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#12 |
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"Daniel Jackson"
May 2011
14285714285714285714
3·13·17 Posts |
If interpreted as either Base-25 or Base-30, it's prime:
M2133277 (Base-25) = 134776610807 (Base-10) M2133277 (Base-30) = 482624813017 (Base-10) Obviously this goes on to infinity, so I won't list anymore. However, 2^2133277-1 isn't prime: https://www.mersenne.ca/exponent/2133277 |
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#13 | |
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"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
31·173 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/base-conversion.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 273 greater than 1. Last fiddled with by kriesel on 2021-05-14 at 17:55 |
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#14 | |
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Apr 2020
15516 Posts |
Quote:
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#15 |
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"Rashid Naimi"
Oct 2015
Remote to Here/There
40078 Posts |
Aren't all Mersenne-numbers with Prime-exponents, Fermat's-Probable-Prime in base 2^n?
![]() Pari-GP: Code:
forprime(n=1,19^1,{
Mn=2^n-1;
print(Mod(2,Mn)^(Mn-1););
print(Mod(2^2,Mn)^(Mn-1););
print(Mod(2^19,Mn)^(Mn-1););
})
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) |
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#16 |
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"Viliam Furík"
Jul 2018
Martin, Slovakia
10728 Posts |
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#17 |
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Undefined
"The unspeakable one"
Jun 2006
My evil lair
11000001101002 Posts |
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#18 | |
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"Robert Gerbicz"
Oct 2005
Hungary
22×7×53 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^p-1)==2^n mod (2^p-1) but we have: 2^p=a*p+2 hence: (2^n)^(2^p-1)==2^(n*(a*p+1))==2^n mod (2^p-1) 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. |
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#19 | |
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
100101000001012 Posts |
Quote:
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#20 |
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"Rashid Naimi"
Oct 2015
Remote to Here/There
40078 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 2021-05-15 at 01:59 |
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#21 |
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Romulan Interpreter
Jun 2011
Thailand
7×1,373 Posts |
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#22 | |
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Jan 2020
22·79 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 2021-05-28 at 17:47 |
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