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#12 |
If I May
"Chris Halsall"
Sep 2002
Barbados
2×5,531 Posts |
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#13 | |
Dec 2008
you know...around...
24×53 Posts |
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#14 | |
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts |
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#15 | |
If I May
"Chris Halsall"
Sep 2002
Barbados
2·5,531 Posts |
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@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... |
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#16 | |
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts |
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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. |
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#17 |
If I May
"Chris Halsall"
Sep 2002
Barbados
2×5,531 Posts |
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#18 |
"Forget I exist"
Jul 2009
Dartmouth NS
841810 Posts |
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If the fact you can get two sequences In effect arbitrarily close apart from odd numbers then what is the sticking point.
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#19 | |
If I May
"Chris Halsall"
Sep 2002
Barbados
2·5,531 Posts |
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#20 |
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts |
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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 2022-11-04 at 21:24 |
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#21 | |
Mar 2019
2×163 Posts |
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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. |
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#22 | ||
"Robert Gerbicz"
Oct 2005
Hungary
32·179 Posts |
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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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