20180129, 12:11  #1 
"Luke Richards"
Jan 2018
Birmingham, UK
2^{5}×3^{2} Posts 
Help with a number theory equivalence
Hi all,
I've found a sequence of numbers which I have submitted to OEIS. Other contributors have taken an interest and found a simpler generating function to mine... ... except none of us know *why* it works. We don't want to include the function on the entry until we can explain why the simpler function outputs the same results. Essentially it comes down to this: Why is: (k>x) equivalent to (k>x) The sequence was defined using the first equivalence, but a pari script, which works in a way equivalent to the second equivalence, generates the same numbers in a fraction of the time. But I'm not enough of a number theorist to ger my head around it! 
20180129, 13:16  #2  
"Forget I exist"
Jul 2009
Dumbassville
10000011010011_{2} Posts 
Quote:


20180129, 14:20  #3 
Dec 2012
The Netherlands
5·353 Posts 
It's not entirely clear what you mean.
The conditions you have given relate to pairs of numbers k & x, and there exist positive integers k,x with k>x for which one condition holds and the other doesn't. So how are you defining the sequence? 
20180129, 14:22  #4 
Jun 2003
5,407 Posts 
I don't understand how this results in a sequence. Can you post both the original code that you used to generate, as well as the faster logic (along with the sequence itself)?

20180129, 14:27  #5  
"Luke Richards"
Jan 2018
Birmingham, UK
2^{5}·3^{2} Posts 
Quote:
This doesn't result in a sequence  it is a part of calculating a sequence. The entry on OEIS in draft (waiting for approval by an editor): https://oeis.org/history/view?seq=A298827&v=34 

20180129, 14:42  #6 
Jun 2003
12437_{8} Posts 
3^k+2 == 3^x+2 (mod 3^x+2)
3^k == 3^x (mod 3^x+2) 3^(kx) == 1 (mod 3^x+2) QED EDIT: In PARI, you would do znorder( Mod(3, 3^x+2)) Last fiddled with by axn on 20180129 at 14:46 
20180129, 14:50  #7  
"Luke Richards"
Jan 2018
Birmingham, UK
288_{10} Posts 
Quote:
And thank you for the patience you afforded me rather than just messaging me effectively saying "why don't you just learn some basic number theory?" and calling me an arrogant newbie. 

20180129, 14:58  #8  
Feb 2017
Nowhere
7×857 Posts 
Quote:
The other congruence implies 3^x + 2 divides 3^k  1. Does that help? 

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