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May 21, 2004, 9:30:01 AM5/21/04

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Hi -

I got an unequality, which i'm unable to analytically to disprove.

Does anyone know useful related material?

The inequality, formulating a restriction, arose whily studying

the question of a certain, primitive assumed loop in the collatz

(3x+1)-problem. Here it goes:

3^L - 1

powceil2( 3^L ) <= 2^L* -------

2^L - 1

where powceil2 ( x) = smallest power 2^S >= x

Empirically the lhs is always *greater* than the rhs (up to L=200)

(Thanks to Otto Forster, Uni Muenchen for his GPL-Freeware num-theory

interpreter-program ARIBAS!)

The benefit of this inequality is, that if it cannot be satisfied,

then a primitive loop in the Collatz-problem cannot exist.

One can reformulate this formulafor instance to:

3^L 3^L - 1

powceil2( ---- ) <= ----------

2^L 2^L - 1

- or many other ways... I didn't find a definitve form, where I could

directly conclude the impossibility of that inequality.

Someone with an idea?

Regards -

Gottfried Helms

May 23, 2004, 7:00:01 PM5/23/04

to

Gottfried Helms <he...@uni-kassel.de> writes:

> I got an unequality, which i'm unable to analytically to disprove.

> Does anyone know useful related material?

>

> The inequality, formulating a restriction, arose whily studying

> the question of a certain, primitive assumed loop in the collatz

> (3x+1)-problem. Here it goes:

>

> 3^L - 1

> powceil2( 3^L ) <= 2^L* -------

> 2^L - 1

>

> where powceil2 ( x) = smallest power 2^S >= x

According to theorems mentioned at the start of chapter 3

of Baker's "Transcendental Number Theory", the following is

true.

Let A1, ..., An >= 4 and B >= 2, and suppose that ...

a1,...,an are algebraic numbers with height(ak) <= Ak

b1,...,bn are rational integers with |bk| <= B.

Write F := b1 log a1 + ... + bn log an. Then either

F=0 or |F| > B^-(C Q log Q), where Q = product(log Ak) and

C depends only on n and on d, the maximum degree of the ak.

In particular, let n=2 and d=1; let a1=2, a2=3; and

write p = b1, q = -b2. Then this says

F = |p log 2 - q log 3|

either F = 0 or |F| > max(p,q)^-C

for a constant C.

C can be found explicitly. Baker's book doesn't give

details of how, but they are out there in the literature.

There have been advances in this area since Baker's book.

(Search Google for "laurent mignotte nesterenko".)

The value of C will still be much larger than you'd

like :-).

Anyway: if the inequality above (the one you want to prove

can't happen) holds then you can find p,q with |F| <= 2^-p

or something like that. And if F is small then p and q

have to be somewhat similar in size.

Now, if p^-C < F <= 2^-p then -C log p <= -p log 2, so

p / log p <= C / log 2. So you can get an upper bound on

the size of p. It may or may not be practical then to

check all actual values of p up to there.

A nasty back-of-envelope calculation using a version

of Laurent-Mignotte-Nestorenko quoted in a paper I've

found on the web suggests that actually you get some

such bound as

|F| >= exp(-17280) / p^(8640+1080 log p)

where those numbers could doubtless be reduced by being

less sloppy than I was, but probably not hugely reduced.

So, that would lead to a contradiction when

2^-p p^(8640+1080 log p) < exp(-17280)

-p + 8640 log p + 1080 (log p)^2 < -17280

p - 8640 log p - 1080 (log p)^2 > 17280

which is true provided p >= 298000 or thereabouts.

So ... if I haven't botched the above in any way (which

I probably have), you "only" need to check a few hundred

thousand values of L. You'd do that in practice by calculating

log 3 / log 2 with sufficient accuracy (plain ol' double

precision should be fine) and looking to see how close to

an integer you can get by multiplying it by an integer

up to a few hundred thousand in size.

--

Gareth McCaughan

.sig under construc

May 24, 2004, 7:30:14 PM5/24/04

to

I wrote:

> According to theorems mentioned at the start of chapter 3

> of Baker's "Transcendental Number Theory", the following is

> true.

..

> So ... if I haven't botched the above in any way (which

> I probably have), you "only" need to check a few hundred

> thousand values of L. You'd do that in practice by calculating

> log 3 / log 2 with sufficient accuracy (plain ol' double

> precision should be fine) and looking to see how close to

> an integer you can get by multiplying it by an integer

> up to a few hundred thousand in size.

