A214850 3x+1 groups: irregular triangle read by rows: row n gives all the numbers p <= A075684(n)+1 such that {T(2n+1,k) /pZ} is a multiplicative finite group, where T(2n+1,k) is the reduced trajectory of the Collatz problem whose elements are all odd.

2, 4, 2, 4, 6, 2, 4, 6, 8, 12, 18, 2, 4, 8, 2, 4, 6, 2, 4, 6, 8, 12, 2, 4, 8, 2, 4, 6, 8, 12, 2, 4, 6, 12, 2, 4, 8, 10, 20, 22, 2, 4, 6, 8, 2, 4, 6, 12, 18, 2, 4, 8, 16, 2, 4, 6, 2, 4, 6, 8, 12, 16, 18, 24, 2, 4, 2, 4, 6, 2, 4, 6, 8, 12, 18, 2, 4, 8, 2, 4, 6

Offset

1,1

Comments

We introduce the structure of a finite group in order to give a possible way to classify the Collatz trajectories.

We see that the study of the classification of the trajectories is dependent on the values p.

The principle of the algorithm is to compute all the products T(2n+1,i)/pZ * T(2n+1,j)/pZ and also the inverse of each element such that if x is in the group, then there exist x' in the group with x*x' = 1.

Rows of triangle:

{2, 4},

{2, 4, 6},

{2, 4, 6, 8, 12, 18},

{2, 4, 8},

{2, 4, 6},

{2, 4, 6, 8, 12},

{2, 4, 8},

{2, 4, 6, 8, 12}, ...

Links

Michel Lagneau, Rows n = 1..400 of irregular triangle, flattened

Example

Row 18 gives 6 groups with p = {2, 4, 6, 8, 12, 18}. The Collatz trajectory T(37,k) = {37, 7, 11, 17, 13, 5, 1}, and if we choose, for example, p=18, we obtain G(37) = {T(37,k)/18Z} = {7, 11, 17, 13, 5, 1} which (as subset of Z/18Z) is a multiplicative group of order 6.
For example 5, or 11, generates the cyclic group:
5^1 == 5, 5^2 == 7, 5^3 == 17, 5^4 == 13, 5^5 == 11, 5^6 == 1 (mod 18).
Other subgroups are {1}, {1, 17} and {1, 7, 13}.

Maple

c:=0:
for n from 3 by 2 to 800 do:
       x:=2:lst:={n}:lst1:={}:x := n:
          for k from 1 to 120 while (x > 1) do:
             a := 0:
             if type(x, 'even') then
             x := x/2:lst:=lst union {x}:a:=a+1:
            else
            x := 3*x+1 :lst:=lst union {x}:a:=a+1:
            fi:
       od:
        n1:=nops(lst):
          for u from 1 to n1 do:
             if irem(lst[u], 2)=1 then
             lst1:=lst1 union {lst[u]}:
            else
           fi:
         od:
         m1:= max( op(lst1)):n1:=nops(lst1):
            for p from 2 by 2 to m1+1 do:
            lst2:={}:
               for q from 1 to n1 do:
               lst2:=lst2 union {irem(lst1[q], p)}:
              od:
              lst3:={}:n2:=nops(lst2):kkk:=0:
                for i from 1 to n2 do:jjj:=0:
                    for j from 1 to n2 do:
                       z:=irem(lst2[i]*lst2[j], p):lst3:=lst3 union{z}:
                       if z=1 then jjj:=1:else fi
                    od:
                    if jjj=0 then kkk:=1:else fi:
                 od:
                 n3:=nops(lst3):iii:=0:
                     for b from 1 to n3 while(iii=0 and n2=n3 and kkk=0)
                 do:
                      if lst2[b]<>lst3[b] then
                      iii:=1:else
                      fi:
                    od:
                    if iii=0  and n2=n3 and kkk=0 then c:=c+1:
                    printf ( "%d %d \n", c, p):
                    else
                    fi:
                od:
                 x:=2:
              od:

Crossrefs

Cf. A075680, A075684.

Sequence in context: A216621 A300448 A214781 * A236186 A143271 A267654

Adjacent sequences: A214847 A214848 A214849 * A214851 A214852 A214853

Keyword

nonn,tabf

Author

Michel Lagneau, Mar 08 2013

Status

approved