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 March 27th, 2013, 03:00 PM #11 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Old King Collatz Ok. You guys are really runnIng with this. Cool. A simple iterative procedure for 5 mod 8 is to reduce it from 8x+5 to 2x+1. Reevaluate it mod 8 and keep reducing from 8n+5 to 2n+1 until you depart the class of 5 mod 8. So 53 is 8(6)+5, And therefore reduce to 2(6)+1. But that is 8(1)+5 and therefore reduces to 2(1)+1. 3 is not 5 mod 8 so it "rejoins" the game. I'm on my iPhone at an Indian restaurant and will try to catch up with the rest later or sometime during my coming 5 day weekend!
 March 27th, 2013, 03:22 PM #12 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Old King Collatz I am envioning a "Collatz cube" power of 2 multiples of odds are the depry dimension. Odds of the form 8x+5 form the height dimension above odds that they are 4x+1 "expansions" of. The bottom row of the face of the cube are odds of the other residue classes. Any infinite increase or loop will critically involve this row, as all other numbers tend back to this row.
 March 27th, 2013, 03:35 PM #13 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Old King Collatz Yes, loops could go back through an 8x+5 number, just as any loop will have to include at least one even. But the logic of these numbers in the system would require that any loop would also have to include at least one number that is neither even nor of the form 8x+5. So we aren't going to miss anything. We have to be careful in other cases that we don't lose critical information. But patterns, yes, this is all about discovering the intricate but still elegant patterns in this Collatz universe!
March 27th, 2013, 04:18 PM   #14
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Re: Old King Collatz

Quote:
 Originally Posted by johnr Ok. You guys are really runnIng with this. Cool. A simple iterative procedure for 5 mod 8 is to reduce it from 8x+5 to 2x+1. Reevaluate it mod 8 and keep reducing from 8n+5 to 2n+1 until you depart the class of 5 mod 8. So 53 is 8(6)+5, And therefore reduce to 2(6)+1. But that is 8(1)+5 and therefore reduces to 2(1)+1. 3 is not 5 mod 8 so it "rejoins" the game.
I can add a bit to that.

If a number can be expressed as 8n+5 but not 16n+5, it will resolve to 2n+1 (and that won't be an 8n+5 number)

If a number can be expressed as 16n+5, 64n+21, 256n+85... (first term increases by 4, second by 4n+1 up through the gateway numbers) then it resolves to the same number as n, in other words 3n+1. For instance 16*2+5=37, which goes to 7 (n=2), 16*0+5 = 5 which resolves to 4 (n=0), 64*7+21 goes to 22->11 (n=7).

I think I've phrased that right, I'll check in the morning, late here.

March 27th, 2013, 04:26 PM   #15
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Re: Old King Collatz

Quote:
Originally Posted by Hedge
Quote:
 Originally Posted by johnr Ok. You guys are really runnIng with this. Cool. A simple iterative procedure for 5 mod 8 is to reduce it from 8x+5 to 2x+1. Reevaluate it mod 8 and keep reducing from 8n+5 to 2n+1 until you depart the class of 5 mod 8. So 53 is 8(6)+5, And therefore reduce to 2(6)+1. But that is 8(1)+5 and therefore reduces to 2(1)+1. 3 is not 5 mod 8 so it "rejoins" the game.
I can add a bit to that.

If a number can be expressed as 8n+5 but not 16n+5, it will resolve to 2n+1 (and that won't be an 8n+5 number)

If a number can be expressed as 16n+5, 64n+21, 256n+85... (first term increases by 4, second by 4n+1 up through the gateway numbers) then it resolves to the same number as n, in other words 3n+1. For instance 16*2+5=37, which goes to 7 (n=2), 16*0+5 = 5 which resolves to 4 (n=0), 64*7+21 goes to 22->11 (n=7).

I think I've phrased that right, I'll check in the morning, late here.
Minor typo, you meant 2n+1, not 3n+1 in the second line of your third paragraph.

 March 27th, 2013, 04:37 PM #16 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Old King Collatz Oh, and for all not immersed in Collatz who may be following this, please note that insofar as Collatz is about ALL positive integers, no number ever really leaves the game. It's just that some numbers will automatically be known to map eventually to 1 by the original 2x+1 -> 6x+4 and 2y-> y rules if certain others do, ie all evens will if all odds do and all odds of the form 8x+5 will if all other odds do. So we don't have to ask any questions per se about numbers of the form 8x, 8x+2, 8x+4, 8x+5 or 8x+6, even though they will be implicated in other mappings and of course are all part of the conjecture in their own right. So IF someone can come up with some abstract inductive or whatever proof that all numbers of the forms 8x+1, 8x+3 and 8x+7 eventually map to 1, the others come for free and we are done. Still a tall order, but I really feel that the rich structure of Collatz (vs, eg, the messiness of Goldbach pairs and where they appear) make this a problem that just might yield to proof, and may do so sooner rather than later once we lock in on the right considerations and conceptualizations. (And yes, this intuition of mine plus $4 will get you a grande at Starbucks, but I'm running with it anyway!)  March 27th, 2013, 05:15 PM #17 Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26 Re: Old King Collatz No, 3n+1 because n goes to 3n+1. This holds, even when n is even for all the 16n+5 etc numbers, which is why I talked about these numbers resolving to the same number as n. For instance 2*16+5 = 37 resolves to 7 (=3*2+1), which is the same number that 2 will resolve to. 2n+1 is as far as you can go for the 8n+5 numbers (that can't be expressed as 16n+5). For 16n+5 numbers you can go down to n or, if n is even, to 3n+1. I think... Pretty sure that is correct though. March 27th, 2013, 05:16 PM #18 Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26 Re: Old King Collatz Quote:  Originally Posted by johnr (And yes, this intuition of mine plus$4 will get you a grande at Starbucks, but I'm running with it anyway!)
You're more optimistic than me, but I like playing with it and seeing where it goes.

 March 27th, 2013, 05:44 PM #19 Senior Member   Joined: Mar 2012 Posts: 572 Thanks: 26 Re: Old King Collatz Oh, I need to explain something in there. 2 would obviously resolve to 1 in the standard way. But 2*3+1 = 7. When I say 2*16+5 resolves t0 2 then 7 (n, then 3n+1), you need to imagine you are treating 2 as an odd number. Probably a clearer way to see it is that 9 (= 4*2 +1) really does resolve to 7 in the standard way, so where the n in a 16n+5 is even, you need to go to the first 4n+1 for the proper resolution. Hope that makes sense.
 March 27th, 2013, 05:50 PM #20 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Old King Collatz I will come back to this tomorrow morning or perhaps Friday when I am off altogether. But yes, if you are treating 2 as if it is an odd number, then you've lost me!

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