I just received some e-mail from a friend of mine who knows

much more about number theory than I do, observing that

(1) it's rather well known (to those who know such things,

I suppose) that the 3n+1 conjecture itself follows from

some sort of bounds on the approximability of log 3 / log 2

by rationals, and (2) that the required bounds are much

tighter than those obtainable by methods of the sort I

mentioned. So if Gottfried Helms's conjecture is strong

enough to do much for the 3n+1 conjecture (I'm not sure

what a "primitive loop" is, I'm afraid) then the argument

I sketched probably has some big holes in it. Contrariwise,

if I've somehow managed to break my perfect record of

never posting anything to sci.math.research without at

least one serious mistake, then Gottfried's conjecture

probably doesn't imply anything very exciting about the

Collatz conjecture. It's after midnight local time and

I should be in bed, so I shan't attempt to determine

which of the three possibilities -- the third being

that Gottfried and I are both right, and that Gottfried

has found a way to make progress on Collatz with weaker

inequalities than others have obtained -- is the truth.

I'm just sounding a note of caution. :-)

--

Gareth McCaughan

sig under construc

May 25, 2004, 12:26:36 PM5/25/04

to

Gareth McCaughan schrieb::

>

> According to theorems mentioned at the start of chapter 3

> of Baker's "Transcendental Number Theory", the following is

> true.

(...)

Hi Gareth -

that's an very interesting material, thank you (hope, I'll

ever learn to apply that myself). I'll study it in more detail

next days, since I'm guessing another connection to that field

here, where my derivations on number-structure possibly could

give additional information - or at least some more insight

- I'll post it another day.

For an explanation for the origin of my question please see my

other post.

Regards -

Gottfried Helms

May 25, 2004, 12:17:31 PM5/25/04

to

Gareth McCaughan schrieb::

Hmmm, the stated inequality results from a "primitive" loop,

let me explain that.

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

Assume the transformation T on an odd x with the parameter A

y= T(x;A)

being

y= (3x+1)/2^A

where A is the highest exponent keeping y integral (which is

actually only a short form for multiple steps of the collatz-

transformation, collecting all subsequent x/2 - operations)

and allowing short form for recursive notation:

z= T(y;B) = T(T(x;A);B) = T(x;A,B)

then a loop of, for instance, length 2 occurs, if

z = T(x;A,B) = x

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

The occuring equations are under investigation in some articles,

that I've come across (mostly via internet), and are obviously

difficult to handle, but of high general interest, as I learned

this way.

My first approach was to deal with any type of loop, the general

form

x = T(x;A,B,C,D,..M)

= T(x;A,B,C,D,...,M,A,B,C,...M)

= T(x;A,B,C...)...

for what I've got some nice results, but still not in the form of a

general formulation, which could, for instance, easily been transferred

to 5x+1 and other classes of the problem.

So I decided, first to investigate a somehow "primitive" loop.

I assumed most primitive loop (besides the trivial one, which is in this

notation 1 = T(1;2) = T(1;2,2) = T(1;2,2,2,2,2,2,...) ) for my

purposes the type of loop, which starts with one or more ascending steps,

and then descends in one step.

These are the assumed loops of the form

x = T(x;1,1,1,1,...,1,A)

where immediately strong restrictions apply on A.

One reason why I assumed that type as somehow primitive, is, that any

eventual loop can be expressed as a collection of ascending steps between

descending steps, if the length 0 is allowed.

So an arbitrary transformation, for instance

y = T(x;1,1,3,1,4,2,1,1,3)

can be segmented in

y = T( T(T(T(T(x;1,1,3);1,4);2);1,1,3)

and eventually I can use my tools, that I developed in the analysis of

the "primitive" loop.

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

If I analyse the transformation of length N

z = T(x;1,1,1,1,...,1,A) where the number of ones is denoted as "L",

and check, whether it is ever possible, that we find a solution in integers,

where z equals x, I come to an expression of the right hand side of

my inequality, that I stated her in my previous postings, which also

reflects an *additional* restriction.

With an arbitray 3-step-transformation, x' = T(x;A,B,C) with the length N=3,

where x' should equal x (thus realizing a loop)

with x,y,z, the intermediate values of each transformation,

we can formulate a strong restriction on the exponents A,B,C:

since y = (3x+1)/2^A z = (3y+1)/2^B x'=(3z+1)/2^C and x=x' assumed,

we can write their product as

(3x+1) (3y+1) (3z+1)

xyz = ---------------------

2^(A+B+C)

Rearranging this we have:

(3x+1) (3y+1) (3z+1) 1 1 1

2^(A+B+C) = --------------------- = ( 3 + ---) ( 3 + ---) ( 3 + ---)

x * y * z x y z

and obviously the range for the rhs is 3^3 .. 4^3 , so the sum of the

exponents S=A+B+C must be an integer, which makes 2^S falling in between

this range.

