Re: A tiny Collatz-exercise

Am 30.06.2008 22:35 schrieb Mensanator:
On Jun 29, 12:56 am, Gottfried Helms <he...@xxxxxxxxxxxxx> wrote:

Yes. take n=7, then 3x=21. Now either 22 or 20 has
exactly one factor of 2,

I was going to ask why this is so, but then
I remembered that the even numbers count of
contiguous LS 0-bits is a 2-adic sequence:

1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,...

so every odd number has one adjacent even
number with exactly 1 factor of 2.


Yes, this simply, because the sequence of even
numbers

[2,4,6,8,10,12,14,...]
is

2*[1,2,3,4, 5, 6, 7,...]

and so each even number with one factor of 2 are followed
by one even number with more factors of 2.

Keeping in mind that this is based on statistics,
which doesn't take certain realities of the Collatz
graph structure into consideration, which can be
seen here:

<http://members.aol.com/mensanator666/Page.htm>

thanks, I'll have another look at it...


In Collatz, only contiguous 1-bits at the LS end
can create continuous increase. But every number has
a finite number of contiguous 1-bits, so, sooner or
later, the Collatz sequence MUST start decreasing
(unless the propagating carry bits can create a new,
unbroken string of 1's at the MS end, but that can't
happen).

The increasing may resume later, but it can never
increase continuously. This does not say anything
about the net result of such alternating
increase/decrease.

Yes, this is the meaning of

2n*2^k - 1 -> 2n*3^k - 1 (k compressed steps)

where 2n*2^k - 1 has k trailing bits

n=1, k=3

7 (= 2*1 *2^2 - 1) -> 17 (= 2*1*3^2 - 1)

and then a decreasing step is required.


This problem sounds very similar to that of the
Collatz-problem, but it seems much easier to
prove.


Well, it's easy to prove in Collatz that every odd 1(mod 3)
and 2(mod 3) number has not just one, but an infinite number
of 0(mod 3) ancestors. Is that true of your system also?
So that any non-0(mod 3) hole in any given sequence must appear
an infinite number of times in the other sequences that
begin with 6K+3? If not, it sounds much harder.

that's why I titled this thread "tiny exercise". ;-)

Proof:

We know, that each odd number has a one-factor-2 neighbour.
This neighbour can be reduced to another odd number by
one D()-transformation.

The D()-transformation stops -by definition- either at 1
for the number n=1
D(1) = 1

or at a number divisible by 3.
D(3*k) = 3*k

Since each step is a decreasing step all trajectories of
iterated D() end at 1 or 6k+3 , k=0,1,2,...

Better formal proof:
??? (I'm not good in formalism, sorry)

Next mail I'll post some other nice observations and thoughts.
I don't think, this shall really help for the Collatz-problem,
but - who knows... and at least it has some niceness.

Gottfried
.

Relevant Pages

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