Re: Status of the Collatz Conjecture?

Am 19.07.2009 06:59 schrieb Mensanator:
On Jul 18, 3:26�pm, Gottfried Helms <he...@xxxxxxxxxxxxx> wrote:

If you define a elementary transformation on odd numbers a>0 only
T(a;A) := a -> (3*a+1)/2^A
where A is the maximal exponent which leaves the result an odd integer
then elementary transformation increase if A=1 and decrease if A>2. (if a=1, then
with A=2 the transformation has a fixpoint, this is the only exception)

(...)
Yepp, we have loops with negative a (a<0), this is known.

But do you have any notion of why?

Yes. See below.

Can you see that it makes a difference to prove, that something
like
2^s =/= (3 + 1/a) (3 + 1/b)..(3 + 1/k)

holds for some a,b,c,...>0 and s with some general restrictions
while the same thing for negative numbers -a,-b,-c,...,-k

2^s =/= (3 - 1/a) (3 - 1/b)..(3 - 1/k)

may possibly *not* hold even with the same general restrictions?

Sure, I'll accept that...as long as a reason is given why
this _must_ be so.

Well, let's do it for a simple case. If we assume a loop of N=2 T()-trans-
formations, we have two members: a and b

a -> b -> a -> b -> ...

where b=(3*a + 1)/2^A and c = (3*b+1)/2^B and to make it a loop, it must c=a
Also a and b are odd, cannot be divisible by 3, cannot be equal and let's
order them that a<b

Leave it open, whether such a loop with positive odd integers a,b>1 actually
exists. But if it exists, then all of the following must also be true:

a * b = (3a + 1)/2^A * (3b + 1)/2^B
2^A*2^B = (3 + 1/a)(3 + 1/b)

let S = A+B then we have my standard notation

2^S = (3 + 1/a)(3 + 1/b)

Now - can a loop exist if a,b>0, odd, not divisible by 3?
We may not find one with small a,b. But what if a and b are some trillion?
Well: this is simple. If a,b greater than some limit, then the rhs is
just very little greater than 3, so if a~1000 and b~1000 then the rhs
is about 3.006 and this is no perfect power of 2.
To arrive at a perfect power a and b must be *small*, we have an *upper bound*.
Ok, let's assume the smallest values, a=5,b=7

2^S =?= (3 + 1/5)(3 + 1/7) = 9 + 3*12/35 + 1/35
= 9 + 1 + 1/35 + 1/35
= 10 + 2/35

The rhs is still smaller than the next perfect power of 2, so for
the smallest allowed a,b there is no solution.

But for increasing a,b the rhs even decreases, so there is no 2-step-loop
with a,b positive odd numbers.



Ok, now let's look at negative a,b.

2^S = (3 - 1/a)(3 - 1/b)
2^S =?= (3 - 1/5)(3 - 1/7) = 9 - 3*12/35 + 1/35
= 9 - 1 - 1/35 + 1/35
= 8
= 2^3

So we have a solution S = 3 and from S=A+B, A,B>0 follows A=1 and B=2

a=-5,
b = T(-5,1) = (3*(-5)+1)/2^1 = -14/2 = -7
a = T(-7,2) = (3*(-5)+1)/2^2 = -20/4 = -5

(btw the same way of arguing is usable for all 3x+C-problems)

The clue is obviously that the product of the N-step-loop on the rhs

2^S = (3 + 1/a)(3 + 1/b)(3 + 1/c)*...*(3 + 1/n)

is very near at 3^N, nearer than the next perfect power of 2, especially
if the members a,b,c,...,n have high (and *positive*) values.

However, for some N the distance to the next 2^S is (relatively!) small.
But we can say, that for a given minimal element "a" we can find a minimal
length N which is required to allow a loop on the above arguments only.
So if T Silveira or E Roosendaal say that no 1<a<2^58 is member of
a loop, thus only a,b,c,..>2^58 is possible (if at all), then we can find
a number N of parentheses such that there is a possibility for a loop:

using "a" = "nearest odd number at 2^58 not divisible by 3"
2^S < ( 3 + 1/a)(3+1/(a+2))(3+1/(a+6))(3+1/(a+8))... (3+1/(a+3N))

and, filling this simply in Pari/GP or another software we find some
N~250000 , so we already know, that no loop N<250000 can exist
for a,b,c,... in the positive domain.




Analoguously we can use the negative signs again:

2^S > ( 3 - 1/a)(3 - 1/(a+2))(3 - 1/(a+6))(3 - 1/(a+8))... (3 -1/(a+3N))

but we see for some small N, small a, that 2^S < 3^N is sometimes near, such
that as loop is possible (by the product-formula criterion only)
and also actually is existent


----------------------------------------
Just for completeness:
for the "1-cycle"-structure the above condition
is much sharper, we get a formula like

2^(A-1) = 2^S/2^N = (k*3^N - 1)/(k*2^N - 1)

which must be satisfied for a 1-cycle of length N.

Here the elements a,b,c are positive, if k is positive. If we want the
negative domain, k is negative, a,b,c are negative too

2^(A-1) = 2^S/2^N = (k*3^N - 1)/(k*2^N - 1)

and N=2,k=-1 gives

2^(A-1) = -10/-5 = 2 -> A=2

which indicates, that a solution is possible. Then
a = k*2^N-1 = -5
b = (3(-5)+1 )/2^1 = -7
a = (3(-5)+1 )/2^2 = -5
and indeed the sequence -5 -> -7 -> -5 has only increasing steps
(in absolute value) and exactly one decreasing step and is thus an
"1-cycle".
-----------------------------------------------------------
What about Lagarias' proof that the minimum bound of
a counterexample is a cycle with 275000 elements?
Obviously, these m-cycle researchers have no clue what
happens since they can't get past 68.

No. The 1-cycle as well as each increasing part of the m-cycle is meant for
*arbitrary* many elements. The notion "primitive transformation" means here,
that the trajectory of a sequence of (*arbitrary many*) elements is strictly
increasing and at the end one time decreasing.

A "1-cycle" is then that this forms a loop.

So Steiner showed: there is no cycle which contains only one decreasing
step of the T()-transformation - no matter how many elements.

Simons/DeWeger arrived at the generalized result:

If a "m-cycle" exists, it must consist of more than 68 segments of
"primitive transformation" (each of arbitrary length!),

or more simple:

it must have more than 68 descendent T()-steps (number of increasing
T()-steps is arbitrary)

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

(concerning Schorer's paper:)

Am I correct in stating no journal has published it?

think so; I didn't come across a reference so far...

Is it concerning his core idea?

Yes. "the i-level value of the anchor tuple of an
(...)

Some lapsus in definitions of limits/bounds?

Well, there's this prattle about counterexamples
(...)

Is there a counterexample when using his arguments?

He also states:
(...)

I told him that this was nonsense and he should refer
to your papers prior to your going off on the tangent
about m-cycles. I take it then, that you agree with me.

Mr. Schorer then says he may not be understanding this
as his Lemma 5.0 contradicts it. Do you accept that his
paper has proven YOUR work is in error? Or do you think
this simply proves his Lemma 5.0 is false?

Hi, thanks for that explanations. I'll need some time to
go through them.

Regards-

Gottfried

.

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