3^N < 2^S <= 4^N (this is valid for any length N of assumed loop)

(for generalizations like the 5x+1-problem this

is equivalently 5^N < 2^S < 6^N)

Even more, this equation shows without any lengthy proof, that whenever the

sum S is in general 2*N, then *all* parentheses on the rhs must take their maximum,

and this is 4 for each of them.

This restricts then all x,y,z to be 1 , which in turn restricts all A,B,C to be 2.

Result: if S = 2*N then x = T(x;A,B,C,D...) only if A=B=C=D=...=2 and x = 1

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

It also shows a widely unknown property of the loop-problem, that the values

of x,y,z have *high bounds* - which is important, since often articles about

the collatz-problem assume loops in "high number" areas, if they don't find a

solution in small accessible values. That is definitely not true: the values

of the members of an eventual loop are much restricted in values.

Since the trivial loop is not of interest, we only have to study the cases

2^S < 4^N

and since 2^S must be at least the next power of 2 after 3^N we have the

inequality

powerceil2(3^N) <= 2^S < 4^N

Now from the expansion of the three transformations into explicit formulas

we get possible S; and I observed, that they were *always* smaller than

powerceil2(3^N) in my cases of primitive loops. The contributions here in

sci.math.research , sci.math and by some private posts showed, that there

is an *extremely* high likelihood, that this inequality never holds.

However - that's still no proof, even not for this simple case.

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

I think, I made the needed progress now by investigating some modular

classes, which exhibited a useful, very simple structure for candidate

numbers x and y.

The above form of the inequality, stated for a "primitive" loop of length

N with L=N-1 ones and one exponent >1

x = T(x;1,1,1,1,1....1,C) with L ones, C>1 and x odd integer>1

This problem can be separated into two:

y = T(x;1,1,1,1,1,1...1) with L=N-1 ones ("iterated transformation")

z = T(y;C) where z should equal x

An iterated transformation like y=T(x;1,1,1,1,1....1) with L ones

restricts x and y to a very specific modular structure.

it comes out, that - with a free parameter i ranging of nonnegative

odd integers - x and y *must* have a very simple structure depending on

the length of the requested iterated transformation:

x = i * 2 * 2^L -1

y = i * 2 * 3^L -1

thus

3* ( i * 2 * 3^L -1 ) +1

z = (3y+1)/2^C = -------------------------

2^C

and if z = x

3* ( i*2*3^L -1 ) +1

x = i*2*2^L -1 = ----------------------

2^C

After rearranging I come to the equation

i*3^N -1

2^C = 2 * -------------

i*2^N -1

and with S = C + 1*L the stated inequality must hold:

i*3^N -1

powerceil(3^N) <= 2^S = 2^(C+L) = 2^L * 2 * --------- < 4^N

i*2^N -1

If this inequality does not hold, then such a type of loop

cannot exist - irrespectively of any assumend length.

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

I stated this inequality for the case i=1 . For growing i the

middle part of the inequality is better written like the following

3^N - 1/i

2^N * ---------

2^N - 1/i

and this converges to 3^N, which is definitevely smaller than the

lhs:

for i->oo the inequality

powerceil2(3^N) <= 3^N

is obviously false.

So the case with i=1 was the most critical case, and I accept

the information and learned the arguments for the extremely

likelihood, that this inequality cannnot be satisfied for any

parameter L and C - but which is still not *proven*....

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

If that's solved, then at least a "primitive" loop can be actually

negated - which is surely no great step in the whole loop-problem or even

the general collatz problem - but... it's a start.

Therefore I'm also more interested in extensible proofs than in

min/max-estimations for separate cases, if they cannot be generalized to

other collatz-type-problems.

Unfortunately, a disproof of this type of collatz-loops is not conversely

saying something on the above inequality, insofar it concerns unproved

assumtions about the 2-log of 3 and the like, what you and others had

derived here - that's bad luck.

But I've got the impression, that some of my equations in this context

could be helpful for progress even in that regard.

I'll post them, if I've time these days.

Hope I answered the questions, that you rose in your post, even if

I had difficulties to really follow your four segments today... :-)

Regards -

Gottfried Helms

May 27, 2004, 8:26:37 AM5/27/04

to

Am 24.05.04 01:00 schrieb Gareth McCaughan:

> Gottfried Helms <he...@uni-kassel.de> writes:

>

>

>> I got an unequality, which i'm unable to analytically to disprove.

>> Does anyone know useful related material?

