20171231, 16:36  #111  
Feb 2017
3·5·11 Posts 
Quote:
If the algorithms are the same...substitute values into the formulas and report what comes out ! Please! I think everybody is crying out for this. May I suggest that you start with the number 1 or any other number that tickles your fancy. That is what we did at primary school be entering numbers into little squares to make the answer true. For Pete's sake. 

20171231, 16:45  #112 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
It's literally a transform of n+2>n in you formula and then a reworking it.

20171231, 16:56  #113  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·29·113 Posts 
Quote:
Suppose you don't understand sheet music at all. Don't expect us to explain it after 3 4 attempts as you clearly need a music teacher and study for >5 years to play this  

20171231, 17:03  #114  
Einyen
Dec 2003
Denmark
3,313 Posts 
Quote:
We claim your algorithm is the fermat pseudoprime test to base 2, so test your algorithm on numbers up to say 10,000. We claim it will find the real primes as well as these composite numbers as "false positives": http://oeis.org/A001567 341, 561, 645, 1105, 1387, 1729, 1905, 2047, 2465, 2701, 2821, 3277, 4033, 4369, 4371, 4681, 5461, 6601, 7957, 8321, 8481, 8911 Here are the list of real primes if you need it: https://primes.utm.edu/lists/small/10000.txt 

20171231, 17:25  #115 
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 
A bit more explicit.
2^n1 == (n+1)/2 mod (n+2) 2^((n+2)2)1== ((n+2)1)/2 mod (n+2) Replacing n+2 with n we get: 2^(n2)1==(n1)/2 mod n 2^(n1)2==n1 mod n 2^(n1)== n+1 mod n Casting out the terms that don't affect the remainder we get. 2^(n1)==1 mod n QED edit if this doesn't register you need to brush up on mathematical laws like the law of replacement, or you need to look up what modular arithmetic is. Last fiddled with by science_man_88 on 20171231 at 17:36 
20171231, 17:55  #116  
Dec 2017
50_{10} Posts 
Quote:
on the left side (in red) are the numbers from "your" algorithm and on the right side (in green) are the numbers from fermat it really does not matter what the exact numbers are therein ... only the fact, that all the numbers that are equal on the left side correspond to some other numbers that are equal on the right side is important here ! can you see, that all the equal signs are in the same positions in both algorithms ? now, the tables are only having 14 lines each but do you think, that the equal signs would always be in the same positions if both tables had 100 lines or even 1000 lines ? you are welcome to make the calculations yourself and you will see that the equal signs always stay in the same positions ? how do we know this without actually having made such huge tables ? well, we use modular math  you should really take your time and learn what this mod function is all about: the "rules" for the modfunction do not equal to the rules of the normal equal sign so you might take a look on https://en.wikipedia.org/wiki/Modular_arithmetic or other pages on the internet to learn what these "rules" are ... and also more important to truely understand WHY they are in that way they are. the "experts" on this forum have spent quite a long time to practice and finally truely understand this WHY this is the reason for their statement that the two algorithms are exactly the same ... using the modfunction is as simple and easy as playing the moonlight sonata on piano for them Last fiddled with by guptadeva on 20171231 at 17:57 Reason: typos 

20171231, 18:35  #117 
Feb 2017
10100101_{2} Posts 

20171231, 18:39  #118  
Feb 2017
245_{8} Posts 
Quote:
Regretfully commentary would not suffice as proof of the identity of the algorithms under the spotlight. By the way, do you know what the algorithm /formula is that awmayer was reducing to? What is his source or the name of this formula/algorithm? 

20171231, 18:46  #119  
Feb 2017
245_{8} Posts 
Quote:
This is more in line with what we need. From what you say indications are that the two algorithms are the same. However, I cannot make out anything on the image you posted, could you perhaps tabulate the results and the algorithms that generated them (as well as the inputs). Thanx, Regards. 

20171231, 18:57  #120  
Feb 2017
A5_{16} Posts 
Quote:
Thanx for indicating the numbers that are the same. Can you give us the EXACT breakdown that each of the Algorithms generate as well please, for the record or at least say the first 30 terms of each. Regards 

20171231, 19:05  #121  
Feb 2017
3·5·11 Posts 
Quote:
Your math is incorrect; LHS of == sign 2^n1....let n=5 Then LHS result is 31 RHS of == sign (n+1)/2 mod (n+2).......substitute for n=5 (5+1)/2 mod (5+2) RHS result is =3 

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