Define a number to be cube-free if there is no prime factor that repeats more than once. Or, $n$ is cube-free if and only if $\frac{n}{\operatorname{rad}(n)}$ is square-free.
I've tested numbers up to $10,000,000$ without finding more than $4$ consecutive not cube-free numbers. Can this be proved to be a maximum or are there consecutive sequences with more than $4$ not cube-free numbers?
Best Answer
Consider \begin{align} 3^3 \mid 18035622\\ 17^3 \mid 18035623\\ 2^3 \mid 18035624\\ 5^3 \mid 18035625\\ 7^3 \mid 18035626 \end{align} This is the smallest example of $5$ consecutive numbers that are not cubefree.
Also, you can now construct other examples from this one. For example, put $K=3^3\cdot 17^3\cdot2^3\cdot 5^3\cdot 7^3$, then we have: \begin{align} 3^3 \mid K+18035622 \\ 17^3 \mid K+18035623 \\ 2^3 \mid K+18035624 \\ 5^3 \mid K+18035625\\ 7^3 \mid K+18035626 \end{align}
Now the hint given in comments gives you much better way to construct arbitrary long sequences of arbitrary powers (not just cubes) of multiples of arbitrary selected primes (for example). I suggest you look at Chinese Reminder Theorem and try to apply it to this problem.
EDIT: Whoops, I accidentally let the searching algorithm running over night, so here are couple more examples (always just first of five numbers): $100942496$, $133799496$, $146447622$, $156406624$, $185966872$, $\dots$.