20200208, 07:45  #1 
Jan 2020
2^{3}·3·5 Posts 
Mersenne numbers in Dozenal base
It seems like when 2^{n}  1 expressions are written in Dozenal base, the repetitive patterns of the ending digits show up with a formula: see attached thumbnail.

20200210, 14:12  #2 
Romulan Interpreter
Jun 2011
Thailand
8868_{10} Posts 
The order of 3 mod 2 is 2 and the order of 5 mod 2 is 4, the order of 9 is 6, etc, so there is no wonder that every second mersenne is multiple of 3, and every 4th is multiple of 5, etc. In your case, the base is multiple of 3 too, so what you see is just an elementary modular property.
Think about why we only test prime exponents, from the perspective of representing Mp in other bases than 10 (like, 2, 3, 6, 12, 15, etc). For example, if you represent them in base 5, or 15, what would happen with the last digit every 4 numbers? Last fiddled with by LaurV on 20200210 at 14:13 
20200210, 18:03  #3  
Jan 2020
2^{3}·3·5 Posts 
Quote:
I prefer to count in Dozenal since they repeat more often, so I have to memorize less digits after identifying the more frequent ending digit pattterns. Any 2^{n * m}  1 will be the multiples of 2^{n}  1, so that's why n has to be a prime in order for the whole result to be a prime too. Last fiddled with by tuckerkao on 20200210 at 18:05 

20200211, 02:57  #4 
Nov 2016
2^{3}×5×59 Posts 
In dozenal, except 3 and 7, all Mersenne primes ends with 27 or X7 (the only two 2digit Mersenne primes), this is a list of first 27 (decimal 31) Mersenne primes in dozenal
Dozenal is my favorite base, it also has these properties: * All squares end with square digits (0, 1, 4, 9). * All primes end with prime digits (2, 3, 5, 7, E) or 1. (equivalently, all primes >=5 end with prime digits >=5 (5, 7, E) or 1). * All numbers k such that the negativePell equation x^2k*y^2=1 is solvable end with all such singledigit number k (1, 2, 5, X). * Except 0 = F0 and 1 = F1 = F2, the only square Fibonacci number is 100 = F10 (100 is exactly the square of 10), thus, 10 is the only base such that 100 is a Fibonacci number (since 100 in a base is just the square of this base, and 0 and 1 cannot be the base of numeral system) * A Fibonacci number can end with any digit but 6, and if a Fibonacci number ends with 0, then it must end with 00. * The period of the final digit of Fibonacci number is 20, that of the final two digits is also 20 (dozenal is the largest base such that the period of the final digit of Fibonacci number is the same as that of the final two digits of Fibonacci number, if there are no WallSunSun primes). For n >= 2, the period of the final n digits of Fibonacci number is 2*10^(n1). (2 followed by n1 zeros, which is an ndigit number 200...000) See https://dozenal.fandom.com/wiki/Properties_of_dozenal for more properties of the dozenal base, I am interested in it :)) 
20200211, 03:02  #5 
Nov 2016
938_{16} Posts 
Problem: In dozenal, 21 and 201 are squares, prove or disprove they are the only squares of the form 2000...0001 in dozenal.

20200211, 03:06  #6 
Jun 2003
2^{3}×19×31 Posts 

20200211, 03:28  #7 
Nov 2016
2^{3}·5·59 Posts 
Well, my problem is whether 2*12^n+1 can be square for n other than 1 and 2
Last fiddled with by sweety439 on 20200211 at 03:28 
20200211, 04:44  #8  
Jan 2020
2^{3}×3×5 Posts 
Quote:
https://www.numberempire.com/numberfactorizer.php 

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