Weighings to find one heavy bean among eighteen

Among $18$ beans, $17$ weigh the same and one is heavier. With a two-pan balance, what is the smallest number of weighings that always finds the heavy bean?

3. Each weighing has three outcomes, so $k$ weighings separate at most $3^k$ candidates; $3^2=9<18\le27=3^3$. Three weighings suffice: split into groups of $6,6,6$, then $2,2,2$, then $1,1,1$.