The problem is not possible.
Suppose there is a term n such that S(n)=0
That means 0 = S(n-2) - 3/S(n-1)
But S(n-1) = S(n-3) - 3/S(n-2)
and substituting in gives
[S(n-3) - 3/S(n-2)]*S(n-2)=3
S(n-3)*S(n-2)=0 <--- Algebra error. This should be 6.
Now S(n-2) cannot be 0, so S(n-3)=0
This means S(n-1)=0-3/S(n-2)
Which is a contradiction.
So there is no n such that S(n)=0.
***** Edit *****
Daniel's solution seems valid and a quick check with Excel seems to confirm this. We posted at nearly the same time so I didn't see his until I hit enter. I think something must be wrong with the above. I wonder what.
Now that I've got S(n-3)*S(n-2)=6 I could probably show S(n-(a+1))*S(n-a)=3a and work backwards to give the same solution as Daniel.
Edited on January 28, 2016, 10:46 am
Posted by Jer
on 2016-01-28 09:34:50