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

-