A265221 Number of steps needed to enter the cycle in the double '5x+1' and '3x+1' problem.

8, 9, 16, 10, 5, 17, 5, 11, 36, 5, 5, 18, 24, 5, 42, 12, 25, 37, 12, 5, 10, 5, 11, 19, 19, 25, 31, 5, 37, 43, 17, 13, 173, 26, 32, 34, 38, 13, 44, 5, 44, 11, 50, 6, 17, 12, 56, 20, 85, 20, 20, 26, 26, 32, 32, 6, 32, 38, 12, 44, 15, 18, 18, 14, 21, 174, 20, 27

Offset

1,1

Comments

See A264811 for the double '3x+1' and '5x+1' problem.

The double '5x+1' and '3x+1' problem is as follows: start with any number n.

Step 1: if n is even, divide it by 2 until the result becomes odd, otherwise multiply it by 5 and add 1. Let r1 denote the result after respective computation.

Step 2: if the result r1 of step 1 is even, divide it by 2 until the result becomes odd, and then multiply it by 3 and add 1. If the result of step 1 is odd, multiply it by 3 and add 1. Let r2 denote the result after respective computation.

Step 3: go to step 1 with n=r2 and continue until stabilization.

It is conjectured that the process always stabilize.

Starting from any positive n, we call the sequence of iterates (n, T(n,1), T(n,2), ..., T(n,k), ...) the trajectory of n. There are three possible behaviors for such trajectories when n>0:

(i) Convergent trajectory. Some T(n,k) = 1.

(ii) Nontrivial cyclic trajectory. The sequence T(n,k) eventually becomes periodic and T(n,k) is different from 1 for any k>=1.

(iii) Divergent trajectory (it is conjectured that this behavior cannot occur).

Links

Michel Lagneau, Table of n, a(n) for n = 1..10000

Example

a(1)=8 because the number 1 generates the trajectory with 8 steps 1 -> 6 -> 3 -> 10 -> 5 -> 26 -> 13 -> 40 -> 20  where the number 20 is the last term of the cycle because the following terms 20/2 = 10 -> 5 -> 26 -> 13 -> 40 are already in the trajectory.

Maple

for n from 1 to 100 do:nn:=1000:T:=array(1..nn):kk:=1:
lst:={n}:m:=n:it:=0:indic:=0:m0:=n:T[1]:=n:
   for i from 2 to nn do:
    if irem(m, 2)=0
    then m:=m/2:kk:=kk+1:T[kk]:=m:lst:=lst union {m}:it:=it+1:
    else
    if indic=0
    then m:=5*m+1:kk:=kk+1:T[kk]:=m:lst:=lst union {m}:it:=it+1:indic:=1:
    else
    m:=3*m+1:it:=it+1:kk:=kk+1:T[kk]:=m:lst:=lst union {m}:indic:=0:
   fi:
  fi:
od:
n0:=nops(lst):ii:=0:
  for a from n0 to nn while (ii=0) do:
    q:=T[a]:it:=0:
      for b from a-1 by -1 to 1 while (ii=0)do:
       it:=it+1:
        if q=T[b] and q=T[b+it]
        then
        ii:=1: printf(`%d, `, a-2):
        else
        fi:
       od:
     od:
     od:

Crossrefs

Cf. A006577, A264811.

Sequence in context: A351048 A292188 A301659 * A286476 A265732 A047393

Adjacent sequences: A265218 A265219 A265220 * A265222 A265223 A265224

Keyword

nonn

Author

Michel Lagneau, Dec 05 2015

Status

approved