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Old 2018-11-26, 15:22   #1
enzocreti
 
Mar 2018

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Default Periodicity of the congruence 1666667 mod 666667

The numbers pg(k) are introduced:
pg(k) is the concatenation of two consecutive Mersenne numbers
pg(1)=10
pg(2)=31
pg(3)=73
pg(4)=157
.
.
.
using Julia software I searched the k's for which pg(k) is congruent to 1666667 mod 666667.
I saw that periodically there are three consecutive pg(k)'s congruent to 1666667 mod 666667.

Infact pg(18),pg(19) and pg(20) are congruent to 1666667 mod 666667...then pg(85196),pg(85197),pg(85198)...then pg(170374),pg(170375),pg(170376).
Is there a periodicity?

Last fiddled with by enzocreti on 2018-11-26 at 15:26
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Old 2018-11-26, 15:31   #2
enzocreti
 
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Default congruences

pg(255551),pg(255552) and pg(255553) are another triple congruent to 1666667 mod 666667

Last fiddled with by enzocreti on 2018-11-26 at 16:28
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Old 2018-11-30, 06:46   #3
LaurV
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You mean numbers congruent to 333333 (mod 666667)


So what?
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Old 2018-11-30, 12:14   #4
enzocreti
 
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Default congruent to 6 mod 7

Quote:
Originally Posted by LaurV View Post
You mean numbers congruent to 333333 (mod 666667)


So what?


I found about 40 primes of the form (2^k-1)*10^d+2^(k-1)-1 where d is the number of decimal digits of 2^(k-1)-1. None of these prime is congruent to 6 mod 7, whereas there are 9 primes of this form congruent to 5 mod 7. What do you think it is due?
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Old 2018-11-30, 15:46   #5
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No idea (I said this before).

There is no reason why these numbers wouldn't be 6 (mod 7).

Maybe it is due to Guy's Law?
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Old 2018-11-30, 16:19   #6
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Default What is the point of this thread?

Numbers of the form 13 mod 666 are my favourite. If only we could discover more of them then we might find some very important significance if we deeply investigate them. Then again, maybe some other random set of x mod y will yield a better chance of a major discovery? Than again again, maybe not. It's probably all just wasted characters on a webpage.
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Old 2018-11-30, 16:33   #7
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for a pair of consecutive Mersennes mod 7, to create 6 mod 7 we have:

(1,0) d=3 mod 6
(3,1) d=4 mod 6
(0,3) d does not exist.
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Old 2018-11-30, 17:20   #8
enzocreti
 
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Quote:
Originally Posted by science_man_88 View Post
for a pair of consecutive Mersennes mod 7, to create 6 mod 7 we have:

(1,0) d=3 mod 6
(3,1) d=4 mod 6
(0,3) d does not exist.



I just conjectured that there is no prime 6 mod 7 of the form (2^k-1)*10^d+2^(k-1)-1.
I think that up to k=800.000 there is no prime 6 mod 7.
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Old 2018-11-30, 17:25   #9
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Moreover the exponents leading to a prime seem to be NOT random at all...for example there is an exponent which is 51456...then an exponent which is 541456...note also that 541456-51456=700^2....
exponents of these primes are NOT random at all and residue 5 mod 7 occur twice than expected
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Old 2018-11-30, 17:28   #10
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i posted the question on mathoverflow and nobody yet has found an explanation
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Old 2018-11-30, 17:28   #11
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If they're not random, predict the next one.
Numerology is not evidence of non-randomness!
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