Finding one heavy bean among eighteen

You have $18$ beans that look identical; $17$ weigh the same and one is heavier. Using a two-pan balance, what is the smallest number of weighings that always identifies the heavy bean?

3. Each weighing has three outcomes (left heavy, right heavy, balance), so $k$ weighings separate at most $3^k$ candidates; $3^2=9<18\le27$. Three suffice: split $6/6/6$, then $2/2/2$, then $1$ vs $1$.