Collatz-Intro - Some general remarks

Workshop
Recreational
Mathematics

Gottfried Helms Univ. Kassel

mailto: helms at uni-kassel

www: math-homepage

 

 

 

This collection of articles is not meant as a general overview on the problem. There are some better articles available in the net, see for instance of [Lagarias],[Schorer],[Conrow].

 

I focus some special topics, prominently the loop-problem, which I even split up into smaller portions, to disprove simpler questions first, like the nonexistence of a "primitve loop" (which are identical to that "1-cycles" or "m-cycles" of [Steiner] and [Simons/deWeger] except for the notation of the parameters).

 

The conjecture can be proven resp disproven in few aspects:

 

CC: (CollatzConjecture) to show that all numbers transform to 1 after a finite number of transformations T() : T(n;A,B,C,...Z) = 1

 

 

CC can conversely be proven by an inversed approach:

 

p1) all numbers can be constructed by the inverse transformation C(), starting at 1

 

 

CC can be disproven if

 

d1) there exists a number which is a member of a divergent trajectory; where the values under transformations T() grow infinitely

 

OR

 

d2) there exist a group of numbers (different from 1) where the transformations T() go into a loop, thus starting at a number n, doing a transformation a'=T(a;A,B,C,D,...) the result a' equals a

 

OR

 

d3) the inverse transformation C() cannot create all numbers if started by 1:
n cannot be expressed by n = C(1;A,B,C,...H)

 

 

CC can be proven if

 

p2) no divergent trajectory exists

 

AND

 

p3) no loop exists.

 

Although I have invested plenty of time in p1 and found some interesting arguments, this is currently not my primary focus. Instead I mainly deal with d2) resp p3), the disproof of a loop.

 

 

 

 

Also in the attempt to *dis*prove I setup some inequalities, assuming some exponents in a transformation-formula like

 

b = T(a;A,B,C,D)

to show, that "a" never can equal "b" - without referring to a special value of "a". This simple remark may be needed for some Collatz-fans, since in such cases some available articles and hommages to the problem discuss it from the view of focusing actual elements of a transformation, instead of focusing the characteristics of a certain type of transformation, which requires a certain possible or impossible structure for its elements a and b.

 

So I won't enter some "sportive" approaches: which is the highest number that has such and such a trajectory, since with the notation of

 

b = C(a;A,B,C,D,...)

 

one can construct many types of examples just by respecting some not too hard modular requirements for the sequence of exponents A,B,C...

 

 

From the definition of my T()-transformation it is obvious, that I only deal with odd numbers as elements of this transformation. For all questions, that I want to discuss this is no restriction, but even a simplification.

 

 

 

For instance, the inverse transformation C() provides a handy tool to generate a very instructive tree, which I have hoped is such simple, that it could prove the CC by p1. [sheet01]

The study of this tree led to many interesting graphs, either as spreadsheet of as graphical picture (my most favorite is a fractal tree in the form of a bottle-brush with whiskers and infinite selfsimilar repetitions) [fractal-graph]. The sheet01, displayed in base-4-number-system exhibits a much simplifying structure [sheet-base4], which is not evaluated finally.

 

 

The notation

 

b = T(a;A,B,C...H)

 

gives raise to display some general formulae.

 

A transform

 

a'=T(a;A,B,C)

 

 

can be written as

 

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

 

 

which can be expanded to

 

a*3^3 + 3^2 + 3^1*2^A + 2^(A+B)
a' = -----------------------------------
2^(A+B+C)

 

 

or more general, where N indicates the "length" of the multi-step-transformation and S may indicate the sum of all exponents

 

 

a*3^N + 3^(N-1) + 3^(N-2)*2^A + 3^(N-3)2^(A+B) +...+ 3*2^(A+...+F) + 2^(A+...+G)
a' = ----------------------------------------------------------------------------------
2^(A+...+H)

 

 

or the canonical form:

 

 

3^N 3^(N-1) + 3^(N-2)*2^A + 3^(N-3)2^(A+B) +...+ 3*2^(A+...+F) + 2^(A+...+G)
a' = a* --- + -------------------------------------------------------------------------
2^S 2^S

 

 

From that canonical form it can also be easily derived that

 

 

3^N
T(a;A,B,C...G,H) = a*--- + T(0;A,B,C...G,H)
2^S

 

 

and thus an "a" can be found easily, if the "standardized" transform T(0;...) is computed and then

 


a'*2^S - a*3^N = T(0;...)* 2^S

 

which can be useful to find a pair of integral a-> a' which satisfy the structure of the specified transformation.

 

The canonical form exhibits some interesting results:

 

         For any combination of exponents we can find one "a", which can be transformed to a valid "a'"

         There are infinitely many solutions for finding "a", namely all the same residue class base 2^S

         The higher the sum of exponents s, the higher must the value of "a" be to result in a valid "a'"

 

 

 

Conversely, if we study the inverse transformation

 

a = C(a',H,G,...C,B,A)

 

we find

 

 

2^S 3^(N-1) + 3^(N-2)*2^A + 3^(N-3)2^(A+B) +...+ 3*2^(A+...+F) + 2^(A+...+G)
a = a'* --- - -------------------------------------------------------------------------
3^N 3^N

 

 

and the Collatz-conjecture CC implies this way, that each odd number can be created by a series of fractions

 

 

2^S 1 2^A 2^(A+B) 2^(A+...+F) 2^(A+...+G)
a = 1* --- - -- - ---- - ------- -... - ----------- - -----------
3^N 3 3^2 3^3 3^(N-1) 3^N

 

 

or

 

with S'=S-N, A'=A-1>=0, B'=B-1>=0 ... d = 3/2

 

2^S' 2^(A'+...G') 2^(A'+...F') 2^A' 2^-1
a = --- - ------------ - ---------- ... - ---- - ----
d^N d^(N-1) d^(N-2) d^2 d

 

 

with appropriate exponents A,B,C,...G,H - which I find a remarkable result in itself.

 

 

 

 

Finally, this canonical form leads on a simple path to the formula, which describes the loop. In that case the lhs and rhs must be equal, thus

 


a'*2^S - a*3^N = T(0;...)* 2^S

with a'=a thus forming a loop

 

a(2^S - 3^N) = T(0;...)* 2^S

 

T(0;...)* 2^S
a = -------------
2^S - 3^N

 

 

and for a loop-candiate of the form


a = T(a;A,B,C,...,G,H)

 

 

3^(N-1) + 3^(N-2)*2^A + 3^(N-3)2^(A+B) +...+ 3*2^(A+...+F) + 2^(A+...+G)
a = --------------------------------------------------------------------------
2^S - 3^N

 

where the rhs must result in an odd integer>1

 

This formula also occurs, if the problem is attacked from an system of linear equations involving equations for all intermediate transforms a,b,c,...g,h

 

 

 

 

 

 

 


last update: 15.8.2004

 

 

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