Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry

From: Mensanator (mensanator_at_aol.compost)
Date: 08/28/04 

Date: 28 Aug 2004 04:48:58 GMT

>Subject: Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry
>From: Gottfried Helms helms@uni-kassel.de
>Date: 8/27/2004 5:18 PM Central Standard Time
>Message-id: <cgoc6u$cgl$02$1@news.t-online.com>
>
>For your interest, I want to add some more notes, relating
>the previous posting to the 1-cycle-problem, as an example:
>
>------------
>
>A 1-cycle is a PC()-transformation with one single descending
>step at its end:
>
> a' = PC(a,N:A) = C(a;1,1,1,1...1,A) with N-1 ones
>
> (in my previous post I omitted the trailing 1 in the
> PC()-notation, should have been PC(a;N:1), sorry)
>
>Now, a loop is, if the outout equals the input, thus
>
>
> a = PC(a;N:A) with N the number of steps,
> and S the sum of exponents = N+A-1
>
>Now we have
> 2^S
> a = a * --- - PC(0;N:A)
> 3^N
>
>which is
>
> 2^S
> a( --- - 1 ) = PC(0; N:A)
> 3^N
>
>That is
> PC(0;N:A)
> a = ------------- * 3^N
> 2^S - 3^N
>
>Because
>
> 3^N - 2^N
> PC(0;N:A) = -----------
> 3^N
>
>the previous becomes
>
> 3^N - 2^N
> a = -------------- (condition for a 1-cycle or "primitve loop")
> 2^S - 3^N
>
>The solution is only valid if a is odd and is integer. For N=1 and A=2 we get
>the trivial loop of one step:
> 1
> a = C(a;2) = --- = 1
> 1
>

Does this only apply to positive integers? The reason I ask is my formulations
span both positive and negative domains and I get two 1-cycle loops:

1 -> 4 -> 2 -> 1

and

-1 -> -2 -> -1

>and for two steps
>
> a = C(a;2,2) = C( C(a;2);2) = C(1;2) = 1
>
>and for arbitrarily many of such steps
>
> a = C(a;2,2,2,2,2....) = 1 .
>
>No other combination of exponents is known to produce an integral a.
>
>-------------------
>
>Back to the primitive loop.
>It can be shown, that for any "a" being an input value for a primitve loop
>of a number N of steps, "a" must have a simple structure:
>
> a = 2^N * i - 1 (where i is a free integer parameter >=0 )
>
>so
> 2^S - 2^N
> 2^N * i - 1 = ----------- - 1
> 2^S - 3^N
>
>and
>
> 1 2^S - 2^N 2^(A-1) - 1
> i = --- * ------------ = ----------------
> 2^N 2^S - 3^N 2^(A-1+N) - 3^N
>
>No integral i is known except in that combination of i,N and A, which
>constitutes
>the trivial loop. I don't want to proceed here, but I think, the example
>shows, what the proposed notation for the collatz-transformation can be
>good for.
>
>Regards -
>
>Gottfried Helms
> 

-- 
Mensanator
Ace of Clubs

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