The following iterative sequence is defined for the set of positive integers:

• $n \to n/2$ ($n$ is even)
• $n \to 3n + 1$ ($n$ is odd)

Using the rule above and starting with $13$, we generate the following sequence: $$13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1.$$

It can be seen that this sequence (starting at $13$ and finishing at $1$) contains $10$ terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at $1$.

Which starting number, under $N$, produces the longest chain? If there are several, enter the smallest one.

NOTE: Once the chain starts the terms are allowed to go above $N$.