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A210483
Triangle read by rows giving trajectory of k/(2n+1) in Collatz problem, k = 1..2n.
4
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
0,2
LINKS
J. C. Lagarias, The set of rational cycles for the 3x+1 problem, Acta Arith. 56 (1990), 33-53.
EXAMPLE
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];
...
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 *)
CROSSREFS
Cf. A210468.
Sequence in context: A199457 A361378 A347208 * A022296 A013069 A096393
KEYWORD
nonn,tabf
AUTHOR
Michel Lagneau, Jan 23 2013
STATUS
approved