Re: The Collatz-Highway

Am 05.10.2006 20:25 schrieb Danny:
Danny -

just take the maximal k, that lets b integer.
In binary representation it is just to cut off
the trailing zeros of the number (3a+1).
Since (3a+1) can have arbitrary size, k can be
arbitrarily large.

Is it that, what you were asking?

Regards -

Gottfried Helms

Ps. in case this was messing your understnding: the article
should be read in fixed-with-font.

Goffried,

My understanding of this first simple part for (k)
could be wrong.

Here is my understanding, but it could be flawed --

Given only this sequence ---

..80,40,20,10,5,16,8,4,2,1

It seems, there is a misunderstanding by mixing
two steps.

With that k I only meant:
Take any odd number: a=13
Then one original step of the Collatz-transofrm is
k
3a+1 = 3*13+1 = 40
Now by 2 20 1
by 2 10 2
by 2 5 3

So (a*3+1)/2^k = (13*3+1)/2^3 = 5
giving the next odd number.
That's all in the first paragraph of my post.

From here I say, it is sufficient (and also
profitable) to look only at odd numbers and
apply the "compressed Collatz-transformation"
b = (a*3+1)/2^k.

Now the next step, involving (odd) numbers of the
form

a = x*2^n - 1

will always transform to

a' = x*3^n - 1

and then, since this is even

x*3^n - 1
b = a' / 2^j = ---------
2^j


Where in the above sequence an (a) --->00
then (k) (could) theoretically--->oo terminating
on the last (a) = 5 but given all other
applicable sequences the starting odd (a)
within that sequence limits the size of the
exponent of (2) which you designate as (k).

When I say applicable sequences I mean
a sequence starting with an odd (a) that is
(2^n-1)/3 which is (not) an applicable sequence
because it never encounters another odd (a)
back to termination of ..8,4,2,1 except (1).

It seems, you are discussing the opposite
(compressed) Collatz-transformation

a = (b*2^j - 1)/3

which is assumed to cover all odd integers,
when started at b=1.

If you are designating a = 1, the terminating
point, then I agree, (k) has no limit!

I am not saying that any odd (a) encountered
before (2^n-1)/3 (higher up the limb) is not
applicable though.

e.g.
Where a = 113

((113*3+1)/2)/2 = 85 = the next (a) where k=2
to obtain 85.

Here it is again the original transformation ...
and yes,

T(113;2) = 85 ; k=2

in my earlier notation:

85 = T(113;2)
1 = T(85;8) = T(113;2,8)

(85*3+1)/2 = 128 = no more (a)'s except 1.
So all sections beginning with an odd (a) that are
higher up the limb from (85) are applicable
where (k) has a limit.
Along with all the rest of the sequences in the
tree.

To avoid further confusion just discount all
sequences that don't include (5) as part
of its' sequence. All others except ...80,40,
20,10,5,16,8,4,2,1 -- (k) has a limit terminating
@ a =(5).
I understand you are talking about the set of numbers

N1 = {1,5,21,85,341,... } where always the m'th

entry of N1 transforms to 1 by one single compressed
step (index m starting at 1):

N1[m]*3+1
T(N1[m];2*m) = 1 = ---------- where k=2m
2^k

All N1[m] have the form (4^m-1)/3

Now one can look at the set N2 = {3,13,53,213,...}
where all entries N2[m] transform to 5 by one step.

N2[m]*3 + 1
T(N2[m]; 2m) = 5 = ------------ where k=2m-1
2^k

While elements of N1 transform to 1, which is also the
first element of N1 itself, we have the recursion
T(N1[1];2) = N1[1]

But this notation is a start for a more general
view:

T(N1[m1];2*m1 ) = N1[1]
T(N2[m2];2*m2-1) = N1[2]
--->
T(N2[m2]; 2*m2-1,4) = N1[1] = 1

and introduce the standard notation

b = T(a; A,B,C,D,...N) where A,B,C,..N are the exponents k,
which occur at each compressed step

a ---> (3*a+1)/2^k




So in the above context, begriming with any
odd integer (a) <> (2^n-1)/3 then (k) has
a limit where the last value of (a) yields
(5) in the sequence.

Please give a counter example if I am wrong!

Hmm, I didn't try this, but may be the question
is already resolved by the above said?

Regards-

Gottfried

.

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