d mod a..............Congruant~63 #6.......c==e.......................True~127(divisor) is Prime[/QUOTE] So after all your talk about n this, n that, your own scribbling show that you suddenly call n-2 "n". Then of course, your "n+2" is n. Congratulations! You proved that all primes are 2-PRP. What is even more impressive is that all composite Mersenne numbers are 2-PRPs (even if you don't know it), so your test will just call them all prime. No false negatives! |
[QUOTE=Batalov;475223]So after all your talk about n this, n that, your own scribbling show that you suddenly call n-2 "n". Then of course, your "n+2" is n.
Congratulations! You proved that all primes are 2-PRP. What is even more impressive is that all composite Mersenne numbers are 2-PRPs (even if you don't know it), so your test will just call them all prime. No false negatives![/QUOTE] Hi batalov That is how I coded the algorithm (in SAGE)....if you like I could(should?) have coded the algorithm as follows as well -I did the above to put the "number" being tested as step #1. #1.......a=(2^7-1)-2.........[COLOR="Red"]Modulo Dividend???[/COLOR]~125 (should be the dividend "seed" value to get "d", the real Modulo dividend as per the algorithm. #2.......b=2^7-1...............Modulo Divisor~(a+2=127)....NUMBER UNDER TEST #3.......c=(a+1)/2............Modulo Target Congruant~63 #4.......d=2^a-1..............Modulo Divident~ As above #5.......e=d mod a..........Algorithm Congruant=63 #6.......c==e....................True~127(Divisor) is Prime To simplify further you could simply call the Divident seed [B]25[/B] in step #1. That would make the divisor in step #2, [B]27[/B]~(n+2). or (a+2) if you like. I only used mersenne notation because jnml was working with M[I]n[/I] notation, which would be more practical if very large numbers are to be considered. I am trying my best to be clear, but bear with me that I am trying to translate my code into words. I will try to type some actual code of a working page as I code the algorithm in SAGEMATH. But everybody would code differently depending on their expertise in SAGEMATH. My coding skills in SAGE are moderate. |
[QUOTE=Batalov;475223]Congratulations! You proved that all primes are 2-PRP.
What is even more impressive is that all composite Mersenne numbers are 2-PRPs (even if you don't know it), so your test will just call them all prime. No false negatives![/QUOTE] No true negatives, either - but hey, who needs all that negativity, it's harshing the holiday spirit, and stuff. |
[QUOTE="old TV cartoon"]That's why we called him Neutron! He is always so positive![/QUOTE]
[COLOR=Wheat].[/COLOR] |
[QUOTE=Batalov;475233]That's why we called him Neutron! He is always so positive![/QUOTE]
Umm... Wasn't that the Protons? At least one of the Electrons refused to play game. |
[QUOTE=Batalov;475223]So after all your talk about n this, n that, your own scribbling show that you suddenly call n-2 "n". Then of course, your "n+2" is n.
Congratulations! You proved that all primes are 2-PRP. What is even more impressive is that all composite Mersenne numbers are 2-PRPs (even if you don't know it), so your test will just call them all prime. No false negatives![/QUOTE] hi batalov I do not understand what you are saying about 2-PRP's ....are you running the algorithm? What are your results? I do not get all "composite mersenne numbers" to be false positives, on the contrary, e.g [B]2^9-1 & 2^15-1[/B], do not register as a false positives but as composite in the algorithm. Please post your results, you could even do the calculation on a calculator or Excel at this level of magnitude. |
SAGEMATH CODE for False Positives
Here is SAGEMATH code to check for false positive (prime) results (for the algorithm) up to 1,000,000. Should only take a few minutes to run.
[1] x=0 [2] for y in range(1,1000000,2): z= next_prime(y) a= y+2; #Modulo Divisor....Number under test b= 2^y-1; #Modulo Dividend c= b%a; #Algorithm Modulo Congruent d= (y+1)/2; #Target Congruent as per Algorithm if c==d: if z>a: x=x+1 print(a,x); # 'a' is the False Positive identified, 'x' is a counter The SAGEMATH software will automatically do the indentation for the code. |
Restatement of Algorithm for Use and Testing
If (2^n-1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME, else COMPOSITE, [I]n[/I] being an element of the set of positive integers =>1
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[QUOTE=gophne;475261]If (2^n-1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME, else COMPOSITE, [I]n[/I] being an element of the set of positive integers =>1[/QUOTE]
Yeah, we heard you the first million-and-six times. Try (n+2) = 2047 = 23*89 in your formula. |
[QUOTE=gophne;475252]
I do not understand what you are saying about 2-PRP's [/QUOTE] A person who walks off a 1km cliff also doesn't understand that he is already dead. But he is. [QUOTE=gophne;475261][B]Restatement of Algorithm for Use and Testing [/B]If (2^n-1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME, else COMPOSITE, [I]n[/I] being an element of the set of positive integers =>1[/QUOTE] This is exactly what is called a 2-PRP test. 1) It is >350 years old. 2) It is false. Try n+2 == 341. (2^n-1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME? Wrong. It is composite. [URL="https://en.wikipedia.org/wiki/Fermat%27s_little_theorem#Pseudoprimes"]F. Sarrus in 1820 found 341[/URL] as one of the first pseudoprimes, to base 2. That was 197 years ago. Try n+2 == 561. (2^n-1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME? Wrong. It is composite. Try n+2 == 645. (2^n-1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME? Wrong. It is composite. |
[QUOTE=Batalov;475266]A person who walks off a 1km cliff also doesn't understand that he is already dead.[/QUOTE]
It's not the fall which kills them. It's the hard stop at the end... |
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