
View Poll Results: Is M2133277 a prime?  
Yes  3  11.11%  
No  18  66.67%  
Dunno  2  7.41%  
Only in some bases  4  14.81%  
Voters: 27. You may not vote on this poll 

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20210514, 16:29  #12 
"Daniel Jackson"
May 2011
14285714285714285714
673 Posts 
If interpreted as either Base25 or Base30, it's prime:
M2133277 (Base25) = 134776610807 (Base10) M2133277 (Base30) = 482624813017 (Base10) Obviously this goes on to infinity, so I won't list anymore. However, 2^21332771 isn't prime: https://www.mersenne.ca/exponent/2133277 
20210514, 17:54  #13  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
1011010010100_{2} Posts 
Quote:
2133277 (Base 25) = 499,267,057 (base ten), which is prime per https://www.alpertron.com.ar/ECM.HTM; 2133277 (Base 30) = 1,484,813,017 (base ten) which is divisible by 61, so the corresponding Mersenne number is composite also, divisible by M61. (base conversions performed with https://calculator.name/baseconversion.php; confirmed with https://www.rapidtables.com/convert/...converter.html) The corresponding Mersenne numbers would have ~150,000,000 and ~447,000,000 decimal digits, respectively, not 12. M499267057 has no factor below 2^{73} greater than 1. Last fiddled with by kriesel on 20210514 at 17:55 

20210514, 18:03  #14  
Apr 2020
3^{2}·5·11 Posts 
Quote:


20210514, 21:02  #15 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·5^{2}·43 Posts 
Aren't all Mersennenumbers with Primeexponents, Fermat'sProbablePrime in base 2^n?
PariGP: Code:
forprime(n=1,19^1,{ Mn=2^n1; print(Mod(2,Mn)^(Mn1);); print(Mod(2^2,Mn)^(Mn1);); print(Mod(2^19,Mn)^(Mn1);); }) Code:
Mod(1, 3) Mod(1, 3) Mod(1, 3) Mod(1, 7) Mod(1, 7) Mod(1, 7) Mod(1, 31) Mod(1, 31) Mod(1, 31) Mod(1, 127) Mod(1, 127) Mod(1, 127) Mod(1, 2047) Mod(1, 2047) Mod(1, 2047) Mod(1, 8191) Mod(1, 8191) Mod(1, 8191) Mod(1, 131071) Mod(1, 131071) Mod(1, 131071) Mod(1, 524287) Mod(1, 524287) Mod(1, 524287) 
20210514, 21:43  #16 
"Viliam Furík"
Jul 2018
Martin, Slovakia
2^{3}×5×17 Posts 

20210514, 23:07  #17 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{3}×5×157 Posts 

20210514, 23:50  #18  
"Robert Gerbicz"
Oct 2005
Hungary
2·3^{2}·83 Posts 
Quote:
For any N if you consider the b bases for that N is a Fermat pseudoprime then these bases form a group in Z_N. For Mersenne numbers this means that 2^n is such a base, because mp is a Fermat pseudoprime for base=2, fortunately these means only p such bases, because 2^p==2^0 mod mp. In an elementary way without group: you need: (2^n)^(2^p1)==2^n mod (2^p1) but we have: 2^p=a*p+2 hence: (2^n)^(2^p1)==2^(n*(a*p+1))==2^n mod (2^p1) what we needed. ps. this is the reason why we are using base=3 for Fermat testing the Mersenne numbers, base=2,4,8 etc is "bad". But you shouldn't fix base=3 to all numbers, because for other type of numbers: N could be a (trivial) pseudoprime for base=3, and you need to choose another base. 

20210515, 00:16  #19  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·3·797 Posts 
Quote:


20210515, 01:57  #20 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×5^{2}×43 Posts 
Well, posts number 18 & 19, and the current result of the pole are the reasons why Science is not a democratic process. I think everything relates.
Last fiddled with by a1call on 20210515 at 01:59 
20210515, 06:20  #21 
Romulan Interpreter
Jun 2011
Thailand
2^{4}·13·47 Posts 

20210528, 17:46  #22  
"Tucker Kao"
Jan 2020
Head Base M168202123
487 Posts 
Quote:
[dozenal] MӾ5,077 M507,Ӿ77 M7,046,577 [/dozenal] All of them got a 0 and a 5 somewhere in the number. Last fiddled with by tuckerkao on 20210528 at 17:47 

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