  
  [1m[4m[31m1. The 3k+1 Problem[0m
  
  
  [1m[4m[31m1.1 Theory[0m
  
  Let  k  in  N be a natural number. We consider the sequence n(i, k), i in N,
  with  n(1,  k)  =  k and else n(i+1, k) = n(i, k) / 2 if n(i, k) is even and
  n(i+1, k) = 3 n(i, k) + 1 if n(i, k) is odd.
  
  It  is  not  known  whether for any natural number k in N there is an m in N
  with n(m, k) = 1.
  
  [1mThreeKPlusOne[0m provides the function [1m[34mThreeKPlusOneSequence[0m ([1m1.2-1[0m) to explore
  this  for  given n. If you really want to know something about this problem,
  see  [W98]  or [34mhttp://mathsrv.ku-eichstaett.de/MGF/homes/wirsching/[0m for more
  details (and forget this package).
  
  
  [1m[4m[31m1.2 Program[0m
  
  In this section we describe the main function of this package.
  
  [1m[4m[31m1.2-1 ThreeKPlusOneSequence[0m
  
  [1m[34m> ThreeKPlusOneSequence( [0m[22m[34mk[, max][0m[1m[34m ) ________________________________[0mfunction
  
  This  function computes for a natural number [22m[34mk[0m the beginning of the sequence
  n(i,  k)  defined  in  section  [1m1.1[0m. The sequence stops at the first 1 or at
  n([22m[34mmax[0m, k), if [22m[34mmax[0m is given.
  
  [22m[35m---------------------------  Example  ----------------------------[0m
    [22m[35mgap> ThreeKPlusOneSequence(101);[0m
    [22m[35m"Sorry, not yet implemented. Wait for Version 84 of the package"[0m
  [22m[35m------------------------------------------------------------------[0m
  
