20220305, 18:49  #1 
Aug 2016
2×3 Posts 
Prime next to multiple of 6
I'm not a mathematician, but I just saw a Tweet by fermatslibrary stating that
every prime is next to a multiple of 6. Isn't this a way faster way of figuring out whether a number is prime or not, instead of attacking it with lengthy tests? https://primes.utm.edu/notes/faq/six.html 
20220305, 19:05  #2  
"Ed Hall"
Dec 2009
Adirondack Mtns
1339_{16} Posts 
Quote:


20220305, 19:12  #3  
Jan 2021
California
2^{2}·5·23 Posts 
Clearly
Quote:


20220306, 15:41  #4 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3577_{10} Posts 
For any positive integer n, all primes not dividing n are coprime to n, and the remainder of them to n must be in the reduced residue system mod n, e.g. for n=15, the reduced residue system is {1, 2, 4, 7, 8, 11, 13, 14}, and for your case of n=6, the reduced residue system is {1, 5}, thus all primes p not dividing 6 must == 1 or 5 mod 6, and if p == 1 mod 6, then p1 is multiple of 6, if p == 5 mod 6, then p+1 is multiple of 6, thus all primes p not dividing 6 (i.e. other than 2 and 3) are next to multiple of 6

20220311, 11:21  #5 
Aug 2016
2·3 Posts 
Thanks for being so polite.
And you just reiterated my statement, however I was looking for an answer to the question whether it would be easier to test first if a given assignment is indeed of the form 6n+1 or 6n1, instead of brutally attacking it with a lengthy test. Assuming that checking the 6n+1 or 6n1 is less brutal off course. 
20220311, 13:37  #6  
"Καλός"
May 2018
5·73 Posts 
Quote:
However, as stated by drkirkby in another thread, within the known sample size there are "more twin primes below Mersenne exponents than above Mersenne exponents", see https://mersenneforum.org/showthread.php?t=26812 and OEIS346645. 

20220311, 16:04  #7  
"Curtis"
Feb 2005
Riverside, CA
17^{2}×19 Posts 
Quote:


20220312, 19:56  #8  
"Καλός"
May 2018
365_{10} Posts 
Quote:
because 2^{p}  1 = (2  1)(2^{p1} + 2^{p2} + ... + 2^{2} + 2^{1} + 2^{0}) = ((31)^{p1} + (31)^{p2}) + ... + ((31)^{2} + (31)^{1}) + 1 and after applying the binomial theorem (see https://en.wikipedia.org/wiki/Binomial_theorem) the terms (1^{p1}  1^{p2}) + ... + (1^{2}  1^{1}) within the parentheses cancel each other. 

20220312, 20:50  #9 
Jan 2021
California
460_{10} Posts 
Well, if you are going to specifically talk about Mersenne primes, the candidates can't have any "small" factors (like 3) because all factors of 2^p1 are of the form 2kp+1, so when looking at Mersenne candidates with exponents over 100 million (which is the low end of the search right now), the smallest possible factor will be over 200 million.
But there are lots of other types of primes that people are looking for, that's just the main project. 
20220313, 02:20  #10  
"Tucker Kao"
Jan 2020
Head Base M168202123
1100001010_{2} Posts 
Quote:


20220313, 03:05  #11  
"Curtis"
Feb 2005
Riverside, CA
17^{2}·19 Posts 
Quote:
Or, please tell us how the last digit allows you to identify whether a number is divisible by 3. Here are numbers not divisible by 3: 10, 11, 22, 23, 34, 35, 46, 47, 58, 59. There's all ten last digits. So, explain. Last fiddled with by VBCurtis on 20220313 at 03:06 

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