20221104, 18:45  #12 
If I May
"Chris Halsall"
Sep 2002
Barbados
2×5,531 Posts 

20221104, 18:50  #13  
Dec 2008
you know...around...
2^{4}×53 Posts 
Quote:


20221104, 19:18  #14  
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
Quote:


20221104, 19:42  #15  
If I May
"Chris Halsall"
Sep 2002
Barbados
2·5,531 Posts 
Quote:
@SM88: Based on your observations, what might you logically work with? Or, more accurately, what can you use to *predict* a statistically significant signal? This is a serious question. I doubt you'll be able to answer. Which is part of the point... 

20221104, 19:48  #16  
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 
Quote:
We get examples: 3 and 5 lead to 3(3)+1=10, 3(5)+1=16, then both fail to 5, and 8 which have difference 3. using that pattern again we can predict 7 will go to 11 as that's 3 away from 8. We then use it again to show 9 goes to 14. 14 and 8 are 6 apart. They are both on division by 2 and yield a distance apart of 3. 

20221104, 19:56  #17 
If I May
"Chris Halsall"
Sep 2002
Barbados
2×5,531 Posts 

20221104, 20:00  #18 
"Forget I exist"
Jul 2009
Dartmouth NS
8418_{10} Posts 
If the fact you can get two sequences In effect arbitrarily close apart from odd numbers then what is the sticking point.

20221104, 20:07  #19  
If I May
"Chris Halsall"
Sep 2002
Barbados
2·5,531 Posts 
Quote:


20221104, 20:47  #20 
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
My point is it's bound by certain quantity difference. I said it got nowhere effectively. The difference has a bound. I helped with semiprime searching, Pari gp use improvements , even attempts to change the very test you to check for primes. I hardly think these are seminal ideas though.
Last fiddled with by science_man_88 on 20221104 at 21:24 
20221105, 00:28  #21  
Mar 2019
2×163 Posts 
Quote:
Reasoning about different \(kn+c\) sequences is all well and good, and you can use a finite number of them to show the conjecture holds up to a certain \(n\) limit. But if you want to use this approach to show the conjecture holds for all \(n\) then you'll need an infinite number of \(k\) and/or \(c\) to do it. 

20221105, 02:08  #22  
"Robert Gerbicz"
Oct 2005
Hungary
3^{2}·179 Posts 
Quote:
Quote:
And not forget that the maximal conjectured number of iterations is O(log(n)^2) if the starter is n, so these are not taking weeks (???). The average number of steps is even smaller, maybe only O(log(n)) but I could be wrong. 

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