A264811 Number of steps needed to reach 1 or to enter the cycle in the double '3x+1' and '5x+1' problem.

11, 12, 6, 13, 13, 7, 5, 14, 5, 14, 27, 8, 5, 5, 13, 15, 27, 5, 39, 14, 15, 28, 34, 9, 15, 5, 46, 5, 8, 14, 58, 16, 22, 28, 28, 5, 8, 40, 14, 11, 20, 16, 23, 29, 29, 35, 28, 10, 41, 16, 64, 6, 8, 47, 12, 5, 53, 9, 53, 15, 15, 59, 17, 17, 88, 23, 23, 29, 29, 29

Offset

1,1

Comments

The double '3x+1' and '5x+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 3 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 5 and add 1. If the result of step 1 is odd, multiply it by 5 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

Table of n, a(n) for n=1..70.

Example

a(1)=11 because the number 1 generates the trajectory 1 -> 4 -> 2 -> 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 90 do:nn:=5000: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:=3*m+1:kk:=kk+1:T[kk]:=m:lst:=lst union {m}:it:=it+1:indic:=1:
    else
    m:=5*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.

Sequence in context: A342076 A329127 A076654 * A352389 A020510 A291521

Adjacent sequences: A264808 A264809 A264810 * A264812 A264813 A264814

Keyword

nonn

Author

Michel Lagneau, Nov 25 2015

Status

approved