OFFSET
1,1
COMMENTS
The problem is as follows: start with any number n. If n is divisible by 3, divide it by 3, otherwise multiply it by 4 and add 1 or 2 in order to find a new integer divisible by 3. Do we always reach the end of a cycle? It is conjectured that the answer is yes.
On the set of positive integers, the orbit of any number seems to end in the orbit of 1, or of another integer.
This problem has a resemblance with the Collatz problem.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 5 because 1 -> 6 -> 2 -> 9 -> 3 -> 1 with 5 iterations ;
a(2) = 5 because 2 -> 9 -> 3 -> 1-> 6 -> 2 with 5 iterations ;
a(3) = 5 because 3 -> 1 -> 6 -> 2 -> 9 -> 3 with 5 iterations ;
a(4) = 7 because 4 -> 18 -> 6 -> 2 -> 9 -> 3 -> 1 -> 6 with 7 iterations.
MAPLE
T:=array(1..2000):for n from 1 to 100 do: T[1]:=n:n0:=n:k:=2:for it from 1 to 50 do: z:=irem(n0, 3):if z=0 then n0:=n0/3:T[k]:=n0:k:=k+1:else n0:=4*n0 + 1:if irem(n0, 3)=0 then T[k]:=n0:k:=k+1:else n0:=n0+1:T[k]:=n0:k:=k+1:fi:fi:od:U:=convert(T, set):n1:=nops(U): printf(`%d, `, n1):od:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Aug 23 2011
STATUS
approved