A194428 Number of iterations of the map n->n/3 if n == 0 (mod 3), n->4*n+a if 4*n+a == 0 (mod 3) where a = 1 or 2, before reaching the end of the cycle.

5, 5, 5, 7, 17, 5, 16, 22, 5, 16, 20, 8, 8, 16, 18, 20, 22, 6, 16, 8, 16, 18, 20, 23, 34, 16, 6, 27, 11, 16, 18, 22, 21, 32, 16, 9, 23, 25, 9, 9, 28, 16, 20, 39, 19, 30, 16, 21, 21, 21, 23, 23, 35, 7, 26, 37, 16, 18, 37, 9, 28, 28, 16, 43, 14, 19, 19, 34, 21, 21, 33, 24

Offset: 1

Author

Michel Lagneau, Aug 23 2011

Keywords

• nonn

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.

Examples

			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.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001281.

Programs

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: