March 27th, 2013, 05:13 PM  #21 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Old King Collatz
In any event, simply taking any number of the form 8x+5 and reversing the 4n+1 rule to get (8x+51)/4 = (8x+4)/4 = 2x+1, dividing by 8 and reapplying the rule every time you get a new number of the form 8k+5 until you finally get a number of some other mod 8 residue class is conceptually as simple as it gets. You will definitely eventually get an odd number that equals 1, 3 or 7 mod 8, guaranteed! 3n+1 either is or isn't odd, depending on n.

March 28th, 2013, 01:05 AM  #22 
Senior Member Joined: Mar 2012 Posts: 569 Thanks: 26  Re: Old King Collatz
The difference is that in my formulation you don't have to keep changing the value of n (or k) and reapplying a rule. There is one rule for 8n+5 numbers (that aren't 16n+5 numbers), and then one rule for 16n+5, 64n+21 etc. So I'd say it is conceptually a bit simpler as it gives an algorithm for the equivalent 8n+1, 3, 7 number* without relying on reiterations. Phrasing it slightly differently as the even/odd thing was a bit confusing: For 8k+5 numbers (that can't be expressed 16n+5) go to 2k+1 (with the same value of k and we know this isn't an 8n+5 number) For 16k+5, 64k+21, 256k+85 etc go to k (if k is odd) or 4k+1 (if k is even) 8k+5 13 = 8*1+5 > 2*1+1 = 3 (>5) 29 > 8*3+5 > 2*3+1 = 7 (>11) 45 = 8*5+5 > 2*5+1 = 11 61 = 8*7+5 > 2*7+1 = 15 77 = 8*9+5> 2*9+1 = 19 16k+5 5 = 16*0+5 > 4*0+1 = 1 21 = 16*1+5 > 1 37 = 16*2+5 > 4*2+1 = 9 (previously I was jumping to the next step here, with 3*2+1 = 7, but it's clearer like this) 53 = 16*3+5 > 3 69 = 16*4+5 > 4*4+1 = 17 85 = 64*1+21 > 1 101 = 16*6+5 > 6*4+1 = 25 117 = 16*7+5 > 7 133 = 16*8+1 > 8*4+1 = 33 149 = 64*2+21 = 2*4+1 = 9 165 = 16*10 = 10*4+1 = 41 181 = 16*11+5 = 11 and so on  using this method, you always reach a number that isn't 8n+5 so don't need to reiterate the operation and can work out the equivalent 8n+1, 3, 7 number for the next step directly. You can also see how it is the 8n+5 numbers that are equivalent to 8n+3 or 8n+7, while the 16n+5 numbers alternate between series of 8n+1 numbers, and another iteration of the 8n+3 and 8n+7 series. *equivalent in terms of the onward path, not the path that went before, obviously. 
March 28th, 2013, 01:16 AM  #23 
Senior Member Joined: Mar 2012 Posts: 569 Thanks: 26  Re: Old King Collatz
The other reason for looking at it this way is that in order to study the pattern it might be useful to break out different series  first for the 8n+5 numbers (3, 7, 11, 15..., then the 16n+5 etc ones (and for 64n+5, 64n+21, 256n+85 etc you get a reiteration of the same series, so above we have started two series of 1, 9, 17 etc, one for the 16n+5 numbers, interlocking with one for the 64n+21 numbers).

March 28th, 2013, 03:42 AM  #24 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Old King Collatz
But you have to do essentially the same amount of calculation to determine which rule to apply out of what is actually an infinity of rules. So even what the next rule is would at times need to be calculated. But the approaches are mathematically equivalent. So in the end, there is no truly meaningful debate here.

March 28th, 2013, 04:10 AM  #25  
Senior Member Joined: Mar 2012 Posts: 569 Thanks: 26  Re: Old King Collatz Quote:
There may be an infinity of rules, but we have already stated the form they all take (because we can define the series 16n+5, 64n+21, 256n+85 etc). On that basis I still think it is helpful to start picking out the different ways that the numbers resolve into series for this reason: 1, 3, 1, 7, 9, 11, 3, 15, 17, 19 looks like quite a messy, chaotic series. but once you start breaking it down it becomes apparent that what you have is several iterations of 3, 7, 11, 15... and several interlocking iterations of 1, 9, 17 (though I need to go a bit further looking at how some of the latter ones work). And we need to understand how the 8n+5 numbers "output" into 8n+1, 3 or 7 numbers if there is to be any chance of dismissing loops. (See following note also). So it's no good just knowing that an 8n+5 number will map to some 8n+1, 3 or 7 number, we need to know more about the mechanics of this mapping.  
March 28th, 2013, 04:12 AM  #26  
Senior Member Joined: Mar 2012 Posts: 569 Thanks: 26  Re: Old King Collatz Quote:
*Because any possible algorithm mapping 8n+1, 3 and 7 numbers to 1 would have to include analysis of how 8n+5 numbers map back to 8n+1, 3, 7 numbers.  
March 28th, 2013, 04:35 AM  #27  
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Old King Collatz Quote:
Exactly the same argument can be made for any number of the form 8x+5.  
March 28th, 2013, 04:57 AM  #28  
Senior Member Joined: Mar 2012 Posts: 569 Thanks: 26  Re: Old King Collatz
Firstly let's say that either approach might have merits depending on the overall approach. However secondly, I basically agree with this Quote:
from 8n+7 to 8n+3 numbers (via other 8n+7 numbers) from 8n+3 numbers to 8n+5 and 8n+1 numbers from 8n+1 numbers back to 8n +3, 5, 7 numbers (via other 8n+1 numbers) From 8n+5 numbers to 8n+1, 3, 7 numbers If you don't have clear rules for the last of these, you can't eliminate loops, and if you can't eliminate loops you can't prove you will get to 1. I think it is feasible to codify all of these transforms, but am not sure there whether or not there is a way of approaching that proof from there, as the rules don't mesh in a way that I can clearly see taking us forwards.  
March 28th, 2013, 05:09 AM  #29 
Senior Member Joined: Mar 2012 Posts: 569 Thanks: 26  Re: Old King Collatz
Another way of putting this  it's easy to codify most of the transforms I mentioned without reference to even numbers, and that's why we can 'ignore them'. But I don't think it is as easy to 'ignore' the 8n+5 numbers, because any transform that goes via them will need to follow two distinct sets of rules  one to get you to a 8n+5 number, then one to get you to the 8n+1, 3 or 7 number that that can be boiled down to. I do see what you're saying, I just think there is a danger of ignoring an important part of the overall picture if you just treat 8n+5 numbers as being 'eliminated from enquiries'. The two easiest transforms to codify are the ones from 8n+1 and 8n+ 7*. The two hardest are from 8n+3 and 8n+5. 
March 28th, 2013, 06:22 AM  #30  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Old King Collatz Quote:
 

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