A210483 Triangle read by rows giving trajectory of k/(2n+1) in Collatz problem, k = 1..2n.

0, 2, 3, 3, 3, 15, 3, 6, 7, 11, 8, 5, 12, 5, 6, 2, 7, 6, 3, 10, 8, 7, 7, 25, 7, 22, 26, 7, 7, 27, 23, 4, 4, 17, 4, 12, 18, 9, 4, 16, 13, 15, 19, 17, 18, 3, 19, 2, 3, 18, 20, 15, 3, 5, 3, 8, 19, 8, 8, 21, 8, 8, 22, 15, 8, 52, 8, 49, 23, 19, 16, 49, 8, 31, 32

Offset: 0

Author

Michel Lagneau, Jan 23 2013

Keywords

• nonn
• tabf

Examples

			The 2nd row [3, 3, 15, 3] gives the number of iterations of k/5 and starts with A210468(2):
k=1 => 1/5 -> 8/5 -> 4/5 -> 2/5 with 3 iterations;
k=2 => 2/5 -> 1/5 -> 8/5 -> 4/5 with 3 iterations;
k=3 => 3/5 -> 14/5 -> 7/5 -> 26/5 -> 13/5 -> 44/5 -> 22/5 -> 11/5 -> 38/5 -> 19/5 -> 62/5 -> 31/5 -> 98/5 -> 49/5 -> 152/5 -> 76/5 with 15 iterations;
k=4 => 4/5 -> 2/5 -> 1/5 -> 8/5 with 3 iterations.
The array starts:
  [0];
  [2, 3];
  [3, 3, 15, 3];
  [6, 7, 11, 8, 5, 12];
  [5, 6, 2, 7, 6, 3, 10, 8];
  [7, 7, 25, 7, 22, 26, 7, 7, 27, 23];
  ...
		

Links

• Michel Lagneau, Rows n = 0..100 of triangle, flattened
• J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.

Crossrefs

Cf. A210468.

Programs

Maple

with(numtheory): z:={1}:
   for m from 0 to 20 do:
             T:=array(1..2*m):k:=0:n:=2*m+1:
      for nn from 1 to 2*m do:
             lst:={nn/n}:x0:=nn: ii:=0:
        for a from 1 to 20 while(ii=0) do:
             if irem(x0,2)=0 then
             x0:=x0/2:lst:=lst union{x0/n} :else ii:=1:fi:
        od:
             x:=x0*3+n: lst:=lst union {x/n}:
               for i from 1 to 6000 do:
            x:=x/2: lst:=lst union {x/n}:  if irem(x,2)=1 then
            x0:=x:x:=x0*3+n: lst:=lst union {x/n}:else fi:
           od:
          n0:=nops(lst):if lst intersect z = {1} then
          n1:=n0-2: k:=k+1: T[k]:=n1: else n1:=n0-1: k:=k+1:T[k]:=n1:fi:
   od:
     print (T):
od:

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[Numerator[#]], #/2, 3 # + 1] &, n, UnsameQ, All]; t = Join[{{0}}, Table[s = Collatz[k/(2*n + 1)]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len, {n, 10}, {k, 2*n}]]; Flatten[t] (* T. D. Noe, Jan 23 2013 *)