[seqfan] Re: corollary to the Collatz conjecture?

[Gottfried Helms]

Am 31.07.2017 um 19:38 schrieb Bob Selcoe:
> Hi Gottfried & Seqfans,
> 
> Thanks.  Here's a link to the Crandall paper.  http://www.ams.org/journals/mcom/1978-32-144/S0025-5718-1978-0480321-3/S0025-5718-1978-0480321-3.pdf.  It doesn't appear to address this issue in particular, but it may be buried somewhere in the text.
> 
> Anyway, it turns out my conjecture is equivalent to the *standard* 3n+1 Collatz conjecture. The reasoning is straightforward: 
> 
> Dealing only with odd terms N in the sequences, let d be the number of times 3 divides N. It takes k-d steps for a successor term to be a multiple of 3^k (obviously if d > k, the number of steps is 0).  Let 1 <= j < k-d. The sequences cannot loop prior to k-d steps, since 3^(j-k+d) is a non-integer.  Once a term reaches a multiple of 3^k, the sequences behave like a standard Collatz sequence "scaled up" 3^k times, so the conjectures are equivalent.
> 
> (Thanks to Jack Brennan for pointing this out and to Jean-Paul Allouche for verifying my analysis).  
> 
> Cheers,
> Bob
> 

let M=3^m with some m >=0

then one step of transformation on the odd numbers (a->b) is

        b = (a*3+M) / 2^A

If we look at the question of cycles, then we should have

          a = (a*3+M) / 2^A
      a 2^A = 3a + M
      a (2^A -3) =  M

and of course, if we have a solution for the problem where M=1, so

      a_0 (2^A_0 -3) = 1

then of course we have the same set of solutions for the case M=3^m
with a_0 * M instead of a_0

      a_0*M (2^A_0 -3) = 1*M

because the parenthese has no common part with M=3^m .

   --------------

Considering two instead of one transformation we have

        b = (a*3+M) / 2^A
        c = (b*3+M) / 2^B

  and looking at cycles we must have c=a such that we can expand to

        a = ((3a+M)/2^A*3+M) / 2^B
          = ((3a+M)*3+M*2^A) / 2^(A+B)
          = (3^2a +M*3+M*2^A) / 2^S     \\ I write S for A+B+... for longer cycles
    a(2^S - 3^2) = M( 3 + 2^A)

 and again, if we have a cycle for M=1 at a_0 then we have accordingly
 a cycle for M=3^m at a_0*3^M

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 I think it is obvious how this generalizes, and I think that answers
 that question so that a full proof can be written easily.


 Gottfried