No. You don't know in advance how many integers there will be for any particular starting value.
Here is the definition, again:
N/2
3*N + 1
And here is a method:
The method prints out the current N
,
then advances to the next by calling itself.
When it hits one, it returns to its caller,
which returns to its caller, and so on up the chain until the first activation.
Fill in the blanks.