>>

>> The inequality, formulating a restriction, arose whily studying

>> the question of a certain, primitive assumed loop in the collatz

>> (3x+1)-problem. Here it goes:

>>

>> 3^L - 1

>> powceil2( 3^L ) <= 2^L* -------

>> 2^L - 1

>>

>> where powceil2 ( x) = smallest power 2^S >= x

>

>

> According to theorems mentioned at the start of chapter 3

> of Baker's "Transcendental Number Theory", the following is

> true.

>

(...)

> So ... if I haven't botched the above in any way (which

> I probably have), you "only" need to check a few hundred

> thousand values of L. You'd do that in practice by calculating

> log 3 / log 2 with sufficient accuracy (plain ol' double

> precision should be fine) and looking to see how close to

> an integer you can get by multiplying it by an integer

> up to a few hundred thousand in size.

>

Gareth et.al -

to continue the discussion I invite to join into personal

communication respecting the current repost of the newsgroup-

charter.

I think I have a definite disproof of the above inequality, but

also it could be faulty.

Thanks for all contributions so far -

Gottfried Helms

Jun 1, 2004, 1:41:08 PM6/1/04

to

Gareth McCaughan <gareth.m...@pobox.com> wrote in message news:<c8radh$4ls$1...@news.ks.uiuc.edu>...

There is a paper by Georges Rhin(1987) in which he proves that

|p log 2 - q log 3| > H^(-13.3),

where H = max(p,q) and H > 2.

After that, just check the convergents of log-2(3).

Regards,

Ray Steiner

Jun 1, 2004, 1:48:53 PM6/1/04

to

Gottfried Helms <he...@uni-kassel.de> wrote in message news:<c94mv1$jl8$05$1...@news.t-online.com>...

I proved the nonexistence of a circuit in the 3x+1 problem in 1977.

Is this what a "primitive cycle" actually is? A circuit is a cycle

with one rise and one

fall.

In fact, B. deWeger has recently shown that there is no cycle in the

3x+1 rproblem

with less than 69 rises and falls in it. I can actually extend this to

70 and 71 rises and falls

with Rhin's inequality which I mentioned in an earlier post.

Regards,

Ray Steiner

Jun 2, 2004, 11:54:37 AM6/2/04

to

Am 01.06.04 19:48 schrieb Ray Steiner:

>>

>> I proved the nonexistence of a circuit in the 3x+1 problem in 1977.

>> Is this what a "primitive cycle" actually is? A circuit is a cycle

>> with one rise and one

>> fall.

>>

>> In fact, B. deWeger has recently shown that there is no cycle in the

>> 3x+1 rproblem

>> with less than 69 rises and falls in it. I can actually extend this to

>> 70 and 71 rises and falls

>> with Rhin's inequality which I mentioned in an earlier post.

>> Regards,

>> Ray Steiner

>>

Hmm, that's interesting. I called it a "primitive", if there

are N-1 steps ascending and 1 step descending, like it is in

(7->11->17) -> 13

which, written as an transfer only noting the occuring

exponents-of-2

13 = T(7;1,1,4)

My question was

"is there any solution in integers x,N,A with

x = T(x;1,1,1,1...1,A) with the 1 (N-1)-times occuring

establishing an N-Step-Loop

".

I found this very tight relation to the 3^N - 2^N properties,

thus my question here.

The case of only one-raise-one-fall, if that means

x = T(x;1,A)

would then just be a special case of my question and can be proven

by enumeration or even modular arithmetic.

But I would like to know, how you accessed the problem of

this specific of rises&falls? I have a formula, how you can

disprove a circle of *any* finite combination of raises&false in

finite number of tests, let say 100 raises&falls by a number

Z of tests, where Z is a combinatorical function of about 42,

which I'm bounding down by optimizing my formulas.

My attempt was to generalize these formulas to *unlimited* Z.

The "primitive" Loop is the most simple structure of that

general problem of unlimited length, and has very tight and handy

relations to the (3/2)^N - structure and that of frac( (3/2)^N),

which I'm currently investigating.

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

Concerning your reference to deWegner:

His literature looks interesting, especially those with

the focus on binomials: that was the next idea, that I wanted

to step in. As I pointed out in a previor post the related

problem can be represented using a rep-unit-form and the

question, whether 3^n-1 div 2^N can ever be a repunit base

a certain power of 2^P with p reasonably greater than N (don't

have it at hand just now). I'm currently studying, what the binomial-

expansion of 3^N = (2+1)^N explains for that problem.

Just have seen your reference to Rhin: could you point to

a reference?

Regards -

Gottfried Helms

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