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

Author

Michel Lagneau, Mar 08 2013

Keywords

• nonn
• tabf

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}, ...

Examples

			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}.
		

Links

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

Crossrefs

Cf. A075680, A075684.

Programs

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: