20181130, 17:35  #12 
Mar 2018
1000001111_{2} Posts 
Not random
It is too difficult to predict which is the next.
Simple questions sometimes require very complex explanations. Even Goldbach conjecture is simple to state, but to proof it it is extremely difficult...so how can I predict which is the next? 
20181130, 17:39  #13 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Using actual math, even Goldbach has heuristic arguments. You have a form of a number prove something about the form it Also has to take to be prime ...

20181130, 17:41  #14 
Romulan Interpreter
Jun 2011
Thailand
2^{3}×5×229 Posts 

20181130, 17:47  #15 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20181130, 17:48  #16 
Mar 2018
527_{10} Posts 
2,3,4,7,8,12,19,22,36,46,51,67,79,215,359,394,451,1323,2131,3336,3371,6231,19179=9*2131,39699,51456,56238,69660,75894,79798,92020,174968, 176006,181015,285019,331259,360787,366770,...,541456
these are the exponents found with Pfwg leading to a prime...look at them...if they are random, I am superman! 
20181130, 17:53  #17  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
Last fiddled with by science_man_88 on 20181130 at 17:53 

20181130, 18:03  #18 
Romulan Interpreter
Jun 2011
Thailand
9160_{10} Posts 
We got that, but where did you take "3 mod 6" from, and generally, where did you take "mod 6" from? Shouldn't that be "6 mod 7" and respective "4 mod 7" there??
Last fiddled with by LaurV on 20181130 at 18:04 
20181130, 18:04  #19 
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 

20181130, 18:07  #20 
Romulan Interpreter
Jun 2011
Thailand
2^{3}·5·229 Posts 

20181130, 18:10  #21 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
10 is coprime with 7, therefore an extension of Fermat's little theorem says that (10^6)^k for arbitrary natural number k will have remainder 1 on division by 7.
Last fiddled with by science_man_88 on 20181130 at 18:11 
20181130, 18:18  #22 
Romulan Interpreter
Jun 2011
Thailand
2^{3}×5×229 Posts 
Right. Thanks. Just now I realized you were talking about d being 3 or 4 (mod 6). I was talking about 10^d being 4 or 6 (mod 7), which is the same thing. I have to remember not to read the forum at 1:15 AM, after a company party... I am going to bed now...

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