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August 30th, 2014, 02:53 AM  #31 
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26  Very frustrating for you. No rush at my end.

August 30th, 2014, 06:16 AM  #32 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Yes, very. In the meantime, would you check that my program is doing what you want? Otherwise the Inetnet plumbers could come and go without us being any further ahead than we are now.

August 31st, 2014, 12:44 PM  #33  
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26  Quote:
Prime Numbers: Attempted proof Bear in mind I'm only addressing the narrow point of whether the outputs from the function tend towards 1 in 2 (for level 1 numbers), 1 in 6 (for level 2 numbers) etc. And I'll acknowledge upfront that an error of even 1 is insufficient if we are attempting a 'counting' proof, but what I'm looking to discuss in the end is whether we could use this method in a different way that could lead to a reductio ad absurdum given the assumption that the conjecture is false. Last edited by Hedge; August 31st, 2014 at 12:51 PM.  
September 1st, 2014, 01:05 AM  #34  
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26  Quote:
 
September 6th, 2014, 06:11 AM  #35  
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26 
I said I'd look at the code Quote:
Quote:
Then, sorry but I can't work out what the last bit below does, which doesn't mean there is anything wrong with it, just that I can't follow the language well enough. Quote:
Firstly it convinced me that the proportion of level n integer outputs from the function do indeed tend up or down to 1 in 2(3^n)/3. (If you don't read it, the main point is that the function involves dividing by 3^n at level n. Every prime power (except powers of 2) has a primitive root; thus the multiplicative group of integers modulo 3n is cyclic. Thus we can arrange the level n inputs as a square/cube/cuboid grid that is cyclic every 2(3^n)/3 places  it's a bit more complicated as the pattern doesn't start out as a cuboid, but it tends to a pattern that can be filled with an increasing proportion of cuboids...) Secondly it became more clear that the region over which this happens is pretty vast. 100 million < 2^21 and at this point the level 3 numbers are only just starting to tend down from about 1 in 19 towards 1 in 18 outputs. And for level 2 numbers we've tended down from 1 in 7 to 1 in 6.001 at about 2^6000. So running tests over 100 million or so wouldn't tell us a huge amount. Last edited by Hedge; September 6th, 2014 at 06:29 AM.  

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approach, collatz, pigeonhole 
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