20181126, 15:22  #1 
Mar 2018
21B_{16} Posts 
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 20181126 at 15:26 
20181126, 15:31  #2 
Mar 2018
21B_{16} Posts 
congruences
pg(255551),pg(255552) and pg(255553) are another triple congruent to 1666667 mod 666667
Last fiddled with by enzocreti on 20181126 at 16:28 
20181130, 06:46  #3 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10100000101001_{2} Posts 
You mean numbers congruent to 333333 (mod 666667)
So what? 
20181130, 12:14  #4 
Mar 2018
1033_{8} Posts 
congruent to 6 mod 7
I found about 40 primes of the form (2^k1)*10^d+2^(k1)1 where d is the number of decimal digits of 2^(k1)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? 
20181130, 16:19  #6 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5×7×191 Posts 
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.

20181130, 16:33  #7 
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
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. 
20181130, 17:20  #8  
Mar 2018
7^{2}×11 Posts 
Quote:
I just conjectured that there is no prime 6 mod 7 of the form (2^k1)*10^d+2^(k1)1. I think that up to k=800.000 there is no prime 6 mod 7. 

20181130, 17:25  #9 
Mar 2018
539_{10} Posts 
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 54145651456=700^2....
exponents of these primes are NOT random at all and residue 5 mod 7 occur twice than expected 
20181130, 17:28  #10 
Mar 2018
7^{2}×11 Posts 
i posted the question on mathoverflow and nobody yet has found an explanation

20181130, 17:28  #11 
"Curtis"
Feb 2005
Riverside, CA
5638_{10} Posts 
If they're not random, predict the next one.
Numerology is not evidence of nonrandomness! 
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